help-23

[fiery junco]

which is 1

aight maybe i got the rest then

i should

y=2x-1 ty

was just that first part tripping me up

i have one more question tho for a different problem

for some reason the answer choices have the slope connected to y, which i dont understand

[median vigil]

what is the relationship between the normal line and the tangent line?

[fiery junco]

well ik its the negative reciprocal

tbh i dont have a clue wtf a normal line is or looks like

1nvm

i do

perpendicular line lmao i brain farted

[median vigil]

so you can make a line using the usual formulas, then see if you can algebraically rearrange it into a form matching the answer choices

[fiery junco]

mmaybe

ill first try to make the equation i usually do

i got y = 1/15x - 40

my tangent slope m = -15

@median vigil

[median vigil]

how did you get that?

[fiery junco]

i mean -15 sorry

taking too long

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[fair goblet]

Alright, can someone help me with number 66

[fair goblet]

I’m not sure how to begin

Would I start by finding the derivative?

[median vigil]

do you know what the normal line is?

[fair goblet]

Yes the negative reciprocal

Of m

Which will be the tangent line

Of this function

[median vigil]

yes

so you should start by finding the normal line

[fair goblet]

And to find that I need to find the derivative first? Since that’s how I get the tangent line

The derivative is 2x

Do I plug in -1 to find that tangent line slope then flip it?

If so, how do I continue from there?

I’m confused on where to go there on

[median vigil]

you do find the tangent line slope that way

what is the relationship between the slopes of perpendicular lines?

[fair goblet]

Where they intersect

They share the same point

[safe radish]

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[little flare]

(ignore the limits) how did it simplify the denominator to that point? it looks like it's derived from (u + 5)^2 but that would be u^2 + 10u + 25 right??

[median vigil]

difference of cubes identity: [ a^3 - b^3 = (a-b)(a^2 + ab + b^2) ]

[flat frigate]

cloud

[little flare]

AHHHHH i like vaguely had that pictured in my head but didnt know what exactly it was

i felt like i just imagined it or something

my bad

thank you very much

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[clever ridge]

$\int_{-3}^{5} \frac{1}{x^2} dx$

[flat frigate]

wakamole

[clever ridge]

can someone just guide me through the steps..

[magic jetty]

2+2=5

[clever ridge]

<@&268886789983436800>

i split the integral into 2

[stiff bobcat]

please don't spam in help channels.

[magic jetty]

ok

[clever ridge]

i got $\lim_{t to 0^-}\int_{-3}^{t} \frac{1}{x^2} dx + \lim_{t to 0^-}\int_{t}^{5} \frac{1}{x^2} dx$

not sure how to make limits

here

[flat frigate]

wakamole

[clever ridge]

final answer is -8/15 , it converges

?

[stiff bobcat]

it doesn't -- you can see that these two limits don't exist separately

think of what the antiderivative of 1/x^2 is and evaluate the corresponding limit e.g. from -3 to t as t goes to 0 from the left

[clever ridge]

ant-derrivative is -1/x ?

[stiff bobcat]

right

and the limit of that as x goes to 0 doesn't exist

[clever ridge]

true. i thought i needed to intput (-1/t) - (-1/3) + (-1/5 + 1/t)

and then put 0 in for t

[stiff bobcat]

I see

well the problem with that is, you can't add limits that don't exist like that

doing algebraic operations with limits requires that both exist

[clever ridge]

oh i didnt know that

ok at least my steps were ok i just wrote it out . so since 1/t = 1/0 and the other 1/0 it does not converge

[stiff bobcat]

right, the limits don't converge so the integral doesn't

[clever ridge]

perfect thank you.

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[lean otter]

Every perfect square is greater than a perfect cube. i am learning discrete math and need to represent this symbolically. i wrote
∀ m ∈ Z, m^3 < m^2

[lean otter]

i can only use symbols, no words

would i have to rewrite this with a negation

[eternal carbon]

you can try to incorporate an existential quantifier

[trim swan]

"Every perfect square is greater than a perfect cube." Is this exactly how the prompt is written?

[lean otter]

yeah would i actually do something like

∀ x, ∃ y ∈ Z, x^2 > y^3

[trim swan]

that looks better, but I'd specify that x is in the integers too

[lean otter]

okay

i think i dont understand how to properly write the notation

another problem i have is

Negate the following statements without using words or the ¬ symbol.

∃ p ∈ R, p/(p-7) ∈ Z

this would mean verbally "there exists p is an element of real numbers such that p/(p-7) is an element of integers"

now i am working on negating and will be back in a moment

∀ p ∈ Z, p/(p-7) ∈ R

i see it as "for any integers, p/(p-7) is an element of real numbers"

i am not sure if this negation is correct

i just flipped the symbol without really thinking deeply about it

for any real numbers p/p-7 is not an element of integers

∀ p ∈ R, p / p-7 ∈/ Z

[safe radish]

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[safe radish]

@lean otter Has your question been resolved?

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@lean otter Has your question been resolved?

