help-39

[wet osprey]

But only 3 different things can fit into the 2nd slot

Now only 2 different things can fit in the 3rd slot

And there’s no choice but to put whatever’s left in the last slot

There are 4! Ways to permute 4 distinct objects, 4 * 3 * 2 * 1

[scarlet harbor]

is it realistic?

I was actually thinking of an election for the government or something

but idk if it will work

is 4•3•2•1 what they want me to do?

[wet osprey]

Well that’s good for the first question, but you can also think of it as

[scarlet harbor]

as?

[wet osprey]

How many different outcomes can you get if you: roll a 4 sided dice, then a 3 sided dice, then a 2 sided dice, then a 1 sided dice

Obviously the 1 sided dice roll doesn’t do anything

But with the set up you can make anything you want really

[scarlet harbor]

so that'll lead to 4•3•2•1?

[wet osprey]

10 * 6 * 3 * 2 simply becomes roll a 10 sided dice then a 6 sided dice then a 3 sided dice then a 2 sided dice

Since there will be 10 different numbers in the first slot

6 different numbers in the 2nd slot

Etc

[scarlet harbor]

where did 10,6,3,2 came from?

[wet osprey]

I just made it up

[scarlet harbor]

ah

So we goods in no. 1

what I'm rlly confused abt is no. 2

[wet osprey]

Well I just gave you a way to make situations for arbitrary numbers

[scarlet harbor]

can I do the same with 2?

like

given

A bakery offers a special promotion where customers can create their own custom cake by choosing from 5 different cake flavors, 2 different frosting options, and 3 different decoration designs. In how many ways can a customer create their custom cake?

[scarlet harbor]

Can this work?

I think it can, right?

5,2,3

[wet osprey]

Yes

[scarlet harbor]

yesss

okay, thank you

[wet osprey]

This is a little weird because the conference has 5 sessions

[scarlet harbor]

okay so not that

[wet osprey]

So somehow you need to permute the 5

[scarlet harbor]

I'll stick with the cake?

[wet osprey]

But we actually only want to select 1 of the 5

Cake was a better example

[scarlet harbor]

okay, thank you!!

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[pearl pond]

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[pearl pond]

@grand valve Has your question been resolved?

[grand valve]

<@&286206848099549185>

[quartz citrus]

uh

have you shown that (a/p) = (a)^((p-1)/2) mod p yet

[grand valve]

Yes!

[quartz citrus]

use that

for (-1/p)

[grand valve]

Sorry, that's wha tthe last part was in reference to. The case of p=6k+1 yields (-1/p)=(-1)^(3k), but it's unknown if 3k is even or odd

Similarly, p=6k+5 yields (-1/p) = (-1)^(3k+2) = (-1)^3k

[quartz citrus]

(-1)^(3k) = (-1)^ k

so two cases

4 cases total

[grand valve]

Hm

I dont think that satisfies what I'm tasked to prove, though

[quartz citrus]

why not

[grand valve]

I'm supposed to find this purely off the value of p mod 6

Aw hell I think the way I handled $\left(\frac{3}{p}\right)$ mgiht have been wrong

[jolly parrot]

cat_food_sounds

[quartz citrus]

p = 6k + 1 case -> 6(2t) + 1 = 1 mod 6, 6(2t + 1) + 1 = 1 mod 6
p = 6k + 5 case -> 6(2t) + 5 = 5 mod 6, 6(2t+1) + 5 = 5 mod6

[grand valve]

of course, but if k is an odd then wouldn't it then flip the returned values of (-1/p)?

[grand valve]

I'm probably just going to abandon this for the time being and come back to it tomorrow.

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[stiff cradle]

Help me pls

[burnt minnow]

do u have a question to ask?m

[stiff cradle]

Yea

I want the formula of surface area of sphere

Pls

[light helm]

google

[burnt minnow]

i second that

[stiff cradle]

Oh yea thank you

[burnt minnow]

if you're done asking you can always close the channel using .close

[pearl pond]

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[pearl pond]

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[fast tinsel]

Problem: Find the volumes of the solids generated by revolving the regions bounded by the graphs of the equations about the given lines.
y = 5x2
y = 0
x = 2

[fast tinsel]

Ive gotten the volume of the region when rotated around the y axis and x axis but my brain cant comprehend how to get the volume of the region when its rotated around y = 20 or rotated around y = 2.

not y = 2, around x = 2

[spark bloom]

y'all are way too smart for me

[fast tinsel]

lol i try my best but this problems making me feel pretty dumb rn

<@&286206848099549185>

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@fast tinsel Has your question been resolved?

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[midnight haven]

I have no idea how to even start this

[pearl pond]

Please don't occupy multiple help channels.

[midnight haven]

Is B the same as P in the diagonlizable formula $A = PDP^{-1}$

[jolly parrot]

Book Reader

[midnight haven]

yes

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[gentle sundial]

yo im cooked

[midnight haven]

I agree

[midnight haven]

a should be easy right?

[gentle sundial]

yeah its four

[vestal tapir]

i get 16

[gentle sundial]

yeah im defo cooked

[midnight haven]

ayo swerrie in help channels

[gentle sundial]

how you get 16?

[vestal tapir]

3 additions

6 "negations"

7 "multiplications"

[gentle sundial]

wait im braindead wtf

[midnight haven]

I counted 10

7 ands and three ors

[gentle sundial]

hmmmmmmmm

[midnight haven]

oh wiat

you need negation gates?

