help-10

[astral ivy]

A polygon is a closed plane figure composed of straight-line segments that meet at their endpoints.

Yes rigid transformations. Imagine you’re cutting a piece of paper and reassembling it

[cobalt shuttle]

so then it's not just edges, you're keeping the "interior" too

[astral ivy]

Wdym by “all edges have to stay edges”? You’re allowed to glue edges together, if that’s what you’re asking

[cobalt shuttle]

that's what we're asking exactly!

It may seem obvious to you what you mean by your question but it may not be to others reading, but that helps

[astral ivy]

Ah

Okay okay

[cobalt shuttle]

https://www.quantamagazine.org/simple-geometry-shows-when-two-shapes-are-equal-20221121/#:~:text=The fact that scissors congruence,them to make the other.

this will be of interest

scissor congruence (which is what you're asking) is very loose (its just area)

and it shows you how to do it too!

[obtuse pebble]

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[topaz echo]

I am pretty confused about a differential equations topic as a whole.
So I am learning about (first order):

1. Exact Equations
2. Separable Equations
   But they seem to literally be the exact same thing, even though they are solved in two different ways

[topaz echo]

For reference, this is what Paul's Notes Says:

[cobalt shuttle]

Not necessarily!

[topaz echo]

wait gimme a sec to put the pictures

[cobalt shuttle]

sure

[topaz echo]

Can't we REWRITE an EXACT EQUATION as a Seperable Differential Equatoin?

[cobalt shuttle]

not quite

a separable one you're separating out what is a function of x and what is a function of y

[topaz echo]

M(x, y) + N(x, y) dy/dx = 0
N(x, y) dy/dx = - M(x, y)

[cobalt shuttle]

an exact your M and N are functions of both!

[topaz echo]

OHHHH

[cobalt shuttle]

yes but you need N(y) dy/dx = M(x)

[topaz echo]

yeah I didn't see that hahah

[cobalt shuttle]

for separable

[topaz echo]

M(x) VS M(x, y)

[cobalt shuttle]

yup, big difference!

[topaz echo]

thank you so much Ossih!

[cobalt shuttle]

no worries!

[astral ivy]

thanks! this is great

"Two polygons are Scissors Congruent if and only if they have the same area."

There is a square of any given area, so the answer to the original question is Yes

[cobalt shuttle]

yup!

[astral ivy]

https://en.wikipedia.org/wiki/Wallace–Bolyai–Gerwien_theorem

ah wow

The theorem can be understood in a few steps. Firstly, every polygon can be cut into triangles. ... Secondly, each of these triangles can be transformed into a right triangle and subsequently into a rectangle with one side of length 1.

damn this is very smart

so cool

[obtuse pebble]

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[timid silo]

Let S be the solid obtained by rotating about the x-axis the region enclosed by the curves y = ln(x) and y = 0, and between x = 1 and x = e.
(1) Sketch the solid S. Label all axes.
(2) Find the volume of S

[timid silo]

how would I draw the solid exactly

This is wht I did

[brave bramble]

That green thing the solid? Looks pretty close

It's going to be like a cone

Consider drawing it, such that it's aligned with your graph. So, on its side

[timid silo]

sorry but how would i do that

[brave bramble]

Draw a cone?

[timid silo]

Oh no that green ting is me crossing out smth lmao

[brave bramble]

But on its side. See how the point is created by (1,0)?

However, yeah that's a sketch of the solid

[timid silo]

how could i show as an arrow that it is rotating about the x-acis

like on my graph

[obtuse pebble]

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[timid silo]

how do i combine this log expression?

[sand oriole]

I'd first distribute the 4

[timid silo]

then it'd be 4logw-16logy+3logx

whats next?

[timid silo]

i appreciate the help!

[sand oriole]

do you know the power rule of logs?

[timid silo]

so logw^4-logy^16+3logx?

[sand oriole]

what about that last log? 👀

[timid silo]

oh

logx^3

[sand oriole]

then are you familiar with your addition and subtraction rules of logs?

if you are, then the easy if trivial!

[timid silo]

thank you man!

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[ripe violet]

Hello I need serious help and a way to do these problems fast

[ripe violet]

My head is going to explote bro

[tardy epoch]

Why fast

[obtuse pebble]

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[vague egret]

I used to do sine rule with two sides but now it has two angles. I can't do further than this. Does anyone have solution?

[static furnace]

cross multiply

[vague egret]

Like this? Doesn't seem right

[static furnace]

yep!

exactly

[vague egret]

Idk about it but the result is negative.

It's supposedly bigger than the other side

[sinful thicket]

I think its supposed to be 11.7

sin rule is a bit weird of how its formated i find it easier to work it reverse ways

7/ sin 35 with ? / sin 105

The main thing your doing is you just solved for the left side

you didnt do the right side at all @vague egret

the awnser at bottom is what your left side equals out to you then convert your sin 105

or .96592etc

then divide that by your awnser at bottom

[vague egret]

So reversing the formula?

