help-0

[blazing ruin]

factorised outside the bracket

[earnest pine]

shouldnt it be 2pi r^2 😭

[blazing ruin]

if its 2pi r^2 then your numerator would be (k 2pi r^2) + (2 pi r^2 sqrt)

[earnest pine]

ohhhhh okok i get it cuz it multiples with the sqrt to give pi r ^2 sqrtk^2 + 16

thank you

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[lucid flame]

can anyone help me

[lone heart]

Please don't occupy multiple help channels.

[lucid flame]

eeeyyyyyy this is occupied

find another channel

can anyone help me

pls

pls

pls

pls

pls

pls

pls

[worn fox]

Stop spamming

[languid belfry]

yea

[lucid flame]

sorry

[worn fox]

Close one of your channels

You have #help-13 , stay there

And close this one

[lucid flame]

.close

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[cerulean talon]

i was getting help from a guy but he had to go

[cerulean talon]

and i just had a quick question

i was checking something and found that m=Mgr/v^2

and it’s around 50g instead

which is correct?

also these results here are based on a fake student in the questions results

is there a reason why when i plug in the different velocities and hanging mass, the swinging mass is different by a few grams?

also i should mention that the question did ask for the values of swinging mass m

and asked in the form of y=mx+b

to find the m

and another thing is that the questions after ask to review the students experiment and point out problems with it

and then the last one was like “the student finds the weight to be 50g from the first experiment” i forget the rest of the question but yeah

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@cerulean talon Has your question been resolved?

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@cerulean talon Has your question been resolved?

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[alpine sable]

How are they calculating P(MnL)

[lament wyvern]

are being a guy and being left handed independent or not

...no wait, it's even easier: they give you P(M n L)

[alpine sable]

wait let me show you the solutions

[lament wyvern]

3/5ths are male, and 1/5th of those are left handed

this gives you P(M n L) over the entire student population

[alpine sable]

Why did they multiply

[lament wyvern]

P(M n L)=P(M)P(L|M)

[alpine sable]

isnt 1/5 P(MnL)?

[lament wyvern]

1/5th is P(L|M)

[alpine sable]

can explain in details

[lament wyvern]

sure

so the probability of P(M n L) is the probability that P(M) and also that given M, P(L). so P(L|M)

P(male and left handed) = P(male) * P(given the student is male, that they are also left handed)

right?

[alpine sable]

wait lemme process it

[lament wyvern]

if they are independent events then this simplifies down to P(A n B) = P(A)P(B) which you might be more familiar with
but fortunately we don't have to worry about that

[alpine sable]

okay, i get what what youre getting at but my question is, isnt 1/5 the same as P(MnL)?

[lament wyvern]

not quite - 1/5th is given that a student is male, they are left handed

let's assume that the chance a student is male is 1% - nice really small number.
then in 1000 students we have 10 dudes. the amount of left handed guys will then be 1/5 * 10 = 2.

but the P(M n L) cannot be 1/5 - because that says we have 200 left handed guys in the school

but we can't have 200 left handed guys when we only have 10 guys total

[alpine sable]

okay i think i get it

[lament wyvern]

black is all students
red is guys
green is left handed

[alpine sable]

so how would it be worded if they wanted to refer to P(MnL)?

[lament wyvern]

this is P(M n L) - chance of picking a student who is both male and left handed

this is P(L | M) - if we only think about guys, what's the chance of picking a lefthander

[alpine sable]

okay i got it

[lament wyvern]

[alpine sable]

i appreciate the effort ><

can i dm if i get stuck at b or c or d ? 💀

[lament wyvern]

you can ping me in here

[alpine sable]

yep

that works

thanks!

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[vocal tapir]

Can someone help me understand basis of a subspace?

[vocal tapir]

so, the definition is

if U is a subspace of some space

The multitude of vectors a1, a2, ..., an is a basis of U

if

span(a1, a2, ..., an)=U

and a1-an are liearly independent

yes?

[frail grove]

Yes

[vocal tapir]

could you please follow it up with an example?

[frail grove]

And if U is a subpace of E
The vectors a1, a2, ..., an must be in E

[frail grove]

The multitude of (0,1), (1,0) is a basis of R²

[vocal tapir]

yes

alright

so (0, 1) and (1, 0) are basis of R^2

[frail grove]

Yes or any other 2 linearly independent vectors

[vocal tapir]

such as

(1, 0), (0, 1)

or for R^3 (1, 0, 0), (0, 1, 0) etc etc

[frail grove]

No for R³ u'll need a basis of 3 vectors

Cause to generate R³ we need at least 3 vectors

[vocal tapir]

yeah

as I said

etc etc

thanks

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[muted rose]

Hey, im not sure if i did this right. Especially the part i marked red is kind of weird to me because im not sure if its possible for the imaginary number to just cancel itself out

[north perch]

in z_2 u missed the negative sign in the front

everything else is correct

[muted rose]

okay tyvm

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[lone heart]

Please don't occupy multiple help channels.

