help-23

[thin bridge]

find the number of trailing 6s?

[wraith yoke]

Find how many trailing 6 are in factorial

[thin bridge]

well once you exceed 5!, the number will end in a 0

so 5!, 6! ... etc will have no trailing 6s

1!, 2!, 3!, 4!
are small enough to check by hand

[wraith yoke]

Oh ok

Thx guys

I have no questions no more

!close

;close

,close

$close

*close

(close

)close

+close

-close

.close

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[lean otter]

hi

[lean otter]

i have a question which i am really struggling at

this one

how to i find the number of moles

n

[ebon rock]

do you have a data sheet?

[neat kiln]

Is this in?

In

[lean otter]

actually its alright

[neat kiln]

Ib

[lean otter]

i get it now

pv = nrt

sub them values in

i forgot that equation existed

unrelated statement : in chemistry i used to divide and multiply some numbers from the given until i got the answer lol

[neat kiln]

Trial and error is the way

[lean otter]

always works

[vestal sedge]

stoechiometric calculations arent necesarily difficult just tedious imo

[lusty widget]

#idealgaseqn

[lusty widget]

oh ok nvm u got it

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[feral cedar]

I need some help with a few complex numbers questions

[feral cedar]

(7-i)^2

[final halo]

can you expand (a-b)^2?

[feral cedar]

Isnt that (a-b) (a+b)

[neat kiln]

No

[final halo]

no its (a-b)(a-b)

[feral cedar]

Whoops

[neat kiln]

That's a^2 - b^2 you are talking about

[feral cedar]

Oh alright sorry

[neat kiln]

(a-b)^2 = a^2 + b^2 - 2ab

(a+b)^2 = a^2 + b^2 + 2ab

[feral cedar]

Alright

[lean otter]

most importantly, i°2 = -1

[feral cedar]

Can I just plug it in that question I sent

[feral cedar]

What does the circle thing mean

[neat kiln]

i^2

Circle is odd choice

[lusty widget]

i#2

[feral cedar]

Oh

Ok hold on

[neat kiln]

Which language?

[feral cedar]

I got 49+i^2-14*-i

Unless I did something wrong

I can combine the first two terms since i^2=-1

So 48-14*-i

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[lean otter]

I know how to solve it

I just dont know how to draw it

[lean otter]

place any point on the paper (the radar)

place a point 4.5cm away with an angle of 42°
another one 3.3cm with an angle of 58°

like this?

@lean otter

yes but use correct scale and correct angles

cause aint no way thats 4.5cm

they said u dont need to draw it by scale

also

when i solve for this

it gives me a dif answer

then what it says on the answer sheet

[lean otter]

if you draw it correctly you can just measure the distance using a ruler

right

so the 2ships are 5.23km apart

it says

its

6 km

in the answer sheet

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[steep sedge]

https://cdn.discordapp.com/attachments/903463355619110912/1040051609926909992/20221109_185315.jpg

[steep sedge]

is this parameterization correct

[safe radish]

@steep sedge Has your question been resolved?

[foggy salmon]

um

it shld b from pi/2 to pi

the t

@steep sedge

[steep sedge]

so to correctly parameterize a clockwise circle you just take what it would be if it was flipped over x axis?

[foggy salmon]

er

idk

i just know this is circle centred about origin with radius r

clockwise so its re^-it

anticlockwise would be

re^it

then from there just try to find the approriate value of r

also generally try to keep it in like

a more normal range

like

(0,2pi]

or

(-pi,pi]

but yea any would work

[foggy salmon]

so this

alternative if we added 2 pi it would work too

so like

5pi/2 <= t <= 3pi

[steep sedge]

Is this a correct parameterization?

Or do i need to conbine it somehow

[foggy salmon]

wot

im so confused with this lol

[steep sedge]

wym

[foggy salmon]

oh its 3 separate stuff

need they connect end to end

[steep sedge]

it starts at 3 and ends and -3

[steep sedge]

wym

[foggy salmon]

like

[steep sedge]

oh so my parameterization of the line should start at 3pi?

[foggy salmon]

do u need the end of l1 be start of l2 or something

[foggy salmon]

well

im saying that works too

[safe radish]

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[foggy salmon]

but the numbers a bit big

.reopen

[safe radish]

✅

[foggy salmon]

n weird

[foggy salmon]

so use like this

generally like

exponential fn repeats every 2pi

so yea

[steep sedge]

is the first parem right at least

for the circle on the right

[foggy salmon]

i didnt even look at it

i was asking about if it needed to connect from end to end

[steep sedge]

Uhhh

Im not sure

[foggy salmon]

ok anyway

um

i just looked at it

idt its right?

like for some circle centred at c or radius r

we have

re^it +c

as the parametrisation

for t in (0,2pi]

ok well sry imma

dip

[steep sedge]

rip

.close

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[wind scroll]

Hey, I was wondering if this makes sense to you, because I'm lost:

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[safe radish]

@lean otter Has your question been resolved?

