help-0

[formal hornet]

find proof that

?

well ok

so

since be is

b is

(-2, -1)

and m is

(4, 2)

the difference is

6 and 3 right

so C is

(10, 5)

[tawdry saffron]

this is the exact triangle btw

to visualise it easier

[formal hornet]

is C 10, 5

-2 + 6 = 4

-1 + 3 = 2

so 4 + 6 = 10

2 + 3 = 5

[tawdry saffron]

wait yea

I was wrong lmoa

[formal hornet]

wait am i right or

[tawdry saffron]

hold on

yea

you are

so proving it's a right angle

[formal hornet]

omg

bro its the first time

im right

for like

10 fuckinf yeras

i was always wrong

iwas in extended maths my entire life

[tawdry saffron]

you're doing really well

[formal hornet]

and i was shit idfk how i even

passed i wanted to

kms so many times cuz

i couldnt getthsi shit

😭

[tawdry saffron]

I'm barely doing anything except tiny hints

[formal hornet]

ok so

to prove its right angle

i can prove that they are

perpendicular

to prove they are perpendicular i need to rpove that

m1 * m2 = -1

right

so slope of AB * slope of AC = -1

AB slope = 1

i just have to prove that AC slope = -1

so

[tawdry saffron]

that's one way to do it

[formal hornet]

A (7, 8) C (10, 5)

[tawdry saffron]

you can also try out pythagoras

[formal hornet]

8-5 / 7-10 =

[tawdry saffron]

and see if it applies here

[formal hornet]

3 / -3 = -1

1 * -1 = -1

so AB is perpendicular to AC

[tawdry saffron]

yep

[formal hornet]

which means it is a right angle

whats the other way?

[tawdry saffron]

that's the first question solved

[tawdry saffron]

pythagoras

and many others probably

pythagoras is slower though

so yours was the fasters

[formal hornet]

ok lets do 2

[tawdry saffron]

on to the 2nd question

[formal hornet]

wait parralelogram is like

whats the symbol for parrarlel again

arrows

[tawdry saffron]

idk about the symbol

[formal hornet]

?

so does that mean

AD = BC?

[tawdry saffron]

yes

[formal hornet]

i mean that doesnt help

cuz 2 variables

and AB = CD

😭

it doesnt help either

fuckkkkkkkkkkkk what do i do

[tawdry saffron]

start with CD

[formal hornet]

k uh

20/-13

20 / -13 = a / b + 1?

[tawdry saffron]

20/-13 is the slope

get the whole function first

[formal hornet]

o ok

21 = 20/-13 (2) + c

21 = 40/-13 + c

273 = 40 + 13c

233 = 13 c

c = 233/13

uhhh

brb

[tawdry saffron]

wrong btw

21=-(20/13) + c

just do

21+20/13 = c

[formal hornet]

ok

uhhhh

back

well

my frined told me

to do this

[tawdry saffron]

do what

[formal hornet]

but then he had to go

:/

i thik he said like

AC = AD or something

ok ineed to do someting for around 10 mins sprry

[tawdry saffron]

that is far from correct

[tawdry saffron]

ping me when you're back

[formal hornet]

ok

@tawdry saffron

okay so i think

he meant

AC = BD

slope of AC is

[tawdry saffron]

yea

[formal hornet]

a - 1 / -16

slope of BD is

b - 2 / -21

a-1/16 = b-2/-21

20/-13 = a/b+1

so uh

tytgfrtyuj

AD = -3 / a - 21

[tawdry saffron]

no

[formal hornet]

fuck how do i do this

😭

[tawdry saffron]

you're getting there

just some tiny mistakes

[formal hornet]

how do i get the

fraction thing here

cuz its rlly annoying to have the /

[tawdry saffron]

where does this 16 come from

[formal hornet]

wait i messed up

AC = BD

a - 1 / - 16 = b - 2 / - 21

there

[tawdry saffron]

the easiest way to do it

is

[formal hornet]

AD = a - 21 / -3

[tawdry saffron]

AB = CD

-20/13 = 0-a/b+1

then you can easily see

[formal hornet]

AB = CD

b - 1 / -a = 13 / - 20

[tawdry saffron]

the solution

[tawdry saffron]

no

[formal hornet]

heres how my teacher did it

[tawdry saffron]

yea

but that's the long way

[formal hornet]

whats the short way

[tawdry saffron]

