help-0

[grave lance]

uh

I dont even know what u did

replace ALL y with u

so we just get an expression with u's inside of it

instead of y

[compact spindle]

14y=196u?

[grave lance]

no still not right

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

this here

all y

replace theme with the u

[compact spindle]

im so confused lol

y^2=4u

3u-14y=-38

[grave lance]

y^2 is = u^4

not 4u

[compact spindle]

im stupid

u^4-u-14y=-38

14y= ?u

[grave lance]

14u^2

[compact spindle]

isnt it 196u^2?

[grave lance]

no

you just replace the y with the u

the number isnt effected

because the squaring only effects the variable

[compact spindle]

that feels weird but ok

u^4-u-14u^2 =-38

s

[grave lance]

yea and now we try to solve for u

[compact spindle]

$u^4-14u^2-u=-38$

[ocean seal]

Drunk

[grave lance]

although I suspect this is gonna get complex really fast

[compact spindle]

of course it is

[grave lance]

it should be +u

not -u

[compact spindle]

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

-14y?

[grave lance]

x + sqrty=11

[compact spindle]

u sure?

[grave lance]

this was ur original term

you replace x

the sqrt(y) was positive

so replacing it with u

shouldnt result in a -

[compact spindle]

ok

$u^4-14u^2+u=-38$

[ocean seal]

Drunk

[compact spindle]

quadratic method to solve it i suppose

[grave lance]

cant

the u at the end blocks you from doing that

but you can factor one u out

[compact spindle]

u(u^3-14u^2+1)=-38

[grave lance]

-14u

so u^3-14u+1

[compact spindle]

my bad

[grave lance]

also not squared

[compact spindle]

$u^3-14u=-39?$

[ocean seal]

Drunk

[grave lance]

no this is actually more tricky than I imagined

it's not = to 0

we're gonna have to bring the -39 back to the original part

and try to factor something else

[compact spindle]

$(u^3-14u^2)=-38-u$

[ocean seal]

Drunk

[compact spindle]

huh

[grave lance]

problem is that isnt really leading to anywhere

[compact spindle]

yep

except

findind a way using quadratic equations

$u^2(u-14)=-38-u$

[ocean seal]

Drunk

[grave lance]

try this

hold on

[compact spindle]

i suppose i simpified it alot

[grave lance]

$u^4-14u^2+u+38=0$

[ocean seal]

lennygo

[grave lance]

try finding any root for this

any will do

if this question is supposed to be humanily solvable

you should have a root at -2 -1 0 1 or 2

just plug it in and see if it works

[compact spindle]

this seems more like a quintic equation

[grave lance]

we dont have to go there

like I said

[compact spindle]

$u^2(u^2-14)+u+38=0$

[ocean seal]

Drunk

[grave lance]

plug in any INTIGER number from -2 to 2 into that equation

and see which one results in both sides being true

[grave lance]

still doesn't lead us anywhere sadly

but lemme just do it for you

[compact spindle]

this is not reasonable

$^even is positive$

[grave lance]

$f(u)=u^4-14u^2+u+38=0 <=> f(2)=0$

[ocean seal]

lennygo

[grave lance]

f(2) is right

so we have a root at u=2

now we do a polynomial division

[compact spindle]

we just guess?

[grave lance]

yea we guess

[compact spindle]

bruh

[grave lance]

but these questions ALWAYS

have a reasonable root

its nothing like 3.45326356434

that's only doable with computers

so if this question isnt done wrong the roots usually hide near 0 with intiger numbers

[compact spindle]

so (2)^4-14(2)^2+38=0

[grave lance]

hold on I think I miscalculated

[compact spindle]

so we got u

18+38-14u^2=0

56-14u^2=0

56/14 =4

u^2=4

u=2

hm

makes quite sense

[grave lance]

no yea

no I didnt miscalculate

its 2 yea

so now we do this:

[compact spindle]

y=4

[grave lance]

$(u^4-14u^2+u+38):(u-2)=$

[ocean seal]

lennygo

[grave lance]

not yet

not so quick there

we have multiple answers

2 is just one of them

there are more

[compact spindle]

whats wrong

[grave lance]

at most 4

[compact spindle]

you said

u = sqrty

[grave lance]

yes it's ONE answer

[compact spindle]

ok

[grave lance]

but for this to be correct

we need to show all of them

all that make this equation true

so anyway polynomial division

you know how to do that?

