help-4

[gaunt fog]

1,296

but with decimals

im gonna

sorry mb

got brain fog

ok now

so ye little bit difficult

so i did

4 divided by 7

since 220 - 216 = 4

and i got the answer of 5.71428

holy sh this is cool

w mental mth

how i do

[muted berry]

your remainder should be 4

[gaunt fog]

yes

[muted berry]

if you want the decimal you can just compute 4/36

[gaunt fog]

wait holy sh

mb

[gaunt fog]

i mixed this up

from our last equation

7 x 100

which we did earlier

so yeah

4 / 36

36.111111111

i think

lemme calculate it

YES

LETS GOOOO

EPIC!

uh

how i close this agin

.close

[vale dock]

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[warped lintel]

hii

How would I calculate the final surface area of an irregular polygon if I was to radial project it onto a cylinder?

Like say I have a weird shape and I project it onto a cylinder, ofc the edges would stretch. How do I calculate this new surface area?

[wintry oxide]

https://tenor.com/view/dexter-shocked-wtf-dexter-morgan-jdooo-gif-3312872781010874916

Do you have the area of the shape or is the an abstract question ?

[warped lintel]

i havent decided on dimensions yet, its more of a "how do i do this and then i can decide on what area i want" thing

[hollow rune]

To be honest, this has a lot of nuance

Where in your cylinder? How big is the cylinder compared to your figure? What is your polygon?

[warped lintel]

projecting directly at the cylinder with no angle, i havent decided on the sizes yet, just wanted to know if this essay topic is doable. my polygon will probably be very weird shapes like this

[hollow rune]

You will probably have a huge problem when trying to do this with quantifying the area.

[warped lintel]

im more of looking at a overview of the method on doing it, and then quantifying the area and solving it with that method ykyk

[hollow rune]

For regular polygons, circles, the more common quadrilaterals, this might just be easy enough, but it can be a huge pain for irregular polygons

[warped lintel]

yeah...

[hollow rune]

The angle does matter btw

[warped lintel]

yeah i didnt clarify that in the original question mb. it's directly perpendicular to the center of the cylinder, so like right at it

[hollow rune]

beyond that: if the angle covered by your figure is relatively small, the resulting area is basically the same as the original

but for "bigger figures", it starts warping more

[warped lintel]

yeah itll be bigger figures

[wintry oxide]

Where is the axis of the cylinder

Is the whole shape projected into the cylinder or do you lose a portion of it ?

How big is the cylinder

[warped lintel]

axis is black line, though i wouldnt mind knowing the blue line too
whole shape yeah
havent decided on dimensions yet, just looking for an overview of the process

[hollow rune]

I could kinda help you on the geometric sense of the problem and prob a numerical method, but im completely stumped on how one would arribe to a generalized formula

[warped lintel]

its oki there doesnt need to be a formula, im looking more for a step to step on e.g. "first you measure area of original polygon, and then abc"

[hollow rune]

Are you good with descriptive geometry and technical drawing?

[wintry oxide]

Let’s take this as a basis for discussion @hollow rune ?

So when you project a portion dx it becomes ds = r dtheta

[hollow rune]

The method im setting relies more on being able to draw out the figure and divide it in "slices"
Disgustingly time wasting

[wintry oxide]

I think integration works, but you need to know what you’re projecting and how

[wintry oxide]

This works if the shape is regular enough

[wintry oxide]

By the way ! For the other problem; i didn’t see the word “interior” !
But the idea of lines is good, the only way to cover a full line is to have 2 opposing parábolas with the same axis that intersect. Since there are infinite lines in the plane, the only way to cover all of them is to have infinite couples of parabols

[hollow rune]

sorry, still drawing things out, its pretty chaotic, i see the idea

For any given slice that describes the width of the figure you chose: You can find from where to where it projects to the x-axis thats situated inside the circle, and use arccos to search for what angle that represents

Based on this you can use a parametric function to integrate over (sint,cost) theta_1<t<theta_2

[wintry oxide]

I would use the arctan to ensure a bijective change of variables but yeah that’s the idea

[hollow rune]

Given we project to the front of the object, 0 to \pi suffices

anyways

[wintry oxide]

You’re Right

[hollow rune]

Yeah, the problem is not finding the projection of a slice
Not even really the fact of integrating over all slices

Its basically defining a function that allows you to get the limits a<x<b for each slice

[wintry oxide]

But there isn’t necessarily a function. If the object is random

[hollow rune]

thats why

[wintry oxide]

That’s Why I explained just the concept of how to approach it. But we need a real case in order to apply it

[hollow rune]

Again, thats the reason i was going for more of a numerical method where you divide the object and measure it manually

[wintry oxide]

The question as is, it’s too abstract

[wintry oxide]

Even so you’ll have the same problem no ?

[hollow rune]

Itll be slow as hell, but pretty easy

For a computer its still fast enough

You could define your polygon as a collection of points in the vertex, measure their distance from the axis of the cylinder projected to that figure, and just lerp between them

[wintry oxide]

Yeah I agree.

[hollow rune]

god forbid you decide to have convex figures along the x axis

[wintry oxide]

🤣🤣

[hollow rune]

that would be fucked up even to code

you could maybe make an algorithm to see if y axis is more numerically stable / simpler

anyways, yeah, the problem is unsolvable for a general basis, it depends a lot on the polygon and its location

[vale dock]

@warped lintel Has your question been resolved?

[wintry oxide]

Is there a way to apply some kind of monte Carlo reasoning here ?

If you solve it numerically. Why not solve it statistically

[hollow rune]

Yeah, you could, instead of mapping as just 0 or 1 and averaging between all elements

You basically have a density map

[wintry oxide]

Project random rays from the axis of the cylinder and see how many land on your shape

[hollow rune]

or that, yeah

[wintry oxide]

Then it’s proportional to the área of the cylinder

[hollow rune]

You would need to be able to project the figure to the cylinder for that anyways

[wintry oxide]

It’s just about solving if a line intersects your shape ?