(0-a)/(b+1) = -20/13

[formal hornet]

so -a = -20

and b + 1 = 13

so b = 12 and

a = 20

[tawdry saffron]

yes

[formal hornet]

but is there like a way to prove it

cuz like

-20/13 can also be -40/26

and then a and b wouldnt work

i mean they would but we cant find the exact values then

[tawdry saffron]

ofcourse

we know how CD looks like

now draw AB

with your points

and see if it's as long as CD

[formal hornet]

ohhhhhhhhh

so like

based off the drawing

we can see that

[tawdry saffron]

it looks like this

[formal hornet]

they must be equal length

[tawdry saffron]

too lazy to connect AD and BC

[tawdry saffron]

yea

[formal hornet]

oh

3 now

[tawdry saffron]

basically same story

start by finding AB and BC

[formal hornet]

cant we just find midpoint of AC

AB = 5/4

BC = 32/11

so

now

AB = y = 5/4x + c

4 = 5/4(3) + c

[tawdry saffron]

one sec

[formal hornet]

4 = 15/4 + c

16 = 15 + c

c = 1

AB = 5/4x + 1

4 = 32/11(3) + c

[tawdry saffron]

the easiest way to do it

[formal hornet]

4 = 96/11 + c

44 = 96 + c

[tawdry saffron]

is finding the midpoint of AC first

[formal hornet]

breh

[tawdry saffron]

sorry

but let me read what you're doing

[formal hornet]

28 + 14 / 2, 24 + 36 /2

[tawdry saffron]

I'm a bit behind

[formal hornet]

42 /2 = 21

21, 30

(21, 30)

[tawdry saffron]

this should be 4/5

btw

[formal hornet]

o

[tawdry saffron]

so you're wrong from the beginning

xD

[formal hornet]

fuc

[tawdry saffron]

but let me introduce you a faster way

[formal hornet]

y = 4/5x + c

[tawdry saffron]

hold on

find the midpoint of AC

first

and use D = (u, v)

to find the midpoint of BD

[formal hornet]

hol up what

3 + u /2 , 4 + v/2????

[tawdry saffron]

everything will become clear

[formal hornet]

so uh

[tawdry saffron]

yes

[formal hornet]

42 = 3 + u

and

[tawdry saffron]

but you skipped this

[formal hornet]

i found it

[tawdry saffron]

oh

where

[formal hornet]

here

[formal hornet]

here

[tawdry saffron]

okay is correct

[formal hornet]

42 = 3 + u
60 = 4 + v

u = 39
v = 54

[tawdry saffron]

yes

[formal hornet]

(21, 30) and (39, 54)

so now

i fidn the equation for

[tawdry saffron]

wasn't that the last part of the question

[formal hornet]

AC and DB

BD*

oh

wait we need the intersection too tho

so to find intersection we need to find

equaiton of

[tawdry saffron]

yea but that's at (21,30)

[formal hornet]

AC and BD

huh

thats midpoint of AC tho

[tawdry saffron]

yea but they intersect in the middle

so it's also M

[formal hornet]

o

so even if we set the

equations of

AC and DB

equal

we still get

[tawdry saffron]

we actually used that fact to find D here

[formal hornet]

(21,30)

?

[tawdry saffron]

yea

[formal hornet]

i. c

ok i might have to gtg

im gonna brush my teeth

and then we can do the rest on phone

[tawdry saffron]

you have mroe?

send picture

[formal hornet]

worksheet 2

worksheet 3

ok thats it theres more but they are more easy ones

wanna do this in dms

[tawdry saffron]

sure

[formal hornet]

so that i dont take this help space?

ok add me

im gonna brb in 5 minis

u might have to graph since ill be on phone

[tawdry saffron]

kk

[lone heart]

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[lone heart]

Please don't occupy multiple help channels.

[royal tapir]

.close.

[lone heart]

@royal tapir Has your question been resolved?

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[lone heart]

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[olive citrus]

i want to learn Karnaugh maps but i dont know how, any good learning resources for it?

[olive citrus]

i understand basic programming, and boolean algebra and how to simplify boolean expressions

karnaugh maps dont seem too difficult, its just that there are some minor things that i dont understand that are blocking me from learning it

[chrome plank]

Karnaugh maps are great, I suggest learning them if you like logic boolean algebra

there are some minor things that i dont understand that are blocking me from learning it
What exactly?

To ask for resources, #book-recommendations may be a better channel (it's fine for non-book resources too)

[olive citrus]

this video states that because all outputs are 1 when A is 1, A should be part of the expression. However that's inconsistent because that would no long be true if every output in the map was 1

[ancient saddle]

Every group that you make represents a term, then the simplified boolean expression is the sum of these terms. Each term is the product of the variables that don't change in its group

[lone heart]

@olive citrus Has your question been resolved?

[chrome plank]

Here's an example with a table with all 1s as output, you can group them however you like as long as you respect the pattern™
(and group all the 1s of course)

[lone heart]

@olive citrus Has your question been resolved?