[gentle sundial]

i dont even care about a i just dont know how to simplify this

[midnight haven]

thats crazy

[midnight haven]

It's like algebra,

$\hat{A} \hat{B}(C+D) = \hat{A}\hat{B}C + \hat{A}\hat{B}D$ distrubutive property

[jolly parrot]

Book Reader

[midnight haven]

There are also other rules you can use to simplify

[gentle sundial]

like absorption law ?

[vestal tapir]

also or gates don't do anything so it could be 13

[midnight haven]

ya

[gentle sundial]

ayt thanks gentleman, im gonna relearn this

[pearl pond]

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[regal sequoia]

"Calculate the double integral. Tip: Be sure to change variables"

[pearl pond]

@regal sequoia Has your question been resolved?

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[pearl pond]

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[sterile pecan]

How do I solve it with substitution method?

[wet osprey]

which one you've sent 5 problems

[sterile pecan]

3 problems but I am stuck with o and p

[wet osprey]

if u let u = x^-1

that sounds like a useful substitution

[sterile pecan]

You are Talking about the p exercise?

[wet osprey]

o

[sterile pecan]

But why x^-1

If there is no -1

[wet osprey]

try it and see

1/x is x^-1

[sterile pecan]

Okey I thought of it a shortcut or sth

It make one x not dissapear

[wet osprey]

show me what you've done

[sterile pecan]

And dt which is missing at the end

[wet osprey]

,rotate

[wet osprey]

but $\frac{e^t}{x} = e^t(x^{-1})$

[jolly parrot]

Frosst

[sterile pecan]

Ohhh

Right

Now I see

But the multiplicstion seems to still be problematic

With the method I was given

[wet osprey]

you dont know how to integrate te^t?

[sterile pecan]

Yes

[wet osprey]

do you know about integration by parts

[sterile pecan]

I mean I know this method But it is not what this exercise is about

It shouldn't be used I guess

[wet osprey]

what's the exercise about?

[sterile pecan]

About using substitution and the basic formulas

Same with the p)

[wet osprey]

is this section before or after IBP

[sterile pecan]

Before

[wet osprey]

well

im not sure then, xe^x is a very common integral

[sterile pecan]

So u suggest i should just know that

And how about the p without using special formula for fractions

[pearl pond]

@sterile pecan Has your question been resolved?

[pearl pond]

@sterile pecan Has your question been resolved?

[sterile pecan]

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[sharp smelt]

so to start I guess the cartesian form will be more relevant here

so let $z=x+iy$

[jolly parrot]

ƒ(Why am. I here)=misery

[sharp smelt]

so this would be

$|x^2-y^2 +2ixy +x+iy+1|=1$

[jolly parrot]

ƒ(Why am. I here)=misery

[smoky musk]

I recommend not using cartesian coordinate here. It's longer

[sharp smelt]

hmm, polar form then?

[smoky musk]

No, just treating complex number like a number

[sharp smelt]

hmm, o

ok

so I have to solve (x^2+x+1)=1 essentially

the equation when real, anyway lies entirely above the x-axis

[smoky musk]

+-, don't forget

Also, when you encounter the determinant < 0, do NOT discard the case.

[sharp smelt]

x^2+x=0 is one case

so x(x+1)=0

so x=0

or x=-1?

how does that make sense

[smoky musk]

That's two of the possible answers

Now, find the other case

[sharp smelt]

but if x=-1

that means |z|=-1, right?

oh

z=-1

right

makes snese

(sense

[sharp smelt]

the second case is

x^2+x+2=0

hmm

this gives two imaginary solutions

[smoky musk]

Now, you just has to plug it out.

[sharp smelt]

why does this work though

[smoky musk]

Because fundamental theorem of algebra

https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

[sharp smelt]

hmm, so option b is right

I guess I substitute and solve for the other too, right?

[smoky musk]

Yes

[sharp smelt]

thanks!

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[jolly parrot]

l'agit

[smoky musk]

+---------------+--------------------+----------------------+
|               | Answered correctly | Answered incorrectly |
+---------------+--------------------+----------------------+
| Prepared      |                  p |                    0 |
+---------------+--------------------+----------------------+
| Not prepareed |                    |                      |
+---------------+--------------------+----------------------+

Fill this table, please?

[pearl pond]

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[restive bluff]

Hello, I have a two part question, firstly, why does the 2n+1 become 2n+3 when doing the ratio test? Shouldn't it become 2n+2 since we are only adding 1?
Secondly, when simplyifing the limit, since n is our term of consideration, we can remove the -16x^2 from the absolute value (as a positive) and solve for whatever is left on the inside correct?

something like this maybe

but it's wrong, and I don't understand why

[trim willow]

Because 2*(n+1)+1=2n+3

And yes you can move 16x^2 outside

[restive bluff]

I see, I didn't assume it was 2(n+1), I assumed 2(n)+1

Makes perfect sense now

[restive bluff]

I moved it outside, and through L'Hopital's rule the inside becomes 1, so I am unsure as to why 16x^2 is an incorrect answer

[trim willow]

It isn't

[restive bluff]

The inside isn't?

[trim willow]

the answer is correct

[restive bluff]

idk it's telling me it's wrong

maybe I need to input -16x^2?

since it's still in the absolute value the way they have it

ok I guess that worked lol 😅

ty for the help

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[restive bluff]

I'm a little confused as to what happens to the limit here

If everything cancels out leaving just |x|/3 and there are no n terms

How can I determine a limit?

[trim willow]

What if I wrote what you had as |x|/3+0*n

could you then find limit?