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[feral raven]

Let's break it down

In first exercise 4rth question

A=5150
B=4635
Find N to make √(5150x4635x N)=y
And y is a natural number
[0, ♾️ [

.

And how do I solve
a^²⁰²³ * b²⁰²⁴-a²⁰²³ * b²⁰²³ +1=b
And b=(√5-1)/2

We only had 1 hour for all that I just entered secondary school ;-;

I allways was excellent at math but this feels wrong for 1 hour
I'm sure I would get all in 2 hours

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[slow crag]

How do I find vector a and b only knowing their lenght and the corner between them?

[near lagoon]

It’s not asking for a,b though, just for some dot product in modulus

[slow crag]

I can do that with their lenght? How?

ok I found formula, but I still need to do addition and scalar multiplication in parenthesis

[near lagoon]

Distribute the dot product

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[slow crag]

I don't see how

[slow crag]

<@&286206848099549185>

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@slow crag Has your question been resolved?

[slow crag]

!close

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[lone heart]

@finite wren Has your question been resolved?

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[finite wren]

<@&286206848099549185>

[remote heron]

You may wanna try #groups-rings-fields

[remote heron]

.

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[desert crystal]

hi, i tried sometimes but i cant solve it

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@desert crystal Has your question been resolved?

[desert crystal]

<@&286206848099549185>

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[alpine sable]

what is this formula to get this?

[alpine sable]

how we got this is the same question

[gray isle]

binomial theorem/expansion

[alpine sable]

you told me everything

or you can say that the sum of the $n$th roots of unity is 0. since this is a third root of unity you have $\omega + \omega^2 + \omega^3 = 0$

[ocean seal]

maxim

[alpine sable]

and $\omega^3 = 1$

[ocean seal]

maxim

[alpine sable]

.close

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[summer hemlock]

A rectangular swimming pool is three times as long as it is wide and is surrounded by a deck 2.5m wide. Find the dimensions of the pool if the area of the deck is 265m^2.

[hot bluff]

area of brown is 265

[errant bough]

(wrong channel, sorry)

[summer hemlock]

And the blue is 3x^2 right

Np

What do I do once I know that

[hot bluff]

build the equation with what you know to find x.

looking at it geometrically, there are two rectangles. one bigger (the deck), and one smaller (the pool). the area of the deck then can be thought of as {area of deck + pool} - {area of pool} = 265

[lone heart]

@summer hemlock Has your question been resolved?

[summer hemlock]

In the equation you wrote what would represent the area of the deck?

[hot bluff]

the entire thing

by getting the area of both pool and deck, then subtracting the pool, you get the area of the deck itself.

imagine the biggest rectangle, then when subtracting the pool's area you are "cutting out" the pool from the rectangle and you're left with just the deck.

[summer hemlock]

So the area of the border minus the area of what's inside = 265

And the pool is whats inside so it's 3x^2 right

Wait so what would the equation be

[hot bluff]

so you said the area of the pool is 3x^2, which makes sense. but what about the area of the biggest rectangle?

we know the deck is 2.5m wide

all we need for the area of a rectangle is length and width

find the length and width of the big rectangle

[summer hemlock]

Sorry for the late reply my mom called me over

I'm not sure how to find it I don't see enough information

[hot bluff]

think about how long those sides are

[summer hemlock]

Is it 6.25

[hot bluff]

?

[summer hemlock]

I mean 18.75

I did 2.5 times 3 times 2.5

[hot bluff]

we need a length and width of the outer rectangle

does this help you?

[summer hemlock]

Oh yes

X + 5 is the width right

[hot bluff]

good

and for the longer side?

[summer hemlock]

3x + 5 right

[hot bluff]

good, now we go back to our equation {area of big rectangle} - {area of pool} = 265

put it all together

[summer hemlock]

(X+5)(3x+5) - 3x^2 = 265 right

[hot bluff]

yes

solve

[summer hemlock]

Wait I have one question

[cerulean parcel]

now the fun part begins

[summer hemlock]

Yes

[hot bluff]

its easy. when you expand you'll see.

[summer hemlock]

Why are those parts of the deck 2.5

[hot bluff]

the width of the deck is given as 2.5m

[summer hemlock]

So the width of the deck is just those parts

Oh that's what I was confused on

[hot bluff]

think of 2.5m x 2.5m squares surrounding the pool

[summer hemlock]

ok I get it now thanks

I'll solve now

I got x is 12

Ok thanks for the help I appreciate it a lot

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[alpine sable]

Can someone give an example of what this graph could show

I don’t get how curved lines work

How could it be curved

And not straight

When there’s a proportional increase

And a set amount it behind with

[tardy stag]

imagine this is the graph of how much a pizza costs ig? it wouldn't be a line

[alpine sable]

Why not

If a pizza costs 20 each

It’s

Y = 20x

Which is straight

😓

[tardy stag]

that would be how much x pizzas cost

[alpine sable]

Oh

What

[fervent timber]

the graph shows how the kostnad increases with more tids

[tardy stag]

but think instead about how much it will cost next year

[alpine sable]

