help-19

[onyx sky]

"
​(Type integers or decimals rounded to one decimal place as​ needed.)"

so 6.7?

now how to find y

[chilly narwhal]

Ah, because I was worried when I got an irrational answer

[chilly narwhal]

same thing, just that now you found what x is. remember use the exact value of x, not just 6.7 in your working

[onyx sky]

giuve me a sec

approximately 9.7.?

@chilly narwhal

[chilly narwhal]

uh, did you solve y(y+6) = 45 to be that?

I don’t have a calculator with me

[chilly narwhal]

sorry, I meant 45, not root(45)

[onyx sky]

correct?

[chilly narwhal]

No

[onyx sky]

whats wrong here

[chilly narwhal]

you have to use the same tangent line, and a completely different secant instead

at the end, you should have an equation for y where you solve it

[onyx sky]

approximately 8.7?

@chilly narwhal

[chilly narwhal]

is it fine if you could show you working this time again?

[onyx sky]

(secant length 1)×(external segment 1)=(secant length 2)×(external segment 2)
(6+6.7)×6=y to the power of 2

12.7×6=y to the power of 2

76.2 = 𝑦 to the power of 2

y= square root 76.2 ≈8.72

@chilly narwhal

[chilly narwhal]

I don’t think this is correct either

just look at the left secant and the same tangent line

And also like I said before, just to have a more accurate answer, use root(45) in your working out rather than 6.7

[onyx sky]

i cant seem to find it, can you provide me with the answer so i can see the explanation from thenm

[chilly narwhal]

You should get the equation y(y+6) = 45. Then just solve for y

[onyx sky]

4.35?

@chilly narwhal

[chilly narwhal]

Looks like it, I don’t have a calculator with me but you should get root(54)-3

[onyx sky]

@chilly narwhal

i have one more try

[chilly narwhal]

You didn’t round this properly it seems

You rounded it down, it should be rounded up to 4.4, now maybe try

[onyx sky]

damn it was 1.5

damn bro now they gave me a whole different one with different numbers to do

[chilly narwhal]

that was the answer?

it cannot be

[chilly narwhal]

just apply the same theorem again

[onyx sky]

yep it was

according to savaas

[chilly narwhal]

whut, no way

[onyx sky]

yh

so what should i do

this is an example they gave to solve likle this

[chilly narwhal]

Omg I’m so sorry, i made a terrible mistake. It was just 6(y+6), not y(y+6)

I’m so clumsy and stupid

My sincere apologies

[onyx sky]

nooo its fine

i have one more attempt but a different numbers

[chilly narwhal]

Alright, this time apply the same thing, and show your working out to me when your done

[onyx sky]

6(y+6) = 45 for y yes?

[onyx sky]

is x 19

[chilly narwhal]

Just remember to apply the theorem properly, and make sure you are doing so, otherwise I might screw you over again. Tbf it’s night over here

[chilly narwhal]

That’s for the previous problem. I should’ve said that instead

[onyx sky]

so here what is it

[onyx sky]

im guessing x is 19 right

[chilly narwhal]

No

x is not 19

[onyx sky]

ahh

its 4.9

[chilly narwhal]

Yes, if that’s what root(24) is

[onyx sky]

wdym by that

[chilly narwhal]

Like square root of 24

[onyx sky]

Ah yes

correct

so now what formula i use for y

[chilly narwhal]

Alright, now find y in the same fashion

[chilly narwhal]

Same formula, use power of a point theorem

[onyx sky]

6.6?

[chilly narwhal]

No

Try again, but this time do not make the mistake I did

[onyx sky]

where my mistake

[chilly narwhal]

The tangent length is x here, and the full secant is the y+4. And remember to write x as the square root of 24, not just 4.9

You should come to a very clean answer btw

[onyx sky]

is it 2?

[chilly narwhal]

Yes

[onyx sky]

YESSS

FINALLY

[chilly narwhal]

yayyy

[onyx sky]

NOW I GOTTA SOLVE 3 more of these

n im done

[onyx sky]

tysm

[chilly narwhal]

No problem :)))

Im sorry for causing a big inconvenience first time round though

[onyx sky]

is this one x≈13.8
y≈14.7

wait lemee double check

[chilly narwhal]

Let me check

[chilly narwhal]

No

Try again properly

[onyx sky]

both incorrect?

is x 10.9

[chilly narwhal]

Yes, both were incorrect

[onyx sky]

?\

[chilly narwhal]

No

Hii, I’m sorry but I have to go now. Just remember to use the Power of a Point theorem correctly, just like last time

[odd edge]

@onyx sky Has your question been resolved?

[onyx sky]

ive applied the pop theorom here, but still incorrect <@&286206848099549185>

?

huh

[whole gyro]

looking at the smaller circle

9 . (9+y) = x^2

and for the bigger circle

6 . (6+15) = x^2

by applying that theorem

[whole gyro]

this one

from this you can get the values of x and y

[onyx sky]

im solcving this one now

[whole gyro]

yeah im talking about that

but you use the same theorem as the one i referenced

[onyx sky]

y x (9+y) = x^2 for x
15 x (6+15) = x^2 for y

?

[whole gyro]

made some edits to the equations

[onyx sky]

for this one

[whole gyro]

check the equations i sent again

[whole gyro]

ik

[onyx sky]

ah

9 times (9+y) = x^2 for x

6 times (6+15) = x^2 for y

this?

[whole gyro]

yup

[onyx sky]

leme try wait

x≈11.2

y=5

[whole gyro]

yes

[onyx sky]

wowww correct

this is a different formula right?

