help-0

[dawn wraith]

So less would be the applicable here.

[marsh valley]

no it said what positive integers

[icy trail]

oh i see

i didn't rtfq

[dawn wraith]

The roots are not real

B is the contrary

[marsh valley]

I dont think I can put "no real roots"

less than 3

b<3

[dawn wraith]

And more than negative 3

[indigo jetty]

your discriminant was wrong

[marsh valley]

it wants pos numbers

[marsh valley]

what how

[dawn wraith]

Ah, yeah

[indigo jetty]

write your discriminant again

[marsh valley]

i think it should be [3,\inf]

oh wait

but it cant be 3

[marsh valley]

ok i guess

4(1)(9)

[indigo jetty]

write it as an inequality

[marsh valley]

hm

so

$\sqrt {b^2 - 4ac} \ge 0$

[ocean seal]

BAAPA|INDIAN ELITES

[marsh valley]

?

would it be that

[indigo jetty]

wrong

[marsh valley]

oh wait

[indigo jetty]

first no square root

second, 2 non-real roots

means no real roots

so it is not greater than or equal to 0

[marsh valley]

$ {b^2 - 9} > 0$

uh

[ocean seal]

querty

[indigo jetty]

no it's not -9

[marsh valley]

how

[indigo jetty]

and it's not the greater than sign

what is your 4ac?

[marsh valley]

4(1)(9)

oh wait

lol

36

[indigo jetty]

correct

[marsh valley]

$b^2 - 36 < 0$

[ocean seal]

BAAPA|INDIAN ELITES

[indigo jetty]

yes, finally

now solve for b and you're done

[marsh valley]

b<6

[indigo jetty]

nope

this is incomplete

[marsh valley]

it wants it in integer form

and they want the positive b's

[indigo jetty]

it's not integer form, it's interval notation

and you're missing part of the interval

[marsh valley]

∞

[indigo jetty]

the question asks for all real values, so it has nothing to do with positive or negative

[marsh valley]

oh lol i read it wrong

[indigo jetty]

anyway b<6 is only part of the interval

you are missing out on another part

[marsh valley]

-6

b>-6

[indigo jetty]

yes

[marsh valley]

[indigo jetty]

combine both into one inequality

[marsh valley]

(-6,6)

[indigo jetty]

yes

[marsh valley]

thanks

[indigo jetty]

👍

[lethal token]

A football player kicks a ball from the ground to a maximum height of 12m at a distance of 18m from the kicker. On its descent, another player hits the ball at a height of 2.2m from the ground.

Write a quadratic fuction to model the situation

Help please, you have to write vertex form first right? If so I keep getting the wrong number for "a"

[old cobalt]

<@&286206848099549185>

I need help to understand what this question is asking

[dark granite]

This kinda sounds like a linear algebra question. Maybe try the linear algebra channel @old cobalt ?

[glass field]

theyre asking what transformations to apply in order to get the parabola you have (pictured somewhere presumably). these transformations are things like stretches/compressions and translations

example, translation x^2 3 units to the right is given by (x-3)^2

[alpine sable]

https://media.discordapp.net/attachments/424404533767307276/867806754603925535/unknown.png

hey, im watching a video on the proof of the harmonic serie divergence

but what i dont get is why the guy made those groups of fractions

like

why 1/3 + 1/4 are in one group

and not

1/2 + 1/3 + 1/4

[vale wigeon]

for convenience

each group is shown to have a sum greater than one-half

if you really really want, you can lump in the 1/2 with 1/3 and 1/4

[alpine sable]

ohh i see

so just to make things easier

wow it makes sense now

thanks^^

[hardy vault]

can someone help me in questions 1

[granite bison]

Can someone help me with the simple formula to figure out if X is <= 20% from A.

[alpine sable]

what theorem/property is this

[gray isle]

consider converse parallel line theorems

[limber quest]

what's the word for describing a graph where as x increases y also increases, except its not a constant factor (so not "proportional")? like a scatter plot where we generally see that a higher x corresponds to a higher y, what would you describe the relationship as?

[gray isle]

increasing?

[limber quest]

i mean in a statistical analysis sense

so like im graphing confidence level vs. accuracy (context is subjects are to select things and convey their confidence level in their choice on a likert scale)

so high confidence + high accuracy = optimal, high confidence + low accuracy = misleading, low confidence + high accuracy = luck, low confidence + low accuracy = ambiguous/generally bad

[alpine sable]

why do we add f'(x)x to f(0)

ngl this is where i got lost haha

[topaz scaffold]

This is a Taylor series

Do you want the derivatives of this series evaluated at x = 0 to match up with the derivatives of your function

p(x) is your series centered at 0 so it's this

p(0) = f(0) and p'(0) = f'(0) is true about the series

I suck at explaining oof

[alpine sable]

so like

what im doing is making the derivatives of f(0) match with p(x)

so its more accurate?

[glass lichen]

You're finding the taylor series of f

[alpine sable]

nah youre good

yea this is the introductory vid to taylor and mclaurin polynomials

the factorial is intriguing tho

[glass lichen]

It's just part of the power series

[alpine sable]

oh and

Why is it f'(0)x and not just f'(0)

[topaz scaffold]

Because when you take the derivative of p(x)

You'll get that p'(x) = f'(0)

[alpine sable]

OHHH

Alright alright

Tysm

[mild pulsar]

Could anyone help me with this it is related to something called Sohcahtoa

[topaz scaffold]

I've never seen someone call it that before

lmaooo

[true crystal]

It is called like that in french

opposé, adjacent, hypothénuse

[mild pulsar]

yah

[alpine sable]

It is called like that in spanish too haha

[topaz scaffold]

Isn't sohcahtoa
sin = opposite/hypothenuse, cos = adjacent/hypothenuse, tan = opposite/adjacent

[true crystal]

Yeah that is it

How do you call it in english ?