Mhm

Inflation

So

[tardy stag]

yeah

[rose void]

The stress you have when a test is coming up

[tardy stag]

and especially how much it cost last year, the year before, etc

[alpine sable]

How did u know

[tardy stag]

lines will cross the x axis at some point. did a pizza have a negative cost in the past? i don't think wo

[alpine sable]

Wait so y = 20 * 1.03^x

Like 3 pro cent inflation

[tardy stag]

sure that's a good equation for inflation

[alpine sable]

Butttt

[tardy stag]

and it gets steeper with time

[alpine sable]

How is it curved

[tardy stag]

wdym how is it curved

the bigger it is, the more it grows

[cerulean parcel]

^

[alpine sable]

OHH so it gets bigger and bigger

[tardy stag]

as it goes up it also grows faster

[alpine sable]

As the number gets bigger

[tardy stag]

yeah

[cerulean parcel]

it gets bigger, faster

[alpine sable]

But percentage change stays the same

OHHHHH

[tardy stag]

bc like at some point it'll be $100 and going up by $3 in a year

ueah

[alpine sable]

so every time there’s squared root it’s curved

Basically

[tardy stag]

square, square root, exponent, pretty much anything that isn't a line is curved

[tribal haven]

you can bake it into anything that grows exponentially

e.g. a bacteria culture doubles in size every day and the cost to purge x amount of bacteria is constant

then the cost to wipe out the entire culture wrt time grows exponentially like such a graph

[cerulean parcel]

yeah when the rate of change isn't constant it will curve

[alpine sable]

Tysm I get it now

[alpine sable]

How do I know if a line is curved by looking at an equation? Should I just be looking for those

[tardy stag]

you should play with these on desmos it's great fun on a rainy day

[tardy stag]

being straight is the exception

if you can't put it into the y = mx + b box then it's curved

[alpine sable]

Ohhh

Okay

[tardy stag]

you'll learn about "parent functions" like x², sin(x), and e^x, and how to stretch and slide and combine those to get new complicated ones

[alpine sable]

So fun can’t wait

But thank you very much

[tardy stag]

yep! 😊 just keep playing with it

[alpine sable]

.close

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[jaunty flume]

Men, Women and Children are employed in the ratio 1: 2: 3 to do a work and their daily wages are in the ratio 6: 3:2. When 50 men are employed then total wages of all the workers amount to Rs. 9,000. Find the daily wages paid to 1 man, 1 woman and 1 child.

[limpid turret]

Three times as many children working as adult men.

[carmine reef]

The children yearn for the mines

[limpid turret]

https://tenor.com/view/baby-cleaning-lady-consuela-free-labor-gif-15076968

[limpid turret]

!status

[lone heart]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[zinc haven]

children working triple as many as adults, yet they get triple as less salary

🤨

[limpid turret]

Do you know how to set up ratio equations at all?

[jaunty flume]

Kinda

[limpid turret]

A good way to start on problems like this is to list your variables, and find the relations between them.

In this case, you would create variable for number of men working, number of women, and number of children being forced to work (wtf kind of problem is this).

[jaunty flume]

Idk it's in my math book

[limpid turret]

You'd also need variables for how much men are getting paid, how much women are getting paid, and how much children are getting paid to work in the coal mines.

[jaunty flume]

50×1:50×2:50×3?

200

Oh wait

300

Yes

Yes but in what way I use their daily wage?

Finding daily wage

Ok

Then it's 11

6/11 × 100

I got somewhere around 54.54 %

Man's wage

Yes

So I need to find other percentage

Ok

4919.5

50

300?

Ok

Ok

I got 24.57 for woman

1636.2

10.90

Thanks for help

.close

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[stuck geode]

my answer is x=3/2 and -1/2

and since 3/2 lies in [0, 2}

such that (such that f'(3/2) = 5)

proving that intermediate value theorem applies

<@&286206848099549185>

could someone check?

[barren portal]

so you get c=3/2?

[stuck geode]

yes

sorry typo

[barren portal]

what did you got ?

[stuck geode]

i got

c=3/2

and -1/2

[barren portal]

What’s f’(x)?

[stuck geode]

is given

[barren portal]

f’(x)= 4x^2-4x+2 right?

[stuck geode]

wasnt f'(x) that?

[barren portal]

yes, my typo

[stuck geode]

we just

[barren portal]

just plug f’(3/2)

[stuck geode]

yea

[barren portal]

and see if you get 5

[stuck geode]

h

i got 5

so its correct right?

[barren portal]

yes, that was your aim

to get a c, such that f’(c)=5

[stuck geode]

ok

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[dry kestrel]

Hello, I'm having problems on this alphametic...
B and R could be: 4,5,6,7,8,9
WB could be: any two-digit number which square gives a three digit number.

[dry kestrel]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

.close

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[dreamy willow]

Can someone help me with this?

[wraith vigil]

with what

[tacit arch]

!da2a

[lone heart]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

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[manic harness]

How do I parameterize for r(t)?

