help-23

[indigo scaffold]

1

[normal moss]

doesn't look like it

[indigo scaffold]

a = 30

[normal moss]

👍

[indigo scaffold]

we cant do that for 19,2 = a*c^1 right

[normal moss]

do what?

[indigo scaffold]

to find a = 30

[normal moss]

Why would you want to find something you already know

You have a system of two equations with two unknowns

[indigo scaffold]

no it was just a question

[normal moss]

You are trying to figure out the values of these 2 unknowns that make both equations true

You have figured out a = 30

You can use that information

[indigo scaffold]

alright

[normal moss]

in the second equation

[indigo scaffold]

19,2 = 30*c^1

[normal moss]

yes

[indigo scaffold]

0,64 = c^1

[normal moss]

yep

hopefully we know what c^1 is

[indigo scaffold]

its 0,64

[normal moss]

well yes

I meant

we usually write just c

😂

[indigo scaffold]

oh alr

[normal moss]

not c^1

Otherwise your equations would look like $7^1x^2y^1 = 18^1x^1$

[flat frigate]

USS-Enterprise

[indigo scaffold]

yeah lol

[normal moss]

Right

So we have figured out a and c

[indigo scaffold]

we found c now so we replace x by 8

[normal moss]

Well first

write the function

[indigo scaffold]

alright

now what

[normal moss]

What did you get

[indigo scaffold]

0,84

[normal moss]

what?

oh

f(8)

yes

$f(x) = 30 \cdot \left(\frac{16}{25}\right)^x$

[flat frigate]

USS-Enterprise

[normal moss]

That is f(x)

16/25 = 0,64

And then just find f(8)

which is approx. 0,84 yes

[indigo scaffold]

yup

[normal moss]

Notice how if you also input x = 0, 1, 2, 3

The y in the table matches the output you get

[indigo scaffold]

cause its exponential right

[normal moss]

I mean, in a way

But because we found the right function

If the values didn't match, that would mean we had made a mistake

easy way to test yourself

[indigo scaffold]

oh yeah

alright tysm

.close

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[normal moss]

no problem

good luck!

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[kind quartz]

Can i get some help with this? I know how to do squeeze theorem but forget what to do with the fact its pi/x

[plucky elk]

-1 <= sin(thing) <= 1

doesn't matter what thing is, even if thing=pi/x

[kind quartz]

So i just put it in as usual? ok

We only recent started it so i was unsure lol

Okay so working it out i have -root(x^3 + x^2) <= root(x^3 + x^2)sin(pi/2) <= root(x^3 + x^2). And working either of those out, they both are equal to zero, so its gotta be 0. Does that seem right?

[plucky elk]

more specifically, the limit as x goes to zero of the left and right functions go to 0

[kind quartz]

yes so its getting squeezed, as the name implies

[kind quartz]

would this be a satisfactory answer to the question though?

[plucky elk]

right

[kind quartz]

Okay nice. i do have one more i fear but its easier, more just making sure im actually understanding

When it comes to this, its l'hopital, yes? i dont quite understand how to do it

[plucky elk]

factoring is better

[kind quartz]

okay ill try that

[plucky elk]

t-1 is a common factor to both numerator and denominator so you do polynomial division

[kind quartz]

im struggling with the polynomial division, its been a while since i last did it. so like. we wanna divide t^4 - 1, by t-1. i lowkey forget how to do that so ill search it up rq i suppose

okay i got t^3 + t^2 + t + 1 for the numerator, and t^2 + t + 1 for denominator

[upper moat]

okay so

Formula for difference of cubes is
(a-b)(a^2+ab+b^2)

[plucky elk]

,w factor x^4-1

[flat frigate]

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[plucky elk]

,w expand (x+1)(x^2+1)

[upper moat]

,w factor x^3-1

[flat frigate]

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[upper moat]

You notice

[plucky elk]

looks right

[upper moat]

The (x-1) can cancel out

[plucky elk]

should just be able to plug in t=1 now

[kind quartz]

yesss cancel the t-1

Okay so plugging in t=1 gave 4/3, which seems reasonable imo.

[upper moat]

Yes

[kind quartz]

okay thankyou! have a good day

.close

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[fringe drift]

can someone help me answer this

do we divide by x

[wild cape]

yes but you did not do it correctly

[fringe drift]

Wwould this be correct.

[stoic dune]

Note it's not asking for the integrating factor, you're doing work you don't have to do

<@&268886789983436800> scammer

Thx you're bae

[fringe drift]

im trying ti find Ax

[stoic dune]

A(x) is just the term multiplying y

You've already found it, it's A(x) = 7

Want to take a guess at B(x)?

[fringe drift]

e^-3x

[stoic dune]

Exactly!

[fringe drift]

stuck after this step

[stoic dune]

Just looking for the integral of tan(x)? Remember that tan(x) = sin(x)/cos(x)

Can finish with u-sub

[fringe drift]

oh so we rewrite it like that

Is sec(3t) *cos(3t)=1

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[indigo scaffold]

how can i find the value of 2 log_8 1

[indigo scaffold]

8^x = 1^2

[quasi timber]

logₐ(1) always = 0 for 𝑎 > 1

[indigo scaffold]

8^x = 0

x = 0

[muted sapphire]

?????

[quasi timber]

❓

[muted sapphire]

everything ok opal?

[indigo scaffold]

no

is x = 0 right

.close

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[sinful citrus]

anyone wanna walk through this with me? I'm like 90% sure I can do it I just wanna make sure I'm doing it right since I don't have a solution for it

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[blazing relic]

i tried 250cos(1/3x)+750and that did NOT work so idk what to do

[neat patrol]

plug in 15

youre missing a factor for the (1/3)x part

dont forget about the period of a cosine function

[blazing relic]

i dont really know how to get those

[sage rain]

your vertical shift of +750 is correct, as is your amplitude of 250

so we have
y = 250 cos(something) + 750

we know that this something takes the form Bx

where B is the period of the function, which is 2pi divided by the frequency

the maximum frequency was heard 6 times on [0,15] so we can tell that 5 full cycles occurred

15 seconds / 5 cycles = 3

Then, B= 2pi/3

[flat frigate]

nc129

[blazing relic]

hmm ok

i think i see how we gotthat

thank u

.solved

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[silk cove]

,rotare

,rotate

[silk cove]

How to find Q1 and Q3?

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[silk cove]

What is critical point?

[light shoal]

did you google it?

[silk cove]

I tried but I didn't understand [e-mail]

In the domain of function the point where it changes its direction

???

Or the derivative is 0

Example |x-1|/x^2

,w d/dx |x-1|/x^2

,w d/dx of |x-1|

Three points

I guess two points because 0 is not in the domain

[plucky elk]

Maybe you should plot the function

[crystal pendant]

do you at least know stationary points for 2D graphs?