[topaz scaffold]

Trig?

[mild pulsar]

basic trigonometry

[mild pulsar]

so your correct

[topaz scaffold]

Well just use cos(θ) = adjacent/hypothenuse

[mild pulsar]

alright

so it would be 7.1?

[topaz scaffold]

Yea

[alpine sable]

If a and b are integers satisfying ab - 3a -2b = 3, find all possible ordered pairs for (a,b)

@topaz scaffold help?

owo

[mild pulsar]

thx

[alpine sable]

someone please put me out of my misery 😩

[topaz scaffold]

That's a line

[alpine sable]

what?

[topaz scaffold]

If I isolate b = 3(a+1)/(a-2)

So a ≠ 2

And b ≠ 3

[alpine sable]

ye thats as far as I wet

idk what welse too

do

[topaz scaffold]

All the other order pairs work

[alpine sable]

huh?

a= 3 +2b/b - 3
b is 3(1+a)/a -2

yeah

[topaz scaffold]

It's a hyperbola right?

[alpine sable]

I got that

ye

I think?

wha howdu solve it so quicklyowo

[topaz scaffold]

Oh

[alpine sable]

also I got b(a-2) = 3(1+a)

wha

[topaz scaffold]

Does ordered pairs also mean it has to be integers?

[alpine sable]

yes

[topaz scaffold]

Ohhhhhhh

This then

[alpine sable]

I got b(a-2) = 3(1+a)

im so confused

idk what i did wrong ffs

[topaz scaffold]

Nothing

Just isolating b instead of a

[alpine sable]

soh cah toa

cho sha cao

what does diophatine mean

is this high school algebra 1

[dawn wraith]

I suppose they don’t just say out broadly that you have to solve a diophantine equation.
You do solve those in high-school yes.

[ancient bear]

is this true, guys?

[warm igloo]

would I need to know what csch and ln mean by the time I take the act test?

[ancient bear]

I have no idea about act

[warm igloo]

end of highchool?

[ancient bear]

but ln is natural log and you do that in high school so I guess yes?

[dawn wraith]

I don’t think so.

[ancient bear]

what is csch^-1 x then?

[dawn wraith]

Is arccsch, 1/arcsinh?

[ancient bear]

no

[warm igloo]

natural log as in log e

[ancient bear]

log base e

[dawn wraith]

Then how did you get the previous result?

Did you assume that:
$$ \csch^{-1}(x) = \frac{1}{\ln(x + \sqrt{x^2 + 1})}$$

[ancient bear]

yes

well, no

there's the ln there too

ln in the denom

[dawn wraith]

Yeah hold up

[ocean seal]

Hunnydrips

[ancient bear]

yes, that was my assumption

[dawn wraith]

Here you go

[ancient bear]

the book doesn't give anything on csch^-1(x), just sinh

and i thought its just the reciprocal

is it not?

[dawn wraith]

You just told me that:
$$ \csch^{-1} ≠ \frac{1}{\sinh^{-1}}$$

[ocean seal]

Hunnydrips

[ancient bear]

yeah, is it not?

[dawn wraith]

You told me it’s unequal.
Though I myself am not sure

[ancient bear]

o

no, its equal

[dawn wraith]

It logically does make sense but I can’t know

[ancient bear]

but i found the formula

yikes lmao, nobody ever showed me this in the book

[dawn wraith]

Okidoki

[humble wadi]

how do i convert a triple integral in cartesian/rectangular coordinates to spherical?

i forgot how to

i'm looking at all the formulas but i'm stumped

[abstract grove]

how do i improve in maths

i got a -5

i rlly wanan get better

[glass lichen]

practice, make sure you understand the concept(s)

[abstract grove]

ok#

should i go over my test sheets

and redo the test after

[ancient bear]

great tip is to really understand the foundations

understanding algebra perfectly is very good going into calculus

[abstract grove]

oh okk

[echo dirge]

I got this for part (a):
$\frac{d\phi}{ds}=\frac{c}{R^2}\csc^2\theta$

[ocean seal]

Sozin

[echo dirge]

does that check out?

quite a long question

[inner sequoia]

why is this true

im p sure this is wrong

i literally put this answer only bcs the other answers were all false too

[echo dirge]

$6k-30<2k+5 \Rightarrow 4k<35 \Rightarrow k<35/4$

@inner sequoia might just be programmed incorrectly in your browser/app whatever you're using

[inner sequoia]

yeah cuz it is not right at all

its probably the school thing

[true crystal]

the answer should be k < 35/4

[inner sequoia]

yes

[inner sequoia]

thats what i got

[true crystal]

Then you're right

[inner sequoia]

ok

[ocean seal]

Sozin

[echo dirge]

oops my bad

[bitter fossil]

Hi everyone, I have a general question, and I'm hoping somebody could shed some light.
I'm new to machine learning, and I'm working on a time series problem. I don't quite understand how a regression problem is different than a time series problem.