[lone heart]

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[alpine sable]

This is my work

the question is

Laura, Henry, and Carlos have a total of 106 in their wallets. Carlos has 2 times what Henry has. Laura has 6 more than Henry. How much does each have

.close

[lone heart]

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[edgy sand]

hey how can I find the local extrema of this function?

I believe local extremas occur when the derivative of the slope is 0 ? but not sure what to do otherwise

[buoyant kayak]

"the derivative of the slope" doesn't make a lot of sense here

you'd just want to refer to either the derivative or the slope

[edgy sand]

oh

I meant when the derivative of the function is equal to 0

would the x value for that be the place at which a local extrema occurs?

[heady pollen]

yes

however, these include turning points

[edgy sand]

the values that I will get for x?

[heady pollen]

yes

did you take the derivative?

[edgy sand]

no not yet

so how will I know which ones are the local extremas from those x values ?

[heady pollen]

if you want to know if one of these solutions is a maxima, you need the second derivative

[edgy sand]

so I need to set the second derivative of the function equal to 0 and solve for x?

[heady pollen]

for turning points the second derivative is 0
for maxima the second derivative is negative
for minima the second derivative is positive

first you need the first derivative

set that to 0

to find all extrema

then you put these in the second derivative

[blissful whale]

youcan also analyze sign of first diervative

maximal points first increase and then decrease

i.e. positive first derivative than negativer

[buoyant kayak]

calculating the 2nd derivative is unnecessary

[heady pollen]

not really

if you want to know if it is a maxima or a minima, then the second derivative is the best way

[blissful whale]

Matter of viewpoimnt

[heady pollen]

i know there are functions where that is obvious but calculating is advised i would say

[buoyant kayak]

only true with simple functions

[blissful whale]

I think first derivative sign analysis is faster

[buoyant kayak]

[heady pollen]

maybe true

i guess im just slow with that because i never do it that way haha

[blissful whale]

to each his own

[edgy sand]

which ones easier to calculate haha?

[blissful whale]

First derivative sign analysis

[heady pollen]

i would say derivative is easy here hehe

🙂

maybe just try both

[blissful whale]

You should definitely understand both

[edgy sand]

okay, how would i do the fist derivate sign analysi?

i get the 2nd derivative method

[blissful whale]

you compute first derivative

,wolf derivate 6x\exp(-13x)

[ocean seal]

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Requested by M8732#6402

[blissful whale]

and then you see when this is zero...

which will be at x=6/78

and then you analyze signs of all factors around this

exp is always positive

6 - 78x is positive before 6/78

nd negative afterwards

-> maximum

I guess the caveat with the sign approach is that you need the function to factor be easily able to tell the sign

Maybe someone else can explain it more throughly if I was not descriptive enoguh

[edgy sand]

I will try it in a few , will let you know if I get it

[lone heart]

@edgy sand Has your question been resolved?

[lone heart]

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[edgy sand]

@blissful whale can I dm you about the question from earlier?

sorry to interrupt here

[blissful whale]

We want to divide $a^2(a^{2n-2} +1)$ by $a^{n+1}+1$.

[blissful whale]

yeah sure, but this is indeed the wrong lace to ask this lol

All I did was substitute b

hoping for more insoriation by seeing it written a nice way

[ocean seal]

M8732

[blissful whale]

ye

try small n... n = 0 would work but I think it might not count as positive enough for you. n = 1 does not work. n=2 also does not with a=1

Latex requires {} for grouping terms

$a^{long\ long\ exponential}$

[ocean seal]

M8732

spraguese

spraguese

[blissful whale]

yeah I think it is a bit contentious but I am pretty sure it is not intended for your task lol

I think it is just naming.

You need positive numbers or zero much more ofdten than strictly positive (which is waht people say that believe in zero being positive)

anyway we are digressing

Number theory is not my area but I think it might be very challenging

actually maybe we can find a reason why starting some n it cannot work anymore

which fraction?

[blissful whale]

btw you are missing a 2 in the exponential

(as I did before I corrected it)

okay so if we consider a prime

then $(a^{2n-2} +1)$ is divisible by $a^{n+1}+1$

[ocean seal]

M8732

[blissful whale]

(because a is not a factor of a^n+1)

hm

Just common factor a²

[ocean seal]

spraguese

[blissful whale]

Just factoring thouh

There are letters are exponentials

German keyboards have keys for these even

he he he

My question is ... are we allowed ot choose a different n for each a?

a^{n+1}+1

just a^{n+1} is divisible by a

but with +1 you will always get reminder 1

so a is not a factor of it

needed to be prime anyhow for what I was saying

[blissful whale]

so just to be clear... what I was saying is that if it is going to work at all then I think this must work as well for a prime

however I think it is definitely impossible to find a single n that works with all a

hm

however let's just consider a=2

it is already not easy!

