help-10

[fiery venture]

If you're trying to get the derivative with respect to the bias in the first layer, you first have to get the derivative of $L(y; \theta) = L(W{2} \sigma(h) + b{2}$ for which you need the derivative of $L(y; \theta) = L(v)$

[warm shale]

yoohoo

[glad sleet]

there's only one layer

[fiery venture]

ok

wait

can you write exactly which function your network is computing?

if X is your input and y your output

I guess if your works it works, so maybe theres no need to bother

[glad sleet]

well

it works but I just did some sneaky way to make it work and now I wonder if I made a mistake in finding the dLdB

[glad sleet]

$L(y; \theta) = L(\sigma(W{1}.X+B{1}) $

odd

this basically

instead of W2 we now have added bias'

so eventually we get the formula in the red rectangle

but for the bias'

the one for the weights worked fine

and my solution for that is: ∂L/∂B1 = (A1-Y) * A1(1-A1) * 1

[obtuse pebble]

@glad sleet Has your question been resolved?

[brisk oar]

Is this the bias in the second to last layer?

I'm getting $$\sigma' \qty(p_1) \odot \qty(w_2^T * \qty(\sigma ' \qty(p_2) \odot \qty(a_2-y)))$$

[warm shale]

RipeOrange

[brisk oar]

$*$ being matrix multiplication and $\odot$ being the Hadamard product

[warm shale]

RipeOrange

[glad sleet]

the bias is just before the sigma function is applied

in this case that's called p

p = wx+b

w being the weights matrix

b being the bias matrix

X being the input

[brisk oar]

Yep

The bias would be $$\pdv{L}{b_1} = \pdv{L}{a_2} \pdv{a_2}{p_2}\pdv{p_2}{a_1}\pdv{a_1}{p_1} \pdv{p_1}{b_1}$$

[warm shale]

RipeOrange

[brisk oar]

Which I think works out to be this

[obtuse pebble]

@glad sleet Has your question been resolved?

[obtuse pebble]

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[woeful jolt]

hi guys, i am a bit stuck on this one

[tired shell]

you'd need a non-saddle point for there to be a max/min on the interior

so the max and min must be somewhere in the C1 boundary

[woeful jolt]

so is the 2nd option implying that f has a max of (1,1) over its entire domain as opposed to S?

[tired shell]

hm? S is the entire domain

[woeful jolt]

oh whoops

i mean im not sure about the difference between 2 and 4

because isnt (1,1) a boundary point of S?

[tired shell]

right, so if 2 is true it's the best choice

but there's not a good reason for 2 to be true, it could be elsewhere

[woeful jolt]

oh yeah
based on what i know up until now, a global extremum of f would either belong to the interior of S or its boundary

so that global extremum may not necessarily be a maximum

[tired shell]

(1,1) doesn't even need to be an extrema

[woeful jolt]

yeah

i would go with option 4 then

[tired shell]

sounds good

[woeful jolt]

thanks!

i also had another quick question

[tired shell]

go for it

[woeful jolt]

for this type of question you wouldn't really need to know any multivariable calculus stuff right? as in you can answer it based off of set theory knowledge

hm actually the first one should not work

because there might be a point in A whose second partial derivative is not 0 inside B

but everything in S would be in A, but may not contain every element of A (so i would say option 2 is true)

[tired shell]

right 2 and 4

multivariable calculus might tell your other options are true potentially

but turns out it doesn't, and you can just go by set theory

[woeful jolt]

thanks so much 😄

now i should free up the channel, but im not sure how to do the command for that

[torn wave]

do

.close

[tired shell]

.close

[obtuse pebble]

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[finite lichen]

I am having trouble finding the most efficient way to prove this

[timid silo]

how about this: odd squared is odd , odd + odd is even , if all a b c are odd, then even = odd which is not possible

[finite lichen]

Like this?

[timid silo]

um no, this proof doesnt need much mathematical calculation or watever that is called

odd numbers when factorised dont have 2 in them so odd squared is odd too

[finite lichen]

That type of reasoning is not allowed in my teacher

*by

[timid silo]

mm k

[finite lichen]

Im confused on how to show it

if that makes sense

[timid silo]

i understand...ill try

[timid silo]

expand c, ull get a long expression with lots of terms being multiples of 4
then u can rewrite it as 4(m) + 1 = 4(n), where m and n are those terms added

4m and 4n are even

but 4m + 1 is odd

odd = even contradiction

[finite lichen]

So the statment is not true?