[restive bluff]

mmmm

oh I get it now

I wrote x/3 instead of |x|/3

ty for the help

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[jolly parrot]

Bean Man

[fluid axle]

they just ignore it

they're just showing w is a sum of $THING1 in ker(g(phi)) and $THING2 in ker(f(phi))

with $THING1 = (uf)(phi)(w) and $THING2 = (vg)(phi)(w)

you can just show these two inclusions independently

[pearl pond]

@rapid bough Has your question been resolved?

[rapid bough]

are they just multiplying by g(phi) or is it some other manipulation?

[fluid axle]

if g(phi) ($THING1) = 0, then $THING1 is in ker(g(phi))

that's what they're doing yes

@rapid bough

it's not multiplication, g(phi) is an endomorphism V -> V

[rapid bough]

doesnt the set of endomorphisms make an algebra with composition so could i view it like multiplication or is that just completely incorrect?

also I’m not seeing the equality

wait we have Ker(fg)(phi) <= Ker(f(phi)) + Ker(g(phi))

and Ker(f(phi)) <= Ker(fg)(phi)
and Ker(g(phi)) <= Ker((fg)(phi))

so it’s good

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[fossil drum]

Does anyone know what this question means?

[fossil drum]

I emailed my professor and he gave me the most vague response

he said "with the question as stated iy should be y’=10000+0.01y."

but in the previous email he said y'=10000-0.01y

[arctic wharf]

it'd be this one i think

[fossil drum]

I'm wondering if he meant withdrawed instead of depsoitied in the start

[arctic wharf]

wait why is it 0.01y and not 0.1y if it's 10% return

every year you gain 10,000 + 0.1*(whatever's in the bank)

[fossil drum]

idk he makes a lot of mistakes and he doesn't provide an answer key so I don't know if my answers to the questions on this practice test are right

maybe he meant .10 but maybe he didn't im not sure

[arctic wharf]

do u get why it would be this

[fossil drum]

no I'm not sure why

[arctic wharf]

so year 0 there's no money in the bank

so we add 10000 + 0.1(0)

and now there's 10000

the next year we add another 10,000, and we also get 10% on the 10000 we put in last year

so the change that year is 10,000 + 0.1(10,000)

and the derivative is the rate of change

so the rate of change per year is 10,000 + 0.1y

where y is how much money is in the bank

[fossil drum]

wait

hmmm

doesn't that equation model how much is in the investment count P(t) instead of P'(t), I'm confused on why it's P'(t)

or actually no it doesnt

oh I see

ok I see thank you

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[arctic wharf]

nice

[pearl pond]

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[fossil drum]

sorry I'm curious about something else

[fossil drum]

what would happen if the balance didn't start with 0

like what if it started with 5,000

oh

would I just add a + 5000 to the equation

well no nvm

I think it would y' = 0.10(y + 5000) + 10000

or no that doesn;t make sense

wait what is y in this equation 😭

it's the amount in the balance right

and y' is the rate of change per year

so that means if we started with 5000

[fossil drum]

yeah I think it would be this equation

[pearl pond]

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[midnight haven]

I drew this diagram, not entirely sure if it's correct

and then for part b what I thought of doing was to find the point A

which i figured would be equal to OA + AB

and in that case, OA + AB = OA + OA - OB

therefore A = 2OA - OB

worked that out though and got (-4,0,9)

which wasn't in the answers

need someone to correct my logic

[thin sigil]

If O is the centerpoint of the coordinate system, then vector OA gives you the coordinates of point A, so you already have the coordinates of point A

OA + AB would give you OB as a result

[midnight haven]

so O would be in the middle of the parallelepiped?

[thin sigil]

No, it would be at one of the corners, but it's treated as the center of the coordinate system from which all coordinates are determined

therefore OA is basically the coordinates of A

if you had a point D, the coordinates of D would be given by the vector OD

[midnight haven]

ok so then wouldn't A be OB - OA?

[thin sigil]

Essentially, you are given the coordinates of 3 vertices and have to find the others

[midnight haven]

oh okay

[thin sigil]

OB - OA would give you the vector BA

[midnight haven]

this is my newly labelled diagram

this question is trickier than I thought, my idea was we'd add and subtract vectors

until we got a corner

but it doesn't seem like we can do that

[thin sigil]

hmmm you're assuming that A, B and C are spread out like that

why is OB a diagonal and not a side?

Try drawing it taking dimensions into account, you might get something else

[midnight haven]

let me do that really quick

not sure if this complicated things or not

[thin sigil]

You're probably supposed to sketch it on paper

Well, I guess in this case you're supposed to assume that each of the point A, B and C is an opposing corner of the parallelepiped from O

so none of them are diagonals

[midnight haven]

this is how it's supposed to look

[thin sigil]

The dimensions in this one are a little iffy, but yeah

so OB is not a diagonal

it's a side

knowing that you can do vector addition to find the other sides

or rather their coordinates

[midnight haven]

O would obviously be our first one

(0,0,0)

and the 3 other points given I assume would be the other ones

[thin sigil]

yes

[midnight haven]

alright will figure the rest out, thanks

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[pearl pond]

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[pearl pond]

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[warm fern]

using integration find the area bounded by y=4 and y=x^2 with the y axis

[warm fern]

would this be 0? as y=+_ sqrt x and that means the two area we get while integrating wrt y will be equal in magnitude but opp in sign

[fast tinsel]

I dont think it would be zero. Your finding the area bounded by those lines so as long as it exists it should be a value.

[warm fern]

i agree, but when I m solving it wrt to y i m getting 0 mathematically

[fast tinsel]

Isn't it just the integral from a to b than the top minus the bottom? I get a number unless I forgot how to get the area. Let me draw it out

[warm fern]

okk

solve it wrt y though

dy not dx

[fast tinsel]

Ok so when I do it from dy I get a number but it's negative. Not getting zero which it shouldn't be but I don't think it should be negative either

[warm fern]

can u show ur work?

u should not get -ve...