For what I know, regression uses the correlation of different features to find numerical data while time-series uses extrapolation to find a likelihood lr a variable.

If that's the case, why do we consider it a regression task and not a classification task?

[echo dirge]

<@&286206848099549185>

[inner sequoia]

how do u do that

[echo dirge]

@inner sequoia it's LaTeX

[inner sequoia]

how do i use it

[echo dirge]

@inner sequoia https://www.overleaf.com/learn/latex/Learn_LaTeX_in_30_minutes

[celest anvil]

how do i do this?

[alpine sable]

just assume that one side is X

if you just have 12 yards as maximum material

you will have X as one and 12 - X as the other side
to calculate the area you do X.(12 - X)

[celest anvil]

so 12x-x^2?

[alpine sable]

exactly

so you solve this equation and you use the correct vallue

the question asks the maximum area

largest area, so you will get the two answer, shift into the first relation x(12 -x) and see which vallue gives the larger area

[celest anvil]

ah ok

i see

oof i got 144 which was wrong

is the answer 0?

[alpine sable]

how come

[celest anvil]

miscalculation

[alpine sable]

how did they get from those first denominators to the second set

[thorn kindle]

A/B * C/D = A/D * C/B

[alpine sable]

ty

How do I factor this by grouping?

[ocean seal]

Christan S

[alpine sable]

My answer is:

[ocean seal]

Christan S

[alpine sable]

Where did i messup

[thorn kindle]

2wv - 2w^2v + 3w - 3 = -2wv(w-1) + 3(w-1)

first of all you forgot to factor out the 2

second of all, you should know how to manipulate the signs to make them match

-(1-w) = w-1

[novel coral]

I got (2wv-3)(1-w)

[alpine sable]

can you show some steps

[thorn kindle]

same thing, you can just factor out a negative 1

[novel coral]

Oh ok

[thorn kindle]

let's start with what you got. wv(2-2w) - 3(w+1)

[alpine sable]

ok

wait no

that isnt the question

[alpine sable]

this is

i only showed my answer

[thorn kindle]

yeah i'm showing you what you did wrong

[alpine sable]

ok

[thorn kindle]

notice that there is still a common factor in the first term

[alpine sable]

uhh

between 2v and 3?

[thorn kindle]

(2-2w)

[alpine sable]

yes

the 2

[thorn kindle]

right

so now we have 2wv(1-w) - 3(w+1)

[alpine sable]

yes

can this still be factored more?

[thorn kindle]

nevermind let's start from scratch. you factored it wrong

[alpine sable]

ok

[thorn kindle]

2wv - 3 - 2w^2v + 3w

rearrange

[alpine sable]

yes

[thorn kindle]

2wv - 2w^2v + 3w - 3

[alpine sable]

ok

[thorn kindle]

2wv(1-w) + 3(w-1)

make sense so far?

[alpine sable]

yes

is this channel available

[thorn kindle]

now factor a -1 out of the first term

[alpine sable]

no sry it in use

nvm sorry

[thorn kindle]

-2wv(w-1) + 3(w-1)

[alpine sable]

ok

[thorn kindle]

= (3-2wv)(w-1)

very nice

[alpine sable]

make sens now

thanks

[thorn kindle]

@alpine sable you can ask now

[alpine sable]

how do i do 1,4, and 5

[cinder owl]

for Q1 there are 2 unknowns of which 1 is the answer. think of how to link them in an equation

[celest anvil]

how?

[thorn kindle]

f'(0) = 0, f(0) = 7, f''(1) = 0, f(1) = -3

that's more than enough to solve

3ax^2 + 2bx + c = f'(x) and f'(0) = 0 so c = 0

d = 7

f''(x) = 6ax + 2b
f''(1) = 0 so 6a + 2b = 0 meaning that 3a + b = 0 meaning a = -b/3

a + b + c + d = -3

-b/3 + b + 7 = -3

b/3 - b = 10
b - 3b = 30
-2b = 30
b = -15

a=5, b=-15, c=0, d=7

[celest anvil]

ok thanks

[frozen cosmos]

hello

is this being used

[thorn kindle]

no you're good

[alpine sable]

Whats wrong with this?

[alpine sable]

the triangle have a 90° degree angle

if they have 2 equal sides, they are congruent

side-angle-side

and as a square, the diagonal is shared, so they have the same size

[alpine sable]

Thank you

[zealous adder]

How do I graph this

[alpine sable]

How did I do this wrong?

[alpine sable]

the external angle is equal to the sum of the two internal that isnt its supplement

60 + 4x + 11 = 11x - 6

for example

you do it with the two

Ohh okay

Thank you again

[alpine sable]

its ok bro

we are all united by the math

[warped phoenix]

how is the solution on the right correct when the graphs on the left show something different?

the graphs are x - 2y < 6, and 2x + y > 6

i simplified them in desmos

but looks like i simplified it wrong

[warm wren]

what was the exercise asking?

