help-19

[woven kelp]

It becomes $2x+4x-6+18=6x+12$

[clever fjord]

Max

[minor bronze]

i thought i had to subract but i wasnt sure

give me one sec

X = 28

i got it

.close

[odd edge]

Channel closed

Closed by @minor bronze

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[vast ravine]

Please help and let me know what I’m missing

[supple mural]

the delta math geometry proofs, the bane of all students.

Show the options

[odd edge]

@vast ravine Has your question been resolved?

[vast ravine]

Idk what am missing

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[void sky]

How do you calculate this vAB?

[void sky]

I don't see any way for this

[hollow lintel]

@void sky where do you think ground is here?

[void sky]

I would think at the left side

[hollow lintel]

why the left?

[void sky]

Due to the current going from the right to the left, but i am not sure of this circuit honestly

Most of the excersises the ground is already defined somewhere and is not so weird of a circuit

[hollow lintel]

one sec let me solve myself so I can make sure I am not misleading you

it is a bit strange

[void sky]

I thought since these were parallel the voltage different would be 0.

But i got it wrong lmao

I tried

[hollow lintel]

still solving it but lfrom looking at it i can tell you a few things

[void sky]

Where do i start with this strange looking thing

Would appreciate it

[hollow lintel]

the current always flows from high potential to low potential, so the spot behind the current source is going to be ground in a circuit like this

and also the potential will be higher on the bottom, not the same, because the bottom has a higher 'drop'

the difference between the two resistors is bigger

4 ohms vs 4 ohms, in series from source to ground, the drop will be the same

4 ohms vs 5 ohms, 5 will drop a bit more than 4

but 10 vs. 15, the 10 will drop ...probalby 2/3 less than 15, i am bad at fractions

so there WILL be a potential difference

if the top two and bottom two resistor setups were identical the voltage drops would be the same, and THEN it would be 0

[void sky]

But doesn't that depend on the current coming in?

[void sky]

Okay this is clear

[hollow lintel]

if the current is the same for the top and the bottom, and coming in from the left (as it is here) then that will be true

[hollow lintel]

how would it depend?

it does, kind of.

[void sky]

As the resistance on the lower part is bigger so there will be less current going there

And just using ohms law v = I * R

Btw does the current source help us in any way here?

[hollow lintel]

eh not other than finding the exact voltage

its a bit more complicated than it would be w/ a voltage source

[void sky]

So at the left dot 4 A in current gets splitted into 2 branches? Should i see it like thatC

[hollow lintel]

yeah it gets split into two

you can kind of rotate it and see it so the two sides are in parallel

and the two resistors are in series

and I will use that to find the equivalent circuit with only 1 source and 1 resistor

and get total voltage and calculate the drops

[void sky]

So basically add the 2 resistors that are in series

9 and 25 ohm

4A coming in

[hollow lintel]

noo add in series

[void sky]

and splits into two

[hollow lintel]

but yes 9 and 25 🤔

yes

[void sky]

mb

But we want the voltage between the two in series right

So we also need to account for that

[hollow lintel]

yeah

this is the same circuit

just rotated and un-kinked

have you done thevenin or norton circuits yet?

if not DW

[void sky]

Yes unfortunately

[hollow lintel]

LOL

[void sky]

We have those

lmao

[hollow lintel]

if you can simplify it to one resistor, then you have the incoming voltage instead of the current

[void sky]

I don't get the current source instead of voltage source honestly

throws me off a bit

[hollow lintel]

yeah you basically just have to use ohm's theorem and pretend its a voltage source

[hollow lintel]

do this

then solve for the voltage in each branch

then solve for x in (incoming voltage - X)/10 = (X - 0)/15

sorry solve for the voltage divided

i have been studying all day

brain melted

[void sky]

So (1/25+1/9)^-1 would be the total resistance right

[hollow lintel]

uhh should be lemme do that rq

yessir should be

man this number sucks

[void sky]

Okay so 4A and that number that sucks

[hollow lintel]

LOL

[void sky]

to calculate the incoming voltage

[hollow lintel]

yep

[void sky]

of the total thing

[hollow lintel]

can you do it w/ this equivalent diagram and knowing the incoming voltage?

the two sides are in parallel

[void sky]

Where would we put the "voltage" source?

[void sky]

At the top?

[hollow lintel]

replace the current source with it

the current creates that voltage when it travels over the resistors

they both travel towards ground

so it is the maximum at the source and the minimum at ground

[void sky]

the voltage of the source is 26.4705..., so the voltage of one total parallel part would be that aswell?

[hollow lintel]

bwuhhhhh the voltage of parallel parts is a ratio of that

wait

current

so voltage yes the same

[void sky]

yeah current would be different

[hollow lintel]

😭

sorry i am literally doing integrals rn

[void sky]

No worries its 1 am here and the deadline of this shit is tommorow cuz i had other tests

So we can use KCL at the node at the top right?