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[gleaming ivy]

This is my question and I am horrible at time management. I'm lost and its due in 14 minutes

please help😢

this is what i have

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[stiff harness]

What steps would I need to take to solve this?

[stiff harness]

I've started out by dividing both sides by 2, but not sure where to go from there.

[pseudo ice]

<@&268886789983436800> spam

[sour verge]

Get a general solution for cos(y)=1/2

Then once you find an expression for y, set y=3x and divide through.

You will get infinitely many solutions, but you can then take the first few before going higher than pi

[stiff harness]

I see, I'll give this a try!

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[solemn urchin]

how is this exact when cos(x+y) dy and cos(x+y)dx give diffrent values?

[last ether]

Well you kinda have to think of it like a vector field

Let's say that F(x,y) = c

Then we can say that $\pdv{F}{x}+ \pdv{F}{y} \dv{y}{x} = 0$ is an exact equation

[ocean seal]

Umbraleviathan

[last ether]

Now let $\phi_x$ and $\phi_y$ be $\pdv{F}{x}$ and $\pdv{F}{y}$ respectively

[ocean seal]

Umbraleviathan

[last ether]

Due to young's theorem, $\pdv{\phi_x}{y} = \pdv{\phi_y}{x}$

[ocean seal]

Umbraleviathan

[last ether]

And that's how you can test if a diff eq is an exact equation @solemn urchin

[solemn urchin]

Thanks

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[torn isle]

,rotate

[torn isle]

,rotate

[torn isle]

May I know where I did wrongly? I checked the answer shouldn’t have -pi/2

Or just like this

For b

,rotate

[last ether]

@torn isle I did the work myself, the book's off

I'm also getting what you're getting

Lemme double check

Yeah no you're correct

[torn isle]

@last ether this is what taught by the tutor which I don’t really agree and I think she did wrongly

[torn isle]

The answer I got from online

<@&286206848099549185>

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@torn isle Has your question been resolved?

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@torn isle Has your question been resolved?

[lone heart]

@torn isle Has your question been resolved?

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@torn isle Has your question been resolved?

[lavish cave]

it should be 1/2 arctan(2y) in the 4th line

u = 2y, du = 2 dy

yeah so your tutor is wrong

[torn isle]

.close

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[lilac thicket]

Hey everyone!
I am struggeling to understand the concept of the vertex space as explained here.
For example if I had like a C4 Graph. What would that Vertex Space even look like ?

[lilac thicket]

Another question would be what that of all function V -> F_2 is supposed to look like. Would that be the functions that map a vertex to a boolean matrix ? And if so why would we wanna do that

or is this just a map that maps each vertex to either 0 or 1 giving us a boolean encoding of a subset of V ?

[mortal trellis]

each vertex to 0 or 1, yes

[lilac thicket]

Oh then the dimension thing would make sense since the basis of the vertex set would consist of n , n x 1 vectors with exactly 1 distinct entry set to 1

[mortal trellis]

note that the vertex space doesnt depend on the edges, so it doesnt really depend on the graph

[lilac thicket]

why is that of importance here ?

[mortal trellis]

well you asked for the specific case of C4

[lilac thicket]

ohh

I see

[mortal trellis]

it doesnt actually matter which graph with 4 vertices you take

the vertex space is the same

[lilac thicket]

OHH

its just about n-Vertex sets I see

[mortal trellis]

(as long as you call the vertices the same obviously, otherwise just isomorphic yada yada)

[lilac thicket]

yes sure

so the vertex space is basically just a set of 2^n , n by 1 vectors encoding every possible subset of n vertices ?

[mortal trellis]

yes

[lilac thicket]

I understand

funny because in programming thats extremely common but here I got super confused

thank you very much for helping me out !!

(once again)

[mortal trellis]

its really not the best description

[lilac thicket]

you mean the way I described it ?

or as in the book

[mortal trellis]

the text

[lilac thicket]

oh yeah I agree lol

thanks !!!

.close

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[worthy marten]

Is there anything wrong here?

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@worthy marten Has your question been resolved?

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@worthy marten Has your question been resolved?

[celest terrace]

you can also factor out a 4 from 4x-4y which makes it (1/2)ln(4(x-y)), then use more log laws to simplify that

otherwise it looks fine

[lone heart]

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[rancid gull]

A

[rancid gull]

An arithmetic series has first term 1 and common difference 4
Find the sum of all terms of the series from the 41st term to the 100th term inclusive.

Is the answer 16740 or 7140

[minor needle]

,w (21+(100-1)4)/2100-(21+(40-1)4)/240

[ocean seal]

Results provided by WolframAlpha

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

Requested by modus7591

[rancid gull]

Tyy

,close

☠️

[minor needle]

.close

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[grizzled trail]

A tree breaks and the broken part bends such that the top of the tree touches the ground making an angle 45° with it. The distance between the foot of the tree to the point where the top touches the ground is 36 m. Find the height of the tree.

