help-13

[lone lagoon]

sure

[static willow]

you know binomial expansion right?

[lone lagoon]

I do.

is

(n k) a^n-k b^k

[static willow]

lets put a as cbrt(x) and b as (-2x^(-2))

and n is 27

[lone lagoon]

sure

[static willow]

and you want to separate the numbers and the variables

,rotate

[lone lagoon]

Yes.

Go on

[static willow]

so that it is in the form of ax^n

you understand every step i wrote, right?

[lone lagoon]

I do.

[lone lagoon]

Wait

[static willow]

so this whole chunk is the coefficient

[lone lagoon]

Lemme think about this

so you did 1/3

so you can then use that as

hm

I mean, express that as

1/1 multiplied 1/2

[static willow]

?

[lone lagoon]

right?

so that way

you have x multiplied root square x? and cancel that with the X from the bottom?

[lone lagoon]

.

[static willow]

i dont really understand what you mean

[lone lagoon]

I mean

how did you end up with

getting rid of this

[static willow]

oh, you know (x^a)^b = x^(ab) right?

[lone lagoon]

No, I didn't

[static willow]

well, it is one of the fundamental identities

this is called “law of indices”

[lone lagoon]

where does this 9 come from

perhaps by dividing that between 27/3?

[static willow]

yes

27-r multiplied by 1/3

-r/3 + 9

[lone lagoon]

so, the answer is?

[cedar kiln]

@lone lagoon Has your question been resolved?

[crimson sedge]

0

[static willow]

yup 0

[static willow]

cuz there isnt a non-negative integer that the power of x is 2

[lone lagoon]

Thank you.

[static willow]

so there is NO term with x^2

[cedar kiln]

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[ancient swallow]

Can a proof by mathematical induction start with
𝑛
+
1
n+1 as a limit in the summation notation before the actual proof begins?

[lusty grotto]

i dont quite understand what u mean

[lusty grotto]

can u send the proof u already have

[ancient swallow]

Before I have to do the n + 1 is in the summation Note

[lusty grotto]

so u show that it is true for n = 1

[ancient swallow]

I am so dumb

[lusty grotto]

then u assume it is true for some arbitrary n

then u deduce that it holds for n+1

[ancient swallow]

Thank you

How do I close the help

[earnest geyser]

.close I believe

[ancient swallow]

.close

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[oblique flare]

(3-2a)*(-1)=(-3+2a)=(2a-3)

[soft lark]

So yes?

[queen ice]

Yes

[cedar kiln]

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[cobalt scarab]

Why is it wrong?

[oblique flare]

at x=

not the value of the minimum value

[cobalt scarab]

Ohh!

My bad thank you

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[slate tulip]

hello! I just wanted to clarify if the "smallest positive coterminal angle" of 300 degrees is still 300? or should I still add 360 degrees?

[coarse kiln]

Add 360deg

[slate tulip]

okay thanks

.close

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[jovial glacier]

What does this say? With respect to t = 1; with t = 1; at t = 1; or something else?

[dawn junco]

compute dW/dt, and evaluate it at t=1

that's what you have to do

this message will self-desctruct in [REDACTED]

[cedar kiln]

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[hexed vortex]

So is E making up for the error in L?

[hexed vortex]

and thats why we can say it equals exactly that formula?

[long swan]

yes

[cedar kiln]

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[sand cradle]

Any idea on what the author is trying to prove here?

Lemma 4 is already proven by those three lines

Is it the last Theorem?

[mighty drift]

long time no see

did you take the linalg exam yet?

[sand cradle]

Hi Bezier!!

[sand cradle]

Actually it will be in january now lmao

[mighty drift]

so that's why I haven't heard from you

[sand cradle]

It was shifted because the prof wasn't in Germany at that time

[mighty drift]

I might be in Germany starting in february for an internship

[sand cradle]

woah

internship for what?

[mighty drift]

probably something somewhere in the intersection of reinforcement learning and some formal methods stuff

[sand cradle]

cool

[mighty drift]

unless I wind up a bit farther north in the Netherlands

or go for something entirely different

[sand cradle]

So you're definitely not staying in academia

[mighty drift]

it is a research internship

[sand cradle]

Oh

[mighty drift]

yeah those are a thing in case you didn't know

[sand cradle]

I didn't know about those

[mighty drift]

researchers often take (underg)grad students as research interns

[sand cradle]

I know there are these REUs

Is that that?

[mighty drift]

that's a US thing

it means "research experience in undergrad" iirc

[sand cradle]

yep

So it's the analogue

[mighty drift]

well I'm not in undergrad, and it's not a short summer thing, it's a whole semester

so it is in fact not the analogue

[sand cradle]

Ah, well then it sounds like an even better version

[mighty drift]

ofc

[sand cradle]

In Germany this is not a thing at all

[mighty drift]

most unis don't do research internships, they're often industry internships

[mighty drift]

you can observe this is indeed a proof of the theorem stated below

[sand cradle]

I did take both of the written exams though

[sand cradle]

What's the induction start?