How do I do the cross multiplication?

[sinful thicket]

you then get 11.3

useing those digits

[vague egret]

Do I also multiply the sin(105) or should I just multiply it by 105?

[sinful thicket]

Thing of it this way 8 divided by 4 equals 16 divided by 8 correct?

both equals 2

depending on how it is you can also reverse it

4 divided by 8 still equals 8 divided by 16

you are solveing for unkown

if you didnt know what 8 equaled for example

4/8 ?/16

you divide 4 by 8 get .5 and multiply it by 16 to get unkown

8

this is why I reverse the formula because it flows quicker this way

otherwise you have to divide so if you reverse formula so sin is on bottom is flows quicker

for example same thing but reverse

8/4 equals 16/8

make 8 the unkown again

8/4 is 2

16/?

divide 16 by 2 to get 8

does that make sense to you?

going back to your sin 35/7 and sin105/?

you can write it either way depending if you want to multiply or divide either way you get correct awnser

[vague egret]

Okay so I was confused. To do the cross multiplication, do I include the sin or should I just multiply it by 105?

[sinful thicket]

just change the sin to decimal format is what I do

[vague egret]

Oh wait that actually make sense lmao

[sinful thicket]

XD okay try it tell me if it makes sense yet

[vague egret]

So it should look like this. Is it correct?

[sinful thicket]

I think you might be overcomplicateing things

step 1 convert sin into decimal

both of em

sin 35 is .573576436

sin 105 is .965925826

you know the value for sin35 is 7

put that above or below sin depending if you want to multiply or divide

[vague egret]

Okay so now I just need to multiply .95692 by 7 and divide it by .5735?

[sinful thicket]

I think so

7 divided by sin35 should get you 12.204

multiply that by your decimal value of sin105 to get correct awnser

[vague egret]

So have I done it correctly?

[sinful thicket]

thats correct

most accurate awnser would be 11.788

althought they would prob ask you to round up to 12

the easiest way is to just divide 7 by sin35 then multiply it by the decimal awnser of sin105 though

I do it this way because it flows easier otherwise with sin on top you have to divide again

@vague egret Do you understand now?

[vague egret]

Yes

But I have another question though. What if in a triangle, the placement of the numbers occupy all the formula?

Wait let me write it

[sinful thicket]

ah

I think i understand what you are confused on

remeber the right triangle rule? at least i assume this is one

what triange is this

[vague egret]

Yes since sine rules are normally done by only two fractions

Non right triangle

[sinful thicket]

k

Use law of cosine i believe

yeah that should be right

[vague egret]

I haven't learn about that since I'm still perfecting my sine law understanding so I might visit again when I'm done with sine law

Thanks tho

[sinful thicket]

No prob its kinda simple though

[vague egret]

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[timid silo]

A triangle has sides of 4 inches and 7 inches. What is a possible length for the third sides?
a. 2
b. 12
c. 10
d. 14

i need help on learning how to solve these kinds of questions.

[tardy epoch]

Use triangle inequality

[timid silo]

so c?

since 7+4 =11 and 11 is greater than 10 ?

ok thanks

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[warm shale]

Ransik

Ransik

Ransik

[obtuse pebble]

@timid silo Has your question been resolved?

[warm shale]

Ransik

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[ripe violet]

Hello my bad for not answering yesterday

[ripe violet]

I am here with a big problem

I need an explanation

And a solution

Iron has a density of 7.86 g/cm". Calculate the volume (in d) of a piece of iron having a mass of 4.91 kg . Note that the density is provided in different units of volume and mass
than the desired units of volume (d.L) and the given units of mass (kg). You will need to express the density in kg/dL (1 cm' = 1 mL) before calculating the volume for the piece
of iron.
Express your answer to three significant figures.

[errant gust]

do you know how to convert the units

[ripe violet]

No

All of these in my mind is confusing

Yesterday I did some unit conversion problems like 1/1000 when a number divided by 1000 got me the result stuff like that

[errant gust]

ok so we need to get 7.86 g/cm into kg/dL

so first step is to do the g to kg

[ripe violet]

We get 0.00786?

[errant gust]

yup

[ripe violet]

1kg/1000g

I want to know how to get the correct conversion factor cuz u can also do 1000g/1kg

[errant gust]

well it depends which way you're going

[ripe violet]

So if I want to go kg to g is the other way around?

[errant gust]

g --> kg is 1kg/1000g
kg --> g is 1000g/1kg

[ripe violet]

I see thank you

Alr let’s keep doing this

The next part I don’t know too much

[errant gust]

so now we need to get from cm^3 to dL

we know that 1 cm^3 = 1 mL

[ripe violet]

So 0.00786 x 1/1

[errant gust]

yes now we have 0.00786 kg/mL

[ripe violet]

It’s the same number but now the unit is different?

