help-28

[tacit siren]

Show the set D \subset C(A, R^m) of functions with norm less than 1 is closed and bounded but not compact

Hint says to show a sequence of D is not equicontinuous

Then use a modified version of arsela-ascoli for R^n

f_n(x) = x^n for A= [0 1]

Next

Suppose $A$ is a compact subset of $R^m$, $f:R^m \rightarrow R^m$, and $\mathscr{B} is a subset of continuous functions from $A$ to $R^m$ such that for each$ x_0$ and $\epsilon$, there exists a $\delta$ such that $d(x, x_0)< \delta$ implies $d(f(x), f(x_0)) < \epsilon$ for all $f$ in $\mathscr{B}$.

Prove Script B is equicontinuous

Ok so

I'm guessing I just take all the B(x, delta_x)

And take a finite cover

Wait hm

Guess there's more to it than that

Ok I dunno

<@&286206848099549185> i could use a hint im kinda tired

Uhh the current problem is the message at 10:06 AM

West coast time

Suppose $A$ is a compact subset of $R^m$, $f:R^m \rightarrow R^m$, and $\mathscr{B}$ is a subset of continuous functions from $A$ to $R^m$ such that for each $ x_0$ and $\epsilon$, there exists a $\delta$ such that $d(x, x_0)< \delta$ implies $d(f(x), f(x_0)) < \epsilon$ for all $f$ in $\mathscr{B}$. Prove that $\mathscr{B} $ is equicontinuous.

I figure we cover $A$ with $B(x, \delta_{x, \epsilon})$ and take a finite subcover of that, but im not sure where to go from there

[glossy valve]

992HallucigeniaEnthusiast

[tacit siren]

Gonna take a nap brb

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[hardy vigil]

I need help
quick dm me
grade 7 linear equation and also inequalities

[wise wyvern]

Why can't a game developer do grade 7 linear equations?

Also, why dm? Just post it here.

[somber niche]

Likely a grade 7 game dev

[shy shore]

maybe the game isnt math related.. somehow

you'd think hp systems and such would require nearly every game to be math related

[somber niche]

Is it possible for a computer game to not be math related in terms of development?

[wise wyvern]

@hardy vigil

[wise wyvern]

Not really, no.

[shy shore]

this seems like a great challenge but we'd need to define math related

no simple arithmetic? absolutely not, no complex functions, probably

[somber niche]

Like, utilizing at least 1 algebraic expression in the code for example

but then logical statements also fall in math

hmm...

[shy shore]

rendering the game would take math, i assume we ignore that factor

is num1++ allowed? are all forms of arithmetic allowed?

[torn jolt]

Y'all arguing over what?

[shy shore]

seems like bored discussion

[torn jolt]

Damn what you upto

[wise wyvern]

More like, I already know the idea of game development. And what it does use, is infact mathematics. I'd call it mathematics all day. So nothing left to discuss lol. But this isn't the right channel for it, so...

[shy shore]

very good point

[hardy vigil]

1. Preeta watches an ant and a beetle crawl in a hole in the ground. The beetle is 3/4 inch below ground level. The beetle climbs 1/3 the distance the ant is below ground level. The beetle is now 2 1/2 inches below ground level.

(a) Let x = the position of the ant relative to ground level. What equation can be written to solve for x?

(b) Solve the equation from Part (a). Show your work.

(c) What is the distance between where the ant is and where the beetle is? Answer the question in a complete sentence. Show your work.

[full forum]

Show your work, and if possible, explain where you are stuck.

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

@hardy vigil Has your question been resolved?

[hardy vigil]

<@&286206848099549185>

[west mist]

?

[hardy vigil]

My questionn

Preeta watches an ant and a beetle crawl in a hole in the ground. The beetle is 3/4 inch below ground level. The beetle climbs 1/3 the distance the ant is below ground level. The beetle is now 2 1/2 inches below ground level.

(a) Let x = the position of the ant relative to ground level. What equation can be written to solve for x?

(b) Solve the equation from Part (a). Show your work.

(c) What is the distance between where the ant is and where the beetle is? Answer the question in a complete sentence. Show your work.

[hardy vigil]

can u help

[west mist]

Ok am gonna see

[hardy vigil]

i have only done till

3/4 + 1/3x = 2 1/2
3/4 + 1/3x = 5/2

[west mist]

Let's solve this problem step by step:

Given information:

• The beetle is 3/4 inch below ground level.
• The beetle climbs 1/3 of the distance the ant is below ground level.
• After climbing, the beetle is 2 1/2 inches below ground level.

Let x be the position of the ant relative to ground level.

(a) Equation to solve for x:
The beetle starts at 3/4 inch below ground level and climbs 1/3 of the distance the ant is below ground level. After climbing, the beetle is 2 1/2 inches below ground level.

So, the equation can be written as:
3/4 + 1/3x = 2 1/2

(b) Solve the equation:
To solve the equation, we need to find x that satisfies the equation:
3/4 + 1/3x = 2 1/2

Convert 2 1/2 to an improper fraction:
2 1/2 = 5/2

Now, solve for x:
3/4 + 1/3x = 5/2
1/3x = 5/2 - 3/4
1/3x = 10/4 - 3/4
1/3x = 7/4
x = (7/4) * 3
x = 21/4
x = 5 1/4 inches

(c) Distance between the ant and the beetle:
To find the distance between where the ant is and where the beetle is, we calculate the absolute difference between their positions:
Distance = |x - 3/4|
Distance = |5 1/4 - 3/4|
Distance = |5 - 3| = 2 inches

Therefore, the distance between where the ant is and where the beetle is 2 inches.