[hollow rune]

I was thinking more of:

Perpendicular to the figure, project rays straight ahead, those at the sides get weighted heavier than those at the center

[wintry oxide]

Oh works too

Yeah I think if you have to solve it numerically. Then Monte Carlo is a good bet

[hollow rune]

Remember the lines get curved when projecting like that onto the cylinder, the figure is pretty warped, being able to find if the line intersects depends on you finding the angles that "represent" the figure

And at that point, just do the slice method

[wintry oxide]

No … I mean no curved lines

[hollow rune]

I know, projecting from the axis straight lines

But the figure projected onto the cylinder is curved

[wintry oxide]

The axis of a cylinder shoots a straight Line, either it touches the object or not

[hollow rune]

If you were to cut it up and stretch it the lines that formed the polygon became curved

[hollow rune]

To know how curved, you need to solve for said projection, and at that point just solve this

[wintry oxide]

Oh I see

[hollow rune]

Thats why i think a density /weighting map would be easier

No clue what that map would look like anyways

[wintry oxide]

You’re absolutely Right

The ray CAN touch the projection without touching the object

So ray weighting is the way to go

Crazy how we solved 2 problems today using rays 🤣

[hollow rune]

lmao

[wintry oxide]

The parábolas and this

This monte Carlo method would actually work for any surface. You just need to figure out how to weigh your “rays”. Also for regular surfaces, it’s not hard to implement and quite easy to compute

[vale dock]

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[compact raptor]

My question is how do I solve this I don't really understand it. I want to understand math. Could someone help me through it step by step of how to solve it. I don't know where to begin in this math problem.

[charred nacelle]

so h(x)=x-5

what is h(1)

[pallid shuttle]

-5?

oh

-6

[charred nacelle]

let velcaria answer it

[pallid shuttle]

ohh mk mk

[compact raptor]

I trying to understand it. I don't know where to start to solve the math problem 😭 I have no clue where to begin to solve

[charred nacelle]

Read the question

[compact raptor]

ok

[charred nacelle]

They are asking for h(1) * k(-2)

You are given h(x) and k(x)

So let's find both h(1) and k(-2) and then multiply

[charred nacelle]

So h(x) = x-5 then what would be h(1)?

[pallid shuttle]

So basically you can try to find hx-x=-5

if x was =1

[compact raptor]

Idk

[charred nacelle]

h(a) = a-5
h(b) = b-5
h(x) = x-5
h(1) = ?

[compact raptor]

y=-5?

[charred nacelle]

What are you going to write for h(1)

[compact raptor]

I have no idea I don't think I'm understand X_X.

[charred nacelle]

So for h(x) = x-5 if they give any value of x , you substitute it

Instead of x

h(80) = 80-5

[compact raptor]

ok

[charred nacelle]

So h(1)=?

[charred nacelle]

?

[compact raptor]

I still don't get it 😭 Like I keep rereading you're messages but I don't understand.

[charred nacelle]

Do you know what substitution is

y= x+3 , find y at x=1

Can you do that?

[compact raptor]

no

[charred nacelle]

Okay so instead of the variable you put a value there

[compact raptor]

okay

[charred nacelle]

Can you do this

[compact raptor]

I can try

[charred nacelle]

Do that

[charred nacelle]

Did you still not get it?!?

Do you understand linear equations?

[compact raptor]

y=4?

[charred nacelle]

yes

FINALLY

TOOK A LONG TIME

Okay so

[compact raptor]

yea I'm sorry I really don't understand math X_X

[charred nacelle]

h(x)=x-5

h(1)=?

This says that

y= x-5 , find y at x=1

[charred nacelle]

Practice more basics I feel your basics and linear equations are not great

[compact raptor]

oki

[charred nacelle]

Use this logic and solve the rest I will leave maybe someone else can help you

[compact raptor]

oki

thanks anyways :D>

h(1)=-4

.close

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[loud jetty]

In this absolute value graph y=5Ix+2I

[flint phoenix]

mhm

[loud jetty]

I'm trying to know the Domain,Range,Increasing&Decreasing

[flint phoenix]

What have you tried in advance?

[loud jetty]

I just know D and R does something with the "inf"

[flint phoenix]

R does not have anything with "inf" unfortunately

Do you know the definition of R?

[loud jetty]

not really

[flint phoenix]

how about D?

[loud jetty]

It's like (inf, inf) or somthing like that

[flint phoenix]

not really, lemme clairfy it for you

Domain means every possible x that can be in a function

Range means every possible y that can be in a function

[flint phoenix]

Check out this cute stuff, is there any point in the function that is below x-axis?

[loud jetty]

No

[flint phoenix]

That's right, so y has no chance to be negative

and that's how we determine range and domain

Can you tell me the range of this function?

[loud jetty]

0-inf?

[flint phoenix]

Try to apply

[loud jetty]

I don't understand what that means

[flint phoenix]

"()"

lol

[loud jetty]

(0, inf)?

[flint phoenix]

Precisely speaking, it's [0, infty)

Because 0 is a possible y value

[] means include
() means not include

Is it alright?

Fantastic, now it's your time to find the domain

it's the exact same logic

Ping me when you're done

Imma go slacking for a minute

[loud jetty]

(-infy, infy)

[flint phoenix]

Alright, anything else?

[loud jetty]

Now increasing & decreasing

[flint phoenix]

So basically, how we determine that is checking whether the y value is increasing or decreasing

from left to right

yk what I mean?

[loud jetty]

Ok?

[flint phoenix]

Alright, based on this concept

Can you find me the answer?

from XX to OO the function is increasing; from PP to QQ the function is decreasing

[loud jetty]

If the x is between -infy and -2, the line in decreasing?