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[runic rampart]

why didn't they use the direct comparison test here?

can't you compare it to 1/n^2?

[solar hazel]

no?

1/(n^2-9) > 1/n^2

[runic rampart]

Ohhh fair enough

Thanks

.close

[solar hazel]

yep ^-^

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[burnt relic]

How many real solutions does the equation x^2 = −9 have? help

[solar hazel]

0️⃣

[burnt relic]

fhow?

[solar hazel]

squares of real numbers are nonnegative

[burnt relic]

i dont understand can u give example

like the i is that it?

[solar hazel]

nono no i

if you take any real number and square it, you get something greater than or equal to 0

so there's no x such that x^2 = -9

[burnt relic]

i dont understand i just use the quadratic formula and got 3i and -3i

so like if the number has i its 0?

[solar hazel]

.-.

i guess

but

you don't need that black magic formula from a more complicated number system for this

[burnt relic]

i dont know any method xd

[solar hazel]

a real solution to x^2 = −9 is a real number x that makes x^2 = −9 a true statement

so like

is 2^2 = -9 true?

[burnt relic]

no

[solar hazel]

so 2 is not a solution

is 5^2 = -9 true?

[burnt relic]

no

[solar hazel]

so 5 is not a solution

is (-2)^2 = -9 true?

[burnt relic]

no

then -3?

[solar hazel]

(-3)^2 is 9

[burnt relic]

oh ye

[solar hazel]

so not a solution

[burnt relic]

forgot

[solar hazel]

and the square of any number is greater than or equal to 0

so no squares are equal to -9

[burnt relic]

oh i understand now

[solar hazel]

yay!

[burnt relic]

.close

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[limber pine]

"Find the value of p, if 3 is a root of 5x^2 - px - 18 = 0

[limber pine]

i feel so embarrassed that i forgot how to solve this

ik its quadratic but

idrk how to find p rn

<@&286206848099549185>

[worthy hemlock]

!15m

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

[limber pine]

ok

"Find the value of p, if 3 is a root of 5x^2 - px - 18 = 0"

what do i do here

i forgot

[sonic spruce]

if 3 is the root

then the value of this polynomial for x=3 is 0

[limber pine]

but its quadratic

wait

[sonic spruce]

so just plug x=3 in

[sonic spruce]

so?

[limber pine]

thats actually what i did

so 5(3)^2 - p(3) - 18 = 0

correct?

[sonic spruce]

yes

[limber pine]

mhm

so 5(9) - 3p - 18 = 0

45 - 3p - 18 = 0

lemme guess

isolate 3p. answer is 9 💀

bro this is easy what

ok i remember how to do it now

thanks! @sonic spruce

[sonic spruce]

ye

np

have a good day

[limber pine]

u too

@sonic spruce yeah sorry but uh

1 more question

given that m * l = m^2 -lm

solve for m givn that m * 4 = -3

' * ' means binary btw

wait yk what

dont answer yet

ima try to solve it again on my own

problem solving skills

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[lean otter]

Let F(x) = [f(x)]^2 + [f'(x)]^2 ; F(0)=6 ; f(x) is thrice differentiable such that |f(x)| =<1 for x in (-1 , 1) ; then prove that there is no point of local maxima of F(x) in (-1,1) ; also prove that for some c in (-1,1) , F''(c)=<0 and F(c) >= 6

this question was done in class by our math teacher and i couldnt understand it whatsoever. i straight up dont know what happened so i have no intial thought process to give [ please explain thru Lagrange's Mean Value Theorem and Rolle's Theorem]

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[lean otter]

Hi guys again at this again

[lean otter]

I have the original sketch drawing

And the circle is touching

[sonic spruce]

What’s the question?

[lean otter]

Find the ?

Actually

Look at text 4

Help

He explains it there

It is unsolvable

But I sent the original one

Where the circle is touching

[solemn ridge]

can u just ask the question here

[lean otter]

Trying to find what the angle ? Is worth

[solemn ridge]

You cant

unless you give some more details

[lean otter]

Well that’s all my teacher gave me

We made a bet

Look

Hol up

[solemn ridge]

do you know if the vertex is on the center of the circle?