[west mist]

Let's continue solving the equation:

3/4 + 1/3x = 5/2

First, let's get a common denominator for 3 and 4 on the left side:

3/4 + 1/3x = 5/2
3/4 + 1/3x = 10/4

Now, combine the fractions on the left side:

3/4 + 1/3x = 10/4
3/4 + 1/3x = 10/4

Now, subtract 3/4 from both sides to isolate the term with x:

1/3x = 10/4 - 3/4
1/3x = 7/4

Multiply both sides by 3 to solve for x:

x = (7/4) * 3
x = 21/4
x = 5 1/4 inches

So, the position of the ant relative to ground level is 5 1/4 inches.

[hardy vigil]

??????

[west mist]

I solve it what?

[hardy vigil]

so this is the answer?

[west mist]

Yeah

[hardy vigil]

sure?

[west mist]

Yeah

[hardy vigil]

cool

thx man

[west mist]

Your welcome

[shy shore]

<@&268886789983436800> this cant not be chatgpt

[west mist]

Bro what is wrong with you?

[spiral vigil]

what is wrong with you

do you think we can't go to chatgpt ourselves?

provide value or leave.

[shy shore]

alot

[hardy vigil]

we need human help

[shy shore]

but no chatgpt

also dont give the answers thx

[hardy vigil]

can you help

.reopen

[shy shore]

since youve gotten the answer from chatgpt i think you can follow along, do you have any specific questions about the solution i can help you with tho

[hardy vigil]

i dont get the answer

i think it contains an error

i would like it if you could resolve it

[shy shore]

im sorry but i cant just provide the answer for you, i could try guiding you im not exceptional at helping people though

[hardy vigil]

no problem just guide me

[shy shore]

alright so you have finished a) right?

[hardy vigil]

yeah

3/4 + 1/3x = 2 1/2
3/4 + 1/3x = 5/2

[shy shore]

alright, now we can start to solve it, do you know how to solve a linear equation?

[hardy vigil]

......

i know but i am bad

REAL BAD

[shy shore]

its alright, just try to write it step by step and ill try to correct you if you go wrong

[hardy vigil]

x = 21/4????

[shy shore]

could you write it step by step?

[hardy vigil]

3/4 + 1/3x = 2 1/2
3/4 + 1/3x = 5/2
1/3x = 5/2 – 3/4
1/3x = 10/4 – 3/4
1/3x = 7/4

i did it till here

i dunno what to do further

[shy shore]

do you know how to go from 1/3x = 7/4 to x = 21/4

or am i misunderstanding you

[hardy vigil]

multiply?

7x3 = 21

4x1 = 4

[shy shore]

yes exactly

[hardy vigil]

21/4

[shy shore]

for multiplication i recommend using * as it doesnt cause confusing with the variable x

[hardy vigil]

k

[shy shore]

great so now you have x = 21/4 so you have solved b)

Preeta watches an ant and a beetle crawl in a hole in the ground. The beetle is 3/4 inch below ground level. The beetle climbs 1/3 the distance the ant is below ground level. The beetle is now 2 1/2 inches below ground level.

(a) Let x = the position of the ant relative to ground level. What equation can be written to solve for x?

(b) Solve the equation from Part (a). Show your work.

(c) What is the distance between where the ant is and where the beetle is? Answer the question in a complete sentence. Show your work.
copying it down for accessibility

[hardy vigil]

What is the distance between where the ant is and where the beetle is? Answer the question in a complete sentence. Show your work.

[shy shore]

do you have an idea how to approach c?

[hardy vigil]

nope

😦

[orchid osprey]

yo

[shy shore]

occupied

or are you here to help?

[shy shore]

alright so how would you go about calculating the distance between for example 2 items where one of them is at 1cm and the other is at 3cm

[hardy vigil]

subtract

but what should i subtract

???

[shy shore]

to calculate the example what do you think the best option is

[orchid osprey]

we substract

the dist would be 2cm

[hardy vigil]

how????????

[shy shore]

if you subtract the 1 cm from the 3 cm you get the difference between the distances, the difference is the distance between them

[hardy vigil]

2

[orchid osprey]

ya

[shy shore]

now you can apply that to the distance of the ant and beetle

[hardy vigil]

are you talking bout my question or the example

[hardy vigil]

bro....

we found x

which is the value of the ant

[shy shore]

knowing the beetle is 2 1/2 inch under the ground, and the ant is 21/4, or 5 1/4 inch under

[hardy vigil]

and the beetle is 2 1/2

[shy shore]

oops yeah

[hardy vigil]

so..

[shy shore]

now we can use the subtract we used in the example with this situation

[hardy vigil]

we would 2 1/2 - 21/4

[shy shore]

almost

[hardy vigil]

5/2 - 21/4

or

21/4 - 5/2

[shy shore]

since the 5 1/4 is bigger we subtract the 2 1/2 from that instead,
5 1/4 - 2 1/2

[hardy vigil]

21/4 - 5/2?

[shy shore]

yes

[hardy vigil]

thanks man you are the best!!

[shy shore]

np, now just formulate it into a sentence and ur assignment should be completed

[hardy vigil]

btw what is ur instgram

[shy shore]

i dont have it, why

[hardy vigil]

nah just wanted to be friends:)

[shy shore]

😂

idk if i can close this channel could you?

[hardy vigil]

.close

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[worthy ginkgo]

what have you tried so far?

[celest zinc]

It's now tan(2arcsin(lsinxl+lcosxl))

If we use tan 2theta

It becomes complicated

<@&286206848099549185>

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[plush egret]

.close @woeful anvil

can you see this?

[plush egret]

oops

haha

.close

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[woeful anvil]

only show up as

suggested

[plush egret]

oh weird

[vestal meadow]

why is that there?