Where does this question come from?

I suspect the answer is (E)

now I am actually not 100% sure IF we allow choosing n depedent of a

but all those fixed choices given DEFINITELY don't work

you can just try with a=2

That was mean to cut away the answer options

[lone heart]

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[blissful whale]

No problem 🙂

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[grim delta]

what is there to "simplify" here to find q and r
only thing would be to factor out and divide both sides by 5 again, which isn't what it's asking

[heady pollen]

35=12 in Z/12Z

[grim delta]

it already said 5m was wrong for q though

[lone heart]

@grim delta Has your question been resolved?

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@grim delta Has your question been resolved?

[oak perch]

He answered you already, you need to close it

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@grim delta Has your question been resolved?

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[valid stream]

Why is this wrong?

the answer weas 3/10 ln(5x-1)

[remote heron]

are you missing the chain rule here

looks like you crossed wires maybe

you picked up a constant from the chain rule but it was the wrong one?

or it was the one as if you had pulled out the 1/2 constant

but youve already multiplied it through

you may want to explicitly write out the u-sub to track the constant error

its only off by that afaict

[alpine sable]

hmm

i dont see the error

,w integral of 3/(10x-2) dx

[ocean seal]

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Requested by pi=e#5213

[remote heron]

the error is in the substitution

[alpine sable]

oh

[remote heron]

in their result, with ln(10x-2), you should have 3/20

[alpine sable]

they are same

tho

[remote heron]

their answer is incorrect

[alpine sable]

u can simplify it gets absorbed in C

[remote heron]

👀

[alpine sable]

ln(10x-2)=ln(2)+ln(5x-1)

ln(2)*3/10+C=C

so simplifying gives 3/10*ln(5x-1)+C

[tall hearth]

(3/10)ln(10x - 2) + C = (3/10)(ln(2) + ln(5x - 1)) + C = (3/10)(ln(5x - 1)) + M

[remote heron]

right but you need to simplify

[alpine sable]

so both answers are correct

[valid stream]

what is the + M?

[tall hearth]

some other constant

[valid stream]

where did that ln(2) go?

[tall hearth]

(3/10)ln(2) gets absorbed, as stated

[valid stream]

absorbed?

[valid stream]

I haven't learnt u-sub

the textbook doesn't go over u-sub at all

[alpine sable]

no need

ur answer is correct

however it can simplified

thats all

I think first time when doing integrals u just guess it out than doing u sub

[valid stream]

but if I were to substitute x in with something, i get different answers

[alpine sable]

like?

[valid stream]

like x = 2

[alpine sable]

u are forgetting there is a +C

[valid stream]

oh so would the c be different for the two answers?

[alpine sable]

yes

[valid stream]

o

[alpine sable]

actually its giving a general for a family of anti derivatives

[valid stream]

so is the c on the right side a different value?

[alpine sable]

yes

but they are both constants

[valid stream]

right

[alpine sable]

so we just use same variable name

[valid stream]

oih

i think i'm getting it

so basically, i can just simplify anything and put it on to the constant?

[valid stream]

why don't we simplify it to 3/10ln(2) + c instead?

[alpine sable]

U can

but its not compulsory

[valid stream]

which would be the 'simpler' way?

[alpine sable]

if I were there I would have probably left it like that

[valid stream]

so is the textbook just being lame?

[alpine sable]

nah its just a simplification of the answer just 2 more steps

really doesnt matter for most part

[valid stream]

cool okay, that makes sense

thanks for the help. that question really confused my sexual identity

❤️

.close

[lone heart]

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[sick notch]

How would I solve this?

[vale wigeon]

use limit laws

and/or algebra

[sick notch]

yeah but how

do i just sub it in or what

[vale wigeon]

do you know any limit laws?

such as limit of a sum, or limit of a product, or limit of a fraction...

[sick notch]

no

[vale wigeon]

so you do not know, for example, that $\lim [f(x) + g(x)] = \lim f(x) + \lim g(x)$?

[ocean seal]

Ann

[sick notch]

now i do

[vale wigeon]

look up "limit laws calculus"

[sick notch]

yep got it

how would i do the q tho

[vale wigeon]

i don't believe that you actually did as i told you in all of 1 minute

[sick notch]

it takes <10 secs to search limit laws

[vale wigeon]

well then apply your newfound knowledge

and if you find that you cant then show what it was that you found so quickly but didnt manage to understand

[vale wigeon]

right

so, yes, apply those here

find the limit of (f(x)+g(x)), then find the limit of 5f(x), then verify that the latter isn't zero, then find the limit of the whole fraction

[sick notch]

with the first rule

in that pic

[vale wigeon]

you mean the second rule?

yes, the limit of the numerator is just 7+3.