[hollow lintel]

yep you can

split it using the ratio

do you know the voltage drop equation

[void sky]

So first add the ones in series together

to simplify

[hollow lintel]

yeah

then you have to split them apart after

[void sky]

then split it later to find exact ones of the first

[hollow lintel]

exactly

[void sky]

and then subtract a -b

[hollow lintel]

yep

if you want to check your answer after I can solve it too

[void sky]

Please do i already lost so many points with my first attempt

lmao

I can get a max of like 7 now

[hollow lintel]

Lol give me an answer and I will tell you if it is right

[void sky]

Alr let me have it a go

thanks

[hollow lintel]

no problem 👍

[void sky]

I hate electronics

LMAO

Just give me math

I am a CSE student so im suffering

[hollow lintel]

im a computer engineering/physics student

i love this stuff

keep your math away from me

[void sky]

I have b being 15.882 V

The current across the 25 ohm shizzle would be (18/17) and on the 9ohm (50/17)

[hollow lintel]

correct for B

idk about the current LOL i didn't find those

but you got voltage right

[void sky]

Okay so i'm not spewing bullshit

good

haha

[hollow lintel]

once you find A I will send my work so you can see how I did it w/o finding current

or if you want me to I can send B

[void sky]

14.70588 V on A?

[hollow lintel]

yep

now the question is... is the answer negative or positive?

(i don't know)

[void sky]

I would assume V(ab) is a-b

Idk

[hollow lintel]

I think so

[void sky]

how do voltage meters work

[hollow lintel]

you connect the positive side to the high voltage

and the negative side to the low voltage

and then it gives you positive

but if you flip it it gives you negative

[void sky]

more like

[hollow lintel]

but the arrow goes from b to a

in the first picture you sent

[void sky]

is Vab Va-Vb or other way around

[hollow lintel]

PROBABLY that but I always stick it in positive and go AW HECK ITS WRONG

does it say magnitude anywhere

[void sky]

Yeah a-b right

No

[hollow lintel]

💀

I'm gonna be honest bro I got you the answer but I can't tell you the direction

I have guesses but they aren't concrete

[void sky]

-20/17 orrrrr the positive one

[hollow lintel]

I would assume its negative

But I don't know

[void sky]

now

should i calculate decimal

[hollow lintel]

So you are gambling because I don't pay attention to details LOL

[void sky]

lmao

[hollow lintel]

What are you inputting it in

ExpertTA? Some freaky pearson hellscape?

[void sky]

Some own uni homework site

[hollow lintel]

I'm so sorry

[void sky]

Nah i don't think so

[hollow lintel]

If you found it in fraction form I would input it in in fraction form

Preciser = better

[void sky]

I can see this just going wrong

lmaoo

[hollow lintel]

probably that but I'm sorry if not 😭

[void sky]

shitty ass sit

[hollow lintel]

at least it isn't pearson

[void sky]

ohh

i got saved

[hollow lintel]

YAAY

[void sky]

it does not like format

no

So decimalll

[hollow lintel]

😭

here is how I solved it

didn't put finding V_s in there but you know how to do that now hopefully

[void sky]

now

how many roundings

lmaoo

[hollow lintel]

i would do 3

[void sky]

so -1.176

[hollow lintel]

but you just know some sophomore CS intern made this site

yeah.. probably

[void sky]

Yep

FInally

FUCKK

THanks

[hollow lintel]

was that the right answer 😭

[void sky]

LETSGOOO

Some cs intern did it right

[hollow lintel]

YIPPEEE

ok now remember

every time you see a scary circuit

[void sky]

i got 0.67 cuz of my stupid guess

[hollow lintel]

find ground and redraw it with proper squares instead of some hexagonal horror

[void sky]

Alright you basically saved me, now i can sleep in peace

[hollow lintel]

you're welcome

[void sky]

So ground is usually where current comes from

[hollow lintel]

noooo

ground is where it goes to

and then it comes back out the other side through the current generator because we draw it like that because its easy

[void sky]

But the current source has an arrow to the left right

[hollow lintel]

ground is where EERYTHNG is going

[void sky]

Or

[hollow lintel]

the current source is bullshit

[void sky]

Oh

LMAO

[hollow lintel]

we have one wire connected into the deep earth

and one wire connected to the positive side of a battery

[void sky]

The current source has made me worse

[hollow lintel]

the side the arrow is pointing to is the positive side of the battery

the side its coming from is the deep earth

[void sky]

Ohhh

[hollow lintel]

they are not touching

they are unrelated

we just draw it like that for the same reason we say electrons are positive

[void sky]

You know the funniest part

[hollow lintel]

what 😭

[void sky]

i guessed a part aswell

from the other excersise

[hollow lintel]

always guess 0 if you are lazy

[hollow lintel]

easiest free points ever

[void sky]

This excerise

how tf they get -20/dec

[hollow lintel]

dude what class is this

[void sky]

from that weird looking shizzle

[void sky]

HAHAH

Yeah I know

Its bad

[hollow lintel]

😭

[void sky]