[sudden spear]

New discord feature

[vestal meadow]

oh

I knew they changed it back recently

didnt know they added new feature

[plush egret]

.reopen

[full forum]

✅

[sudden spear]

You probably have to sign up to see all the help channels in here

[umbral dome]

use id:customize and select "i need help with math" to get access to all help channels

[woeful anvil]

wroked

[woeful anvil]

🫡

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[nova topaz]

Is there a quick way to work this out without having to do a lot of working out?

[nova topaz]

My first thought is let A be the first matrix. 2 * column 1 of A + x * column 2 of A -3 * column three of A = column 7 of A

and solve for x

[fathom saddle]

I'd just reduce "as much as needed" to get the final answer

You don't need to reduce the bottom 3 rows. You don't need to include any of the numbers outside of the first 3 columns

You do want to keep the -13 in the ? spot, and keep track of it, though

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[torn jolt]

The first statement is wrong

For n >= 2 .....

[elder cape]

how so?

[vestal meadow]

how did you get the last inequality?

[torn jolt]

Say n is 5

[elder cape]

We r assuming log n to be base 2

[vestal meadow]

oh

[torn jolt]

Oh

[elder cape]

yeah

is it correct now then? lol

[vestal meadow]

i still don't see how the last inequality is true

@elder cape you gonna explain how you got the last inequality?

[elder cape]

I'm figuring it out, one sec

nvm

.close

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[vital spruce]

i have a vector and an axis

[vital spruce]

how can i get the magantude of the vector on the axis

[torn gust]

dot product the vector with the unit vector along the axis

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[tropic grotto]

Hey there! Got a physics question on wave interference

[tropic grotto]

This is the question

and this is our work so far

Using desmos, we found that the answer should be 2, but we're having trouble finding the answer symbolically

(that is, that the position should be at 3or 7 meters from the left source, which is 2 away from the midpoint between the sources)

You can also ignore the "k" part

The main confusion I'm running into is calculating the phase difference deltaPhi. Honestly the whole thing is really confusing too

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@tropic grotto Has your question been resolved?

[tropic grotto]

<@&286206848099549185>

[torn jolt]

Let's say the wave by source1 is Asin(kx-wt)

What would the wave equation for the wave released by source 2 be ?

[tropic grotto]

Asin(kx-wt+phi_0), no? But we're taking a picture of the wave at an arbitrary time t, so can't we just get rid of the wt term by assuming we're starting our time at t = 0?

Though i guess the wt terms would cancel?

[torn jolt]

And why phi_0?

I am not sure about removing w

Give me a min

[tropic grotto]

well technically both waves have the same initial phase, so it wouldn't be phi_0, it would be delta_phi, right?

that is, the phase difference. Since both waves have the same initial phase, I just assumed phi_0 = 0

Anyway, that's what I've written on the whiteboard, after dividing out the amplitude

the issue is I'm stuck on how to represent delta_phi symbolically, since the equation relies on delta_D, which relies on the distances from x to the sources

So the expression I get at the very end is circular, like its saying "x depends on x"

Thanks for replying btw, been here a minute lol

[torn jolt]

Trying to make sense of what you are saying

Thought this was going to be easy

[tropic grotto]

I wish lol, this has taken me hours

I mean, if it makes it easier, we could ignore my work and start from the initial question

The idea is we want to solve the equation 0 = Asin(kx - wt) + Asin(ks - wt + delta_phi) for x, right?

[torn jolt]

Yes

Wait a min

The wave from the second source is travelling towards negative x axis

[tropic grotto]

Yeah

[torn jolt]

So it would be kx+wt

[tropic grotto]

That makes sense, and would explain why it cancels if we move it to one side and take the arcsin

So we should be able to get rid of the t term

[torn jolt]

What you are saying is right,but will that help us in anyway?

[tropic grotto]

It reduces the equation to one variable. Unless we just let t=0, which also takes care of that

[torn jolt]

What should be the phase difference ?

[tropic grotto]

That's what I'm trying to find

it should be equal to k times the path difference delta_D

0 = sin(kx) + sin(kx + delta_Phi)
-sin(kx + delta_Phi) = sin(kx)
-kx - delta_Phi = kx
- delta_Phi = 2kx

[torn jolt]

Mb

[tropic grotto]

no worries, its a dense problem lol

does the above look correct to you so far?

i forget big delta notation lol

[torn jolt]

Yes

[torn jolt]

?

[tropic grotto]

$$x = - \frac{\Delta \phi}{2k}$$

[glossy valve]

Rafzekael

[tropic grotto]

there we go

[torn jolt]

Oh

[tropic grotto]

so since delta_Phi = k*delta_D, and delta_D = |d_1 - d_2|

we should get

$$x = - \frac{\abs{d_1-d_2}}2$$

[glossy valve]

Rafzekael

[tropic grotto]

the issue with this is that both d_1 and d_2 depend on x...

and that's where I'm stuck

(also it depends on where you define your orgin which makes things even more confusing lol

idk lol. this phase difference stuff is hella confusing to me

[torn jolt]

Say origin is at source 1

What would the wave equation be?

For source 2

(I don't know)

[tropic grotto]

I think it was this

to simplify, sin(kx + delta_Phi)

[torn jolt]

Why does delta phi keep showing up

They said same phase difference right?

[tropic grotto]

That's the phase difference between two waves

[torn jolt]

^

[tropic grotto]

The waves have the same initial phase, but that's different from phase difference

because they are offset by position

[torn jolt]

Man,I am getting confused

[tropic grotto]

haha, yeahhhh...me too

Here's a desmos graph with the actual values

[torn jolt]

Sorry for intervening

This was supposed to be easy

[tropic grotto]

Nah, no worries at all! I appreciate any help

[torn jolt]

Simple

[tropic grotto]

how so?