[mighty drift]

dim V = 1, A necessarily has an eigenvalue and therefore it is trivial

[sand cradle]

Thanks!

.close

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[empty spear]

,rotate

[empty spear]

What an I doing wrong

[craggy geode]

how do i find the number for 30 years and how do i calculate interest using the annuity method?

[cedar kiln]

@empty spear Has your question been resolved?

[craggy geode]

.close

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[pearl talon]

Could someone help me with this exercise? soles is the money. I is the revenue C is the cost and U is the profit

[hidden mural]

profit = revenue - cost

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[halcyon glacier]

Not a math question rather a question about my future to do with math

[halcyon glacier]

I'm a 9th grade student in a school that I consider pretty difficult

From what I've heard grades 11 and 12 are really hard due to academic pressure with SAT and APs

I am thinking about starting AP's now with physics AP 1 to ease off pressure later on

My father is an engineer and has experience teaching all types of physics and maths

I heard that precalculus is heavily recommend to be mastered before physics AP 1

So where and what source should I use and how many hours of studying would it take to master precalculus

[dire geode]

help channels are for math questions. maybe try #discussion or #serious-discussion

[steel plank]

Got some answers for GP - send me a note when you move your discussion.

[halcyon glacier]

GP?

[steel plank]

Glorious Phantom!

[halcyon glacier]

Done

@steel plank

.close

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[meager vapor]

is this correct?

[dire geode]

and where did -1.37 come from

[cedar kiln]

@meager vapor Has your question been resolved?

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[midnight linden]

how can you do this?

.close

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[coarse atlas]

can someone check my work?

[dull oxide]

Subtracting $\frac{20(20+1)}{2}$ would remove all numbers1...20 from the sum 1...1000

[wraith dagger]

SWR

[queen ice]

But we want to keep the 20, so we only want to remove all numbers 1…19

[coarse atlas]

ah yes

[cedar kiln]

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[ancient vector]

Unsure what to do from here on.

[livid hound]

sub in x=9

[ancient vector]

How do I know to sub in 9

[livid hound]

because that's what you're being asked for

y'(9)

[crimson sedge]

LOL

[livid hound]

derivative at 9

[crimson sedge]

I WAS WONDERING WHERE U GOT 9 FROM

[ancient vector]

Input 9 for x or y?

[livid hound]

x

[livid hound]

and you are also given the value of y when x=9 from:

y(9) = 81

[ancient vector]

Kind of confused what the information given actually means. Cuz no where does it state x = any value

That's why I did the work up to a point and I'm stuck

[livid hound]

have you done basic type differentiation questions?

e.g. given
$$f(x) = x^3 + 7x$$
find $f'(7)$

[wraith dagger]

ℝαμOmeganato5

[livid hound]

and given
$$g(x) = x^2 + k$$
determine the value of $k$ if $g(3) = 10$

[wraith dagger]

ℝαμOmeganato5

[ancient vector]

154

[livid hound]

and how did you arrive at that?
subbed x with 7 right?

same idea applies here

[ancient vector]

Yea

So y is the function?

[livid hound]

yes

[ancient vector]

9 is the x input then?

[livid hound]

yes

[ancient vector]

Honestly don't know why it was made this confusing.

[livid hound]

its not supposed to be confusing

[ancient vector]

My brain hurts tho

So f(x) is y

y(9) is x???

[livid hound]

in the basic example, you have the derivative in terms of one variable
with implicit you have it in terms of more than one

y(9) is the value of y when x=9

[ancient vector]

I'm not sure how I'm suppose to know what to plug in.

Aside from that I got the derivative part.

[livid hound]

again refer back to the example with f(x)

would it be better if it were something like
$$\sqrt{x} + \sqrt{f(x)} = 12, f(9) = 81$$
find $f'(9)$

[wraith dagger]

ℝαμOmeganato5

[ancient vector]

So in this problem how am I to determine the x and y values?

[livid hound]

its implied from the notation

[ancient vector]

So x I put 9 and y I plug 81?

[livid hound]

yes

[ancient vector]

Ty. I'm gonna go jump up a building now .....

[livid hound]

please refrain from making comments relating to self-harm here

[ancient vector]

thanks

.close

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[vapid grotto]

Can I have help idk how to start

[vapid grotto]

Also is this correct

[tender scroll]

Is some confusing notation here, does it help if I phrase a(n)=a(n-1)-4, as "the next term is the current term minus 4"?

[tender scroll]

And this does look correct, just you dropped a ^3 on the 2nd one but you reported it right

[clear umbra]

i think the person who wrote it doesnt know how to type subscripts

[tender scroll]

Yeah or somehow thought it was easier pretending they're functions idk

[clear umbra]

actually writing it as functions works too

now i think of it

[last cipher]

consider a(1) to be the first term of one of those options, then check a(2)

[cedar kiln]

@vapid grotto Has your question been resolved?

[cedar kiln]

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[vapid grotto]

What do u mean?