[flint phoenix]

keep going

[loud jetty]

If the x is between -2 and infy, the line is increasing

[flint phoenix]

If the x is between -infy and -2, the line in decreasing?
✅
If the x is between -2 and infy, the line in increasing
✅

Very Good

Alright, time to bye. Anything else?

[loud jetty]

Why does an absolute graph have a change in line direction compared to a linear equation?

by mean, there's a vertex

but why

[flint phoenix]

I'm not sure how to explain that, but it's the result after poltting the points on it

[tawny moon]

consider the definition of absolute value

in an absolute value, we don't have negative numbers, so if whatever inside an absolute value is negative, it comes back out positive.
so when the expression inside an absolute value goes from positive to negative, the graph "bounces" sharply off 0

(not formal terms, but i hope it's intuitive enough)

[loud jetty]

.close

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[tawny moon]

i don't think it's correct

[copper stump]

,w tan(10pi/6)

[rocky lotus]

Results provided by WolframAlpha

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

Requested by bacter10fr4g

[verbal elk]

.close

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[proper hearth]

Simple matrix operations (addition, scalar multiplication)

Determinants (2×2, 3×3)

Chain rule & product rule

Derivatives of standard functions (polynomial, trig, exponential, log)

[proper hearth]

wondering what resource would be best to revise these prerequisites before my uni course

[hollow rune]

To be honest, if you are just revisiting, just search up a few problems on whatever plan you got for uni

These are relatively simple concepts about algebra / calculus

[proper hearth]

true but i honestly forgot everything haha

[proper hearth]

ah alright

thanks a lot bro!

[hollow rune]

If you have a particular problem, wouldnt mind helping

[proper hearth]

oh nah i js wanted to know how to go about revising these concepts

[tidal terrace]

pretty much just go to YouTube or answer some problems

[hollow rune]

From these, id advice you practice chain rule the most

All the others are formulae or really intuitive operations

[proper hearth]

ohh ight imma go thru yt and look up some problems online tysm

[proper hearth]

oh alright will focus on that

[tidal terrace]

Bro became green 🔥🔥🔥

[proper hearth]

ohh

tysm guys for the help

[hazy pivot]

How am I still not 😭

I was gonna congratulate him on active, and boom he's helpful too now

(on a serious note, congratulations @hollow rune)

[hollow rune]

Thanks, no clue why i just got it, lmao
wrong reply 🔥

[proper hearth]

congrats lmaop

tysm again guys ig ill close it now

.close

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[normal temple]

Hey guys, I’m currently working on a question from my maths assignment. I’ve already derived the Maclaurin series for cos(x) and cos(x^2), and I know a bit about the Lagrange Form of the Remainder (it wasn’t covered in class, but a tutor suggested I use it).

The question asks me to find the minimum value of n such that the error is less than 0.2.

My main issue is that I don’t know how to handle the numerator in the Lagrange Form of the Remainder. If I were dealing with cos(x), the maximum value of the derivative is always 1. But with cos(x^2), it doesn’t seem to have a fixed bound. What can I do here? Do I need to modify the Lagrange Form of the Remainder, or is there a trick I can use?

[ebon glade]

its still bounded by 1

its still cos(something)

[normal temple]

yes but lets say we chose the first derivative of cos ie n=0. Would the max value be still 1 or 2?

[ebon glade]

|cos(anything)|<=1

always

[steady charm]

well the derivatives of cos(x^2) would be annoying and not nicely bounded. but also the taylor series for cos(x^2) should be alternating...

[ebon glade]

whether thats the tightest bound depends on the anything

oh wait ok sry I misread

[normal temple]

is there a trick with it being alternating?

[steady charm]

there is a separate error bound for alternating series

[normal temple]

maybe this helps with the context

[normal temple]

thats very good to now thanks

[vale dock]

@normal temple Has your question been resolved?

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[wintry cape]

I know how to factor but I have no idea what a circuit is. Can someone look at the directions and see how I advance here

[runic scroll]

you have two factors. look at the "answers" of each box in the sheet. the one with an answer that matches either factor is your next cell.

the "circuit" here is not mathematical. it's merely to tell you to follow a path, cell by cell, by using the factors as a guide.

[wintry cape]

I should go ahead and factor the next one and look at those answers for a factor that goes with #1 ?

[stark wedge]

i think the idea is that you factor each question and then find one of the factors in the Answer: line of another question

[wintry cape]

Line of another question ?

[runic scroll]

notice how each box has an "Answer:" line?

[wintry cape]

Yeah

[runic scroll]

from your first question, look around the "Answers" of the other boxes.

[wintry cape]

So from number 1 I’m going to the bottom left box?

[stark wedge]

so you did #1 and now you need to find which question says "Answer: x+2" or "Answer: 2x+1" at the top of the box

yes, bottom left.

put #2 in there

[wintry cape]

What would I put on the line

[wintry cape]

Ah ok

[runic scroll]

then keep proceeding in this manner until you're done.

[wintry cape]

Thank you 😊

[runic scroll]

don't mention it. might want to thank the other helper as well.

[wintry cape]

@stark wedge thank u to you as well

[runic scroll]

please remember to close the channel when done.

[tawny moon]

hi OP, are you done with the channel?

[river shale]

.solved

[vale dock]

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[graceful island]

confused as to why the answer is not 274

[steady charm]

can you show your work?