[lusty hearth]

try drawing a diagram of the tree and ground

the legs are the ground and unbroken tree

[grizzled trail]

okayy

[lusty hearth]

have you learned about special right triangles

[grizzled trail]

no

[lusty hearth]

sin, cos, tan?

[grizzled trail]

yes

[lusty hearth]

use sin, cos, and tan to find the other 2 side lengths

[grizzled trail]

tan or sin?

what to use

[lusty hearth]

you use tan to find AB, and cos to find AC

[grizzled trail]

i kinda did not listen in class when this was being taught

oki

[lusty hearth]

add the two side lengths together to get the full tree

[lusty hearth]

sin means opposite over hypotenuse, cos is adjacent over hypotenuse, and tan is opposite over adjacent

[grizzled trail]

some options have 1+ root 2 and all

[lusty hearth]

do you know how to use tangent to find AB

[grizzled trail]

no

[feral raven]

I think it's the only way tho

[lusty hearth]

$tan\theta=\frac{o}{a}$

[ocean seal]

Ruby

[lusty hearth]

we want to find the opposite side

[feral raven]

Yeah ab

[lusty hearth]

so multiply both sides by a

giving adjacent times sin(angle)

[feral raven]

I'm not the question asker

[lusty hearth]

or 36 times sin45

[grizzled trail]

ooh

[lusty hearth]

,calc 36sin45

[ocean seal]

The following error occured while calculating:
Error: Undefined symbol sin45

[lusty hearth]

,calc 36sin(45)

[ocean seal]

Result:

30.632526883228

Mathematics (268882317391429632)

​ ​ ​ ​ ​ ​ ​ ​ Owner: mniip (207092805644845057)
​ ​ ​ ​ ​ ​ ​ ​ ​ Icon: Icon Link
​ Verification: High | Must be a member for more than 10 minutes
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​ ​ ​ ​ ​ ​ ​ ​ Roles: 70
​ ​ ​ ​ ​ ​ Members: 180871 humans, 8 bots | 180879 total
​ ​ ​ ​ ​ Channels: 152 text, 10 voice, 16 categories, 1 stage, 5 forum | 184 total
Server Boosts: Level 3 | 99 boosts total
​ ​ ​ Created at: <t:1484176940:F>

[lusty hearth]

hold on

,calc sin(1)

[ocean seal]

Result:

0.8414709848079

[lusty hearth]

yeah thats in radians hold on ima get the right calculation for u

[grizzled trail]

oki

[grizzled trail]

should i take decis?

[lusty hearth]

no

[grizzled trail]

oki

[lusty hearth]

hold on somethings wrong

[grizzled trail]

uhm

[lusty hearth]

OHHH

i swapped tan and sin along the way

it should be 36tan45

so side AB is 36ft

@grizzled trail

but theres more

we need to find AC now

same idea but with cosine

$cos\theta=\frac{a}{h}$

[ocean seal]

Ruby

[lusty hearth]

so to find the hypotenuse, we need to multiply both sides by h, then divide by $cos\theta$

[ocean seal]

Ruby

[grizzled trail]

i see

[lusty hearth]

leaving us with $h=\frac{a}{cos\theta}$

[ocean seal]

Ruby

[lusty hearth]

or 36 over cos45

[grizzled trail]

hm

[lusty hearth]

but we need an answer in terms of square roots

[grizzled trail]

yes

[lusty hearth]

cos45 is always $\frac{\sqrt{2}}{2}$

[ocean seal]

Ruby

[lusty hearth]

when dividing by a fraction, you can mult by reciprocal

so $\frac{36(2)}{\sqrt{2}}$

[ocean seal]

Ruby

[lusty hearth]

multiply by $\frac{\sqrt{2}}{\sqrt{2}}$

[ocean seal]

Ruby

[lusty hearth]

you get $36\sqrt{2}$

[ocean seal]

Ruby

[lusty hearth]

so the total height of the tree is $36+36\sqrt{2}$

[ocean seal]

Ruby

[lusty hearth]

@grizzled trail

[grizzled trail]

yes i saw it

i got it

tysm

.close

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[serene moon]

hi

[serene moon]

i dont get how that simplifies to that can any1 explain

is it right?

[lusty hearth]

you distribute the x/|x|

and the sqrt can be written as |x|^1/2

exponent rules

|x|^1/2 times |x|^1 becomes |x|^3/2

[serene moon]

so which x goes on the numerator and which goes on den

[lusty hearth]

the numerator(x) of the fraction stays on the numerator of the bigger fraction, and the denominator(|x|) goes to the denominator of the bigger fraction

[serene moon]

aight hm

[serene moon]

which rule is this?

oh wait

ig im getting kinda confused cause its in the denominator

[subtle birch]

,tex .exp rules

[ocean seal]

B-eard

[lusty hearth]

product rule

[serene moon]

cause of the rule that states with same base, you can just add the exponents right

[lusty hearth]

yes

[serene moon]

ok thanks

just gonna go over it

.close

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[vocal tapir]

for

[vocal tapir]

-7+7i

[void nymph]

yes

[vocal tapir]

r=7sqrt(2)

sin=sqrt(2)/2

cos=-sqrt(2)/2

what's the argument?