[lean otter]

Nope

Nothing

Only angle b

As far as we got

Was this

But we made a mistake

The circle is touching though

[solemn ridge]

oh now youre telling me XB=BY

And MX=MY

[lean otter]

We’re not sure

[solemn ridge]

wdym by not sure

is it given

[lean otter]

No

[solemn ridge]

so its wrong

[lean otter]

My teacher said

That we have to find out

If it’s

Equal

[sonic spruce]

So you are saying your teacher gave you this?

[lean otter]

Yea

[sonic spruce]

Drawn like this?

That’s it?

[lean otter]

Yep

The original sketch

Just like that

I asked for more

[solemn ridge]

tf

its a geometrical prove

[lean otter]

Apparently I asked some of my math teachers

And they said

That it’s solvable

Just take a a lot of time

I have till next Friedan

Friday

[sonic spruce]

I know not the most rigorous way

[solemn ridge]

well then I wanna make an assumption that the angle is 52

lmao

[sonic spruce]

But wtf else are you to do with rigour when all they give in the question is this sloppy picture

[solemn ridge]

Youre supposed to say it's unsolvable.

[sonic spruce]

…

[lean otter]

Welp

I got this from another person

In a channel

[solemn ridge]

did you send them the exact same thing?

[lean otter]

I sent the sketches

From my friend

Which were bad

Because the circle wasn’t touching

[solemn ridge]

show them

[solemn ridge]

lmao they mention that MY and MX are equal as I said before

so that means its given?

[lean otter]

Ik

But we can’t prove it

[solemn ridge]

so your friend just said "oh I assume" MX=MY?

[lean otter]

Well

We had no other way

If we were only giving

Given

B

[solemn ridge]

thats what I said. But it doesn't mean that you can just add stuff and said they are given

[lean otter]

Yea

But ok back to the guy

That

He told me

From those sketches

unsolvable

go back to the first drawing

by moving the xy line closer to the b point, but without rotating it, you change angles x1, x2, y1, y2 and the one you're looking for but not those already given

so the given angles are not enough because you can change the outcome without changing the starting variables, depending on the drawing

unless the xy line is touching the circle

in that case it should be solvable, but it's not touching in any of the drawings

@solemn ridge

[solemn ridge]

I dont understand what u are trying to say

[lean otter]

This part

How can solve it if it is touching g

[solemn ridge]

You mean if X was still on BC but touching the circle?

[lean otter]

Yes

[solemn ridge]

well then its easy

[solemn ridge]

btw im looking at this drawing

[lean otter]

Yea that was a sketch

But remember the circle is touching

[solemn ridge]

like so?

@lean otter

[lean otter]

No

[solemn ridge]

ok

[lean otter]

Look back at the original one

[lean otter]

This is the one my teacher gave

[solemn ridge]

Exactly like that is unsolvable

theres basically nothing given

[lean otter]

Yea

[solemn ridge]

ok so thats it

[lean otter]

ig so

so it isnt possible?

[solemn ridge]

no

[lean otter]

then?

@carmine glen

my teacher sent me this

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[summer lark]

how do I factor 5x^2+23x-10 step by step?

[sudden birch]

1. What are some factors of 5 * -10 that add up to positive 23?

2. put them into the form (5x - factor1)(5x-factor2)

3. are some of the factors divisible by 5? If so, take out the 5

For example if u have (5x-10)(5x-3), then make it into (x-2)(5x-3)

BOOM u have ur factors

@summer lark

[summer lark]

so the answer will be 5(x+5)(5x-2)?

is my answer correct @sudden birch?

[sudden birch]

Just take the five out of the entire expression, like forget it. Including it will result in an incorrect expression

[summer lark]

why do I need to take the five out?

[sudden birch]

Firstly, if u take out the five, then do u see how it will multiply out to be the original form 5x^2 + 23x + 10?

[summer lark]

yes using the foil method

[sudden birch]

And notice that in (x+5)(5x-2), the coefficients of the x’s have to multiply to 5 in order to get 5x^2. If both of the coefficients of the x’s were 5, then we would have 25x^2 when we multiply them

That’s all the insight I have left. But it has something to do with the fact that x^2 has a 5 in front of it

And that does something

[summer lark]

ohh so taking the five out is just needed to complete and simplify the given?

[sudden birch]

Yup! Very necessary indeed

[summer lark]

ahh ok thx for the help😁

.close

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[umbral solar]

Guys if I had 130$ and now i have 108.2$, how much percent I lost?