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[wet plume]

how do i find the range of (lnx)^2 -1

[wet plume]

and domain

[quiet plume]

For the domain, assume it's R to begin with and look for functions that restrict the domain (like a square root or a log) or denominators that might be zero somewhere.

[wet plume]

how about the range?

[quiet plume]

Well we know (ln(x))^2 is always >= 0 right?

[wet plume]

yeah

wait wouldnt ln0 be invalid

[quiet plume]

And so we should have (ln(x))^2 - 1 >= -1.

[wet plume]

wait nvm i thought u meant x

[quiet plume]

Is there any value of x you can find that yields (ln(x))^2 - 1 = -1 ?

[wet plume]

i dont think so

[quiet plume]

Say x=1 does that work?

[wet plume]

ohh right

[wet plume]

oh yeah

[quiet plume]

Great. So that means that the function actually attains a global minimum of -1 at x=1.
Now ln(x)^2 will be as big as we want since ln(x) is increasing.

So that means that every value bigger than -1 is achieved also

[wet plume]

ohh that makes sense

awesome

thanks so much

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[hexed widget]

help.

[hexed widget]

why do i keep getting decimals

i dont think i should be though

[mighty spoke]

uhh i got
6 -3y = -4.5x
10 = -2.5x
x = -4
y = 8/3??

i havent done simultaneous in a while tho

[hexed widget]

im so tired i just got y = 8 and x = 4

after multiple tries

[safe radish]

@hexed widget Has your question been resolved?

[sinful iris]

hi

what is this

system?

[hexed widget]

yes

system of equations

[sinful iris]

i got x=12 and y=20

[hexed widget]

guys this is so confusing

[sinful iris]

lol where are you struggling?

what i did was

[hexed widget]

might as well turn into a parliamentary debate at this point

[sinful iris]

multiply the whole second equation by 4

[hexed widget]

yes

[sinful iris]

so u have linear on both

then what method did u choose

to solve

[hexed widget]

3x-2y+4=0
2x-3y+16=0

elimination

[sinful iris]

me too

what did you do

[hexed widget]

x3 and x2

up and down

[sinful iris]

u eliminated x?

[hexed widget]

yah

6x -9y +48

[sinful iris]

i think u eliminated y

oh nono

im dumb

[sinful iris]

huh

this is top?

[hexed widget]

no

bottom

6x -4y +16 ithink

[sinful iris]

no

ìts +8

multiply by 2 not squared

[hexed widget]

oh wait oops yeah i mean 8

let me check what i did

here −9y+6x=−48
6x−4y=8

[sinful iris]

[hexed widget]

gosh i have no idea what im doing

my head is so fried

[sinful iris]

i think you just messed up the multiplication

try this instead

[sinful iris]

multiply the top equation by 2, and the bottom by 3

[safe radish]

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[verbal fable]

Hey 👋
I need some help with Sinus theorem
I would also appreciate if they could also explain the steps they took to resolve the task

[verbal fable]

I'm confused where to start to start resoliving the task

[pure agate]

You are given angle 1 which you can use to find the angle of elevation of the tree relative to the ground.

[verbal fable]

do you mean this?

that 79° is from taking away 11 from 90 as the old angle was square

and if im correct there was something to do with the triangle now

[pure agate]

Yes.

You are also given angle 2 which you can use to find the third angle from the top of the tree.

[verbal fable]

oh so the top of the tree angle would be 45° , because the sum of all of the angles should be 180

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[long harbor]

linear equations cant be shifted horizontally right, just vertically

[peak estuary]

well I would argue both have the same effect visually

[long harbor]

in this form y=f(x-h)+k

whatever its called

what does the f represent, wouldnt "a" usually be there

like this y=a⋅f(x−h)+k

or is that something different

[tepid summit]

Well.

if you want to translate some function f(x) (a,b)

then you do f(x-a)+b

think of how that works

you literally change how x functions, so when you input say x=2 but you translated it by 1 it's basically like you inputted x=1 normally

think of it like siding the plane over

[long harbor]

wait what

what is a and b here

[tepid summit]

just general coordinates

variables

https://www.desmos.com/calculator/c04femhoun

Here, I made this

[long harbor]

wait how is the blue curve parabolic

i dont really get how the f works

like why make the red line f(x) = 3x^2+3x instead of y=3x^{2}+3x

oh does the blue thing only do the job of translating the red parabola

[tepid summit]

the black dot represents (a,b)

the blue curve is the red curve but moved via the black dot

if you look at it, if a=0, b=0 then they are the same thing

[long harbor]

i still dont get this tho

[tepid summit]

isn't that the same thing

f(x) = y

f(x) is just a function that allows you to change the input

normally it is defined that y=y(x)

[long harbor]

input being x?

[tepid summit]

yeah

[long harbor]

oh

okay thanks

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[lean otter]

$$6 \times 4^x - 13 \times 6^x + 6 \times 9^x = 0$$

Guys can you help me solve this equation

[flat frigate]

Edmund Cloudsley

[lean otter]

I have come to a solution wherein $(\frac{2}{3})^x = \frac{3}{2}$

[flat frigate]

Edmund Cloudsley

[lean otter]

and one more solution would be

$$(\frac{2}{3})^x = \frac{2}{3}$$

[flat frigate]

Edmund Cloudsley

[lean otter]

Now my professor thinks that there is only one valid solution to this equation and that being 1 and the other solution -1 is only a conicidental equation

can you guys help me please

[sinful iris]

can’t u plug both in to verify?

