help-27

[queen dirge]

It's equal to 0

[cinder bobcat]

in the limit yes

[queen dirge]

Soo if lim g(x) - 8 = 0

Can that mean lim x->5 g(x) = 8?

[cinder bobcat]

yes it does

[queen dirge]

That's it?

[cinder bobcat]

yup lol

[queen dirge]

My god

[cinder bobcat]

I feel that

[queen dirge]

Okay, thanks so much

[cinder bobcat]

np lol

[queen dirge]

I really didn't know you could do that < wud be <= in that context

[queen dirge]

Alright, have a good one

[cinder bobcat]

now ya know, you too 🫡

[queen dirge]

.close

Oh nvm

[plucky kernel]

uhhh

help isnt this my channel what happened

anyways

why did my messages get deleted?

...

:/

.close

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[weak coral]

Help

[weak coral]

Two classmates poorly copy the text of an exercise written on the blackboard by the teacher. The exercise consists in determining the point of intersection of the line of equation y = 2x with another line, of which the teacher provides the equation. The teacher also provides as a solution 3/4 3/2 The second straight line has an equation of the type y = mx + g, m and q being two integers. Luke makes a mistake in transcribing m and Mark makes a mistake in transcribing q. Luca gets the solution (-3, -6) while Marco gets the solution (1, 2). Both correctly solved the exercise that they had mistakenly transcribed. Would you be able to determine the equation of the second line assigned by the teacher and the equations erroneously transcribed by Luke and Mark, knowing that the original value of q given by the teacher was a positive integer equal to and not exceeding 10?

[stiff ledge]

Someone please help

[weak coral]

Can you solve this only using linear equation

[stiff ledge]

Explaiin to me

[weak coral]

Who did that

[stiff ledge]

What?

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[dire frigate]

I need help solving F

[dire frigate]

so plz help me out anyone

[restive river]

you retrieved the equation y = 7x +13 yes?

[dire frigate]

yes

[dire frigate]

yes

[restive river]

so

what are y and x representing

[dire frigate]

they mean i forgot....

[restive river]

what

[dire frigate]

x and y are representing the points on the graph.

[restive river]

no they have an associated meaning

[dire frigate]

um...

[restive river]

that's why u set up the equation to begin with

read back the question n ur answet

[dire frigate]

oh!!!! x represents the lines and y represents the cost!!!

[restive river]

yeah

so you know the y

and you want to find the x

[dire frigate]

my y is 97.00 so it's 97.00=7x+13 right?

[restive river]

yeah!

so solve for x

[dire frigate]

yay can youhelp me out on something else.

[restive river]

sure

[dire frigate]

I help on letter H

-17=17/3x-170 right?

[restive river]

yes

[dire frigate]

ok

thank you!!!!

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[haughty yacht]

can someone provide examples of these 2

[stone stump]

$A=\begin{pmatrix} 1 & 0 \ 2 & 0 \end{pmatrix}, B=\begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}$

[woven radish]

Denascite

[stone stump]

the row space of A and B are both spanned by (1,0)

the column space of A is spanned by (1,2)^T, while the column space of B is spanned by (1,0)^T

[haughty yacht]

ohh thank you

[devout snow]

@haughty yacht Has your question been resolved?

[haughty yacht]

I keep getting the wrong answer for this

[devout snow]

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[haughty yacht]

.reopen

[devout snow]

✅

[haughty yacht]

can someone explain this in simpler terms

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[haughty yacht]

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[soft peak]

Help

[soft peak]

Shouldn't it just be mirrored around the y axis?

Then the answers would be (0,9) (8,0) (4,0)?

Or is it around the x axis

x axis seems about right

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[crimson niche]

If I was asked to evaluate the integral using area. When do I say A1+A2 and A1-A2

[crimson niche]

Not for these specific questions but overall.

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[quiet hull]

find Dn

[quiet hull]

no matter what I do I cant get a triangular matrix 😦

I tried develpoing by R1,Rn,C1 and Cn but I always get stuck

[restive river]

r u finding the determinant

[quiet hull]

ye Im trying to

<@&286206848099549185>

[devout snow]

@quiet hull Has your question been resolved?

[uneven notch]

what about multiplying 1/x to every row except first one?

nvm I was wrong

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[sturdy yew]

How do i show without just substituting values that
f(x+y) = f(x)f(y) reepresents the exponential function?