[fast tinsel]

You know the formula right?

[warm fern]

ya

[warm fern]

u took only + sqrt y

wait a sec...4

[fast tinsel]

So u don't know the formula?

[warm fern]

ok so like ya, u have to consider both right?

x=+- sqrt y

[fast tinsel]

That shouldn't matter, let me look at what I'm doing wrong it's been a while. But you do understand why I have the integral from 0 ro 4 right

[warm fern]

it would, ur missing half the area. yes i get that

[fast tinsel]

I see what I did wrong, hold on

[warm fern]

ok

[fast tinsel]

Ok so the left graph isn't y=4 our left graph is the y axis which is just zero. I'll send you the picture if u don't get something just tell me.

[warm fern]

ok

[warm fern]

ok um... wht did u do??

also with dx i m getting -ve 16/3

[fast tinsel]

What did u do?

I got 16/3

Let me write down an explanation hopefully it'll ne better than typing it

[warm fern]

ok

[warm fern]

i will just write my work clearly n share it

ok um... ur ans is wrong

[fast tinsel]

Welp I don't know what I did wrong lol mb.

That should be the formula though, im not sure what I'm putting in wrong.

[warm fern]

this is with dx

I want it with dy

[fast tinsel]

Ye I understand that. Let me see ur work. What u got there is correct right?

[warm fern]

Ya

Should we ask someone else to help?

Fresh eyes might help

[fast tinsel]

Ye I don't mind lol I'm in the same boat as u now

[warm fern]

lol, um whom do we ask though?

[fast tinsel]

Idk lol I just joined yesterday

[warm fern]

<@&286206848099549185>

i m tryin this

[fast tinsel]

Let me look at ur work though maybe I can remember how ti do it dy

[warm fern]

mhm ok

[fast tinsel]

But why are u doing the area outside the curve and not the inside, aren't we getting the area of the region bounded by y=x^2 y =4 and y axis

[warm fern]

i m doing area inside only

see my eq

[fast tinsel]

Ye but u shaded the outside, so u just ignored the shaded region right

[warm fern]

i subtracted that

[fast tinsel]

Okay just confused me a little usually I shade in the region im finding, I get it now

Are u sure my answer was wrong? I mean should it be the same area as what u got when doing it dx. 16/3 is the area and if u want the negative side as well than it's just 16/3 timed 2 because both sides are equivalent leading to 32/3 unless that answer is wrong.

[warm fern]

ok then ur ans is right

but i dont want u to do that

write an integral for the -ve side

and u will see that u get a -ve area

uk wht nvm

i have lost my mind

[fast tinsel]

Yes because it's on the negative side u get -16/3

[warm fern]

ur right

i got the ans

[fast tinsel]

If u were to add it u get 0 but that's not logical therfore u just times it by two

Ok

[warm fern]

thanks for the help

[fast tinsel]

Np

[warm fern]

i bothered u for such a long time, sorry abt that too

.close

[pearl pond]

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[fast tinsel]

Lol np I was just trynna help

.close

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[warm fern]

Where am i going wrong

[torpid thicket]

take mod value of the -ve x-axis part

the areas are cancelling out

[warm fern]

But we cant just take mod like that?

[torpid thicket]

but the area of parabola on the left will give negative

in these cases we give logic that, the figure is symmetric, so we calculate the positive side and double it, since the negative side will also have same area

[warm fern]

Ya but i read somewhere area is only -ve under x-axis

[torpid thicket]

that is true, if u are calculating wrt to x-axis

here ur axis is y

[warm fern]

Ok if we have the same ques but (x-1) ^2 instead of x^2

There wont be symm.

[torpid thicket]

yes they wont be

then u have to use mod value for negative area

[warm fern]

But how can we just apply mod like this

It doesnt make sense

[torpid thicket]

but area cant be negative man

we are just take absolute value

[sterile tusk]

for area wrt y, we would have to do the right minus the left function

[jolly parrot]

y0shi

[warm fern]

Ya but of we were in the 3rd quad we would take it as -ve no?

[sterile tusk]

just like how you would do wrt x, you do top minus bottom

[warm fern]

Um, please explain that again

[sterile tusk]

when you find area enclosed between two functions with respect to x, we would integrate from lets say a to b, and our integrand would be the top minus the bottom function

same applies to when we are finding area with respect to y, but instead of doing top minus bottom, we would do right minus left

think of the integrand as the height of the infinitely small rectangles

[warm fern]

Thank you so much!

.close

[pearl pond]

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[pearl pond]

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[warm osprey]

The first image is a common diagram to show where the formula of a determinant comes from

I have rotated this diagram, and I get a different formula for the area.

= ab + ad + bc + cd - cd - ab
= ad + bc```
What has gone wrong here?

[limber ravine]

In the first diagram, all the values are positive, so they can be interpreted as the lengths

In the second diagram, c is negative, so the length of the red and green line is not a + c but rather a + |c| or alternatively a - c

[warm osprey]

Thanks

.close

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@long nest Has your question been resolved?

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[tidal vine]

i dont get how i got it wrong

[rough forge]

Bonjour, can you translate it

[shut perch]

its just saying martin's dad wants to build a thing for his horses

and it wants you to find the length

the notes say to use a period as the decimal symbol

use pi on your calculator

idk wtf the third thing says

[grim fractal]

Since this is a prism, just find the area of the base

[shut perch]

and don't round your answer

[grim fractal]

then area of base * length = volume, so <310

[pearl pond]

@tidal vine Has your question been resolved?

