help-23

[lone arch]

{7, 19, 26}

[runic pine]

So x can be all these till 27 ryt

[lone arch]

Should be the set C

[lone arch]

x is a^3 - b^3

Where a^3 and b^3 are anything out of that list

[runic pine]

Got it

[lone arch]

Oh, plus minus everything

[runic pine]

Yeah

So u solved it

[lone arch]

Yep

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[last wren]

I'm not gonna be incredibly rigorous here, but does anyone know of a proof that

$$\int_{t_1}^{t_2} dt = \int_{x_1}^{x_2} \frac{dt}{dx} dx$$

I understand it intuitively, but I despise hand-waviness

[flat frigate]

doaby

[last wren]

the whole "treat it like a fraction" thing is nice and all but is not always justified

so i'm wondering why this is technically allowed, could you just use a substitution?

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[dawn sierra]

How can I solve this? <@&286206848099549185>

[last heath]

!occupied

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[brazen dragon]

I have an integral like 1/f(x) what is best way to integrate this

[brazen dragon]

by parts?

or would DI method work?

and is this considered algebraic if i use di table or just by parts

f(x) is polynomial

[lean otter]

no easy way to say

oh polynomial?

[vast obsidian]

Usually Partial fraction until you have something linear or a quadratic

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[vital wing]

Solving an indefinite integral with substitution and parts method

[vital wing]

I don't understand the way to solve these with substitution

the definition I have : /

I should send this actually

[glass flame]

did u get this by solving for ques and got stuck here?

[glass flame]

cos integration for cosx*sinx is easy

[vital wing]

no, it's the initial problem

[glass flame]

why are you using this then ??

[vital wing]

The prompt asks me to solve using substitution , then parts if those are possible

[glass flame]

for Integral of cosx sinx
use u*v

cosx int(sinx dx) - int( d(cosx)/dx int(sinx dx) dx)

that will give us, cosx (-cosx) - int( (-sinx) (-cosx) dx)
and lets consider the starting eqn as A,
now we have,

A = -cos^2x -A

A+A = -cos^2x
2A = -cos^2x
A = -1/2 cos^2x

[glass flame]

and after that just put limits of A into the answer

[vital wing]

alright I get how to do the other one then

thanks friend

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[gaunt garnet]

u=sinx
So you just get
0.5 * (sinx)^2 |1->4

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[clear crest]

Can anyone figure out why this isn't correct?

[stoic dune]

How did you get to line 3?

[stoic dune]

It's (2x + 3)ln(1 + x) in the original question?

Or is it the sightly different 2x + 3ln(1 + x)?

[clear crest]

Oh

Right

So 3 Is actually the exponent of ln

[clear crest]

Yes

[split ether]

1. Looks like you forgot to put lim_{x -> 0} on the right hand side
2. That is true when a is a constant; Is 2x+3 a constant?

[clear crest]

Oooh ok makes sense then

Thanks all

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[lunar lagoon]

mb

[shrewd topaz]

why doesnt the degree symbol work

sorry i didnt see you typing

[lunar lagoon]

dw

[rough storm]

there is degree mode in desmos

[shrewd topaz]

i already put it in degree mode

[rough storm]

you don't need to put degrees symbol then

[rough storm]

oh

you're trying to plot polar coordinates?

[shrewd topaz]

yea

[rough storm]

desmos only allows you to plot in cartesian, so use the conversion formula x=rcos(theta),y=rsin(theta)

so instead of (r,theta), plot (rcos(theta),rsin(theta))

and you don't need degree symbol for theta if you have it in degrees mode

[shrewd topaz]

oh i see

thanks

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[thick holly]

Can someone please help me with this question the first part

I know I have to integrate (Mx^2/L)dx but I don’t know what my bounds are thanks

[steel stag]

it's pinned at L/5, so from that pin it goes -L/5 to the left and 4L/5 to the right

since x means distance to the fixed point

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[modest bison]

hello

[modest bison]

can someone help me with
trigonometric equations with multiple angle functions

e.g. sin(2x + 1) = 0,3

i went for this

$u = 2 \cdot x + 1 \ u_1 = arcsin(0,3) = 0,3046 \ u_2 = -0,3046 \ x_1 = \frac{0,3 - 1}{2} = -0,34 \ x_2 = \frac{-0,3 - 1}{2} = -0,652 \ x_{1,1} = \frac{0,304 + 2 \cdot \pi - 1}{2} = 2,793 \ x_{1,2} = \frac{0,304 + 4 \cdot \pi - 1}{2} = 5,935$

[flat frigate]

Kraut

[modest bison]

and how do i calculate x_2,1

[shadow glade]

i'm a bit confused about what youre doing, it's just:
sin(2x+1) = 0,3
2x+1 = arcsin(0,3)
2x = arcsin(0,3) - 1
x = (arcsin(0,3) - 1) / 2
so just solving for x after you take arcsin of both sides

[pliant egret]

your notation is very confusing

not really sure what x1,1 x1,2 etc means

you just got x = (arcsin(0,3) - 1) / 2 as said above

and other "x" are just (arcsin(0,3) - 1) / 2 +kpi

[modest bison]

its like solving trigonometric equations with multiple angle functions in the area [0; 2pi[

[pliant egret]

are your angles radians?