[lean otter]

I tried plugging both of them in

and the values do indeed match

but my professor belives that the x = -1 is a conincedental solution

[sinful iris]

both of the answers verify the equation?

[lean otter]

and not a valid slution

[lean otter]

yeah they do

but according to my teacher this is only a coincidental solution and not a general solution

I would like to know why

<@&286206848099549185>

[sinful iris]

I think it most definitely is an equation that has 2 answers

[lean otter]

i have answer.

so there is no confusion in x = 1 being a solution

[sinful iris]

2 solutions

1 and -1

[lean otter]

but for x = -1,my teacher belives that it is not a solution for the following reason

$$(\frac{2}{3})^x = \frac{3}{2}$$
$$\frac{2^x}{3^x} = \frac{3}{2}$$
$${2^x \times 2^1} = 3^x \times 3^1$$
$$2^{x+1} = 3^{x+1}$$
$$(x + 1) \ln{2} = (x + 1) \ln{3}$$
$$\cancel{x + 1} \ln{2} =\cancel{x + 1} \ln{3}$$
$$\ln 2 = \ln 3$$
Which would be untrue

[spark zinc]

1

[flat frigate]

Edmund Cloudsley

[spark zinc]

3/2≠2/3... EDIT: I misread it lol i'm wrong here

[lean otter]

guys so would -1 be a solution to this equation or not?

and if not why so?

[sick scarab]

it is, why not

it satisfies it

[lean otter]

it is a coincidental solution

[lean otter]

see this explanation

this would imply that ln 2 = ln 3

and that would be untrue

is there something that this explanation is perhaps missing?

[craggy sedge]

umm your prof cant just cancel x+1

[spark zinc]

yeah it is anything ^-1 is just it's reciprocal

[snow ridge]

He took ln of both sides but you can't always do that can you

Logs aren't defined for inputs 0 and 1 between

[sinful iris]

is this because you can’t divide by 0?

[craggy sedge]

yes

[sinful iris]

since the proof starts assuming x = -1

[spark zinc]

plug it into a calculator

see what happens

Your teacher is trying to troll or something

[sinful iris]

x+1/x+1 is only possible when x ≠ -1

[spark zinc]

true

[lean otter]

Guys I just got a message from my professor

and they agreed to -1

Thanks so much for your help

I sent them the graph

[spark zinc]

I'm not even trying to troll or be a dick but how is this not obvious?

I learnt this in basic indicies in yr9

[lean otter]

.close

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[gleaming ginkgo]

can someone help me w this ?

[gleaming ginkgo]

😭

i have to construct a line here but idk where

im confused af rn

its like circle theorems and stuff

[gloomy scaffold]

I think 15 degree is the answer

[gleaming ginkgo]

ooo how tho?

[velvet nebula]

BC looks good

[gloomy scaffold]

use the relation of angle of circles

[gleaming ginkgo]

huh

what

[gloomy scaffold]

angles in the center of the circle is twice of the angle in the circumference

[gleaming ginkgo]

ohhh yeah

tht one

thru the BC line

ok so theres BC as the line right

[gloomy scaffold]

through the arc DC

[gleaming ginkgo]

yup

this is what ive done as per ur idea

but in order to find DEC

wouldnt i need to know DOA

wait no

[gloomy scaffold]

you don't need to know DOA

[gleaming ginkgo]

i figured 😭

i got DOC

wait no

ODC

65 degs

[gloomy scaffold]

yes

[gleaming ginkgo]

and BAO and OBA is 70 degs

[gloomy scaffold]

sorry my approach was rather incorrect. But the method is still same

find CAD and BDA

[gleaming ginkgo]

cod is 50

[gloomy scaffold]

add them to get desired angle

CAD is will be 25 and BDA will be 20 . Adding these you will get CED = 45 degree

[gleaming ginkgo]

OHH CAD

okk 👍

tysm for ur help ❤️

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@gleaming ginkgo Has your question been resolved?

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[lean otter]

bruh i checked the LHD and the RHD and its coming out to be 1

but also at x=1 there's a sharp turn

so it shouldnt be differentiable

but how do i show it

[rich elm]

LHD will not equal to RHD

it should be quite clear LHD is negative whilst RHD is positive

[lean otter]

do you want me to send pic of how i did it?

[rich elm]

yes please

[quasi ember]

Best is to split up the function in how it behaves when x>1 and x<1

[lean otter]

bruh i just did how summi told me to

and im still getting 1

lemme send

[quasi ember]

When you take x < 1 then x -1 will always be a negative function

But f(x) = |x-1|

[lean otter]

when x >1, y = x-1
when x<1, y = -(x-1)

differentiating you get 1 and - 1 respectively

[quasi ember]

Exactly like that

[lean otter]

you always have to seperate the mod function even in integration

[quasi ember]

The absolute value ia always positive, so to acommodate for that you have to work with -(x-1) for x<1

[lean otter]

the nature of the function changes completely because of mod

omg i was using the wrong formula to calculate the LHD and RHD this whole time

im so stupid

thanks anyway

.close

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[desert pasture]

This doesn't make an ounce of sense to me

[desert pasture]

so N and M are arbitrarily large positive numbers

Thanks!