[tropic grotto]

https://www.desmos.com/calculator/36kgfllh2x

[torn jolt]

I thought

Assumed

XD

[tropic grotto]

hahah, yeah if only lol

I think I'm missing an equation

cuz otherwise i feel like i'm really close to the answer

[torn jolt]

Ahhh

Y2 I.e wave from second source

Is 0 at x equals 10

We get delta phi from that won't we

[tropic grotto]

We can, but the issue is that the question wants us to solve it symbolically first for full credit

so without any numbers lol

I suppose we could say it equals zero at x = d...

[torn jolt]

I am taking origin at source 1

[torn jolt]

And I am done

[tropic grotto]

lmao

[torn jolt]

Out of my league

[tropic grotto]

that's fair

take care, friend. thanks for the help ^^

.close

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[deep robin]

Is this correct?

[brisk kestrel]

Absolutely corrrct

But u need to do one more thing

The surface area for the top is (21x12)-(8x4)

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[inland silo]

how should i proceed with question 2?

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[buoyant yew]

How to find the average speed for the whole jounrye?

[tiny terrace]

Average speed is total length travelled over time

Average velocity is total displacement over time

[desert trout]

Well for distance you just need to add how much it went up and how much it went down

[buoyant yew]

Man you a lifesaver fr

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[quaint mortar]

i dont understand how to do this question

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[quaint mortar]

<@&286206848099549185>

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[quaint mortar]

<@&286206848099549185>

[pallid hatch]

you can use trigonometry for abd and e.

[quaint mortar]

how tho

[pallid hatch]

for the first one, BC is adjacent to the angle of 40 degrees and the length of the hypotenuse is given (18m). as we have the hypotenuse and are looking for the adjacent we must use cos(θ) = adjacent / hypotenuse. substitute in the values: cos(40) = adjacent / 18 . Now rearrange to have the adjacent equal to 18 * cos(40) = adjacent. This is BD

[quaint mortar]

ahh

i get it now

thank you

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[lost raptor]

how would i do this question? I cant seem to apply the approximation

[open igloo]

For a quantity $$Z= AB,$$
$$\frac{\Delta Z}{Z}= \frac{\Delta A}{A} + \frac{\Delta B}{B}$$

[glossy valve]

Sulphur

[open igloo]

(Delta A)/A, for example, is the relative change of A
Similarly for the other things

[lost raptor]

would that make it -1%?

but answers say -9

[open igloo]

Yeah coz you have to find change in V, not v

[lost raptor]

so I rearrange to V=v/(kA)

and because its a division would it become -5%-(+4%) = -9%?

[open igloo]

You could do that, yes

[lost raptor]

okay thanks alot

[open igloo]

Hmm yes

Np

[lost raptor]

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@royal kettle Has your question been resolved?

[deep robin]

Ohh, got it, thanks!

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[upper flower]

Hi, sorry I don't know how to use the math bot

but I have this problem here

I solved this by doing ln to the expression

and after simplifying and all I get 4

but by doing it the way outlined in the other solution, I obtain e^4, which is the correct solution

was just wondering if anyone could let me know how I could get e^4 using the method i solved by (and why)

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[unborn pike]

You are taking ln both sides

So let y= the limit

So you got lny= 4

y= e^4

[upper flower]

yeah that's true

guess it was that simple

didnt think about other side

thanks for your help !

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[dim pumice]

"Linearize the function
...
around the point A = (6, 5). Then use it to calculate an approximate value of f(B), where B lies on
distance 0.17 l.e. from A in the direction (8, −15). (Please compare with f(B) calculated with a mini calculator).
Also state the equation for the tangent plane to the surface z = f(x, y) at point A."

[dim pumice]

i dont really get how this is solved

i start by putting in A(6,5) into sqrt(228+x^2+y^2) and get 17

[viral jasper]

Now you need the normal of the tangent plane of the curve at that point

[dim pumice]

hmm okay can i not calculate b first by doing a + h?

[viral jasper]

Sure

[dim pumice]

okay then we get (6.08, 4.85) right

but then my teacher does f(a) = f(7,5) = 342 + ln(1) * 17 = 342

and i dont understand why

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[fiery sun]

How do i solve this

[fiery sun]

part (b)(ii)

[spare dust]

65

[fiery sun]

How?

[umbral dome]

there are only two possible arguments for an imaginary number (which ones?)

[fiery sun]

Based on the argand diagram?

[umbral dome]

yes

[fiery sun]

Sry i dont really understand this part

[umbral dome]

a purely imaginary number must lie on the imaginary axis

[fiery sun]

Ohh

Ohhhhh

Since arg(z) = -5pi/9 lies in the third quadrant

Yea its -pi/2

[delicate patio]

||+- pi/2 ?||

[fiery sun]

As it has to lie on the imaginary plane?

Is it positive or negative pi/2? now

Well based on the argument it is in the third quadrant

So i assume it is negative

??

[delicate patio]

idk cus its purily imaginare it can be positive or negative axis right ?

[fiery sun]

Yea but z = -5pi/9

[ashen crescent]

From i) did you get positive or negative i?

[delicate patio]

my imaginary numbers are not my best so im asking as well

[fiery sun]

i is a show qn

[umbral dome]

well just from the fact that it's purely imaginary it must be either +/- pi/2. and we know that z has a negative imaginary part, therefore its conjugate has a positive imaginary part (which we are subtracting)

[fiery sun]

The qn has alrdy given the argument

ohh

So its - pi/2?

[umbral dome]

yes

[fiery sun]

Thank you

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[royal crest]

Hi I need help setting up this double integral
The double integral of (x-y)DA over the region R who is defined by the inequalties y <= 2x -1, y >= -x, and y <= 4 -x^2

[viral jasper]

Try drawing the region R and figure what your bounds should be

[royal crest]

So where i'm getting at is that the domain of x ranges from 1/3 to (1+sqrt(17))/2?

and y is bounded by the 2x- 1 and -x? but I don't think that is correct of the parabola of 4-x^2

[gritty robin]

Can a number 44...41, whose decimal representation consists of an odd number of digits 4 followed by a digit 1, be a perfect square?