[gritty sable]

Critical points are points where function becomes undefined, functions derivative become zero or undefined

[silk cove]

,w d/dx |x-1|/x^2

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[gritty sable]

Critical points where derivative becomes zero are called stationary points too!

[silk cove]

And what is critical point?

Where is first derivative is 0 no?

So this one is 0 at

[crystal pendant]

for one variable functions, its where the first derivative is either 0 or undefined

for multivariable functions, its where all first partial derivatives are either 0 or undefined

[sage rain]

1 and 2

[silk cove]

so it is 0 at 1,2 yes

And undefined at x=0

But 0 is not in our domain

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[daring flume]

just dropping by to say that you should be very careful writing exponents at near full font size.

[fickle mantle]

Oh

[quasi bison]

yeah also line spacing should be like double what you got

[sage rain]

lwk multiply everything through by 2^x ?

[quasi bison]

you're on the right track though

[sage rain]

2^2x + 16 -10(2^x) = 0 and solve the quadratic?

[quasi bison]

do this and also call 2^x by a new name like z

[quasi bison]

let this sink in for op now

[fickle mantle]

Huh

[daring flume]

try multiplying the entire equation by 2^x.

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[quasi bison]

welp

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[fickle mantle]

I’m did

I did it

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[sage rain]

um

what did you do here

[median vigil]

,rccw

[daring flume]

preferably, when asked to show {expression 1} = {expression 2}, you should try to manipulate just one side to achieve the other.
but if your syllabus allows manipulation of both sides simultaneously via reversible operations, then your methodology could work.

[fickle mantle]

Oh ok

I always get confused w this

But how do I do the second one

[daring flume]

note that the question starts with the word "Hence".

normally, when questions start with that word, it is an indication to use the information you have or derived from the previous subquestions.

in this case, you have shown that the LHS of the equation is identical to the RHS in subquestion (1).
you are then given that the LHS of that equation is equal to some number.
what can you replace the LHS of the equation in (2) with to make it simpler?

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[latent knoll]

hii. so this is a graph and i just need a confirmation whether or not this is correct. if you need me to translate something lmk. va is vertical asymptote, ha is horizontal and ka is oblique as.

'ekstremi' means extremes and 'nema rj' means no result

[safe radish]

@latent knoll Has your question been resolved?

[slender steeple]

what are we looking for here?

this one checks out

[latent knoll]

im so sorry i didnt get a notification

i am not entirely sure. i am just supposed to calculate a graph and to draw it?

[slender steeple]

you're asked to graph it?

[latent knoll]

i basically need a domain, zeros, asymptotes and extremes and then draw it

[slender steeple]

is it also asking to label asymptotes and/or intercepts?

[latent knoll]

i believe so

[slender steeple]

i see

[latent knoll]

yeah which is why i drew the zero point intercept or whtv it's called on 2 and all the other lines which are just vertical asymptotes

[latent knoll]

so is the graph drawn well?

cuz if it's not i have this variation as well

[slender steeple]

better than how i would've graphed it

that's for sure

[slender steeple]

huh

this is peculiar

[latent knoll]

and is it correct?

[slender steeple]

you suddenly dip down on an x intercept?

[latent knoll]

yeah this is another one i've drawn if the first one isn't correct

[latent knoll]

yeah idrk how to read the VA that well. so im taking it that the 2nd picture is supposed to be wrong

[slender steeple]

this variation is almost correct

this part is

wrong

but aside from that you should use your latest variation

cause that's the correct one

[latent knoll]

what's wrong there

[slender steeple]

as per why, well it's not continuous

[latent knoll]

what does that mean

[slender steeple]

it has a sharp turn

instead of being a smooth curve

[latent knoll]

i think they wont look at that that much tbh

[latent knoll]

is this one correct?

[slender steeple]

the calculations check out

but the graph is wrong sadly

[slender steeple]

i got deducted 2 points because i made a sharp turn in my graph

[latent knoll]

oh

well i've calculated that the extreme has no solutions but got root72 insteas of -40 so that sucks. that's 1 point despite an accurate result. the other part where i've failed is the graph ig

istg if i fail math i'll die

[slender steeple]

just keep doing practice problems and you'll be fine

[latent knoll]

chatgpt told me it was wrong as well and proceeded to draw my graph. oh wait.. he for some reason made it go through 0?? huh??? but my professor told us and chatgpt that if there is a zero point that it should go through it?.. i'm so confused

[slender steeple]

this is because..

HAs only describe the END behaviour

in between the HAs almost anything is fair game

i'd reccomend using test values to check

[latent knoll]

yeah idk how to use that

[slender steeple]

basically

values between the horizontal asymptotes

[latent knoll]

wait, so is my problem is that i'm too close to x and that i did a sharp turn?

[slender steeple]

yes

you're supposed to transition smoothly through the x intercept

if you know what i mean

sorta like this ig

[latent knoll]

oh..

i hope my teacher doesn't minus me for that

[slender steeple]

he most likely will

[latent knoll]

but how was i supposed to know ts

i think that i did give it a little more space on my exam haha

[slender steeple]

definition of a function

[latent knoll]

but i dont rmb if i drew the first or the 2nd graph though lol

[slender steeple]

one of them is that it's continuous

[latent knoll]

i swear if i lose like 4 points and i have 3 extra points and not enough for a passing grade i'll die

[latent knoll]

so if this entire task is 20 points. how many points do u think i'd lose from only that part? cuz the other parts of my graph and calculations are correct

[slender steeple]

i'd say

around 3-5

depends on how important the graph is

[latent knoll]

i did have to calculate the domain, zero points, asymptotes, extremes, write all that correctly and draw it on the graph which is did except for the 2 lines in the middle. do you rly think i'd lose so many points for that?

[slender steeple]

thats how much i normally lose

[slender steeple]

yeah see

graph is smooth an continuous as it should be

[latent knoll]

but why is the left graph going down

instead of up

did i miscalculate smth or what

[slender steeple]

you mustve messed up

on the end behaviour

[latent knoll]

let me calculate it again

i got the correct drawing now

the thing is.. i dont remember how i drew it

i think i might've been drew it this way bcz i rmb paying attention to the 3,99 4,01 -3,01 -2,99... and that is how i drew this graph correctly.. if only i could remember

[slender steeple]

still wrong sadly

[latent knoll]

sharp turn?

[slender steeple]

no

random shift that is not needed

[latent knoll]

eh whatever i cant remember. let's just pray i drew it correctly

only God can save me now

thanks for ur time. i appreciate it

[slender steeple]

ok

[latent knoll]

.close

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[tired kernel]

Ankita travels 14 km to her home partly by rickshaw and partly by bus She takes half an hour if she travels 2 km by rickshaw and remaining distance by bus . On the other hand , if she travels 4 km by by rickshaw and takes 9 mins longer Find the speed of the rickshaw and of the bus .