[whole gyro]

no

[onyx sky]

this

look

[whole gyro]

i mean this is for angles

and that theorem is for lengths

[onyx sky]

yh ik

what formula do i use here

[whole gyro]

lemme have a look

[onyx sky]

okok

@whole gyro

[whole gyro]

ok igot it

sory was away

first

label the centre of the circle as point O

then label the point the 72 is on as say M

then we know

ROS = 90

and VOU = 82

and ROV = 2.RMV (angle at centre is twice angle at circumference) = 2 . 72 = 144

thus we can calculate SOU = 360 - ROS - VOU - ROV

= 360 - 90 - 82 - 144

[onyx sky]

44?

[whole gyro]

YES

sorry cap[s

[whole gyro]

then we get that MRS + MVU = 1/2 MOS + 1/2 MOU

no thats not thw answer bro

[onyx sky]

oh thats not the final

[whole gyro]

im not done

[onyx sky]

lol

its ok

i have

3 more

tries

[whole gyro]

by the same theorem i stated

thus MRS + MVU = 1/2(SOU)

but we know SOU

44

so MRS + MVU = 22

now notice the quadrilateralRMVT

sum of angles is 360

so MRS + MVU + reflex RMV + RTV = 360

=> 22 + 288 + RTV = 360

thus RTV = 50

BUT

Try to understand before putting in the answer

[onyx sky]

Ok

[whole gyro]

and i may have made mistakes too

[onyx sky]

ur right

i double checked

[whole gyro]

did u understand the answer

[onyx sky]

how does one acheive this smartness

[whole gyro]

its not smart bro

its monkey brain trial and error

try to explore these questions a bit further

[onyx sky]

i got one more

to solve

lemee see

if its different

look

[whole gyro]

bruh

u can do it

recall the first problem we did

[onyx sky]

this formula?
y x (9+y) = x^2 for x
15 x (6+15) = x^2 for y

but diff numbers

right

[whole gyro]

yup

[onyx sky]

which numbers should i inject into this formula

[whole gyro]

use the theorem stated in the first pic u sent here

[onyx sky]

lemee see one sec

x×(x+21)=y to the power of 2

[whole gyro]

close

but why x * (x+21)?

if u look at the line the x is part of

external sement (the length of the line outside the circle) = 5

full secant length (total length of the line) = x + 5

so it should be 5 * (x+5) = y^2

[onyx sky]

oh

this is for x?

[whole gyro]

sorry i meant 5 * (x+5)

this formula is between x and y

then u can look at the y line and the 3rd line

and write the formula

[onyx sky]

one sec

so now tha ti have this

for x and y

whats the next step

[whole gyro]

ok

so how you wrote this formula for the 1st and 2nd line

now write a formula that concerns the 1st and 3rd line

[onyx sky]

6×(6+15)=x ^2

[whole gyro]

first of all

x is not a tangent

2nd of all

where did u get 15?

3rd of all

i was talking about the 1st (the one with y) and the 3rd (one with21 and 6) line

cus the formula only applies to a tangent and a secant\

Okay, ill walk you through the steps

[onyx sky]

y(y+9)=162
?

[whole gyro]

.

[onyx sky]

yh maybe thats better

[whole gyro]

ok so

we are considering the 1st and 3rd line

the 1st line is thee tangent = y

for the 3rd line

the orange part is the external segment = 6

and the entire brown part is the full secant = 21 + 6

so in the formula, (external segment) . (full secant length) = (tangent length)^2

we susbtitute these values

and 6 . (21+6) = y^2

@onyx sky

[onyx sky]

Okay

[whole gyro]

did u get it?

[onyx sky]

this is for which value

[whole gyro]

wdym which value

[onyx sky]

what are we finding here

y correct?

[whole gyro]

ok we need to equations to get x and y

[onyx sky]

yes

[whole gyro]

the first equation we got was 5 * (x+5) = y^2

and now with the second equation 6 * (21+6) = y^2

[onyx sky]

correct

[whole gyro]

we can solve for x and y

[onyx sky]

lemee try

x≈27.4

y≈12.7

[whole gyro]

see if it works

[onyx sky]

yessss

correct

[whole gyro]

important thing is you understood

[onyx sky]

yes thank u

i screenshoted all and will continue practicing

thanks broski

[whole gyro]

[onyx sky]

.close

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[gray hare]

How do I calculate the bounds for polar integrals?

[gray hare]

I understand where 4 cos theta and 2 cos theta are coming from, but I don't understand where 0 and pi are coming from

[olive needle]

It depends on the region being integrated. What was the original question?

[gray hare]

And here's what I've got so far

<@&286206848099549185> ?

[mystic saffron]

what?

[gray hare]

Trying to figure out how to calculate the bounds for polar integrals

Specifically, 0 and pi

The example problem I'm working through simply states what they are and doesn't show how they're obtained

[potent folio]

think about where the functions r = 2cos ϴ and r = 4sin ϴ are when ϴ = π/2

they both correspond to r = 0

oops

4cos ϴ*

so you could also integrate from 0 to π/2 and double your answer

[gray hare]

Because the value of cos at pi/2 = 0... but if that's the case, why aren't we using 3pi/2?

And where is 0 coming from.. because cos (0) = 1, which means our values are 2 and 4

[potent folio]

this is the graph when ϴ ranges from 0 to π/2

and this is from π/2 to π

so from 0 to π covers the entire region

[gray hare]

ok.. that makes sense...

my only question at this point is.. how do I figure out what the graph is going to look like? Because come test time, all we have is pencil and paper.. no calculators, no desmos

[olive needle]

You don't need to worry about what the graph looks like. Just worry about the region being intergrated looks like.

[gray hare]

Replace my words with yours, then...

I still don't know what the region is going to look like

[olive needle]

Have you learned how to find the intersections of polar curves?

[gray hare]

(which is why in my original question, I stated I'm trying to calculate the bounds)

[gray hare]

Yep.. got 2cos and 4cos..

[olive needle]

It's the r^2 that gets overlooked.

[gray hare]

I didn't overlook it.