[modest bison]

yes

[shadow glade]

yeah, arcsin will give you the values in the range [-pi/2, pi/2] so if you get negative solution then add 2 pi to it to get something in [0, 2pi], then also consider (pi - your solution) since that has the same sin value (same y on unit circle because of symmetry over x axis)

[modest bison]

i need to get all the radiant value in the region [0 ; 2pi[

[modest bison]

so if negative value just add 2pi and then pi - this new value?

the other two solutions i dont know how to get are 0,918 and 4,060

[shadow glade]

so you got arcsin(0.3) = 0.304692654 right? that's some small angle in the first quadrant

[modest bison]

yes

[shadow glade]

theres also an angle in the 2nd quadrant that has the same y value

[modest bison]

cuz of this right

[shadow glade]

on the unit circle, that angle will be pi - 0.304692654

that gives 0.3 also

[modest bison]

yes understandable

[shadow glade]

so you have 2x+ 1 = 0.305
and also have 2x + 1 = pi - 0.305 equations, now solve both for x

that's it

[modest bison]

system of equations?

[shadow glade]

no, theyre just separate solutions

one gives one solution, the other gives another

[modest bison]

sry for shit handwriting

[modest bison]

should i then just add PI to the solutions?

i need to find 4 degrees which have the same thing in the are of [0; 2pi[

yk?

hello?

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[old basalt]

hi! im confused on this question, particularly

this part, i don't understand why it rounds out the way it does....

[shadow aurora]

Because there is a common factor

which is 4 and the square root of it is 2

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[wispy granite]

im too dumb guys, sry if anyone can help me please i would love to know because i think i was born different 😭

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[novel sage]

any idea to get MP

[novel sage]

I know it should be possible through midsegment theorem

But I can't see where to apply it

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[left gyro]

1. prove that NQFM is a ||parallelogram||
2. prove that NR is parallel to ||MP||
3. draw segment MQ, then use similar triangles to find MP

[stark stratus]

is that 8y? or 8 and like french

or whatever

[novel sage]

That is 8

y is just "and" in spanish

[novel sage]

I'll try it, thanks

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[tardy lark]

if we use x=cos+1

y=sin

why is r not r=2cos(t), is this a glitch in maths ?

<@&268886789983436800>

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[steel stag]

it's different coordinate systems where the letters mean different things. in the first you have r as a constant and t draws a circle, for the second in desmos it's polar coordinates where theta and r are actual variables

[tardy lark]

hmm

[tardy lark]

then how could i convert the parametric equation into a polar funciton ?

[steel stag]

you can do r=sqrt(x^2+y^2) and cos(theta)=x/sqrt(x^2+y^2) and do a bunch of reorganizing

it'll turn into this then you can complete the square to get (x-1)^2+y^2=1 which is basically the first equation you had

[tardy lark]

2cos(t) ?

[steel stag]

yea starting from r=2cos(theta) and plugging in the polar->cartesian transformation

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[crimson creek]

Can someone explain this step?

[crimson creek]

how did we do the integration, which type is this?

[drowsy karma]

the integral on the right side results immediately from the formula, while the integral on the left side requires some transformations and substitutions

right side is solved with such a formula:

[crimson creek]

ya I don't get the left side

I'm confused

[drowsy karma]

so it comes imemdiately

i can give you a beinning of the solution fo r the left integral

2 minutes plz

[crimson creek]

sure, take your time.

[drowsy karma]

so you get integral of the rational function, which is standard

[crimson creek]

I do not understnd why we have 2 root(f(y)) here and root(f(y)) in denominator.

[drowsy karma]

compute the derivative to check my formula

[crimson creek]

got it

[drowsy karma]

kewl )

[crimson creek]

I already got that but thanks!

[drowsy karma]

yw )

[crimson creek]

Just a question, what does int. f'(x)/f(x) give?

[crimson creek]

nvm. got it

[drowsy karma]

but i do that automatically in my mind at once

[crimson creek]

log(f(x)) + C?

[drowsy karma]

no

you talk ab similar but diff formula

[crimson creek]

int. f'(x)/f(x) dx = log | f(x) | + C right?

[drowsy karma]

write them in your exercise book, i always give them to my students

[crimson creek]

thanks, will do.

[drowsy karma]

smiles

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[crimson creek]

.reopen

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✅

[crimson creek]

hey, I get
1/2 int. 2u . u/ u^2 - 1
what do I do afterwards?