[compact spindle]

never heard of it

but i can learn

[grave lance]

it's basically this

https://cdn.discordapp.com/attachments/490557019623915520/1014939413047955587/182418252830343170.png

take the original function

and divide it by the root

Im gonna have to write a bit for this division

so bear with me

[compact spindle]

division?

oh making 2 roots?

huh

i dont get it

[grave lance]

no we are simplifying

I will show you in a sec

[compact spindle]

we got u already

[grave lance]

what happens

no like I said

u has multiple answers

it's not just one

you'll see

[compact spindle]

i kind of dont get why is there multiple possible answers

[grave lance]

I will show you in a sec

it has to do with thee powers

[compact spindle]

never done this before

alright

[grave lance]

$u^3 + 2u^2 - 10u - 19$

alright this is what we get

[ocean seal]

lennygo

[grave lance]

so if we take this

[compact spindle]

i know the four main ways

[grave lance]

and multiply it by (u-2)

[compact spindle]

but power and sqrt really confuses me off

[grave lance]

we are back at the old equation

[compact spindle]

you mean x(u-2)

[grave lance]

so just for explaning this

with the root we sorta guessed

we cracked the equation a bit

and managed to get a term that's a bit more simple to solve

so now we continue with u^3+2u^2-10u-19

let that be = to 0

and solve for the rest of the answers

but I can already tlel

we are stuck

[compact spindle]

it was a mistake for making (u)

[grave lance]

not really

[compact spindle]

but we get y=4 anyways

[grave lance]

anyway lemme show you what I mean

with that we have multiple answers

this is the graph we have for the f(u)

or a little more compressed

do you see how the graph goes through the y=0 line multiple times?

and once exactly through 2?

[compact spindle]

4

[grave lance]

yea 4 times

so we have 4 roots

we found one of them

which is 2

[compact spindle]

you typed this in the graph?

$u^3 + 2u^2 - 10u - 19$

[grave lance]

no

the full one

https://cdn.discordapp.com/attachments/490557019623915520/1014937461085962250/182418252830343170.png

this one

it just sorta depends now

do you only NEED one answer?

or do you want all of them

if this is a question from your school they usually want all of them

but the problem is that we're gonna get really weird numbers

and not intigers anymore

[compact spindle]

just the answer

[grave lance]

well the answer would be all of them

[compact spindle]

no i mean

1

[grave lance]

ok so u = 2

[compact spindle]

the answers of both questions are integers

means its only 1 answer oof

[grave lance]

u = sqrt(y)

which means

[compact spindle]

y=4

[grave lance]

sqrt(y) = 2

[compact spindle]

sqrtx+y=7

[grave lance]

and that means we get 2 answers again

[compact spindle]

x=9

[grave lance]

hold on

just a minute

it gets a bit odd again

[compact spindle]

????

[grave lance]

to get the answer to sqrt(y) = 2

we obviously do .² right

or ()²

[compact spindle]

ye

[grave lance]

that gives us 2 answers

• and -

[compact spindle]

u=sqrty

[grave lance]

so we have

[compact spindle]

+-2?

[grave lance]

$y_1=+4$

[compact spindle]

whats wrong

[ocean seal]

lennygo

[grave lance]

$y_2=-4$

[compact spindle]

y=-4

wtf is u?

[grave lance]

ok I keep messing it up

there

[ocean seal]

lennygo

[grave lance]

no it's just a natural consequence of squaring something

you get 2 answers

[compact spindle]

i do not think you can sqrt negative numbers

[grave lance]

I didn't square a negativen umber

we just got a negative and a positive number from squaring

that's the difference

[compact spindle]

ignore it

xD

[grave lance]

not really

lemme show you

x²=4

right?

pretty simple question

here are the answers

so us doing the sqrt on both sides

results in + and - 2

not just +2

[compact spindle]

ok

what is $sqrt-4$

[ocean seal]

Drunk

[grave lance]

oh wait no ur right

[compact spindle]

you ok bro?

[grave lance]

ok no I got a bit confused myself now

no yea just ignore this rambling

[compact spindle]

you should sleep lol

[grave lance]

that's my bad

anyway

y = 4

[compact spindle]

sqrtx=3

x=9

[grave lance]

yea

that's it

for one of the roots atleast

[compact spindle]

why not just bring it up

#help-0 message

both works

x>y

[grave lance]

yea that was odd on my part

with the + and -

I forgot it only happens when you take the sqrt of both sides

but anyway

if your teacher is happy with that one intiger answer

and only one solution to this system of equations

then that's the question done

[compact spindle]

i dont get what you're saying lol

isnt there only 1 solution?

sqrtx + y =7
x + sqrty=11

[grave lance]