[errant gust]

yes cuz 1 cm^3 is exactly 1 mL

[ripe violet]

Ahhh alr

[errant gust]

now for mL --> dL it's 1dL/100mL

[ripe violet]

So 0.00786 x 1/100

[errant gust]

yup

[ripe violet]

Now we get 0.0000786

[errant gust]

perfect

[ripe violet]

So that’s our answer

[errant gust]

now we need to calculate volume

[ripe violet]

Dam

So much stuff

The calculations aren’t even difficult is the units are confusing

[errant gust]

it said the mass is 4.91kg

[errant gust]

yeah it be like that sometimes

[ripe violet]

Crazy man

Volume over mass

[errant gust]

almost

mass over volume

[ripe violet]

Ah I see because u trying to find volume?

[errant gust]

yes

[ripe violet]

I see

[errant gust]

the equation is D=m/V

[ripe violet]

We get this 0.000385926

[errant gust]

so rearranging it you get V=m/D

[ripe violet]

Ah

I see

[errant gust]

so plug in the values you know and see what you get for volume

[ripe violet]

Alr

This is the volume I got 0.624681933842239

4.91/7.86

[errant gust]

use the value we converted earlier

[ripe violet]

Alr I believe if I use that one I get We get this 0.000385926

[errant gust]

hmmm that's not what i got

[ripe violet]

I will do again

[errant gust]

4.91/ 0.0000786

[ripe violet]

Ye I will calculate that one again

62,468.19338422392

I got that number

[errant gust]

same

[ripe violet]

Alright so that’s the volume

[errant gust]

yes

and it's in dL

[ripe violet]

Ah and also dl

Btw do u also know about ranking stuff from low density to high density?

[errant gust]

yeah

[ripe violet]

I will ask rn btw how do u write this like shorter

62.5?

I got one more attempt and can’t fail it lol

[errant gust]

it says 3 sig figs

so you have to use scientific notation

[ripe violet]

A true

62.5 x 10^13

Its prolly wrong

[errant gust]

62.5 x 10^3

where did you get 13?

you only go to the dot

[ripe violet]

I counted 13 decimal places for some reason

I need to also rank from lowest to highest density butane chloroform ice gold naci and copper

How would I do this

Nvm bro already did it

Alr panda thank u man🙏🏻

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[turbid bane]

this integral isn’t matching up with the calculator’s answer, anything wrong here?

[fathom flicker]

, rotate

[fathom flicker]

what's the derivative of cos(x)

[turbid bane]

-sinx

[fathom flicker]

okay so if u=cos(x) du=...

[turbid bane]

so I should reverse the signs?

[fathom flicker]

?

[turbid bane]

-sinx dx

[fathom flicker]

yea that is your mistake

[turbid bane]

this look good?

[fathom flicker]

you said yourself you had an integral calculator

does it?

[turbid bane]

the calculator did it differently, it turned the sin/cos into a tanx

that’s why I’m asking here

[fathom flicker]

If you redo your subsitution to have u=cos(x) du=-sin(x)

then continue from there

I can recheck your work if you show it

[fathom flicker]

okay but

why did you hcange the isng

change the sign*

but do the rest of the integral the same?

you want to replace sin(x)dx

that is equal to -du

not just du

[turbid bane]

I multiplied the outside and the inside by -1

[turbid bane]

what does that change?

[fathom flicker]

the sign?

[turbid bane]

that’s why I multiplied by -1 right

[fathom flicker]

I don't see you doing that in the first step

[turbid bane]

I put the -1 at the start of the second line

after I substituted u in

it kind of blends in with the integral sign

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[timid silo]

How does $-\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}$ simplify into $-\frac{\sqrt{6}+\sqrt{2}}{4}$

[warm shale]

Matt

[timid silo]

there we go

how does the plus get there

[static furnace]

nice

[timid silo]

where tf does the subtraction sign go

[static furnace]

well -sqrt6-sqrt2 is the same as -(sqrt6+sqrt2)

they just factored out the negative to put it in front of the whole fraction

[timid silo]

the other negative was already there

[static furnace]

$\frac{-\sqrt{6}}{4}+\frac{-\sqrt{2}}{4}$

[warm shale]

Wumbo

[static furnace]

thats the same thing as $-\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}$

[warm shale]

Wumbo

[static furnace]

right?