[timid silo]

yes

proof by contradiction

[finite lichen]

Can you show me your work

I think I udnerstand you

but not really

[timid silo]

(2p + 1)^2 + (2q + 1)^2 = (2r + 1)^2
4p^2 + 1 + 4p + 4q^2 + 1 + 4q = 4r^2 + 1 + 4r

4p^2 + 1 + 4p + 4q^2 + 4q = 4r^2 + 4r

4(p^2 + p + q^2 + q) + 1 = 4(r^2 + r)

even * odd is always even

so 4(p^2 + p + q^2 + q), and 4(r^2 + r) are even

but even + 1 is odd

4(p^2 + p + q^2 + q) + 1 is odd

odd = even

contradiction

[finite lichen]

I seee

[timid silo]

so assumption that all a b c are odd is wrong

[finite lichen]

I see

I need to go rewatch the lecture

Thank you so much

🙂

.close

[obtuse pebble]

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[timid silo]

which grade is this if u dont mind

[finite lichen]

College

[timid silo]

mmk

[finite lichen]

Like first year

[timid silo]

what course?

[finite lichen]

Discrete Math

[timid silo]

that comes in first year!?

thought that was a last year thing

[finite lichen]

hypothetically

I think calc is the only pre req

[obtuse pebble]

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[obtuse pebble]

@finite lichen Has your question been resolved?

[tired shell]

.close

[obtuse pebble]

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[stone phoenix]

Not sure what to do for this question

Tried starting off with

$$(x-0.15x)$$

[warm shale]

Gr00by

[stone phoenix]

<@&286206848099549185>

[elder helm]

(1-0.15)x = 1

[stone phoenix]

oh k

[stone phoenix]

does it matter if the x is before the bracket or after

[elder helm]

yeah

you want 0.85x in this case

[stone phoenix]

ok

ye

so how do u find the answer

0.85x = 1

do u divide

[elder helm]

x = 1/0.85

[stone phoenix]

by 0.85

yeah

ok

[elder helm]

and u get like 1.17 something

[stone phoenix]

x = 1.17

[elder helm]

so u need to increase by 17%

[stone phoenix]

yeah

the answer is 17.6

bits 17.65

but the question didnt say to do it one decimal point

or two

but answer is 17.6

[elder helm]

cool

[stone phoenix]

k

thanks

[elder helm]

👍

[obtuse pebble]

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[empty jungle]

help

[empty jungle]

How to find the range?

for y=f(IxI)

[royal basin]

you mean f(|x|)?

[empty jungle]

yes

[royal basin]

you used capital i's to replace absolute value bars. i don't recommend doing that.

anyway, do you know what the graph of y = f(|x|) looks like?

[empty jungle]

yes

i drew y = f(|x|)

[royal basin]

you can even have desmos plot it for you, if you put f(x)= at the beginning of the first line and write y=f(|x|) on another

y = f(|x|) can be created by taking the part of y=f(x) to the right of the y-axis and mirroring it

[royal basin]

yes, like this

[empty jungle]

but how to find the range?

I know how to find the domain

but the range is different

[royal basin]

the domain is what values your graph covers on the x-axis

the range is what values your graph covers on the y-axis

it should be clear that the graph goes down to -3 on the y-axis

and the peak is at y=0

[empty jungle]

that means Range: {-3<y<0}

right?

[royal basin]

≤

-3 ≤ y ≤ 0

[empty jungle]

it should be right

but

the answer is

x<0

x≤ 0

[royal basin]

can you show the original question?

[empty jungle]

Question b

i think my teacher got it wrong

thats the only think i can think of

[royal basin]

no, it sounds like your teacher only corrected you on the upper boundary

but wrote it in a somewhat confusing manner

[empty jungle]

?

so what is the answer?

what is the range?

.close

[obtuse pebble]

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[empty jungle]

.reopen

[obtuse pebble]

✅

[royal basin]

the answer is {y : -3 ≤ y ≤ 0}.

[empty jungle]

nice

thxs

.close

[obtuse pebble]

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[swift mortar]

my current answer is (x * 85/100) * 125/100 : y

[swift mortar]

I don't know the answer just feels wrong

[elder helm]

looks goo

d

just simplify

[obtuse pebble]

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[agile turret]

problem number 5. I attached my work, the original problem, and the book’s answer. is there something I am doing wrong here? why is my equation so much different?

[agile turret]

(obviously, it’s problem #5)

i literally get it up until that last step

also realizing i just made a -2x magically disappear, my bad, i’m very tired. regardless, my answer would still be very different

[timid silo]

you actually got the ans

just a small mistake

[timid silo]

in your answer, the place u boxed ln(2) - 2

before that

(8ln(2) - 2))/8

so ans = ln(2) - 1/4

u just cancelled 8 and 8 by mistake

[agile turret]

ohhh. so you’d factor out an 8 from the -2, which gives you the -1/4?

[timid silo]

yes

(a+ b) /c = a/c + b/c

[agile turret]

gotcha, just didn’t keep my quantity together. thank you so much!

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[elder pilot]

quick question

[elder pilot]

when doing range of a function

${\mathbb{N}$

[warm shale]

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

[elder pilot]

or $\geq1$

[warm shale]

Fadanglebang

[elder pilot]

for this type of function

[novel knoll]

$x \in \bN$ means $x$ is in ${1,2,3,\dots{}}$ while $x\geq 1$ (without other context) means ${x \in \bR : x\geq 1}$. So here $x$ can also take values such as $1.212121$ or $\pi$ for example.