1 sec lemme show you

I'll just throw it into an equation solver

do you see what I mean

it has some more answers

[compact spindle]

lets say

u is 3.131313

[grave lance]

yea Im gonna say it now

technically

"technically"

we would have to check for the validity of ALL possible answers

[compact spindle]

then y=9.8051211

[grave lance]

but because we aren't computers we can't really get solutions with those other answers

[compact spindle]

and

[grave lance]

now I am telling you this just because of one reason

because you will encounter other questions

where you cant just take the first intiger solution and be done

there will be more intiger solutions

[compact spindle]

sqrtx + y =7

[grave lance]

and you have to solve for all of them

but for this one just going on with u = 2 is fine

[compact spindle]

(possible solutions)

sqrtx+9.8051211=7

huh

im just weirded out

[grave lance]

its really odd yea

I got off on the other answers

because I thought they would be intigers aswell

but obviously not

just to recap all I've said:

make sure you try to solver for every answer you get not just the first one that you get.

other than that we are done

unless you need further help

[compact spindle]

the ways that we solved this question are so random and unreliable

bruh

[grave lance]

yes and no

you mean the part where we guessed a root right?

[compact spindle]

yep

10/10

[grave lance]

that is

funnily enough

a common method

for solving polynomials higher than the power of 2

[compact spindle]

u srs

[grave lance]

yes

you can ask your teacher if you don't believe me

but like I said before

the questions are usually arranged in a way where one of the roots is always an intiger

that is close to 0

so -3

or -2

or -1

or 0

or 1

or 2

or 3

and so on

[compact spindle]

what if it's isnt

[grave lance]

then

1. the question isn't solvable for you
2. the teacher did a mistake
3. u just did something wrong

usually it's 3.

there is one way of getting roots aswell

[compact spindle]

i mean

[grave lance]

but since ur still solving equations at this level and not doing derivatives yet you probably haven't encountered it

it's called Newtons Method

[compact spindle]

the answer is with a few decimal points

[grave lance]

but yea that's just a sidenote

[compact spindle]

ok

[grave lance]

more actually

way more

I think those numbers might even be irrational

[compact spindle]

so messy with power and sqrts

[grave lance]

yea

anything else that might still confuse you?

( other than my ramblings )

[compact spindle]

haha

there's still alot but thanks

have a good day

.close

[lone heart]

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[dreamy jolt]

I am trying to decompose the force F2 into X and Z axis i have found the correct answer on the X-axis (F2x=10kN) and in my head the equation circled tells me that F2z will be 5 times larger (50kN). But apparently it is wrong, What am i missing?

[grave lance]

Ok first of all this is a math server so a lot of people here have never seen classical mechanics

but I have seen them

so all you're trying to do is split F2 into x and z values?

do you not have an angle at which F2 attacks the body?

[dreamy jolt]

no angle

only ratio

[grave lance]

thats a weird question

I guess what you could do

no that wouldnt work aswell

huh

I mean you're looking for F2x

which depends on F2z

but you dont have z of F2

or any equation for it

[dreamy jolt]

i have calculated F2x

[grave lance]

lmao how

without an angle?

is there any missing equations?

[dreamy jolt]

i can show you

[grave lance]

sure I have never seen this

[dreamy jolt]

using the ratio given i know F2z is 5 times larger then F2x. With that i calculated the angle (taking a pic... 2 sec)

[grave lance]

and this is correct?

[dreamy jolt]

F2x is correct
But F2z came out wrong according to the homework software

[grave lance]

well instead of using the angle

you have F2x

just let F2x * 5 = F2z

but that would bee 50 aswell

maybe an error on the software?

[dreamy jolt]

and thats 50 right

[grave lance]

bcs if F2x is correct

[dreamy jolt]

..... maybe

[grave lance]

then F2z has to be 5 times bigger per equation

[dreamy jolt]

that is what i am thinking as well... i am just not 100% sure i read the equation correct

[grave lance]

no what you did is entirely correct

it has to be an error on the software

[dreamy jolt]

hec.... i guess i have to ask the Prof, and he cannot speak Norwegian or English that well.... thanks for the help

[grave lance]

huh thats interesting?

where do you go to university

[dreamy jolt]

NTNU in Norway. Like... he is understandable, its just his typical French way of constructing his sentences and how he forgets to use certain contextual filler words that confuses the hec out of me at times.

[grave lance]

that's a first time I heard of a French prof teaching at a norwegian school

kinda odd

[dreamy jolt]

I have more foreign profs then Norwegian. the struggle is real.

