help-36

[lapis raven]

yes

[shell mountain]

Now, to find the answer, we need to know how R relates to r

[lapis raven]

We cant just assume it is half or something

But there is no direct equation

So im not sure

[shell mountain]

Its not guaranteed to be half

[lapis raven]

I dont know then

[shell mountain]

Lemme think

[lapis raven]

k

im thinking that they have the same radius

[shell mountain]

They can't, otherwise it won't be inscribed

[lapis raven]

oh

nvm

[shell mountain]

I don't know an easy way to do this, but I know a way that's probably more complicated than necessary

The slant of the cone can be modeled as a line passing through the points (-R, 0) and (0, 6R)

[lapis raven]

ok

[shell mountain]

Which means the height of the cylinder will be the value of that line at -r, which we can set = 3r

[lapis raven]

ok

[shell mountain]

Or, to ignore the negatives, it passes through (R, 0) and (0, 6R). That's neater

So what's the equation of the line through (R, 0) and (0, 6R)?

[lapis raven]

y=6x+6R

sorry

[shell mountain]

That's through (-R, 0), which I said originally

[lapis raven]

y=-6x+6R

The negative didnt press

[shell mountain]

And so what's the height at x = r?

[lapis raven]

0

[shell mountain]

r

Not R

[lapis raven]

oh

6R-6r

or 6(R-r)

[shell mountain]

Yes. But the height should be 3r

So 6R - 6r = 3r

[lapis raven]

6R=9r

R=1.5r

[shell mountain]

So we can replace R with 1.5r

[lapis raven]

ok

[shell mountain]

The cylinder, 3πr³, remains unchanged, but what about 2πR³?

[lapis raven]

That is now 3πr³

[shell mountain]

Not quite

[lapis raven]

Wait

1.51.51.5

1.5 times 1.5 times 1.5

3.375

So multiply by 2

6.75πr³

[shell mountain]

Yes

What percent is 3πr³ of 6.75πr³?

Or, what fraction

[lapis raven]

3/6.75

6/13.5

12/27

4/9

44.444%

[shell mountain]

4/9, yes

[lapis raven]

ok

thats the answer

ok

thank you so much

you were a great help

hope to see you around

bye

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[dry light]

What have you gotten so far?

@tranquil pine

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@tranquil pine Has your question been resolved?

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[south jasper]

.

hello

[south jasper]

Im solving question (e)

i know lambda will equal to .25

but what value would i but in as X

P(X = 0) or P(X = 1) as one call is required i think but unsure off

[burnt fable]

How did you arrive at 0.25?

[south jasper]

5 calls per hour .2 of them are printer problems, which mean there is 1 printer problem call per hour

and in a duration of 15 mins, the mean of the being a print call is .25

[burnt fable]

Oh ok good I didn't read the 0.2 properly lol

[south jasper]

oh ok

[burnt fable]

So on this question, what does X represent?

[south jasper]

There being a printer call in less than 15 mins, Which im "guessing" X = 1

but that im unsure off

[burnt fable]

No, X is a discrete random variable (a certain possible number), not an event (i.e. a thing happening)

In Poisson distributions, your X usually represents the number of occurrences (phone calls in this case)

[burnt fable]

^ Note that X represents the same thing your mean represents

[south jasper]

So would X = .25?

[burnt fable]

Does this make sense so far?

Not the number, but more the concept

X isn't fixed to a number, but when you do P(X = 0) for example you're "checking" how likely X is equal 0

So X itself is actually the number of calls about a printer problem in 15 minutes, but you have no way of knowing what X is equal to exactly

[south jasper]

So would the answer to my question be 1 - P(X = 0)

[burnt fable]

Yup ^^

Figuring out what the variable tells you about the problem helps :)

[south jasper]

Yessir, ty for the explanation.

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[karmic elm]

Hi, What do you call the expression in the denominator? Also, isn't it possible to reshape it into: for instance, lets say: (s+3)(s-5)? What do i call "that" method?

[karmic elm]

https://www.wolframalpha.com/input?i=+2.5e-06+s^2+%2B+0.25+s+%2B+25

Wolfram solved it lol

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[turbid marsh]

Hey, I'm kinda stuck here on how I can solve this integral:

[turbid marsh]

,tex $\int 5x\frac{1}{x^2+4}-\frac{1}{x^2+4}-\frac{3}{x+1}dx$ \newline\newline
I know that \newline\newline $\frac{1}{x^2+1} = arctan(x)$ \newline\newline
but how do I use this here?

[soft zealot]

Eric B.

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@turbid marsh Has your question been resolved?