[timid silo]

yes

that makes sense

[static furnace]

which turns into this $\frac{-\sqrt{6}-\sqrt{2}}{4}$

[warm shale]

Wumbo

[timid silo]

yes

[static furnace]

and then you can factor out a -1 from the numerator

[timid silo]

oh and you can just factor out the -

[static furnace]

yayaya

[timid silo]

yes that makes sense now

thank you

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[static furnace]

no problem

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[vague ibex]

can someone help me with this problem

idk why is it wrong

[true pond]

sorry

1/7^x

[vague ibex]

so do I take the intergeral of y^1/7x

like 7^1/7x / ln7

[true pond]

instead of (+)

it should be (-)

otherwise its fine

[vague ibex]

why is it minus because I thought you seperate the intergeral with the plus sign

[true pond]

because it is 7^(-x)

[vague ibex]

so if its a negative exponent for the x as a power do you change it to an minus

[true pond]

if there is negative sign yes

if here it would have -x as exponent then ans is -e^(-x) +c

[vague ibex]

ok

thx

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[dawn bane]

Will send work in a second

[dawn bane]

Worked through the whole problem and then I got it wrong

No clue what I did wrong

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[hollow ridge]

im not sure how to do 7

[obtuse pebble]

@hollow ridge Has your question been resolved?

[hollow ridge]

<@&286206848099549185>

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[grave quartz]

Would someone mind explaining this proof to me?

It seems that they've let x = a + h, and y = b + k, otherwise how do they substitute them?

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@grave quartz Has your question been resolved?

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@grave quartz Has your question been resolved?

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@grave quartz Has your question been resolved?

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[subtle skiff]

(repost from the discrete math channel, sorry if this counts as spam)
ive encountered a problem while solving this question, i got that the answer is
b = -19,
c = 90
d = 112,

but when my friend tried doing it using almost the same exact method but differing ever so slightly, he couldnt get the answer, even though logically we did the same thing

[subtle skiff]

in my work, i decided to multiply the 2's from the nC2 terms into the 56 and reached the correct answer

but here my friend decided to fully evaluate the nC2 instead of multiplying the 2's from them it into the 56, and found that the value for b changed every time he tried to use different numbers of i for his elimination

is there any reason why this happens?

[obtuse pebble]

@subtle skiff Has your question been resolved?

[subtle skiff]

<@&286206848099549185> sorry for the ping, but my exam is tomorrow 😰 and i would love to know why this happens so i can avoid doing it

[lost otter]

honestly bro

ive no idea

its over for you

JUST KIDDING JUST KIDDING

a helper will help you soon

[subtle skiff]

😭 😭

[candid yarrow]

You might try to find the error when you substitute the values of b and c you got back into the equations you get

[subtle skiff]

how would i do that if the values for b and c arent constant?

[candid yarrow]

Aren't they constants you have to find, but that you keep getting different answers for?

[subtle skiff]

yup

[candid yarrow]

So substitute them back into the initial equations you get to solve for b and c, then see which ones are right and which ones are wrong

this will help check if there's a problem with the final algebra

and if there isn't a problem with the algebra, the problem may be the method with which you got the equations to solve for b and c

[subtle skiff]

i have actually tried plugging in all the various numbers for b and c using code and none of them produced the correct answer

since the correct answer would be using b = -19 but by doing it how my friend did it, the closest value for b we could get is -18

[candid yarrow]

how about retracing the steps of your argument?

[subtle skiff]

just not sure why we get completely different answers even though seemingly we use the same method

[candid yarrow]

maybe you double counted something

[candid yarrow]

if you use the same method, then where do your steps differ?

[subtle skiff]

i dont fully evaluate nC2 while he does

i leave mine as a fraction

the steps after that are the complete same though

[candid yarrow]

are your values for nC2 the same?

[subtle skiff]

yup

[candid yarrow]

so the steps that are completely the same can't be the problem, if one of them produced the right answer

if they really are completely the same

How about comparing the values you got for all the constants at each of the steps and finding the error that way

[subtle skiff]

at least i would like to believe theyre the same since its just simple substitution and elimination to find b and c

[candid yarrow]

isn't E[X] a constant?

[candid yarrow]

and then you're required to express it as a sum?

[subtle skiff]

yes

ok ive checked that leaving the nC2 as a fraction and fully evaluating the nC2 both give the same answer so it shouldnt be a problem there

its getting way too late for me i think i might just have to give up/hope i get it right for this one question during the exam 💀

thanks though @candid yarrow, appreciate the help

[obtuse pebble]

@subtle skiff Has your question been resolved?

[obtuse pebble]

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[sly flicker]

test

[sly flicker]

why can’t I just equate 448k^5 to 2240k^4 instead and find out that k = 5

divide both sides by k^4

why they got another solution (ik it still got canceled after but why’d they go out of their way to get that solution first?)

[timid silo]

that's what they did

[timid silo]

you won't be finding all solutions doing that

dividing by k^4 is working under the assumption that k is never 0

but, problem is, k = 0 happens to be a solution

take for example $x^2 = x$

[timid silo]

if you had divided both sides by x, you will restrict the possible values of x to exclude x = 0

and you get x = 1

which is a solution, but x = 0 also is a solution you would have neglected here

[obtuse pebble]

@sly flicker Has your question been resolved?

[obtuse pebble]

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[dim quail]

Hello, I'm trying to find the value of pi with only using a square and a circle and apparently, things don't seem great with my work, can anyone guide me how to do that?