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[neat knoll]

Question 3

[neat knoll]

the triple variable systems of equations is driving me insane

Hello

[worldly chasm]

Here's how I approach problems like this:

I don't like fractions if I can help it, so I try to find equations that will lead to no divisions if possible. I look and notice that there are two equations where we do not have coefficients on the x, y, and z. Next, I think about solving for each variable and what will be leftover when I try to substitute. It looks like if I solve for y in the top equation and do a substitution in the second, then I'll have -3z + 2z, so I can solve for z without causing a division

So my plan is to solve for y in the first equation, substitute into the second equation, solve for z, then substitute into the 3rd equation both y and z, and solve for x.

Once I have x, I can solve for y and z.

Does that make sense? Or was your worry a bit lower level and mechanical rather than planning the smoothest route through the problem?

[neat knoll]

ik the concepts but i dont know for which variables i need to plug also. I also only know how to solve triple variable s.o.e's when one of the equations is x + y + z = any real number

[worldly chasm]

Generally, you solve for one variable in terms of the other two.

Substitute into some other equation so now you have two variables, solve for one of them

Substitute into the third using both substitutions you found, after doing this you get an equation in a single variable, finally, and you can solve that for a value

Then you use that value to solve the other two variables

[neat knoll]

lets say i want to solve for y. how would i alter the first equation to make it a value to plug in for the other 2 equationd to find the other solutions?

[worldly chasm]

You isolate the y on one side of the equals sign

[neat knoll]

turn it into y = ?

[worldly chasm]

Yes

By moving the 2x and z to the other side

[neat knoll]

thats what i did originally

now what should i do

[worldly chasm]

In the second equation, everywhere it says y, replace it with your right hand side of the equation y = ?

[neat knoll]

k

wait hold on

[worldly chasm]

This is the substitution step

[neat knoll]

wait

so i take the first equation. pick the variable i want to solve for, isolate it. plug that into the second equation

[worldly chasm]

Yes

[neat knoll]

and then when i plug it into the second equation

would i take the answer of the 2nd and move it to the third?

[worldly chasm]

You'll have two variables instead of three

[neat knoll]

or not quite

[worldly chasm]

So you'll make another equation, z=?

This time in the question mark is only x and constants

[neat knoll]

wait so lets pretend i have a whole new example

[worldly chasm]

Ok

[neat knoll]

2x + y + 3z = 15

4x - 2y + z = 32

5x - 6y + 4z = 50

hypothetically the numbers dont make sense

[worldly chasm]

So first I would solve for a nice variable, y looks most nice

[neat knoll]

but i would isolate the first equation to y = -2x - 3z + 15

then plug that into the 2nd equation and get

[worldly chasm]

Yup

[neat knoll]

4x + -2(-2x + -3z + 15) + z = 32

[worldly chasm]

Exactly

[neat knoll]

4x + 4x + 6z + -30 + z = 32

8x + 7z - 30 = 32

8x + 7z = 62

thats where i get stuck

what do i do with this

take the 3rd equation and do z = ?

or x =

[worldly chasm]

So solve for either x or z

Probably better with x

[neat knoll]

so isolate x in the 3rd equation to get x = ?

[worldly chasm]

No second equation

[neat knoll]

oh ok

[worldly chasm]

This one

[neat knoll]

8x + 7z = 62

8x = -7z + 62

x = -7/8z + 62/8

so then what do i do with this?

[worldly chasm]

Now use the third equation

[neat knoll]

plug this into the third equation?

[worldly chasm]

Plug y first

[neat knoll]

plug y into the equation first

simplify

then plug x?

[worldly chasm]

Yes!

[neat knoll]

oh ok!

i think i get it now

ima try the original problem and check in a little bit

ty

(dont close tho)

[worldly chasm]

Cool! Good luck!

I won't close, but I might not still be around

Late here

[neat knoll]

yeah its late the late night cram is real

can u stay like 10 more minutes?

[worldly chasm]

No promises

[neat knoll]

i have to take notes on my history textbook too 😦

[neat knoll]

ok thats fine

[worldly chasm]

But I'm not leaving right right now

[neat knoll]

i can probably finish in 5 minutes

OMG TY. I got an answer

i got x = 1

so now i can just plug it into the equations to find the other ones now

[worldly chasm]

Yup

[neat knoll]

tysm

you can close now

have a good night!

[worldly chasm]

Try instead of starting from the beginning

Try starting from z = ?

[neat knoll]

i mean does the order matter though?

[worldly chasm]

Because that depends only on x

So it's less work for you

[neat knoll]

i feel like u can start from any equation

[worldly chasm]

You can, less work though

[neat knoll]

i mean i would rather just do it in order to keep my work organized

[worldly chasm]

Sure as you wish 🙂

Just trying to help you keep your work to a minimum

[neat knoll]

ok im done now but i have one tiny little problem but i wont bother u anymore

i learned substitution

now i needa do elimination with this 😭

[worldly chasm]

What's the problem?

[neat knoll]

but i can teach myself

you helped me enough

ty

🙏

[worldly chasm]

Yw

If you're done with the channel you can type in .close to close it

[neat knoll]

ok

[worldly chasm]

I can also close it, but I'll let you do the honors.