[void nymph]

what's an argument again 💀

im so shit with terms

[vocal tapir]

is it this?

[void nymph]

oh for normal form?

yes i guess

[vocal tapir]

I don't need guessing

tysm

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[serene moon]

if you use lhopital rule and it still evaluate to 0/0 at the limit

[serene moon]

do u just use l'hoptial again?

[mortal trellis]

yes

[tawny crown]

Yes you can

[serene moon]

or does that mean that the limit there doesn't exist

i'm confused

[serene moon]

it wants me to evaluate the limit at 0 does it not exist here? cause i graphed it on desmos and used lhopital and its still indeterminate

[mortal trellis]

well use lhopital again

[serene moon]

yeah but shouldnt there be some limit on the graph at 0?

[mortal trellis]

well the limit according to the graph is apparently -1

[serene moon]

or is this the limit or something

oh

yeah

[lone heart]

@serene moon Has your question been resolved?

[lone heart]

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[rotund crater]

why is 6-x?

[snow cobalt]

what?

[carmine reef]

f(x) is 4-x

So f(x)+2 is 6-x

[rotund crater]

Because I can't apply the formula directly

Why should you do that?

[rotund crater]

why

Wait

i see

[rotund crater]

"I see now, correct me if I'm wrong: if it were y = 4-x and
y=−2, and you take their difference, you would have the complete area, but that's not what you want. What you want is for the area to be above the x-axis or the axis of rotation, which is why you add 2 to make it above, right?"

[carmine reef]

i mean you are taking their difference

you're subtracting -2

if you now just integrated the difference it would give you area, which isn't what you want

so instead you take the difference and do the volume formula thing with it

[rotund crater]

Sooo I m good?

[lone heart]

@rotund crater Has your question been resolved?

[carmine reef]

probably?

it's hard to tell what you're doing

[lone heart]

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[floral jacinth]

How do I find the angle?

P+q = 315i+196j

Do I just use
Tan ∅= 196/315??

This results in the angle 31.9⁰

Although I'm not sure if this is the answer

<@&286206848099549185>

[mighty robin]

<@&286206848099549185>

[lone heart]

@floral jacinth Has your question been resolved?

[floral jacinth]

<@&286206848099549185>

[mighty robin]

<@&286206848099549185>

[lone heart]

@floral jacinth Has your question been resolved?

[lone heart]

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[tough badge]

more limits 🤓

[tough badge]

I think I got a) but Im not sure

all of the limits are 3

and k = 3

[echo socket]

That's correct

Have you tried part b so far?

[tough badge]

okay, Im not rly sure why k = 3

its just makes sense xD

[echo socket]

You showed that $\lim_{x\to1^-}f(x)$ and $\lim_{x\to1^+}f(x)$ are both equal to $3$, right?

[ocean seal]

A Lonely Bean

[tough badge]

My understanding is that its because both of them 1- and 1+ are 3

so it like joins them together

idk how to explain it

[echo socket]

Just say that therefore $\lim_{x\to1}f(x) = 3$ and we would like to choose the value of $f(1)$ so that it is equal to $\lim_{x\to1}f(x)$, i.e., we need $f(1) = 3$ and $f(1)$ is defined to be $k$. We hence choose $k = 3$.

[ocean seal]

A Lonely Bean

[tough badge]

yes indeed

ok ye but part b kinda stumped me

[echo socket]

Have you evaluated the one-sided limits in terms of a so far?

[tough badge]

idk how to do that

as in g(1) = a+1 ?

ok no thats cant happen

cuz x isnt < 1

[echo socket]

No, I am talking about the limits, what do you think $\lim_{x\to1^-}ax^2 + 1$ is equal to?

[ocean seal]

A Lonely Bean

[tough badge]

a + 1

[echo socket]

Right

And what about $\lim_{x\to1^+}2x+a$?

[ocean seal]

A Lonely Bean

[tough badge]

a + 2

[echo socket]

Will a+1 and a+2 ever be equal?

[tough badge]

nope

oooooh

[echo socket]

Neither will the limit exist

[tough badge]

:}

yes indeed

I see now

and if the 2 limits arent ==

it cant be continous right

[echo socket]

$\lim_{x\to a}f(x)$ exists if and only if $\lim_{x\to a^+}f(x)$ and $\lim_{x\to a^-}f(x)$ exist and are equal

[ocean seal]

A Lonely Bean

[echo socket]

If $\lim_{x\to a}f(x)$ does not exist, then $f$ is automatically not continuous at $a$

[ocean seal]

A Lonely Bean

[echo socket]

Uh assuming it is also defined on a neighbourhood around a but that's too much of detail

[tough badge]

alright

U wanna help me with another?