[warm shale]

ScapeProf

[elder pilot]

ohh tru

thanks

.close

[obtuse pebble]

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[desert gorge]

Hello. So first, I deed to say I'm German and my English skills might not be the best. Im gonna try to translate the exercise with google translate.

So I have to do that:
Determine the mapping matrices of the rotation R ^ 3 -> R ^ 3 around the coordinate axes by an angle φ.

I almost have the solution but I've always put the minus wrong

[desert gorge]

and the solution is

I get for example :
( cos(phi) sin(phi) 0
sin(phi) -cos(phi) 0
0 0 1)

Can anyone help?

[obtuse pebble]

@desert gorge Has your question been resolved?

[desert gorge]

<@&286206848099549185>

am I in the wrong help section?

[dawn parrot]

you can ask here or you can try #linear-algebra both works

[desert gorge]

ok

.close

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[obtuse pebble]

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[timid silo]

how do I do this?

this is what I did so far then I’m stuck

[carmine hawk]

Looks good, from the last equation you get 20+9x = m+mn+n^2 x

Now, match the constants with the constants, and the x from the left side with the x from the right side

rather, 20 = m+mn as you have written

but also, 9x = n^2 x

Does that make sense?

[timid silo]

yes

[carmine hawk]

So from here solve for n

[timid silo]

n=3

Ohh

Substitution method

I get it now thank you !!

.close

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[wary zenith]

How might one do this:

$\lim_{h \to 0}f(x+h)-f(x)$

[warm shale]

HappyAlt

[mossy bane]

Need help with this please:

Positive angles x and y are such that:
x+2y=300 degrees and tan y = 2 cos160 degrees
Find x and y.

[wary zenith]

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[unique solstice]

Whats f(x)

[wary zenith]

Oh I'm stupid

I didn't see that in the question header

That's why I got confused too lol

[unique solstice]

Reasonable

[wary zenith]

Apparantly, they do give you f(x)

Ight, sorry for my stupidity

c ya

!close

.close

[obtuse pebble]

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[radiant cobalt]

can anyone help me with this?

[timid silo]

do you know impplicit differentiation?

[radiant cobalt]

yes

[timid silo]

differentiate wrt r on both sides

left hand side we have to use the power rule

right hand side chain rule product rule

[radiant cobalt]

okay

so my left hand side will be 2r dr/dx

?

[timid silo]

no

we are differentiating with respect to r

not x

[obtuse pebble]

@radiant cobalt Has your question been resolved?

[radiant cobalt]

oh

so it will be 2r d/dr

?

[obtuse pebble]

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[opal linden]

Hello, I'm kinda new to field and i wanted to know how do we find the primitive element of a power of 2 galois field ( like GF(8)) cause, 2**n is not prime so alpha (the primitive element) doesn't exist ?

[opal linden]

For any number n in GF(8)\{0}, n**i doesn't construct GF(8) for any i, so how it's possible

<@&286206848099549185>

[obtuse pebble]

@opal linden Has your question been resolved?

[opal linden]

No one ?

[timid silo]

I think you'd be better off asking in an advanced mathematics channel

[opal linden]

I'll do that, thank you

.close

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[midnight oak]

how do i find the angles if the sum of angles inside the polygon equals to 100

[green flame]

hi can you help me?

[midnight oak]

yea hi

whats your question

[green flame]

we have math test now

[midnight oak]

i dont understand that language sorry

[green flame]

ou sorry

[midnight oak]

<@&286206848099549185>

[midnight oak]

^

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[tired shell]

what do you get for the 1/4 part, -2 part, and the 2/3 part?

[dense dirge]

[16-4] . [4/9]
= 16/3 or choice b

[obtuse pebble]

@coarse abyss Has your question been resolved?

[nocturne minnow]

Don't give out answers

[obtuse pebble]

@coarse abyss Has your question been resolved?

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@exotic sparrow Has your question been resolved?

[exotic sparrow]

hellooo

<@&286206848099549185>

[forest sinew]

for which?

also 👋

[exotic sparrow]

3e and 3f

e first

[forest sinew]

what have you tried for e part one?

can you tell me any numbers you think work or dont?

[exotic sparrow]

its true when x is bigger than -1

or equal and bigger than 0

nvm

equal and bigger than 1

[forest sinew]

not equal, right?

[exotic sparrow]

yeah

so 2 and over

[forest sinew]

so should be a similar process for the other part

just looking for pathological integers

should be a fairly small set

[exotic sparrow]

how would I write the answer

for the first part

[forest sinew]

as a set

you can use set builder notation

[exotic sparrow]

can you show me

[forest sinew]

can you express it in words?

i guess you said 2 and over

[exotic sparrow]

x has to be equal or bigger than 2

for it to be true

[forest sinew]

so its gonna be like this

${ x \in \bZ : x \geq 2}$

[warm shale]

jan Niku

[forest sinew]

read: all integers, such that x is greater than or equal to two

try the second part

[exotic sparrow]

x has to be less than 1

or equal

[forest sinew]

no

its a finite set

its actually super small

[exotic sparrow]

-1 0 1

[forest sinew]

yea

so write it in set notation

[exotic sparrow]

{1,0,1}

just that?