[neat knoll]

g'bye

.close

[cedar kiln]

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[worldly chasm]

Gnight

[cedar kiln]

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[weary granite]

how to work this out? answers says 4C2*6C2 but i dont get how that works, could someone explain?

[cedar kiln]

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[flat tapir]

this is What I have done but I have no idea if the direction is correct and what's the next step:

Assume n is square

(n^2-3)/4 = k where k is any integer

[dreamy void]

If n is a square you can write n = k² in the form.

Now you wanna prove k²-3 is not divisible by 4

[cedar kiln]

@flat tapir Has your question been resolved?

[flat tapir]

yes but the problem is I have no idea or any theorems can prove

for any integer k to test divisibility

[dreamy void]

Suppose k is odd or even

Consider
(2m-1)²-3 and (2m)²-3

[opaque root]

Contraposition is contrapositive?

[dreamy void]

You can then derive it's not divisible by 4 aka in the form 4q where q is some integer

[solid juniper]

mod 4 we have {0^2, 1^2, 2^2, 3^2} = {0,1}. so k^2 = 3 mod 4 has no solutions

[flat tapir]

thanks for all , let me take a look

[cedar kiln]

@flat tapir Has your question been resolved?

[cedar kiln]

@flat tapir Has your question been resolved?

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[carmine cloud]

Pouvez vous me aider dans cet exercice?

[carmine cloud]

I did this and I got stuck

[lusty grotto]

i think u expanded (p+1)^3 wrong

its not (p-1)^3, its (p+1)^3

and btw its probably easier to factor out a (p+1) first

[mental trail]

tu n'as pas besoin de récurrence il me semble, p^3 - p divisible par 3 ne va pas t'apprendre grand chose par rapport à (p+1)^3 - (p+1)

il te suffira de factoriser p^3-p

et d'expliquer pourquoi au moins un des facteurs est divisible par 3

[carmine cloud]

Ohhh

Mais la factorisation est p(p^2-1)

Comment je peux connais si c'est divisible par 3

[mental trail]

il te reste encore quelque chose à factoriser

[carmine cloud]

P^2 - 1 est une identitée remarquable

P (p+1)(p-1)

[mental trail]

exactement

[carmine cloud]

Mais comment je peux connais si c'est divisible par 3

[mental trail]

est-ce que tu remarques quelque chose sur les facteurs?

p-1, p, p+1

[carmine cloud]

Ohhh, 3 nombre consécutifs ?

[mental trail]

oui

bon pour prouver formellement que y en a un des trois divisible par 3

il suffit de faire une division euclidienne

le reste c'est soit 0, 1 ou 2...

[carmine cloud]

Je divise p et p+1 sur p-1?

[mental trail]

nononon

tu veux montrer que y en a un divisible par 3

donc la division euclidienne c'est de p (par exemple) par...

[carmine cloud]

Je les divise sur 3?

[mental trail]

oui

mais prends juste p par exemple

p = 3q + r, r est 0,1 ou 2

maintenant, on distingue les cas

[carmine cloud]

Ohhh je comprends

Si c'est 0 donc c'est divisible par 3

Si c'est 1 donc p-1 est divisible par 3

Si c'est 2 c'est p+1 qui est divisible par 3

Oui oui je pense que je comprends

[mental trail]

exactement

[carmine cloud]

Merciii

[native heath]

est-ce que parles-tu anglais ?

wait did you get it already

[carmine cloud]

Yeah yeah I did

Thanks!

[native heath]

Oh neat

de rien mdr

[carmine cloud]

:))

[native heath]

je ne pas parle en beaucoup francais

mais j'aurais essayé mdr

[carmine cloud]

We can talk in English 😂

[native heath]

it's good practice tho

[carmine cloud]

Okayy

.close

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[slate garden]

how did sin^23x become the 1-cos6x

i do the addition formula

and i gret sin2xcosx +sinxcos2x

oh its squared tbf

how did this person spoit its 1-cos6x instantly tho

i dont get it

[cold mulch]

double angle formula

[slate garden]

they use this one right?

[azure stirrup]

not necessarily

[azure stirrup]

where n= 3

[slate garden]

oh what

we are only given these

[cold mulch]

double angle formula is equivalent to half angle formula, they can be directly deduced from each other

[slate garden]

oh right

ok thanks

.close

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[ocean marsh]

I need help

[crimson sedge]

Which part mate?

[ocean marsh]

I think I did it right but I dont know the answer so Im not sure If i did it corrrectly

all of it

for 10 a I got 0.60

I mean -0.60

[crimson sedge]

The formula for PED is as follows

$$\gamma = \frac{% \Delta Q_d}{% \Delta P}$$

[wraith dagger]

Edmund Cloudsley

[ocean marsh]

right

[crimson sedge]

and this can be further expanded to

[ocean marsh]

is it not

[crimson sedge]

what?

go ahead

[ocean marsh]

(1.25-0.75)/1 =0.5 =>50%

[crimson sedge]

nah mate

I dom't thing that is correct

have a look at this

your answer is correct btw :)

[wraith dagger]

Edmund Cloudsley

[crimson sedge]

@ocean marsh are you there?