[echo socket]

Sure

[tough badge]

part a is cuz its a polynomial I know that

unless my TA was capping

[echo socket]

For part b it looks like you are expected to apply intermediate value theorem

[echo socket]

If you have so far shown that sum of continuous functions is also continuous and that the power functions are continuous, you should be fine to just say that f is a polynomial and hence continuous

[echo socket]

Can you recall it?

[tough badge]

no idea what that is tbh

[echo socket]

Given a continuous function $f$ and two real numbers $x_1, x_2$, for every $k \in [f(x_1), f(x_2)]$ or $c \in [f(x_2), f(x_1)]$ (depending on which one of $f(x_1)$ or $f(x_2)$ is greater), there exists $c$ between $x_1$ and $x_2$ with $f(c) = k$

[ocean seal]

A Lonely Bean

[echo socket]

This theorem allows you to find intervals where a root of a function lies

Simply show that your function is negative at some point and positive at another point

Then it must follows that it is equal to 0 somewhere between those points

Assuming the function in question is continuous, of course

[tough badge]

ooh that makes sense actually

for once

so I sub in numbers lets say until I get an answer that is less than 0 and one that is greater than 1

and that proves those 2 points

[echo socket]

Not necessarily greater than 1

As long as the value of f is negative and positive at two points, there will be a root inbetween

[echo socket]

So try calculating f(0) and f(1) in this case

And see if they have different signs

If they do, it means there is a root of f in the interval (0, 1)

[tough badge]

is it because subbing in points as in like, lets say u get -1 and 1

it means its going up and down

as in

u know what I mean?

[echo socket]

Yes but it may go down somewhere there

Overall it still increases though

And it will hit every value between -1 and 1 as a consequence of continuity

A fun exercise regarding ivt would be something like this

Let f be a real continuous function with f(3) = 2 and f(5) = 7, show that there is a real number x with f(x) = x

[tough badge]

huh

now u gone and confused me lol

[echo socket]

Where exactly?

[tough badge]

if f(3) = 2

how can f(x) = x

[echo socket]

f(3) = 2 and f(5) = 7

Try plotting the points (2, 3), (5, 7) and the line y = x

You will see that you can't go from one point to another without intersecting the line

[echo socket]

This exercise asks you to prove that using ivt

And I didn't say f(x) = x for all x just in case

The conclusion should be that there is one such x

[tough badge]

ye but ur question legit contadics itself xD

[echo socket]

Where?

[tough badge]

if y = x

y = 3

x = 5

3 != 5

[echo socket]

I never said y = x is the graph of f

[tough badge]

u say f(x) = x

[echo socket]

I say there exists at least one x such that f(x) = x

I didn't say that x is 3

or 5

[tough badge]

ye alri

U still there?

@echo socket

[echo socket]

Yeah

[tough badge]

idk if im not supposed to @ you

how do I prove if a limit doesnt exist lol xD

[echo socket]

I'm fine with pings

You don't need to deal with limits, you are given that f is continuous

[tough badge]

oh Ive moved on now

Im doin this now

[echo socket]

Looks like just factorizing

[tough badge]

mhm

honestly

Ima be honest

I just read the question again...

I didnt see this

xD

[echo socket]

So what did you get for parts a and b?

[tough badge]

0/0

:I

bro I remember I got this before and I did somthing but now I forgot

[echo socket]

Factor the quadratic expressions and cancel out some terms if possible

Do you know how to factor quadratics?

[tough badge]

ye

but it takes me a while xD

[echo socket]

There is a trick of just guessing two numbers whose product is the constant coefficient and whose sum is minus the coefficient of x

For example, when I see x^2 - 7x + 12, I immediately see that 3 and 4 are roots because 3 *4 = 12 and 3 + 4 = 7, meaning the quadratic can be factored as (x - 3)(x - 4)

[tough badge]

x-2/x-4

[echo socket]

Yeah

The fraction simplifies to (x - 2)/(x - 4)

What's the limit of that as x -> 3?

[tough badge]

-1

[echo socket]

And what about x -> 4?

[tough badge]

🙂

does not exist

u are smart boi

[echo socket]

Right

[echo socket]

Thanks

[tough badge]

Ok ima let u rest from my brain

ty

fr

[echo socket]

You are welcome

[tough badge]

.close

[lone heart]

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[tardy fern]

Can I do this problem with degrees instead of radians?

[tardy fern]

dtheta/dt = 1080 degrees

[jagged cobalt]

why would you want to

[tardy fern]

im curious if it's possible

and if not, why?

[tardy fern]

so substituting that into the equation, i get dx/dt = 5760 miles/min which is clearly not the same

[jagged cobalt]

derivatives get dodgy if you dont

[tardy fern]

?

[jagged cobalt]

the derivatives of trig functions you know are made on the basis that the angle is in radians

[tardy fern]

really?