I need help pls msg me

[sullen hinge]

this is also an in use channel, you want to pick one from the Math Help (Available) section

[gritty robin]

ah ok

[royal crest]

Sorry to bother you again but how else could i approach this problem?

[viral jasper]

Consider dividing you region by this line AB here

[royal crest]

Ohhhh alright that made so much sense thanks

c:

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[craggy plinth]

A 9200 USD investment grows to 12000 USD in 4 years. What is the annual interest rate?

[craggy plinth]

So i did 12000 = 9200(1+- r)^4

then 12000/9200 = 1.304347826

1.304347826 = (1 +- r)^4

square root both sides by 4

therefore

[hallow walrus]

are you sure it's not simple interest

[craggy plinth]

i get +-1.068681656 = 1 +- r

[craggy plinth]

nah it's for math 2 extended

10th grade math class

so listne

[hallow walrus]

so?

what does that matter

[craggy plinth]

+-1.068681656 = 1 +- r

right

so

what if

R =

2.07

[craggy plinth]

idk i dont study econ

because basically

if we can have it as -1.068681656

= 1 - r

we subtract both sides by 1

and then divide both sides by -1

[hallow walrus]

that isn't a valid solution

[craggy plinth]

ok

so what do we do

[hallow walrus]

why 1+-r?

[craggy plinth]

because that's from the original formula

A = A0 (1 +- r)^t

[hallow walrus]

$\frac{12000}{9200}=(1+r)^4$

[glossy valve]

kheerii

[craggy plinth]

t being years in some cases, such as this

[craggy plinth]

correct

[hallow walrus]

you're not given any information about how the interest compounds

whether it's per annum, twice per annum or so on

you're assuming it compounds once per annum

[craggy plinth]

yes

because that's how the information is given here

again

[craggy plinth]

we're following this formula

[hallow walrus]

the formula isn't the holy grail

think about how you're applying the formula

[craggy plinth]

but how else do we solve

[craggy plinth]

isn't this correct already

except it has to be a +-?

[hallow walrus]

that formula assumes that the interests compounds annually. do you KNOW that that assumption lies behind that formula?

[craggy plinth]

not sure

im guessing yes

[hallow walrus]

the +- doesn't really matter, r will just come out to be negative

[craggy plinth]

ok

we are coming out to the fact that there is an annual interest rate, and we are trying to find it after

[hallow walrus]

and you shouldn't take the negative 4th root here

that gives you a ridiculous value for the rate

[craggy plinth]

im not doing the -4th root, but doesn't it come out to be +- whatever the 4th of the root that is

[hallow walrus]

the rate of interest can't be -2.07, that means you lose 207% of your money every year

i.e you go into debt

[craggy plinth]

exactly

so we're thinking

that

[hallow walrus]

so we don't take that solution

[craggy plinth]

we are given and before and after numbers

[hallow walrus]

just 1.06....=1+r

[craggy plinth]

9200 before, 12000 after

[hallow walrus]

so r=0.06...

[craggy plinth]

but if it's under 1, doesn't that make it go down

[hallow walrus]

if it's negative, that is what makes it "go down"

[craggy plinth]

but isn't it exponential decay

if the rate is under 1

[hallow walrus]

no, it is exponential decay if the rate is under 0

you're confusing two different things

(1+r)<1 implies exponential decay yes

in other words r<0 is exponential decay

[hallow walrus]

yeah. and in this case the base is (1+r)...

[craggy plinth]

so the rate

can be 0.06?

and stay like that fine?

[hallow walrus]

of course it can

yes

[craggy plinth]

wait a second

oh wait

nvm

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[cold sail]

i got the answer but

i dont understand how differentiating g1 with respect to t finds f_xy

[neon crystal]

you could write it with (y) instead of (t) as [g_1 \colon y \mapsto f_x(0, y)]

[glossy valve]

Invariance

[neon crystal]

does that help?

[cold sail]

yes but i dont understand why differentiating wrt to y would still find f_xy

cz like

if you took the partial derivative of f_y wrt to x then u wold get f_xy

well i i dont understand the significant of making the x component 0 basically

[neon crystal]

it doesn't matter

[cold sail]

*significance

[neon crystal]

you can set x = 0 before or after differentiation

either way you treat it as a constant

so it doesn't make a difference

[cold sail]

why tho cz wouldnt setting x = 0 get rid of some y terms too

wouldnt that give u an entirely different expression

eg the 4x^2y^2

in f_x

[neon crystal]

You'd get rid of those anyway. Using that example, if we set (x = 0) first, we get
[\frac{\partial}{\partial y} 4 x^2 y^2 \to \frac{\partial}{\partial y} 4 (0)^2 y^2 = \frac{\partial}{\partial y} 0 = 0]
and if we set (x = 0) after, we get
[\frac{\partial}{\partial y} 4 x^2 y^2 = 8 x^2 y \to 8 (0)^2 y = 0]

the order doesn't matter because partial differentiation by y treats x as a constant

so if we have a specific constant in mind, it doesn't matter if we set it before or after partial differentiation

it would matter if we were differentiating by x, though, for the reasons you state

[neon crystal]

it's only because of this property that we get away with it

[glossy valve]

Invariance

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[drifting vapor]

hi

[drifting vapor]

i have issues

with the formula

with y-axis and x-axis

on a rotation

what the diffrent and whats the formula for both when its about the volume

and rotation area?

[viral jasper]

Where are you getting volume?

[drifting vapor]

what do u mean?

uhh the figure volume

[viral jasper]

What volume?