[tired kernel]

Need guidance how to solve such kind of question

[hard crest]

probably easiest thing to do here is to say the rickhsaw travels R km/h, and the bus travels B km/h

and set up a system of equations

[tired kernel]

took x and y respectively

[hard crest]

actually you might choose to say the rickshaw travels at R h/km instead, probably easier to work the numbers

[hard crest]

okay, that's fine. can you set up the equations?

[tired kernel]

i tried making 2 equations but got stuck at point which makes me think my equations are wrong

[hard crest]

hmm ok

what were your equations and reasoning?

[tired kernel]

i am sharing just wait a moment i am on laptop with no camera so it's little hard to share my progress

equation 1

eq 2

i took the time on one side and added the time taken my both vehicals

after converting it into distance/speed

[hard crest]

right ok yea so

you kinda rocked yourself a little here

but if you write it like this

[tired kernel]

Are the equations correct?

[hard crest]

$4(\f1x) + 16(\f1y) = \f{13}{20}$ etc

[flat frigate]

hayliänus austrǎlis

[hard crest]

then you can solve for 1/x and 1/y

yea they seem right to me

13/20 is 39/60

[tired kernel]

ahh i am confused how do i solve it further? taking 4 common?

[hard crest]

well clearing denominators by multiplying by 20 could be nice, but the point is that you take "1/x" as a Thing

and solve for 1/x and 1/y like you would any system of equations

you can even replace 1/x with X (capital) if you want

[tired kernel]

oh alr got it

let me solve it further like that then

after trying to solve it further i am getting q(1/y) = 7/320 by using elimination methord

which i think might be wrong i think

i took 1/7 as q it is not multiplied up there

oh wait i might have realised by mistake

[hard crest]

q?

[tired kernel]

i took 1/x as p

and 1/y as q

i did 12*2 = 32 by mistake let me correct it

but still getting 7/160

as q

i did this , is this correct?

i tried to elimination of p in this

by multiplying 2p+12q=1/2

into 2

[hard crest]

sure

(maybe do -2)

[tired kernel]

but after doing that and then using elimination

i am getting

8q=7/20

trouble is on the rhs side i think

[hard crest]

what's wrong with it

[tired kernel]

how will i solve 8q=7/20 further?

[hard crest]

well

[tired kernel]

it will become very complex

[hard crest]

seeing as you're multiplying 8 by q, how about try dividing by 8

[tired kernel]

ahh i did'nt understand that

you want me to do it like q=7/160?

[hard crest]

sure

and so now we have 1/y = 7/160

so we can do a little upsy daisy and say y = 160/7

[tired kernel]

yep

[hard crest]

now what units were those?

what was y supposed to mean?

[tired kernel]

speed of bus

x=speed of rickshaw and y as speed of bus

[hard crest]

in what units?

[tired kernel]

km/h

hmm i think i did it wrong

when i subsitute value of y=160/8 into equation

i get x=-80

and it's impossibel for speed to be in negative

i am doing mistake in elimination i think

[hard crest]

no i get the same from your system

let me check over what you did

oh 16q is wrong

should be 10q

[lean swan]

This isn't that hard want me to solve it for you in paper?

it's my first time here im new here

in this server

[tired kernel]

hmm actually i wanted to learn step by step how to solve such problem i do not just want the solution

oh

[lean swan]

k so im in 10th grade so we were thought 2 ways of solving this problem One is the elimination method and the other is substitution method and they both work perfectly fine in these simultaneous pairs of linear equations

which grade are you in by the way

I mean I'm not really a good teacher but I can try to make you understand

[tired kernel]

i got it

in the equation

i wrote 16q

the remaining distance was 10 km not 16

that why i was getting stuck

[lean swan]

This one This one is easy we did in class

[tired kernel]

so how are preparations for boards?

[hard crest]

hey shado i appreciate your help but i think we're pretty much there

[lean swan]

it's from a combined sample paper Ik

[tired kernel]

i am in 9 th i am studying 10 th material

[lean swan]

why tho?

[lean swan]

k well...

[tired kernel]

wait let me just solve the question first

then talk about this

[lean swan]

kk take ur time

[tired kernel]

y=40 km/h speed of bus

x=10 km/h

finally solved

@hard crest Thank you very much for helping

[hard crest]

[tired kernel]

so basically bhaiya i want to go for iit

and there are some class mates of mine who are literally preparing for jee since class 6 th

now they have completed the whole syllabus of 10 th

and heading to study for 11 th instead of preparing for 9 th final

since they have already study all that long ago

so i also started studying 10 th material as soon as i finished 9 th

so i can also head on 11 th

and start preparing for jee as soon as possible

seeing the compition

ahh you here to read all this or should i close the channel ? @lean swan

.close

[safe radish]

Channel closed

Closed by @tired kernel

Use .reopen if this was a mistake.

[lean swan]

yeaaa

correct

[lean swan]

sorry i went out emergency

nice ambition u aiming JEE?

[tired kernel]

yep

to get an iit somehow

[safe radish]

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[lean swan]

me too planning to take PW jee arjuna batch in 11th

[stiff magnet]

ok

[tired kernel]

ahh we opened a help channel by mistake

let's talk in personal dm

[stiff magnet]

offline batches are better

[tired kernel]

.close

i cannot even close it bruh

well nvm btw

what batch should i take

when i go in 11 th?

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[granite flower]

Here you are

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[mellow tide]

So I'm self-studying math right now in order to prepare for my math test next week and I'm currently stuck on a problem in my math book. Since it's in Norwegian I will try to translate closely as possible over to english.

A company got a bill of 47500 kr (Norwegian kr to clarify), to repair a roof of one of the city farms.
The company has 20000kr in their account and bills out around 2500kr for every apartment.

So I solved problem A to make a model being K(x) = 2500x + 20000kr

Now I'm stuck on problem B to find the area of validity. I thought I managed to solve it by doing 2500x/2500 which ends up being 0 since X, and then 20000kr/2500 which is 8. So the X would be between 0 and 8.

The issue here is, once I looked at the answer sheet in the book, it says it's supposed to be that X is between 0 and 16. So I don't quite understand how it became like that. I feel like I'm missing something in order to solve it and I been wrapped around this problem for a while now.
I cannot use any assisted tools like calculator and etc. I can only do it written.

[mint plume]

send the norwegian problem too

!xy

[safe radish]

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.

[mellow tide]

In the first picture is the one I’m doing, the second picture is doing similiar problem in a different section of the book.