[drowsy karma]

have you ever integrated rational functions?

[crimson creek]

2u / u^2 -1 is log(u^2 -1) right?

[drowsy karma]

in numerator, you have got: 2u^2

not 2u

[crimson creek]

ah yes

[drowsy karma]

🙂 so you must

thikn

about

decomposition on partial fractions

but before decomposition, you need to notice that the degree of the numerator is the same as degree of polynomial in denominator

hence you should divide them

or other trick giving you proper rational function

proper means: degree of numerator is less then degree of denominator

such trick helps you:

plz continue

[crimson creek]

got it

so 2 int. 1 + 2 int. 1 / (u^2 -1)

[drowsy karma]

yes!! ))

and

the second one on partial fractions

[crimson creek]

2u + 2ln(u^2 -1)

[drowsy karma]

no

[crimson creek]

and half outside?

[drowsy karma]

the second one you wrote wrong

[crimson creek]

so cancels out

[drowsy karma]

that wil be difference of logarithms ..

[crimson creek]

2 int. 1 / (u^2 -1) -> 2 int. 1 / (u -1)(u+1)
which property do we apply here?

to get log u-1 - log u+1

[drowsy karma]

diffeence of logarithms is logarithm of quotient

[crimson creek]

how did we change that?
I don't get what we've done in the 4th step.

[drowsy karma]

i said: decomposition on partial fractions

then you obtain values of A and B, I made it at once, so i wrote it above

but normally you have to find them

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[crimson creek]

Thanks, got it.

Had to revise partial fractions.

[safe radish]

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[steel spire]

Hi I'm doing a section in Real Analysis called cardinal numbers... The first page is the original quiz questions I took down and the are the answer I wrote and the last one is everything I wrote for the solution with the question above the solution

[steel spire]

I just want someone to correct my last part properly without just saying why🙂

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<@&286206848099549185>

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[burnt wyvern]

i proved this through stars and bars
is there any other way we could do it

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[burnt wyvern]

<@&286206848099549185>

anyone

WAIT i got it

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[gloomy wren]

Messing around with complex numbers in python. If i is definitively equal to sqrt(-1), shouldn't (-1)^(1/2) return i as well? Pretty much every other calculator I found that could do complex numbers also returned i, but for some reason python returns this monstrosity. Why? What the heck does l mean? And where are 6.123233995736766 and -17 coming from?

[gloomy wren]

Okay my bad. Turns out that the "e-17" in this context is a weird form of notation that python uses, the l is actually a 1 and j is another way of writing i. It actually means 1.000000000000000000000000000000001874699728327322. It's a floating point error.

Python can be so annoying sometimes

[edgy fjord]

There's a seperate module for complex numbers I reckon

[gloomy wren]

probably

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[buoyant pond]

assuming them x + iy is a lot of work

any easier method?

hint me pls

[eternal carbon]

hint: ||x^3 + y^3 can be factored||

[buoyant pond]

oh dang

Imma try tyty

yoo got it thanks

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[kindred slate]

im a bit confused on this question

[kindred slate]

it says here the first order partial derivatives have to be continuous

im doing 1a f(z) and showing wheres its differntiable

but the ms says nothing about them being continuous at a point

im confused

its only shown that the cauchy riemaan equations are satisfied

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[inner sand]

But the equations aren't satisfied

I think the image got cut off before stating everything

There's an if floating there

[kindred slate]

they didnt mention anything about continuity

which is mentioned in the original proposition

[inner sand]

I guess because the derivatives are expressed as linear eqs which are already continuous for x,y R but idk tbh

[kindred slate]

could it be that

since they r infinitely differentiable

they r continuous everywhere?

automatically

since if something is continuous it is diffable at that point

[inner sand]

I mean they look pretty much for example the first one as Ux = f(x) = 2x + 3, which with the restriction that x is real then we know f(x) = 2x + 3 and that would be continuous idk if i explained myself still not sure

[kindred slate]

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[hoary mirage]

okay guys, I dont know how I got #4 wrong

I looked over my equation and Im still puzzled

[safe radish]

@hoary mirage Has your question been resolved?

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[hoary mirage]

.help

[safe radish]

Commands:

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[delicate zinc]

Why is this line of line of reasoning invalid?

The theorem is true for N = 0, and if the theorem is true for N+1, then it is true for N.

I can't seem to wrap my head around it, but to me it seems as if if it is true for N = 0, and true for N + 1, then surely that'd span all N.

Assuming N is a non-negative integer.

[median vigil]

for example if n = 1, then it's true only if we assume it's true for n = 2, but it's only true for n = 2 if we assume it's true for n = 3, etc. if we proved it for n = 100, then that would imply n = 99, 98, etc all the way down, to 1, but it would not prove for any n past 100

[delicate zinc]

But if you were to prove it for the general n+1, without picking a specific value to do the proof for, shouldn't that span all N?

it's true for 0

it's true for n+1

so it's true for 0+1 = 1

so it's true for 1+1 = 2
etc

What am I missing?