[ocean marsh]

yeah

[crimson sedge]

did you understand why this answer works

?

[ocean marsh]

ohhh

ok

yeah

[crimson sedge]

next question

at what price is the total revenue maximum

well the formula for revnue is

$$\text{Revenue} = \text{Price} \times \text{Quantity}$$

[wraith dagger]

Edmund Cloudsley

[crimson sedge]

for every single price and quantity, let us find the respective revenue

and try to find where the revenue is the highest

,w calculate {200, 250, 300, 340, 400} * {50, 45, 40, 35, 30}

R = {10000,11250,12000,11900,12000}

this is the array of revenue

[ocean marsh]

hmmm

[crimson sedge]

the highest revenue is at $400 or $300

since the revenue at both the prices is exactly the same

next question

[ocean marsh]

ohhh ok

[crimson sedge]

why don't you try the next one

now that you know the formula for PED

[ocean marsh]

I understand the banana one

yeah

[crimson sedge]

cool

[ocean marsh]

I did it already thanks

[crimson sedge]

that's nice

have a good day

bye

[ocean marsh]

THank you

[crimson sedge]

Do close the channel using .close

if you are satisfied with the help

[cedar kiln]

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[upbeat pike]

Can we write a - b = tan c as tan^-1 a - tan^-1 b = c

[mental trail]

not really no

[upbeat pike]

Fuck me

[mental trail]

first of, if you suppose c is between -pi/2 and pi/2

then a-b = tan(c)

is the same as tan^-1(a-b) = c

[upbeat pike]

Ahh

[mental trail]

but tan^-1 is not linear

[upbeat pike]

I get it

[upbeat pike]

What will be the trigonometric formula for tan^-1 x

Like tan x = opposite/ adjacent

[mental trail]

tan^-1 is the inverse of tan, so there aren't really things like that

tan^-1 takes the ratio opposite/adjacent, and finds the appropriate angle smaller than pi/2 with that ratio

[cedar kiln]

@upbeat pike Has your question been resolved?

[upbeat pike]

Yes

[cedar kiln]

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[raw kernel]

I’d like to get some help with radicals

[raw kernel]

With exercise 101

The goal in the exercise is to find the conditions of existence of the radical in question

Problem is that im not sure that when im doing them with |x| i have to make 2 conditions with negative and positive x

Because the book makes it seem like that

But when i make two conditions and proceed with the study of the signs (+/-) the outcome is very similiar to the result but theres a small “mistake” and i dont understand why its like that

The small “mistake” is the one with “???” And an arrow

Sorry if i used some weird terms but i am in italy so i dont know the terms well in english i just translate word for word from italian

Because the result in the book is 1/2<=x<=0 but my outcome is 1/2<x<=0

[wide pollen]

1-2*(-1/2)=1+1=2

that means x could possible be -1/2

cuz thats doesnt make a/b, b=0 which would result in math error

[raw kernel]

Yes but thats through trying to give x the value -1/2 if i didnt have the result in hand how would i know that its -1/2<=x and not -1/2<x

[wide pollen]

god knows

cuz i dont

[raw kernel]

Bruh😭😭

Thanks anyways ill wait for someone else to explain it to me

[cedar kiln]

@raw kernel Has your question been resolved?

[dense kayak]

you are correct and the book has a typo 😃

[cedar kiln]

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[woven ruin]

any general tips for solving second, third, fourth or fifth degree 2x2 systems

[dire geode]

do problems

[woven ruin]

because i feel like there's a more or less general method of solving such things

ok

look

[dire geode]

go

[woven ruin]

(x^2+y^2)/xy = 5/2
x^2-y^2=3

this is a simple one

yet i get stuck at some point

[pastel vault]

probably try adding $x^2 + y^2 = \frac{5}{2} xy$ and $x^2 - y^2 = 3$

[wraith dagger]

south's secret twin brother

[pastel vault]

so that you can solve for y

[woven ruin]

$x^2=\frac{5xy+6}{4}$

then what

[wraith dagger]

eightzeroeightsix

[pastel vault]

so then rearrange for y

$\frac{4x^2 - 6}{5x} = y$

[wraith dagger]

south's secret twin brother

[woven ruin]

ok so lesson learned, do not rearrange for 2+ degree terms

cuz then sqrt

ugly

does it make sense to sub x^2 with something

because i get a biquadratic

[pastel vault]

perfect, then use the quadratic formula or factor or whatever
to find what x^2 must be

,w expand x^2 (25x^2) - (4x^2 - 6)^2 = 3(25x^2)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by sooouth.

[cedar kiln]

@woven ruin Has your question been resolved?

[cedar kiln]

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[neat solar]

I need help with the 1st and 3rd bits

[dire geode]

write all trig functions in terms of sines and cosines

[neat solar]

Hmm

One sec

All of them???

@dire geode

[dire geode]

?

[neat solar]

Can i just cancel out the tan theta, sec theta, with their reciprocals in the other bracket

[dire geode]

i mean if you want to do it a different way go ahead

[neat solar]

Or is this a pemdas nono

[neat solar]

I dont know how i would write tan or sec in terms of sin or cos😭

[dire geode]

$\tan(x) = \frac{\sin(x)}{\cos(x)}$

,tex .rocket trig

[neat solar]

Oh yea this exists

Mb...