[amber holly]

Do a u-sub in the first fraction (let u = x^2 + 4)

You can rewrite the second fraction as 1/4 * 1/((x/2)^2 + 1)

Let t = x/2

So that you have 1/(t^2 + 1)

Integral of which is just arctan(t)

Notice that you can generalize the rule into $\int\frac{1}{x^2+a^2}dx = \frac{1}{a}\arctan{\frac{x}{a}}$

[soft zealot]

A Lonely Bean

[turbid marsh]

interesting, ok, will try to understand that =D

thanks alot

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[tidal jolt]

how do I change this function into standard form

[abstract monolith]

is f(x) an quadratic?

[tidal jolt]

yes

[abstract monolith]

$f(x) = a\cdot(x-x')(x-x'')$

[soft zealot]

sopinha

[abstract monolith]

where x' and x'' are the roots

[tidal jolt]

i don't know how to answer that

[abstract monolith]

You have x' and x''

[tidal jolt]

what does that mean

[final saddle]

@tidal jolt Has your question been resolved?

[hollow basin]

What's this?

Guys?

I'm here

What do I do next?

[tidal jolt]

.close

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[fallen valve]

because the surface integral is taken with respect to dS' (Where S' is a vector and is equal to the the UNIT normal vector times dS (where S here is just surface area). dS is equal to the magnitude of the normal vector times dA. so the magnitudes of the cross products i.e. the normal vectors cancel, and we are oeft with a NORMAL vector dot the curl of the field dA. but then i dont get the same answer as my tracher for some reason

[fallen valve]

<@&286206848099549185>

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[loud sundial]

You know that's a bot ... right?

[fallen valve]

I think?

[fallen valve]

waitl do you know the answer to my question

[loud sundial]

If you're going to act entitled like that, honestly, I don't want to help

Plus you already posted the question twice, which makes me even less incentivized to help you.

[fallen valve]

ok, im sorry

i ddint know I was acting entitled, i apologize, its just slightly frustrating when the channel keeps on closing when the question hasnt been answered. I understand it is to free up space, but all I wanted was just an answer, not a bot to just close my reply, only forcing me to post it again, and again

[fallen valve]

I truly do not believe I was acting entitled. If I have hurt anyone please let me know, and that was clearly not my intention. I am just trying to clarify a misconcpetipj I am havong, and I ask you to please keep your comments about others to yourself. Its just really frustrating when people dont understand the full situation and then make fun of others for it.

[loud sundial]

I was just trying to be honest since you asked me to help you, but I see where you're coming from

But again, since you reposted this in #multivariable-calculus, I'll leave it to the people there to answer.

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[fallen valve]

oh alright, i am sorry, thanks again

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[cedar venture]

Why is the derivative of 100+50x - (7x^2/350) = 50 - x/25 ? How do I solve this?

[severe dawn]

so $f(x)= 100+50x - \frac{7x^2}{350}$\
$f`(x)= 50 - \frac{x}{25}$?

[cedar venture]

100 is a constant and therefore 0
50x becomes just 50
but deriving the fraction I get to 350*14x - 7x^2/350 and I am not sure it is correct

[severe dawn]

(7x^2/350) 50 whats that 50 here?

[cedar venture]

oh sorry like this

the derivative is suppsosed to be 50 - x/25

[severe dawn]

ah

[cedar venture]

of $f(x)= 100+50x - \frac{7x^2}{350}

[severe dawn]

I see

[cedar venture]

I am just not sure how to get there

[soft zealot]

Jigglyproff

[cedar venture]

yes

[severe dawn]

alright, so like you said, $100\times x^0$ becomes 0

[soft zealot]

Jigglyproff

[severe dawn]

$50x^1$ becomes $1\times 50\times x^0$, so just $50$

[soft zealot]

Jigglyproff

[cedar venture]

yes

[severe dawn]

and lastly

[cedar venture]

I just seem to be having trouble with the fraction

[severe dawn]

$\frac{7x^2}{350}$

[soft zealot]

Jigglyproff

[severe dawn]

you can write it as $\frac{7}{350}\times x^2$ if that helps

[soft zealot]

Jigglyproff

[severe dawn]

whats the derivative of x^2?

[cedar venture]

x?

[severe dawn]

mhm not quite

wait

[cedar venture]

wait no

2x

[severe dawn]

yes

$x^2$ becomes $2 \times x^1$

[cedar venture]

yeah I thought of that too then

[soft zealot]

Jigglyproff

[severe dawn]

so for that term, you end up with $\frac{7}{350}\times 2x$\
now you can just shorten that a bit, since 7 and 2 both divide 350

[soft zealot]

Jigglyproff

[cedar venture]

so i would get 2x/50 and then x/25?

ah alright, i think I got it

thanks!

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[pulsar cliff]

"Let Σ = {0, 1}, and let L be the language over Σ consisting of all strings of 0’s and 1’s of length 4 with an equal number of 0’s and 1’s. What would the elements of L be"
I'm having a hard time understanding what it's asking for. This is in my discrete math textbook and its on one of my homework questions. Bit of a crunch time now and I know this is going to be on my exam yet I have no understanding of it

[strange rune]

they ask you to write all the words of the language

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@pulsar cliff Has your question been resolved?