[drifting vapor]

of the rotation figure

Let Ω denote the area in the first quadrant between the line y =
16 and the curve y = x^4. When the curve Ω is rotated one revolution around y-
the axis forms a body of rotation. calculate about the volume of the body.

like this

[viral jasper]

Ah so early calculus

[drifting vapor]

yea

or kinda

its mixed

[viral jasper]

There's different formulas based on the axis over which you are rotating

[drifting vapor]

yes i know

i want all of them

but the rules

is typed

diffrent

online

iam confused lol

[viral jasper]

Where?

[drifting vapor]

this is how my preffoser have done

and on google

it say like only pi

[viral jasper]

Where

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[fallow aurora]

Can someone please explain to me the Tan, Sin, Cos A and B things? How do I know for future reference how to know which side is the opposite/adjacent?

[fallow aurora]

This is taken directly from yesterdays notes from my class.

[fathom saddle]

A is one of the angles on your triangle. You can see it notated in black ink

The "opposite of A" is the side that isn't touching A.

The "Adjacent of A" is the side that is touching A, and isn't the hypotenuse

The hypotenuse is the longest side of the triangle.

[devout valley]

[in your picture, the red is in reference to the angle at A, and the blue is in reference to the angle at B]

[fallow aurora]

Oh, alright. I was a little confused because in the previous page it was just explaining how to use Sin Cos and Tan, and then it just immediately jumps to this. It felt like a huge jump.

Thank you :D

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[odd rose]

I have a vector in R4 and I want to find the basis of this vector.

The vector is {[a b c d] | a+b+c+d = 0, a = -c-d}

[odd rose]

So, Ive been having some problems checking on how to find the span of this vector. I did b = -a-c-d and for a = -c-d, i plugged that into [a b c d]. But when i try verifying with other values i get different outputs

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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.

[odd rose]

It was a quiz so i cant really show it... i dont have the paper

[ember shadow]

what was the original question though?

[odd rose]

.

[balmy lake]

Looks like a subspace of R4 and asks for a base for the subspace, ig?

[odd rose]

yea

how i do it is:
apply the constraints, seperate the variables, factor out the variables, write the span and then check for linear independence

But im getting stuck on applying the constraints. Not sure if im doing it right

[ember shadow]

you just row reduce the matrix for those equations

[odd rose]

Yea i know but im stuck on applying the constraints like i said

[ember shadow]

what constraints?

[odd rose]

a+b+c+d = 0, a = -c-d

i have to apply those to the vector right?

[ember shadow]

no, you just solve the system of those 2 equations

[fathom saddle]

Those are rules that the vector needs to follow, in order to be a member of the subspace

[odd rose]

yea thats what im doing no?

[balmy lake]

Gnah. Sideways

Always

[odd rose]

yea i got the same thing

[balmy lake]

Lemme fix that

[ember shadow]

that's not right

[odd rose]

but then i somehow got 0

[balmy lake]

It is isn't?

[odd rose]

so ik i made a mistake

[ember shadow]

no, b = 0

[odd rose]

-(-c-d)-c-d = 0

[balmy lake]

Oops

True that

[odd rose]

so we should have 3 vectors right?

[balmy lake]

No, it's still two vectors

Just (-1, 0, 1,0) and (-1,0,0,1) instead of what I wrote

[odd rose]

if we do c = -a-b-d and a = -c-d and then expand and simplify a we get a = a+b

so we have 3 variables a,b,d

[balmy lake]

Thanks for catching the miss btw, much appreciated

[odd rose]

what am i doing wrong here

[balmy lake]

Well considering b is 0

a=a+b is still true ig

[odd rose]

why is b 0 for this case

[balmy lake]

It is. You cannot find a vector satisfying all constraints with nonzero b

[odd rose]

im just not sure what im doing wrong

[balmy lake]

If you solve the first restraint for b

And than plug in the second restraint

It will give you b=0

[odd rose]

yes im talking about if we solve the first one for c

[balmy lake]

Well you have to chose 2 of the components as parameters, since with two constraints in R4 the subspace is 2-dimensional

[odd rose]

Can i actually show you how I would do it when solving the first one for c?

[balmy lake]

It doesn't matter which two you chose, as long as it isn't b (because it's zero)

Sure

[odd rose]

So c = -a-b-d and a = -c-d

Is the correct that the vector now will look like:
[ -c - d
b
c - b
d]

[balmy lake]

Where did you get the c-b from?

[odd rose]

so we had c = -a-b-d

and we know a = -c-d

so i plugged a into c

[balmy lake]

I see

But that immediately gives you b=0

Since in c=c-b you can simply subtract c from both sides

[odd rose]

well could i not also just write the augmented matrix of the 3 varibales and solve to see linear independence

[balmy lake]

Sure

[odd rose]

all 3 cols have pivots

unless im doing something wrong.. it seems as we have different answers

[balmy lake]

Can you show it?

[odd rose]

sure but it might not make sense. I didnt document it well

[balmy lake]

I'm not sure if I know what you are trying to do here. Where did you get the matrix from?

[odd rose]

You can try solving it your way because it’s the same way you did it when solving for b

[balmy lake]

I can't follow

[odd rose]

Like just try solving it yourself and see what you get

[balmy lake]

I get the exact same thing as above with a different matrix.

[odd rose]

Could you show your work?

[balmy lake]

It'll probably be sideways again, tho... : <

[odd rose]

It’s all good lol

[odd rose]

I’m not sure what ur solving for here

could you maybe solve it in this manner

[balmy lake]

Since c and d don't have pivots I can choose them freely.

b does have a pivot and must be 0 since the last line is 1b = 0

a does have pivot, so it isn't free either it is determined in the second to last line by the chosen values for c and d and the calculated value for b:

a+b+c+d=0, where b=0 and c and d are a chosen.