[tidal vine]

wait but how

[tidal vine]

Here is a sketch of a drinking trough that Martine's father would like to build for his horses. He wants it to have a capacity of 310 litres. Work out for him how long the trough should be.

Notes:

Use the "." for the decimal symbol.
Use the π on your calculator.
Use your calculator's memory wisely.
Don't round off your final answer. Write the first three decimal places.
The length of the trough should be

Translated with DeepL.com (free version)

[rough forge]

Ok so we are long for the longeur

[tidal vine]

yes

[rough forge]

The whole thing is 310 volume wise

[tidal vine]

yes

[rough forge]

310 = V_cuboid + V_half_cylinder

310 = width x length x height + πr²h/2

ok?

[tidal vine]

abc?

[rough forge]

width length height

Let's collect what we have

[tidal vine]

r= 25 and largeur= 50 cm

[rough forge]

ok

[tidal vine]

and volume= 310 L/dm

[rough forge]

notice we are looking for the height

we are given the length and width

[tidal vine]

yes

but shouldnt i convert 310 dm to cm

[rough forge]

yes

[tidal vine]

so that would be 31 cm

[rough forge]

no

[tidal vine]

oh 3100

[rough forge]

no

310 liters = 310 dm³

[tidal vine]

ohhhh 310 000

[rough forge]

yes

[tidal vine]

ooppsies

so that would be 310 000= 25 times 50 time height

[rough forge]

hold on

,,310.000cm^3 = 25cm \times 50cm \times h + \frac{h \times (25cm)^2 \times \pi}{2}

[jolly parrot]

𝔸dωn𝓲²s

[rough forge]

half cylinder😅

[tidal vine]

so taht would be 310 000= 1250h + 981.747... h

[rough forge]

I wouldn't calculate pi yet!

[tidal vine]

oh why?

[rough forge]

,, = 310.000cm^3 = 1250cm^2 \times h + \frac{625}{2} \pi cm^2 \times h

[jolly parrot]

𝔸dωn𝓲²s

[rough forge]

because of the rounding thing

I would first solve for h

[tidal vine]

ohh so then i do 625/2

then +1250?

[rough forge]

We can factor h

,, = 310.000cm^3 = \left( 1250cm^2 + \frac{625}{2} \pi cm^2 \right ) \times h

[jolly parrot]

𝔸dωn𝓲²s

[rough forge]

And then if we divide by this get the decimal number

310000/(1250+625π/2) = h

,w 310000/(1250+625π/2)

[jolly parrot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by adoniszgks

[tidal vine]

oh thats gives the answr

i get it now

[rough forge]

Actually I think we could have done pi before

but then we would have to be cautious about the decimals

,w 310000/(1250+981.747)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by adoniszgks

[rough forge]

ok good gives the same!

[tidal vine]

yay!

[rough forge]

like as long as we took the first 3 decimals it's alr

[tidal vine]

yes

thank you!!!

[rough forge]

no problem

[tidal vine]

.close

[pearl pond]

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[pearl pond]

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[frigid jasper]

Is this notation ok?

[white juniper]

i havent seen notation like that

but if you define it well i dont see the problem

maybe give some more context?

[fossil jewel]

Mm I’ve only seen that when x goes to infinity

For a specific p it’s a bit weird

[river marten]

In my uni we use this notation sometimes but we put x->p under the ->

[pearl pond]

@frigid jasper Has your question been resolved?

[frigid jasper]

yep, I've been studying variable substitution in limits and I wondered if I can be a bit more rigorous instead of just substitute the value of the tendency of a variable just to find out the other one. E.g., given that y=5x:

[jolly parrot]

Daisuke

[frigid jasper]

Most text books I use are specific for problem solving of pre-University math questions for applications for military carrier, so calculus isn't that rigorous at all.

[pearl pond]

@frigid jasper Has your question been resolved?

[frigid jasper]

<@&286206848099549185>

[frigid jasper]

.close

[pearl pond]

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[pearl pond]

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[vivid pilot]

kind of stuck

[vivid pilot]

i started by finding the points of intersection with

20+10cos(2theta)=sin(2theta)

but i have no idea how to find theta here

wait, the graphs dont intersect

the area of A_2 is $\int^{\frac{\pi}2}_0{\frac12 (\sin(2\theta))^2\mathrm{d}\theta}$

[jolly parrot]

talk_less
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vivid pilot]

which is

pi/8

as for the big thing

i found the area of the big thing to be 225pi i think

but 225pi-pi/8 is wrong

oh wow

the limits were from 0 to 2pi for the big thing

.close

[pearl pond]

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[calm chasm]

XD

[pearl pond]

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[fossil drum]

struggling with these questions, I think I got 10a right, but not sure about 10b, let me show what I have

[fossil drum]

is that right?

[pearl pond]

@fossil drum Has your question been resolved?

[rough forge]

Yes

[fossil drum]

and is this one correct also for the second part

[rough forge]

Yes what i meant

Wait this 10')

[fossil drum]

yeah

[rough forge]

hold on

,, (-x^2)^n = (-1)^n x^{2n}

[jolly parrot]

𝔸dωn𝓲²s

[rough forge]

So you actually have

x^(2n)/(n+2)!

Differentiated would be

(2n-1)x^(2n-1)/(n+2)!

[rough forge]

It's because you have (-1)^n(-1)^n which is (-1)^(2n) = 1

Imma go sleep hope it helps

[fossil drum]

ok thank you

[rough forge]

Just noticed

You are squaring somthing

Negative is gone

10a) and b)

[pearl pond]

@fossil drum Has your question been resolved?