[delicate zinc]

What makes it go backwards?

[median vigil]

it says that if the theorem is true for n+1, then it is true for n. so if the theorem is true for n=2, then it is true for n=1, not the other way around

[buoyant shadow]

it's valid, i don't see any problem

[delicate zinc]

so it's basically induction but backwards?

in induction you'd instead prove that if the theorem is true for n, then it is true for n+1

[delicate zinc]

it's from a list of invalid techniques of proof, which makes me doubt that

[median vigil]

yes. but the problem is that if you prove in the "backwards" direction, you'd need to pick some high integer as a base case and then it would prove for all n lower than that. but since there's no "highest integer", you could never possibly cover all integers like that

[buoyant shadow]

there's no such problem here

they start with 0, and they say negatives don;t matter

[delicate zinc]

does "reverse recurrence" mean "reverse induction"?

[delicate zinc]

well, I say N is a non-negative integer

bc it's not specified what N is at all, but I feel it most fair to assume it to be that

but for all we know N is a complex number

but that makes it obv nonsense

[buoyant shadow]

i don't know what they could mean by saying it's invalid

[delicate zinc]

this is labelled "Proof by reverse recurrence"

the n=0 base case doesn't actually do anything

is that correct?

[buoyant shadow]

cloud misread it

oh i see

yeah, cloud is right, it's backwards

[median vigil]

yes, if we do consider negative integers then it does prove for all negative integers, but if we only consider positive integers then it accomplishes nothing

[buoyant shadow]

if the theorem is true for N+1, then it is true for N

this is going down

[delicate zinc]

same as "if the theorem is true for N, then it is true for N-1"

?

[buoyant shadow]

yes

i misread it bigly

[delicate zinc]

Okay I think I get it

Thanks everyone!

.close

[safe radish]

Channel closed

Closed by @delicate zinc

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[safe radish]

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[crystal atlas]

hi

[crystal atlas]

i had a question on smth

so the question is (8x-5) divided by (2x+1)

and i don’t know how we would do this

@helper

<@&286206848099549185>

we’re tryna figure that out

it’s a polynomial

[hardy lion]

!original

[safe radish]

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

[crystal atlas]

ok

its question 10

@lean otter sorry here’s the original question

<@&286206848099549185>

[hoary wind]

The question is to divide the polynomials and establish if the divisor evenly divides into the dividend - basically, do you get left with a remainder when you do the long division

Have you done Qs 7-9? The method should be the same

[crystal atlas]

we’re only directed to do 10

they like pick questions

that we have to do

[hoary wind]

Oh I see, that's fine

[crystal atlas]

alr

[hoary wind]

So set up the long division, divide the leading term of the numerator by the leading term of the denominator (8x by 2x)

What do we get?

[crystal atlas]

ok

so

4x

right

[hoary wind]

Not quite -> It's x^1 dividing x^1

What is $\frac{8x}{4x}$?

[flat frigate]

TayBee

[hoary wind]

2x even

[crystal atlas]

2

x

[hoary wind]

Just 2 in that example which was meant to be the same but I mistyped lmao

Because x^1 divided by x^1 is x^0

I.e. the x's cancel

[crystal atlas]

oh

[hoary wind]

So Dividing 8x by 2x just gives 4

[crystal atlas]

ohh

ok

[hoary wind]

So now we've got the 4, what's the next step of long division?

[crystal atlas]

see how many times 2 goes into -5

[hoary wind]

Long division works slightly differently so we ignore the -5 for now, we multiply the divisor (the denominator) by what we get

I.e. we want to work out 4(2x+1)

So what does that give us?

[crystal atlas]

8x+1

wait no

4

+4

sorry

8x + 4

[hoary wind]

Yes good

So now we take that

And we subtract it from our dividend

(the numerator)

What are we left with?

[crystal atlas]

-9

and that’s the

remainder

[hoary wind]

Exactly

[crystal atlas]

so the result

is

4

?

[hoary wind]

So the result is 4 and then the remainder divided by the divisor

So putting it all together

$4-\frac{9}{2x+1}$

[flat frigate]

TayBee

[crystal atlas]

ohh ok

and

the divisor doesn’t evenly divided with the dividend right

[hoary wind]

Exactly - if there's a remainder then it hasn't evenly divided, so that's the answer to the question

[crystal atlas]

alr

[crystal atlas]

also

for 12

would it be -6

[hoary wind]

It would indeed

[crystal atlas]

ok

and i’m having trouble doing 17

it’s in the first image

[hoary wind]

The first thing you'll want to do with that one is rewrite the numerator so it has descending powers, then long divide the same way

The only difference is, now the degree of the polynomials are different

So if you're dividing x^2 by x, you are still left with x^1

But then it's the same as before, multiply and subtract the result, then divide again if it's possible

[crystal atlas]

ok

wait so

on the top

would we have -7x+8

[hoary wind]

-7x is spot on, shouldn't be +8 though, what do you get left with when you subtract -7x(-2x-5) from your dividend?