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[hexed river]

Can someone help me with this

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@hexed river Has your question been resolved?

[dry light]

Find when r = 0

[hexed river]

R = 0 at 0/2pi

Sorry 0 or 2pi

[dry light]

You sure?

[hexed river]

Not anymore

Hold on

[dry light]

Do you know what kind of polar function 1 - 2cos(θ) is?

[hexed river]

It's a inner loop limacon

[hexed river]

Is it -1

[dry light]

No

[hexed river]

I have no idea how to find it then

[dry light]

$$1 - 2\cos(x) = 0$$

[soft zealot]

Umbraleviathan

[dry light]

For convenience's sake, x = θ

You know how to solve trig stuff right

[hexed river]

Yes

Do I solve for x

[dry light]

Yes

[hexed river]

Cos(x)=½

[dry light]

No

Wait yes

Sorry yes

cos(x) = 1/2

Isolate x

[hexed river]

How would I do that

[dry light]

You remember your inverse trig?

arcsine, arccosine, arctangent?

[hexed river]

Yea

[dry light]

Use that

arccos(cos(x)) = x

[hexed river]

Ohh

X=arccos(½)

What would I do next

[dry light]

Know your unit circle

what would arccos(1/2) be

Or use your calculator

[hexed river]

Pi/3

[dry light]

What else

(And just give answers between 0 and 2Pi)

[hexed river]

Pi/3 and 5pi/3

Would I organize the interval of the domain like (r,theta)

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[serene patio]

Show that there are infinitely many linear maps f : R^4 → R^2 that satisfy the conditions f(v1) and f(v2).

How do I solve this? I was thinking about making a system of linear equations, but I don't really know how to do that in this case

[worldly vale]

you know a linear map from R^4 to R^2 can be represented by a 2x4 matrix so use those two conditions to get relations on the elements of the matrix

[serene patio]

@worldly vale
Is this correct? And then I can use f(v1) and f(V2)?

[worldly vale]

no you want a generic 2x4 matrix applied to $v_1$ to equal $\begin{pmatrix} 1 \ 1 \end{pmatrix}$

[soft zealot]

ΣAC

[worldly vale]

and similarly for v_2

[serene patio]

@worldly vale like this?

[worldly vale]

yeah exactly

plug in v1 and itll give you conditions on a,b,...,h

same for v2

and then argue there are infinite ways to satisfy those conditions

[serene patio]

Show that such maps take on the same value at the point v – which value is it?

v = (1,1,1,1)^T

@worldly vale do you have an idea how I can solve this?

[worldly vale]

apply your generic matrix to (1,1,1,1) and use the relations to show it doesnt matter which one you pick

[serene patio]

@worldly vale generic matrix * (1,1,1,1) = (a+b+c+d, e+f+g+h), but how do I show that?

[worldly vale]

well what were the relations you got between the a,b,c,...,h before

[serene patio]

with v1 I got: a + 2b + 3c -d = 1 and e +2f + 3g -h = 1

with v2: 2a +b + 4d = 1 and 2e + f + 4h = 2

[worldly vale]

okay so somehow those relations should tell you that a+b+c+d = e+f+g+h

not immediately clear to me how to extract that information

btw your 4th relation shouldnt have a g in it

[serene patio]

@worldly vale idk whether that helps but: 1/3 *(v1+v2) = v

[worldly vale]

oh that helps massively

dont need to bother with this relation business

you know what f(v1) and f(v2) is so just use that

[serene patio]

@worldly vale so the value would be 1/3*(f(v1)+f(v2)) ?

[worldly vale]

right and you know f(v1) and f(v2)

[serene patio]

@worldly vale ok thank you for your help 🙂

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[hard mist]

How would you solve this using l'hopitals rule?

[hard mist]

Guess the first step would be write as a fraction?

[oblique portal]

whats up with the < in the limit

[hard mist]

Only evaluate limits smaller than π/2

[oblique portal]

ohh

oki

id take ln of that to get rid of the exponential

wait you cant really do that

i mean e ^ ln(whole limit) would cancel it and still be valid

like

the bit inside

and yes split tg x into a fraction

[final saddle]

@hard mist Has your question been resolved?

[hard mist]

Ill try in a min

[hard mist]

$... = cos(x).\frac{ln(sin(x))}{ln(cos(x))}$

[soft zealot]

rainy

[hard mist]

So like this?

Which becomes

$... = \frac{cos(x).ln(sin(x))}{ln(cos(x))}$

[soft zealot]

rainy

[oblique portal]

$ln \left( \frac{x}{y} \right)\neq\frac{ln(x)}{ln(y)}$

tf

how did u make the long brackets

[soft zealot]

Kel.plush

[oblique portal]

there

[hard mist]

So how should I rewrite the ln(tan(x)) then

as a frac

🧐

[oblique portal]

do u have to use lhopital?