[sick notch]

@vale wigeon

[vale wigeon]

your limit is (7+3)/(5*7) yes

bad notation but your value is correct

[sick notch]

got it

and using the first rule

it would just be the fraction right?

since theres no x?

[vale wigeon]

...??

[lone heart]

@sick notch Has your question been resolved?

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[delicate lake]

help pls

[delicate lake]

i dunno how to do this

[alpine sable]

Take -1 out ig

Rootx-3)i +2

[delicate lake]

it says this when i do it on a website

[alpine sable]

Idk

[serene junco]

If you factor out the -1, you'd get $i \sqrt{x+3} + 2$

[ocean seal]

tatpoj

[serene junco]

which, yeah, that's as "simplified" as any other expression

[alpine sable]

That's what I said before

[serene junco]

you said x-3, not x+3

[delicate lake]

i dont understand

[alpine sable]

Oops my bad

[serene junco]

$$\sqrt{-x-3} + 2$$
$$=\sqrt{-1(x+3)} + 2$$
$$=\sqrt{-1}\sqrt{x+3} + 2$$
$$=i\sqrt{x+3} + 2$$

[delicate lake]

i understand the math

but do they graph to be the same thing?

the original and the simplified equation

because my calculator is not graphing the simplified version for some reason

[alpine sable]

Do u know complex nos eddie

[ocean seal]

tatpoj

[delicate lake]

yes

but why is my calculator not graphing it?

thats what i dont understand

[serene junco]

Graphing calculators don't really like complex numbers

[delicate lake]

oh

but if u actually decided to graph it would it probably be the same thing?

[alpine sable]

Lol

[serene junco]

You need three dimensions to graph a real input and complex output

so a calculator with a 2d screen can't do it

but yes

[delicate lake]

damn

so anytime i get an equation like this and i am asked to graph it, i shouldn't simplify it??

[serene junco]

you were asked to graph y = sqrt(-x-3)+2 ?

[delicate lake]

yes

[serene junco]

Then, forget the complex numbers. Just take the real numbers

[delicate lake]

wdym

[serene junco]

It sounds like your domain is just the real numbers, not complex

So, in that context, sqrt(-x-3)+2 is only defined for x <= -3

If the problem asks you to graph, you can trust your calculator.

[delicate lake]

oh

ok

i think i understand now

[serene junco]

Your calculator doesn't like complex numbers, but your teacher isn't going to ask you to graph complex functions

[delicate lake]

alr

im in precalculus btw

started 2 days ago

[serene junco]

Yeah, definitely just real numbers for now then

Until your teacher says otherwise

[delicate lake]

ok thx bro

[serene junco]

No problem

[lone heart]

@delicate lake Has your question been resolved?

[delicate lake]

yes babe

[lone heart]

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[alpine sable]

I’m new to college and haven’t been in school for a very long time and I’m struggling with putting nouns into equations such as 1 of something - 9 of something = 2 of something as well as 1 of something - 23 of something = 60 of something

[alpine sable]

<@&286206848099549185>

[naive sorrel]

I'm not really sure what you mean (maybe someone else is?) but if you have 5 of one thing like apples minus two of another thing like oranges it doesn't really work cuz u can't compare apples and oranges

[alpine sable]

So like 1-90=10 is 1dollar-90cents=10cents

.close

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[simple idol]

Question: Let the graph of g be a translation 2 units left and 3 units up, followed by a reflection in the y-axis of the graph of f(x)=x2−2x. Write a rule for g.

I am confused on how to add the 2 units left and 3 units up to this given equation.

[simple idol]

Ive tried converting it to vertex form

but i get y=x^2

so when i add the rest of the translations

i get y= (x+2)^2 +3

I changed x to -x

but when i simplify i get y=x^2 -4x +7

i think im converting it wrong

but idk how to convert

<@&286206848099549185>

nvm i figured it out

.close

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[hollow dock]

is g'(x) the same thing as g(x)'

[blissful whale]

likely

the second is slopy notation though

[hollow dock]

yeah it looks very weird

[blissful whale]

I think those are all right

alright

[alpine sable]

Well that , isn't '

[hollow dock]

o:

[alpine sable]

maybe they wanted to wrote $h(x) = \frac{1}{g(x)}$ , then h'(1)

[hollow dock]

does that mean it's not g prime

[ocean seal]

Shihab

[alpine sable]

you see ?

[hollow dock]

OHHHHH

I get you

like it's a comma separating the definition of h(x) and then it's telling me to find h'(1)

[alpine sable]

yes

[hollow dock]

thank you!