In text format:

Et borettslag med 16 leiligheter har fått en regning på 47500kr for reparasjoner av taket på en av bygårdene. Borettslaget har 20000kr på konto og sender i tillegg ut en faktura på 2500kr til hver leilighetene.

Problem A) Lag en modell, K(x), som viser hvor mye borettslaget har på konto etter at x leiligheter har betalt fakturaen. So K(x) = 2500x + 20000kr.
Problem B) (Which I'm currently stuck at), Bestem et gyldighetsområde for modellen.
So I attempted 2500x/2500 which is 0 due to X, then do 20000/2500, which becomes 8. I don't know if the answer sheet is wrong or if it's correct, but the supposed answer is that X is between 0 and 16. I don't really know how they got that answer at all. It feels like I'm missing something to solve it.

[safe radish]

@mellow tide Has your question been resolved?

[mellow tide]

<@&286206848099549185>

Gonna attempt the other problems by myself, but I can't find out how the X is between 0 and 16. I've been trying to brainstorm any possible solutions to find the part that I'm missing in order to solve it.

Nvm, I think I managed to solve it. Since X is between 0 and 8, then the y has to be 16. (0, 16).

.close

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[safe radish]

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[timid echo]

i know the method to solve this but kinda bad at it

[opaque fern]

Elaborate

[daring flume]

$\frac{5^2}{625^5}$?

[flat frigate]

Nicole

[timid echo]

yes

[daring flume]

so what is your confusion with this?

[timid echo]

so um 5^2 ÷ 625^5 = 5^2 ÷ (5^4)^5

so now i cant find the answer, it shows some random ass letters and symbols on cal

[flat frigate]

scoob

[daring flume]

use the power of power rule to settle the denominator first.

[timid echo]

ok

[daring flume]

then it should be clear how to proceed.

[timid echo]

so (just powers) it is 2-20 right?

cause 5^2 and 5^20

[daring flume]

correct.

[timid echo]

ok so the answer results 5^-18

[daring flume]

looks good.

[timid echo]

ok, thanks i couldnt verify, cal shows something weird i havent learnt

[opaque fern]

Good time to stop using the division symbol by the way

Or things will get annoying (and confusing)

[timid echo]

oh yeah but it was hard to do it in the #bots

[daring flume]

prefer using a fraction to express division.

[timid echo]

ok

got it

[opaque fern]

,, \frac ab

[daring flume]

#latex-testing is a good place to experiment too!

[opaque fern]

Thats the LaTeX command if you are curious

[timid echo]

hm ok

[daring flume]

isn't this with a preamble?

[opaque fern]

no

You can try that

[timid echo]

thank you guys, now should i close this or its alright

[daring flume]

,, \frac ab

[flat frigate]

Nicole

[opaque fern]

Yeah close it

[daring flume]

oh, interesting. I'm more used to math mode.

[timid echo]

.close

[safe radish]

Channel closed

Closed by @timid echo

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[timid echo]

so i would wanna right it like this right?

[daring flume]

yes, preferably.

[safe radish]

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[restive turret]

The drawing in notebook is mine with the auxiliary construction, the other with a zoom and that hasn't the line CD, is the original from the original problem.

[restive turret]

Hypothesis or Data: 1. AD is perpendicular to AB, 2. BC is perpendicular to AB. 3. AD ll BC. Auxiliary Construction: It draws a perpendicular from vertex C to D this perpendicular will be called CD. Thesis: Prove that ABCD is a rectangle.

I want someone to check if my proof is correct

The right is the original drawing, the left is mine with the auxiliary construction

[safe radish]

@restive turret Has your question been resolved?

[restive turret]

<@&286206848099549185>

<@&286206848099549185>

[sour bolt]

don’t constantly ping helpers

[sudden seal]

yess

[restive turret]

Srry

But can you help me pls

[sudden seal]

give me 5 minutes, im helping one right now, will be here in 5

[restive turret]

Ok

[safe radish]

@restive turret Has your question been resolved?

[safe radish]

@restive turret Has your question been resolved?

[restive turret]

<@&286206848099549185>

[restive turret]

<@&286206848099549185>

<@&286206848099549185>

Pls i need help

Well a check

<@&286206848099549185>

<@&286206848099549185>

[fickle monolith]

Yo chill

[dreamy meadow]

bro

dont spam

[restive turret]

But pls i need you to check me pls

[dreamy meadow]

If you need help ping once

theres another guy coming to help

[restive turret]

Oh ok

[dreamy meadow]

if you spam ping any more we need to ping a mod

dont spam cuz everyone with that role gets absolutley cooked

[restive turret]

So what do i put?

[dreamy meadow]

read the ticket info, it says "ping helpers once after 15 mins if no one comes"

[restive turret]

Bro it happens 2 hrs

[dreamy meadow]

ping us once if no one comes in 15 mins

ok so dont spam ping then

[restive turret]

Ok

But pls help, i've been wairing 2 hrs

[sudden seal]

ok give it

what u want us to check

ohh fckk my bad

i was working over smtjh

[restive turret]

Hypothesis or Data: 1. AD is perpendicular to AB, 2. BC is perpendicular to AB. 3. AD ll BC. Auxiliary Construction: It draws a perpendicular from vertex C to D this perpendicular will be called CD. Thesis: Prove that ABCD is a rectangle.

The right is the original drawing, the left is mine with the auxiliary construction

[sudden seal]

its only a bit correct

[restive turret]

Why only a bit?

Why not entirely correct??

[sudden seal]

Your written proof does not explicitly justify:

Why CD ∥ AB

Why having two pairs of parallel sides + right angle ⇒ rectangle

Since AD ⟂ AB and BC ⟂ AB, we get AD ∥ BC.
Construct CD ⟂ AD. Since AD ⟂ AB, CD ∥ AB.
Thus, opposite sides are parallel and one angle is right.
Hence, ABCD is a rectangle.

use this it would be good

[restive turret]

Oh so the form that i write is wrong but the purpose is good we can say??

[naive dragon]

It's insufficient, it can only show ABCD is a right trapezoid

[restive turret]

Well no, because all 4 right angles, in a right trapezoid 2 are right and 2 obtuse

[naive dragon]

How can you show there're 4 right angles

[restive turret]

Well by the hypothesis CB is perpendicular to AB, and AD is perpendicular to AB and as CD is a perpendicular, CD is perpendicular to CB and AD

[sudden seal]

here

i got it on ss

[naive dragon]

Now you have to show why CD is perpendicular to CB

That would be all you need to complete the proof

[naive dragon]

Lol no

[restive turret]

Yeah but CD is a perpendicular from vertex C to D, so CD is perpendicular og CB and AD

[sudden seal]

bro we can construct CD perpendicular AD and since AD is already perpendicular to AB, we willl have CD parallel to AB

[naive dragon]

Yes we will but your job is to show why

I meant OP's

[sudden seal]

yea i got i

it

[restive turret]

Emm, what is OP's?