[hard mist]

Yes

[oblique portal]

oh

well $ln \left( \frac{x}{y} \right) = ln(x)-ln(y)$

[soft zealot]

Kel.plush

[oblique portal]

so u can split $e^{cos(x)ln(tg(x))} = e^{cos(x)(sin(x)-cos(x))} = e^{cos(x)ln(sin(x)} - e^{cos(x)ln(cos(x))}$

hello????

latex??????

so u can split $e^{cos(x)ln(tg(x))} = e^{cos(x)(ln(sin(x))-ln(cos(x)))} = e^{cos(x)ln(sin(x))} - e^{cos(x)ln(cos(x))}$

there

jesus christ

[soft zealot]

Kel.plush

[oblique portal]

or maybe just the approach is wrong

[hard mist]

Well yeah first step is surely to bring the exponent down, then I have lim ln(f(x)) = cos(x).ln(tg(x))

Its just the cos(x).ln(tg(x)) part that im having problems with rewriting as a frac

[oblique portal]

you need to put that in e^ln(f(x)) tho right

otherwise ur taking the wrong limit

[hard mist]

Can do that in the end

[oblique portal]

but ur gonna be takin the whole function to the e

mesning even if u write that as a fraction youll get e^fraction

[hard mist]

Okay i see

Thx

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[ruby sorrel]

I'm doing factorization currently and I've came across this question in the practice test. keep in mind we never worked out with factorization combined with measurement. I'm overall confused all around with this question (ignore the writing)

[ruby sorrel]

the teacher did explain that the volume is important to the question however I don't see the importance as well

[final pewter]

should i start watching dragon ball

[ruby sorrel]

lmao that's completely irrelevant to the question

if you have the time on your hands go ahead

<@&286206848099549185>

...

[final saddle]

@ruby sorrel Has your question been resolved?

[final saddle]

@ruby sorrel Has your question been resolved?

[copper chasm]

you're given a value for the volume, and the volume is the length * width * height (length is 4z + 3, width is 4z - 3 and height is 3 or whatever way you want it to be). You can use the volume to find z

[ruby sorrel]

when you find z then what

[copper chasm]

the surface area is basically adding the areas of six rectangles which make up the faces of the cuboid if that makes sense

[ruby sorrel]

mhm

[copper chasm]

use your value of z to find the area of each face and then add them all together

[ruby sorrel]

for example the length with the newly founded z would be (4 X {Z}) +3

[copper chasm]

yes

[ruby sorrel]

i'll add things up and see what I get

did I find z correctly here?

[copper chasm]

not exactly

it should be a whole number i believe

unless i did something wrong

[ruby sorrel]

go check that then

[copper chasm]

pretty sure I did it right

i checked over it

[ruby sorrel]

are you allowed to share how you found it?

[copper chasm]

anyway if the question is about factorization then it's likely that you'd have to get a whole number

[copper chasm]

3(4z-3)(4z+3) = 2325

if you expand that and solve for z

you should get a whole number

it should expand to 3(16z^2-9) = 2325

bruh texit is offline

but anyway it expands to that

[ruby sorrel]

to this? 3(16z^2-9)

[copper chasm]

yes

which is still equal to 2325

[ruby sorrel]

why would 3 still be outside the brackets

[copper chasm]

I just decided to leave it there

[ruby sorrel]

didn't you just multiply everything inside

[copper chasm]

yeah I did but I expanded the (4z+3)(4z-3) first then multiplied by 3

[ruby sorrel]

oh

yeah

realistically 3(4z+3)(4z-3) would be the same as (12z+9)(12z-9)

actually hold on

[copper chasm]

it wouldn't be

actually nvm it would be

but in this case i wouldnt write it like that

because 2325 is divisible by 3

[ruby sorrel]

when you got 3(16z^2-9) = 2325 I think you might have multiplied wrong i'll write out what I did

actually nvm

you did it correctly

im still confused how to write out the length for example

it would be a mess like ({4 X 3(16z^2-9)} +3) y'know

[copper chasm]

it'd be easier to solve for z first in that case

since you know that 3(16z^2-9) = 2325

it will turn into a difference of two squares if you get one side equal to 0

[ruby sorrel]

jeez this suff is difficult to remeber since I did it about 4 months ago

i'll see what I could do

in order to get z alone wouldn't I do 16+9?

making the entire thing 25z

[copper chasm]

you can either multiply out the 3 to get 48z^2 - 27 = 2325 or divide by 3 on both sides to get 16z^2-9 = 775

for now you're looking for an equation which has z^2 in it that is equal to 0

so like 4z^2 - 16 = 0 for example (this is not the actual solution)

[ruby sorrel]

after I multiplied by 3

48z^2 - 27 = 2325

then what lmao

[copper chasm]

you can subtract 2325 from both sides

[ruby sorrel]

the problem is it isn't just one big whole number on the left side

[copper chasm]

the right side will become 0 and the left side will become 48z^2 - 27 - 2325

[ruby sorrel]

continue..

how exactly would 2325 be subtracted though.