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[celest spruce]

The last row is a total guess...

[versed remnant]

didnt u get the correct answer?

[celest spruce]

Nope, the homework program said it was wrong... I just left the answer I entered...

[versed remnant]

yeah wheres the inverse coming from?

shouldnt it just be sin(8x)?

[celest spruce]

Sorry for the clutter...

This is part of what the answer is based on...

[celest spruce]

If anyone told me the answer to this madness was 🐟 I wouldn't question it.

[versed remnant]

the rule u say has 1-k^2x^2

so u need 1 in the first place to apply that

the 3^2 is still under sqrt root i forgot to put it there

also there should be a "k" in between sin^-1 and x

[celest spruce]

There is, it's in red.

[versed remnant]

in ur answer

so ur answer would become
$$\frac{1}{8} \sin^{-1} \big{(} \frac{8x}{3} \big {)}$$

[jolly parrot]

JustToPro

[versed remnant]

+c

[celest spruce]

I'm so confused...

[versed remnant]

ok ill help wait

$$\frac{1}{8} \int \frac{1}{\sqrt{9-u^2}}$$

[jolly parrot]

JustToPro

[versed remnant]

till this , all is correct

now lets solve the integral

[celest spruce]

Here's what I have so far...

[versed remnant]

$$\int \frac{1}{\sqrt{3^2-u^2}}$$

[jolly parrot]

JustToPro

[versed remnant]

u have a formula / rule that states
$$\frac{1}{\sqrt{1-(kx)^2}} = \frac{1}{k} \sin^{-1}kx +c$$

[jolly parrot]

JustToPro

[versed remnant]

lets convert ur integral into that form first before using the rule

[celest spruce]

How did the 3^2 under the u disappear?

[versed remnant]

from here

im starting from there

[celest spruce]

Ready

[versed remnant]

is the rule correct ?

yeah its correct nvm

[celest spruce]

I just rewrote what was circled down below where there's more room. I don't know what to do next...

[versed remnant]

i forgot

wait dont do anything

$$\int \frac{1}{\sqrt{3^2-u^2}}$$

[jolly parrot]

JustToPro

[versed remnant]

just tell me what u dont understand btw

u have a formula / rule that states
$$\frac{1}{\sqrt{1-(kx)^2}} = \frac{1}{k} \sin^{-1}kx +c$$

[jolly parrot]

JustToPro

[versed remnant]

we have to convert ur integral into this form

so to get 1 we need to divide by 3^2

but if we divide we also have to multiply

[celest spruce]

Pretty much everything... I've been guessing my way forward...

[versed remnant]

$$\int \frac{1}{\sqrt{3^2-u^2}\cdot \sqrt{\frac{3^2}{3^2}}}$$

[jolly parrot]

JustToPro

[versed remnant]

everything as in this question or integrals in general?

[versed remnant]

simplify this we get
$$\int \frac{1}{3\sqrt{1-\frac{1}{3^2}u^2}}$$

[jolly parrot]

JustToPro

[versed remnant]

take out 3 and we get

$$\frac13 \int \frac{1}{\sqrt{1-\frac{1}{3^2}u^2}}$$

[jolly parrot]

JustToPro

[versed remnant]

now here k^2 = 1/3^2

[celest spruce]

I was only being slightly sarcastic. 😅
I don't know what to do after the part that was originally circled....

[versed remnant]

using the formula

[versed remnant]

yeah im sending the solution with a bit of explanation

just reply to the part u dont understand

$$\frac{1}{\sqrt{1-(kx)^2}} = \frac{1}{k} \sin^{-1}kx +c$$
in ur integral $k^2 = \frac{1}{3^2} \implies k = \frac{1}{3}$ so
$$\frac13 \int \frac{1}{\sqrt{1-\frac{1}{3^2}u^2}} = \frac{1}{3} \big{(}\frac{1}{\frac{1}{3}} \sin^{-1} \frac{1}{3}u\big{)} + c$$

[jolly parrot]

JustToPro

[versed remnant]

simplify the answer we get

$$\int \frac{1}{\sqrt{3^2-u^2}} = \frac13 \int \frac{1}{\sqrt{1-\frac{1}{3^2}u^2}} = \sin^{-1}\frac{1}{3}u$$

[jolly parrot]

JustToPro

[versed remnant]

use this in here $$\frac{1}{8} \int \frac{1}{\sqrt{9-u^2}}$$
We get $$\frac18 \sin^{-1}\frac{1}{3}u$$

[jolly parrot]

JustToPro

[versed remnant]

use value of u

We get $$\frac18 \sin^{-1}\frac{1}{3}8x$$

[jolly parrot]

JustToPro

[versed remnant]

$$\frac{1}{8} \int \frac{1}{\sqrt{9-u^2}} = \frac18 \sin^{-1}\frac{1}{3}8x$$

[jolly parrot]

JustToPro

[celest spruce]

Stuck on this part... I'll show you what I get in a bit...

[versed remnant]

basically ur rule/formula only works when u have $\sqrt{1-(kx)^2$ , the $1$ needs to be there

[jolly parrot]

JustToPro
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[versed remnant]

so to get 1 we just divide and multiply the entire square root with 3^2 because 3^2/3^2 is 1

[celest spruce]

total guess...

[versed remnant]

uh

thats correct

next step

[celest spruce]

How does the 3 get to the outside of the radical...? I don't know what's going on..