[dire geode]

top 4

[neat solar]

So uh one sec lemme write it

[hot crag]

jeez riemann that's a lotta tex

[dire geode]

credit roketto

,tex .recip trig

[wraith dagger]

riemann

[dire geode]

ah there we go

[neat solar]

Now what

[dire geode]

algebra ✨

[neat solar]

Distribute it????

Im throwing rn arent i 😭🙏@rie.mann

[dire geode]

,rotate

[neat solar]

@dire geode ???

[dire geode]

what

you haven't finished

[neat solar]

No 😭🙏

[dire geode]

keep simplifying

[brisk dagger]

@drusje the final answer can be represented as an integer. you won't be done until then

[neat solar]

Alright i guess

[brisk dagger]

i don't follow your factoring work here

go back to the first line and foil it

@neat solar

[neat solar]

Alr

,rotate

[neat solar]

@brisk dagger is this the right direction?

[brisk dagger]

yes but it should be cos not cot

[neat solar]

Oh mb rookie mistake

[brisk dagger]

you could have directly foiled btw instead of finding least common denominator but this will work

[neat solar]

@brisk dagger this might ne a dumb question but i cant just cancel the cos theta in the num and denom of first bracket, and sin theta in num and denom of second bracket righ

Yea I cant

[brisk dagger]

yeah no worries

[neat solar]

@brisk dagger im stuckkk

[brisk dagger]

multiply across the top

cos*sin+cos^2-cos.......etc

@neat solar

[neat solar]

What

Wait one sec

Sorry for ignoring what u said but i might have just found a very tedious step to do this

[brisk dagger]

@neat solar and btw you should have a list of trig identities handy if you don't know them offhand

i have to go now

[neat solar]

Kk bye, and no i dont need a list, i have google

🤣

I DID IT

HOLY FUCK THIS EQUATION WAS TEDIOUS

[brisk dagger]

@neat solar ya good job
Q3 is much quicker

[cedar kiln]

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[charred socket]

the base case isnt working what should i do

[charred socket]

the pattern is (2n)(2n+1) where n increases by 1/2

i tried plugging in a positive value for n and the statement doesn't hold up

[cedar kiln]

@charred socket Has your question been resolved?

[covert canyon]

You can prove this with induction

[charred socket]

yea i just realized the base case works

but to do the substitution

would it be 1(2) + 2(3) + 3(4) + ... + (2k)(2k+1) + 2(k+1/2) (2(k+1/2) + 1) + (2k+2)(2k+3)?

[queen ice]

I suppose that works? It’s a strange way to look at the series though. Avoid the 1/2. Read the series as gaining 2 elements for each increase of n, rather than as gaining 1 element for each half increase of n. n is natural, after all. There are no half increases of n

[charred socket]

what do you mean 2 elements

[queen ice]

2 Summands

[charred socket]

how would the pattern work, cause for the first one its 2(0.5) (2(0.5) + 1)

[queen ice]

That’s precisely not the case

n is Natural

It cannot be 1/2

Nor 24.5

It’s one of 1, 2, 3, …

[charred socket]

then what would n have to be to produce 1(2)

[queen ice]

It can never produce 1(2)

[charred socket]

but its in the question

[queen ice]

It can produce 1(2)+2(3)

But never 1(2)

[charred socket]

what value would you plug in to get that

[queen ice]

n=1

[charred socket]

yea true

but then how did they get 3(4)

or 3(4) + 4(5) if thats how you think of it

cause at n = 2 its 2(2) * 2(2) + 1

[queen ice]

$\sum_{k=1}^{n} [(2k-1)(2k)+(2k)(2k+1)]$

[charred socket]

do you only get the second part

[wraith dagger]

FirstNameLastName

[queen ice]

That’s the sum on the left hand side

[charred socket]

Ohh I see

Would I have lost marks if I made k start at 1/2

[queen ice]

Yes absolutely

[charred socket]

So k can only be whole numbers

[queen ice]

k must be Natural

Yes

And not negative either

[charred socket]

Ok that makes sense thanks

Might be harder to notice the pattern tho

[queen ice]

You mean now that we have only natural k?

[charred socket]

Yea idk why the first and only idea I had was k= 1/2

[queen ice]

Well, as a rule of thumb, you will generally consider natural numbers only when applying induction. n is also generally defined to be a natural number. If you remember that I think you won’t have much trouble in that regard in the future. And practice will really aid you in recognizing patterns like this

[charred socket]

Sounds good, thanks 🙏

.close

[cedar kiln]

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[oblique prawn]

how to use kinetic energy to solve for final velocities in collisions

[dire geode]

!original

[cedar kiln]

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.

[oblique prawn]

A 5 kg ball moving at 2 m/s makes a totally elastic collision with a 3 kg ball at rest. What are the speeds and directions of each ball after the collision?