[grave lance]

Yea I get that I had some foreign profs aswell and it makes it quite difficult to understand them sometimes

but look at it with the positive side

in the industry you will have to meet foreign actors aswell that you will have to talk to about your project

so it prepares you for the worst

[dreamy jolt]

that's thru

thanks for the help. 🙂

[lone heart]

@dreamy jolt Has your question been resolved?

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[runic wigeon]

i need help

pls

[lone heart]

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[uncut torrent]

What have you tried?

[runic wigeon]

everything

-x=-4

not right

[uncut torrent]

well, try finding the equation of the line

[runic wigeon]

nvm i got it

[uncut torrent]

o ok

[lone heart]

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[runic wigeon]

.close

,close

[lavish delta]

Anyone know whether u use sine or cosine rule here coz I’ve tried both and neither have gotten me the right answer

[alpine sable]

What is here

There are two questions

None of them seem to be right triangles

Oh read your question wrong

[lavish delta]

It’s the top triangle

[alpine sable]

Of course cosine rule

$10.5^2 = 7.6^2 + x^2 - 7.62x*\cos{61^\circ}$

Solve this quadratic

[lavish delta]

There is an answer sheet at the back and it says the answer is 11.8cm to 3dp but every time I do it keeps giving me 9.52 to 3 dp

[ocean seal]

Endou Mamoru

[alpine sable]

It is 11.8126 (https://www.wolframalpha.com/input?i=10.5^2+%3D+7.6^2+%2B+x^2+-+7.6*2*x*\cos{61+deg})

You are solving the quadratic wrong

[lavish delta]

What am I supposed to add

[alpine sable]

?

[lavish delta]

Like what do I actually do

[alpine sable]

Solve this equation correctly

[lone heart]

@runic wigeon Has your question been resolved?

[lone heart]

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[lone heart]

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[alpine sable]

hello,
how do i find the cos(n)th root of something

[lone heart]

@alpine sable Has your question been resolved?

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[hot bison]

Can someone help me with 2 and 3?

[keen pasture]

The channel is already being closed due to you deleting your message

[hot bison]

Oh thanks

.close

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Oh ok

[keen pasture]

It takes some time to reopen

[lone heart]

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[sleek cradle]

can someone solve this

[pulsar seal]

try plotting those points on a piece of paper

you can define a linear function with two points

[fierce prairie]

yeah.. or you could find it algebraically using point slope form

(after finding the slope m)

[sleek cradle]

it says i got it wrong and the question changed but whts the answer

[fierce prairie]

well what answer did you put?

[sleek cradle]

-1

[fierce prairie]

how did you find that..?

that would be a straight line at y=-1

[fierce prairie]

first of all, do you see how the question gives you 2 coordinates?

[sleek cradle]

-2 and 3 right

[fierce prairie]

that's one coordinate, yes

properly written as (-2, 3)

and the other one?

[sleek cradle]

(-2,2)?

[fierce prairie]

no

the form is f(x) = y

so what is x and y if f(2) = -2?

[sleek cradle]

2,-2

[fierce prairie]

yep

now can you find the slope between these 2 points?

[sleek cradle]

yes got it lemme try the other question now

thank you

[lone heart]

@sleek cradle Has your question been resolved?

[fierce prairie]

you know the rest of the steps? finding the y-intercept?

[lone heart]

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[sleek cradle]

yes the textbook had them i just had to read thro

through*

ty tho

[fierce prairie]

oh okay np

[lone heart]

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[slow hound]

g is a function

it maps a scalar to a triplet

[weary wyvern]

Function and variable are used almost interchangeably

A line refers to a set

[slow hound]

yeah

$\color{lime} g: \mathbb{R} \to \mathbb{R}^3$

[alpine sable]

i guess a line can be represented as both, either a set or a function

[slow hound]

bot's still dead rip

it's just implied

[lone heart]

@thick lynx Has your question been resolved?

[sleek cradle]

how do i solve this?

[potent garnet]

You're not supposed to post your questions in someone else's channel if they haven't closed it yet.

Please use an unoccupied channel.

[sleek cradle]

oh my bad

i didnt know that

[potent garnet]

That's okay, now you know. 🙂

[lone heart]

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[cosmic shard]

Hi, i need help with this problem please

[keen plinth]

what have you been able to do

[lone heart]

@cosmic shard Has your question been resolved?