[dim quail]

So, I was trying to find the value of pi and ended up finding the (half of AC - the radius of circle)

[ancient nacelle]

ok

[dim quail]

Do you have any idea what to do. If yes then please help

[ancient nacelle]

i am analysing the figure

[dim quail]

Ok

[ancient nacelle]

what do you mean by len(ride)?

[dim quail]

The length of the side of the square

[ancient nacelle]

i got confused as len () is also used as log sometimes lol

brooooooooo

💀

[dim quail]

Sorry for that

[ancient nacelle]

you just cant do that

[dim quail]

Uhhhm what?

[dim quail]

How?

[ancient nacelle]

like

what was your thinking

like what did you plan on how to find pie

[dim quail]

Well I read about it in an article and wanted to prove it myself

[ancient nacelle]

uhm can you send the article?

[dim quail]

Just a random start, was gonna make some things for my game and ended up finding the value of pi

[dim quail]

Lemme search for it

[ancient nacelle]

what is "y" though?

[dim quail]

The length of ( half of length of AC - r)

Couldn't find it @ancient nacelle

[ancient nacelle]

ok

[dim quail]

If I can't do that then I have another doubt up my sleeve

[ancient nacelle]

whats that

[dim quail]

How to find the tangent of the curve of equation y = x³ + 4x² - 3x + 6 at co-ordinates (-3,24)

@ancient nacelle

[ancient nacelle]

sorry brother

[dim quail]

Its ok, thanks for taking out some time to help a 13y old. @ancient nacelle

.close

[obtuse pebble]

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[dim quail]

.reopen

[obtuse pebble]

✅

[dim quail]

.close

[obtuse pebble]

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[obtuse pebble]

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[coarse summit]

Pythagoras.. im confused

[dim quail]

Pythagoras theorem? @coarse summit

[coarse summit]

gotta figure out x

what I did

the books correct answer

[dim quail]

Lemme check it, give me a second

[coarse summit]

okay

bro is as cooked as me 💀

[dim quail]

Bro, I was so confused when I saw that comma. Now lemme try

[coarse summit]

💀

[dim quail]

Bro I'm getting the answer as an imaginary number @coarse summit

[coarse summit]

dawg what

[dim quail]

I found their dot product and did the math and got the answer as an imaginary number

[coarse summit]

what is an imaginary number

[dim quail]

xi where i is the square root of -1 and x is any real number

[coarse summit]

ah

the hell then bro

[carmine bear]

commas mean a decimal point in european things so 11.2^2 is not 22.4

also don't round the numbers until the very end

[dim quail]

Wait what l

[coarse summit]

how isnt 11,2^2 same as 22,4

[coarse summit]

so 3,64 is same as 3.64

[dim quail]

I thought it was matrix @carmine bear

[carmine bear]

11.2^2 means 11.2 x 11.2 = 125.44

[coarse summit]

ohhhh shit yeah

[dim quail]

Bruh

[coarse summit]

Ahhhh I messed up bro

Now i understand

give me a moment let me try again

[dim quail]

Sorry @coarse summit I couldn't help you

It fin thought it was matrix

[carmine bear]

if it was itd be put into brackets or whatever

to make it clear

[dim quail]

I see thanks for the help

[carmine bear]

that its a vector

or coordinate

[coarse summit]

Mb I didnt understand what you didnt understand

I messed up on such an easy thing

my fault

but lemme try now again

[dim quail]

It happens.

[coarse summit]

books answer is 10,7

so probably just a

round up error

[carmine bear]

don't round the 3.45 to 3.5

[coarse summit]

good enough for me ill submit it 💀

[carmine bear]

until the very end

otherwise that happens

[coarse summit]

hmm aight lemme fix

[carmine bear]

if you use 3.45 you'll get 10.65 which rounds to 10.7

[coarse summit]

what

why does it look like that

is

A = 165

B = 235

C= x

[carmine bear]

yes

[coarse summit]

thank you

[carmine bear]

again don't round any numbers cuz itll affect ur final answer

[coarse summit]

i got it

How are people good at math

Its legit hard

[carmine bear]

practice

[coarse summit]

like rn u could tell whats A, whats b and whats c in just a moment

[carmine bear]

everyone who is good at maths had to do this at some point so yh

[carmine bear]

c is always the longest side thats all you need to remember

[coarse summit]

bro interacting others abt math is making me like it

[carmine bear]

the other two can be swapped around if you wanted

if you had A = 235, B = 165 and c = x youd still get the same answer with pythagoras