[rigid inlet]

What's giving you issue?

[umbral solar]

im trying to calculate but I cant

[rigid inlet]

How have you tried to calculate it?

[rigid inlet]

Forget percents for a bit

How much money did you lose?

[umbral solar]

21.8$

[split ether]

(108.2 - 130)/130 turned into percentage, no?

[lean otter]

130 -108.2 = x
then get the percentage of x wrt 130

[split ether]

Oh wait, how did we lose

Then 130 - 108.2 on the top there

[umbral solar]

hmm

[split ether]

Just turn (130 - 108.2)/130 into percentage

[umbral solar]

very nice thanks

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[safe radish]

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[timid blaze]

guys this means 8+4 right

[safe radish]

@timid blaze Has your question been resolved?

[timid blaze]

!close

[sand minnow]

i have no idea how you come to that, my best guess would be that its $\sqrt{2+3}=\sqrt5$ but its not clear

[flat frigate]

Duh Hello

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[crisp plaza]

hello

[crisp plaza]

so I was solving this problem

and I'm not sure if I'm in the right path...

[crisp plaza]

derivative...

[lusty widget]

what’s the question

kind of hard for me to read haha

[crisp plaza]

oh

[crisp plaza]

y = 7(2x+3) / 6 (x+3)^1/2 (2x-1)^1/3

here...

[lusty widget]

so you have $\frac{7(2x+3)}{6(x+3)^{\frac12}(2x-1)^{\frac13}}$

[flat frigate]

Springsskateboard

[crisp plaza]

yes

btw is my solution on my paper clear?

[lusty widget]

no

I’m not sure what ure doing

u can look at it as $\frac{7}{6}\cdot{\frac{d}{dx}}(\frac{2x+3}{(x+3)^{\frac12}}\cdot{\frac{1}{(2x-1)^{\frac13}}})$

[flat frigate]

Springsskateboard

[lusty widget]

and use product + quotient + chain

or double product + chain

I would just take it as $\frac{7}{6}(2x+3)(x+3)^{-\frac12}\cdot{(2x-1)^{-\frac13}}$

[flat frigate]

Springsskateboard

[crisp plaza]

oh I think that would be easy...

it's kinda shortcut right?

[lusty widget]

ig it’s so you can allocate f(x) and g(x)

then follow the formula

[crisp plaza]

however...

we are not allowed to do such thing...

but anyways...

I'm solving now and I'll send the solution to you...

[safe radish]

@crisp plaza Has your question been resolved?

[lusty widget]

what

[crisp plaza]

yeah

[crisp plaza]

this... we do not do this...

[safe radish]

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[fresh widget]

why is finding the roots of a polynomial helpful/ important?

[fresh widget]

in other words, why do we set a polynomial equal 0 and solve for x?

[sudden forum]

To see where the graph of that polynomial hits x axis

[lusty widget]

🛹

[fresh widget]

yes i can visualize it graphically but whats the point

[lusty widget]

to find roots

[fresh widget]

how does that help?

[sudden forum]

Why would you solve x+7=0

[fresh widget]

idk

[fresh widget]

how does roots help

[lusty widget]

to get x

[fresh widget]

roots help to find x?

[lusty widget]

it helps u to see if it had roots in the first place

has

and if it does

u can find it

[lusty widget]

real/complex

[fresh widget]

i still dont see how knowing what x value equal the equation to 0 is going help to solve for x

[lusty widget]

for a graph

the x intercept

is on the x axis right

and at the x axis

y is 0

[fresh widget]

yes

[lusty widget]

usually u have an equation in the form y= something in x

[fresh widget]

then u set y=0 then for solve for x

[lusty widget]

yes

[fresh widget]

then thats how u find the roots

[lusty widget]

yes

[fresh widget]

but how does finding the roots of a polynomial equation is going to help me find x

[lusty widget]

the roots are x

roots is another way of saying x-intercept

so if u have roots 1 and 2

that means x=1, x=2

[fresh widget]

oh ok but why do the find the roots of a polynomial expression?

[lusty widget]

why don’t we need to

if we don’t

then what would u do

[fresh widget]

well i would set it to any y

[lusty widget]

then that won’t be the root

because y is always 0 at the x intercept

[fresh widget]

but how is the root of a polynomial expression helpful/ important

[lusty widget]

it helps us roughly know what the graph looks like

combined with other factors ofc

[fresh widget]

oh so the point of finding the roots of a polynomial equation helps us to find how the graph roughly looks like

[lusty widget]

Yes

[fresh widget]

so we can graph it easier

[lusty widget]

in order to sketch a graph, u need x, y intercepts, turning pts etc

so it’s a manual way

[lusty widget]

depending on the type of graph

[fresh widget]

how do you find turning pts?