[cosmic shard]

Not too much, i only know that there can only be 70 coins of 5 dollars and 35 of 10

I need the other equation

[keen plinth]

yes that's one of the equations you'll need

except

you have not 350 dollars

you have 2500

[cosmic shard]

Yes

2500 dollars

[keen plinth]

so 5x + 10y = ?

[cosmic shard]

350

[keen plinth]

what does the quantity 5x + 10y represent

[cosmic shard]

Wait what

[keen plinth]

why do we calculate 5x + 10y

5 times the number of 5 dollar coins + 10 times the number of 10 dollar coins

[cosmic shard]

Because there are only coins of 5 and 10 dollars

[keen plinth]

what does the quantity give us

[cosmic shard]

The thing is that i dont know the amount of each coin, i only know the total of coins

[keen plinth]

that's why we have x and y

we need to determine them

if we have x 5 dollar coins and y 10 dollar coins

then calculating 5x + 10y has a specific meaning

5x is how much money in 5 dollars you have

10y is how much money in 10 dollars you have

in total it's how much you have altogether

this means 5x + 10y = 2500

[cosmic shard]

yeah but how do i find the x and y values

[keen plinth]

well you need another equation

we've used the fact that you have 2500 dollars to get 5x + 10y = 2500

there's still the fact that you have 350 coins in total

how do you represent that as an equation

[cosmic shard]

5x + 10y = 350
so 350 = 2500?

[keen plinth]

no

how do you count how many coins you have

x is the number of 5 dollar

y is the number of 10 dollar

[cosmic shard]

i wrote that ecuation in geogebra and it says that x = 70 and y = 35 but if i multiply 5 x 70 its equal to 350, and 10 x 35 is equal to 350 so idk 😭

[keen plinth]

okay first of all do you understand why 5x + 10y = 2500

[cosmic shard]

yeah, because the amount of dollars is 2500

[keen plinth]

okay good

so then 5x + 10y should not be 350

that equation is wrong

we have to count how many coins we have

if we have x 5 dollar coins

and y 10 dollar coins

the total number of coins is?

[cosmic shard]

Well the problem says that there are accumulated 350 coins and 2500 dollars with 5 dollar and 10 dollar coins. So i need to find for each coin, how many coins are there

[keen plinth]

yes that's what x and y are

you can add the number of each coin

to get the total right

x + y is the total number of coins you have

[cosmic shard]

Ohhhhhhhhh i found the answer tysm

I was making something wrong lol

.close

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[mint tundra]

why is the derivative of the second pic equals zero

while the first doesn't

whats the difference

[lone heart]

@mint tundra Has your question been resolved?

[rose sigil]

the line tangent to the first curve at (x_0, 0) would look more like a \ and the line tangent to the second curve at (x_0, 0) would look like a - (a horizontal line)

[mint tundra]

Alright thank you sm

.close

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[rose sigil]

np!

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[jovial tusk]

I’ve got the equations down correctly (I hope lol) I’m just stuck on the last one

[alpine sable]

solve for one variable and then plug it in to anotehr equation

so you can do w = c/2

[jovial tusk]

I’m trying to solve it using matrices in the calculator

I’m confused on how to plug 2w=c into matrices

If that makes sense

[sudden hinge]

the 4th one should also be the other way around: w = 2c

[raw fractal]

You can write w = 2c as 2c + 0s - 1w = 0 in a matrix. So the row would be 2, 0, -1, 0.

[jovial tusk]

Oh shoot my bad

[jovial tusk]

Ohh thank you

[raw fractal]

You needa make sure all the variables are on one side and in the same order. So since the first 2 equations have c first, then s, then w, when you rearrange w = 2c, you needa keep that order.

[jovial tusk]

Okok thanks man

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[lone heart]

@nocturne ridge Has your question been resolved?

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@nocturne ridge Has your question been resolved?

[lone heart]

@nocturne ridge Has your question been resolved?

[nocturne ridge]

Who reacted

Come help me

[rustic coral]

Reposted here: #help-6 message

[nocturne ridge]

.close

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[tired star]

hi

[silver onyx]

what to find

[tired star]

well idk if its ok to post here cause its math channel

[silver onyx]

well idk

[tired star]

but i need the to find the Intensity before it goes in the resistors

so basically i1 i2 i3

is it ok if i do 25-(i2*5)=0

wait no -25+(i2*5)=0

[wary stream]

,rotate

[wary stream]

Intensity? Never heard current referring as that

[tired star]

yeah sorry im french speaking haha

tried poorly translating it

[wary stream]

But I suggest asking in the physics server, linked in #old-network since this is more physics than math

[tired star]

yeah ik but i need verified phone number to post lol

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[maiden hill]

hello! I am working on the difference quotient in algebra -> how does 2/(x+h)+5 - 2/x+5 work out?