[alpine sable]

nope

[hollow dock]

I mean it makes sense since I have to use a table

[alpine sable]

yes

[blissful whale]

oh woah

[hollow dock]

thank you so much I appreciate it

.close

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[tulip gazelle]

Can someone help explpain this? 🙂

[tulip gazelle]

Why do 1/(2-x)<3 become 1<3(2-x)

[lilac nest]

they multiplied both sides by (2-x)

[tulip gazelle]

Ah thanks for help 🙂

Why is there 2 cases?

I would have come to (-∞, 5/3)

and stopped there

.clos

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[boreal stag]

Need help with ai and aii

I don't understand what the question is asking

[lone heart]

@boreal stag Has your question been resolved?

[boreal stag]

<@&286206848099549185>

[lone heart]

@boreal stag Has your question been resolved?

[boreal stag]

Anyone help?

[potent garnet]

Could you please zoom in a little bit - my eyesight is very bad.

[fickle owl]

what dont you understand

to get to X after one throw, do you need to go counterclockwise or clockwise from W

(clockwise)

given the rules, you need to throw a 1 OR a 2

therefore since the probability of throwing a 1 OR a 2 is 2/6 = 1/3, the probablility is 33.33% or 1/3

[boreal stag]

I don't understand the second one

I got 4/9

But ans is 5/9

[fickle owl]

to get to Y after two throws

what are the two options you have

go counterclockwise on both or go clockwise on both, correct?

the probability of going clockwise on both is (1/3*1/3) = 1/9

the probability of going anticlockwise once is 4/6 (can get 3,4,5, or 6) = 2/3

so the probability of going anticlockwise on both is (2/3*2/3) = 4/9

1/9 + 4/9 = 5/9

is this good explanation? @boreal stag

[boreal stag]

Let me see

I think i get it

.close

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[vernal dragon]

hi, the trig idenity says sin^2 u + cos^2 u = 1

[vernal dragon]

i have sin^2 (t/2)

can i write it as 1-cos^2 (t/2)?

[alpine sable]

Yep

[lilac nest]

Yes, in that case u=t/2

[lone heart]

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[vernal dragon]

how do i int e^2t * sin(2t) dt

[lone heart]

@vernal dragon Has your question been resolved?

[nocturne dove]

Integration by parts

[lone heart]

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[lilac nest]

What have you tried to do?

$n^{-1}=\frac 1n$

[ocean seal]

Andrea276

[tight herald]

wait

what if x^-1 in the denominator is used as the numerator like this

(-1)+1 = 0

x^0 = 1

[lilac nest]

You can't bring x from the denominator to the numerator like that

[tight herald]

why?

[lilac nest]

It's a sum, not a product

[tight herald]

so it has to be a number?

[lilac nest]

No, you can't do that in general

If you have $\frac{2+5}{3+6}$ you can't bring the 3 up and do $\frac{2\times \frac 13+5}{6}$

[lilac nest]

Try to rewrite all x^-1 and y^-1 this way

[tight herald]

yea but the ^-1 becomes ^1

[ocean seal]

Andrea276

[lilac nest]

Doesn't change

You can't do that

You have sums, not products at the numerator & denominator

[lilac nest]

.

$\frac{\frac 1x + \frac 1y}{\frac 1x - \frac 1y}$

[ocean seal]

Andrea276

[lilac nest]

Can you see how you could simplify numerator and denominator even further?

Try to find the common denominator between the fractions

[ocean seal]

Andrea276

[tight herald]

hmm

[lilac nest]

Try to rewrite the fractions at the numerator with a common denominator, then do the same for the fractions at the denominator

[tight herald]

like rationalizing?

[ocean seal]

Andrea276

[lilac nest]

No I'm just talking about adding two fractions

[lilac nest]

Add those like you would add 1/2 and 1/3 in this example

[ocean seal]

Andrea276

[tight herald]

ohh

[lilac nest]

It's not correct

How did you find the numerator?

[lilac nest]

Of each fraction

It shouldn't be 1

[tight herald]

alr wait

[lilac nest]

xy at the denominators is correct though

[lilac nest]

Correct

Do you see how you could further simplify it now?

[tight herald]

i guess

[lilac nest]

Correct👍

Now you're done

[tight herald]

alright thanks

.close

[lilac nest]

Yw

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[fleet edge]

$5sin\left(5t\right)-\frac{5}{2}Q=\frac{dQ}{dt}$

[ocean seal]

Alkose

[fleet edge]

Hi all, need help finding a solution to this differential equation using the integrating factor method

[abstract fractal]

What have you tried

[fleet edge]

I'm not really sure where to begin, since all the other examples I've done so far were in the form

f(t) + g(t)Q = F(t, Q)

However I'm having trouble finding where the g(t) is, since it doesn't appear to have a g(t) and to me looks like it's in the form:

f(t) - Q = F(t, Q)

[abstract fractal]

g(t) is just the thing multiplying Q, no?