I know why just how

[copper chasm]

like any other equation

-27 - 2325 = -2352

so 48z^2 - 2352 = 0

[ruby sorrel]

okay

so thats z itself?

or is there more steps

[copper chasm]

one more step before you get the difference of 2 squares

you have to find a number that goes into both 48 and 2352

so that you can simplify the equation

[ruby sorrel]

okay i'll get my calculator and see if I could find it

[copper chasm]

nvm i made a small mistake

[ruby sorrel]

lets hear it

[copper chasm]

should be 2352

not 2342

[ruby sorrel]

good to know

[copper chasm]

now it should work well

[ruby sorrel]

its divisible by 48 itself

[copper chasm]

yup

so divide the entire equation by 48

then what do you get

[ruby sorrel]

z^2 -49

[copper chasm]

perfect

now that can be factorised

[ruby sorrel]

so 4(z^2 -49) right?

[copper chasm]

not quite

we're solving z^2 - 49 = 0

it's a difference of 2 squares

if that makes sense

[ruby sorrel]

so theres even more steps in order to find z

[copper chasm]

yes, technically 2 more

[ruby sorrel]

since that -49 is sticking out like a sore I asume you +49 on both sides?

[copper chasm]

in this case that would work but you probably shouldn't for other equations because you might end up losing 1 solution. For example z^2 - 16 = 0 -> z^2 = 16 but some people miss that z = + 4 or -4

[ruby sorrel]

what do you do instead then

[copper chasm]

well z^2-49 factorises to (z+7)(z-7) = 0 if that rings a bell

[ruby sorrel]

not necessarily

[copper chasm]

well it's a useful thing to know anyway

a^2-b^2 = (a+b)(a-b)

like z^2 - 64 = (z+8)(z-8)
9z^2 - 49 = (3z + 7)(3z-7)

[ruby sorrel]

so we would break it down into (z+7)(z-7) then?

[copper chasm]

yeah so (z+7)(z-7) = 0 and since they're equal to 0 then that means either z + 7 = 0 or z-7 = 0

because any number multiplied by 0 is 0

so either (z+7) or (z-7) must be 0

then hopefully that shows the 2 solutions for z which are z = 7 and z = -7

but in this case only z = 7 works

because length here isn't negative so z cant be -7

[ruby sorrel]

so my solution would've worked from the start because I actually did round the 6.9 from the start

by luck I asume

[copper chasm]

by luck yes

but in any other case you would've got the wrong answer

[ruby sorrel]

this information is more complicated than it should for me

lots of steps in order to find z

[copper chasm]

well a lot of them can be compressed

[ruby sorrel]

might be easier if it was actually written out

[copper chasm]

that's the way I did it

but i divided by 3 at the beginning

then divided by 16 later on

[ruby sorrel]

lemme take a quick look at it

[ruby sorrel]

when you said this

how so

[copper chasm]

well going through the z + 7 and the z -7 bit, you can just skip straight to z = + or - 7

rather than going z + 7 = 0 and z - 7= 0 then solving

when you get used to it, it's easier to just jump certain steps

like going straight from 3(16z^2 -9) to 48z^2 - 2352 = 0

instead of multiplying out then subtracting and showing that

[ruby sorrel]

our schooling systems are likely much different and that's probably why I don't 100% understand it

our school does online tutoring in 3 hours so i'll see if I understand better then

if I ever want to view this conversation again it'll be saved to help-36?

[copper chasm]

well it will be in this channel at this time yes

but youd have to scroll up till you found it

[ruby sorrel]

with the search bar at the top right that'll be easy

[copper chasm]

mhm

[ruby sorrel]

anyways whats the command to end this?

[copper chasm]

.close

[ruby sorrel]

.close

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[slate prism]

second part

about interquartile range

???

[rare quarry]

okay so

whats an interquartile range

[slate prism]

subtract the left line by the median from the right

i know for the dog its 15 and the cats is 12

is it 33%?

[rare quarry]

no but ur on the right track

[slate prism]

how would i solve it

[rare quarry]

assuming the ranges are correct, you want to find the %

that's (15-12)/12

[slate prism]

25%?

[rare quarry]

yep

[slate prism]

TYSM

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[south jasper]

hi

[iron needle]

hi

[south jasper]

Any tips about approaching part(i)

[iron needle]

||:).||

[south jasper]

As i got no clue how to solve for it

[iron needle]

i forgot statstics

i have to revise it

[final saddle]

@south jasper Has your question been resolved?