[cinder bobcat]

well first you need the assumption f(0) = 1, I assume that's implied here

[sturdy yew]

oh yea

they said f(0) = 3 but it doesnt make a diff

[cinder bobcat]

oh true nvm

that should make a difference tho, wait

you don't need any assumptions here, f(0) = 1 is clear

just replace x and y with 0 and solve for f(0)

generally to do these kinds of problems, you need to show 2 things: it is exponential for integer arguments and it is exponential for rational arguments. then if you assume f is continuous, you can show it is true for all real numbers

[sturdy yew]

i just want to show it for integer arguementss

[sturdy yew]

wait iif u do that u get f(0) = 1 but in the question it mentions f(0) = 3

[cinder bobcat]

well that can't be true, 3a^(x+y) =/= 9a^(x+y)

can I see the exact question?

[sturdy yew]

oh nvm

iit says f(1) = 3

not f(0)

mb

[cinder bobcat]

oh that's good, well just so yk, if you wanna only show it for integers, induction is the way to go

assume f(x) = 3^n

[sturdy yew]

i havent learnt induction

is there no other way

[cinder bobcat]

not really that I know of, usually if you wanna prove something about integers or natural numbers, induction is THE way to go

[sturdy yew]

ok

thanks you

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[onyx loom]

What did I do wrong? For part b the answer is -0.326 and I got -0.5

[meager yarrow]

f’(x)=2x+4/x^2+6

[onyx loom]

nvm it was bc f'(x) is +4/x^2 not -

[meager yarrow]

Right

[tawdry wing]

you did $-0.4^2$

[woven radish]

proofAd

[tawdry wing]

should be $(-0.4)^2$

[woven radish]

proofAd

[onyx loom]

i had it as ANS it was bc i diffiencated wrong

[tawdry wing]

oh alr

let me check it rq

[tawdry wing]

yeah it was differentiated wrong

did you correct it while calculating the values?

because the term $\frac{4}{x^2}$ shouldn't be negative

[woven radish]

proofAd

[onyx loom]

i got the correct answer when i did +4/x^2 instead of -4/x^2

[tawdry wing]

ah alr

If you dont need help anymore, do .close

[onyx loom]

.clsoe

.clodsds

.closke

.close

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[celest elk]

can someone find the nth term

its cubic so ax^3 + bx^2 + cx +

d

but i dont know how to actually find it

[lost laurel]

what makes you think it's a cubic series?

[celest elk]

i checked an the difference was 12

[twilit comet]

so...?

and how is the difference 12

[celest elk]

can you just do the question

[twilit comet]

the difference changes

[twilit comet]

!nosols

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[candid lance]

Yo I think I found a pattern

[twilit comet]

?

what is it

[candid lance]

u want me to say it

ok here

[twilit comet]

i mean i kinda see one

[candid lance]

the differnces between the numbers

[twilit comet]

powers of 2?

[candid lance]

no no

[twilit comet]

accha ok

[candid lance]

differences bw the numbers

[candid lance]

have differences

in the order of 12

[celest elk]

12

[candid lance]

yeh yeh yeh

[twilit comet]

oh i see

[celest elk]

how do you use cubic

the formulas

[twilit comet]

that was fast

[candid lance]

I usually do that stuff

whenver I see a non progression sequence

[winter patrol]

solve a system of equations

[celest elk]

i dont know those equations

[candid lance]

wel well well

I think

I found it

wait nvm

uhh

u see he mentioned cubic

[winter patrol]

you know that
t_x =

ax^3 + bx^2 + cx + d

and the numerical values of the first 5 terms

[candid lance]

yo yo yo

I found the sequnce buddy

[celest elk]

what is it

[candid lance]

should I answer him?

or nah

[winter patrol]

!noansw

[devout snow]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[candid lance]

cuz theres a pattern

I found

[candid lance]

well then how do u want me to explain it

[winter patrol]

describe the steps involved (and also don't do the work required for them either)

[celest elk]

dude

[candid lance]

Uhm I did all the stuff in mind reading the numbers and trying to find the general ter

I didnt evn do a single step

[winter patrol]

.use that to set up 5 equations (which is more than enough to determine the values of a,b,c,d)

[celest elk]

how do i find equations for this

i know the equations for the normal

like ax^2

[winter patrol]

same idea

[celest elk]

okay

so

i would get

[candid lance]

thats tuff

[celest elk]

3a = a3 ??

instead of 2

[winter patrol]

where's that coming from

[celest elk]

since ax^2 is 2a for the formula

it would make sense that ax^3

is 3a

[winter patrol]

you have deduce that your terms are part of a cubic sequence

t_x = ax^3 + bx^2 + cx + d

to get your first equation, replace all x in that with 1

and you also know the first term of the sequence t_1 is 3

show me what you get after doing that

[celest elk]

a + b + c + d = 3

[winter patrol]

then repeat for x=2,3,4,5

[celest elk]

ohhhh

wait that makes sense

8a + 4b + 2c = 14?