[sudden seal]

both pairs of opposite sides are parallel, so ABCD is parallelogram and because DAB = 90*, ABCD IS rectangle

[naive dragon]

Yeah I have no idea what you meant here ngl

[naive dragon]

Original Poster, in this case you

[sudden seal]

im just gonna send the whole solution

[naive dragon]

!nosols lol

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[sudden seal]

oopss

[naive dragon]

Why do you think I'am doing this for

[sudden seal]

it slipped from my mindd

[restive turret]

I'm gonna write it

And you check me

[sudden seal]

i have been awake from quitee too mcuh time

my head is goinn

WIII

[restive turret]

Gimme 10 min

[sudden seal]

hm

[restive turret]

And i'll write it and check

Me pls

To finish this odyssey

@sudden seal @naive dragon

Only check me

Pls

And we finish this odyssey

[naive dragon]

The wording in this problem is weird, so CD perpendicular to AD is what given

Yeah that seems good

[restive turret]

Ok

Finally

So

Thanks!!!!!!!!!!!! @naive dragon @sudden seal @dreamy meadow @sour bolt @fickle monolith

.close

[safe radish]

Channel closed

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[safe radish]

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[verbal flare]

hello

[slate pebble]

what is your question

[verbal flare]

i need help understanding this

trigo

[open junco]

You were on the right track to solve the question tho?

[verbal flare]

was i?

[open junco]

Im 90 percent sure

[verbal flare]

alr

uhh

cos like im really fing clueless after bi done that 😭

[open junco]

Like assume the case where tanx is not 0 and cancel out the numerator?

[open junco]

@verbal flare

[verbal flare]

uhh

yea im still here

im racking my brain

what d you mean assume the case where tanx is not 0 bruhhhhhh

not like im criticizing you but i cant seem to understand

[slate pebble]

if tan(x) is zero you would be dividing by zero

[open junco]

I mean divide both sides by tanx yk and you cant divide by 0

[slate pebble]

but because it is stated in the original equation that tan(x) is in the denominator we can assume it is not zero

[verbal flare]

ohhhh ok

i understand why its not zero

[open junco]

We dont have to assume ig its just given because of the original equation

[verbal flare]

ok

[slate pebble]

that's true

mb

[verbal flare]

alr so

$tan 2x / tan x$

[flat frigate]

isa.

[verbal flare]

so do i use the double angle idenities in the numerator?

like

[icy aspen]

Yes that would help

[open junco]

Yes

[verbal flare]

so

hold on ill be back

i need to reset my ipad

because of some keyflicks

so

$(2tan x/1-tan^2x)/tanx$

[flat frigate]

isa.

[open junco]

Correct

[verbal flare]

alr thx

[open junco]

Youve got it from here?

[verbal flare]

yes ttsm

tysm

i undertsand

tjhxxxxxxx

.close

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[vocal stirrup]

000.33333333330000*10⁹

[vocal stirrup]

I understand I have to multiply, but I don't understand how

I haven't done any math in three years because I dropped out of middle school, so I was trying to refresh my memory before getting back into school but everything looks scary and confusing, I asked around for help and I still couldn't understand a lot of it

[winged flare]

shift the dot right 9 times

[vocal stirrup]

Like this?
000333333333.30000

[winged flare]

remove the 10^9 here but yes

just 000333333333.30000

[vocal stirrup]

Ok got it

That's the answer?

[winged flare]

yeah

when multiplying by 10^n shift the dot right n times

for positive n

[vocal stirrup]

Ohh I see thank you!! 🩷

.close

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[solar sand]

Are there any alternative ways to find the matrices A^2 A^3 A^4 etc to compute Bn matrix while solving the questions of path matrix? Manually counting the paths causes a lot of error in exams, so was wondering if there are alternative ways to find those matrices easily

[mortal sandal]

Well, there's matrix multiplication

Especially if you have access to a graphing calculator during exams

[solar sand]

my calculator can only hold upto only 3*3 matrix 💀

[mortal sandal]

what kinda calculator does that 💀

ig manual matrix multiplication then

[solar sand]

ugh there are no other ways?

[mortal sandal]

Well you don't need to find A^3 to find A^4

You can just square A^2 again

But if you don't have a calculator that can handle them then as far as I know it's manual matrix mult or count paths

What kind of calculator do you even have

[solar sand]

a basic scientific calculator

[mortal sandal]

oh I didn't know those had matrices at all

welp if you're allowed to use a graphing calculator you can buy one or see if you can borrow one

[solar sand]

alright

thanks

.close

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[mortal sandal]

np

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[quiet hollow]

I’m stuck on this question bc I can’t find any common factors between 2 & 9

[lean otter]

what does the question ask to do

[open wedge]

factor?

[open wedge]

-# completing the square's the best shot

[lean otter]

the problem is theres no RHS

unless you assume all this is 0

[open wedge]

then it has to be factorizing

[lean otter]

mid-term?

[open wedge]

ay ay

don't spoil the answers yet

[lean otter]

oops

[open wedge]

i got several methods to introduce

3 specifically

[lean otter]

do you know what mid-term factorization is?

[open wedge]

this is 1 method

but i don't really use that

[burnt notch]

Me neither, actually in Italy we aren't even taught it

[lean otter]

I do completing square if one value of x is a fraction

otherwise its mid term

its a matter of preference really

[open wedge]

that's the second method, but i only use that in 1-variable quadratic inequalities questions and some other questions

the 3rd method is the one i use the most

it's to find the roots of the polynomial

[quiet hollow]

It’s Triomal factoring

[lean otter]

then you are looking for mid term split i assume

[open wedge]

multiply 2 with -9. the 2 numbers you need to find should add together to get 3 and multiply to get -18

[lean otter]

have you factorised quadratic before?

[burnt notch]

Do you have your class notes on that topic? If so, can you send them?