[pearl glen]

This one is a really neat problem

Have you tried out Bayes theorem?

That way you can switch it over from probability that it's a male given that the subject is x cm tall to probability that the subject is x cm tall given that he's a male

Which is 10 times easier to find out

[south jasper]

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[boreal spruce]

I am reading the solution of the problem. I don't know how they derive the yellow part particularly.

[hearty zephyr]

pretty sure it's a typo
the 2 shouldn't be there

[boreal spruce]

Okay thank you. It makes sense now.

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[fathom stream]

what am I doing wrong

[fathom stream]

I find what f(5) and f(5+h) is

and then plug in

(f(5+h)-f(5))/5+h-5

im an idiot

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[fathom stream]

I find the inverse

which is

(-x-1)/(x-1)

and then find the derivative of that

which was

2/(x-1)^2

and then plug in

x-1/x+1 for x

and then got

[terse siren]

it want's the answer not the derivative

[fathom stream]

yes I gave the answer

I plugged in 3

[light wren]

You didnt evaluate it

[terse siren]

from the picture you just put 3

[fathom stream]

it did it for me doe

[light wren]

[terse siren]

🤔

[light wren]

wait

you expanded (x-1)^2 into x^2+2x+1

instead of x^2-2x+1

although for a q like this I would just leave it in factored form and plug 3

to get 2/(3-1)^2

[fathom stream]

yes

oh wai

fuck

[light wren]

:DD

[fathom stream]

omg

why am I so

shit

ty

sm

[light wren]

[fathom stream]

Ive been lookin at my work for the past 10 mins confused asl

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[fleet pelican]

can someone explain to me how to strech linear functions (transformations)

[fleet pelican]

pls

like for example, strech the line 2/5x + 5 from the line y = x by a scale factor of 5

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[ruby nebula]

help pls

[ruby nebula]

nvm

i figured it out

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[limpid pike]

Hi I am trying to understand the answer to this question. I understand step 1 (marked in red) but don't see how they got step 2 (marked in red) from step 1. I was under the impression that after you transpose the matrix you just negate the numbers inside it to get -A^t. Unless there is some property I might have forgotten. Thanks in advance.

[desert mantle]

well that's step 1

on the left is A, on the right is -A^t

and then you compare each entry, they have to be the same

you need a=-a, which means a=0

then b=-d

c=-g

and so on

[limpid pike]

oh wow

Thank you so much

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[grand rose]

4!(9n-1) divided by (9n)!

[tranquil pine]

What did u try

[grand rose]

Sorry wrong problem

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[violet rune]

There's something that i didn't get about the definition of a limit at a point.
Why (on the photo) the delta is define with a "it exists" instead of a "for all"

[ocean lintel]

It can get smaller, but it can only get so large

[strange shore]

for each epsilon, there is only one value of delta

[violet rune]

So when epsilon is very small it exists only a delta very small?

[strange shore]

thats what that means

[ocean lintel]

The smaller epsilon, the smaller the delta you have to take

Try to visualize it graphically

If delta0 works, any delta < delta0 works as well, but there's no guarantee that a delta > delta0 will

[violet rune]

Since all delta< delta0 work then there's more than 1 value that works

[ocean lintel]

Yes

[violet rune]

Ah never mind I think a get it

It's a it exists not a it exists only

[ocean lintel]

Yes

[violet rune]

So it works

[ocean lintel]

That's why you have $\exists$ and $\exists !$

[soft zealot]

mateo713

[violet rune]

Yeah

[ocean lintel]

The second one is "there exists a unique"

[violet rune]

Thx guys

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[ruby nebula]

<@&286206848099549185>

[shell mountain]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[ruby nebula]

i posted this in another channel like an hour ago with no response and it closed so i opened a new channel and pinged the helpers

[shell mountain]

When you open a new channel, you still need to wait 15 minutes

[ruby nebula]

why

i already waited an hour

[shell mountain]

A new channel is a new channel

[ruby nebula]

well no one helped me so i pinged

an hour is a lot more than 15 mins

[shell mountain]

You shouldve pinged in your original channel

[ruby nebula]

ok i will rn

[shell mountain]

? But your old channel is closed

[ruby nebula]

someone else used it after it closed

so my question is still there

and i just pinged

happy?