[warped phoenix]

find the solution set of both inequalities

for x - 2y < 6, i subtracted x from both sides and divided both sides by -2, giving me y < 1/2x - 3

for 2x + y > 6, i subtracted 2x from both sides, giving me y > -2x + 6

[warm wren]

I think x - 2y < 6 is supposed to be y > x/2 - 3 not smaller

in your graphs on the left you have y < x/2 - 3

[alpine sable]

is this even solvable? how can there be an empty natural log

not sure where to even start on my calculator

[austere bridge]

$\ln x$ is shorthand for $\ln(x)$ because math people are lazy and too many paranthesis gets hard to read

[ocean seal]

cgodfrey

[alpine sable]

im still learning about this

and i see

thanks for that

so i input in the calculator

y= e^x - ln(x)

y2 = 3

[austere bridge]

if there's more than one term in the log like $\ln(x^2+1)$ or something, the parenthesis are mandatory, but if it's just $\ln x$ they're not necessary

[ocean seal]

cgodfrey

[alpine sable]

yeah this hw is already beginning to frustrate me

and i see

[austere bridge]

and yeah, you'd graph those two functions and use the calculator to find their intersection

[alpine sable]

so i have two x intercepts

though the answer choices only wants one

[austere bridge]

so you have to be careful, because there's two ways you can find the answer

one way is to graph both curves and find their intersection

the other way is to subtract one from the other, graph that, and find its x-intercept

because if e^x - ln(x) - 3 = 0, then e^x - ln(x) = 3 and that's a solution

[alpine sable]

i think i understand

so i find where x = 3

and where the function curve is

that's my solution?

[austere bridge]

you find where those two graphs intersect

[alpine sable]

yes i see two x intercepts

so if i understood you

the difference between the value of those two intercepts are

or is the solution?

[austere bridge]

those aren't x - intercepts

it's only an x intercept when it crosses the x-axis

[alpine sable]

oh wait

you're right

ugh im burned out

sorry

[austere bridge]

it just looks like that because the graph is just a flat line

all good

so those are your two solutions

[alpine sable]

yeah

[austere bridge]

idk which one it wants you to pick

[alpine sable]

so far the solution you're currently seeing is uh

1.14

the first is

.161

[austere bridge]

cool, so 1.14 it is, then

[alpine sable]

ah okay

is there a reason why

.161 is omitted

or is it not a solution

[austere bridge]

it is a solution

I guess because it's single-answer multiple choice they picked one

idk

[alpine sable]

i see, ugh thats a bit annoying

[austere bridge]

yeah

[alpine sable]

i overthink

thank you!!

[gaunt magnet]

hi

You are given a polynomial function and two of the known roots.  You must determine the values of m and n and then use polynomial division to determine the other x-intercepts and then write the function in factored form.  Use polynomial division to determine the other x-intercepts.  Show your work!

I got

so for this question
I got n=-42 and m=-11
and then using these value we need to find x intercept using long divison
and then x intercepts i got are x=2, - 1, 3/2

**can someone check if i did this right **

bruh use another channel

#❓how-to-get-help

channel is taken

[gaunt magnet]

<@&286206848099549185>

[hexed bloom]

it is given that 7 and -4 are the roots of the equation

hence we know that if we replace x with the 7 and -4, it will give us 0

thus first put x as 7 and then as -4, both these equations will be equal

now u have 2 equations with 2 variables

u can find out values of m and n easily

[hexed bloom]

o wait u just want to check if its right, my bad i didnt read the messages below the image

i'll check and let uk in a sec

[warm wren]

I found m = 11 and n = 46 🤔

[gaunt magnet]

🤔

how did you get them?

[warm wren]

never mind

you're correct

I missed a -

:p

your answers are right

[gaunt magnet]

and my x intercepts are good too?

[hexed bloom]

yep i just checked too

values of m, n and x intercepts are all correct

[thorn tapir]

I did the whole sum, seems correct - if you didn't do any errors in the process it should be perfect

[warped phoenix]

hm..why would you change the inequality sign?

oh wait that seems to be a rule

[warm wren]

These were the originals x - 2y < 6, and 2x + y > 6, so if you take the first one, you have 2y > x - 6 -> y > x/2 - 3

[warped phoenix]

yeah

i just figured that out

that when you divide or multiply an inequality by a negative

you have to flip the sign from > to < for example and vice versa

i think 😅

[dusk smelt]

yes

[warm wren]

well, yeah, I mean, it's basically just swapping sides, just like when you solve an equality, if you bring something on the other side you have to change their sign, same here, but instead of moving everything over we just say "if we swap the sign invert the inequality", it's just a shortcut

[velvet condor]

uhh apparently im stuck on an inequality problem

prove that $\sum_{cyc}{\sqrt{a^3+3b}} \ge 6~ \forall~ a,b,c>0,a+b+c=3$

[ocean seal]

Minh Baka

[velvet condor]

bro use another channel

basic thales stuff and why cant you do it?:)

[alpine sable]

?

I didnt notice you just asked something.

Calm down buddy

[velvet condor]

<@&286206848099549185>

[warped phoenix]

refer to the rules please, you've tagged helpers at least twice now, and you're ignoring guidelines

[velvet condor]

have i?:)

[zealous adder]

Help pls

[warped phoenix]

yes you clearly have...

[velvet condor]

show me the evidence and i will admit it 🙂

[warped phoenix]

[velvet condor]

i tagged once?:)

[gaunt magnet]

I did 7m+n+119=0 and 4m-n+2=0
that how i got my M and n value

[warped phoenix]

you tagged once before, and deleted it

[zealous adder]

U know they have logs right

[velvet condor]

then show me the logs

:v

[warped phoenix]

doesnt matter if you deleted it because for those who didnt see it, they got host pinged

not an admin, i cant

[velvet condor]

tell them to show the logs then?