[sly abyss]

u dont ned kinetic energy huh

oh nvm

[oblique prawn]

what do i do after $10=5v_1+3v_2$

[wraith dagger]

Jash

[dire geode]

Use definition of elastic

[oblique prawn]

K_i=K_f

1/2mv_i²=1/2mv_f²

[oblique prawn]

do i use the kinetic energy of the two ball system

[dire geode]

Yea try it

https://www.khanacademy.org/science/ap-college-physics-1/xf557a762645cccc5:energy-and-momentum/xf557a762645cccc5:representation-of-changes-in-momentum/v/elastic-and-inelastic-collisions

More examples there

[oblique prawn]

ty i got it

.close

[cedar kiln]

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[dusky current]

Good evening everyone! I'm brushing up on fractions and clearing them from equations and could use some clarity.

In equations with a fraction only on one side, is it the same as clearing them on both sides?

[violet flume]

you can think its a subcase, yea

if i understand what you mean

like, if you have $x = \frac 54$

[wraith dagger]

jan Niku

[violet flume]

we can just rewrite it as $\frac x1 = \frac 54$

[wraith dagger]

jan Niku

[violet flume]

proceed as normal

[dusky current]

turn 10x and -8 into a fraction, distribute the 2/5 and create whole numbers, correct?

[violet flume]

its up to you at this point, how uncomfortable you are with fractions being present

we actually have an exact form of the solution to this problem

so i dont think that clearing the fractions you could say is "necessary"

[violet flume]

but no, i would multiply both sides by 5

[dusky current]

my teacher's preferred method is to eliminate fractions entirely, that is her first step

[violet flume]

okay

then multiply both sides by 5

[dusky current]

so not using distributive property, instead using multiplication property (multiplying by the denominator) to eliminate the fractions. Is that right?

[violet flume]

yes if the goal is to get rid of fractions why would you distribute and create more fractions

is the reasoning

[dusky current]

good point lol, I'm overthinking as usual

[cedar kiln]

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[turbid isle]

Did I set up this correctly?

[cedar kiln]

@turbid isle Has your question been resolved?

[cedar kiln]

@turbid isle Has your question been resolved?

[cedar kiln]

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[umbral crow]

i need help with part c)

[cedar kiln]

@umbral crow Has your question been resolved?

[umbral crow]

<@&286206848099549185>

[cedar kiln]

@umbral crow Has your question been resolved?

[umbral crow]

<@&286206848099549185>

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@umbral crow Has your question been resolved?

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[cedar kiln]

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[odd dagger]

Hi could someone please help me with my maths investigation

[odd dagger]

This is my working out so far but I’m stuck on what to do for the graphs and sign diagrams

[cedar kiln]

@odd dagger Has your question been resolved?

[topaz pewter]

For the graph, you can plot it on desmos using your surface area as a function of the radius

So you'd use your $S = \frac{2V}{r} +2\pi r^2$

[wraith dagger]

Katrro

[topaz pewter]

Then the graph will show you your optimal values of r.

[odd dagger]

Ive already tried putting that into desmos but a graph hasn’t shown up?? Do I need substitute anything?

[topaz pewter]

try using this

[rancid moss]

hey im a 9th grader in Utah what are some good tips to write better notes pretest on y intercept from a graph and how to graph using Y=mx+b ?

[odd dagger]

.close

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[thin bane]

guys, im stuck on this, its a trapezoid, im not sure what else to do with it but i need to find ABED

[cedar kiln]

@thin bane Has your question been resolved?

[stark river]

i'm trying to

if we find h of the triangle ABC

[thin bane]

like, trace a line from the A base to the C point?

[stark river]

can we say that h of DEC is the double of h of ABC

[stark river]

yep

cuz u can use (B+b)*h/2

[thin bane]

i'm not sure.. it doesnt state that the area of DCE is 10, thats why im stuck

but it that's the case then

[stark river]

nah

[stark river]

@thin bane

U DON't have any other information?

[thin bane]

nothing

[stark river]

dang

[thin bane]

i think we just have to assume 10 would be the area of the other triangle right?

i was thinking those triangles were similar and i did arrive at an answer but it does assume a few things

[stark river]

nah u will have to justify it

[thin bane]

right

[stark river]

bro we relly stuck

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[jaunty hemlock]

Hello, I was trying to solve this exercise but when I compared my result with my teacher’s solutions, I was a bit confused. To get the acceleration he subtracts not only the static friction force, but also the normal force from the gravitational force of the left package. In my understanding you would only subtract the static friction force. So could someone please tell me if I’m in the wrong or if his solutions are wrong? Thank you very much for helping me
Here is what is given:
Calculate the acceleration and the force on the connecting cable. Data: Coefficient of friction µ=0.3

[crimson sedge]

Where did he subtract the normal force?

Could you mark it?

@jaunty hemlock

[jaunty hemlock]

in the equation for the acceleration in the denominator. There he subtracts the the right mass * g * sin(alpha) which I think is the normal force) and the µ * right mass * g * cos(alpha) is the static friction force.

[cedar kiln]

@jaunty hemlock Has your question been resolved?