[lusty widget]

dy/dx =0

[fresh widget]

oh true

thank you for helping me understand this better

[lusty widget]

np

pleasure

[fresh widget]

have a nice day!

.close

[safe radish]

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[lean otter]

One thing i always was confused about regarding polar coordinates, why exactly would those points be equal?

It is pretty clear that the point would just be a 180 degrees difference, and it would have the same distance, but it is physically not in the same place as the other one no?

[foggy salmon]

well

whenever u spin 2pi

u end up in the same location

take a simple example

e^it

with t = 0 and t=2pi

e^i(0) = 1

e^2pii = 1

same for any number of rotations

anticlockwise or clockwise

clockwise would be -2pi

[lean otter]

I was more talking about the ones with the pi difference in angle

[foggy salmon]

so we get

hm?

oh well

e^it with t being pi

is the same as

-1

they added a - in front

the 2 - cancel

normally we look at positive length

aka

re^it

with r>0

[lean otter]

(5, pi/3) is equal to (-5, 4pi/3) just was confused about that because aren't those two points completely opposite how can they be the same

[foggy salmon]

but if u consider r<0 as well, yes theres that

well its just like

the same

-1 = e^ipi

so just an extra rotation of a bit

thats how they get it

unless im misunderstanding smt idk

[lean otter]

Didn't really understand that because I don't understand Euler's formula all that much, but I'm just going to assume that polar coordinates care more about the length and direction of the point rather than its actual place

.close

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[keen pebble]

I need help with my homework tho

[keen pebble]

They say the 3 extremas wich i calculated already

equals an triangle

wich is true

And they want us to calculate the interior angle

But at extremas the gradient is 0 right?

Like when id put a tangent on that point

It would pe parallell to the x-axis

right?

So what do i exactly do?=???

And the gradient equals the angle right? and if its 0 the arctan of 0 is 0 aswell

Im confused

really

[drowsy moss]

it sounds like they want you to use the three extremes to make a triangle, and then find the angles of that triangle.

[keen pebble]

Yea

Thats right

But the thing is

Gradient= arctan (gradient)

And the gradient at extremas is 0

wich makes the angle equals to 0 aswell

[safe radish]

@keen pebble Has your question been resolved?

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[compact egret]

bruh

[compact egret]

The slope is -3/(2^9)

But I can't find right calculation for tangent

-3/(2^9) * (x - 16) - 1/8 ?

[polar lynx]

You have the slope and you know the x coord for your original curve y you want to find a tangent at

That gives you a point that should also be on the tangent line

[compact egret]

@polar lynxyeah I wrote equation for that?

[polar lynx]

So the rest is just y=mx+b stuff you know m and you can find (x,y)

[compact egret]

Yeah

Did you even read what I wrote?

[polar lynx]

Yep

Did you read what I wrote

[compact egret]

-3/(2^9) * (x - 16) - 1/8 ?

this is

y = mx+b

no you didn't lol

[polar lynx]

Ok?

[compact egret]

Why are you telling me I need y=mx+b when I already have it?

[polar lynx]

I'm sorry I didn't read your random equation and immediately guess what you meant by it

[compact egret]

random

when I said

tangent line

[polar lynx]

I'll be sure not to help you in the future

Bye

[compact egret]

Sure is random

cancer

<@&268886789983436800>

He doesn't even read my full question and gets mad for it

[polar lynx]

You posted a random equation and got mad that I didn't immediately know what you meant by it after I gave you a good faith effort explaining what you needed to do?

[warped roost]

please dont call people "cancer"

[compact egret]

So ask?

[warped roost]

anyway you did not present an equation in y = mx + b form, you presented an expression and called it a "calculation"

[compact egret]

"Oh he wrote some equation let's assume it's irrelevant"

[polar lynx]

Maybe the burden should be on you to explain your work?

[warped roost]

you did not write an equation

[compact egret]

big brain stuff there

[polar lynx]

Ah true nami

[compact egret]

I wrote the calculation without y = as prefix

[warped roost]

the confusion is understandable, i dont understand why you escalated it like this

[compact egret]

which I assume is obvious since imtalking tangetnt

????????

[polar lynx]

It's really nt

[compact egret]

The confusion is obvious when I ask for tangent?