[merry pelican]

Arbitrary-ish in the sense where the choice of N depends on the value M.

[fast gazelle]

this is pretty similar to the definition of $\lim_{x \to a} f(x) = b$ if you've seen that

[flat frigate]

bee [it/its]

[desert pasture]

$\varepsilon-\delta$?

[flat frigate]

Veni, vidi, perii

[fast gazelle]

yeah

$M$ is sort of like $\varepsilon$ and $N$ is sort of like $\delta$

[flat frigate]

bee [it/its]

[fast gazelle]

the idea is just that close to $\infty$'' means very large''

[flat frigate]

bee [it/its]

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[brave wolf]

I managed to prove i -> ii, how do I do ii -> iii?

[brave wolf]

And btw, my idea was doing i -> ii -> iii -> i, is that good or is there something simpler?

[acoustic flower]

help

[safe radish]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[lapis shadow]

thats a circle theorem, right

angles in the same segment

the last one

[viscid garden]

If you do i <-> ii <-> ii <-> iii <-> i

It's better

Cuz otherwise, you are not considering every case...

[brave wolf]

But I dont know the circle is there

or do I?

[brave wolf]

I actually am.
A -> B -> C -> A
is equivalent to A iff B iff C

[viscid garden]

If you know ii hold, then you can prove that the 4 points lie on the circle

Hint: consider the center

And how angles double when moved to the center

[brave wolf]

The center?

[viscid garden]

Oh ok... But usually A -> B mean A implies B

[brave wolf]

How do I know there is a center if I dont know if there is a circle?

[brave wolf]

Yes, it does

I am saying that A implies B, B implies C, C implies A is equivalent to A iff B iff C

[viscid garden]

Oh sorry sorry mb

[lapis shadow]

a cyclic quadrilateral is one whose vertices lie on a circle right?

[brave wolf]

Yes

[lapis shadow]

if uve shown its cyclic

then u know that right

or rather if u assume its cyclic

[brave wolf]

Yes, if I assume it's cyclic I can prove ii quite easily

[lapis shadow]

i -> iii

[brave wolf]

and I think I could also prove iii

Oh yes, that makes sense

it's just inscribed angle theorem twice

so now I have i -> iii and i -> ii

[viscid garden]

You can also prove iii -> ii

[brave wolf]

that sounds interesting

if I manage to do that, then I only need ii -> i

[viscid garden]

Its trivial...

[brave wolf]

Is it?

[viscid garden]

I mean... if you just try to see what iii means...

I mean, if you just draw a circle of ABD

You can prove easily that C lies on the circle...

Cuz that's just the property of circles

You get this?

[brave wolf]

what property of circle?

[viscid garden]

Any point X on the arc AD is gonna have the same angle AXD = ABD

And it's iff

[brave wolf]

the iff part is essentially what I need to prove

[viscid garden]

It's just angles subtended from the same arc...

[brave wolf]

or the only if part

I only proved the if part till now

[viscid garden]

So here is what we are proving right now:
Let X and Y be on the same arc AB <- AXB = AYB

right?/

Do you get what I am saying?

(I have to go in 10 mins...)

We were originally trying to prove iii -> ii

[brave wolf]

sorry, im back

[viscid garden]

oh right!

[brave wolf]

yeah, thats essentially what I need to prove

maybe I could just construct circle around ABD

then assume C is outside

and possibly get a contradiction

[viscid garden]

Yeah that's what I was thinking

[brave wolf]

imma try that

[viscid garden]

By considering the angles of those arcs from the center

I mean, if C is on circle, then the angle from center should double... (This is and iff property surely)

[brave wolf]

well, the iff part is unfortunately still unproven

[viscid garden]

no... You dont have to prove it...

I mean... Its a property..

It's how circles are defined...

I mean, if you just look at the proof of this property, it's obvious that the point should lie on the circle for the property to hold...

[brave wolf]

circles are defined as a set of points that are equidistant from its center

[viscid garden]

.

ok, gtg

[brave wolf]

ill try to do the contradiction thing

i think I have an idea

so assume C lies outside

then we can construct the blue thing as I did

the blue angle must also be strictly greater than the green one

or at least I fucking hope so

but ABD-blue is cyclic quadrilateral

and so the blue angle should equal green

but blue > green

contradiction

similar can be done if C is inside

at least I hope so

god I hate geometry

its just a ton of handwaving

.close

[safe radish]

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[brave wolf]

I have i -> ii and i -> iii

i also have iii -> i

and so I only need ii -> anything

and that sounds easy

[safe radish]

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[storm spruce]

why cant i open a help channel

[icy lance]

you just did

[pine horizon]

u just did

[storm spruce]

bruh

alr then

ima ask my question now

[lean otter]

u just did

[storm spruce]

if i do a y'=___ and the solution i find is y'=something is it different if i solve it and find the X instead?