[forest sinew]

[exotic sparrow]

for all values for n, n squared is bigger or equal to 0

which is true

[forest sinew]

sure

sounds reasonable

[exotic sparrow]

so the answer is just "True"?

[forest sinew]

yea

each one will just be T or F

[exotic sparrow]

so ii would be false

for some values of n, n^2 = 2

[forest sinew]

sounds false, yea

kinda infamously false

[exotic sparrow]

and iii false

[forest sinew]

why do you say that?

[exotic sparrow]

nvm its bigger or equal to

if it was bigger i was gonna say n=1

[forest sinew]

youll get 4 pretty fast

but you should notice that 4 is the negation of 1

and think about what that means

[exotic sparrow]

false

n^2 cannot be negative

[forest sinew]

yea

but you should definitely notice what i said

if its given that $\forall x : P(x)$ is true, then $\exists x : \neg P(x)$ is false just directly @exotic sparrow

[warm shale]

jan Niku

[exotic sparrow]

yep

[forest sinew]

being equivalently like "this is always true" tells you something about "this is sometimes false"

anyways ez pz

[exotic sparrow]

yeaa

[forest sinew]

lol you have a lot of homework

i do too

[exotic sparrow]

ok can you just help 4(I)

I know the rest

<@&286206848099549185>

[sour mortar]

you can say that iff a = b = n, then you can show a . a = n^2 = a^2 (reflexive)

a . b = n^2 = b . c = a . c (transitive)

a . b = n^2 = b . a (symmetric)

I think

[exotic sparrow]

.close

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[supple cypress]

how do i find the sum of this

i tried and got 6/(pi - 1)

but that was wrong

[mild ocean]

@supple cypress what you put is correct

maybe the system didn’t like the way you formatted your answer

[supple cypress]

oh

guess ill ask my teacher tmrw then

[mild ocean]

replace x with -61 and 11

[obtuse pebble]

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[hollow yoke]

I don’t understand something about derivatives. The equation e^((x^2)(y)). Depending on the question it makes me use chain rule or power rule, however power rule would be 2xy(e^((x^2)(y))). If I use the chain rule it would be 2xy+x^2(e^((x^2)(y)). I don’t get how this works? If I use implicit differentiation I do chain it seems, if I just solve for that equation it would be power rule? I just don’t get when to do what.

[mild ocean]

@hollow yoke i can’t think of any situation where you would use the power rule for that

you would use the chain rule on that function

[hollow yoke]

I just plug it in on something like Photomath and it would give me the power rule answer. Are you saying chain rule should always be applied though when possible?(e.g. a number having an exponent)

[mild ocean]

no, there’s a number that’s not a variable in the base

i know it’s e, but there’s still a function in a function

use chain rule on it

[hollow yoke]

From what I’m seeing Photomath uses chain rule then power rule on the x^2 instead of using chain rule then chain rule on x and product rule. But if I made the question a implicit differentiation question, it would then use the chain rule, chain rule, product rule. Which I don’t get how you can use both of these when they give different answers it seems

[mild ocean]

you only need to use the chain rule once

[hollow yoke]

But why is it used twice in implicit differentiation? How can it be?

[mild ocean]

unless y is a constant, implicit differentiation is required

because implicit differentiation uses the chain rule

it’s pretty much just a rewrite of the chain rule

[hollow yoke]

Okay, it seems I just don’t understand implicit differentiation enough then. It seems to use the chain rule to a good extent? Thank you

[mild ocean]

dy/dx = dy/du * du/dx, du/dx = du/dy * dy/dx @hollow yoke

the du/dx part is implicit differentiation

where u = something in terms of y

[obtuse pebble]

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[split latch]

I need help please

[muted flame]

Ask your question

[split latch]

What is the maximum number of solutions for a system of a hyperbola and an ellipse?

1, 2, 3, or 4

?

[unique solstice]

What do you think

[split latch]

I have no idea

I don’t know

[unique solstice]

Do you know what both look like

[split latch]

Yeah

[unique solstice]

Try drawing a picture

[split latch]

Ones like a circle and the other one is two half circles facing opposite sides

Okayyy

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[ember rune]

2x – 3y = 8
–3x + 2y = 8
How do i classify the system? (consistent and dependent, inconsistent, consistent and independent)

[elder helm]

reduce it to rref

[ember rune]

what is rref

[elder helm]

Reduced row echelon form

Have you been taught Gauss Jordan elimination?

[ember rune]

no i haven't

im in algebra II

[elder helm]

have you been taught any ways to simplify the system?

or maybe just solve for x and y

[ember rune]

i think putting it into slope intercept form but that' sit

that's it*

[elder helm]

yeah

the intersection is the solution

might be helpful to graph them

[ember rune]

ah ok

is there a faster way to do it though?