This one was easy

the limit

[hollow lintel]

practical circuits for inventors is a good book if you can get it

[void sky]

But the second part was chinese

[void sky]

Fundamentals of electronics

at a uni in the Netherlands

[hollow lintel]

and youre a CS major??

man this is a crime

[void sky]

Yeah i had physcis but they stopped giving it to cs people

[hollow lintel]

my CS school makes you take programming only

oh

[void sky]

but now

[hollow lintel]

that is your replacement physics then

that makes sense

[void sky]

i am the old bachelor program

[hollow lintel]

this is my same book in my circuits class right now

btw

LOL

[void sky]

so i have to do another course that replaces it

[hollow lintel]

i am using this exact book

man you can't upgrade

[void sky]

Wdym upgrade

[hollow lintel]

well good luck, hopefully your semester is almost over

[void sky]

uhh

[hollow lintel]

upgrade your uh

degree plan

[void sky]

5 weeks left

im dead

[hollow lintel]

I can change to the current years degree plan at any point

nice

i have my first final on Thursday

🫡

digital logic

[void sky]

Discrete math is worse tho

[hollow lintel]

and no one makes an A in that class

not even someone who programmed a mini-GPU in verilog

in a month

HEY I LOVE DISCRETE MATH

[void sky]

We have in-class tests of 45 minutes using proofs you just have to memorise

i love discrete

but not the way they test us

[hollow lintel]

THAT IS NOT DISCRETE MATH

ours were 'hey heres some stupid thing. now evaluate it with only a list of theorems and postulates'

[void sky]

Let me have some time

[void sky]

yepp inclusion esclusion theorem

yay

[hollow lintel]

hated those thangs

i need to go back to studying 💀

have a good night

[void sky]

What is digital logic about

fast

[hollow lintel]

uhhh boolean logic

programming motherboards

state diagrams

[void sky]

oh in that sense

[hollow lintel]

it's theoretical computer engineering

[void sky]

We had a whole class about logic and proofs

it was nice

best course ever

Flag-style derivations

[hollow lintel]

what in gods name is your discrete math class then

ours was just boolean logic and proofs using them

[void sky]

Yeah that

was our logic course

[hollow lintel]

interesting

[void sky]

lol

Discrete math is just some shit they made up

because they needed another course or sum

[hollow lintel]

yeah considering you LITERALLY ALREADY HAVE IT AS A COURSE

that sounds pretty made up to me

[void sky]

Oh

[hollow lintel]

they should just make you do a hands on programming class

[void sky]

We have data structures next semester

yay

[hollow lintel]

haven't taken that im CpE never will

im taking embedded systems, electronics, operating systems, and signal processing

[void sky]

4??

Damn

okay ima stop using this channel

/solve

or sum

.close

[odd edge]

Channel closed

Closed by @void sky

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[cold sage]

Am i just going crazy

[night rapids]

my bad

[cold sage]

Or is d wrong there

Nw

[brittle beacon]

,rccw

peak

[brittle beacon]

,w rref [[1, -1, 2, -2], [2, -2, -1, 1]]

[clever fjord]

Results provided by WolframAlpha

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

Requested by chartbit

[brittle beacon]

Seems fine to me I think, how comes you think it's wrong?

[cold sage]

im just not quite seeing how the span of (1,-1,0,0) and (0,0,1,-1)
and the span of (1,1,0,0) and (0,0,1,1) are the same
unless you split the first 2 vectors into (1,0,0,0),(0,-1,0,0),(0,0,1,0),(0,0,0,-1) then put them back together

[brittle beacon]

I mean, assuming it's supposed to be a kernel they're working with, then if you had $\pmqty{x_1 \ x_2 \ x_3 \ x_4}$ in there, that RREF tells you that $x_1 - x_2 = 0$ and $x_3 - x_4 = 0$, so then you get it being in the form they say $\pmqty{x_1 \ x_1 \ x_3 \ x_3}$

[cold sage]

ah right, thats makes sense, thanks chart

[clever fjord]

@brittle beacon

[brittle beacon]

[cold sage]

.close

[odd edge]

Channel closed

Closed by @cold sage

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[median cedar]

no idea what to do

i tried proportions and stuff

dont work

[odd edge]

@median cedar Has your question been resolved?

[median cedar]

<@&286206848099549185>

[verbal loom]

WHATS 2+BQ

[odd edge]

@median cedar Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[rotund fjord]

hello

[rotund fjord]

i just wanted someone to double check my work

the question was find the taylor series for x/8 when a = -2

this one is find the power series when x = 0 and the intervals for the series

<@&286206848099549185>

[whole pumice]

Sry Taylor series is bit too much for me but I would have liked to help u

[rotund fjord]

so do i wait for someone else?

[rotund fjord]

can i ping helpers again? or just keep wiating?