[versed remnant]

btw multiplying and dividing isnt necessary if u are kinda good at factoring

$$(3 - u) = 3(1 - \frac{u}{3})$$

[jolly parrot]

JustToPro

[versed remnant]

laws of indices states
$$(ab)^n = a^n b^n$$

[jolly parrot]

JustToPro

[versed remnant]

and square root is just powered 1/2

so $$(ab)^\frac12 = a^\frac12 b^\frac12$$

[jolly parrot]

JustToPro

[versed remnant]

keep practicing u would understand

usually these can be tricky to understand

[celest spruce]

It's all been an eldritch horror to me... I'll try rewriting it as (stuff)^1/2. That should make it a little less frightening...

[versed remnant]

yeah u can do that

it can be horror to ppl new to calc

[celest spruce]

My final is in 2 weeks.... 😐
Not sure I'll pass at this rate...

[versed remnant]

u will

good luck

[celest spruce]

I still don't understand what's going on... How did it get to be (1/3^2)u^2??? I'm even more lost than when I started.

How do I get to the next step from here?

Found a clue here... I'll keep stumbling forward...

Never mind... totally lost...

Don't know what I'm doing wrong...

<@&286206848099549185> ?

[midnight haven]

I guess you should end up with an ln, bc you have to do u=g(x) and du=g'(x)

And taking that 1/8 out isn't correct, you have to get it from the 9 too, which you can't cause you didn't change anything abt that 9

[celest spruce]

So... How do I solve this...?

[midnight haven]

I'm not really sure

Let me look into it

[celest spruce]

I give up... I'll do the walk through on this problem... Maybe there will be a hidden clue there. 🤷‍♂️

[midnight haven]

I'm sorry, but I can't help much now

It's 5am and I haven't slept yet

Gn

[celest spruce]

I don't understand how the 9 disappears from the final answer...

.close

[pearl pond]

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[halcyon blaze]

Does anyone know what the runtime for this is?

[marble sigil]

seems like it's usually always true

[halcyon blaze]

My bad I mean the runtime complexity

O(n) right?

[marble sigil]

oh woops I read it wrong

[halcyon blaze]

Haha all good I think it is O(n)

Tyty tho 🙏🏻

[fossil jewel]

O(n^3) methinks

[halcyon blaze]

.close

[pearl pond]

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Use .reopen if this was a mistake.

[pearl pond]

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[delicate dock]

What would the bounds for the integral be for the one on the right? I'm trying to compute mass moment of inertia

[delicate dock]

can u @ me pls if u know

i thought it would be x to x+R

[pearl pond]

@delicate dock Has your question been resolved?

[delicate dock]

<@&286206848099549185>

[pearl pond]

@delicate dock Has your question been resolved?

[pearl pond]

@delicate dock Has your question been resolved?

[pearl pond]

@delicate dock Has your question been resolved?

[dry yacht]

Use the parallel axis theorem. Dont integrate

integration wont work for your purpose i think

[pearl pond]

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[smoky solar]

yo im kinda stuxk on this question, I dont know how to solve it

[smoky solar]

<@&286206848099549185>

[wanton latch]

So you are given the angles of <C, <A, <DE and <CB -‘d you have also been given the length of DE+AC

What you want to find is the length of AB

[smoky solar]

yeah but like idk how to go from there

[wanton latch]

With that given information, you look at the sides of the shapes of your image

essentially it’s a triangle divided into a smaller triangle and a trapezoid. (Ik u know that just bear with me)

[smoky solar]

umm what next?

[wanton latch]

Do you know what the side splitter theorem is?

Idk if my internet is letting me send or receive messages

[smoky solar]

i do not know what that is

[wanton latch]

what side splitter theorem says is if a line is parallel to another (like DE being parallel to BC) and perpendicular to the another (DE is perpendicular to AC) then the triangle split by that line all be directly proportional

The pain in the butt part is that they gave us too much info to go off of

[smoky solar]

so...?

what does the side splitter theorem do?

[wanton latch]

Did you learn about that in class?

It helps you find the length of any missing side, but this problem isn’t giving us the necessary information that I thought could help. And if you didn’t learn anything about it in class I doubt this is what this is on

[smoky solar]

nope

im in year 9

[wanton latch]

Alright then, looks like we are going to the basics, do you know sine cosine and tangent?

[smoky solar]

yeah

SOHCAHTOA?

[wanton latch]

great!

We are going to utilize sohcahtoa to our advantage for this kind of problem

even though we don’t have many exact values, we know DE+AC= 12 and the angle of A is 30° and angle C is 90°

with that information we can isolate AC and use which trig function to solve for AB?

[leaden wadi]

Would you like some help?

[smoky solar]

um I think im kinda dumb

I dont know how to do it

is it cosine?

[leaden wadi]

There are three relationships in the problem that are "given" to you". Two are explicitly given, DE + AC = 12 and AD = BC, and the other is implied from the given information, ACB is similar to ADE.

[smoky solar]

is it because acb and ade share the same angles?

[leaden wadi]

Yes.

Because they are similar triangles, you can make the relationship, AD/AC = DE/BC.

So you have DE + AC = 12, AD = BC, and AD/AC = DE/BC.

Let's simplify things a little bit and get rid of the segment names. Let's rename them with something easier, easier for me anyways.

[smoky solar]

i see

so if we cross multiply ADxBC is 12?

[leaden wadi]

No. AD + BC =12 which is not the same as ADxBC.

[smoky solar]

oh

yeh

so ad is 6?

[leaden wadi]

No.

So given two of the relationships previously mentioned and using the renamed segments, you have the two relationships a + b = 12 and c/a = b/c.

These are two equations and three unknown variables.

That is one too many unknown variables. You need to winnow it down to two unknown variables.