[pine horizon]

what?

[storm spruce]

alr one sec

i have x-2\sqrt{x+3}

[lean otter]

.

this

question

[storm spruce]

now if i do a y' to the question its gonna look like this: 1-2\cdot \frac{1}{2\sqrt{x+3}}

how do i make this look normal

[icy lance]

$$y=x-2\sqrt{x+3}$$
$$y'=1-2\cdot \frac{1}{2\sqrt{x+3}}$$

[storm spruce]

$$1-2\cdot \frac{1}{2\sqrt{x+3}}$$

[flat frigate]

IMry_

[storm spruce]

ok

[flat frigate]

AℤØ

[icy lance]

then what?

[storm spruce]

one sec im trying to type the thing

$$\frac{2\sqrt{x+3}-2}{2\sqrt{x+:3}}$$

[flat frigate]

IMry_

[storm spruce]

makes sense right?

[icy lance]

sure?

[storm spruce]

so now you can remove the same numbers and get -2=y' right?

[icy lance]

thats not how fractions work

[storm spruce]

should i solve it like there is zero on the other side and then find x?

[icy lance]

i dont get what it is youre trying to do

[icy lance]

you have y' right here, though you can do some simplifying

[storm spruce]

$$0=\frac{2\sqrt{x+3}-2}{2\sqrt{x+:3}}$$

[flat frigate]

IMry_

[storm spruce]

im saying like is this the thing i need to solve?

find the x?

[icy lance]

i dont know because you havent said anything to indicate that

you havent showed what question youre trying to solve

youd set y'=0 and solve if maybe you were trying to find a stationary point, is that your aim?

[storm spruce]

im trying to find the like top or buttom point

idk how you say it in english

the min or max point

[icy lance]

then you need to solve y'=0, yeah

[storm spruce]

ok so if i just get -2 like i did here i cant say thats the point?

[icy lance]

because the way you got -2 makes no sense at all

[storm spruce]

the two numbers are cancelling each other

[icy lance]

(a+b)/a is not equal to b

thats not how cancelling works, the entire numerator would have to have a as a factor

not just part of it

[storm spruce]

alr

so i did the rest of the excersize

let me show the steps

first i removed the bottom by multiplying it

so its

wait

alr i think i know what i did wrong

but just to make sure: if i find the y' its not the solution. i have to find the x right?

[icy lance]

if youre aiming to find a stationary point then yeah, solve y'=0 after

[safe radish]

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[safe radish]

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[soft matrix]

when is a triangular number a perfect square

[soft matrix]

elp

[eternal carbon]

do you want an answer, or is this something you are trying to prove?

[soft matrix]

i want a answer ( im trying to prove smthing else )

[eternal carbon]

https://en.wikipedia.org/wiki/Square_triangular_number

[hearty egret]

can you recall what is a triangular number for me ?

[soft matrix]

sum of n natural numbers

[soft matrix]

let me check thx

nice so all sol to the pells eq x^2 - 8y^2 are square triangular no nice

.close

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[hollow coyote]

checking for convergence

[safe radish]

@hollow coyote Has your question been resolved?

[safe radish]

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[deep steeple]

There is error right?

[deep steeple]

1+0=0 xd

[desert swallow]

Yeah that's the error

[deep steeple]

thanks for confirming

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[jolly bay]

Can someone help me with this one

[jolly bay]

I combine fractions, but then get lost on what to do

[fathom jewel]

,, \lim_{t \to 0} \frac{1-\sqrt{1+t}}{t\sqrt{1+t}} ?

[flat frigate]

bacc

[fathom jewel]

are you at this step?

[jolly bay]

why in the denominator is only t and not t^2?

[fathom jewel]

the lcd is t sqrt(1+t)

[true veldt]

they already have a common factor of t in the bottom

[jolly bay]

and why one?

[fathom jewel]

anyway with squares you can multiply by the conjugate

[fathom jewel]

the other fraction...

[jolly bay]

when combining them is it t - t sqrt(1+t)?

[fathom jewel]

no

,, \lim_{t \to 0} \frac{1}{t\sqrt{1+t}} - \frac{1}{t} \cdot \textcolor{cyan}{\frac{\sqrt{1+t}}{\sqrt{1+t}} }

[flat frigate]

bacc

[fathom jewel]

that's what was done

[jolly bay]

ohhhh

I see my error

wait no

i still dont understand

[plucky elk]

did you simplify the fraction bacc showed you

[jolly bay]

I didnt

Like I now understood how to get to that expression since its LCD

but now that I have that expression I dont know what I could do

[fathom jewel]

multiply by the conjugate of the numerator

[jolly bay]

let me try it out

so its 1/t?