[elder helm]

this lol

[ember rune]

oh lmfao

okay thank you

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[ember rune]

.reopen

[obtuse pebble]

✅

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[boreal crest]

If I have the function f(x, y) = sin(pi/2 * xy) and I'm asked to graph this on the line y = x, how do I go about doing this? I'm not exactly sure what it means to graph a function on the line y = x

[last comet]

normally you see a function f(x) and are asked to graph that on an x-y coord system
this implicitly already makes you graph the function on a line, a very trivial one

this case is pretty much the same
the function f(x,y) generates points in 3 dimensions but if you are asked to graph this function on just a line, then it simplifies back to a 2 dimensional plot

and how do you go about it? very simple:
take a look at the information that oyu are given about the line and see how that helps you
(remember, you want to "reduce" the function to something that only depends on 1 variable, not 2)

[boreal crest]

That makes perfect sense--thank you!

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[thick torrent]

Oh interesting this seems like the python server help system

We're learning drawing graphs in class from an equation using calculus. One of the steps is to find the x and y intercepts. Y intercept is easy as x can just be set to 0, but finding roots is difficult. We've been taught 2 ways to find roots:

1. Intermediate value theorem, do this many times to approximate (I don't like this as it's time consuming)
2. Rational zero theorem. I was wondering, if the constant of the polynomial is 0, if this can still be used, and how?

[thick torrent]

• Also, are there any better/clever ways to quickly find roots of a polynomial?

[unique solstice]

Newtons method is good for certain things

[thick torrent]

Ok, cool. What about 2.?

We were told to use rational zero theorem but i'm not sure if it's possible when the constant term = 0

[unique solstice]

Youd prob just factor out as many xs as you need to get a new constant

[thick torrent]

Oh yes good point!

Thank you

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[glacial mesa]

hello I don't really understand what's wrong in here - can someone give pointers/hints of what i am doing wrong? thank you!

[glacial mesa]

oh

nevermind I found it the moment I posted it

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[glacial mesa]

~~I forgot to differentiate the 9x ~~

.close

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[azure scarab]

someone help me with triganomerty

[azure scarab]

trig*

[obtuse pebble]

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[ripe flare]

(x-3)^2-3(x-3) how to factor

[ripe flare]

i got out (x-3) wuts the other one?

[obtuse pebble]

@ripe flare Has your question been resolved?

[obtuse pebble]

@ripe flare Has your question been resolved?

[ripe flare]

nvm, i solved it

[obtuse pebble]

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[ripe flare]

lol

[obtuse pebble]

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[topaz harness]

how to solve this

[obtuse pebble]

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[woeful jolt]

hi guys, since (0,0,1) is the top of the sphere you would need a flat vector right?

[woeful jolt]

so its z coordinate 0

would the third option suffice then?

also for this i am unsure how to find dimension

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[marble juniper]

hello i just am having trouble understanding jordan decomposition

[marble juniper]

i know how to find the geometric multiplicity and algebraic multiplicity of any eigenvalue. so i can determine the jordan normal form up to reordering of the jordan blocks on the diagonal

[proven breach]

better ask in linear algebra or abstract algebra channel

[marble juniper]

ok!

im going to leave this open for a minute.

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[timid silo]

how do i find the domain for this function (part a) ?

[carmine hawk]

Do you know what a domain of a function is in general?

[timid silo]

yea

who far the function goes in the x axis

[carmine hawk]

That can be the graphical intuition for it, but a domain is the set of numbers that you can input in the function

For example, if I had a square root function, I could put only non-negative numbers in it if we are working with the set of real numbers

So it is the set of numbers that you can put into that x

Do you think that you can put any value in e^(-x^2 /2) and get a result?

[timid silo]

in the x, yes

[carmine hawk]

Great, that means that the whole set of real numbers is your domain, since the function will be well-defined there

[timid silo]

does that means its an even function hence, reflects on the y axis?

since its squared

[carmine hawk]

Again, what does it mean for a function to be even?

(algebraically, not geometrically)

[timid silo]

power of x to be even

[carmine hawk]

That isn't enough, there are also functions without powers of x that are even, and yet the whole function can be even

The function is even when f(x) = f(-x)

[timid silo]

and odd when -f(x)

[carmine hawk]

Yes, now considering that f(x) = e^(-x^2 / 2), is f(x) = f(-x) true here?

[timid silo]

it is

what about question e)

[carmine hawk]

How would you approach part e)?

What do you know about squares first in general

A property that is so obvious for squares

[timid silo]

u get a positive number

[carmine hawk]

Yes, (more accurately non negative)
We write that as x^2 >= 0

Now, multiply that inequality with -1

Remember how signs flip when multiplying with -1?

[timid silo]

yea

[carmine hawk]

-x^2 <= 0

[timid silo]

if u multiply both sides, the sign flips

[carmine hawk]

Hence, what is the maximum of that?

[carmine hawk]

Something is smaller or equal to zero, so the highest value is?