A function can be a constant

[fleet edge]

g(t) is normally another term which is dependant on t

so g(t) can be -(5/2)?

[abstract fractal]

Yea

Why not?

Nothing wrong with horizontal lines

[fleet edge]

Yeah okay, so that means the next step should look something like this?

[ocean seal]

Alkose

[abstract fractal]

It shouldn't be I(t) - 5/2 Q

You're multiplying everything by I(t)

[fleet edge]

So what would it be instead?

[abstract fractal]

You'd multiply it instead of adding it

I(t) times -5/2 Q is not I(t) - 5/2 Q

[ocean seal]

Alkose

[fleet edge]

Like this?

[abstract fractal]

Sure

But what's I(t)?

[fleet edge]

I(t) is the integrating factor, which is used to solve differential equations via the integrating factor method

[abstract fractal]

I know, but what is it?

Like, what is the function itself?

[fleet edge]

Not too sure sorry, I'm pretty new to this

How would you figure it out?

[abstract fractal]

It's a linear differential equation

In a differential equation dy/dx + f(x)y = g(x), the integrating factor is e^∫f(x)dx

[fleet edge]

This is taken from my textbook, would it be this?

[abstract fractal]

That's not the integrating factor, no

But you'll use it later

https://youtu.be/gd1FYn86P0c

[fleet edge]

So in my case it would be?

dQ/dt + f(t)y = g(Q), the integrating factor is e^∫f(t)dQ

[abstract fractal]

No

You have to write the differential equation in the form dQ/dt + f(t)Q = g(t)

In the video, y was the function you were trying to solve for

In your problem, it's Q

[fleet edge]

Okay thanks, I'll rewatch the video and see If i can do it

[lone heart]

@fleet edge Has your question been resolved?

[lone heart]

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[languid bolt]

$\sqrt{x^2}$ should be $\pm x$ no?

[ocean seal]

help

[languid bolt]

for example

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

[ocean seal]

help

[rigid smelt]

no

[alpine sable]

its |x|

[rigid smelt]

sqrt(4) is 2

[alpine sable]

no

[languid bolt]

why isnt it -2 tho

[analog basalt]

^

[vale junco]

in a square root you take the positive

[alpine sable]

$\sqrt{x}$ is principal square root

[rigid smelt]

the square root that you are using is called the principal root, this is defined to output a positive value

[alpine sable]

4 has two roots, 2 and -2, but if you do it as a term, a term always has just one value

[ocean seal]

Analytic Continuation of zeta

[rigid smelt]

this is why i said earlier sqrt(x^2)=|x|

[languid bolt]

yeah and that got me confused

[alpine sable]

principal square root over reals is defined to always take the non negative value

[languid bolt]

if i want it to get a + - what is the square root called

[alpine sable]

if youre solving x^2=4 it might look a bit different

its called splitting into two terms

[rigid smelt]

well, unfortunately, there isn't explicitly such a thing. If we do use it, we will specify that we are using a multi-valued square root

[alpine sable]

x^2=4 <=> x=2 or x=-2

[rigid smelt]

this is especially useful when working with complex numbers, where square root can be ill-defined

[languid bolt]

oh ok i think i get it

i just remembered that i explained this thing to someone 3 months ago

lel

my brain is lagging from learning quadratics

like 2 months ago

anyways..

thanks

.close

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[tulip gazelle]

What did I do wrong? Supposed to be: (0,1]

[analog widget]

Send the problem

[vale wigeon]

is there a reason why you refused to simplify x+1-2x into 1-x?

[analog widget]

What the question asks

[vale wigeon]

assuming solid means + and dashed means -, you have the signs the wrong way around for (1-x) @tulip gazelle

[vale wigeon]

the question asks to solve the inequality (x+1)/x >= 2

[tulip gazelle]

@vale wigeon So minus should be positive?

@stark ermine Cause im a bit tired. Didnt get much sleep when I fucked your mom all night

@stark ermine What your mom also told me

@stark ermine To bad this question isnt as easy as your mother

[vale wigeon]

@stark ermine please stop harassing other users.

[keen plinth]

calm down yall

[vale wigeon]

you are insulting other users for no reason. you are being rude.