[maiden hill]

as a reference

i can do the others, but since this one is a fraction, I cant quite wrap my head around it

[rustic coral]

common denominators

[maiden hill]

@rustic coral so that would be ((x+h)+5)(x+5)?

[rustic coral]

yeah, that's your common denominator.

[maiden hill]

@rustic coral question, how do i treat the +5 in ((x+h)+5)?

does it get multiplied as well?

[rustic coral]

yeah, it's a term like any other in the factor.

[maiden hill]

@rustic coral so it evaluates to:
x^2+5x+hx+5h+5x+25?

[rustic coral]

,w expand (x+h+5)(x+5)

[ocean seal]

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Requested by messy circle creation#1465

[maiden hill]

so the next step would be to
multiply what I just did in the bottom to the top?

like, 2(x+5)

so 2x+25?

@rustic coral

[rustic coral]

yeah, rewrite both fractions to have this common denominator

[maiden hill]

so, 2x+25 and 2x+2h+10 are the numerators

@rustic coral

[rustic coral]

should be 2x+10 and 2x+2h+10

[maiden hill]

ooooohhhhh, right

all that over h

@rustic coral

@rustic coral now i subtract the numerators?

[rustic coral]

yeah, the difference of those 2 fractions over h

[rustic coral]

mhm

[maiden hill]

is the second numerator negative?

because the negative sign gets distributed?

of the original funtion

[rustic coral]

you mean $(2x+10)-(2x+2h+10)$?

[ocean seal]

messy circle creation

[rustic coral]

yeah that's equal to -2h

[maiden hill]

so I have $-2h/hx+5h-x^2+10x+25$

[ocean seal]

Biblicalwizardx

[maiden hill]

and now i cancel all h's, because im dividing by h?

[rustic coral]

$\frac{-2h}{hx+5h-x^2+10x+25}$

[ocean seal]

messy circle creation

[rustic coral]

you can't divide the numerator and demominator by h since they both don't have a common factor of h

well theoretically you could

but it would make it more compliated unnecessarily

[maiden hill]

@rustic coral so whats the next step?

or is that the answer?

[rustic coral]

that should be the answer

[rustic coral]

oop

,w \frac{\frac{2}{x+h+5}-\frac{2}{x+5}}{h}

[ocean seal]

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Requested by messy circle creation#1465

[rustic coral]

hm

oh ik what we forgot

the h in the denominator

$\frac{-2h}{h(hx+5h-x^2+10x+25)}$

[ocean seal]

messy circle creation

[rustic coral]

so yeah you can cancel the h lol

[maiden hill]

so why do we now multiply by h?

[rustic coral]

it wasn't here so I forgot lol

💀

[rustic coral]

wait what?

[maiden hill]

how do i go about canceling the h's?

or is that the answer?

[rustic coral]

what you said here, divide the numerator and denominator by h

[maiden hill]

so the answer should be
$\frac{-2}{x+5-x^2+10x+25}$?

[ocean seal]

Biblicalwizardx

[rustic coral]

well you don't cancel all the h's

since the common factor is h, not h^2 or whatever

you should get what's under 'expanded form'

[maiden hill]

niiiiiiice!

omg thanks so much, damn that's complicated

[rustic coral]

👍

[maiden hill]

youre good

[rustic coral]

we had a brain dead moment but we got there eventually lol

.close

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[foggy onyx]

hi

[foggy onyx]

i have a stupidity question

i dont understand the meme lol

[foggy onyx]

OH

LOL

ok

thanks

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[normal holly]

how do i solve this?

[normal holly]

tysm i rly appreciate it

one more question

how do i solve part b

i did it twice and it’s still wrong

2 whole?

but it’s 2/5

yh but the denominators aren’t the same

[ornate condor]

remaining*

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@normal holly Has your question been resolved?

[lone heart]

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[west brook]

im supposed to find the domain and range of this... but I don't understand why range is [0,sqrt(3)) U (sqrt(3), inf)

[ornate condor]

um

idk

didnt see the whole thing

but

[0

is defn wrong

oh wait range

oops

ok yea lol

so 4/x can essentially take on all values except 0 right

[west brook]

yes

[ornate condor]

so cuz of that

u cant get sqrt3 exactly from the func

also u cant get neg from sqrt

[west brook]

so where is the [0 from then

or is that just wrong

[ornate condor]

when 4/x=-3

[west brook]

oh i see

[ornate condor]

nah its right

i thoguht they asked for domain mb earleir

[west brook]

ok that helped ty

[ornate condor]

np

<3

[west brook]

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[limber hill]

How to integrate log(xy) with respect to x

[limber hill]

?