[open wedge]

then split the midterm into (first number)x + (second number)x

that's midterm split

[lean otter]

-# too many people i back out

[burnt notch]

<@&268886789983436800>

[raw pivot]

<@&268886789983436800>

[quiet hollow]

Yeah that’s where I’m stuck. I used 2(-9) to get -18 but idk what I’m supposed to do with that 3 afterwards? I never saw my teacher use the C. The next equation would be smth like:

(2p^2 -9) (2p-9)

[open wedge]

you need to find 2 numbers that multiply to -18

and add to 3

[lean otter]

what were trying to do with split up the 3

into 2 new factors/things

so when you split it, the numbers should multiply to -18 and add up to 3

[quiet hollow]

You got like a visual I can follow

Bc I don’t get I

It

[open wedge]

search up on google for midsplit

[lean otter]

uh wait let me doodle it for you

[open wedge]

$x^2 - 5x + 6$ for example

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

two numbers that times to $1 \cdot 6$ and add up to $-5$ is $-2$ and $-3$

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

so i'd split $x^2 - 5x + 6$ into $x^2 - 2x - 3x + 6$

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

$= x(x - 2) - 3(x - 2) = (x - 2)(x - 3)$

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

done factorizing

[lean otter]

aaand I cant paste images

great

[open wedge]

lol

[lean otter]

nvm it works now

my handwriting is a bit messy

[open wedge]

oh my doodle

[lean otter]

but you get the gist of it

WAIT WAIT LET ME MAKE A BETTER ONE

Stop roasting 😭

[open wedge]

online lagrida if you know $\LaTeX$

[flat frigate]

1 divided by 0 equals Infinity

[quiet hollow]

I understood everything but what that last row is supposed to say

[open wedge]

well idk

u + y are 2 numbers that add up to 3

and times to make -18 i suppose?

@quiet hollow

[quiet hollow]

I already have my -18 by doing 9(-2)

[lean otter]

.

[open wedge]

you need to find

2 numbers

[open wedge]

that add up to -18

and times to make 3

[lean otter]

I know $0$ LaTeX

[flat frigate]

noonebtw

[open wedge]

9 + (-2) is not equal to 3

[open wedge]

you just knew a lil bit

[lean otter]

yeah

actually

try and find the number yourself

[open wedge]

refer to this example @quiet hollow

[lean otter]

hint: both are divisible by 3

[open wedge]

that's why you put u and y

smart

i have 2 more ways to introduce dw

[lean otter]

thats x

[open wedge]

complete the square

[lean otter]

but close enough 😆

[open wedge]

and find the roots

[lean otter]

you can disregard completing the square here

since theres no RHS

[quiet hollow]

I used an X method

[open wedge]

eh?

[quiet hollow]

Ok

.close

[safe radish]

Channel closed

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[lean otter]

whar

[safe radish]

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[open wedge]

i can still do it without the RHS

[open wedge]

.close

[safe radish]

Channel closed

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[lean otter]

ahh

looks "complex"

compared to something like a mid term

[open wedge]

imagine accidentally opening a channel...

[lean otter]

no need to

[open wedge]

if you send a message into an empty channel => you claimed the channel

[lean otter]

I am aware

[open wedge]

ts rarely happens

it's like the 1st time happening to me or smth

[safe radish]

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[earnest mirage]

Hello, could someone check if this proof looks good please?
⁨```latex
\begin{Theorem}
Assume $n \in \bZ$. If $5n^2 + 3$ is even, then $n$ is odd.
\end{Theorem}

\begin{proof}
We use the contrapositive.
Suppose $n$ is even.
Then, $n = 2k$ for some integer $k$.
So, $5n^2 + 3 = 5(2k)^2 + 3 = 20k^2 + 2 + 1 = 2(10k^2 + 1) + 1$.
Since $k$ is an integer, so is $10k^2 + 1$.
Then, $5n^2 + 3 = 2l + 1$, where $l = 10k^2 + 1$.
Thus, by the definition of odd integers, $5n^2 + 3$ is odd.
Therefore, by the contrapositive, if $5n^2 + 3$ is even, then $n$ is odd.
\end{proof}

[flat frigate]

Mor Bras

[tardy mango]

Maybe explicitly state the contrapositive? But this is fine.

[safe radish]

@earnest mirage Has your question been resolved?

[earnest mirage]

Thanks!

[safe radish]

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[earnest mirage]

Hello, could someone check if this proof looks good please?
⁨```latex
\begin{Theorem}
Assume $n \in \bZ$. If $8 \nmid (n^2 - 1)$, then $n$ is even.
\end{Theorem}

\begin{proof}
We use the contrapositive.
Suppose $n$ is odd.
Then, $n = 2(2k) + 1 = 4k + 1$ for some integer $k$.
So, $n^2 - 1 = (4k + 1)^2 - 1 = 16k^2 + 8k + 1 - 1 = 8(2k^2 + k)$.
Since $k$ is an integer, so is $2k^2 + k$.
Thus, by the definition of divisibility, $8 \mid n^2 - 1$.
Therefore, by the contrapositive, if $8 \nmid (n^2 - 1)$, then $n$ is even.
\end{proof}

[flat frigate]

Mor Bras

[stoic dune]

Suppose n is odd: I'll choose n = 3.

You suggest that 3 = 4k + 1 for some integer k. What's k?

I see that n is not just any odd, but a specific odd. An odd kind of odd

But ultimately you've now proven that if 8 doesn't divide (n² - 1), then n is not of the form 4k + 1

[earnest mirage]

Yes, I see

[visual kraken]

After fixing the part Kaynex mentioned you'll probably have just one extra step, but otherwise the proof looks nice

[earnest mirage]

I'm writing a fix for this

Here's the fix:
⁨```latex
\begin{Theorem}
Assume $n \in \bZ$. If $8 \nmid (n^2 - 1)$, then $n$ is even.
\end{Theorem}

\begin{proof}
We use the contrapositive.
Suppose $n$ is odd.
Then, $n = 2k + 1$ for some integer $k$.
Now, either $k$ is even or odd.

\begin{case}
Suppose $k$ is even.
Then, $k = 2l$ for some integer $l$, and then $n = 2(2l) + 1 = 4l + 1$.
So, $n^2 - 1 = (4l + 1)^2 - 1 = 16l^2 + 8l + 1 - 1 = 8(2l^2 + l)$.
Since $l$ is an integer, so is $2l^2 + l$.
Thus, by the definition of divisibility, $8 \mid n^2 - 1$.
\end{case}

\begin{case}
Suppose $k$ is odd.
Then, $k = 2l + 1$ for some integer $l$, and then $n = 2(2l + 1) + 1 = 4l + 3$.
So, $n^2 - 1 = (4l + 3)^2 - 1 = 16l^2 + 24l + 9 - 1 = 8(2l^2 + 24l + 1)$.
Since $l$ is an integer, so is $2l^2 + 24l + 1$.
Thus, by the definition of divisibility, $8 \mid n^2 - 1$.
\end{case}

In both cases, we have that if $n$ is odd, then $8 \mid n^2 - 1$.
Therefore, by the contrapositive, if $8 \nmid (n^2 - 1)$, then $n$ is even.
\end{proof}

[flat frigate]

Mor Bras

[earnest mirage]