[timid silo]

wait why did u multiply by -1

[carmine hawk]

The question asks about -x^2

[timid silo]

oh right

[timid silo]

x=0?

[carmine hawk]

yeah

[timid silo]

wouldnt it be $x > 0$

[carmine hawk]

It wouldn't, it is asking you about the maximum value of -x^2

[timid silo]

but like x doesnt equal to zero

its approaching it

[carmine hawk]

Ok, let me rephrase what I said

[warm shale]

Bread

[carmine hawk]

You are required to find the maximum value of -x^2

We showed that -x^2 <= 0

This makes sense, a square is always non-negative, put a negative sign in front and it's always non-positive

Since we want the maximum value, we see that the maximum value is reached when -x^2 = 0

If you are having difficulty visualising that, draw the -x^2 graph, the parabola open downwards

you will see that the highest point is reached when x=0

[timid silo]

oh

[carmine hawk]

Also, where did you get that x>0? The domain is specifically all real numbers

[timid silo]

the vertex basically

[carmine hawk]

Yup, that will be the highest point

[timid silo]

i thought it related to the function

then u just sub zero to the function to get the y point for the coordinate

ty for clearing things up

[carmine hawk]

np

[timid silo]

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[timid silo]

The area for this is 104, could someone tell me why my method is wrong? Wouldn't the base have to be the distance between the bottom 2 points?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Oh ye, I realise my mistake there, I understand it now
Thanks

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[tulip vigil]

may i ask about the question's meaning? does this mean n ={1,2,3,4,5} but not {0}? thanks

[tulip vigil]

ah

.close

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[dim basin]

Hello, I don't understand this, is there really a way to find the slope without the function? With only the given points of a curve? This has got me stuck for an hour now. It is item 19-20

[warped shell]

slope : f(b)-f(a)/b-a

[timid silo]

that's for a straight line

this function isn't straight, so the slope is always changing

[warped shell]

yes

[timid silo]

what would a value be then?

[warped shell]

youre right mb

[timid silo]

mm

yeah this is tricky actually

[warped shell]

alrigjht

you can guess that the function is 3x^2

and calculate it from that

maybe there is a easier way but idk

[dim basin]

I tried to guess the function and find the derivative of the function to find the slope but I don't think we are allowed to do that

[timid silo]

i see no other way tbh

[warped shell]

im sry maybe im wrong but i dont think that with discrete values you will find the slope

if its not a line

[dim basin]

Ok. Thank youu, I'm just gonna guess the function and go from there

[warped shell]

yeah its 3x^2

btw

[dim basin]

Thankss

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[solid kernel]

I need a hint as to how i would go on about solving this, please dont solve it tho

[timid silo]

have you learnt lhopital rule

[obtuse pebble]

@solid kernel Has your question been resolved?

[solid kernel]

yep sorry i alrdy solved it thank u

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[hazy ridge]

Solve the first order differential equation, i need help idk how to do this one! This is what i've tried so far:

[rigid pine]

Hint.

Actually it doesn't work. 😢

[hazy ridge]

oh

i dont know what to do

[rigid pine]

There's another way though.

[hazy ridge]

is it involving substitution?

[rigid pine]

Yes.

I may have another way which doesn't need to use substitution, I just need to check if it works here.

[hazy ridge]

I think i'm meant to do a substitution as i haven't learnt the other methods yet for special cases

but which substitution is what i dont know

[rigid pine]

Yes. The substitution should work. It looks like you have it. I hate working in other variables than y and x.

[hazy ridge]

v = y/x ?

I mean

v = H/P ? should work?

With the H^2 it doesn't work

[rigid pine]

Let $x = p$ and $y = H$ then the question becomes:\
$x^2\dv{y}{x}= 25x^2 - 9xy+y^2$.

[warm shale]

stabulo

[hazy ridge]

ok

[rigid pine]

My alternate trick won't work here. Probably have to just use v = y/x to do it. It should definitely work.

[hazy ridge]

But

It isntworking

It is not i tried the h^2

[rigid pine]

$\dd{y} = \left( 25 - 9\frac{y}{x}+\left(\frac{y}{x}\right)^2\right) \dd{x}$.

[warm shale]

stabulo

[rigid pine]

Let $v = \frac{y}{x}$ then we also have that $y = xv$ so we have that $\dd{y} = v\dd{x}+ x\dd{v}$.

[warm shale]

stabulo

[rigid pine]

Using this we can get:

$v\dd{x} + x\dd{v} = (25 - 9v + v^2)\dd{x}$.

[warm shale]

stabulo

[rigid pine]

Simplifying. There results

$x\dd{v} = (25 - 10v + v^2)\dd{x}$.

[warm shale]

stabulo

[rigid pine]

We can divide by x(25 - 10v + v^2) for the values of x and v which it doesn't equal zero to obtain the separable equation

$\frac{\dd{v}}{25 - 10v + v^2} = \frac{\dd{x}}{x}$.