[jagged cobalt]

if you do it in degrees you have sin(x) becoming sin(xpi/180) for a degree angle then when you differentiate youd get pi/180 cos(xpi/180)

etc

it would be a bunch of chain rule thats unnecessary

[tardy fern]

oh ok ty

.close

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[jagged cobalt]

i wrote that wrong oops

[tardy fern]

.reopen

[lone heart]

✅

[jagged cobalt]

there

[tardy fern]

ok

.close

[lone heart]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[serene moon]

can anyone help me on this question

[serene moon]

im completely lost

[keen karma]

why are you doing this is multiple channels

its purposeful disruption

@brisk inlet

[velvet veldt]

diagram?

[serene moon]

there's no diagram, but there's this one online

i think youre supposed to find the right balance of when the cable has to go underwater and when it's underground

but i just have no idea where to start even given that diagram

[velvet veldt]

hmm

maybe like a quadratic and try to optimize stuff

[serene moon]

yea but idk where tf to formulate the quadratic lmao

[velvet veldt]

3(sqrt(16 + x^2)) + 14 - x

and try to manipulate

idk tho i could be wrong

if you are allowed to use that IM path

[serene moon]

where'd you get that from

[velvet veldt]

let’s just say underground is 1 dollar and underwater is 3

[serene moon]

yea

[velvet veldt]

i used pythagorean theorem to get the distance of IM and then added the rest of the way MB

[serene moon]

ight thanks

.close

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[lone heart]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cursive edge]

can someone help with what test to prove series is convergence or divergent? (5^(-n)+10n)/(6n^2+4^(-n))

[lament forge]

i don't know how to prove it formally but intuitively it feels like the 5^-n and 4^-n should be pretty much irrelevant for large n

and if you just ignore those then you get 10n/6n^2 which is a lot easier to work out

[slow haven]

Uhm

[cursive edge]

its a series

[slow haven]

are you familiar with the "algebra of limits"? @cursive edge

oh

oh

Sry Boss

Hehe

[foggy pecan]

the series diverges because the necessary condition for convergence for your series is not met

[cursive edge]

it is

its lmit is 0

[foggy pecan]

no it i snot true

[cursive edge]

it is

its limit is 0

[foggy pecan]

you are wrong, the nominator

is a way greater then donimantor

[cursive edge]

10/6n

that limit is 0

its a series

[foggy pecan]

yo wrote 5^n

[cursive edge]

(5^(-n)+10n)/(6n^2+4^(-n))

yes

10n and 6n^2

ominates

[foggy pecan]

this $\sum_{n}^{\infty }\frac{5^{-n}+10n}{6n^{2}+4^{-n}}$ ?

[ocean seal]

Joanna Angel

[cursive edge]

yes

10n and 6n^2

dominates

[foggy pecan]

yes ok, i did not see it lookign at yor notificaiton )

so oen can use the comparison test in limes version

$\sum_{n}^{\infty }\frac{5^{-n}+10n}{6n^{2}+4^{-n}}=\sum_{n}^{\infty }\frac{4^{n}+10n\cdot 5^{n}\cdot 4^{n}}{6\cdot 4^{n}\cdot 5^{n}\cdot n^{2}+5^{n}}$

[ocean seal]

Joanna Angel

[foggy pecan]

$\text{then }b_{n}=\frac{1}{n}$

[ocean seal]

Joanna Angel

[foggy pecan]

$\lim_{n \to \infty }\frac{a_{n}}{b_{n}}>0\text{ finite limes}$

[ocean seal]

Joanna Angel

[oak chasm]

You can prove that f(n) - 1/n > 0 for all positive n, so f(n) > 1/n, so it diverges.

[foggy pecan]

yes that is the same almsot method

[cursive edge]

wait

it converges

i compareid t iwth

1/n^2

u got it mixed up

an < bn

if bn converges then it als oconverges

[oak chasm]

What did you find when you compared it to 1/n^2?

[cursive edge]

if bn < an

and bn diverges

then an converges

1/n^2 converges

so t must also converge

cause an < bn\

[oak chasm]

No, your series doesn't converge.

[cursive edge]

how

[oak chasm]

Because it sums to infinity.

It's greater than 1/n for all positive n.

So, since 1/n sums to infinity, yours must sum to that or higher.

[foggy pecan]

divergence is also proved base don comparison test in lim version

[cursive edge]

its not

greater than 1/n

how is it

its smaller than 1/n

the limit is 0

[oak chasm]

Prove that your expression minus 1/n must be greater than zero for all positive n.

[cursive edge]

because

10/6n

thats after simplifying it

with the greatest powers

[foggy pecan]

In the limit comparison test, you compare two series Σ a (subscript n) and Σ b (subscript n) with a n greater than or equal to 0, and with b n greater than 0. Then c=lim (n goes to infinity) a n/b n . If c is positive and is finite, then either both series converge or both series diverge.

[cursive edge]

1/n < 10/6n

[oak chasm]

The limit of 1/n is also zero, but it diverges.

[foggy pecan]

yes that i shwoed above

[cursive edge]

1/n < 10n/6n^2

?

[foggy pecan]

$lim\frac{a_{n}}{b_{n}}=\frac{10}{6}>0$

[ocean seal]

Joanna Angel

[foggy pecan]

since series b_n is divergent

then also a_n