[crystal atlas]

x + 14x^2 - 75

[hoary wind]

In that case it looks like you might have expanded the bracket wrong or cancelled the wrong terms

Taking it back a step

What do you get when you expand the bracket?

[crystal atlas]

wait i think i see what i did wrong

ok so

i got 14x + 35x

from the brackets

when i dust sri herd it

distributed

[hoary wind]

35x is right, there's something wrong with the 14x, what's x times x?

[crystal atlas]

oh

14x^2

[hoary wind]

Yes much better

So now subtract 14x^2 + 35x from your dividend

(as a sanity check when doing polynomial division, if you haven't eliminated the leftmost term of your dividend, something's gone wrong)

[crystal atlas]

50x-40

[hoary wind]

15x - 35x ?

Check your signs

[crystal atlas]

oh wait

yeah

we’re subtracting

-20

x

[hoary wind]

Yes good, -20x -40

So now we divide again

How many times does -2x go into -20x

[crystal atlas]

10

[hoary wind]

Exactly, so now you rinse and repeat, multiply -2x-5 by 10 and subtract from what you've got left

[crystal atlas]

-20x - 50

and then

wait that’s right tho rught

[hoary wind]

Hmm something has gone a bit wrong there

[crystal atlas]

after multiplying

[hoary wind]

You've got -20x -40 left

[crystal atlas]

yeah

[hoary wind]

Oh sorry

Yes

That's what you get when multiplying

[crystal atlas]

ohh ok

[hoary wind]

So now we subtract that

[crystal atlas]

ye lemme substract biw

subtract

now

mb i hate fast typing

i got

-100

wait no

10

[hoary wind]

Yes much better, 10

So we can't divide 2x into 10 (there's no x left), so that's our remainder

[crystal atlas]

so it’s -7x with a remainder of 10

[hoary wind]

Don't forget the +10 you worked out as well!

You should be left with

[crystal atlas]

ohhh ye

[hoary wind]

-7x + 10 and a remainder of 10

And generlaly I'd then write the remainder over the divisor

So we end up with

$-7x + 10 + \frac{10}{-2x-5}$

[flat frigate]

TayBee

[crystal atlas]

oh alr

[hoary wind]

That all make sense?

[crystal atlas]

yeah

thank you so much

[hoary wind]

Excellent, no problem!

[crystal atlas]

wait so

can i just write a remainder of 10

or do i have to add that

[hoary wind]

Writing remainder 10 is correct, honestly I think it depends a) whether the question specifies, and b) who's marking

The most important thing is the -7x + 10

How you deal with the remainder I think is up to you

[crystal atlas]

ah alr

ohh ok

ok thank you

[hoary wind]

But if you want a pure algebraic solution, the fraction is best

No problem :)

[crystal atlas]

aight

.close

[safe radish]

Channel closed

Closed by @crystal atlas

Use .reopen if this was a mistake.

[crystal atlas]

@hoary wind hey are you still online

.reppen

.reopen

[safe radish]

✅

[crystal atlas]

alr

so

we’re doing 32

and idrk what it is

ik how to do synthetic division

tho

i think it might be

9x - 2 - 10/x+3

<@&286206848099549185>

<@&286206848099549185>

[severe pond]

https://youtu.be/gUPJjRPYA5g?si=Ez2p9UwbM3Wp8iGX

[crystal atlas]

yea

i watched fhT

that

[severe pond]

did you what confused you

[crystal atlas]

wait so

if we’re adding

in the parenthesis for the divisor

would the k value be negative

[severe pond]

yes

[crystal atlas]

ok

[severe pond]

if the divisor is

x+3

then you use -3

[crystal atlas]

it’ll be -3?

ok

[severe pond]

yes

it’s of the form x-k

[crystal atlas]

so i got 9x-2-10/x+3

is that correct

[severe pond]

what was the original

so strt with 9 25 -4

[crystal atlas]

32

[severe pond]

yes urs is correct

[crystal atlas]

oh ok

[severe pond]

oh wait

-10?