[shell mountain]

It's not your channel anymore

[ruby nebula]

well i already pinged

all i want is help with that question

im not here to argue

[limpid pike]

2x + y^2 = 6
y^2 = 6 - 2x
y = +/- sqrt(6 - 2x)

two answers because +/- so y is not defined as a function of x

https://www.youtube.com/watch?v=1uykTqTYDHM

[ruby nebula]

thank you

[grim badger]

Don't do the work for people or give answers

[limpid pike]

Sorry

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[fervent scroll]

A continuous increase of 37% is the same as an annual increase of ___ %

[fervent scroll]

I don't know how to convert continuous percentages to annual increases

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[rocky remnant]

can someone explain

how professor gets residue values here without laurant series

because dont we need the find the laurant series around z=1 and z=-1

[vital crag]

You don't need laurant series for simple poles

[rocky remnant]

oh there is something about poles?

is it the f(z) =g(z)/z-z0

and then if g(z) is analytic at z0 then res of f(z,z0) = g(z0)?

[vital crag]

Yes

[rocky remnant]

ok tyty

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[fathom stream]

I dont rlly understand

this question

[supple mantle]

Ok firstly

Do you know what your second derivative represents?

[fathom stream]

over here?

[supple mantle]

Yep

[fathom stream]

uh second deriv is to find inflection points right

where the concavity changes

[supple mantle]

Yep

[fathom stream]

so if its > 0

that means ?

[supple mantle]

And the question tells you it never goes to 0

It means it's always increasing

It never decreases

[fathom stream]

oh so it cant be under 0 and -1.4 is under?

[supple mantle]

Wait

Let's go step by step

Then we find conclusions

[fathom stream]

ok

[supple mantle]

Now we're done with second derivative

[fathom stream]

yes

[supple mantle]

Let's move on to the next condition

Tells you tangent line at a=1 (Idk what a is doing here) y= -2x +1

[fathom stream]

mhm

[supple mantle]

And the tangent line's slope represents the derivative's value

At that point

[fathom stream]

the slope is -2

[supple mantle]

Alrighty

[fathom stream]

and since its neg

its going down?

[supple mantle]

Yep

The earlier condition btw is about the first derivative

The first derivative is always increases

[fathom stream]

the

f''(x)?

oh the one u mentioned

so if f''(x) > 0 that means our first deriv is always increasing?

[supple mantle]

Yep

At x=1, f'(x) = -2

[fathom stream]

oh so slope is -2 so we know f'(x) = -2

[supple mantle]

We should predict that f'(x) at x = 1.2 should be somewhere like -1.6 or -1.4

Since it increases

[fathom stream]

alright

[supple mantle]

That means there will be an error when we find the linear approximation from -2x + 1

And since the slope increased, our linear approximation will always be an underestimation

[fathom stream]

wait

isnt -2x+1 like a decreasing line so isnt the slope constant

[supple mantle]

Yeah but that's the tangent line

The actual function isn't a line

[fathom stream]

how do we know that doe

what the actual function looks like

[supple mantle]

The fact that the f''(x) > 0 tells us that it is never a straight line

Take the derivative of a straight line twice and see what its value will be

[fathom stream]

0

[supple mantle]

Yep

And our function has f''(x) > 0

So it won't be a straight line

[fathom stream]

wait

so we know

at x = 1

f'(x) = -2

right

[supple mantle]

Yep

[fathom stream]

and they are askin us for the estimate

[supple mantle]

Yep

[fathom stream]

at x = 1.2 for

f'(x)

[supple mantle]

No

[fathom stream]

oh

[fathom stream]

.

[supple mantle]

Yep

[fathom stream]

isnt that what u meant

[supple mantle]

I was trying to explain why the tangent line estimate we have for the function will be an underestimation

It will be lower than the actual value

[fathom stream]

oh so like -2(1) +1 will be more then -2(1.2)+1?

[supple mantle]

More accurate

[fathom stream]

shit im so dumb I acc dont get it

[fathom stream]

oh im right?

on that

[supple mantle]

The value itself isn't that way

But accuracy at x=1 should be 100% because the tangent line for it touches the function at exactly that point

But the accuracy starts to go off once you go away from x=1

Lemme try to demonstrate it with a drawing

[fathom stream]

my friend sent me this

doe I dont understand

is it sayin that it will always be greater for point under zero if f"'(x) > 0?

[supple mantle]

You have the actual function, and you have its tangent line

Tangent line touches the function at exactly 1 point, so it approximates the points around it

But when the slope of the function (the derivative of the function) keeps increasing, it means that the tangent line at a single point will be below the function at other points

[fathom stream]

wait so at 1 it will be -1 then wont that mean at 1.2 it will be a little lower than -1

[supple mantle]

Exactly

That's just the tangent line

But the actual function will be higher than what the tangent line will give you

[fathom stream]

ohh

OHHHH I SEE

[supple mantle]

All because of the first sentence saying that the second derivative of the function is higher than 0

[fathom stream]

if it

was less than 0

concave down

[supple mantle]

Mhm

[fathom stream]

then our point would be loweer than

what the tangent

line gives us

[supple mantle]

Yep

[fathom stream]

around the 100% accuracy point

I see

now

[supple mantle]

The tangent line is gonna be above the function

Think of the function as a ball

And a paper going over it as the tangent line

When the ball is over the paper, that means that the ball's points are above the paper

Or the function's points are over the tangent line

[fathom stream]

I get it now

tysm

[supple mantle]

Glad I helped :)

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[tranquil pine]

I'm admittedly not even too sure to start with this

[tranquil pine]

im assuming it doesnt mean the literal dimensions like 3x4. Would it be asking for what Rn is equal to ?