Here's where I calculated 2cos and 4cos already

(hence why my original question is asking about 0 and pi)

[olive needle]

2cos(theta) only equals 4cos(theta) when cos(theta) = 0.

Does that make sense?

[gray hare]

Is that the same 0 that becomes the lower bound 0?

Yes, it does

the only way to eliminate coefficients is for the variable to be 0..

at least in a case like this

[olive needle]

Happens a lot of with polar equations.

[gray hare]

So.. that's the 0?

[olive needle]

Yes.

[gray hare]

Awesome. And pi?

Because cos pi = -1, and... the equations aren't equal at that point

[olive needle]

I'll be honest with you, they picked a weird bound. It wouldn't be my first choice.

[gray hare]

I mean.. looking at the graph, it's where the circles are complete

it makes perfect sense, graphically

I just don't know how to calculate where that "completeness" is

Normally a circle is complete at 2pi

[olive needle]

The tricky party with what they did is that the upperbound is when r is negative.

Which is why I said they picked on odd boundary.

[gray hare]

Is it?

[olive needle]

One moment and let me make a graph that will visually show what I mean.

[gray hare]

Wait.. This isn't true

[olive needle]

https://www.geogebra.org/classic/dk3tmcx4

[gray hare]

cos(0) = 1... which means we'd have 2 = 4 at cos(0)

[olive needle]

And this is where the r^2 comes into play.

[gray hare]

cos(pi/2) = 0, which is where cos = 0 and 2cos = 4cos

[olive needle]

What r should equal is ±2cos(theta) and ±4cos(theta).

[gray hare]

How do you figure? Since we never took the square root...

one of the rs cancels out.. no square root is taken

[olive needle]

r^2 = 2r cos(theta)

[gray hare]

hence, no +/-

Divide both sides by r

r = 2 cos

not +- 2cos

[olive needle]

You're right. My bad.

I had trouble with polar regions as well. 😛

Especially when you had to consider negative radii.

[gray hare]

At any rate, the 2cos and 4cos are pretty easy to obtain.. we're just converting the rectangular formulae to polar ones

I'm sure the 0 and pi are easy, too.. I just don't know what route to take

[olive needle]

I would have chosen pi/2 and 3pi/2 for bounds because it avoids the negative radius.

[gray hare]

what negative radius?

[olive needle]

That negative radius.

In the red segment, the radius is negative. In the green segment, the radius is positve.

Theta points in the green direction, but because the radius is negative, it ends up on the red direction.

[gray hare]

So how do you calculate pi/2 and 3pi/2?

[olive needle]

That's just from when cos(theta) = 0.

[gray hare]

So.. if you do that...

.. you're going to get a different answer

(at least.. looking at it, it seems like we would get a different answer)

Which.. actually no.. it's the same

[olive needle]

I just realized I'm showing you incorrectly. You're doing a double integral. 😛

[gray hare]

Either way, the bounds are coming from the same equations

[olive needle]

Anywhos, the r values you found will be the upper and lower bounds.

[gray hare]

Seems like that works.. thanks

I'm wondering if they didn't just take the difference between pi/2 and 3pi/2 and "slide" it back so the lower bound is 0

Easier to calculate, since it's just (upper bound integrand) - 0

Any ways.. thanks

[olive needle]

yw, all roads lead to Rome and we got there eventually. 🙂

[gray hare]

.close

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[lethal galleon]

Siggi the carpenter undertakes to do work for ISK 382,500. with value added tax which
is 24.5%. What is Sigga's salary minus the tax?
Answer: ___________________

[lethal galleon]

we have to add 24.5%

x * 0,775=382.500 kr

[grizzled tide]

,calc 100-24.5

[clever fjord]

Result:

75.5

[grizzled tide]

but anyways it's x(1+0.245) = 382000

when you pay tax you pay the original (the 1) and the tax (the 0.245)

[odd edge]

@lethal galleon Has your question been resolved?

[lethal galleon]

.close

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[next mica]

The limit when directly evaluating is 1^inf

therefore, I know I can use l'hopitals rule, but I just dont know how to apply the natural log to solve this

I get ln(y) = (1/x)ln(1-4x)

How would I find the limit of this then?

[viscid flint]

hmmm

really is a shame that the exponent is 1/x and not 1/4x

[next mica]

yeah

[viscid flint]

you know, 1/4x is equal to 1/4 * 1/x

[next mica]

yes

how exactly will that help me

[viscid flint]

i guess maybe it would be better to say that 1/x is 4 times 1/4x

$\lim_{x\to0} (1-4x)^{\f1{4x}\cdot4}$

[clever fjord]

hayley is stateside!!

[sterile saddle]

how do u solve gradient

[viscid flint]

can you manipulate that expression to make a piece of it match the form for e?

[viscid flint]

alternatively you can just l'h this

[odd edge]

@next mica Has your question been resolved?

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[feral bramble]

just need to know where to start w these

[whole gate]

for first one you can show b1 and b2 as multiples of d

and show that d divides b1 + b2

[feral bramble]

okok

[whole gate]

for second one maybe use odd * even = even thing? idk if u can use that

[sand horizon]

For second one do two cases

One where x is even and another where x is odd

And use product with congruence

[feral bramble]

sorry what is product with congruence lmao

[sand horizon]

Sorry for that goofy gandlaf top right

[whole gate]

lmao

[feral bramble]

ohh ok

tysm guys for helping

[sand horizon]

Yw

[feral bramble]

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[crisp flume]

Find the largest possible ( n ) for which it is possible to divide the set ( {1, 2, \dots, n} ) into 5 non-empty subsets such that the numbers in each subset are pairwise coprime.