[coarse summit]

aah

a question here

is it

x^2 = (-27225) - 55225

[carmine bear]

youve used 165 instead of 162

but once youve squared both you add them together

[coarse summit]

na im just thinkin

on how do i put the x on left

[carmine bear]

you don't

[coarse summit]

i think i did sm wrong

[carmine bear]

you square root it

[coarse summit]

whaat

okay

[carmine bear]

you add the two numbers together then square root it

to find x

[coarse summit]

ok now

[carmine bear]

i mean the reason why u didnt get 285 is cuz u used 165 instead of 162 which is whats in the question

but the method is correct

[coarse summit]

wha

ohh shi

hhhhh 💀

its fine

what the hell man

https://tenor.com/view/grandson-of-ottoman-dirilişertuğrul-diriliş-ertuğrul-bey-ertuğruş-gif-25296850

Calculate the length of the altitude of an isosceles triangle, h

[carmine bear]

its a bit of a trick you only need half the triangle

[coarse summit]

imma go eat eat bro i aint dealin with this

ill do it later

[carmine bear]

the dashed line splits the bottom line in two so you have a = 4 b = h and c = 14.6

[obtuse pebble]

@coarse summit Has your question been resolved?

[coarse summit]

the length of the base of an isosceles triangle is 42 cm. What is the length of a side of a triangle when its area is 588 cm2?

wharra do

[carmine bear]

use the area to find the height of the triangle then use pythagoras to find the side length

[obtuse pebble]

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[limber quartz]

!status

[obtuse pebble]

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

[winged estuary]

1

[limber quartz]

Otto gave you a formula

that's a Kinematic Equation

(the first, iirc)

that one is the formula for displacement with respect to time. It requires a starting displacement x_0, a starting velocity v_0, and constant acceleration a

the problem gave you the starting speed and the constant acceleration.

It also implicitly tells you that the beginning displacement is 0, since this person is at the starting line

@winged estuary can you start now?

[nimble copper]

melvin help pls u seem like u could help with my vector problem in help-32

[winged estuary]

Otto?

[nimble copper]

https://tenor.com/view/hello-hi-friend-penguin-waving-gif-17013804

can he not see my messages lmao

[obtuse pebble]

@winged estuary Has your question been resolved?

[obtuse pebble]

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[solar leaf]

bro what is something/something

[rough dawn]

Just rewrite the function

[timid silo]

Yeah Lol

[rough dawn]

Get the common denominator

[timid silo]

But its coverg bro

[timid silo]

@solar leaf Use this

[solar leaf]

idk the steps for that

do i just

multiply

like the terms together

or smth

so they get the same denominator

idk

[rough dawn]

Think as if you were adding up fractions

[solar leaf]

okay i multiplied 3x + 1 by x/x and i have 3x^2 /x + x/x + 5/x

and im gonna add them up

is that the answer

(3x^2 + x + 5)/x

idk

can it be simplified

[obtuse pebble]

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[timid silo]

Can anyone help me out for this

I got 2x/sqrt(9-4x^2)

[plain stag]

Seems fine

Now divide numerator and denominator by 2

[timid silo]

How would I divide the denominator by 2 amd simplify it

[spiral knot]

$a\sqrt{x}=\sqrt{a^{2}x}$

[warm shale]

Chixen

[timid silo]

Yea I got it thanks

[obtuse pebble]

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[obtuse pebble]

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[worthy cargo]

wait so if we have v(t) as a velocity function for a particle moving in the x axis is the arc length just the distance travelled

[brazen gorge]

no

[worthy cargo]

so why works for parametric functions

[brazen gorge]

the distance travelled is $\int |v(t)|dt$

[warm shale]

artemetra

[worthy cargo]

oh yes

so the fact that distance for parametric equations is arc length is like irrelevant for the 1d stuff

?

[brazen gorge]

uh

show example

not sure what you are talking about

[worthy cargo]

random website

https://socratic.org/questions/how-do-you-find-the-distance-travelled-from-t-0-to-t-3-by-a-particle-whose-motio

[spiral knot]

Well, arclength in 1d is the same as distance.

[worthy cargo]

so it is?

[brazen gorge]

oh yeah then it's arclength

[spiral knot]

(about. It’s a bit more complicated if the velocity changes sign, but yeah.)

[brazen gorge]

your parameters are coordinates, not velocity

so yes it's arclength

[alpine verge]

only for parametric functions i believe

[brazen gorge]

yes

[worthy cargo]

oh wait i see

for the parametric stuff the function gives the coords

not velocity components

[alpine verge]

since in a normal 1d function we have x axis as time

[worthy cargo]

make sense

[alpine verge]

whereas the x and y axes there are just the corresponding x and y directions

[worthy cargo]

i gotcha

ok thanks

.close

[obtuse pebble]

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[alpine bison]

me too

[timid silo]

what r u "solving"

[alpine bison]

ok now i m not confused anymore

yes then when you got $c^n>b$ just use $\log_{c}$ to both sides

[warm shale]

everg

[alpine bison]

if you know the word then you know what it is

[obtuse pebble]

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✅

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[quasi apex]

when square rooting anything is it always + or -

[odd stag]

That really depends on the context of the question

[quasi apex]

ok so the equation is x^2 + y^2 = 4 and i need to sub in 0 for x

[spark bramble]

if you take the square root of a positive number it always returns a positive answer, you cant take the square root of a negative number

[quasi apex]

would it = +-2?