[rare cradle]

Do you know what span means

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[white night]

Is this correct

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<@&286206848099549185>

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<@&286206848099549185>

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[austere pumice]

How do I do this

[grim badger]

Do you know what f(1) means?

[austere pumice]

Nvm got it

What about this one

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[fringe kite]

$My'' + (2y * \pi a^2) * \rho g = 0$

[soft zealot]

Zim

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[simple parcel]

is this correct?

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[snow fox]

Polybius square to be deciphered

Any suggestions on what the threes and twos on the sides could mean?

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[oak kraken]

Do I understand this correctly? If so, why is the third step allowed?

Would the same rule apply for 2/e?

[grand bronze]

No because $2^{kt}\neq 2$

[soft zealot]

Duh Hello

[grand bronze]

But $1^{kt}=1$

[soft zealot]

Duh Hello

[grand bronze]

Hence why the first is allowed

[oak kraken]

Hmmm.. OK so this only works for e/1 to begin with

[grim nebula]

this doesnt make sense

you need parens around the whole thing

and then a couple more exponents

[oak kraken]

Oh right that only brings e to the denominator, not 2 to the numerator

The way I wrote it

I keep thinking negative exponent is reciprocal to the entire fraction but it’s not

It’s only reciprocal to the base with the negative exponent

@grim nebula

[grim nebula]

last line is a

[oak kraken]

This is incorrect too right?

The 2 does not belong in parentheses

[grim nebula]

it only works with 1 because 1^blah = 1

[oak kraken]

OK.. so that would be incorrect let me fix that

This would be OK?

@grim nebula

[grim nebula]

ye

[oak kraken]

Is there any situation where 1^n can be anything other than 1?

[grim nebula]

uh

probably when you do complex analysis

but uh

thats a whole can of worms

[oak kraken]

Lol .. so nothing to worry about in Calc 1 or Calc 2?

[grim nebula]

probably not

unless you're doing limits ig

indeterminate form stuff

but then in that case it isnt truly 1^blah

its a limit

[oak kraken]

OK Ty!

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[tranquil pine]

the function is positive when x < -3, negative when -3 < x < 1, and positive when x > 1

[tranquil pine]

I dont understand this question

[sour geyser]

Parabola Equation?

[tranquil pine]

yes!

please i really need help

ive been spending the last hr trying to understand

[oblique portal]

what exactly dont you understand

[tranquil pine]

how this equation should be graphed

[oblique portal]

what do you think of the function being positive and negative

[tranquil pine]

well

for the first qs

i thought it was this

[oblique portal]

i think you mixed increasing and decreasing

[tranquil pine]

"The function is increasing when x > -1 and decreasing x < -1"

[oblique portal]

wait no its right

thats correct yeah

[tranquil pine]

u sure?

[oblique portal]

yeah

its right

[tranquil pine]

the second question

the function is positive when x < -3, negative when -3 < x < 1, and positive when x > 1

[oblique portal]

the graph is kinda not that visible

ill do u one in desmos and u can check if thats right

[tranquil pine]

how do i go to desmos

[oblique portal]

nvm that desmos doesnt really help

[oblique portal]

when x<-3, f(x) is positive means any value of the function to the left of -3 should be pointing up

[tranquil pine]

can you do a demonstration

[oblique portal]

this should be the function graph

ugly sketch in paint

but

[tranquil pine]

yay thanks1

okay ty ty

so its the same with the first question right

[oblique portal]

kinda yeah

[tranquil pine]

TY TY

[oblique portal]

here i drew it as a parabola but you can do it with a modulus liniar as well

[tranquil pine]

can u accept my friend reuqest pls

[oblique portal]

just remember the bit under the x axis

[tranquil pine]

oki!

[oblique portal]

thats when -3<x<1

is negative

#❓how-to-get-help

[tranquil pine]

.close

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[plucky ermine]

Can somone help me

[tranquil pine]

i will close this channel

since u have another one open

.close

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[unique vapor]

can anybody help me with a maths equation/

[paper zodiac]

The give says that AD is a bisector of BAC

It means it cuts it in half

[unique vapor]

yes

[paper zodiac]

So the bottom angle is also 19 degrees

[unique vapor]

oh right

[paper zodiac]

Using soh cah toa to find line DB

[unique vapor]

tysm

[paper zodiac]

Oh I got this wrong

@unique vapor

Use line DB for later

We know that angle DAB is also 19 degrees

19 + 19 = 38