[warped glacier]

yep, first one is correct

yeah 2nd one is correct

next time you can use Wolfram Alpha and Symbolab to check your work

[rotund fjord]

ah ok ty

[warped glacier]

https://www.wolframalpha.com/input?i=8%2Fx+taylor+series+around+x+%3D+-2

for instance

[rotund fjord]

its ccause i wanted to check my steps also

[warped glacier]

ah ok ic

[rotund fjord]

not just my answer

cause i want to be sure im doing it correctly

[warped glacier]

wait there's a tiny error in the 1st one

(-2)^(2 - n) would be alternating signs

you mean $-1 \cdot 2^{2 -n} (x + 2)^n$

[clever fjord]

south

[rotund fjord]

yeah

i think i can just move the negative sign outsideright?

[warped glacier]

nah, cause $(-2)^{2 - n} = (-1)^{2 - n} 2^{2 - n}$

[clever fjord]

south

[warped glacier]

(-1)^(2 - n) is alternating between 1 and -1

[rotund fjord]

i meant like -(2)^2-n

would that work or no?

[warped glacier]

oh yes that works

yeah the brackets really help

you put the - inside the bracket though

[rotund fjord]

ok thank you for catching my mistake

[warped glacier]

no worries

[rotund fjord]

yeah i moved it outside

[warped glacier]

you didn't lol

[rotund fjord]

sry i meant on my paper

[warped glacier]

oh ok

[rotund fjord]

is that the only mistake i had?

[warped glacier]

yes

[rotund fjord]

ok im going to close this then

thank you for your help

.close

[odd edge]

Channel closed

Closed by @rotund fjord

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[desert marlin]

Okay

I'm having trouble understanding your question about the proof

It is proving that differentiability implies continuity

Yes?

We didn't show f(c+0)=f(c) and then claim that this completed our proof

[whole vine]

not c + 0.
its c + apporaching 0

its not the same when you are talking about limits

[desert marlin]

we showed that, given a function is differentiable at a point, then we can use this definition of differentiability to show that it also satisfies the definition of continuity at a point

[whole vine]

yes. becaus if h = 0, then its just c+0. which is just c

you cant prove anything from this

can you?

[desert marlin]

We are taking the limit as h->0, we never have to actually deal with or care about when h=0

write down the definition of f being differentiable at a point c

and also write down the definition of f being continuous at a point c

they should look somewhat similar

so then just with playing with them a little bit

they made this

yes

f is differentiable at c, if that limit exists

and all you need to say for the second part (the bottom) is,
if lim x->c f(x) = f(c)

because if the limit or f(c) didn't exist then that statement couldn't be true

yes

What

No?

the limit does not exist

the right and left side limits disagree

the right side limit is 1

the left side limit is -1

the overall limit cannot exist if the right/left limits are not equal

[quasi sparrow]

You're confusing yourself

[quasi sparrow]

Read this again

[desert marlin]

no I meant the difference quotient limit that you wrote

if the limit as h->0 of [f(x+h)-f(x)]/(h) exists

no

[quasi sparrow]

The derivative itself is a limit

[desert marlin]

The derivative is that limit

[quasi sparrow]

You wrote it yourself here

[desert marlin]

if that limit exists, the derivative does

Well

you have two limits going on

involving the same point

like can you see how they are somewhat similar

no

I am referring to that the definition of differntiability, involves a limit and the point c

and likewise the definition of continuity, involves a limit and the point c

so they're like atleast somewhat similar at first glance

agree?

$f$ is differentiable at $c$ if the limit $$\lim_{h\to 0} \frac{f(c+h)-f(c)}{h}$$ exists. And, $f$ is continuous at $c$ if the limit $$\lim_{x\to c} f(x) = f(c)$$

[clever fjord]

Austin

[desert marlin]

like somewhat similar

Okay so there really is no explanation more to the proof rather than just following their steps again

I mean all they did was take the top piece

the part that was given to you

and they did some algebra to make it into the bottom piece

the piece that you wanted

So just like go through the steps a few times more

they don't ever say that

you need to read it more carefully

write down the steps 1 at a time

At every step of the way, understand why they are "allowed" to do what they do to get to the next line

This question doesn't make sense

I am having trouble helping you because your questions aren't full sentences

Looking only at this line

let x=c+h okay?

then the limit as h goes to 0

is meaning that x goes to c

so that is why they are equivalent

And the part I blurred out, is part of the next sentence

it doesn't matter in that sentence

No one said h isn't 0

that isn't stated in this line

we aren't just saying things willy nilly without proof

and what are you even asking?

say that what IS f(c)?

h?

we aren't saying h=f(c) either

I genuinely don't know what you are confused about

again because you aren't really asking in complete sentences

That is literally in the next line

I blurred it out for a reason

It doesn't matter

this is the sentence you are asking about

so stop reading ahead and confusing yourself

No

who cares what h is

it doesn't matter at all

it is just a variable

...........

they are saying

to show that the limit as x goes to c of f(x) is equal to f(c) [the top line]

is equal to showing the bottom line

they aren't claiming that either are true yet

just that both statements are equivalent

In this line, they defined h this way. They said (without actually writing it) Let x=c+h

then the top line becomes

lim c+h->c f(c+h) = f(c)

agree or disagree?