[wide raven]

Log(xy)=log(x )+log(y)

And then ig simple enough

[limber hill]

Yeah but then what would be the integration of log y with respect to x

[serene junco]

If you're integrating with respect to x, then you can treat y as a constant

@limber hill

[limber hill]

That means log y will be xlogy?

[serene junco]

yes

[wide raven]

exactly

[limber hill]

So the answer would be xlogx - x + xlogy

[wide raven]

i think so

[serene junco]

+C

[wide raven]

oh yea ofc

[limber hill]

Final answer would be x(logx+logy) - x + C

And logx+logy would be logxy again so....

xlogxy - x + C would be the final answer

[serene junco]

Looks good to me

[limber hill]

Lol then why I was doing so much hard work

🥲

[serene junco]

Lol

beats me

It happens

[static osprey]

shut up nerd

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[signal scroll]

Can someone explain me why we say is not continuous at [-3,3]? Why do we get that -3? I get that the function is not define at 3 but why do we chose the interval [-3,3]

[tacit arch]

can you finish your interval

[signal scroll]

What do you mean?

[tacit arch]

"[-3" is not an interval

[signal scroll]

Fixed it

[tacit arch]

are you questioning this?

[signal scroll]

I honestly when I first see it I think the domain is (-infinity, 3) and then (3, infinity)

[signal scroll]

Yeah

[tacit arch]

they mention that because the function $f(x) = \frac{x^2 - 9}{x-3}= \frac{(x-3)(x+3)}{x-3}$ and $g(x) = x+3$ are equal almost everywhere except $x=3$. $g(x)$ is defined for all real numbers, but $f(x)$ isn't

[ocean seal]

riemann (eric tao for honorable)

[signal scroll]

Oh okey I just realized that the interval was something given and not something we calculated out of the problem sorry

But I just got a question out of that

If 3 isn’t included why do we put a parenthesis sign after

[tacit arch]

[a, b) = {x | a <= x < b}

more generally
https://andymath.com/wp-content/uploads/2019/06/Interval-Notation.jpg

[signal scroll]

Oh I see thank you

I just feel burned out right know I’ve been study material for 4 hours

Test tomorrow, thanks for the help again, good night

What’s the command to close this?

[tacit arch]

.close

[signal scroll]

.close

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[fathom wedge]

why is x = y² not a function?

[fathom wedge]

if someone cares to explain this can you tell me the reasons why it is not a function

[proud pike]

it is a function of x in terms of y, but it isn't a function of y in terms of x, because there are several values of y for a single value of x.

For example, if f(x) = y, then what is f(4)? It could be either 2 or -2, but functions can only have one value as an output.

[fathom wedge]

so what u said is gonna be the same thing as this statement "If x is the independent variable and y is the dependent variable, then x=y2 is not a function"?

[inner viper]

You could express it as a root if roots were defined 2-ways

[proud pike]

yes, because a function has to give exactly one result for each value in its domain

y = ±√x isn't a function

[fathom wedge]

okay thank you

that helped me

[lone heart]

@fathom wedge Has your question been resolved?

[vale wigeon]

.close

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[naive sorrel]

Do you know what a maximum of a function is

?

oh sorry

I didn't look at it all

do you know what happens when the x is divided by 2

[sudden lagoon]

umm

is it that the graph is stretched by scale factor2?

@naive sorrel ?

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[feral juniper]

How does one prove a function is positive within a certain range

[tacit arch]

Depends on the function

[ornate condor]

do u mean domain?

[feral juniper]

Basically I need to prove that 1/2 * cos(x) is positive within the range (-pi/2, pi/2), and ik I can plot it but is there a mathematical proof

[keen plinth]

domain and range man

[ornate condor]

wot

anyway

[feral juniper]

This is a part of a larger problem where I need to prove a probability density function is valid, which I know one of the conditions is that it's positive

[ornate condor]

u can find the values of cos at the start n end points

[feral juniper]

So in this specific case that function and domain is given

[keen plinth]

nu

[ornate condor]

then u can check gradient to see it increases

also there are no other zeroes

[ornate condor]

wot

[keen plinth]

cos is concave on that interval

it's just mean value theorem

use mean value theorem to prove that it's positive

[keen plinth]

gradient>0 implies increasing is a consequence of MVT

so is concave function lying above the secant

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[lone heart]

@rapid ingot Has your question been resolved?