[spark bramble]

yes

when you take the square root to get rid of the power you have to add a +-

so y^2=4

[quasi apex]

ah ok

[spark bramble]

y=+-sqrt(4)

y=+-2

[quasi apex]

thx

[spark bramble]

but if the question is y = sqrt(4)

then y = -2 is not a valid answer

[quasi apex]

.close

[obtuse pebble]

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[quasi apex]

.reopen

[obtuse pebble]

✅

[quasi apex]

got another question, why isnt "the relation between distance and time if john walks 5km/h" a function?

wouldnt it just be a stright line going up and to the right

[obtuse pebble]

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[obtuse pebble]

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[tame sorrel]

Given the vector function, determine its expression in cylindrical and spherical coordinates, that is, determine f(ρ,θ,z) and f(ρ,θ,φ) with their units expressed correctly.

pls can anyone help me with this?

well my teacher just gave us the formulas of the spherical coordinates and told us to try this

yeap

yes

i have no clue

no problem thanks 🙂

it is in spanish, but this is what i´ve got for the cylindrical coordinates

and this is for the spherical

ok 🙂

another question

the x, y and z is just by doing the bottom formulas in both coordinates right?

here the (x +y) for example

the way to transform it from cartesian to spherical is to using the bottom formulas from the picture above?

Ok i´ll see

thank u a lot

mmm i see

what i have to do is transforming the scalars to the other coordinates using the formulas and the unit vectors using the other?

do i have to do anything with rho?

😮

mhm

mhm mhm

i see

nope

I'm supposed to know them this course in mathematics, but we're not there yet.

sorry if i have bad english btw

and thank u for ur help it really helps me with the documents

yeah

yeap

yeah i see it

mhm

is it

something squared + something squared = r squared?

oh this: (x-a)²+(y-b)²=r² ??

i have searched it in google

mhm

aww right

mb

one coordinate for the length and one for the radius?

i mean

the angle*

mhm

yeah tell me :))

yeah that's the rudimentary way

its easier to describe it as it length and just an angle to show full motion of the circle

mhmhm

yeap

i see it

its nice

ok so i gotta go to do some chores , but i´ll be back for sure, thank u I erally appreciate the effort :))

do u mind if i talk to u private?

just curiosity

sure :))

yeap

mmmmmmmmmm yeap

its like if u were calculating a volume

the "height"

the radius

and the angle where it goes

i see it

they add another angle

mhm

cool :))

thank u so much

yeap

u are great teaching

thank u for sure

[obtuse pebble]

@tame sorrel Has your question been resolved?

[obtuse pebble]

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[jolly sparrow]

I don't know where to even start to get that solution.

[jolly sparrow]

1/x

1/f(x) *f'(x)

OHHHHH

i just use the denominator as u?

yeah

thats so smart

thank you bro, that's all I needed

.close

[obtuse pebble]

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[obsidian scarab]

can someone explain the answer

[obsidian scarab]

specifically the 6!

[obtuse pebble]

@obsidian scarab Has your question been resolved?

[obsidian scarab]

<@&286206848099549185>

[obtuse pebble]

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[upper void]

Does anyone here specialize in helping with projects? I have a trigonometry project that is due on the 15th and I really need help because I have no idea what to do. No matter how much I pay attention in trig I continue to have no clue what we are doing.

[stiff thicket]

Is this answer correct? This is my third attempt on this test 🥲

<@&286206848099549185>

[obtuse pebble]

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

[unreal musk]

And the channel is already occupied by someone else, so get a new one

[stiff thicket]

oh sorry

[upper void]

Do you mind deleting your comments so that my message can be seen?

<@&286206848099549185>

[obtuse pebble]

@upper void Has your question been resolved?

[obtuse pebble]

@upper void Has your question been resolved?

[obtuse pebble]

@upper void Has your question been resolved?

[obtuse pebble]

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[forest lake]

Calc 2 improper integrals

[tardy epoch]

is sqrt(2) > 1 ?

[forest lake]

Number 6

[forest lake]

Yes

1.41 …….

[tardy epoch]

Great now use improper integral to find the integral

[forest lake]

Idk where to start I tried using u sub but idk if u sub will work here

[tardy epoch]

Use integral of x^n dx = x^(n+1) / (n+1)

[forest lake]

Omg

I’m so slow

🤦‍♂️

.close

[obtuse pebble]

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[obtuse pebble]

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[obtuse pebble]

@chilly swan Has your question been resolved?

[obtuse pebble]

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[sharp anchor]

Is there another way I should have written this? Is it even correct? I think it is, but I feel like I'm missing something.