[polar lynx]

Anybody can write word salad and random expressionsy

You need to communicate your work?

[compact egret]

You need to maybe ask what it means instead of assuming its irrelevant?

[polar lynx]

Not get mad when people don't magically understand your random string of symbols

You are the one asking for help I already passed calculus

[drowsy moss]

anyway, you're equation is wrong. -1/8 is the y for the curve, not the y-intercept of your tangent

[compact egret]

You fucking just jump to conclusion and it's my fault you miss something

[drowsy moss]

if you follow what Doot said and plug the point into y=mx+b with the slope you found you'll find b

[polar lynx]

I didn't jump to any conclusion

[drowsy moss]

are keep bickering like children. you do you

[compact egret]

Like it has absolutely nothing to do with understanding, I wrote -3/(2^9) * (x - 16) - 1/8 instead of y = -3/(2^9) * (x - 16) - 1/8

And even said tangent

[warped roost]

i will say that, when i read that, i assumed it was a derivative computation and not meant to be the equation of the line

take that as you may

this is probably because you called it a "calculation"

[compact egret]

The slope is -3/2^9

Which is the derivative

[warped roost]

in any case, this was a really sudden and hostile escalation

things are hard to communicate over text and our helpers are volunteers

sometimes they get confused

[compact egret]

Read with context to my original question, he's asking for derivative of x^3/4, then wants to know what the tangent is at x = 16

[compact egret]

Yeah it's hard when helper jump to conclusion and autopilot answer what a tangent is instead of asking what my "random" term at bottom meant

[polar lynx]

I literally gave you the exact method you needed

[warped roost]

anyway, i'd recommend you read what dootdoctor said initially and correct your answer

[compact egret]

I already had done that method, my question was if my calculation for the tangent (y = mx+b) was correct

[warped roost]

no, your value of b is incorrect

[polar lynx]

Fwiw I don't think most people will want to check your calculations for you.

[compact egret]

16^(3/4) = 1/8

[warped roost]

also fyi your presented equation isnt even of y = mx + b form

its of y = m(x-a) + b form

[polar lynx]

Getting into the habit of checking your work on your own is a great skill to work on.

[warped roost]

traditionally this needs to be simplified by distributing in the m (then it becomes part of the constant term)

idk if your calculus class cares about that or w/e

but yeah

[drowsy moss]

caugh since we're complaining about people not reading cough

[compact egret]

why would I not distribute the m?

[warped roost]

idk, you tell me

[compact egret]

???

sudden random derail okay

[warped roost]

youre the one who wrote -3/(2^9) * (x - 16) instead of distributing

[compact egret]

Yeah because writing it distributed is tedious af, I want to know if it even is the correct form, which it isn't according to you since b (1/8) is wrong

So would be waste of time distributing a wrong answer to begin with

[warped roost]

it just seems like youre taking a weirdly adversarial stance here rather than trying to cooperate to get help

i hope you dont behave like this with your teachers/professors

communication errors happen, they dont have to devolve into insulting each other

[compact egret]

Hahaha

Teachers don't usually jump to conclusions

[warped roost]

i have held many office hours and i have definitely misunderstood what students were asking initially, sometimes due to fault of my own

usually it became clear within a quick exchange that, no, they know that already, what they need help with is this

[polar lynx]

Teachers won't care if finding the correct answer is too tedious for you

[warped roost]

if that makes me a bad ta/professor, then i guess i was a bad ta/professor

[compact egret]

yeah if youdid it like doot you are prob terrible ta/prof

if not then no

[polar lynx]

Fwiw I don't really like the accusation that I jumped to an incorrect conclusion.

I disagree with that

[warped roost]

regardless of whether its unreasoanable for someone to instantly and immediately recognize that your "calculation" of the form m(x-a)+b was actually meant to represent an equation of the form y = mx + b

[polar lynx]

I'm not going to check your calculations for you even if you do them right and actually ask for that.

[warped roost]

this is the kind of thing that can be quickly resolved by just saying "yes, i have the equation right there, im wondering if my numbers are correct"

[polar lynx]

I gave you the method to use. You do the calculation.

[warped roost]

also yeah this isnt an answer-checking server

why not plot it on desmos and see if it actually forms a proper tangent?

[compact egret]

It was literally the first reply I gave to him

[polar lynx]

"Yeah I wrote equation for that" no you didn't

[compact egret]

It didn't work :)

[warped roost]

i dont see any equations

i have definitely had students say "i did that already" referring to work they'd done on paper and not shown off

[compact egret]

oh right you can't put y = in front

[polar lynx]

You barely even let me finish my explanation before getting mad about your "equation".