[jaunty hemlock]

<@&286206848099549185>

[crimson sedge]

isnt that a physics question

[jaunty hemlock]

yes, I'm sorry. I'm new here. Do I have to post it somewhere else?

[crimson sedge]

i guess all the people here have a mathematical expertise

i m not sure if we can send physics question too

😊

[jaunty hemlock]

ok, thanks you for your help.

.close

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[sly tulip]

help me factor dis: y²-7y-18

[vivid radish]

(y-9)(y+2)

[south tundra]

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[sly tulip]

help me and full explanation

<@&286206848099549185>

ALEKS

[vivid radish]

For statement 3 there is only one line left to fill

[sly tulip]

which line

rep

bro

[vivid radish]

There are only 5 lines and 4 are used in statement 1 and 2

[cedar kiln]

@sly tulip Has your question been resolved?

[sly tulip]

reason pls

BD=BD

?

[vivid radish]

Yes

[sly tulip]

reason

reflexive or transitive

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@sly tulip Has your question been resolved?

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[unique crane]

Can I factor out 1/25 to make this easier to integrate

[dire geode]

Yes

,tex .int rules

[wraith dagger]

riemann

[dire geode]

First and second

[unique crane]

I see, thank you

.close

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[tame sleet]

how to get the f of n

[tame sleet]

I know the pattern but idk whats the equation

[queen ice]

Maybe you can express it as a sum first end turn it into its closed form after?

[tame sleet]

?

[queen ice]

Do you know how to work with $\sum$?

[wraith dagger]

FirstNameLastName

[tame sleet]

no

[queen ice]

Hmm okay

[tame sleet]

it starts at 10 and n + 4 every time

[tame sleet]

what do I do?

[queen ice]

It seems the sequence corresponds to the pictures above it, correct?

[tame sleet]

I think?

[queen ice]

Can you see in what way they correspond?

[tame sleet]

1 cube is one ?

[queen ice]

Yep

[tame sleet]

???

[queen ice]

So for the first element in the sequence

You can count the cubes to obtain 6

But you can also calculate it. Can you see how?

[tame sleet]

how

[queen ice]

Maybe it becomes more obvious if you think about how to easily calculate the number of cubes in the 4th element of the sequence

[tame sleet]

bruh idk 😭

[queen ice]

Hint: 2*3 =6
6*8 =48

[tame sleet]

?

[queen ice]

Well, where may the 2 and 3 be coming from? Where may the 6 and 8 be coming from?

[tame sleet]

number of terms?

[queen ice]

Look at the pictures. They are the sequence

[tame sleet]

they add 4 more each time

[queen ice]

It’s not about that yet

What you need to figure out is how 6 corresponds to the first element in the sequence

And how 48 corresponds to the 4th element

[tame sleet]

how

the only thing i see is they start + 10 and + 4 more each time

[queen ice]

You see these rectangles? At the top?

[tame sleet]

multiply by 8?

[queen ice]

Yes or no

[tame sleet]

yea

[queen ice]

You see what 6, 16, etc are described as in the table?

[tame sleet]

add 10?

[queen ice]

No

[tame sleet]

6 squares

[queen ice]

Yes

And 16 squares etc

[tame sleet]

but how is the picture even related

[queen ice]

That’s what you have to figure out

[tame sleet]

bruh I tried for 30 minutees

Im too dumb to

[queen ice]

You see how it says „number of squares in sequence“?

[tame sleet]

i see

???

is there any formula

for this

[queen ice]

The picture shows the first 4 elements of the sequence meant here

Yes there is

[tame sleet]

what is

I dont think I can figure out the pattr

[queen ice]

You could if you tried to follow what I’m saying. You’re always rushing to the next part instead of actually comprehending what I’m saying

[queen ice]

The first rectangle has how many squares?

[tame sleet]

6

[queen ice]

The second has how many?

[tame sleet]

16

[queen ice]

The third how many?

[tame sleet]

30

[queen ice]

Do you see how they are connected now?

Each number and its corresponding rectangle?

[tame sleet]

even number?

[queen ice]

.

[tame sleet]

?

[queen ice]

What has the 6 got to do with the first rectangle

[tame sleet]

is the pattern addtion

?/

[queen ice]

No

[tame sleet]

whay form

[queen ice]

Why are you at the pattern again

[queen ice]

Just answer

[tame sleet]

1 retangle has 6 squares?

[queen ice]

Yes

And second has 16 squares and third 30 and so on

That’s how you know what the next number in the sequence is

By looking at how many squares are in the next rectangle

Do you understand that?

[tame sleet]

but whats the formula

it said F(n)=

[queen ice]

Mate I’ll tell you this one more time. Stop rushing on and instead answer my questions. If you don’t I’ll dip and you can find it yourself.

We will get to the formula

[tame sleet]

yes i understand

[queen ice]

Good

Now instead of focusing on how the numbers increase, let’s instead focus on the rectangles

What are the side lengths of the first rectangle?

[tame sleet]

3

2

[queen ice]

Correct

How about the second rectangle

[tame sleet]

4

[queen ice]

4
4

Yess

How about the third

[tame sleet]

6

5