[graceful island]

to find t, I did sqrt((-100 times 2)/-9.8)

which got me 4.5175 for time

then I did ((4.5175)(24.5)) + (0.5 times 4.5175^2 times 16)

to get 274

[steady charm]

24.5 is the roadrunner's speed, not the coyote's

[graceful island]

uhh

I think it is the coyote's tho

because 24.5 is the bare minimum for the road runner

[steady charm]

but the coyote is constantly accelerating. he starts out much slower than the roadrunner but by the time he reaches the cliff he is going much faster

[graceful island]

ahhhh I see

[hollow rune]

first and last time this sentence will be said btw

You have to consider the speed in which the coyote launched off the cliff after accelerating for a few secs (based on the previous part)

[hazy pivot]

Is that fucking greens theorem

[hollow rune]

And you use the limit time for it to fall to the ground from 100 m to calculate the position he'll end up in

[graceful island]

got the answer

385

[hollow rune]

yeah

[graceful island]

alr thx

I do need help on the next (and last) question but lemme do part c of this one first

alr part c I did

i am pretty lost

[hollow rune]

Okay, for a) youll first have to find the acceleration the ball feels horizontally.

[graceful island]

I thought it would just be 16 times (189/1000)

but that was wrong

[hollow rune]

Lets go over it. We know the following;
It starts from rest, aka 0 m/s
its accelerated for 0.189 seconds
it ends up with a speed of 16 m/s

Since it has been stated that the acceleration is uniform, we know it must equal some constant.

[graceful island]

ohhh

I misunderstood the question

acceleration is 84.7 m/(s^2)

[hollow rune]

yes

Based on that, and assuming it also starts from pos = 0 m, you could find the distance it moved during those 0.189 seconds

[graceful island]

Okay i got a correct, 1.51m

[hollow rune]

Okay, do you know how to do B?*

[graceful island]

no

[hollow rune]

Well, lets go over it, what is the unit of force?

[graceful island]

newtons

[hollow rune]

and what do those equal?

[graceful island]

so mass time acceleration over time

ik acceleration

wait

no I dont

[hollow rune]

just mass.acceleration

[graceful island]

really?

one sec

[hollow rune]

#help-4 message

[graceful island]

right but that's just in his hand until it's thrown

acceleration changes after that right?

[hollow rune]

The problem asks for the force the thrower exerts

[graceful island]

ah

[hollow rune]

Now, objects under gravity's influence suffer a force that pulls them downwards. We were given a fact: The grav. force the ball experiences is 2.25 N

[graceful island]

ah yes and acceleration is in distance/(time squared)

I remember now

[graceful island]

right and acceleration downwards is -9.8 m per s^2

so mass is

one sec

0.230 kg

[hollow rune]

Yeah

So now we know the mass of the ball, and the acceleration it experienced by the pitcher

[graceful island]

got it

oh

[hollow rune]

The "direction" part is more of a purely theorical thing, it has almost nothing to do with our math

[graceful island]

weird thing one sec

I did (2.25/9.8) times 84.7 and got 19.4

but that was wrong

[hollow rune]

the 84,7 m/s^2 registered as correct?

[hollow rune]

Hm, strange

[graceful island]

I mean it should be?

16/(189/1000) should 100% be the acceleration

which is 84.7

then 84.7 times (2.25/9.8) should be the newtons

[hollow rune]

Yeah, it should

Try completing the second part with the direction

Btw, since we were given the fact that it leaves with fully horizontal speed, that means its 0 degrees above the horizontal

[graceful island]

it's not 0 above the horizontal

I have an idea

[hollow rune]

Oh, maybe it consider the 2.25 to be instantly applied

[graceful island]

yep

[graceful island]

that's what it was

for the magnitude at least

so 19.6 newton magnitude

[hollow rune]

Oh yeah, i figured why

[graceful island]

and lemme do angle

[hollow rune]

Its cause they pitcher has to excert upwards force to make the ball not fall through his hand

(normal force)

[graceful island]

question rq

why is it not -6.6 degrees

[hollow rune]

Then the force is slightly upwards, just do atan(2.25 / 19.4)

[hollow rune]

The force the pitcher is opposite of the gravity

[graceful island]

but the gravitational force is down

so the 2.25 should be negative

[hollow rune]

Its the normal to avoid the ball falling through his hand

Since we know it left horizontally, something had to cancel the force from gravity

[graceful island]

I've been struggling with like when it's above a horizontal or not

because sometimes I'm meant to include the negatives in the tan

and get a negative angle

and sometimes not

and I don't get it

[hollow rune]

generally dont, its easier, at least in my experience

[graceful island]

well quizzes and all that

[hollow rune]

just accomodate for your negatives angles yourself

[graceful island]

I got a 69% on my last quiz because of an issue like that (it's a 3 question quiz)

[graceful island]

but I mean like

ig you're not my professor

idfk

like this is negative angle

and it makes sense

[graceful island]

and this shouldn't even be above the x axis

the ball should start dropping immediately

so it should be -6.6 degrees

not positive 6.6 above the horizontal

or am I crazy

[hollow rune]

Its asking for the force of the pitcher, not its trajectory

imma try to draw it 🥀

the red force is gravity, its because we are in a gravitational field

The green is the normal the hand of the pitcher causes to avoid the ball falling while being held
The purple is the horizontal force

The forces that the pitcher exert are the combination of the green and purple

[graceful island]

OH

thank you

it makes sense now

that's awesome

[hollow rune]

You simply got the opposite angle cause the weight** and the normal are opposites themselves here

[graceful island]

gotcha

alr imma close out

it's 5am and I didn't sleep tonight

but i also procrastinated that assignment a ton

shoulda been like a 1 hour assignment max

it's been 5 hours

anyways

thank you a ton

.close

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[autumn spade]

i got stuck at the green highlight, i cant find my mistake. i been stuck for 2 hours on this

[zealous pendant]

I don’t remember the method but should there be a - sign on all 3 of the 1 0 2 equations?

Like $f_{x}+ag_{x}+bh_{x}=0$ instead of $f_{x}=ag_{x}+bh_{x}$?

[rocky lotus]

BBMaths

[vale dock]

@autumn spade Has your question been resolved?