[warped roost]

well, your equation is not an equation for "that", if "that" is meant to be responding to what doot was talking about (ie finding a point using the tangent)

like i get the impression that you feel like doot was insulting your intelligence or something by explaining it to you in detail

and thats why you got so offended and escalated

im pretty sure he just wanted to see what was tripping you up

whether it was a fundamental misunderstanding or whether you just wanted your numbers checked

¯_(ツ)_/¯

[polar lynx]

Fwiw I was absolutely not trying to insult you with the method I gave you initially.

[compact egret]

it's the over-explaining autopilot google answer of tangent equation when he's not even sure what the full question is that's annoying

;)

[warped roost]

in any case, please dont call users "cancer" for trying to help with good intentions, even if you think their help is inadequate

[polar lynx]

If I came across that way I apologize.

[warped roost]

you can ask them not to help you and block them

[compact egret]

good intentions lol

alr

good point

I'll try in a new fresh channel

.close

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[manic juniper]

Please help me with basic geometry

[sonic spruce]

What do you think you should use?

[manic juniper]

I think slope

?

.close

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[faint saffron]

how to solve this, the answer is 5% but i didnt get why

[safe radish]

@faint saffron Has your question been resolved?

[vale pulsar]

Let income before tax =y, let income tax =x and let net income = a. You can then form 2 simultaneous equations. Income before tax-tax paid on income =net income

[faint saffron]

and what to do next?

[vale pulsar]

So you get something like y-xy=a

Then it says when income tax is increase by 19% net income falls by 1%
So y-1.19xy=0.99a

[faint saffron]

but why inc tax is xy

[vale pulsar]

Income tax paid = income tax x income before tax

[faint saffron]

but like the solution was like 0.19x + 0.01x

and then 19+1=20

then

1/20 . 100

= 5

but WHHYYY i didnt get that

[vale pulsar]

If you subtract first equation from second

You get 0.19xy=0.01a

a=y-xy

[faint saffron]

AAA like we already got the percentage rate

but then we again finnd that

[vale pulsar]

If you sub in for a you end up with 0.19xy=0.01y-0.01xy

[faint saffron]

i mean i didnt even get what the question asks

[vale pulsar]

It’s asking for what the original income tax rate is

[faint saffron]

oooo

now it is clearer

[vale pulsar]

Just realised I overcomplicated this but you then get 0.2xy=0.01y so x=1/20 =0.05 so 5% tax

[faint saffron]

yep thank u sm!!

[vale pulsar]

np

[faint saffron]

.close

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[gusty lodge]

hi I have a physics calculation question

[gusty lodge]

how do I work out the currents through the components

[glass carbon]

u know KCL and KVL?

[gusty lodge]

Yes

[glass carbon]

so

u can do system of simultaneous equations using these laws

then solve

I've got 1.5 A for 27V emf, 3 A for 24V emf and 4.5 A for the 4 Ohm resistor

[gusty lodge]

those are right but for 27v the mark scheme has -1.5

How do I create the simultaneous equations

[glass carbon]

in fact directions of currents can be chosen arbitrarily

but - sign tells us current should be in opposite direction

[glass carbon]

u have 2 nodes, so u can make (2-1) = 1 equation using KCL

which is let's say I1 = I2 + I3

[gusty lodge]

Yep I got that one but I’m not sure what to do with the emfs

[glass carbon]

then u need to choose two loops and close them

[gusty lodge]

Will they combine to make 3v?

[gusty lodge]

Ohh I see

So two loops with each one having a different emf and both having the 4ohm resistor?

[glass carbon]

it doesn't matter at all, but in fact there are only 3 possibilities in this case

I mean 3 loops

[gusty lodge]

Is this what I was meant to do?

Actually the labelled part on the top graph is wrong

[glass carbon]

you've redrew loops, but you've assumed current is same which is wrong apparently

do e.g. this

for your notation

for upper loop

it should be 24 = 4I3 + 2I1

and for lower loop: 27 = 4I3 + 6I2

[gusty lodge]

thank you for helping

sorry I was a bit slow

[safe radish]

@gusty lodge Has your question been resolved?

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[calm mason]

ODE

[calm mason]

can anyone help me prove this theory?

[safe radish]

@calm mason Has your question been resolved?

[safe radish]

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[royal berry]

The area of a circle is decreasing at a rate of 110 square inches per minute. At the time when the area of the circle is 4\pi4π, what is the rate of change of the circumference of the circle? Round your answer to three decimal places (if necessary).