@tulip gazelle the (1-x) thing should have its signs reversed

[heavy otter]

in the future maybe it's best to ignore things like that

[tulip gazelle]

Yeah im sorry about the jokes, just had to

[analog widget]

I inverted mb

wait

\begin{equation}
\frac{x+1-2x}{x} \geq 0 \Leftrightarrow \frac{-x+1}{x} \geq 0 \Leftrightarrow -x²+x \geq 0 \Rightarrow x≠0 \Rightarrow [1;+∞]
\end{equation}

[ocean seal]

everytimecrusader

[analog widget]

\begin{equation}
-x²+x \geq 0 \Leftrightarrow x \geq 1
\end{equation}

[ocean seal]

everytimecrusader

[analog widget]

X cannot be 0, since division by 0 does not exists

and X assumes values greater or igual 1,

[keen plinth]

it isnt x>=1 though

[analog widget]

how

[keen plinth]

it should be 0 < x <= 1

[keen plinth]

this was said when the question was asked too

[analog widget]

The inequalitie was $\frac{x+1}{x}-2 \geq 0$

[ocean seal]

everytimecrusader

[analog widget]

X cannot assume 0

[keen plinth]

-x^2 + x >= 0 gives you 0 <= x <= 1

then you remove 0

not that the method is particularly great

[analog widget]

-x²+x>0 then x²>=x then x>=1

[keen plinth]

other way round

x>=x^2

[analog widget]

Oh mb, i see what i done, i have not inverted the direction

Ur right

[keen plinth]

|| (x+1)/x = 1 + 1/x >= 2 implies 1/x >= 1, this can only hold for positive x so then 0 < x <= 1 ||

thats a much more direct method

[analog widget]

X need to be different then 0

[keen plinth]

well i didnt include it

[analog widget]

X>0 do not include negative values

[keen plinth]

i dont know what you mean

0 < x <= 1 doesnt include negatives

[analog widget]

\begin{equation}
\frac{x+1-2x}{x} \geq 0 \Leftrightarrow \frac{-x+1}{x} \geq 0 \Leftrightarrow -x²+x \geq 0 \Rightarrow x \leq 1 / x≠0
\end{equation}

[ocean seal]

everytimecrusader

[analog widget]

$[1;0) \cup (0;-∞)$

[ocean seal]

everytimecrusader

[keen plinth]

?

[analog widget]

X need to be different than 0

not greater

The only condition we have is x<=1

[keen plinth]

the answer is (0, 1]

[analog widget]

Yeah

[keen plinth]

you can graph it out and check for yourself

[analog widget]

Wich is this

[keen plinth]

im not sure why you have -infinity there

[analog widget]

Becuse it can assume all negative values

[keen plinth]

it cant

[analog widget]

How

[keen plinth]

im saying the answer is only (0, 1]

[keen plinth]

this is the explanation

what doesnt make sense to you

[analog widget]

We got x<=1 right?

And x need to be different from 0 right?

[keen plinth]

no my solution gets 0 < x <= 1

thats all there is

[analog widget]

My doubt is this x>0

[keen plinth]

|| "1/x >= 1, this can only hold for positive" ||

[analog widget]

x needs to be ≠ 0

[keen plinth]

yes x cannot be 0

but the inequality is not satisfied when x is negative

hey there my rotund friend @ornate condor

[analog widget]

You are right, i was forgetting to relate the conditions both nominator and denominatior.

gonna review this topic

[ornate condor]

大家好！

[keen plinth]

sorry i dont speak chinese

[ornate condor]

same

[lone heart]

@tulip gazelle Has your question been resolved?

[lone heart]

@tulip gazelle Has your question been resolved?

[lone heart]

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[compact spindle]

how to start with

[compact spindle]

sqrtx + y =7

x + sqrty=11

[grave lance]

you wanna solve both of these?

for what variable?

[compact spindle]

find x and y

hm

[grave lance]

sure that shouldn't be too hard

2 equations and 2 variables works

[compact spindle]

idk where to even start solving this simultaneous equation

[grave lance]

do you know how to solve for example sqrt(x) + y = 7?

well lets solve that first

[compact spindle]

ok

[grave lance]

with square roots you always wanna try to isolate that term on one side of the equation

so in this case you can just do -y on both sides

and end up with sqrt(x)=7-y

[compact spindle]

how about

(7-y)^2+sqrt y = 11?

idk if this works

[grave lance]

ok I mean that works

I was gonna simplify a bit before but it works too

[compact spindle]

49-14y+y^2 = 11 - sqrty

14y-38-y^2 = sqrty

ok so

what now??

[grave lance]

well try putting all the y on one side

and all the values that don't have y attached on the other

[compact spindle]

-14y+y^2-sqrty = -38

^2 everything?

[grave lance]

the sqrt is annoying yes

no

well you can

but also

u = sqrt(y)

[compact spindle]

ok

y2-14y-u=38

more pleasant

[grave lance]

replace the other y with the u aswell

and then we will solve for u

and then supplement back in the end

[compact spindle]

does y2=4u?

hm

[grave lance]

yes

thats correct

[compact spindle]

soo

3u-14y=38

u=sqrt(y)

now im even more confused

oof