[clever fjord]

Slowaq

[crisp flume]

is it 11?

because for n greater than 11 in some subset there are always going to be at least two numbers that are divisible by 2

[fallen surge]

i think that's right

just need to find an example for 11

i think ||(1,2,3,5,7,11), (4), (6), (8), (9,10)|| works

[crisp flume]

Yea thanks

[odd edge]

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[plucky coyote]

Can someone explain q)d with info given

From first image

Help

[median bear]

If its a square then AQ is parallel to PB so a line that passes through AQ would have the same gradient as l1 since that passes through PB

then to find the y intercept i guess youd need to work out either point P or B

[plucky coyote]

How would i find p

[median bear]

i cant lie the only way i can think of right now is by using the discriminant because l1 touches the circle at only one point so if you sub the equation mx + b (for line l1, you already know m so you should sub that in) into the equation for the circle then expand and find an equation for the discriminant of the quadratic you get

y = mx+b*

[plucky coyote]

Calm that makes sense

[median bear]

the discriminant will be equal to 0

[plucky coyote]

So i have to find b

[median bear]

because theres only one solution

since theres one point

[median bear]

nah this will get you b directly

[plucky coyote]

Yh

[median bear]

its the alternative if you dont know P or B coordinates

When you expand the equation of the circle with y = mx+b substituted in

[dim sandal]

Do we not know the gradient m already?

[median bear]

yeah this is to find b

[plucky coyote]

The gradient is 1/3

[median bear]

the y intercept

[dim sandal]

AQ is parallel to PB

[median bear]

i feel like im missing something really obvious to find the y intercept

[plucky coyote]

(0,y)

[median bear]

but thats the only way i can think of

[plucky coyote]

Sub into l1 to find y intercept

[median bear]

yeah

but theres probably a faster way

i just cant think of it

do u have the coordinates for E?

[plucky coyote]

No but ur way makes sense

This is the markscheme for it

I dont get this way

[median bear]

oh right its because its a square so the vector AQ is perpendicular to AP

if youre going 6 to the right and 2 up to go from A to Q then going from A to P is going 2 to the left and 6 up

thats the thing i felt like i was forgetting

i mean you can still do it the way i suggested but itd take way longer

[plucky coyote]

So i flip the coordinates for point Q

[median bear]

yeah the diagram helps to picture it

Q to B would be the same vector as A to P

so technically you couldve also done it that way and found B

and then found the y intercept

[plucky coyote]

Yh that makes sense

With your way i got a weird answer idk if i made a mistake

Nvm

Thanks

.close()

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[reef zealot]

this is apractice problem and the answer is 47.729435 and im im not sure what process they took, so Ik the anti derivative is 0.6 arctan(t)

[odd edge]

@reef zealot Has your question been resolved?

[nimble blaze]

you'd want to first convert R(t) to
million gallons / hour

[wanton bison]

Why would we not get the same amount if we said t = 4 × 60 min = 4 hours instead?

[nimble blaze]

the function uses the value of time in hours

to use that approach, you would keep the rate as gal/min
but change the input to use minutes instead of hours

[wanton bison]

I see thx

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[opaque wasp]

whats the best way to solve this using the inverse laplase

[opaque wasp]

doing this and solving to get the a's

is it the best solution or is there a simpler way because feels like a bunch more work than previous exercises so I think i might be approaching it wrong

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[opaque wasp]

<@&286206848099549185>

[mystic saffron]

what?

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@opaque wasp Has your question been resolved?

[grand swift]

Hi

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[pearl smelt]

D?

[cursive field]

yes

[pearl smelt]

why

[inland rover]

profit is revenue-cost

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[mystic saffron]

How to prove a process has independent increments?

[low locust]

context?

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[lament estuary]

This is one of the question in my teacher previous exams
The answer of it is b) which is e^1/5
But I dont know solve it
My problem is I dont know how to get ride of the multiple n as exponent and even how the heck do I even find the constant e in it
Can someone whitout telling me exactly how to do it (I still want to try on my own) which theorem to use for this question?

[viscid flint]

well first thing i see is that n^n 5^n = (5n)^n

[warped glacier]

ah try using the limit definition of e^n

[viscid flint]

not sure how helpful that is but it's something

[sand horizon]

Write it as (a/b)^n

[warped glacier]

you can simplify $a_n$

[clever fjord]

south's secret twin brother

[sand horizon]

Bro got the totally spies helpers fr

[lament estuary]

wait please slow down you guys confusing me ;-; but thank you very much

[sand horizon]

Look at first message of each others

[rapid kestrel]

I personally was taught to do it like this

[lament estuary]

is this correct?

i think i got it

[rapid kestrel]

Yeah

[lament estuary]

oh i didnt knew we could use hospital theorem

ty

Big thank you for you all

it help me a lot

.close

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[sick spire]

used sophie-germain and made them into partial fractions (because the previous parts of the question guided you through that) and im stuck on this final part now

[sick spire]

you get this as the partial fraction expansion of this

[chilly narwhal]

I think you can simplify the right picture even further

[sick spire]

the left is the simplified version of the right

sorry i pasted it in the wrong way

[warped glacier]

try expanding 4(k + 1)^2 + 1

[sick spire]

i know thats the denominator but i dont see how that helps

[warped glacier]

and similarly 4(k + 1)^2 + 8(k + 1) + 5 becomes 4(k + 2)^2 + 1

it telescopes!