[crystal atlas]

and also

what

[severe pond]

hmm

9x-2 is correct

[crystal atlas]

yeah

wait so

i did -2 times -3

[severe pond]

yes

[crystal atlas]

and then that’s 6

[severe pond]

and that’s 6

[crystal atlas]

do we -4 - 6

and that’s -10

[severe pond]

u add

don’t subtract

[crystal atlas]

oh

wait but for the others

[severe pond]

it’s -4+6

[crystal atlas]

i did

[severe pond]

well you got -27 correct

then 25+ (-27)

not 25-(-27)

[crystal atlas]

oh wait

ohh

ok

so it would

be

2/x+3

instead

of that

[severe pond]

yes

[crystal atlas]

ok and also

for 36

would the leading coefficient be -7

[severe pond]

yes

then 0

then 12

[crystal atlas]

wait

where is 0

[severe pond]

0x

you go in descending order for the degree

you can’t exclude the degrees that have a coefficient of zero

[crystal atlas]

what has a coefficient of 0

[severe pond]

x^1

[crystal atlas]

where is that

[severe pond]

it’s not there

so you put a zero

[crystal atlas]

oh

[severe pond]

you go in descending order

[crystal atlas]

so the answer would be

-7+7+5/x+1

i meant

-7x

[severe pond]

yes

[crystal atlas]

ok

and also

can u check some of my other work

for 6 i got -x-7, and for 10 i got 4-9/2x+1 and it doesn’t evenly divide since we have a remainder

for 12 i got -6 and it does indeed evenly divide

for 17 i got -7x+10+10/-2x-5

is that all right @severe pond

[severe pond]

lemme check

[severe pond]

these r correct

[crystal atlas]

aight

and the rest

[severe pond]

true

[crystal atlas]

ok

[severe pond]

true

all right

[crystal atlas]

aight thank you so much brother

.close

[safe radish]

Channel closed

Closed by @crystal atlas

Use .reopen if this was a mistake.

[tulip warren]

oh good work

[safe radish]

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[flat frigate]

moneymagnetlol

[obtuse plover]

@lean otter

[brazen ravine]

798

you have to press each number in it’s prime factorising forum

🤫🧏🏻‍♂️

[safe radish]

Channel closed

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[half berry]

help

[obtuse plover]

@half berry

what have u tried

[half berry]

Uploading work

I wrote this down idk if this is right but idk what to do next anyways

[obtuse plover]

Ok so for the first term in numerator

(x^2 + 1)(e^3x)

u didnt differentiate

the e^3x

oh wait

uh its kinda confusing the way u write it

it should be (x^2 + 1)(d/dx e^3x) - (e^3x)(d/dx x^2 + 1) /(x^2 + 1)^2

for ur first line

It reduces confusion

[obtuse plover]

good

now do the necessary differentiatoon

[half berry]

i think i forgot how to do this

do i just distribute

[obtuse plover]

just take the derivatives of the terms with d/dx in fromt of them

[half berry]

Then what

[obtuse plover]

now just distribute

[half berry]

do i write it like this

[safe radish]

@half berry Has your question been resolved?

Channel closed

Closed by @half berry

Use .reopen if this was a mistake.

[half berry]

Oops

.reopen

[safe radish]

✅

@half berry Has your question been resolved?

[half berry]

@obtuse plover

[leaden glen]

not sure why u have a d/dx in there

u should be getting

e^3x(3x^2+-2x+3)/(x^2+1)^2

well thats what i got atleast

@half berry

[half berry]

thats just the answer

how did u get there

this distrubted/

*/

[half berry]

how tho

steps

[leaden glen]

did you use quotient or product rule

im not a fan of quotient rule so i moved the (x^2+1) to the top and did a product rule

you can still get the same answer though

[half berry]

im just stuck on where the other person left me

[leaden glen]

im confused on your thought process

why did u leave a d/dx

[leaden glen]

in here

[half berry]

cuz im lost

i was asking if i was supposed to do that

or take it out

and then like do the distribution and stuff

[leaden glen]

once u take the derivative, you dont need it

[half berry]

ok

[leaden glen]

alr let me walk u step by step from the beginning

[half berry]

so then it looks like this

i think

[leaden glen]

from right there

yeah ur right

but its a fraction

so over (x^2+1)^2

[half berry]

yeah

[leaden glen]

and there u go thats the answer

[half berry]

oh

[leaden glen]

if u need to simplify, then u can factor out the e^(3x)

and then u will have the same answer i got

[half berry]

yeah can u show how to simplify it

do i just like

[leaden glen]

first take out the e^(3x)

and show me what u get

[half berry]

like leave it like this and take it out?

[leaden glen]

it should just be multiplying it out once u factor the e^3x

[leaden glen]

factor out the e^3x

so like:

e^(3x) [ 3(x^2 + 1) - 2x ]

and then multiply everything out in the inside

e^3x [ 3x^2 + 3 - 2x]

and thats the final answer

e^3x [ 3x^2 + 3 - 2x] / (x^2 + 1)^2

[half berry]

ohh

i understand

okay

[leaden glen]

yep there u go

[half berry]

i see how u got it now

ty

sm

fr

[leaden glen]

np

[half berry]

.close

[safe radish]

Channel closed

Closed by @half berry

Use .reopen if this was a mistake.