[flat vale]

bruh just stop

[velvet condor]

once you have the evidence of me tagging twice

[warped phoenix]

LOL right

[velvet condor]

i will admit that i tagged them twice

:()

you just cant argue

[warped phoenix]

[velvet condor]

you have no evidence of me tagging them

🙂

[warped phoenix]

mhm

[velvet condor]

and neither have i

so

[warped phoenix]

whatever floats your boat bud

[velvet condor]

i will just shut up

[gaunt magnet]

Yes

!

[sour spindle]

Q. Evaluate and simplify 4/3 ÷ 5

so confused

[velvet condor]

4/15

[sour spindle]

u sure

fraction division confuses me

[velvet condor]

u mean

[gilded kiln]

DAYUM YOU IS MAD ANNOYING DAWG

[velvet condor]

$frac{4}{3} \div 5$?

(4/3)/5 or 4/(3/5)?

[gilded kiln]

WTF IS HIS PROBLEM 😂😂😂😂😂😂😂😂😂

[velvet condor]

still, you have no evidence
and i am not gonna admit that i did that because clearly i didn't 🙂

if i did that i would've admitted already

[gilded kiln]

SURE DAWG

YOU IS A LYIN BITCH

[velvet condor]

sorry for being a bitch to you 😛

you can just block me if u want

i dont care smh

[gilded kiln]

“You can just block me if I want”

[warped phoenix]

can someone help me in channel 1

[thorn tapir]

uh, I mean, you could eliminate one variable - did you not do Systems of linear equations with two variables before?

Actually you can just even add the two equations, n gets cancelled from LHS

[velvet condor]

nice alt though 🙂

i made that typo with purpose

[warped phoenix]

alt?

bruh??? 😹

[flat vale]

also, if ur mini-modding u should abide by the rules as well..

[warped phoenix]

daddy 👶

[alpine sable]

I-

How would I solve this?

[thorn tapir]

..uh

[velvet condor]

rsq = tsq:/

[thorn tapir]

You can see it's a rhombus and so, it's also an parallelogram. Try to use all the properties a parallelogram has

[velvet condor]

notice the isosceles triangle

[alpine sable]

Is it just 74 @velvet condor @thorn tapir ?

sorry for the pings

[thorn tapir]

you mean the SRQ angle?

if yes then you're correct

[alpine sable]

im so confused

If all triangles add up to 180 and we have 74 then itll be 106 right

and since 2 sides are the sam

you divide that by 2 and get 53 no?

[velvet condor]

that works though

technically qts+tsr=qts+tsq+sqt=180 degrees

bisector thingy

[thorn tapir]

yeah 53 is the correct answer

wait what

wat.

Do they want to put the degree sign or something?

[alpine sable]

I dont believe so

help someone please

[thorn tapir]

I think ask your teacher where is it wrong

[thorn tapir]

It streches horizonatlly by a factor of 3

So

in normal y = sin theta gives y = 1 at x = pi/2 right?

As it's streched by factor of 3, now it will give y = 1 at x = 3pi/2

[alpine sable]

uh yeah?

[quick lake]

sin 1/3 theta

[thorn tapir]

but y = sin x never gives 1 at 3pi/2, it needs to be pi/2

So you need to divide the input by 3 before you plug it in sin

so

$y = \sin{\frac{\theta}{3}}$ is your answer

[ocean seal]

rikusp2002

[alpine sable]

alr tysm

[thorn tapir]

uh.. I guess you;re welcome

[alpine sable]

that's equivalent to this right?

[thorn tapir]

yeah

[alpine sable]

alr

[next pumice]

Just add all the numbers in each answer up and divide them by four

Mean = sum of all numbers / amount of numbers

[bleak basin]

ah

[alpine sable]

can anyone help me with this epsilon proof?

[next pumice]

You're welcome

[alpine sable]

Let $f(x)=x\cos(x)$ Prove that for every $\epsilon >0$ that If $0<|x-0|<\epsilon$ then $0<|f(x)-0|<\epsilon$

[ocean seal]

jswatj

[alpine sable]

my proof so far

[ocean seal]

jswatj

[alpine sable]

er

i should have done |cos(x)| when multiplying x>0

i made a mistake

that makes no sense

im stuck

Let $\epsilon >0$ and suppose that $0<|x-0|<\epsilon$, multiplying both sides by $|\cos(x)|$ we get that $0<|\cos(x)||x|<\epsilon\cdot|\cos(x)|\implies 0<|f(x)-0|<\epsilon\cdot|\cos(x)|$

[ocean seal]

jswatj

[alpine sable]

im confident up to here

[austere bridge]

what properties do you know about cos(x)?

specifically, what's the largest value it can be?

[alpine sable]

it can be 1

[alpine sable]

What number do you see the most of?

its between -1 and 1

[alpine sable]

.

oh

does that mean that $\epsilon\cdot{|\cos(x)|}\leq{\epsilon}$?

[ocean seal]

jswatj

[alpine sable]

since $|cos(x)|\leq{1}$

[ocean seal]

jswatj

[austere bridge]

yep

[alpine sable]

but

is that still okay?

[austere bridge]

why wouldn't it?