Okay

but what is lim c+h -> c

well that is the limit of h -> 0

agree or disagree?

Great

so equivalently, after substituting x=c+h, the top line is equal to lim h->0 f(c+h) = f(c)

which is all that sentence is saying

it hasn't proven anything yet

it hasn't proven that those limits mean anything for your proof

But this

is what you want to prove

eventually

right?

because that is proving continuity at c

by definition

so this first line is saying, we can equivalently prove

this instead

since we have shown they are equal

both of them imply continuity

and then it continues with the proof from there

does that make more sense?

Great

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[winter rapids]

With regards to this Boolean algebra, is there any way to simplify it further using any theorems, I used De Morgan's and I'm not sure what should I do after that:

F = xy + !z + !(xy)z
F = xy + !z + (!x + !y)z

[odd edge]

@winter rapids Has your question been resolved?

[odd edge]

@winter rapids Has your question been resolved?

[winter rapids]

.close

[odd edge]

Channel closed

Closed by @winter rapids

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[wispy vine]

How do you solve C? I don’t know where to start

[forest sky]

you just sketched the graphs of the two functions, so look at the graphs and find where the graph of 6/x is lower than the graph x+1

[restive delta]

You could solve it as a quadratic independently too

Since it says "hence or otherwise"

[wispy vine]

i did bfr

i got x=2,x=-3

[restive delta]

Yeah

Oh no wait did you multiply both sides with x?

[wispy vine]

yuh

part a is where i got the x values

also ill need help with part d later (the one above part a)

[restive delta]

$\frac{x^2 + x - 6}{x} \geq 0$

[clever fjord]

Lorentz

[craggy matrix]

x + 1 -6/x

[wispy vine]

where did the x in the denominator come from

[restive delta]

Which becomes
$\frac{{(x+3)}{(x-2)}}{x}\ge 0$

[restive delta]

I subtracted 6/x from both sides instead of multiplying x

In case of inequalities, multiplying x would lead to loss of solution(not always, but at times)

[clever fjord]

Lorentz

[wispy vine]

ohh right right

[wispy vine]

cus x cant be 0

[restive delta]

Yeah

[wispy vine]

so what do i do from here

[restive delta]

You can find the intervals that x belongs to

And that satisfies the inequality

[wispy vine]

in which cases are these two solutions

[restive delta]

You can find the critical points(the values of x at which each term becomes 0)

And then take intervals

For ex, 2 is one such critical point
So two such intervals would be x>=2 and 2>x>(the critical point just lesser than 2)

And so pn

On

[wispy vine]

alrlr

[restive delta]

And then see which intervals satisfy

Ye

[wispy vine]

so is d correct

the working that is

any steps that i should fix or add

[odd edge]

@wispy vine Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[sterile wadi]

should i just use discriminant tas input that in the function or is there an easier way to do this

[mystic saffron]

LMAO

the drawing

[sterile wadi]

sums up my feelings abt this

[mystic saffron]

range of a quadratic

[sterile wadi]

kind strangers, i summon you

[mystic saffron]

☠️

[mystic saffron]

yk

you can solve this using a parabola

[ember oak]

<@&268886789983436800>

[mystic saffron]

is the range -oo to 53/5

[runic swift]

quadratic functions have the shape of a parabola, and parabola's have a maximum/minimum point, you just need to find that point and you have it

[mystic saffron]

so it goes from -oo to the vertex right

[runic swift]

yes

[odd edge]

@sterile wadi Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[cinder perch]

what am i supposed to do here?

[cinder perch]

i know with a quadratic equation you use b^2-4ac for discriminant thing

but how for x^5

what do you do

[terse roost]

So first, based on your understanding, what does it mean for a root to be real?

[cinder perch]

its a real number?

i have no idea

isnt it just a value of x that works in the equation

[terse roost]

Yep it’s a solution that’s real

So whenever the graph crosses/touches y=0

You have a real solution

You can also make a couple of conclusions about the general shape of the graph

[cinder perch]

is there like a method?

for bigger equations like this

[terse roost]

A couple, mostly logic based

Alright simplify the problem for yourself and remove the y translation

[cinder perch]

why translation?

[terse roost]

So try and factor $x^5+3x^3$

[cinder perch]

okay

[terse roost]

And you’ll have a better understandinging*

Then sketch it

And apply a why translation

And you can see the num of sols

[cinder perch]

what is a why translation

i can kind of see the logic for 3

[terse roost]

Y*

[cinder perch]

x^3(x^2-3) so theres 2 solutions at least

[terse roost]

Sorry autocorrect is sxfeweing me so much

[cinder perch]

thats fine

[terse roost]

Diff of squares!

[cinder perch]

im just using desmos

[terse roost]

You can factor further using difference of squares

[cinder perch]

oh

[terse roost]

And then you can see 3

[cinder perch]

wait

i forogt that existed

doesnt that only work for quadratics?

like x^2+x+6

wait im thinking of complete the square

i have no idea what that is

so its like this?
how does difference of squares work for x^5 and x^3?