[autumn spade]

nah

[gritty sinew]

what you did looks alright

the issue is you missed a case while solving your lagrange equations

[zealous pendant]

Perhaps b is complex lol

[gritty sinew]

2y(2a+b)=0

y=0 ain't the only situation where this is 0

[zealous pendant]

$b=-2a$

[rocky lotus]

BBMaths

[zealous pendant]

Actually it doesn’t matter which way around it is since it just flips the signs of the Lagrange constants

[vale dock]

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[gaunt mantle]

im lost

dont know where to start

[next snow]

try u = x^10

[vale dock]

@gaunt mantle Has your question been resolved?

[gaunt mantle]

just did

can't do any meaningful substitutions after that

differentiating the outside function gives the same x powers as the one inside the radical, but the coeffs are all messed up so can't really do anything with that approach

<@&286206848099549185>

[charred nacelle]

the ^1/10 is only for the (2x^40+5x^10+10) part right

[gaunt mantle]

yup

makes life a whole lot harder

[charred nacelle]

You sure? I feel this makes it a lot complex

[gaunt mantle]

i am sure

[charred nacelle]

From your other integration question this feels 10 steps ahead

[gaunt mantle]

here's the full question

can differentiate the other side and get a system of equations but i wanna do the integral

[gaunt mantle]

tbh this is very lengthy also

[hardy coral]

Next time post this first

Try b = 11

[gaunt mantle]

but i want to integrate it though

[hardy coral]

Good luck

[gaunt mantle]

😭

[zealous pendant]

Sometimes the integral doesn’t have a closed form, take what you can get from the answer sometimes

[hardy coral]

That has nothing to do with it

It obviously has a closed form, it's given up to the values of a and b

It's just awful to derive by hand

[zealous pendant]

Sure but what I meant is that sometimes integrating is hard

[gaunt mantle]

hmm

alrighty then

💀

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[sick lynx]

Hello I am having great trouble with this problem (its probably very easy)

So far what i understand is that the contours of a map are basically how high it gets for each line. Here i have 8 contour lines. and i have to find out the height for each contour line. i have tried counting out each line between 800 and 400 and dividing it by that. (800-400=400 real height then 400/6 but gives me 66.66... which doesnt work because i counted it out.)

So then i moved onto part E because maybe that would help me get contour line ect, so i did 800/2.8 but got weird number that doesnt work.

I am incredibly confused and a bit frustrated because i feel like im missing something here very obvious. So please explain to me like im 5 how to go about solving this.

[mild kernel]

you don't get any assurances that it will be a nice even number on the spacing

I don't think you should be using 6 though

Normal off-by-one error

Consider it like: there's 400m, then theres 5 steps (4 unmarked lines, and then the marked 800m line). We need to go from 400 to 800 in these 5 steps, one step for each line.

[sick lynx]

so then 400/5?

[mild kernel]

Yea.

,calc 400/5

[rocky lotus]

Result:

80

[sick lynx]

so then each contour line is an increase of 80 meters

[mild kernel]

Yea

[sick lynx]

which would mean point b is 880 and point a is 320

[mild kernel]

I believe so, yea

[sick lynx]

880-320=560 then 560/2800=1/5

[mild kernel]

sorry for which question?

[sick lynx]

for e

[sick lynx]

for E mb

[mild kernel]

they confusingly make you use different units

[sick lynx]

yeah..

[mild kernel]

so you want the rise / run, in m/km

you got the m right, its 880-320=560 m rise

but you don't want to use 2800m

because we want the denominator to be in km

[sick lynx]

so turn meters into kilometers?

[mild kernel]

well, they do give you the run in km already

we want rise/run

you have to calculate the rise from the contour plot

but they tell you, 2.8km horizontal distance, thats run

[sick lynx]

right, so then we would turn meters into kilometers for the rise right?

[mild kernel]

nope

"the units of gradient are m/km"

[sick lynx]

oh

[mild kernel]

this tells you, if you are calculating the gradient, you need to divide meters by kilometers

[sick lynx]

so then it would be 560/2.8

[mild kernel]

yea i think so

,calc 560/2.8

[rocky lotus]

Result:

200

[mild kernel]

a nice round number

not that we know for sure it will be

[sick lynx]

wow what a beautiful number

ok i think you helped me tremendously with this. i was very confused but now i understand mostly. so thankyou very much

[mild kernel]

no worries.

[sick lynx]

❤️

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[tidal timber]

I need help with Karnaugh maps

What should I do with unknowns? Should I ignore them or treat them as 1s?

[vale dock]

@tidal timber Has your question been resolved?

[viscid spade]

Do what you want with it iirc

You can set it to 0 or 1 depending on how its better for you

[tidal timber]

The whole point is to minimize the formula as much as possible

[viscid spade]

For the reduction

[tidal timber]

Yeah I realized that

So umm 110 = 1 and 101 = 0 for the shortest result?

[viscid spade]

101?

Oh yeah this works well

But can't you set both to 1 and do two groups ?

[vale dock]

@tidal timber Has your question been resolved?

[tidal timber]

If I do the bottom group (111 and 101) then it would be a & c so yeah that one works too but picking both unknowns to be 1 doesn't work because I'll have to include all 3 variables and that doesn't make the formula shorter

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[magic grove]

Stuck on this problem, idk if it is ordered or unorded and ai isnt helping that much

[magic grove]

is it ordered or unordered

[mortal temple]

i think it's saying two arrangements will be considered the same if you can rotate one to look like the other

[viscid spade]

"Chairs don't matter" means what in terms of order ?

[mortal temple]

i would say order matters

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[rotund vale]

is there a quick way to get the distribution of all possible sums of rolling n sixed dice? in other words, is there an easy mathematical implementation for finding the probability of any sum n given i roll some number of dies

[steady charm]

https://math.stackexchange.com/questions/2290090/probability-that-the-sum-of-k-dice-is-n

[rotund vale]

ty!