[warm shale]

stabulo

[hazy ridge]

o

[rigid pine]

$\frac{\dd{v}}{(v-5)^2} = \frac{\dd{x}}{x}$.

[warm shale]

stabulo

[rigid pine]

(x cannot be zero and v cannot be 5.)

Hopefully you can continue now. 🙂

[hazy ridge]

Hey man@rigid pine

I uhhhhh

Realised i wrote my first line wrong i forgot a p^2 on the last term

[rigid pine]

Oh. So you found where you went wrong initially?

[hazy ridge]

Yeaa

[rigid pine]

Perfect. 🙂

[hazy ridge]

Ty for the help tho

[rigid pine]

No problem. 🙂

[obtuse pebble]

@hazy ridge Has your question been resolved?

[hazy ridge]

Yes

[obtuse pebble]

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[hard meteor]

@split adder

[obtuse pebble]

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[celest glacier]

How does someone prove or refute this claim?

[celest glacier]

I mean this one is true. But can I just write true?

[vivid valley]

A. Find the class mark for each interval.

B. Is it possible to determine the number of cars that
did they exceed the speed limit? In case
yes, what is the corresponding percentage?

[celest glacier]

Sorry what?

[vivid valley]

i am latin

i speak spanish

but i need help

[celest glacier]

You need to go for a channel which is free

[vivid valley]

i don't find a another mathematics spanish server

aaa i just understand

sorry

[obtuse pebble]

@celest glacier Has your question been resolved?

[obtuse pebble]

@celest glacier Has your question been resolved?

[obtuse pebble]

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[obsidian isle]

what's the original question

[small kestrel]

just to check, I can't see it properly - is the polynomial

$T_{3}(ln(2x)+\frac{1}{6}x^{-1},1)$

[warm shale]

Smethisko

[small kestrel]

oh, sorry, didnt notice you posted it already

When I was learning Taylor polynomials, we had instructions written differently, so I'll just ask - the 1 after the comma at the end, right in front of the closing parenthesis, it is the point around which we approximate?

I hope the question makes sense

yup, okay

I'll take a look at it

but no miracles granted

[obsidian isle]

T_n means the nth degree taylor polynomial?

so T_3 means 3 terms

not much left to do but apply the formula

[small kestrel]

watch out! @regal sierra you made a mistake calculating the derivatives

that's it

you wrote it was

$1/{2x}$

[warm shale]

Smethisko

[small kestrel]

but it's

$1/x$

[warm shale]

Smethisko

[regal sierra]

Oh,oh my gosh, thank you, I will take a look

[obsidian isle]

(ln(2x))' = 1/(2x)

[small kestrel]

nope, you have to multiply it by 2

because it is a composite function

the natural log has 2x for the argument

so you have to differentiate the argument as well

and multiply these two

[small kestrel]

take a look at this

how did you come up with this? @regal sierra

[obtuse pebble]

@regal sierra Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[wind kettle]

tell me what I have to do

[spiral maple]

Choose situations/activities that you do in your house this time of pandemic that you will represent or see the concept of direct and inverse variation through video presentation.

[wind kettle]

what activities

[spiral maple]

idk

Im not you

[wind kettle]

how about sweeping the floor

how will that represent variation

[spiral maple]

No clue, can't really do thought for you.

[wind kettle]

ok

.close

[obtuse pebble]

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[obtuse pebble]

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[heady briar]

Henlo all, I need help lul

[heady briar]

The 1/3 one

This is how it works

You have to make a square thing which I don't know how to translate it, and you put those on graphs to make a straight line

<@&286206848099549185>

[split scaffold]

Could you perhaps be more specific?

[heady briar]

Ok so

Uhh

Idk how to explain it cuz I don't know to translate it properly

There y and x

[split scaffold]

I can give you a formula that might help

[heady briar]

Okay I can see the formula

Can you send it

[split scaffold]

g(x) = f(x - h) + v

g(x) is the new function

h is the units you translate horizontally

v is the units you translate vertically

[heady briar]

Okay that's kinda off track from what I'm talking about

I meant translate as uhh

The words I can't put in english

So there's y and x, let's say y = 2x

[split scaffold]

Alright

[heady briar]

So you draw your square thing like the photo above, you can put any number but it will have to match on the graphs to get a straight line

And say I put -3 on the first one, and you multiply it with 2

So it's - 6 if I'm not dumb

[split scaffold]

Yeah it is lol

[heady briar]

This is kinda long to explain lol

Ok so then

You draw your graphs

You match put a dot between - 3 and - 6

Here I'll just show you a Pic instead

[split scaffold]

Right so you wrote an equation and drew a line with it?