[safe radish]

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[quick crater]

find the minimum value of n if n is a positive integer and $3>=\frac{n}{x}>=7$ where x is also an integer and has exactly 35 possible values

[quick crater]

i dont even know where to stsrt

[flat frigate]

Skill_Issue

[solar hazel]

wait

what

are you sure this is stated right

how can n/x be greater than or equal to 7 and less than or equal to 3

[dusk gate]

They're both greater-than symbols..

[solar hazel]

i know

do you see the 3 >= 7

[dusk gate]

I think something in there's backwards

That sort of notation works with less-than symbols..

[dusk gate]

No, yeah.. I'm agreeing with you..

seems like OP ditched, anyways, though

[safe radish]

@quick crater Has your question been resolved?

[quick crater]

sorry, i had to do something

[quick crater]

oops, sorry i switched them around, it supposed to be 3 <= n/x =< 7

[solar hazel]

ok

i would start by putting 3x <= n and n <= 7x

or equivalently x <= n/3 and x >= n/7

so now fix an n

and ask how many integers x satisfy those

ceil(n/7) is the smallest, right?

ceil(n/7) + 1 is the next, and so on, until ceil(n/7) + k = floor(n/3)

for some k

working with ceiling might be a little tough

maybe ignore it to start

[quick crater]

uh then what

[solar hazel]

btw i haven't done the problem yet, i'm just giving ideas
write n/7, n/7 + 1, n/7 + 2, ... as
(n+7)/7, (n+7)/7, (n+14)/7,...

and we want to know how long this list is without going over n/3 = (7/3*n)/7

so in other words what's the largest k such that n + 7k <= 7n/3

which is... floor(4n/21)?

so i think if we fix n, there should be approximately floor(4n/21) + 1 satisfactory x's

[solar hazel]

with it only being approximate because of this

smallest n with floor(4n/21) + 1 = 35 is 179

i conjecture the answer is close to that

[gray sentinel]

for polynomial A = 3x^4 - 5x^2 - 3x + 7 - 2x^4 + 4x^2
a. Collapse and find the degree of A
b. Find polynomial B such that A+B= 2x^2 +1
c. Calculate the value of polynomial A when x=1/2

help

[solar hazel]

!occupied

[safe radish]

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

[quick crater]

ok ngl my brain is not at the capacity to do this rn, so i think ill close this and ask later at a diffrent time

[solar hazel]

wait i wanna do a little more

[quick crater]

k

[gray sentinel]

help me

[quick crater]

!occupied

[safe radish]

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

[solar hazel]

let's check it exactly for 179 just for fun...
how many integers x satisfy 3x <= 179 <= 7x?
x = 26 is smallest and x = 59 is largest i think

hmmmm so 34 x's

that's pretty close

[eternal carbon]

34*

[solar hazel]

oops yea

[solar hazel]

ok so actually this did as much as i was hoping for from my silly method

the answer should be just a little bigger than 179

guess and check to find it is reasonable probably

for 180...
largest is 60 and smallest is 26?

so that does it i think

answer is 180

[gray sentinel]

So Is that accurate

[solar hazel]

wdym

[gray sentinel]

What do you mean?

[solar hazel]

what are you asking?

[gray sentinel]

This

[solar hazel]

dawg

nothing i said is related to your problem

[solar hazel]

@quick crater if you have the answer available i'm curious if this is right

[quick crater]

i dont sadly

[solar hazel]

ok the idea was: estimate the answer to be near n = 179, check how many x's there are for n = 179, see it's 34, go up until you find an n that makes 35 satisfactory x's

and 180 did it

[quick crater]

oh ok

[solar hazel]

you can close if you want tho

[quick crater]

thank you btw for your time

.close

[safe radish]

Channel closed

Closed by @quick crater

Use .reopen if this was a mistake.

[solar hazel]

[solar hazel]

where is the problem from?

[quick crater]

grade 8 olympiad

[solar hazel]

phew, i can feel at least as smart as the 8th graders

[quick crater]

.reopen

[safe radish]

✅

[quick crater]

is there a good way to find the distance between a line and a point

[delicate apex]

projection

[eternal carbon]

many exist

[quick crater]

my way from logic is to find the ewuation for the perpendicular that passes trough the point, then find the intersection point then find the distance with the sqrt((x1-x2)^2+(y1+y2)^2)

[eternal carbon]

if ur given point and equation of line, there is formula

[quick crater]

can you tell me it?

[eternal carbon]

,

[quick crater]

uh ok

thanks

.close

[safe radish]

Channel closed

Closed by @quick crater

Use .reopen if this was a mistake.