Fixed 👌
⁨⁨```latex
\begin{proof}
We use the contrapositive.
Suppose $n$ is odd.
Then, $n = 2k + 1$ for some integer $k$.
Now, either $k$ is even or odd.

\begin{case}
Suppose $k$ is even.
Then, $k = 2l$ for some integer $l$, and then $n = 2(2l) + 1 = 4l + 1$.
So, $n^2 - 1 = (4l + 1)^2 - 1 = 16l^2 + 8l + 1 - 1 = 8(2l^2 + l)$.
Since $l$ is an integer, so is $2l^2 + l$.
Thus, by the definition of divisibility, $8 \mid n^2 - 1$.
\end{case}

\begin{case}
Suppose $k$ is odd.
Then, $k = 2l + 1$ for some integer $l$, and then $n = 2(2l + 1) + 1 = 4l + 3$.
So, $n^2 - 1 = (4l + 3)^2 - 1 = 16l^2 + 24l + 9 - 1 = 8(2l^2 + 3l + 1)$.
Since $l$ is an integer, so is $2l^2 + 3l + 1$.
Thus, by the definition of divisibility, $8 \mid n^2 - 1$.
\end{case}

In both cases, we have that if $n$ is odd, then $8 \mid n^2 - 1$.
Therefore, by the contrapositive, if $8 \nmid (n^2 - 1)$, then $n$ is even.
\end{proof}

[queen ingot]

looks good besides that one issue

[flat frigate]

Mor Bras

[earnest mirage]

Thanks for checking it!

.close

[safe radish]

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Closed by @earnest mirage

Use .reopen if this was a mistake.

[safe radish]

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[vagrant steppe]

How come we don't times the ^-2 or 2 to the imaginary ^1 the 1 has

[normal moss]

what?

[vagrant steppe]

Or is that not a rule in these kind of situations

[solar hazel]

what 1?

[normal moss]

I am sorry I just don't understand what you meant

So I can't answer that

[vagrant steppe]

Wait I think I understand

[normal moss]

Ah

[vagrant steppe]

Right?

Or should I have flipped the 1 and exponent first

[solar hazel]

are you talking about like (a^b)^c = a^(bc)?

[vagrant steppe]

Because like I know 1^2 is just 1
But don't that 1 really have a ^1 to it too

[solar hazel]

i guess but it doesn't make a difference

[vagrant steppe]

I guess my question now is how would you have worked it out

Bc I flipped it last

[burnt notch]

(stuff² = stuff × stuff)

[vagrant steppe]

But maybe I should have done that first

Should I have flipped to get rid of the ^- first

[burnt notch]

<@&268886789983436800>

[burnt notch]

You can choose the order you prefer, because $(a^b)^c = (a^c)^b$, since they both equal $a^{bc}$

[flat frigate]

Alberto Z.

[burnt notch]

(That happens because b•c = c•b, since multiplication is commutative)

[vagrant steppe]

I'm just confused because -1^-2 is -1

And then the exponent would be -1^-2

Because that 1 has an ^1 and the ^-2 times it which then again just gives you negative one

So that's why I'm wondering if flipping first just idk

Because if I flip it now I still get stuck with a -1

WAOT

No

[burnt notch]

$$(-1)^{-2} \neq -1^{-2}$$

[vagrant steppe]

Bc -1*-1 is 1

[flat frigate]

Alberto Z.

[vagrant steppe]

Oh

So here I wouldn't ex knowledge how the -1 technically has a ^1?

[burnt notch]

$$(-1)^{-2} = \frac{1}{(-1)^2} = \frac{1}{(-1) \times (-1)} = \frac11 = 1$$

[flat frigate]

Alberto Z.

[burnt notch]

Wdym, sorry?

[vagrant steppe]

Doesnt the 1 have an imaginary exponent of 1?

[normal moss]

Almost everything has an exponent of 1

"imaginary" is unfortunate wording though

[vagrant steppe]

She sorry

Ah*

[normal moss]

4^2

is just (4^2)^1

[vagrant steppe]

Is that an example

[normal moss]

Yes

[vagrant steppe]

It's just my fault I over think and complicate everything lol

I think I'm okay

Thank you guys I'm sorry that was stupid

[normal moss]

Well but do you get it now?

No need to apologize

[vagrant steppe]

I think

[safe radish]

@vagrant steppe Has your question been resolved?

Channel closed

Closed by @vagrant steppe

Use .reopen if this was a mistake.

[vagrant steppe]

.close

[safe radish]

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[earnest mirage]

Hello, could someone check if this proof looks good please?
⁨⁨⁨⁨⁨```latex
\begin{Theorem}
Assume $m,n \in \bZ$. If $m^2(n^2 + 5)$ is even, then $m$ is even or $n$ is odd.
\end{Theorem}

\begin{proof}
We use the contrapositive.
Suppose $m$ is odd and $n$ is even.
Then, by the definition of odd and even integers, $m = 2k + 1$ and $n = 2l$ for some integers $k$ and $l$.
And so,
\begin{align*}
m^2(n^2 + 5) &= (2k + 1)^2((2l)^2 + 5)\
&= (2k^2 + 4k + 1)(4l^2 + 5)\
&= 16k^2l^2 + 20k^2 + 16kl^2 + 20k + 4l^2 + 4 + 1\
&= 2(8k^2l^2 + 10k^2 + 8kl^2 + 10k + 2l^2 + 2) + 1\
\end{align*}

Since $k$ and $l$ are integers, so is $8k^2l^2 + 10k^2 + 8kl^2 + 10k + 2l^2 + 2$.
Thus, by the definition of odd integers, $m^2(n^2 + 5)$ is odd.
Therefore, by the contrapositive, if $m^2(n^2 + 5)$ is even, then $m$ is even or $n$ is odd.
\end{proof}

[earnest mirage]

Typo: fixed 👌

[wild cape]

the claim is supposed to say 5 instead of 1?

[earnest mirage]

The theorem states m^2(n^2 + 5)

[wild cape]

ok so typos in conclusions too

[earnest mirage]

Fixed

[wild cape]

,w expand (2k+1)^2((2l)^2+5)

[flat frigate]

Results provided by WolframAlpha

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

Requested by rokettojanpu

[woven hound]

I think you've way overcomplicated this

[flat frigate]

Mor Bras

[woven hound]

Like the approach is technically correct but a bit overkill I think

I would do assume m^2 (n^2 + 5) is even, and assume m is odd.
Goal: Show n is odd

[wild cape]

the algebra is right so proof is good

[earnest mirage]

For sure, nevertheless, it was the only way I knew how to proceed

[earnest mirage]

m odd and n even are a give by the contrapositive

[woven hound]

That way doesn't use the contrapositive

if you want to try that approach it's a direct proof

[earnest mirage]

The exercise stated to use the contrapositive

[woven hound]

oh I see

[earnest mirage]

Otherwise, by direct proof, then only given is that m2(n2 + 5) is even

[woven hound]

and assuming m is odd

[safe radish]

@earnest mirage Has your question been resolved?