[mellow tusk]

5

[rapid ingot]

how

[mellow tusk]

factorize

[lone heart]

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[rigid tundra]

Hi, I'm trying to split this fraction into a sum of powers of x and partial fractions but have exhausted the methods I know.

It's top heavy so tried to do long division and ended up taking away unlike terms.

I also couldnt factorise.

[thin pivot]

you can't simply this any further

[rigid tundra]

You can't split into partial fraction and a sum of powers of x?

[gray isle]

what do you have after long division

and/or show your attempt at it

[rigid tundra]

7x - 21x-1/x+3

7x seperate

My work is messy so this Easier soz

[gray isle]

not for me

rewrite more nicely and show it

[rigid tundra]

Ok one sec

[thin pivot]

is the original equation 7x^2 - 21x - 1/x+3?

[rigid tundra]

No one sec

Soz for rotation, also have slow speeds

[gray isle]

,rotate

[gray isle]

you didn't complete your long division

[rigid tundra]

How so

[gray isle]

-21x-1 has the same degree as x+3 so you can continue dividing

[rigid tundra]

Ah I see

Until I get to a point where the top has a lower degree than bottom?

[gray isle]

poor way to view it

until the degree of the current remainder is less than the divisor

you don't really want to explicitly write the result after each stage to check whether you should continue

[rigid tundra]

Ah fair makes sense

I got 7x - 21 +62/x+3

Would that satisfy a question asking to simply expression as "a sum of powers of x and partial fractions"

Also when doing long division, I thought you couldn't take Away unlike terms, but you can I'm assuming if it's a remainder?

[thin pivot]

what'd the question be?

[rigid tundra]

.

[thin pivot]

well personally, i'd just show the long division

[rigid tundra]

7x^2-1/x+3

[gray isle]

7x - 21 +62/(x+3)
but yes

[rigid tundra]

Yh

[gray isle]

parentheses are important when communication in plain text

[rigid tundra]

Thanks both for help

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[languid bolt]

forgot to ask this so ima open a new channel

[languid bolt]

wait nvm

i forgot the question

brb a min

nvm i found the answer

.close

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[languid bolt]

.reopen

[lone heart]

✅

[languid bolt]

wait nvm

P(A or B) = P(A) + P(B) right?

if i can try 5 times

the probability of me getting either a or b once is

5(P(A) + P(B)) right?

wait no

yeah im confused

[sudden hinge]

if A & B are mutually exclusive, then P(A or B) = P(A) + P(B), otherwise P(A or B) = P(A) + P(B) - P(A and B)

[languid bolt]

what do you mean by mutually exclusive

independent?

[vale wigeon]

mutually exclusive = cannot happen simultaneously

[languid bolt]

ohh

[vale wigeon]

mutually exclusive events are never independent save for some trivial cases

[languid bolt]

oh ok i think i get it

what do i do if i can try 5 times tho

if the probability doesnt change every try

[vale wigeon]

you will have to be more specific in what kind of experiment you are conducting and what event you are really talking about

[languid bolt]

hmm

ok

probability of getting a black ace out of a deck of playing cards

with replacement

but i get 5 tries

[vale wigeon]

ok so you are drawing cards from a deck with replacement and want to draw at least one black ace

do i understand that correctly

[languid bolt]

yes

[vale wigeon]

it must be noted that "draw black ace on 1st try" is not the same event as "draw black ace on 2nd try", so denoting them with the same letter (as you appear to have attempted earlier) would be wrong

anyway

this is most easily calculated as 1 minus your probability of never drawing any black aces

so 1 - (25/26)^5

[languid bolt]

oh yeah i forgot about that

are there any other way tho?

[vale wigeon]

well you could write down 5 events A_i = "draw black ace on i'th try" (for i=1,2,3,4,5), note that these events are independent, then compute the probability of their union via incl-excl...

[languid bolt]

ok

What about if there is no replacement

[vale wigeon]

gonna be painful tho

[languid bolt]

yeah, ima just use the first method, the easy one for the first question

[vale wigeon]

if you are doing it without replacement you will have to get into lots of combinatorial hypergeometric bullshit

[languid bolt]

ok then ima just skip this

ty btw

.close

[lone heart]

Channel closed

Closed by @languid bolt

Use .reopen if this was a mistake.

[languid bolt]

.reopen

[lone heart]

✅

[languid bolt]

forgot to ask again

what is a half life

google said it's the time for something to become half of it's initial value

ok im dumb nvm i already found my answer