[royal berry]

.close\

.close

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[lusty gust]

can you not

so fucking wait your turn

[broken yew]

What the fuck

<@&268886789983436800> spammer

[wild cape]

ty

.close

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[hidden steppe]

Hello, could someone explain how to solve this?

[safe radish]

@hidden steppe Has your question been resolved?

[lean ruin]

well what is h(v) equal to?

try isolating that variable @hidden steppe

[hidden steppe]

depends on what the input is

[lean ruin]

well in general if the input is x

what is h(v)?

its u where -2u + 6v = 9

h(v) = u. define u by moving around terms

[hidden steppe]

how do you know it equals to u? I don't really understand the concept of this

[lean ruin]

read the first line of the question

[hidden steppe]

yes

[lean ruin]

for a given input v, the function h outputs a value u

so h(v) = u

now how are v and u related?

-2u + 6v = 9

[hidden steppe]

h is function v is input and u is output

as I undestand it

but I don't know what they mean with the formula part

[lean ruin]

we are supposed to define h(v) in terms of v

what is h(v)?

h(v) = u

how is u related to v?

-2u + 6v = 9

now move the 6v over

-2u = 9-6v

divide by -2 on both sides

u = 3v - 9/2

u = h(v)

so h(v) = 3v - 9/2

[hidden steppe]

I don't follow but it's this part that I don't understand

[lean ruin]

how is a general function written

in terms of its input

for eg

[hidden steppe]

f(x)

[lean ruin]

f(x) = x2

if i say f(x) = y

and y = x2

and i say write it in terms of x

so f(x) = y

and y = x2

f(x) = x2

[hidden steppe]

yes that makes sense but I still don't understand my problem

[lean ruin]

is it a language problem bro?

ill try simplify

we are given two equations

one says x = y

and the other says y = z^2

[hidden steppe]

don't think so

[lean ruin]

now write x in the terms of z

[hidden steppe]

yeah don't understand what this means

[lean ruin]

writing x in terms of z

means writing x as a function of z

not y

right now, x = y

but we also know the relation between y and z correct?

y = z2

and y = x

so x = z2

the variables are making it confusing

take out a pen and a paper

h(v) = u

-2u + 6v = 9

write these two down

now try to write u as a term of v

in that, isolate u

[hidden steppe]

no I understand this

just don't see how it relates to my problem

[lean ruin]

can u try this

[hidden steppe]

I don't know what you mean by term of v

[lean ruin]

in that, defining it in terms related to v

not to u

we want to eliminate u

you know how acceleration is velocity/time and velocity is displacement/time

but what is acceleration IN TERMS OF displacement?

[hidden steppe]

not familiar with this

[lean ruin]

youre not familiar with speed?

bro think about it this way

you know your sister

your sister knows your brother

but how do you know your sisters brother?

hes also your brother

correct?

thats his relation in your terms

not your sisters term

idk how to explain this its 3am im an idiot sorry

basically write h(v) as a function of v instead of a function of u

[hidden steppe]

I don't really undersatnd the point

[lean ruin]

there is no point

youre overthinking it

just write down the equations and solve

it means write it as a function of v

not as a function of u

[hidden steppe]

i'm trying to understand functions in general and how they work

[lean ruin]

we know h(v) = u

[hidden steppe]

I don't understand the question they ask

[lean ruin]

ok

do you not understand the words IN THE TERMS OF?

it means as a function of or with only this variable

only with v

[hidden steppe]

well not what they mean by it

[lean ruin]

not with u or x or b or a

it means what must i do to v to get h(v)

does that help?

[hidden steppe]

but v is the input

[lean ruin]

yes

so what do i with the input to get the output?

[hidden steppe]

h needs to have some functionality

to get output

[lean ruin]

well what is the output? "U"

we have input

we have output

[hidden steppe]

yes

[lean ruin]

we have relation between input and output

yes?

[hidden steppe]

what relation?

output is just the result of the function

[lean ruin]

-2u + 6v = 9

u is a result of h(v)

correct

[hidden steppe]

yes

[lean ruin]

-2h(v) + 6v = 9

right?

now can u solve for just h(v)?

[hidden steppe]

you want me to calculate this?

[lean ruin]

just solve for h(v)

yes

i want u to calculate that

just put h(v) on one side

and everything else on other side

basic algebra

you got this

@hidden steppe what did u get

[hidden steppe]

h(v) = -4.5 + 3v

[lean ruin]

good

thats the answer

what did they ask you

they asked you what is h(v)