Give me a minute

[odd rose]

ok

[balmy lake]

You went from (-c-d, b, c-b, d) as a vector, ig?

[odd rose]

yea

[balmy lake]

I think the problem here is, that b shouldn't exist at this stage anymore

[odd rose]

how tho

[balmy lake]

It's zero.

Since the equation c=c-b forces it to be zero

[odd rose]

Ok, im gonna leave this for now tbh

Thank you

[balmy lake]

Hm! Sorry : (

[odd rose]

ita ok hah

[balmy lake]

: )

[odd rose]

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[royal crest]

Using Rieman Sum approximate the double integral of x^2 + 4y Da

Where the integral D is region inside the upper unit semi circle centered at the origin (inside the circle, above the x-axis). I want you to approximate this using a polar Riemann sum, dividing theta into 6 sub divisions, and dividing r into 3 subdivisions (so there will be a total of 18 pieces). Use the point with minimum theta and maximum r in each subdivision piece as the point you put into the function.

So far i got:
delta theta: pi/6
delta r: 1/3

lim (m,n goes to infinity) i= 1 sigma m = 6, j= 1 sigma n=3 of F(rj, thetai)

I'm not sure what to do for this part of the question:
"Use the point with minimum theta and maximum r in each subdivision piece as the point you put into the function."

[royal crest]

I'm assuming that I go from pi/6 to 5pi/6
and for r I go from 1/3 to 1 for every pi/6 upto 5pi/6?

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@royal crest Has your question been resolved?

[fathom saddle]

I'd consider drawing the division. It's definitely not easy to keep track

[royal crest]

Hm wdym by drawing the divisions?

Suggesting to draw each individual square?

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@royal crest Has your question been resolved?

[royal crest]

<@&286206848099549185>

[royal crest]

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[torpid perch]

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[dusky drift]

can someone tell me why or how this is even wrong?

[plain fox]

can u show working?

[quaint prawn]

Remember to use absolute values where appropriate

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[dusky drift]

👍

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[torn jolt]

Hii js a quick intro - I've started calculus and I've realised how abysmal my algebra skills are 💀 could you please help 🙏

How in the world do you get

(1/2)x^(-1/2) = 1(2sqrt(x))

[woeful nova]

its 2sqrt(x) or 2/sqrt(x)?

[torn jolt]

2sqrtx,

[woeful nova]

this?

[torn jolt]

Ohh ok Ty

I’ll defo be back for more 💀

How do I differentiate this

[woeful nova]

give me a second

You can do this because there is nothing above in function of x, because otherwise you should use the formula for divisions in the derivatives, I don't know if you understand me because I am using a translator.

nothing as a function of x in the numerator,It would also work with constants since you can take them out from the derivatives.

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[modern geode]

.close

How do I calculate a sin wave?

[modern geode]

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[torn jolt]

Ty

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[warped needle]

The polynomial P(x) = ax+ 3 satisfies the equation P(P(P(x)))−3P(x) = −2x for all real values
of x, find all polynomials that satisfy this equation.

[warped needle]

not sure if im correct

but i got x = -3?

[stuck storm]

x?

the question is not looking for x

[warped needle]

as in P(x) = x-3

+*

[stuck storm]

oh ok

any value of a would work in this case

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[torn jolt]

im really confused

if the following equation doesnt have an x intercept

[torn jolt]

like clearly it doesnt touch the x line

then what did i just calculate

[clever sluice]

in abs only positive value are the outputs

[torn jolt]

wha

[clever sluice]

absolute value

[torn jolt]

yea...?

[clever sluice]

so how a function that has only positive output values have a negative value?

[torn jolt]

OHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

[torn jolt]

omg i love you so much thank youuuu

[clever sluice]

welcome

[torn jolt]

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[stray rock]

Given is a rectangle ABCD. Let a = AB, b = AD. On the side AB is a point E, on the side AD is a point F and let AE = 2/3a and AF = 1/2b. Where on the line FE should we choose the point G so that the rectangle inscribed in the pentagon DFEBC with the intersection at G and the sides parallel to the sides of the original rectangle, will have the maximum area?

[stray rock]

yeah so I alr took this channel

@torn jolt

[clever sluice]

i think its c

[stray rock]

...

[ashen crescent]

Draw a diagram

[stray rock]

yeah I wanted to send but this guy sent his question like its his channel

[ashen crescent]

Looks like he’ll be banned soon anyway

[stray rock]

so like this

Im guessing I need to form a function for something and then find the maximum right

I thought maybe it has something to do with ratios of the sides

[ashen crescent]

Yeah I believe you need to describe the area of the inner rectangle in terms of a and b

[stray rock]

the answer it "for G we have to choose a point (..., ...)

how do we find the point

[ashen crescent]

The BC side of the inner rectangle will be the difference in y component of G and E subtracted from b

You can do something similar for the adjacent side

[full forum]

@stray rock Has your question been resolved?

[ashen crescent]

@stray rock are you still stuck?

[stray rock]

I am

I had to eat in between also, so thats why I was gone sorry

but yeah idk how to do this

[ashen crescent]

Can you do me a favour and redraw the diagram with A and C swapped

[stray rock]

you want me to flip the whole thing?

[ashen crescent]

Yeah

In fact, just leave it

[stray rock]

oh

alright then

[ashen crescent]

Lets call the area you want to maximise A=L*W

[stray rock]

yep

[ashen crescent]

L is gonna be the horizontal side

We can describe L in terms of G and a

Since L is just a - Gx

Where Gx is the x component of G

Right?

[stray rock]

yeah alright I understand

and then the same for the width

[ashen crescent]

Yeah, which would be?