[wispy vine]

let u = x^3

[terse roost]

Difference of squares is where if you have x^2-a^2 it factors to (x-a)(x+a)

[cinder perch]

yeah ik about it just didnt know the name

[terse roost]

Nw and then since the 0 solution has a multiplicity of 3 so not an even one

[cinder perch]

then x^5 becomes u^5/3

[wispy vine]

no it becomes u^2

[cinder perch]

am i stupid

wait

u=x^3 then u^2 = x^6? no?

[wispy vine]

yes

but to simplify it

substitute u = x^3

that way it forms into a quadratic

[cinder perch]

x^5+3x^3+1=0

you cant simplify it with u=x^3

[wispy vine]

u get u^2 - 3u + 1

[cinder perch]

no

[cinder perch]

this\

doesnt work

[wispy vine]

hold up

i might be dumb

is it B?

[cinder perch]

its c

[cinder perch]

look here

[wispy vine]

i think this has something to do with cubic curves

[cinder perch]

https://www.youtube.com/watch?v=3Z2BGDpHYRY

think its like this

but it doesnt work on mine

[wispy vine]

thats true

i actually dont know this

sorry man

ping a helper if you'd like

[cinder perch]

sure

<@&286206848099549185>

[wispy vine]

the guy bfr who helped seems like he knows what he's doing

better than me for sure

so try to get him back

again, sorry about that

[cinder perch]

https://www.youtube.com/watch?v=Iaq7z7reznM

oh

it might be this

[wispy vine]

wait does ur question fall under polynomials?

[cinder perch]

i think?

[wispy vine]

oh then i cant do anything at all

im not up to that stage yet

[cinder perch]

yeah it does i think

[wispy vine]

i learn that next year

[cinder perch]

i mean

i am in undergraduate

[wispy vine]

me too

but have you learnt polynomials

do you konw the laws and theorems and all that

[wise idol]

maybe descartes rule of signs can help

[cinder perch]

i dont tihnk so

im going insane

decrates rules of signs

you need to actually solve for x

to check it

[odd edge]

@cinder perch Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[modest badge]

Need help with 179

[odd edge]

@modest badge Has your question been resolved?

[modest badge]

<@&286206848099549185>

[odd edge]

@modest badge Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[mortal veldt]

if anyone understand lodev raycasting tutorial(https://lodev.org/cgtutor/raycasting.html#Textured_Raycaster) i want to ask him about this H: hitpoint of the ray on the wall. Its y-position is known to be mapY + (1 - stepY) / 2 yDist matches "(mapY + (1 - stepY) / 2 - posY)", this is the y coordinate of the Euclidean distance vector, in world coordinates. Here, (1 - stepY) / 2) is a correction term that is 0 or 1 based on positive or negative y direction, which is also used in the initialization of sideDistY.

[mortal veldt]

i cant prove it

[odd edge]

@mortal veldt Has your question been resolved?

[odd edge]

@mortal veldt Has your question been resolved?

[odd edge]

@mortal veldt Has your question been resolved?

[mortal veldt]

.close

[odd edge]

Channel closed

Closed by @mortal veldt

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[gusty blaze]

how can i figure out the upper limit of integration

[gusty blaze]

sorry i did not include the equation

[grizzled tide]

notice that the two bounds are when r = 0 and r = 5

use the formula to solve the bounds in terms of theta

[gusty blaze]

ohh

so i need to solve

5 cos 5theta = 5

cos(5theta) = 1
acos(cos(5theta)) = acos(1)
5theta = 0
theta = 0?

[grizzled tide]

thats one of the bounds

whats the other?

[gusty blaze]

5 cos 5theta = 0

cos(5theta) = 0
acos(cos(5theta)) = acos(0)
5theta = pi / 2
theta = pi / 10

i was thinking 5 cos 5theta = 5 would give me the upper bound tbh

[gusty blaze]

but this is it

thank you

[grizzled tide]

[odd edge]

@gusty blaze Has your question been resolved?

Channel closed

Closed by @gusty blaze

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[carmine tree]

how to obtain the reference angle: 30 degrees

[green ledge]

whats the question

do you have to calculate sin(5pi/6)?

[carmine tree]

the third quadrant formula doesn't work

[green ledge]

well it's not in the third quadrant

how many degrees is 5pi/6?

[carmine tree]

150

[coarse bane]

150

[green ledge]

so thats which quadrant?

[carmine tree]

aight mb

2nd

[green ledge]

ye

should work if u apply that

[carmine tree]

got it

thankss

/close

how

.close

[odd edge]

Channel closed

Closed by @carmine tree

Use .reopen if this was a mistake.