[sick spire]

wait acc

[warped glacier]

yeah

[sick spire]

oh i didn't spot that cause i made an addition error 💀

tysm

[warped glacier]

no worries

[sick spire]

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[spark vessel]

What is the result of $\int {0}^{1}\lim{ n \to \infty } nx^{n}\mathbb{1}_{[0,1)}(x) , dx$

[clever fjord]

Rootsyl

[spark vessel]

oof the mbb is broken i see

its the indicator function for x

[lavish kite]

$\mathbb{1}$

[clever fjord]

bee [it/its]

[lavish kite]

$\bold{1}$

[clever fjord]

bee [it/its]

[lavish kite]

there

mathbb is for stuff like $\mathbb{N}$ and i have no idea how you'd do that to $1$ so it kind of makes sense that it does something else instead, although i don't know why $\mathbb{1}$ in particular

[clever fjord]

bee [it/its]

[spark vessel]

you know that its not the same xD

but ok i use bold

[lavish kite]

...oh huh yeah idk how to latex that symbol

[spark vessel]

$f_{n}=nx^{n}\mathbb{1}_{[0,1)}(x), n \in \mathbb{N}$

[clever fjord]

Rootsyl

[spark vessel]

it works in my obsidian

welp

What is the result of $\int_{0}^{1}\lim_{ n \to \infty } nx^{n}\bold{1}_{[0,1)}(x), dx$

[clever fjord]

Rootsyl

[lavish kite]

$\int_0^1 \lim_{n\to\infty} nx^n \bold{1}_{[0,1)}(x) \dd x$

[clever fjord]

bee [it/its]

[lavish kite]

...wow this is a weird integral

[spark vessel]

the integral isnt the issue

the limit is weird

[lavish kite]

i mean i think that just gives you 0?

[spark vessel]

i would say so as well since x is defined on 0 to 1 and the ratio test for the series version of $nx^n$ also gives this

[clever fjord]

Rootsyl

[spark vessel]

but anywhere else on the real line the limit is not defined

its not even divergent

directly UNDEFINED

[lavish kite]

well no the limit would still be zero actually, because the indicator function would give zero

[spark vessel]

i mean yea thats why i said anywhere else on the real line

[lavish kite]

if you ignore that then for negative x it does crazy things (for even n it goes to +inf and for odd n it goes to -inf so overall it doesn't converge to anything) and at x >= 1 it goes to +inf

...so yeah i'm pretty sure the answer is just 0?

[spark vessel]

but the ratio test tell us that the terms converge to abs(x)

if x != 0

then wouldnt the sum be different?

[lavish kite]

well the ratios between the terms converge to |x|

the terms still converge to 0

try taking like x = 0.9 and compute 1000 * 0.9^1000

you get about 10^-43, it's clearly approaching 0

[spark vessel]

yea thats basically 0

ok

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[shadow pilot]

im having trouble drawing sin and cosin functions in radians

[shadow pilot]

should i do two drawings? a first with A and C, and a second where i add H and K? or other way around

[steady tide]

!xy

[odd edge]

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.

[shadow pilot]

for my period

would it be 4 pi

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[frail nova]

someone help me with this

tomorrow exam

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[rancid agate]

Hello! This is an old question from a previous exam and I need help understanding the problem. I understand Newton's Method but I don't understand the problem and I'm not really sure how to approach this type of question. I'm also not really sure what to look at in the graph itself. If someone could explain a step by step solution and approach to the question I would be really grateful.

[odd edge]

@rancid agate Has your question been resolved?

[brittle beacon]

Notice how the tangent to y = h(x) at x = f seems to intersect the x axis at x = a, and it appears the tangent to y = h(x) at x = a seems to meet the x axis at x = b…

Can you state Newton’s method and what it says?

[rancid agate]

ok the equation for Newton's method is xn+1=xn-f(xn)/f'(xn) and in my understanding its a way to approximate the zeros by repeating the formula untill you get closer

is it that the tangent line will cross the x axis closer to the root so when the tangent at x=f crosses the x axis that means that its a new "guess" for the root but at x=a the tangent line crosses at x=b so would that mean that the tangent line at x=a is closer than at x=f?

because I assume they guessed the point at x=f first which would be the x1 then they solved for x=a which would be the x2

so then if

x1 is f
x2 is a
the x3 would be b?

so then would it converge at c because thats where the zero is?

ayyy I got it tysm @brittle beacon ! I understand it now

[brittle beacon]

thought that would “instantly” make it clear

[rancid agate]

: D

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[odd edge]

@steady dirge Has your question been resolved?

[wispy cliff]

Honestly i dont understand english very well but i believe the line P is where it meets with y that s why the points where it meets it are the solutions which are x= -1 and x=5 so take the point where the curve is at these 2 points and link them, then u ll get the y coordinates and be able to calculate m and so be able to calculate c

[robust wyvern]

wait so a line P means a curve P too?

[wispy cliff]

No the line meets the curve at these 2 points

[steady dirge]

u have to create a straight line using the co ordinates

[wispy cliff]

That s why we call them solutions

[robust wyvern]

ohhh

[robust wyvern]

way to go, leroy

[olive needle]

I know English and the wording on that question was confusing even to me.

[robust wyvern]

me too

[steady dirge]

so basically im finished 😦

[wispy cliff]

....

U got 1 question at least

[steady dirge]

this is the video to it

[wispy cliff]

B cannot be solved without a

[steady dirge]

but idk what part a is

[olive needle]

The only issue that there are an infinite number of quadratic equations that can go through two points. You need a third constraint to determine a partial equation.

[wispy cliff]

That s what i told you 🤷🏻‍♀️

But it s not the same

[wispy cliff]

They changed the numbers

[wispy cliff]

I dont think they study that?

[steady dirge]

i have no clue what the answer is

it doesnt make sense at all

[robust wyvern]

bruh

[wispy cliff]

What do you not understand?

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[remote hare]

i am confuse about the third step of the function

[boreal crag]

I didn't even know functions had steps

[remote hare]

i mean the note's third step

2/3+x*ln2=0

how to get this

[olive needle]

It looks like they mutiplied by x^(1/3).

[remote hare]

oh

i see

than you

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[meager monolith]

Apply menelaus theorem

[chrome stump]

bro i have no clue what that is

[meager monolith]

Search it up if you want to learn more, but in this case, it means (BD/BC)(CE/EA)(AZ/ZD)=1

[chrome stump]

okay thanks

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[scarlet pond]

I need

[meager monolith]

you need?