[stray rock]

b - Gy

[ashen crescent]

Yeah

Now as for G, it lies on the line connecting F and E

So we can describe it as (Et, Ft) = (2/3 at, 1/2 bt)

[stray rock]

oh so like multiplied by a factor t

[ashen crescent]

Yes

[stray rock]

alright yeah

[ashen crescent]

then we rewrite A with the new G

[stray rock]

so we replace Gx and Gy with those?

[ashen crescent]

So you have A = (a-2/3at)*(b-1/2bt)

Yeah

Then, multiply it out and factor ab

[stray rock]

okay wait I have a slight doubt

[stray rock]

with this step

so you just described G with those coordinates?

[ashen crescent]

Yeah though, you may be right in spotting something dodgy there

[stray rock]

shouldnt it be

oh nvm...

[ashen crescent]

You are right, it is wrong

[stray rock]

well

[ashen crescent]

Give me a second to think about this

[stray rock]

in your case

t just has to be <=1 then right

for it to work? or is it completely wrong

ohhhhh

no but is it really wrong?

[ashen crescent]

Yeah, it wouldn’t be lying on the line, it would be a region

[stray rock]

oh yeah true

[ashen crescent]

I think instead it should be G = F(1-t) + E(t), taking a and b as vectors

Then when t = 0, G is at F

t = 1, G is at E right?

[stray rock]

okay but cant we do it without vectors?

if EB is 1/3a

and FD is 1/2b

that means both of those are being stretched right

one is being stretched, and one the opposite

[ashen crescent]

I’m not sure I can see where this is going

[stray rock]

well when G is at F

the sides of the rectangle are 1/2b and a

when its at E the sides are b and 1/3a

maybe something with that? sorry im just kinda bad with vectors

[ashen crescent]

Yeah that is the L and W we had earlier

It may be possible to get the equation of the line FE in terms of a and b

Under the assumption that A is fixed at the origin

Then l_FE would be given by
y = mx + c
y = (1/2b * 3a/2)x + 1/2b

[stray rock]

okay so you got m how

ohoh okay

sorry

yeah I understand

[ashen crescent]

Hang on

[stray rock]

okay wait no just quickly I actually dont understand, how did you get m?

[ashen crescent]

It would be just rise over run -> F/E

Then its negative too clearly

[stray rock]

okay

[ashen crescent]

[ y = - \left(\frac b2 \cdot \frac 3{2a}\right)x + \frac b2 ]

[glossy valve]

shsgd

[ashen crescent]

It’s a bit hard to describe G with this though

Which is why vectors work better i’d say

[ashen crescent]

Because we have this as G= (b/2)(1-t) + (2a/3)t

= 1/2 b - 1/2 bt + 2/3 at

Hmm

Vectors may make this a bit weird since a and b now have 2 components

[stray rock]

there has to be a way to just do it geometrically

[ashen crescent]

The only problem is describing coordinates of G

[stray rock]

well

what about what we did before, with the t right

if you stretch 1/b with t

if t is 1

then you stretch 2/3a with

like it has to be something like (a - 2/3a*ct)(b - 1/2t) where c is some number?

yk so that then it wont be a region like you said

[ashen crescent]

That would just be this, no?

Where t is in the range 0-1

[stray rock]

yeah but lets do that for this

[ashen crescent]

Well what would the coordinates of G be in terms of a, b, c and t then?

[stray rock]

(a - 2/3a*ct, b - 1/2t) this?

yk what I mean like if you stretch 1/2b you ahve to stretch 2/3a by the same amount no?

[ashen crescent]

You would stretch by the opposite amount

Which is where i got t and 1-t

[stray rock]

so can we use that then?

1-t as c

or I mean like as

a - 2/3a(1-t)

and then b - 1/2t?

if not then im out of ideas

[ashen crescent]

That would be this

[stray rock]

yeah so why cant we do that there

[stray rock]

here

[ashen crescent]

It changes a bit since length and width are scalars

G = b/2(1 - t) + 2/3 at
= 2/3t a + (1-t)/2 b
A = l*w = 2/3t * (1-t)/2 = t - 3t^2

This is why I initially said it may be better to flip the diagram, since then A = Gx * Gy

A’ = 1 - 6t = 0 iff t = 1/6

Then we would get G = 1/6 * 2/3 a + (1-1/6)/2 b

= 1/9 a + 5/12 b

Possibly

[stray rock]

I have the answer here if you want it

[ashen crescent]

Yeah, go on

[ashen crescent]

Can you send the diagram again

[ashen crescent]

I need to go get some paper, can’t do this in my head give me a min

@stray rock can’t seem to get it and don’t see where i’m going wrong

[stray rock]

it's alright

[cinder robin]

i didnt get the question

[ashen crescent]

t would be 1/4

And it works

[cinder robin]

oops ok

[stray rock]

really?

[ashen crescent]

Yeah

Since G = 2/3ta + ((1-t)/2 )b

Then t = 1/4

G= (2/3)(1/4)a + (3/4)/2b = 1/6a + 3/8b

So A’ = 1-4t

[stray rock]

okay tysmm

[ashen crescent]

A = t - 2t^2 + c

If you backtrack a bit and don’t mess up the vectors, you’ll be able to get it

Again, I think it would help to flip the diagram

Sorry I couldn’t fully help, but hopefully you get the idea

[stray rock]

No dont worry thank you for your time

@ashen crescent

I appreciate it

🙂

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[honest lark]

how calculate cos and sin into that form

with calculator

[lofty vine]

i dont understand your question

how to calculate cos(2pi/3) ?

[honest lark]

yes using calculator machine

[grand hatch]

Hit the cos button and type in the angle

[lofty vine]

almost all calculators can calculate cos(x) for any x

just type it in ur calc

[honest lark]

I use this casio

[lofty vine]

you see that cos button in the middle

[honest lark]

oh

[lofty vine]

also make sure your calc is in radians mode

[honest lark]

ohhh