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[mental island]

im failing at understanding how to solve this specific problem can anyone help me

[fickle rose]

Write:
[\frac{1}{x^2(x-1)^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}+\frac{D}{(x-1)^2}]

[mental island]

heres my work so far

[rocky lotus]

PajamaMamaLlama

[fickle rose]

ah

[mental island]

idk how to find A or C

[fickle rose]

well from (1=As(s-1)^2+B(s-1)^2+Cs^2(s-1)+Ds^2)

[rocky lotus]

PajamaMamaLlama

[fickle rose]

expand out each guy write it as (a_3x^3+a_2x^2+a_1x+a_0)

[rocky lotus]

PajamaMamaLlama

[fickle rose]

then equate coefficients, i.e.: (a_3=0,; a_2=0,; a_1=0,; a_0=1)

[inner abyss]

if you don't want to compare coefficients, you can also evaluate at more points

[rocky lotus]

PajamaMamaLlama

[mental island]

what points would specifically get me A or specifically C tho

[inner abyss]

try 2

or -1

[mental island]

I tried 2 but it gets me both A and C

[inner abyss]

so do -1

then you have simultaneous equations in A and C

[mental island]

OHHH okay

wait so then i get A = -1, B = 1, C = -1/2, and D = 1

does that look right

[inner abyss]

hmm, I don't think so

[mental island]

😓

[inner abyss]

what equations did you get

[mental island]

ima screenie one sec

[inner abyss]

where did the stuff after the green star come from?

[mental island]

i subbed in the C

[inner abyss]

A + 2C = -2 is right

oh you did that first, I see

I don't see any A's in your equation for s=-1 though

[mental island]

yes i subbed in the A = -2 - 2C there

for A

and then i went back to the other one and subbed in C = -1/2

[inner abyss]

I think you may be missing a minus sign in the s=-1 equation

[mental island]

where

😟

[inner abyss]

the coefficient of A is s(s-1)^2

so -4

not 4

[mental island]

omg

i forgot ab the s

tysm 🙏

[inner abyss]

rather than sub in A immediately, I would write out the two linear equations in A and C first

makes it easier to spot where mistakes happen

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[rotund vale]

this is the formula for the number of ways to reach sum x with n D sided dice. I'm struggling to figure out how to alter this function, so that it returns all the ways minus any combination of dice outcomes that include a 1. can someone please help?

[rotund vale]

obviously i can just subtract all combinations of dice rolls including 1, but thats not the most efficient way

from a glance, i would have to do n-1 choose k. but idk how to alter the other combination

https://math.stackexchange.com/questions/2290090/probability-that-the-sum-of-k-dice-is-n

link to equation if needed

<@&286206848099549185>

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[ivory mulch]

im meant to put this on the graph and i have no genuine idea what im doing i know almost nothing (im very bad at math)

[river shale]

Mathematical tools available?

[ivory mulch]

wdym? i can use tools yes

i just dont know how to go about it

[river shale]

What tools specifically?

Is desmos allowed?

[ivory mulch]

yes

[hazy pivot]

That trivialises it so let's not

I imagine they want three things from you here, maybe four

1. The points where the function is zero, these are easy to find since the function is already factorized for you

2. The sign of the function at other points. This can be easily found by substituting values in between consecutive zeroes

Are you familiar with derivatives

@ivory mulch

[ivory mulch]

no not really

[hazy pivot]

Okay in that case this is really all you need

I don't imagine they expect you to give a completely accurate graph

Probably just want signs and zeroes

Do you know how to do the two things I said

If not, we'll go over them

[ivory mulch]

no i dont know

[hazy pivot]

Okay

Well for the first one

[rocky lotus]

Xavier 🌺

Your function is $f(x) = (x-1)^3(x-2)^3(x-3)^2(x-4)^2$

[hazy pivot]

Can you tell me when this is zero

Just by looking at it

[ivory mulch]

umm no im not sure

[hazy pivot]

Well when is a product zero

If you have a, b, c, d and abcd is zero

What can you say about the individual variables

[ivory mulch]

that one is zero?

[hazy pivot]

Yup

At least one of them is zero

So now to back to the function

For what values of x does the function multiply to zero

Bella you still here?

[ivory mulch]

yes

im just having trouble understanding

[hazy pivot]

Then say it lol, I'm happy to explain

Essentially the function is a product right

And as we just said, if a product is zero, one of the things being multiplied is zero

With me so far?

[ivory mulch]

yes

[hazy pivot]

Good

So now look at the function again

If it's zero, one of the terms in it must be zero

Now tell me, for what values of x can that happen

[ivory mulch]

itd have to be zero straight up for all of them no???

[hazy pivot]

No no

Let's say x = 0

Then f(x) is not zero

Since (x-1) isn't zero

Neither is (x-2)

And so on

So when is f(x) zero

You want (x-1) or (x-2) or one of the other terms to be zero

[ivory mulch]

well none of these r zero then, do we have to make one to be zero?