[earnest mirage]

Why m odd can de assumed?

[woven hound]

An equivalent statement to (P or Q) is (not P) implies Q

Direct proof is:

1. Assume m^2(n^2+5) is even
2. Show m is even OR n is odd

You can do step 2 by assuming m is odd implies n is odd

[earnest mirage]

Assuming m odd makes "m odd or n even" true automatically, isnt it?

Oh I see

Yes

[earnest mirage]

Thanks for the advice!

.close

[safe radish]

Channel closed

Closed by @earnest mirage

Use .reopen if this was a mistake.

[safe radish]

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[earnest mirage]

Hello, could someone check if this proof looks good please?
⁨```latex
\begin{Theorem}
Assume $m,n \in \bZ$. If $(m^2 + 4)(n^2 - 2mn)$ is odd, then $m$ and $n$ are even.
\end{Theorem}

\begin{proof}
We use the contrapositive.
Suppose either $m$ is odd or $n$ is odd.

\begin{case}
Assume $m$ is odd.
Then, by the definition of odd integers, $m = 2k + 1$ for an integer $k$.
So, $(m^2 + 4)(n^2 - 2mn) = (4k^2 + 4)(n^2 - 4kn) = 2(2k^2 + 2)(n^2 - 4kn)$.
Since $k$ and $n$ are integers, so is $(2k^2 + 2)(n^2 - 4kn)$.
Thus, by definition of even integers, $(m^2 + 4)(n^2 - 2mn)$ is even.
\end{case}
\begin{case}
Now assume $n$ is odd.
Then, by the definition of odd integers, $n = 2k + 1$ for an integer $k$.
So, $(m^2 + 4)(n^2 - 2mn) = (m^2 + 4)(4k^2 - 4mk) = 2(m^2 + 4)(2k^2 - 2mk)$.
Since $k$ and $m$ are integers, so is $(m^2 + 4)(2k^2 - 2mk)$.
Thus, by definition of even integers, $(m^2 + 4)(n^2 - 2mn)$ is even.
\end{case}

We showed that if $m$ is odd or $n$ is odd, then $(m^2 + 4)(n^2 - 2mn)$ is even.
Therefore, by the contrapositive, if $(m^2 + 4)(n^2 - 2mn)$ is odd, then $m$ and $n$ is even.
\end{proof}

[flat frigate]

Mor Bras

[hard crest]

it's probably fine but you've posted a lot of these very sledgehammery proofs so far. Do you not have something that says if an odd number is equal to a * b, then both a and b must be odd?

[earnest mirage]

Not really, as far as I can see

[hard crest]

wait it's not fine, i don't get this part --- it looks like you've assumed m = 2k?

wait this isn't even true

m = n = 1
(m^2 + 4)n(n-2m) = 5 * 1 * -1 = -5

[earnest mirage]

I made a mistake typing the theorem and the proof, let me fix it

[austere goblet]

it also seems like you may find it useful to check your own proofs

while we can give you second opinions, it's always good to be able to do so without external assistance (or with minimal assistance)

esp. on typos and algebra mistakes

[earnest mirage]

⁨```latex
\begin{Theorem}
Assume $m,n \in \bZ$. If $(m^2 + 4)(n^2 - 2mn)$ is odd, then $m$ and $n$ are odd.
\end{Theorem}

\begin{proof}
We use the contrapositive.
Suppose either $m$ is even or $n$ is even.

\begin{case}
Assume $m$ is even.
Then, by the definition of even integers, $m = 2k$ for an integer $k$.
So, $(m^2 + 4)(n^2 - 2mn) = (4k^2 + 4)(n^2 - 4kn) = 2(2k^2 + 2)(n^2 - 4kn)$.
Since $k$ and $n$ are integers, so is $(2k^2 + 2)(n^2 - 4kn)$.
Thus, by definition of even integers, $(m^2 + 4)(n^2 - 2mn)$ is even.
\end{case}
\begin{case}
Now assume $n$ is even.
Then, by the definition of even integers, $n = 2k$ for an integer $k$.
So, $(m^2 + 4)(n^2 - 2mn) = (m^2 + 4)(4k^2 - 4mk) = 2(m^2 + 4)(2k^2 - 2mk)$.
Since $k$ and $m$ are integers, so is $(m^2 + 4)(2k^2 - 2mk)$.
Thus, by definition of even integers, $(m^2 + 4)(n^2 - 2mn)$ is even.
\end{case}

We showed that if $m$ is even or $n$ is even, then $(m^2 + 4)(n^2 - 2mn)$ is even.
Therefore, by the contrapositive, if $(m^2 + 4)(n^2 - 2mn)$ is odd, then $m$ and $n$ is even.
\end{proof}

[flat frigate]

Mor Bras

[austere goblet]

maybe on your next attempt, you can show what you think is mistaken that warrants a second opinion from us

helps build your self-confidence too

[earnest mirage]

I understand that, when writing, it's useful to check one's writing before requiring assistant to check the writing, specially for long writings

[austere goblet]

mhm. the idea is that you can produce and verify proofs

I don't see anything wrong on a skim but I'm outside atm

[earnest mirage]

I also understand the work volunteers put in checking my work whenever possible, and I appreciate it whenever someone reads and point out the errors, or give their approval, so thank you for your time!

[safe radish]

@earnest mirage Has your question been resolved?

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[lean otter]

is there a way to simplify general solutions for linear diophantine equations? when finding a solution, using the euclidian algorithm is kinda annoying sometimes as it gives rise to larger numbers, which create problems when dealing with stuff like finding the smallest positive integer solution,

for example i was doing this problem:
Mr smith cashed in a cheque in a bank, the teller mistook the number of cents for number of dollars and vice versa. Unaware of this, he spent 68 cents and then noticed that he had twice the amount of the original cheque. determine the smallest value the cheque couldve been written

let x=$, y=¢, then i got the general solution as x=2254-98t and y=4554-199t, which is pretty much impossible to do cuz there is a solution but its from a non integer value of t which is impossible to guess

[quasi bison]

wdym non integer value of t

can you show all your work please

[lean otter]

well its js messy rough work, but u can understand, the minimum value is $10.21, which cant be obtained the general solution i have with integral values of t

[quasi bison]

if you can't get this with integer t then your general solution must not be as general as you claim

so you must have screwed up somewhere

but diagnosis requires seeing all of your work laid out clearly

[lean otter]

yeah, im asking HOW to make it more general (this is my alt )