[carmine tree]

.reopen

[odd edge]

✅

[carmine tree]

no.6

this was actually the one i got lost

-150-180 = -30

[carmine tree]

idk how the reference angle become 30

formula gives -30

<@&286206848099549185>

[green ledge]

well your angle is specified as -150

you can rewrite that as 210 degrees

210-180 is 30

[odd edge]

@carmine tree Has your question been resolved?

[green ledge]

that's if you really want to use the formulae otherwise you can just visualize it using the trig circle and calculate it

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[odd edge]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[nocturne belfry]

specifically whats your confusion

[strange drum]

end points

[nocturne belfry]

$\sum _{n=1}^\infty \frac{(x-2)^n}{3^n \sqrt n}$

[clever fjord]

jan Niku

[strange drum]

im gotta submit soon

[nocturne belfry]

what do you mean end points

what the radius is?

[strange drum]

3

[nocturne belfry]

i mean what do you mean you are confused about the end points

what about them

[strange drum]

idk if the diverge or converge

[nocturne belfry]

hmm well you should be able to check each one right

lets see the endpoints are 6 and -1?

5 and -1

so if i had to guess

-1 is going to be fine

and 5 is going to be bad

can you see why?

ping if u reply

[strange drum]

no

[nocturne belfry]

so when x=-1, the summand becomes $\frac{(-1)^n}{3^n \sqrt n}$

[clever fjord]

jan Niku

[nocturne belfry]

this is term-by-term less than the alternating harmonic series

and the alternating harmonic series converges

clearly $3^n > n$ i hope

[clever fjord]

jan Niku

[nocturne belfry]

so $\frac{1}{3^n} < \frac 1n$

[clever fjord]

jan Niku

[nocturne belfry]

since $3^n \sqrt n > 3^n$ we get $\frac{1}{3^n \sqrt n } < \frac{1}{3^n} < \frac 1n$

[clever fjord]

jan Niku

[nocturne belfry]

hows this

[strange drum]

ok

[strange drum]

so diverges at -1?

[strange drum]

lim n-> inf | (x-2)^n+1 / 3^n+1 Vn+1 * 3^n Vn / (x-2)^n

lim n-> inf | (x-2)Vn / 3V(n+1)

|-2+x|/3

|-2+x|/3 < 1

-3<-2+x<3

-1<x<5

R=5- -1 /2 = 6/2 =3

check endpoints

x=-1

(-1)^n /Vn

converges via ast absolutely

x=5

1/Vn

harmonic diverges

[nocturne belfry]

well we showed its term by term less than the alternating harmonic series

[strange drum]

interval (-1, 5)

R=3

[nocturne belfry]

and the alternating harmonic series converges...

[strange drum]

converge abs (-1, 5)

conditionally at -1

[nocturne belfry]

,w Sum[ (x-2)^n / ( 3^n Sqrt[n] ), {n,1,Infinity}, VerifyConvergence -> True, GenerateConditions -> True]

[clever fjord]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by janniku

[nocturne belfry]

no offense if you just want the answer im not interested im busy with this paper and not looking for additional homework atm

[strange drum]

idk what it is

[strange drum]

do you?

[nocturne belfry]

do i what?

[strange drum]

know the answer

[nocturne belfry]

yes

,w SumConvergence[(x - 2)^n/(3^n Sqrt[n]), n]

[clever fjord]

Results provided by WolframAlpha

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

Requested by janniku

[nocturne belfry]

there ya go

[strange drum]

it doesnt tell me abs or conditional

[odd edge]

@strange drum Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[rose quail]

Hi

[rose quail]

Okay so

Its a question in durch so ill translate it to english

Given the curve with equation 8x^3 + y^3= x^5y^4 give an equation of the tangent to the curve in the point p(1;2)

What i did was this

Ignore the big text

So basically i took the derivative

[fervent pewter]

okay..then?

[rose quail]

So yh now i jave this

[fervent pewter]

perfect!

then?

[rose quail]

And when u fill in the points (1;2) u get -56/29

And thats where im stuck

How do i make an equation out of that

[fervent pewter]

okay

so can you define tangent for a curve first?

[rose quail]

Y=f(a)+ f’(a)(x-a)?

This is the formula we use

[fervent pewter]

no forget the formula

what is the definition of tangent

like how will you explain meaning of tangent to a common man

[rose quail]

Its the rico of the first derivative

Wait no rico is in dutch

Lemme look up in english

So the tangent is the direction coefficient of the first derivative?

[fervent pewter]

wait

have you heard of tangent before? like when you were young in middle school maybe?

[rose quail]

I know what a tangent is

I just gotta find a way to get that equation

With my point

[fervent pewter]

no I get your problem

[rose quail]

And the number i became (-56/29)

[fervent pewter]

for that I will guide you

[rose quail]

Thank u

[fervent pewter]

so now can you tell me the what kind of curve tangents are?

[rose quail]

Do u mean what sorts there are?

[fervent pewter]

no like can tangents be anything? can it be a circle, straight line or parabola?

[rose quail]

I’d say a straight line

[fervent pewter]

are you sure?