[alpine sable]

even tho im trying to prove '<'

strictly less than

[austere bridge]

you already have that $|f(x)-0|<\epsilon\cdot|\cos(x)|$

[ocean seal]

cgodfrey

[austere bridge]

in the 'worst case', $\epsilon\cdot |\cos(x)|=\epsilon$

[ocean seal]

cgodfrey

[alpine sable]

ohhhh

right

yes

[austere bridge]

as long as you have < anywhere in the chain of inequalities, you're good to go

if they're all $\leq$, then you have a problem

[ocean seal]

cgodfrey

[bitter sage]

is this chat occupied, can someone help me?

can anyone help me solve this?

[rigid smelt]

you can establish a system of equations by plugging those points into h(x)

[thorn tapir]

And then solve for m and n

you'll get exact forms of f(x) and g(x)

[bitter sage]

once i have h(x)

how do i solve for m and n

[rigid smelt]

system of equations

[bitter sage]

can u explain further ?

[rigid smelt]

if you plug the two points into h(x)

you will get two equations in m and n

[bitter sage]

if i plug it in for the y and x values how do i get m and n sepearte

seperate*

i expanded f(x) x g(x) and now i just plug in one of the points?

[thorn tapir]

yes one, and then do the second one as a seperate case

those two cases would give two equations for you to solve as a system

[bitter sage]

ok

[alpine sable]

why is it equal 6?

[gray isle]

by definition

[thorn tapir]

yep

$a^{\log_a x} = x$

[ocean seal]

rikusp2002

[thorn tapir]

because $\log_a x = y$ if and only if $a^y = x$

[ocean seal]

rikusp2002

[alpine sable]

thats weird

nvm

thanks

yo

im back

cuz

;_;

[alpine sable]

Let's think about this for a second

You know what the degree of a polynomial is, yes?

yes

uh

So, we know the degree of the dividend is 4

i think i got the answer

Let's hear it

the hard way

1

nope

:/

Let's start simple

sure

The degree of the dividend is 4, we know that

Let's say the polynomial was divided by a regular number, say 3

$\frac{x^4-2x^3+4x-5}{3}$

[ocean seal]

Cicopath

[alpine sable]

Would this change the exponent of the highest degree variable?

nope

Exactly, so the degree would stay as 4

$\frac{1}{3}x^4$

[ocean seal]

Cicopath

[alpine sable]

Now, let's kick it up a notch. Let's say it was divided by x-3 instead.

$\frac{x^4-2x^3+4x-5}{x-3}$

[ocean seal]

Cicopath

[alpine sable]

I presume you've been working with dividing polynomials

So, would this change the exponent of the highest degree variable?

How many ways to site 70 people across 7 tables, 10 people per table. The people must be sorted in pairs.

@alpine sable Channel taken

[alpine sable]

The highest degree would change! Because it's being divided by a polynomial with a degree of 1, the quotient would have a degree of three.

Now, let's take the next divisor.

$\frac{x^4-2x^3+4x-5}{x^2-4x+2}$

[ocean seal]

Cicopath

[alpine sable]

A polynomial with a degree of 4 divided by a polynomial with a degree of 2.

So, an easier way of thinking of this is imagining the divisor as factored.

So, let me make up a hypothetical example

yes...

And this isn't going to be accurate

Just an example

Let's say we have a polynomial $x^4+2x^3-5x^2+3x-10$

[ocean seal]

Cicopath

Cicopath

Cicopath

[alpine sable]

So you see how dividing the original polynomial with a degree of 4 by two different binomials has a quotient with a degree of two less than the original degree.
Basically, the original polynomial has a degree of 4. The quotient has a degree of 2.

Does that make sense?

but the degree of the divisor is 2

What is the product of two binomials, each with a degree of 1?

(x-3)(x+5) for example.

$(x-3)(x+5)=x^2+2x-15$

[ocean seal]

Cicopath

[alpine sable]

$\frac{x^4+2x^3-5x^2+3x-10}{(x-3)(x+5)}=\frac{x^4+2x^3-5x^2+3x-10}{x^2+2x-15}$

[ocean seal]

Cicopath

Cicopath

[alpine sable]

You see the resemblance?

So essentially, a trinomial with a degree of 2 is the product of two binomials, each binomial with a degree of 1

Another way you can think of it is you take the divisor's degree, which in your case is 2, and subtract it from the dividend's degree, which is 4.

And you get 4-2, which equals...?

2

Exactly!

So, I just want to make sure

Did any of that make sense to you?

it did

but i had a presumption in my mind

that

When I hear remainder, I think of the part of the quotient where it has a remainder, e.x. $\frac{19}{x^2+3x-5}$

[ocean seal]

Cicopath

[alpine sable]

did not get it

I might be thinking of this wrong

Is that from Google?

Eh

The degree can be any real number, doesn't necessarily have to be equal to bigger or smaller than the dividend

@alpine sable To mention, the quotient and the remainder are two separate things

ik

For example if the dividend has a degree of 1 and the divisor has a degree of 3, the remainder will have a degree of 3 only

omg

im dumb

very

im sooo sorry for wasting ur time

i regret this

forgive me

There's no need for apologies lol

didnt even care to read the question

properly

smh

sorry

[vapid seal]

Please provide full answer

[tough hatch]

read. the. rules. on. #rules and on #❓how-to-get-help
you'll more likely to get help if you ask how to get the answer instead of the answer
absolutely no need to post on more than one questions- channel

[tight locust]

this is a CS question but i'm sure you can also put it in pure math language

[vapid seal]

Yes sir

[tight locust]

i give you a function fac(n) defined as followed:

fac(0) = 1

fac(1) = 1

[tough hatch]

factorial?