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[gaunt zephyr]

i need to check my moms answer for the question "a cafe makes 275 dollars by selling 60 drinks, lattes are 4.25 and cappucinos are 5.5. how many of each did they sell?". my mom got 16 cappucinos and 44 lattes. if this isnt the answer can someone give me an indepth guide on how to get the real answer?

[steady tide]

this is correct

[gaunt zephyr]

can you still give me the breakdown on how?

[steady tide]

let x be the number of cappucinos sold, y be the number of lattes sold

the cafe sold a total of 60 drinks, so x + y = 60

the cafe made 275 dollars, so 5.5x + 4.25y = 275

this is now a system of linear equations. solve for x and y

[gaunt zephyr]

i multiply the top one by 5.5 and the other by -1

right?

[steady tide]

yeah

thats the general procedure in solving for system of equations; get rid of 1 variable

[gaunt zephyr]

either that or the infinitely harder substitution

thanks

[steady tide]

!done

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[timber fiber]

time for my hourly desperate scrambling for help about calculus

[timber fiber]

information:
dr/dt = 2 cm/h
dh/dt = 2 cm/h
r = 6
h = 15
V = (1/3) x pi x r^2 x h
dV/dt = ?

[obtuse nymph]

what can you do to both sides of V = (1/3) x pi x r^2 x h to have a lot of the stuff you know appear?

[timber fiber]

should i derive it?

[obtuse nymph]

yes, differentiate (i think that's what you meant)

[timber fiber]

yeah mb

how would i notate it? bc now it's in terms of r and h

[obtuse nymph]

wdym notate it?

[timber fiber]

dV/d?

what would replace the ?

[obtuse nymph]

oh, it would be t

in these types of problems, you always differentiate with respect to time

[timber fiber]

ah ok

so for the derivative of r, that's 2r * dr/dt?

[obtuse nymph]

yes

[timber fiber]

i think dV/dt would then be (1/3)pi x 2r * dr/dt * dh/dt?

does that look right?

[obtuse nymph]

you need to apply the product rule

the derivative of f(x)g(x) is not just f'(x)g'(x)

[timber fiber]

yes

on which term do i need to do that? r^2? i thought i did

[obtuse nymph]

imagine your functions are r^2 and h

[timber fiber]

yes

[obtuse nymph]

when you differentiate r^2h, you want to have two terms

[timber fiber]

ohhhh right

got it

i think

[obtuse nymph]

alright, try it again and lmk what you get

[odd edge]

@timber fiber Has your question been resolved?

[timber fiber]

⅓π x 2r x h + r² x dh/dt?

<@&286206848099549185> sorry, cant get this ahh

[olive needle]

What is confusing you?

[timber fiber]

derivative of formula for volume of a cylinder in terms of r and h

[olive needle]

The first thing to pay attention to is that neither r nor h are constants.

[timber fiber]

yes

[main pike]

find the first partial derivative with respect to r?

[timber fiber]

i

[olive needle]

So you have to treat them as variables that change. As such, when you find the derivative, you need to use the product rule.

[timber fiber]

those are big words 💀

[timber fiber]

yes

im not entirely sure if i'm applying it correctly

[timber fiber]

^ this is what i think it is, but i'm not sure

[olive needle]

So here, you derivated r^2 but you neglected the dr/dt.

[timber fiber]

so, ⅓π x 2r x dr/dt x h + r² x dh/dt?

[olive needle]

If you meant

pi/3 (2rh dr/dt + r^2 dh/dt)

then yes.

[timber fiber]

why is h not multiplied on the left side?

ah ok

[olive needle]

My bad, left that out.

[timber fiber]

i think that makes sense then

thank you!

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[crisp wadi]

Try integral test

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[vast quartz]

I'm struggling a little with Taylor Polynomials in my calc 3 class, to be specific multivariable (x,y) ones.

What exactly is Lagrange's general formula for the remainder looking like here? Ofc I understand the singlevariable one, but I'm staring at the one for n=2 here and I'm not sure why for Fxxy and Fxyy the pattern changes

[vast quartz]

I asked ChatGPT and I got this back

the 2 in Fxy is what I don't understand

meanwhile the 1/2 in fxx and fyy I do because that's just part of the remainder formula where the function's over (n+1)!

in this case n=1

[olive needle]

You don't have a book?

[vast quartz]

got stewart's multivariable analysis but it just mentions for single variable (lol). my profs gave me formulas for 1st and 2nd grade polynomials but i want to know why the numbers are there themselves if i wanted to say construct remainder formulas for n=3 or 4 or 5

[olive needle]

f_xxx f_xxy f_xyy f_yyy, the pattern is similar to Pascal's Triangle.

f_(3x 0y) f_(2x 1y) f_(1x 2y) f_(0x 3y)

[vast quartz]

hm. so, for n=2 it'd be something like this?

one sec

[olive needle]

,rotate

[vast quartz]

q a point between (a,b) and (x,y)

[olive needle]

xxx , xxy, xyy, yyy would be for the third derivative, or the error if that is what you are looking for.

[vast quartz]

yeah, i got it now though. that's more or less what i was looking for

for R1 you just go down one level of derivation since its k+1, alright

i guess i just answered my own question hahah

[olive needle]

🙂

[vast quartz]

thx

.close

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[strange bobcat]

hello how do I do this..?

[strange bobcat]

?

[warped glacier]

how much interest is gained in 12 months?

if it's too confuisng find how much interest is in 1 month first

[strange bobcat]

alr

[warped glacier]

then you can do (interest gained in 1 year)/400

times 100%

[strange bobcat]

alr thanks

i got it

[warped glacier]

cool

[odd edge]

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[weak kiln]

Can someone help with this question

[steady tide]

what have you tried so far?

[weak kiln]

I’m not sure what to do

I know inverse is y=k/x and direct is y=kx

Would it work if I solve for k and apply it for 2 columns in the table to see if it works?