[fathom jewel]

the numerator should be -t and the denominator tsqrt(1+t) + t(1+t)

[jolly bay]

what... 😭

[fathom jewel]

,, \lim_{t \to 0} \left ( \frac{1-\sqrt{1+t}}{t\sqrt{1+t}} \cdot \textcolor{cyan}{\frac{1+\sqrt{1+t}}{1+\sqrt{1+t}} } \right )

[flat frigate]

bacc

[jolly bay]

ohh

[fathom jewel]

then you should have it

if you simplify the terms

[flat frigate]

bacc

[jolly bay]

im sorry

i just dont get it

[fathom jewel]

what

at which step are you

[jolly bay]

how did you get this

im a bit dumb

[plucky elk]

$(a+b) (a-b) = ?$

[flat frigate]

[fathom jewel]

[jolly bay]

a^2 - b^2?

[fathom jewel]

yes

see now

[fathom jewel]

i put the squares just for you

[safe radish]

@jolly bay Has your question been resolved?

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[remote flint]

https://cdn.discordapp.com/attachments/1082836993593049169/1283190769397665852/brave_tvYOXHdHVQ.png?ex=66e21807&is=66e0c687&hm=f45635d33f0c5451cc2c40aa194c14560ad5d4820ace4c5d273d2e3fe3ed5dce&

[remote flint]

how is my answer wrong?

am i missing smth?

[burnt notch]

Yeah the period, because your expression has a period of 2π, but it's got to be 6 according to the picture

[remote flint]

oh

so its just

3sin(pi/3 x) - 3

fuck im actually stupid

[burnt notch]

Yep, exactly

[safe radish]

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[noble bane]

How does one do 24 and 25

[marsh walrus]

not allowed to use lhopital?

[noble bane]

nah i dont think so

im pretty sure at least

[marsh walrus]

so you can use it?

[noble bane]

no i don't even know what it is

[marsh walrus]

okay

[hearty egret]

for the 25 I think it could be useful to rationalize the numerator

[marsh walrus]

you could try 25 for the power series

however id use an identity

[noble bane]

so what i square it so i can use Pythagorean identity?

[marsh walrus]

https://en.wikipedia.org/wiki/List_of_trigonometric_identities

youll find it here

its gonna end up being half angle

or, think of it like

$\sin ^2 a = \frac 12 \qty( 1 - \cos (2a))$

[flat frigate]

jan Niku

[marsh walrus]

now you have $\lim _{x \to 0} \qty(\frac{\sin ^2 x }{ \sin (2x) })$

[flat frigate]

jan Niku

[noble bane]

so what? is it 0/0?

[marsh walrus]

try double angle

$\sin 2x = 2 \sin x \cos x$

[flat frigate]

jan Niku

[marsh walrus]

our substitution creates a factor of 1/2 but it doesnt impact the limit

[noble bane]

so what would the answer be here?

[marsh walrus]

what do you think? did you try to simplify using this?

[noble bane]

correct me if im wrong but (sinx)(sinx)/2(sinx)(cosx)

[marsh walrus]

yea

[noble bane]

then sinx/2cosx

[marsh walrus]

simplify more

[noble bane]

0/2

[marsh walrus]

yea

[noble bane]

so 0?

[marsh walrus]

yup

[noble bane]

alright thanks

[marsh walrus]

24 lets see

[noble bane]

im not sure where to check my answers in this text book so imma take your word for it

[marsh walrus]

just use desmos

$\lim _{s \to 0 } \frac{ \qty( \frac{1}{\sqrt{1+s}}) -1 }{ s}$

[flat frigate]

jan Niku

[marsh walrus]

what have you tried

[noble bane]

i multiplied by the conjugate for 1-sqrt(1+x)/x(sqrt(1+x)

wait

no sorry

i set all the denomenators equal for that

[marsh walrus]

try doing it one step at a time

[noble bane]

nvm I got it i think

is it -1/2

[marsh walrus]

use desmos

[noble bane]

how do you do limits on desmos?

[marsh walrus]

type in the function, and look what it approach

[noble bane]

oh its negative 1

lemme try redoing this

[marsh walrus]

it is?!

[noble bane]

unless I did something wrong

[marsh walrus]

but the limit is to 0

isnt it

[noble bane]

nvmnvm

I was right

I forgot what I was looking for 😭

alright thank you

[marsh walrus]

[safe radish]

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[last heath]

fellow canadian

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[jovial scaffold]

1

[last heath]

have you learned exponents rules?

[jovial scaffold]

Yes

[last heath]

can you try apply these rules for number

for example what is 3^2

etc

[jovial scaffold]

Ok

I just have a problem with this

[paper rover]

what have you tried

[jovial scaffold]

I tried (-5^12)

[paper rover]

why?

[jovial scaffold]

Because I thought that was the summation

Or equivalent to the answer I’m looking for

[paper rover]

$x^a * x^b = x^{a+b}$

[flat frigate]

albert einstein

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@jovial scaffold Has your question been resolved?

[safe radish]

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[latent compass]

I forgot which property this is, how to solve "ab + abc + ab + abc = ?"

[safe radish]

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[west pumice]

Why can't this be K? Number 30

[west pumice]

Nvm I got it

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[meager bane]

can someone help me find the x value

[merry swift]

Write pythag theorem out for the triangle and you'll have an equation you can solve for x

[meager bane]

.close

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[leaden wasp]

Not sure how to start 4b…..