[hazy pivot]

Yes

We want values of x such that f(x) = 0

Hullo Michele

[ornate nacelle]

gday

[ivory mulch]

okay i dont know how to get there though

[hazy pivot]

^^

When is x-1 zero

[ornate nacelle]

if we have a bunch of terms $(x-a)(x-b)(x-c)(x-d)\dots=0$ then at least one of those terms must be zero. so, we can set each product individually to zero and solve for the required value right? because if any of them are zero then the entire function becomes zero

idk if tyhat makes sense

[rocky lotus]

Michele

[ivory mulch]

ohhhhhhhhhh

[hazy pivot]

I was trying to not overwhelm her with too much information, but if it helps it helps

[ornate nacelle]

yeah me too thats why i was typing for so long lmao and gave up 😭 im so bad at explaining things

[hazy pivot]

Also \textbf{} for bold in latex

[hazy pivot]

Well your explanation worked so

[ornate nacelle]

ye ik

https://tenor.com/view/latex-my-beloved-latex-overleaf-heart-locket-locket-gif-12308574570829297726

[hazy pivot]

Pfft

[ivory mulch]

i understand it a bit more but im not sure on why i need to do this to put it on the graph, im sorry you are working with someone who has never learned an ounce of math i was homeschooled unfortunately

[hazy pivot]

It's okay

Well tell me now, at what points is the function zero

I'll come to why this is important in a moment

[ivory mulch]

any point?

cus if we need to make it zero any point could be zero

[hazy pivot]

No no

You have a function

You plug in a value

For what values does the function give you zero

Read what Michele said as well

[fleet burrow]

I use {\bf ...} 😈

[hazy pivot]

Hullo 0, we are looking for you

[ivory mulch]

at x-1?

[hazy pivot]

When x-1 is zero yes

That's one of the points

When is x-1 zero

[ivory mulch]

when x is equal to 1?

[hazy pivot]

Good

That's one of the points where the function is zero

What are the others

[ivory mulch]

i dont know, wouldnt it be the same?

[hazy pivot]

You have more terms in the function

[hazy pivot]

X = 1 makes the first term zero

But the function can be zero due to the other terms as well

So what other values make the function zero

[fleet burrow]

Computing derivatives of this 😨

[ivory mulch]

x-2 is also zero then?

[ornate nacelle]

Yup!

now use the same logic for the last two terms

[hazy pivot]

And when is that

[ivory mulch]

wehn x = 2

[hazy pivot]

I mean it isn't that bad

[hazy pivot]

Good

Now the other two terms

When are they zero

Not together, just on their own

[ivory mulch]

x-3 is zero when x=3 and the same for x-4 x=4

[hazy pivot]

Yup

[ornate nacelle]

holy peak

[hazy pivot]

So now those are the "zeroes" of your function

Hey Michele be nice she's a victim of homeschooling

[ornate nacelle]

huh

i wasnt being mean im saying this is good

[hazy pivot]

Ah

I thought it was sarcastic

My bad

[ornate nacelle]

ohh mbmbmb

[hazy pivot]

So you can mark these on the graph as zero

[ivory mulch]

yeah i got "homeschooled" which just means in my case neglect lmfao so im trying to get my highschool diploma myself online, which im almost done with i just struggle with the math portions

[ivory mulch]

okay

[hazy pivot]

It's okay, good luck to you!

And we're happy to help you out

[ivory mulch]

thank you :)

how do i mark them on the graph as zero though? im not sure what that means for me to do

[hazy pivot]

Just put a dot at x = 1 and y = 0

And the other values of x that we found

[ivory mulch]

ohh i see ok hold on

[hazy pivot]

After you do that

I want you to convince yourself about something

[ivory mulch]

like this???

[ivory mulch]

huh?

[hazy pivot]

Yup

[hazy pivot]

The function is not zero anywhere except these points

[ivory mulch]

okay

[hazy pivot]

Do you see why

[ivory mulch]

i dont know

[hazy pivot]

Well we said that a product is zero only if one of the terms is zero

[ivory mulch]

yes

[hazy pivot]

Can any of the terms be zero at any other points

For example, can x-1 be zero anywhere except x = 1?

[fleet burrow]

(there's a name for this for easy reference, it's called the zero-product property)

[hazy pivot]

No I know, I just wanna spell things out as much as possible for her

[fleet burrow]

i too know that you know, i was saying it mainly for her in case she wanted to refer to what you said easily

[ivory mulch]

no

[hazy pivot]

Exactly

So the function can't be zero anywhere except 1,2,3,4

[ivory mulch]

yes i understand

[hazy pivot]

So everywhere else it's either positive or negative

[ivory mulch]

mhm

[hazy pivot]

And now this is going to be a bit difficult to digest so take your time

It can only change sign at 1,2,3,4

So if you find what sign it is for a random point in between

It will have that sign in that entire region

Feel free to ask for more explanation

[ivory mulch]

what is a sign exactly?

[hazy pivot]

Positive or negative

Are you familiar with positive and negative numbers

[ivory mulch]

YES LOL

[hazy pivot]

Just had to confirm lol

No offense intended, genuinely

[ivory mulch]

i know LOL just funny

ok so i have to figure out if its positive or negative?

[hazy pivot]

Well now do you understand what I said

Yes that you do

[ivory mulch]

i assume its positive?

[hazy pivot]

But do you understand why the sign doesn't change

[hazy pivot]

There will be regions

[ivory mulch]

no i do not

[hazy pivot]

It's not gonna be the same sign everywhere

[hazy pivot]

If f(x) is positive somewhere and negative somewhere

It has to be zero in between

So if the region we are considering doesn't have a zero point

[ivory mulch]

yes that i understand

[hazy pivot]

It cannot change sign

[ivory mulch]

ahhh

ok i see

[hazy pivot]

So we split the x axis into regions

(-inf, 1)

(1,2)

(2,3)

(3,4)

(4,inf)

And we find the sign in these regions

Are you familiar with interval notation

If not, (a,b) is just all numbers that are between a and b

[fleet burrow]

At a zero point (formally known as a root), the function can go from positive to negative, negative to positive, or keep its sign.

[hazy pivot]

More importantly, it cannot change sign anywhere else

[ivory mulch]

this i understand

[ivory mulch]

i am not

[hazy pivot]

Does this make sense to you

[ivory mulch]

umm kind of? i dont know what a and b are representative of

[hazy pivot]

Random numbers

Well arbitrary, not random

[fleet burrow]

so like if we have 1 < x < 2, we can say x is in the interval (1, 2)

[ivory mulch]

i see

i understand that a bit more then

so how do i find the signs in this

[hazy pivot]

Well

When x < 1

What is the sign of x-1

[ivory mulch]

positive?