[split scaffold]

On a graph

[heady briar]

So for example here I wrote - 3 to 3 on the square thingy, which I multiplied each of them with 2 to find y

So you just put a dot between x and y

And a line appears

This is kinda complicated to explain

[split scaffold]

Okay

Also when you say “square thingy” I think you mean coordinate plane

Or graph

Just try and be familiar with that term

[heady briar]

Yeah so I'm basically stuck on the 1/3 part, I'm assuming you divide x with 3 to find y

[split scaffold]

Yes exactly

Remember $\frac{1}{3}x= \frac{x}{3}$

[warm shale]

Crimson Blitz

[heady briar]

Mmm

[split scaffold]

Yes

So here it would be quite hard to make a line if you picked any number

[heady briar]

I think I just found the solution myself instead

[split scaffold]

Haha

[heady briar]

If I have to divide it with 3

I just have to find the specific number to match

[split scaffold]

Yes

Basically any number that’s divisible three (so you end up with a whole number) would be easier to use

Like -6,-3,0,3,6

And son

[heady briar]

Mmm

Okay thanks

[split scaffold]

One more thing

[heady briar]

That's what I was supposed to do all along

[split scaffold]

I’m pretty sure it says $-\frac{1}{3}$

[warm shale]

Crimson Blitz

[split scaffold]

Just remember that it’s negative

Other than that I think your good

[heady briar]

Yeh

Thank you so much

[split scaffold]

No problem, even though I wasn’t much help 😅

[heady briar]

Have a nice day

.close

[obtuse pebble]

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[latent panther]

Please I need help with this

[flat anvil]

that a test?

[latent panther]

no

[flat anvil]

homework?

[latent panther]

Its not even for credit

But there will be a test for it

[flat anvil]

okay well

[latent panther]

and the video lecture is absolutely useless

[flat anvil]

$\tan(\theta) = \frac{1}{\sqrt{3}}$

[latent panther]

I know 30 is an answer to it

[warm shale]

not discordmod

[flat anvil]

right so what satisfies this equation

i can think of 2 things

$\frac{\frac12}{\frac{\sqrt{3}}{2}}$

[warm shale]

not discordmod

[flat anvil]

so this is one of the casese

$\frac{\frac{-1}{2}}{\frac{-\sqrt{3}}{2}}$

[warm shale]

not discordmod

[flat anvil]

and this is the other one

so what angles satisfy those values?

[latent panther]

30 is good

apparently 210 is I have no idea why

i assume -210 does not satisfy it?

[flat anvil]

no it wouldn't

so 30 works because

sin(30) = 1/2

and cos(30) = sqrt(3)/2

[flat anvil]

that is what we had here right?

210 works because

sin(210) = -1/2

and cos(210) = -sqrt(3)/2

[flat anvil]

that was this caes

[latent panther]

Alright

Do you know this one?

[lethal bone]

Secant is never 1/2

[latent panther]

So 0?

And how about this monstronsity?

[timid silo]

first write sin^2 (x) in terms of cosx

[latent panther]

uhmmmm

how do I do that?

[lethal bone]

sin^2+cos^2=1

sin^2=1-cos^2

[latent panther]

Alright

[timid silo]

yeah

its an identity you should know

[latent panther]

So now what?

[lethal bone]

2(1-cos^2) - cos - 1 = 0

Solve

[latent panther]

uhmmm

2-2cos^2-cos=1

-2cos^2-cos=-1

then what?

@lethal bone

[lethal bone]

No

You can’t move it

[timid silo]

you have it in the form of a quadratic

[lethal bone]

Factor it

Think of cos x as x

-2cos^2 - cos + 1 = 0

[latent panther]

-2x^2-x=-1?

[lethal bone]

Don’t move the 1

[latent panther]

So what are you supposed to do?

[timid silo]

just do it like how you'd solve a quadratic

[lethal bone]

Factor

[latent panther]

How do you factor it?

[timid silo]

first make it = 0

[latent panther]

It already does = 0???

[timid silo]

according to this it does not

[latent panther]

-2cos^2 - cos + 1 = 0

according to this it does

[timid silo]

well then factor it

just like how you would factor any other quadratic

[latent panther]

how

this makes no sense

[timid silo]

what

[latent panther]

how do you factor it?????????

[timid silo]

do you know how to factor a quadratic

i cannot do everything for you

how would you factor x^2 + 4x + 4

[latent panther]

this makes no sense doing it with cosine and all these other random numbers

[timid silo]

no

let cosx be x

[timid silo]

then you can say -2x^2 - x + 1 = 0

then factor it

then replace back to cosx in the end

[latent panther]

So the answer is 2 then?

[timid silo]

answer to what

the question or the equation?

[latent panther]

the question

its 2

[timid silo]

yes it would be 2

[obtuse pebble]

@latent panther Has your question been resolved?

[obtuse pebble]

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[ivory geode]

@void arrow Do not interrupt other people's questions please. Stay in your own channel.

[nocturne minnow]

@timid silo If you're still here, could you close this channel? I believe this person has been banned and helpers can close channels

[timid silo]

let me try

.close

[obtuse pebble]

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[timid silo]

ah nice

[obtuse pebble]

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[rotund cloak]

I'm stuck on this question from Spivak's book. I've tried to use the limit proof for lim (f+g)(x) for the problem, but I still don't understand how I'm supposed to find the min for delta