[serene frost]

I think it’s asking you to find the largest number that will get you an error bound less than 0.1 when using the first order Taylor polynomial

So if you want you can just graph the given inequality and go from there

[leaden wasp]

wait huh? how would you graph an inequality w/ x and t1?

[serene frost]

Like you can graph $|f(x)-T_1(x)|$

[flat frigate]

za

[serene frost]

And look at where it is less than 0.1

[leaden wasp]

ohhhh damn ur right

ok tysm

[leaden wasp]

pls help..... got stuck again......

😭

[serene frost]

Why don’t you plot them separately

As two different y equations

And look at the intersection

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[magic ermine]

hi i decided to plot the convolution of 2 functions but it seems to be scaled by a factor of 2. does anyone know what i did wrong?

[magic ermine]

i just used the standard integral formula but it doesnt seem right

[plucky elk]

!original

[safe radish]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[magic ermine]

well i just made it up myself

[sleek sentinel]

what formula

what's your goal

[plucky elk]

Then how do you know it doesn't seem right

And yea what formulas

Nobody can answer anything so far from the little information you've presented

[magic ermine]

https://www.desmos.com/calculator/pm2syk7ane

[plucky elk]

You're plotting using a discrete tool

[magic ermine]

also well bc the area between those two is obviously 2 and not 4

[magic ermine]

it should still work if increase the number of points

[plucky elk]

Then do that

[magic ermine]

still looks the same tho

[plucky elk]

Why should this be 4

[magic ermine]

because thats what the convolution is at that point according to this graph

[plucky elk]

Between what 2

Green and purple?

[magic ermine]

yes

[plucky elk]

Why do you think the convolution gives area between two functions

[magic ermine]

because thats what it does

or is supposed to do

at least if the input isnt a matrix

ig ill just solve to integral to find the error

.close

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[tiny frost]

I asked this yesterday but couldnt get an aswer to b, its basically the probability going from A to B, first question asks what if you have to pass C, and the answer to that is 0.6. b question says what is the probability from A to B, and the book says its 0.7 but I cant figure out how it got to that number.

[safe radish]

@tiny frost Has your question been resolved?

[tiny frost]

I dont know what you mean?

P(A to B) ?

yea

you mean like $A\cap B$ or $A\cup B$

[flat frigate]

Totalani

[tiny frost]

I assume this would be $A\cup B$

[flat frigate]

Totalani

[tiny frost]

but im not sure

Right, im not sure..

is this something like $P(A\mid E_{1})$ ?

[flat frigate]

Totalani

[tiny frost]

or no

yea fair

I dont know how to get A and B

E_1 and E_2

AB or ACB ?

I have the probability for ACB

$P(B\mid A)$ right?

[flat frigate]

Totalani

[tiny frost]

A has to happen for B to happen

Im really lost

P(E_1)

?

P(E_2) times P(E_3)

Sure

yes

Yes, im with you so far

It says P(E_1) is 2/5 right?

and I know from previoius question P(E_2 and E_3) is 3/5

Thats where I get lost, do I need to use something like $P(A\cup B)=P(A)+P(B)$ ?

[flat frigate]

Totalani

[tiny frost]

becuase that becomes just 1

oh

the difference

0.4/0,6 ?

I mean it gives me the correct answer but I dont know if that was the correct way

2/5 and 3/5

I assume you wold round up but gguess not

the idea was if E_2 and E_3 didnt happen then E_1 had to happen

Ok, please excplain how I get to 0.7 cuz I dont know

so I need the summetric differens?

Well the union is 6/25

10/25+15/25-6/25=19/25 ?

2/5 times 3/5 ?

What is P(A | B) ?

aight im just gonna give up

thanks anyway

.close

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[safe radish]

Please don't occupy multiple help channels.

[outer mesa]

Why not if it isn’t

<@&286206848099549185>

[viscid garden]

I think 3rd one is correct

[viscid garden]

Because we are given the cost of production

The cost of product is 5.

So we have to find what the initial cost would be...

Such that C(D(x))=5

If the initial cost is x, then our cost of production would be 5

[outer mesa]

Yes but that isn’t total cost

That’s just per item

You were wrong

I was right

Idk why I changed my answer

Now I got it wrong

[viscid garden]

Yeah I was also thinking about the 4th one... But I thought you asked cuz you got it wrong...

lol

[outer mesa]

Nah

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@outer mesa Has your question been resolved?

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[rare vapor]

I remember learning for poisson approximation to binomial, the conditions were N >50 and np < 5

[rare vapor]

However when searching it up, Its giving me a totally different answer

Can someone clear up where exactly the N > 50 and np < 5 conditions apply

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[gleaming sinew]

Why the union of empty sets is empty but the intersection of empty sets is universe set?

[frozen marlin]

it's not?

the intersection of any set with an empty set is an empty set

where did you get that information from

please don't ping me

[toxic stratus]

[toxic stratus]

if you have a definition of what unions and intersections of sets are, you can verify this by working through the definitions

[gleaming sinew]

I had tried

[toxic stratus]

what are your definitions

[gleaming sinew]

Axiom of unions. For every collection of sets there exists a set that contains all the elements that belong to at least one set of the given collection.