[safe radish]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[main mural]

bruh

lol

claim a new channel

[hollow iris]

oops

ok

[safe radish]

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[lean otter]

help

[lean otter]

the stuck is gemeotry

after i draw the vertical 90 degree line from G straight down

tehn what

then**

<@&286206848099549185>

[lone arch]

What does 25 mean

Area?

Length?

[lean otter]

no

the length of the perpindicular

its a trapezoid

[normal harness]

info is not clear

can you articulate all the availabe infos?

[lean otter]

there is a trapezoid defined by GHAF

AFGH I am asked to find side GH

the perpindicular length of the trapezoid is 25

it is too hard

ya H = 25

[normal harness]

to clculate the length of GH in the trapezoid GHAF
height of trapezoid is 25m
and H is 104 20 10

we can use Cosine of angle
and we are dealing with external right triangle. therefore
theta = 180 - 104 20 10
and I get roughly 100.97

[lean otter]

sorry if that makes it more clear

[normal harness]

question

[lean otter]

the angle is in dms

[normal harness]

literraly GH = 25 * cos(theta)

[lean otter]

im getting 6.190

[normal harness]

show me how you got there work?

[lean otter]

you said 25* cos ( theta)

[normal harness]

the given angle is 104 degree 20 min 10 seconds

correct?

[lean otter]

ya

no its

180 - 104 degree 20 min 10 seconds

that is how i got theta

[normal harness]

yes I did that
and cos(theta) = GH/25
and GH = 25 * cos(theta)

[lean otter]

you got 100

[normal harness]

that is range of caculator

yes

I got roughly 100.97 meters

[lean otter]

is theta 75 degrees 39 mins 50 seconds

[normal harness]

yes

[lean otter]

why did i get 6.19

[normal harness]

what is your calculator mode?

[lean otter]

degrees

[normal harness]

can you try radian?

or I can try degree

equation is right tho. I still get 100.97

[lean otter]

can you show me your work

again

[normal harness]

first we need to find the theta
theta = 180 - (104 + 20/66 + 10/3600)
so its 75 39 50
height of trapezoid is 25M
and with these 2 evidence we can use cosine of Angle
which GH / Hypotenuse
since cos(theta) = GH/25 under rule of cosine of Angle
GH = 25 * cos(theta)

[lean otter]

but cos theta is

0.247

you mean 25 divided by cos theta

anyway that number is too big

[normal harness]

this is just my calculator Issue my bad

I dont know how it got 100.97 m at the first place but its 6.19

[lean otter]

can you explain

how to find that interior angle

[normal harness]

are you asking interior angle of H ?

[lean otter]

ya

[normal harness]

if you asking me back tracking

you need theta arc cos(GH/25)

[lean otter]

shouldnt the lenght be technically longer then the height

since its a hyp

[normal harness]

I assume 25 is the hypotenuse

assume GH as adjacent

[lean otter]

i want the other assumption

GH is hyp

[normal harness]

let me claify given information.
external angle is still 104

[lean otter]

yes

[normal harness]

ghaf is still trapezoid

[lean otter]

ya

[normal harness]

and just gh is hyp

im finding adjacent ?

[lean otter]

ya

[normal harness]

I will leave this to someone its 4 am to me tho, if nobody solve it by tmrw, Ill show the work for you

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[summer dock]

I get a negative value when I calcualte SA of revolution

[summer dock]

This is the integral

3.3365.94-0.0821515x3+0.859522x2-2.93095x+7.55275 1+(-0.2464545x2+1.719044x-2.93095)2 dx=35.7664

bruh

pi * Integral from 5.94 to 3.336 of ((0.0821515 x^3 + 0.859522x^2 - 2.93095x + 7.55275)* (1+ (-0.2464545 x^2 +1.719044x-2.93095)^2)^1/2 dx

Yeah its a pain in the ass, I solved it using a GDC

But i'm getting a negative value

[sleek hornet]

[summer dock]

Not sure why, anyone help me out

[summer dock]

I know.

[safe radish]

@summer dock Has your question been resolved?

[safe radish]

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[opaque pelican]

question about dy/dx

is it fair to say that dy/dx is the same as d/dx(y)

(d/dx)[y]

[umbral swan]

ys

leibniz notation

[opaque pelican]

ok so

in this problem

do i solve for y?

[umbral swan]

you want to find dy/dx

you derive both sides by d/dx

then solve for dy/dx

and insert x, y to get the value

[opaque pelican]

Ok so when i take these derivatives, i will get a dy/dx spat out at me because of the -3y in the sin function correct?

[umbral swan]

ys

[opaque pelican]

and then i just solve for dy/dx and im done

[umbral swan]

ys

[opaque pelican]

thank u so much

[umbral swan]

np

leave it open if you want to have it checked

[opaque pelican]

all good, seems legit

[safe radish]

@opaque pelican Has your question been resolved?

[summer dock]

.reopen

bruh