[tight locust]

for n >=2 the function returns n*fac(n-1)

how many calls to the function are made for a given input n?

[tough hatch]

so it's just the factorial.

[tight locust]

yeah

what's the math version of calls to a function

[tough hatch]

ah

the function is called until you get the last value of n as 1 right?

since fac(1) = 1

if n=3, then you get
fac(3) = 3*fac(3-1) = 3*fac(2)
which at this point would call the function again to find fac(2)
fac(2) = 2*fac(2-1) = 3*fac(1) = 3
then it stops calling the function since you already know what fac(1) is.

this should give u an idea @tight locust

[tight locust]

faccalls(n) = 1 + faccalls(n-1)

[tough hatch]

what is faccalls(n) defined as in words, exactly?

[tight locust]

yep

yes

shouldn't it just be n+1?

unless i'm being dumb

ah yeah let me modify the defn of the function

[tough hatch]

only expection is when n=0,1

[tight locust]

fac(0) = 1, fac(n) = n*fac(n-1)

fac(1) not explicitly defined

[tough hatch]

wasn't it?

[tough hatch]

.

[tight locust]

yeah let's change that i think it's a more natural construction

[tough hatch]

hmmmm does faccalls include the time when fac(1) is called?

[tight locust]

yes so you have to go to fac(0)

[tough hatch]

u need to define exactly what faccalls(n) is
as in the number of times fac(n) = n*fac(n-1), NOT fac(0) = 1 is called?

[tight locust]

yeah it is just n+1

yeah faccalls(n) is the number of times fac(n) is called

[tough hatch]

seems legit

[tough hatch]

yeah, make sure to include except fac(0) = 1 there

[tight locust]

right yeah. faccalls(0) = 1 and faccalls(1) = 2

n+1

i thought that'd be more interesting

the fibonacci series has a cooler recursive construction

[tough hatch]

yep

[wanton spoke]

why did they divide the 2 out

[tough hatch]

that is a disgusting demonstration of Divide BOTH SIDES by 2

[tough hatch]

you are trying to find the values of x (solutions) for which |x-5| = 4

[wanton spoke]

ohhh ic, ic

[grave egret]

Can anyone explain me why critical value is 1.96?

[tight locust]

z_alpha/2?

[alpine sable]

hello

ur computer has a virus

get a 500 dollar target gift card

and ill assist u further

[tight locust]

there's no target in india

[alpine sable]

but ya all have target

-_-

[tight locust]

then why would you need a target card?

are you going to sell it to someone else?

[mild trail]

Ayo no off topic in questions

[ionic jewel]

ironic

[topaz scaffold]

Lmaoo

[tight locust]

Lol!

[umbral pulsar]

ur computer has multiple viruses

[topaz scaffold]

I like Bunny's shade of blue in his name tag

Guess I just gotta grind it out

[alpine sable]

sir

r u playing games with me

[umbral pulsar]

give me 500*10^x(x tending to infinity) dollars

[alpine sable]

i did not recieve the gift card

i have many people waiting in line

[topaz scaffold]

Sir

[alpine sable]

make it quick

[topaz scaffold]

I have a Mac

[alpine sable]

microsoft

bought apple]

from all the scam money

i-i mean-

hard earned money

yes

[topaz scaffold]

And now you're a monopoly

And the government has a problem with that

[umbral pulsar]

https://tenor.com/view/ramdev-baba-to-indian-government-pawpaw-business-money-cash-gif-18905018

[alpine sable]

ye boii

[ionic jewel]

i don't even know why I have it I'm not even that active

i feel like ur on more than me

[alpine sable]

hello mr bunny

i think u have a virus

[thorn tapir]

What is this.

[stark grail]

may i know how to solve this?

[noble temple]

Try to write a formula for the volume of the container and set that equal to 28pi

[stark grail]

pi r square h = 28 pi?

[noble temple]

Yeah

So there are 2 variables, r and h, you only want one because it will result in a cubic equation which is easier to solve.

In this case I think expressing the formula in terms of r is the easiest to do

[stark grail]

r= square root 28 over h ?

is this dy/dx ques?

[noble temple]

Yes

But you have to account for material cost

So you dont have the function yet

Btw I would write h = 28/r² because it will be a bit more easy to use

So you get the equation:
pi•r²•(28/r²) = 28pi

[stark grail]

okie ~

whats should I do after done this?

[topaz scaffold]

You've been here for a longer time

[glossy forge]

Can someone help me with a math problem

[stark grail]

me stuck TT

[vale wigeon]

@glossy forge until you post the problem, we can't

[noble temple]

You now have the formula for the volume but you want the area

[tough hatch]

[noble temple]

Whats wrong?

[tough hatch]

what are they solving for here?

dimension of the container sounds vague

[noble temple]

I know it just says 28pi = 28pi

[noble temple]

Height and radius

[tough hatch]

so the question is actually asking What are the dimensions of the container

[noble temple]

Yeah

@stark grail can you write a formula for the total area?

[tough hatch]

but what are RM5 and RM2

[noble temple]

Im guessing some kind of currency

But it doesnt matter it could be the amount of cows in your backyard.

[tough hatch]

by surrounding they mean the surface area 2pi rh?

[noble temple]

So like €5 and €2

[noble temple]

Yes

@stark grail you there?

Ping me when you are back

[alpine sable]

This is going to for a parabola right? Did i do this properly?