[polar fossil]

confused about what you're trying to prove

where did those numbers come from? the 17, 12, 5, -7

[sharp anchor]

It is just an example that works.

What I'm trying to prove is the first line in the picture.

gcd(m,n) = 1, if and only if there exists some integers x,y, such that mx +ny = 1.

So, I used Bezout's Identity and and an example I knew that worked.

pls help

[obtuse pebble]

@sharp anchor Has your question been resolved?

[sharp anchor]

<@&286206848099549185>

Can anyone help?

[obtuse pebble]

@sharp anchor Has your question been resolved?

[sharp anchor]

hello?

[obtuse pebble]

@sharp anchor Has your question been resolved?

[raw crypt]

can anyone help with this ? https://hips.hearstapps.com/hmg-prod/images/math2-1569526052.jpg?crop=1xw:1xh;center,top&resize=980:*

[obtuse pebble]

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[cursive mirage]

I saw multipule answers to this what is the correct one?

[cursive mirage]

.close

[obtuse pebble]

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[wicked ice]

how do i do this?

[obtuse pebble]

@wicked ice Has your question been resolved?

[obtuse pebble]

@wicked ice Has your question been resolved?

[static marten]

you need to manipulate the expression to fit the definition of a rational number
e.g. rationalising so denoms are integers

||after combining both fractions into one fraction, notice that the numerator contains a surd that must =0 for the expression to be rational
therefore, set it =0 to get final result||

[obtuse pebble]

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[obtuse pebble]

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[uneven wren]

How to solve

[uneven wren]

I understand how to do chain rule for the left side

But

I don’t get how

To

Do

Product rule for the left side

Actually

I understand that

What I don’t get is

[dim quail]

Uhhm please explain again

[uneven wren]

Why is it d / dx

And not

Dx / dy

[dim quail]

Wait a bit

[uneven wren]

Why dx / dx

[dim quail]

Uhhm Imma just solve the sum, this is my first time helping so I don't know from where to start

[uneven wren]

Ok

But i just need to understand why it’s dx / dx

[flat geyser]

Product rule

[uneven wren]

Idk

I know product rule

But why is it dx / dx and not dy / dx

[flat geyser]

Okay

Do you know the derivative of xsin(x)?

[uneven wren]

cos

[flat geyser]

x*sin(x)

I meant that

[uneven wren]

non

No

[flat geyser]

Product rule

Apply it here

[uneven wren]

Ok

I got

x cos (x) + sin (x)

[flat geyser]

Yeah

You got that by doing
x (d/dx sin(x)) + (d/dx x)sin(x)

see that d/dx x?

That’s where it came from

[obtuse pebble]

@uneven wren Has your question been resolved?

[uneven wren]

I don’t get it

like the theory behind it

[obtuse pebble]

@uneven wren Has your question been resolved?

[uneven wren]

<@&286206848099549185>

[outer hedge]

I think

The reason for using dx/dx instead of dx/dy is bcz the derivative is with respect to x

[last pilot]

dx/dy?

[outer hedge]

dx/dx

[last pilot]

you just said "dx/dy".

that is not a thing

@uneven wren d/dx f(x) refers to d(f(x))/dx, aka infinitesimal change in f(x) divided by infinitesimal change in x

$\frac{d}{dx}f\left(x\right)$ refers to $\frac{d\left(f\left(x\right)\right)}{dx}$

[warm shale]

ren

[last pilot]

the same way that we say $\frac{du}{dx}$

[warm shale]

ren

[obtuse pebble]

@uneven wren Has your question been resolved?

[uneven wren]

I meant the other way around

[obtuse pebble]

@uneven wren Has your question been resolved?

[obtuse pebble]

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[dense tinsel]

Help with functions

[dense tinsel]

How do i do 4 A

textbook says b = -2 and c = 4 but i have absolutely 0 clue as to how to get there

[glacial obsidian]

First, write which point is the vertex of f(x)

In terms of b and c

[dense tinsel]

??

not sure

well i know c is the y-intercept

that's about it haha

[glacial obsidian]

But in general, how do you find the vertex of a quadratic function?

[dense tinsel]

-b/2a

[glacial obsidian]

That is the x coordinate

[dense tinsel]

OH so do i set it to

1 = -b/2?

so then b = -2?

[glacial obsidian]

Yes

[glacial obsidian]

And what about the y coordinate?

[dense tinsel]

and from there i just plug in 3 for f(x) and 1 in for x

so

3 = (1)^2 -2(1) + c

or no

oh wait yeah im on the right track

so

1-2 = -1

then add 1 to both sides and c = 4!

:DDD

[glacial obsidian]

Yes, that is correct
I would have chosen -b/2 for x but it would have been more difficult

[glacial obsidian]

Yes

[dense tinsel]

wooah

one sec

im doing 7B

do i do x-int form?

and just plug everything in?

[glacial obsidian]

What is x-int form? f(x)=c(x-a)(x-b)?