[hazy pivot]

No

Think of a number less than 1

Let's just say 0

What's the sign of 0-1

[ivory mulch]

nothing it keeps it

[hazy pivot]

Huh

[ivory mulch]

im confusing myself arent i........

[fleet burrow]

the sign can be thought of as + or -

e.g. the sign of -52 is -
the sign of 93 is +

[hazy pivot]

Thanks for the help 0

[ivory mulch]

i understand that

[hazy pivot]

So what's 0-1

[ivory mulch]

• then

[hazy pivot]

Write it as one number first

What does 0-1 equal

[ivory mulch]

1?

[fleet burrow]

that's zero minus one btw

[ivory mulch]

-1 then omg

im stupid bruh

[hazy pivot]

Lol it's ok

So what's the sign

[ivory mulch]

negative

[hazy pivot]

Good

[fleet burrow]

so you know that the sign never changes in the interval between two roots

[hazy pivot]

Are you familiar with how signs work with multiplication

[ivory mulch]

uhhh no?

[ivory mulch]

yes

[hazy pivot]

If you multiply two negatives you get a positive

If you multiply two positives you get a positive

If you multiply a negative and a positive you get a negative

[ivory mulch]

okay i see

[hazy pivot]

Using this information

What's the sign of f(0)

Also another thing that would help

If you raise a negative to an odd power it stays negative

If you raise a negative to an even power it becomes positive

And positives always stay positive no matter what power you raise them to

[ivory mulch]

okay

[hazy pivot]

Now tell me this

[ivory mulch]

positive?

[hazy pivot]

I believe you're right yea

[ivory mulch]

also we can leave it here for now i need to go do some stuff

[hazy pivot]

Sure sure

[ivory mulch]

thank you for ur guys help i understand this stuff a bit more

[hazy pivot]

But the idea is to find the sign in between the roots

[ivory mulch]

mhm

ill research it a bit when im back to it

[hazy pivot]

Sure sure

[ivory mulch]

tyty

[hazy pivot]

Good luck!

[fleet burrow]

do you want a summary of everything so far?

[ivory mulch]

only if you want to do that, up to you but id appreciate it a lot

[hazy pivot]

Do you wanna do it 0 or should I

[fleet burrow]

up to you

[hazy pivot]

Hmm I'll do it

So we started with the problem of graphing a function.

Since the function is already factorized, we can find the roots (zeroes) using the zero-product rule.

Then we observe that the function can only change signs if it hits zero.

So now we want to find what sign the function has in the regions between the roots, since it cannot change there.

This gives us enough information to do a rough graph of the function.

[ivory mulch]

thank you :D

[vale dock]

@ivory mulch Has your question been resolved?

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[fluid blade]

I lowkey got no idea what to do after this or if I should do it a different way

[fluid blade]

Idk how to rotate

,rotate

[fluid blade]

bang

[worn hemlock]

Substitute t=tanx/2

[fluid blade]

why

[worn hemlock]

Integrals of the form 1/(a+bsinx) get reduced to something simpler using that

sinx = 2tan(x/2)/(1+tan² x/2)

[fluid blade]

So a is -1? And b is sqrt7 sintheta?

[worn hemlock]

a is -1 and b is sqrt7

[fluid blade]

Yea

Mb

That’s what I mean

Meant

But like where does the t go or the tanx/2

[worn hemlock]

Inside

[fluid blade]

Where

[worn hemlock]

tan(x/2)

[fluid blade]

where’s tan x/2

[worn hemlock]

You can write sinx

In terms of

tan(x/2)

[fluid blade]

With this?

[worn hemlock]

Yes

[fluid blade]

That seems mad complicated 😭😭

[worn hemlock]

Trust me

It will get simplified

[fluid blade]

what method is this

[worn hemlock]

Substitution?

[fluid blade]

It’s js a regular substitution

[worn hemlock]

Its very standard tbh

[fluid blade]

I didn’t know if it was like some special one

[fluid blade]

My bad bruh I’m only in hs 😭😭

So would it go to (sqrt7)(2t/(1+t²))-1

All that over 1

I mean under

[worn hemlock]

Yes and dont forget to change dx

[fluid blade]

dtheta

To Dr

Dt

[worn hemlock]

t = tan(x/2)
So whats dt

[fluid blade]

sec^2(x/2)/2

oops

/2

[worn hemlock]

Yea so replace dx with 2dt / sec²(x/2)

[fluid blade]

Ok

One sec

[worn hemlock]

You can proceed from there on your own

[fluid blade]

Hopefully 😭✌️

Yea idek where tf to go

My sec isn’t even in terms of t

I js have 2 variables now bruh 😭

[vale dock]

@fluid blade Has your question been resolved?

[vale dock]

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• Show your work, and if possible, explain where you are stuck.
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[tawny arrow]

finite and union solutions help pls

[tawny arrow]

never done these before

@fleet burrow

you still up?

[golden gate]

!noping

[vale dock]

Please do not ping individual helpers unprompted.

[tawny arrow]

they asked me to ping them

.close

[vale dock]

Channel closed

Closed by @tawny arrow

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[frank path]

Random q but would it be ok if I like put v=dx/dt=d(function) instead of d/dt(function) on my physics test today

[frank path]

Or would I lose pts

[steady charm]

that would probably lose you points because in physics especially it is important to specify what variable you are differentiating with respect to

[pallid oasis]

Why do you want to do that? Also those can mean different things

[vale dock]

@frank path Has your question been resolved?

[frank path]

But I had already established that for instance velocity was equal to the derivative of displacement w respect to time

And then weren’t rlly any other variables

[steady charm]

d(function) also just isn't standard notation for the derivative of a function

[frank path]

So I’d lose pts even if I’d established that?

It was just bc of the time pressure and everything