[vivid quartz]

try to brute force it

trial and error

input the values in each case and see if the condition stands

[weak kiln]

so the correct answer is one where k is equal for all of them?

[vivid quartz]

yes

its obviously not A or C

or B

the rest u can try

[weak kiln]

.close

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[weak kiln]

Ty

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[topaz lantern]

looks like tedious IBP to me

anyone have any other ideas

[warped glacier]

well to save on the calculations use the DI method of by parts

tabular integration by parts

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@sly drift Has your question been resolved?

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[last citrus]

what does this mean

why should the nth derivative be differentiable on the open interval

[forest sky]

the last term involves the n+1th derivative

[last citrus]

yes but how does that relate to the interval

it makes sense to be closed

so that it converges to a function

but what about the nth derivative being open

[forest sky]

the mean value theorem requires continuity on the closed interval and differentiability on the open interval. this theorem probably depends on that

[last citrus]

ohh

yea that makes sense

thanks alot

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[craggy jasper]

I was wondering about a something:

which numbers, when multiplied by 9, get permuted?

for example, 1089 * 9 = 9801

I wrote a program to print such numbers but I am not able to figure out any patterns or common properties to help me answer my question

The only thing I have come up with is that the number should be divisible by 9, because its permutation will also be divisible by 9

[craggy jasper]

here are some of the numbers that I generated

one more thing I noticed is that all of them start with 1

the program has reached 9 digit numbers and not a single number yet has started from something other than 1

more

[proven beacon]

it coudl be because if it were anything higher than 1 it would have more digits than the original number

2000*9=18000 which has 5 digits compared to 4 in 2000

[craggy jasper]

youre right

[proven beacon]

so then one condition would be that it has to start with 1 and have a 9 somewhere

[craggy jasper]

and it must be less than 111111...

the number of 1s will depend on the number of digits

[craggy jasper]

because otherwise it will give a number with more digits

[subtle atlas]

this could be something interesting in number theory ngl

[craggy jasper]

Im assuming that the logic behind needing to have a 9 is that its the only way to get 1 as a digit in the product

If that is correct, then we could make cases

like if 2 is present in the product, 8 must be there in the original number

similarly, n corresponds to 10 - n like that (for example 3 corresponds to 7)

[subtle atlas]

the first step maybe to plot all the numbers you get on a graph?

and then see if there's any relation?

[craggy jasper]

idk how to automate that

[subtle atlas]

i can code it up in python, add me on discord ill get back to you in a few hours

[craggy jasper]

I should get these numbers in a text file

[craggy jasper]

ok

[meager juniper]

Consider a sequence of digits, (a, b, c, d, ...) and the number that they form when concatenated a + b*10 + c*100 + ...

If we have a permutation of this set, for instance (b, a, d, c, ...) then we can consider the difference of these two numbers.

We can notice that when we subtract powers of 10 from each other we are left with numbers that look like 99990000, these are all divisible by 9.

So the difference between the two numbers is also divisible by 9.

So you have N - M = 9k and also N = M * j

So N - M = 9k, Mj - M = 9k, M(j-1) = 9k, so either (j-1) is divisible by 9 or M is. But j cannot be 10 or larger.

So M is divisible by 9

[craggy jasper]

ill send the code I currently have over here

from itertools import permutations

n: int | str = 1
while True:

    n2 = str(n * 9)
    n = str(n)

    if int(n) % 9 == 0 and len(n2) == len(n) and tuple(n) in tuple(permutations(n2)):
        print(n, "* 9 =", n2)

    n = int(n) + 1

[meager juniper]

And because digital roots are invariant under permutations of digits, and the only multiples of 9 are guaranteed to do that as well, j must also be equal to 9

So that explains both of your observations.

[craggy jasper]

ill update the condition to include the ones we have discussed so far

[craggy jasper]

nice

[meager juniper]

Well, I guess that's sufficient but not necessary.

Let me see if I can tighten that up a little bit.

Everything up to the digital roots stuff is solid. I just don't have a reason why j must also be 9 yet.

Obviously j can trivially be 1.

[craggy jasper]

from itertools import permutations

n: int | str = 1

while True:

    n2 = str(n * 9)
    n = str(n)

    flag1 = int(n) % 9 == 0

    """
    n must be less than 1111..., otherwise n * 9 will have more digits than n.
    the expression on the right is (10^k - 1) / 9, which is 1111... [sum of a finite gp]
    """
    flag2 = int(n) < ((10 ** len(n)) - 1) / 9

    if flag1 and flag2 and tuple(n) in tuple(permutations(n2)):
        print(n, "* 9 =", n2)

    n = int(n) + 1

from itertools import permutations

n: int | str = 1

numbers = open("numbers.txt", "w")

while True:

    n2 = str(n * 9)
    n = str(n)

    flag1 = int(n) % 9 == 0

    """
    n must be less than 1111..., otherwise n * 9 will have more digits than n.
    the expression on the right is (10^k - 1) / 9, which is 1111... [sum of a finite gp]
    """
    flag2 = int(n) < ((10 ** len(n)) - 1) / 9

    if flag1 and flag2 and tuple(n) in tuple(permutations(n2)):
        output: str = f"{n} * 9 = {n2}"
        print(output)
        numbers.write(output)
        numbers.write("\n")

    n = int(n) + 1

this one writes the output to a txt file

Im uploading the file on pastebin

[meager juniper]