In other words: for each collection e, other than 0, there exists a set V such that x E
V if and only if x E X for every X in e, here E is belonging sign

.close

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[serene hound]

hi

[dense wadi]

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[smoky nacelle]

That is not readable

at all

[dense wadi]

Also your teacher should invest in a blackboard

[smoky nacelle]

Yes

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@serene hound Has your question been resolved?

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[wind laurel]

Is this correct?

[wind laurel]

I feel like this contradicts the fact

but i'm not sure if its correct

I agree that in an open neighborhood of t, f must be concave. but what about outside that neighborhood, why cant the curvature change?

I hope what i showed is wrong cause thats actually a proposition that i'm trying to prove and already made a good amount of progress on

and this contradiction just came to mind and im scared now that i wasted my time...

[hearty egret]

what ? the preporsition talks about the existence of a function ...it s hard to give a couter example in these cases

[finite yacht]

I guess that f is concave in the whole interval

It does not specify the point

[wind laurel]

wdym? f is a function that satisfies all the properties but still isnt concave

[finite yacht]

No, f does not satisfy the condition at some points

Like at the local minimum next to the t you chose

[hearty egret]

d oh ok , in that case you have to prove that for every t you can find such a g ?

[finite yacht]

You can't find any g there

[wind laurel]

ohh so these properties need to apply to every point on f?

[finite yacht]

^

[finite yacht]

^

[wind laurel]

okayy thanks!!

[hearty egret]

woomy in fact in your example, in a very little neighbourhood of t, f is going to be concave

[wind laurel]

Yea

I already proved the proposition

but more hand-wavy

and intuitive

so for practice im trying to write a rigurous proof about it

#help-0 message

[finite yacht]

But I'm not really sure that works in general
There may be functions f such that g exists for one t\in(a,b) but f is not concave in any open neighborhood of t

[wind laurel]

I proved it here

maybe theres a mistake in the proof but basically

[hearty egret]

yes you are right

[wind laurel]

for any f that is convex you can chose a small enough epsilon such that g < f and that contradicts the first property

[wind laurel]

since epsilon is arbitrary and the curvature of g depends on epsilon

just chose a very small epsilon

such that g "becomes" flatter than f

"in the limit" g becomes a straight line and thus f must be concave (ofcourse g will never be a straigh line since epsilon is strictly greater than 0)

[hearty egret]

ok so just lets use f with f''(t)=0, but still convex

for example x^4

[finite yacht]

https://en.m.wikipedia.org/wiki/Weierstrass_function here at t=0 you can find g for all epsilon but f is far from being concave

So the fact that it is for all t must be central

[wind laurel]

here i chose g = x^2 and now g < f

[hearty egret]

right and g''(0)=0<< eps

[wind laurel]

f must be C2 differentiable

(its second derivative must be continous and exist)

[hearty egret]

I m not sure you can find such a g here

[hearty egret]

i don t think so

[finite yacht]

Oh
That's not in the original question

[wind laurel]

Oh wait no

[wind laurel]

yea no my bad g must be C2 differentiable

[finite yacht]

You can, choose g constant and equal to the maximum of f

[hearty egret]

? g must be equal to f in t

so your point t cannot be that minimum

[finite yacht]

I mean the maximum

At t=0

[wind laurel]

sorry im back

[hearty egret]

oh right! it works i think

[wind laurel]

But do the properties work for any t?

[finite yacht]

No

Just for t=0

And a few other points

[wind laurel]

So i guess it still doesnt contradict the proposition since these properties need to hold for any number in the domain of f

[finite yacht]

^
And not only that, the fact that it works for all t must play a role in the proof

Because if you only use one point you don't even get local concavity

[wind laurel]

Yess!! Thanks a lot, this was actually my original doubt lol

so just to make sure

is saying "for any" equivelant to "for all"?

[finite yacht]

I think it is

In this case it is

Maybe in some contexts it's not

[wind laurel]

maybe its better i change it to for all?

to make it clear

[finite yacht]

Yes

[hearty egret]

and also c^2 assumption is not enough for this ...as X^4<X^2 is another counter example

[wind laurel]

Heres what i have so far btw

[wind laurel]

Why wouldnt it be?
The main point of the proof is that for any convex f(x), theres a g(x) thats "less" convex that f, thus f > g, contradicting the first property

[finite yacht]

But you could choose δx^2 as g(x) instead, and it would work

[hearty egret]

yeah they seem quite similar

f is not assumed to be convex ... so g cannot be less convex than f .. and g is just greater than f everywhere but in a point..and it has a restriction on the second derivative

[wind laurel]

(the lemma is to show that if at t, f'' < 0 (concave), then it cant "suddenly" become negative (convex)

[finite yacht]

But f'' does not exist in general

f is just continuous

[wind laurel]

i guess if f'' doesnt exist then asking about the curvature of f doesnt mean anything as it has no curvature

do i specify in the proof that f must also be C^2

[finite yacht]

But there is a definition of concave for general functions

Which definition are you using?