[worthy verge]

Helpo

[velvet condor]

DE=?

bro are you kidding me

DE/BC=?/?=?/?

haven't you studied the properties

[worthy verge]

wait

4

[velvet condor]

so?

dne and cnb are ? triangles

so their ratio is (?/?)^2=?

fill in the blanks

[worthy verge]

4sq 12 sq

16/144

[velvet condor]

=1/9

[worthy verge]

ty

[velvet condor]

anyone----

[upper pebble]

:)

[reef hare]

Hello does somebody know if I can use the quadratic optimization method for quadratic function with two variables? e.g in the form of f(z) = 0.5 * x^T * H *x + g^T * x + y^T * H *y + y^T

[vale wigeon]

yes

[reef hare]

how would that work?

[vale wigeon]

you'd have $\bmqty{H & 0 \ 0 & H}$ as your quadratic coefficient matrix and $\bmqty{g \ \mathbf{1}}$ as your linear coefficient vector

[ocean seal]

Ann

[vale wigeon]

assuming you put all x's before all y's

...wait, y^T?

that's a vector and everything else there is a scalar

are you sure you didn't mean g^T y

in any case, you're gonna need to construct the parameters of your function as block matrices and block vectors

[reef hare]

oops yes of course i mean g^T

[vale wigeon]

yeah then it's $\bmqty{g \ g}$

[ocean seal]

Ann

[reef hare]

wow thank you! Did not expect help that quickly. I'll try that

[ocean seal]

Jerom

[alpine sable]

help!

the question is above

[alpine sable]

this one

[light mango]

how do i deal with the squared d?

[tough hatch]

LMAO

[light mango]

ive never seen this and idk how this would work

[true crystal]

This just means the second derivative of u with respect to y

[light mango]

hmmmmmmmmmmm

i see

[true crystal]

Check partial derivative

[light mango]

btw

is there any difference between that symbol and the standard d?

[velvet condor]

no

[light mango]

okay thx

[tough hatch]

there is.

there is!

[velvet condor]

what.....

[tough hatch]

.....

[true crystal]

You use d when you only have one variable

f(x) = x^2 for exemple

[velvet condor]

people have been telling me they are the same

[light mango]

well i assume in practice theyre the same?

[true crystal]

But for f(x, y) = x^2 + y^2 you would use partial

[light mango]

essentially

yea honestly ive use the d before for partial too

guess thats wrong if you strict about it

[true crystal]

I suppose it depends your education

[light mango]

yea well i assume engineering is not as strict about it

[true crystal]

But you should use partial for multivariable function

[glass lichen]

who?

[tough hatch]

[light mango]

whats the practical difference between the square above the d and above y?

on the right side

[glass lichen]

it's just the notation....

how are you in multivariable and havent seen a 2nd derivative?

[light mango]

me?

[velvet condor]

my math teacher at school:)

[glass lichen]

yes you

[vague coral]

d/dx is the same as the sign of partial derivative, its just that the notation for partial derivative is mostly used when we are dealing with more than one variable

[light mango]

ive seen multivariablse a lot , im just confused with this square writing bcs this is new to me

[glass lichen]

again, you've never seen $\dv[2]{y}{x}$?

[ocean seal]

Mosh

[light mango]

no

[tough hatch]

this @light mango

[vague coral]

bruh

[glass lichen]

. . .

[light mango]

?

[light mango]

thank you

it is more simple than it looks

[thorn tapir]

I just did C by myself, I can give you a hint about C - Find out the ratio $\frac{b}{a}$, square on both sides, and then apply componendo-dividendo - and finally take reciprocal. Simplifying the right hand side should prove your equality.

[ocean seal]

rikusp2002

[tough hatch]

it's similar to how $\frac{\diff^2 u}{\diff x^2} = \frac{\diff}{\diff x}\left[\frac{\diff}{\diff x} y\right]$

[vague coral]

$\partial$

[ocean seal]

Herels

[vague coral]

$\frac{\partial}{\partial x} = \dv{x}$

[tough hatch]

not always.

[ocean seal]

Herels

[vague coral]

I know

[tough hatch]

take for example f(x,y) = xy

[vague coral]

I know

we use partial derivative when we are dealing with more than one variable

But if you have a function with one variable

using partial derivative or d/dx is just the same

[light mango]

if i mulitply by dy^2 . And d^2*u remains on the right side. Does that mean that integrate the right side two times?

thats where i get confusded

(im saying d for ease of writing)

[tough hatch]

u mean "multiply by $\partial y^2$ ?

[light mango]

yes

[thorn tapir]

@alpine sable Also btw, If I suppose that the statement C is correct, which I think it is as it has been froved deductively already - using that statement gives us

$\sin (\theta + \phi) = \frac{2ab}{a^2 + b^2}$ Instead of what is given. I'm not sure why that is given and I have no idea about the equivalence of those two fractions.

[ocean seal]

rikusp2002

[light mango]

i need to integrate left side two times by y

thats why i multiply with that

[tough hatch]

there's no integral sign anywhere though.

[light mango]

uhhhh

[vague coral]

why are we talking about integral xddd

[light mango]

thats what im supposed to do

hmm

[tough hatch]

is this a PDE?

[light mango]

whats that

[tough hatch]

so it's not

what is the actual original problem

[vague coral]

what do you want to get at the end btw ?

[light mango]

its an engeenering problem

[vague coral]

Not the problem, what do you want to get by solving this