[rose quail]

In dutch its called raaklijn which means touching line

So yeah im sure…

[fervent pewter]

ok nice!

so you are correct on that aspect

so now we need to find the equation of the tangent line for the given curve at the point (1,2)

so now tell me to get equation of any straight line what informations do you need?

[rose quail]

U need the equation itself with the x of ur point filled in it

And the same with the derivative of ur equation

[fervent pewter]

forget derivatives, calculus for now

think just coordinate geometry

[rose quail]

Ion understand what ur asking me then 😭

[fervent pewter]

what are the requirements for finding equation of any straight line?

[rose quail]

U mean like

The gradient mx and its height b?

Like y=mx+b?

[fervent pewter]

yes good!

[rose quail]

God thats so long ago im surprised i even got that

[fervent pewter]

what is m exactly?

[rose quail]

Ur gradient? Are u asking me what a gradient is?

[fervent pewter]

no this is fine!

Actually it's slope but gradient is fine as well for now

[rose quail]

Aaah stijging of daling in dutch

Alr i get it

[fervent pewter]

now tell me what is this value? Is this gradient/slope?

[rose quail]

So earlier i talking about a “touching line” my teacher told me that number is the touching line of my tangents

[fervent pewter]

okay

umm I think the language difference is confusing you a bit

but anyways..

[rose quail]

Translated “fill in every x=1 and in y=2 then u have with this number the touching line of the tangent

Rico is touching line

[fervent pewter]

this value is the slope/gradient of the line

[rose quail]

Wait yh nvm rico is not touching line

Tangent is touching line

My apologies

[fervent pewter]

no worries

[rose quail]

So this is m?

[fervent pewter]

now in y = mx + b

yes

m = -56/29

[rose quail]

Whats my b then

[fervent pewter]

now all you need to find is b

[rose quail]

Or how do i get it

[fervent pewter]

now think about it..there is some information already given in the question which can be used to solve for b

[rose quail]

Our point?

[fervent pewter]

just think about that for a few mins and then let me know again

[fervent pewter]

are you sure?

[rose quail]

😭😭 no man i really have no idea

[fervent pewter]

no problem

[fervent pewter]

you guessed it correctly!
so the point lies on the tangent line
so in y = mx + b put x= 1 and y = 2..m you have already found out
solve for b

then you are done!

is this fine?

[rose quail]

If ive done it correctly then yes

I have my answer

Just no correction sadly but it’s cool

Ty for this amazing rollercoaster

.end

Ah doesnt work like that

[fervent pewter]

Just one thing

Try to understand the steps of the solution.. I just have a feeling you are trying to solve the problem by trying to remember the steps which is not the correct way tbh.. try to understand why the steps are being taken..

you might face difficulties if a little variation of problem is introduced which ideally should not be the case.. so try to have a good understanding of the topic

[fervent pewter]

it's "close"

[rose quail]

I try that but see the problem is, my math is dookie. Working something out i can do that perfectly fine but indeed when faced with questions like these i freeze

But still ty

Hopefully we’ll see each other on another prob of mine

“Close”

Close

[fervent pewter]

no

haha

[rose quail]

.close

[fervent pewter]

wait

[odd edge]

Channel closed

Closed by @rose quail

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[sly wedge]

s

[sly wedge]

i need help

for this limit

[odd edge]

@sly wedge Has your question been resolved?

[sly wedge]

<@&286206848099549185>

[stable robin]

The trick is to get the stuff inside the parentheses to be (1+h)

Because this will ultimately allow you to use the definition of e. Try substituting h = -3x

[sly wedge]

Ok

if h=-3x then lim x->0 (1+h)^-2/x

so x = h/-3 ?

then lim x->0 (1+h)^6/h ?

then we do [lim x->0 (1+h)^1/h]^6

@stable robin like this ?

[stable robin]

Yea exactly

[sly wedge]

so its = e^6

[stable robin]

Don’t forgot to substitute the x in the lim x -> 0 as well

[sly wedge]

yes lim h->0

[stable robin]

Yes

[sly wedge]

thank you

but the substitue work for every lim ?

[stable robin]

You’re welcome

[stable robin]

Usually it works when the limit almost looks like the definition of e

[sly wedge]

its the limit looks like e it always work ?

[stable robin]

I can’t say “always” because I’m not sure. But usually yea

[sly wedge]

okey thank you

[stable robin]

Np!

[odd edge]

@sly wedge Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[rigid bramble]

I'm not too good with function questions, but this one came up on an exam I did recently and I wanna get to the bottom of it.

After some experimenting, I found that f(x) = 1 - x would work, and more generally, f(x) = 1 - g(x), where g(x) is just some injective (not too used to this kind of terminology too, so it may be wrong) odd function of x which maps integers to integers, it satisfies the properties in the question. Even if this is correct, how would I go about proving that functions in this form are the only ones that satisfy the properties of the question?

I was just thinking of functions that are their own inverses, and this and f(x) = something/x are the only ones, but the latter doesn't always map onto integers, so I excluded it.

Thank you in advance! :)