There are a lot of examples that don't multiply by 9

1035 * 3 = 3105
1089 * 9 = 9801
1359 * 7 = 9513
1386 * 6 = 8316
1782 * 4 = 7128
2178 * 4 = 8712
2475 * 3 = 7425
10035 * 3 = 30105
10089 * 9 = 90801
10350 * 3 = 31050
10449 * 9 = 94041
10890 * 9 = 98010
10899 * 9 = 98091
10989 * 9 = 98901
12375 * 3 = 37125
13590 * 7 = 95130
13599 * 7 = 95193
13860 * 6 = 83160
13986 * 6 = 83916
14085 * 6 = 84510
14247 * 3 = 42741
14724 * 3 = 44172
14859 * 6 = 89154
15192 * 6 = 91152
17604 * 4 = 70416
17802 * 4 = 71208
17820 * 4 = 71280
17982 * 4 = 71928
18027 * 4 = 72108
19728 * 4 = 78912
19782 * 4 = 79128
20178 * 4 = 80712
21780 * 4 = 87120
21798 * 4 = 87192
21978 * 4 = 87912
23751 * 3 = 71253
23958 * 4 = 95832
24147 * 3 = 72441
24714 * 3 = 74142
24750 * 3 = 74250
24876 * 3 = 74628
24975 * 3 = 74925

from itertools import permutations

n: int = 1
while True:
    n2 = n * 9
    for i in range(2, 10):
        n3 = n2 * i
        s2 = str(n2)
        s3 = str(n3)
        if len(s3) == len(s2) and tuple(s3) in tuple(permutations(s2)):
            print(f"{s2} * {i} = {s3}")

    n += 1

[craggy jasper]

upto what number should I upload currently?

ive noticed that these numbers seem to come in clusters

from the way that the program stutters at high numbers

it stops for a while and suddenly outputs a bunch of numbers at once

and those numbers are close to each other

[craggy jasper]

well need to wait for him to make the graph to see if this is true

[craggy jasper]

I might just be looking for a pattern where there isnt one

well see

[meager juniper]

I saw something about 5 being impossible?

[craggy jasper]

I was guessing

but I need to prove that

[meager juniper]

I don't think you'll be able to

Because I have an example of 5

[craggy jasper]

this is what Im talking about

it froze for more than 10 seconds after 10599345

then it printed the next number and forze again

then it printed all of the other ones all at once

and the difference between those numbers is less than 1000

I saw a bigger cluster before this

[meager juniper]

142857 * 5 = 714285

[craggy jasper]

isnt that the decimal expansion whenever something = 1mod7 is divided by 7

[meager juniper]

Actually 142857 is super interesting because a bunch of different numbers multiply to permute it

Yes

[craggy jasper]

and any natural number / 7 is just a permutation of 142857

[subtle atlas]

i've got this running lets see what a's length comes out to

[subtle atlas]

from 0 to 1*10^13

[craggy jasper]

make a a set

it might run better

[subtle atlas]

eventually we're gonna plot it and gonna need it in ascending order

also i've let it run for a while i might as well just let it run

[craggy jasper]

https://pastebin.com/7cqQXmZx

[craggy jasper]

this is the latest iteration of the program from my side btw

[meager juniper]

This version is much faster than the previous

[craggy jasper]

should we move this discussion somewhere else?

because its going to get buried soon and the channel will get closed automatically when I go to sleep

[meager juniper]

n: int = 1
while True:
    n2 = str(n * 9)
    for i in range(2, 10):
        n3 = str(n2 * i)
        if sorted(n3) == sorted(n2):
            print(f"{n2} * {i} = {n3}")

    n += 1
    if n > 1111111:
        break

[craggy jasper]

nice

[meager juniper]

This gets all of the numbers up to 10 million in a few seconds

[craggy jasper]

sorting will be faster than permuting

W

[meager juniper]

Actually you can clean up the intermediate values now that they're no longer being used.

Fixed

[craggy jasper]

this is so much faster

[subtle atlas]

wait i looked at the pastebin some of the numbers dont hold

like 10449

[meager juniper]

Which one does not hold?

[subtle atlas]

10449

just because the sorted numbers are equal doesnt mean its a reverse

[craggy jasper]

my question wasnt about the numbers being reversed

it was any permutation

[subtle atlas]

oh

mb

[craggy jasper]

its ok

[meager juniper]

B)

[craggy jasper]

I can still see this behaviour

in the version that uses sorting

[subtle atlas]

cause these are the only ones that hold for a reverse

uptil 10M

[craggy jasper]

it might be a pattern or just the program lacking in optimization at certain numbers

[meager juniper]

@subtle atlas I haven't posted a pastebin because I'm remoting into my server to run this stuff

[craggy jasper]

is the next one 10999989

[meager juniper]

But have you tried the code I posted above?

[subtle atlas]

yes

[proven beacon]

10999989*9 = 98999901 ,yes

[meager juniper]

All of the values above are of the form 11*99*9, 111*999*9, etc

[craggy jasper]

if were seeing only the ones that get reversed then I think the pattern is clear

we just need to prove it

[subtle atlas]

do you want me to run the thing and provide output?

cause im running it on a collab atm

[meager juniper]

Nah, I have output, just no super easy way to exfiltrate it

[subtle atlas]

ohk

[meager juniper]

Easy to browse, pita to copy paste

[craggy jasper]

.

[subtle atlas]

probably

[meager juniper]

In other words, these values are related to 99*99 and 999*999 and so on

I don't know if it is possible to prove that these are the only reversed ones, because I'm not entirely convinced they absolutely must be.

[craggy jasper]

I thought it was 10999...989

[subtle atlas]

a help forum should be a good place right

[craggy jasper]

the forum in this server or a forum website?

[meager juniper]

And that number is 111...111 * 999...999

[dense stone]

Maybe it is always starting with 1

[subtle atlas]

forum in the server

[craggy jasper]

yes

[subtle atlas]

im gonna run the thing upto a trillion and check if it always starts with 1

[dense stone]

Or does something like 00109 count

[craggy jasper]

we proved that

[meager juniper]

Only when multiplying by 9 does it need to start with 1.

[craggy jasper]

no need to check it

[subtle atlas]

ohk

[meager juniper]

I guess actually 9, 8, 7, 6, and 5, all need to start with 1.

[craggy jasper]

im making one