[smoky solar]

so we can turn umm

[smoky solar]

a into b-12?

so c is b-12

[leaden wadi]

a + b = 12

a = 12 - b

[smoky solar]

i see

[leaden wadi]

So you have these two equations, what else can you do?

[smoky solar]

we can substirute a?

[leaden wadi]

Yes.

[smoky solar]

so b x (b-12) is equal to c?

c squared?

[leaden wadi]

c^2 = b x (12-b)

c^2 = 12b-b^2

Now look at this image. Can you think of a way to solve for b or c in terms of the other variable?

Think of SOH-CAH-TOA.

[smoky solar]

so umm

should be use sine law?

wait no

um

[leaden wadi]

What are the two sides in relation to theta?

Are they adjacent, opposite, or hypotenuse?

Is b adjacent, opposite, or hypotenuse?

[smoky solar]

opposite

[leaden wadi]

Is c adjacent, opposite, or hypotenuse?

[smoky solar]

c is adjacent

[leaden wadi]

So you have an adjacent and opposite side, which of the trig functions should you use?

SOH-CAH-TOA

[smoky solar]

tan right?

[leaden wadi]

Yes.

[smoky solar]

but does that not find the angle

[leaden wadi]

You know the angle, it was given to you by the problem.

[smoky solar]

but if we plug it into the equation do we not get tan30=b/b-12?

[leaden wadi]

tan(30) = b/c

c ≠ b - 12

a = 12 - b

[smoky solar]

oh

shoot

so we use b/tan30 to get c?

[leaden wadi]

Yes!

Or you could solve for b and get b = c x tan(30).

Either way eliminates another unknown variable.

[smoky solar]

wait I thought we dont know what c is

oh so then we can substitute that into the equation

[leaden wadi]

Yes.

[smoky solar]

so (b/tan30)^2=12b-b^2

[leaden wadi]

Yes.

[smoky solar]

but how do we solve this equation?

[leaden wadi]

Have you taken algebra?

What is tan(30) equal to?

[smoky solar]

1/2

[leaden wadi]

No.

[smoky solar]

oh wait its umm

1 right?

[leaden wadi]

No. Do you know the hand trick for the standard angles?

[smoky solar]

its like 123,321 rooted , and then divide by 2

and then take the top 2 for tan?

oh so its 1/root3

[leaden wadi]

Yes.

[smoky solar]

so then we do (b/root3)^2 = 12b-b^2

which is

[leaden wadi]

No.

tan(30) = 1/sqrt(3)

If you divide by tan(30), you end up with sqrt(3) * b.

b/(1/sqrt(3)) = sqrt(3) b

[smoky solar]

and then after its squared it will b 3xb^2?

[leaden wadi]

Yes.

[smoky solar]

so 3xb^2=12b-b^2

[leaden wadi]

Yes.

[smoky solar]

so 2xb^2 =12b

so b is 12?

[leaden wadi]

No.

[smoky solar]

wait no

i meant

6 right?

[smoky solar]

b^2=6b?

[leaden wadi]

What math operation do you perform to move the -b^2 to the left-hand side?

[smoky solar]

wait forgive me for being so dumb

i thought it was +

so 4b^2 is 12b

so 3b =b^2

b = 3?

[leaden wadi]

b^2 = 3b

b = 3 and b = 0

[smoky solar]

but it must be 3 right?

[leaden wadi]

But for this problem, b = 0, does not make any sense so you can ignore that value.

[leaden wadi]

Correct.

So b = 3. Now you can solve for c and then a.

[smoky solar]

I see

so using the a+b =12

we get that a is 9

[leaden wadi]

Yes.

[smoky solar]

and we use pythagorus theorem to get line ab?

[leaden wadi]

You need to find c first before using the Pythagorean Theorem.

[smoky solar]

c is umm

12x3-9

[smoky solar]

root 27?

[leaden wadi]

Yes, or 3sqrt(3).

So now you have the two sides of the right triangle, 9 and root(27).

What does AB equal?

[smoky solar]

its root9^2+27

root 108

[leaden wadi]

,calc sqrt(108)

[jolly parrot]

Result:

10.392304845413

[smoky solar]

Thank you So much!!

you are the best!

[leaden wadi]

yw. LPT : Try to solve this problem again from scratch and refer back to this chat, or screen save it, as a reference.

[smoky solar]

Okay thank you!

.close

[pearl pond]

Channel closed

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[pearl pond]

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[sharp smelt]

$\lim_{x\rightarrow7}\left(\frac{18-\left[1-x\right]}{\left[x-3a\right]}\right)$

[sharp smelt]

let this limit exist

where [x] is teh floor function

find a

[jolly parrot]

ƒ(Why am. I here)=misery

[sharp smelt]

$\lim_{x\rightarrow7^+}\left(\frac{18--7}{\left[7-3a\right]}\right)$

[jolly parrot]

ƒ(Why am. I here)=misery

[sharp smelt]

$\lim_{x\rightarrow7^-}\left(\frac{18--6}{\left[7-3a\right]}\right)$

[jolly parrot]

ƒ(Why am. I here)=misery

[sharp smelt]

these must be equal

so I have to find an "a" such that

$\lim_{x\rightarrow7^+}\frac{25}{\left[7-3a\right]}=\lim_{x\rightarrow7^-}\frac{24}{\left[7-3a\right]}$

[jolly parrot]

ƒ(Why am. I here)=misery

[sharp smelt]

so I was thinking, if a=1

or rather 2

we have the right hand limit=25

nah, that doesn't work

<@&286206848099549185>

I have to find a

just a hint please

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

Use .reopen if this was a mistake.

[limber oasis]

Did you solve it?

[sharp smelt]

no

but I think I'm close

is it -6