[winter patrol]

how are you getting that

[celest elk]

a x 2^3

2^3 = 8

so 8a

then same thing for the rest

[winter patrol]

where's 14 coming from

where's d

[celest elk]

from t_2

[winter patrol]

t_2 isn't 14

[celest elk]

17?

[winter patrol]

the second term given to you is 17
also

where's d

[celest elk]

so it'd be 8a + 4b + 2c + d = 17

i dont see how this helps me though

[winter patrol]

well you'll get more equations from x=3,4,5

[celest elk]

yeah, but i need to find the nth value

[winter patrol]

have you solved systems of / simulatenous equations before?

[celest elk]

yes

[winter patrol]

this is just that

[celest elk]

simultenous yeah

oh

to find x?

[winter patrol]

no

the goal here is
to find a,b,c,d

[celest elk]

okay okay

i think i understand

so

7a + 3b + c = 14

wait

no

no yeah

[devout snow]

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[devout snow]

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[halcyon kernel]

How do I get going with this problem?

[late rose]

do you know the integral of sec^2x

[supple knot]

where even is b

[arctic field]

a) b) ...

[candid wing]

a would be 2

because we can put 3x-5=t
from where we get dx=dt/3

after substbuting it in the equation we get the intergral of 2sex^2 t

[halcyon kernel]

is it not tan(x)+C

[candid wing]

who said that its answer is tan (x) +c

[halcyon kernel]

I meant integral of sec^2

U sub ? 3x-5

[candid wing]

in we integrate sec^2 (t)
where t=3x-5
then we get tan(t) in which if we substitute t then we got

tan(3x-5)+c

[halcyon kernel]

I'm just getting started on integrals

[candid wing]

you will be better in it

a=2

[halcyon kernel]

ok thanks

.close

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[bitter halo]

,,\lim_{x \to 0^+} e^\frac{1}{x} , \quad \lim_{x \to 0^-} e^\frac{1}{x}

[woven radish]

レナト (renato , ping if reply)

[acoustic leaf]

what about them?

[bitter halo]

why

the left side of this limit is different from the right side?

,w lim x to 0 e^(1/x)

[woven radish]

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[acoustic leaf]

because the left and right limits of 1/x are different

and e^x has very different limits when approaching +infinity vs -infinity

[restive river]

1/x goes to minus infinity when approaching 0 from -ve side
and positive when approaching from minus side

a^(something minus) means 1/a^(something)
so , when approaching from positive , limit is infinity
and negative , limit is 0

[devout snow]

@bitter halo Has your question been resolved?

[bitter halo]

what is ve side?

[devout snow]

@bitter halo Has your question been resolved?

[bitter halo]

,w plot 1/x

[woven radish]

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[faint zinc]

@bitter halo "-ve" I believe was intended to be short for negative,

But anyway, not rigorously speaking, e^-inf = 0, whereas e^inf = inf, this is why the limit is different in the two cases.

[devout snow]

@bitter halo Has your question been resolved?

[restive river]

to calculate the limit from the left of e^1/x you can work backwards. 1/x becomes large when x is small (you can see this in the graph) but when we approach from the left we look only at negative values of x so 1/x will also be negative

1/x will therefore go to -infinity when x goes to 0 from the left

so e^1/x will go to e^(what 1/x goes to from the left) = e^(-infinity) = 0 when x goes to 0 from the left

[bitter halo]

however we are talking about e^(1/x) not e^x , I believe

[faint zinc]

yes.

but if you go to 1/0 from the negative direction you get negative infinity

and you get positive infinity when you to go 1/0 from the positive direction

so knowing the behavior of $e^{\pm \infty}$ is useful

[woven radish]

全能の存在

[devout snow]

Channel closed

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.reopen

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[calm belfry]

help pls how do i solve the 2 blanks

[hardy zenith]

so ur trying to find the velocity when its at height zero

and if you know what the height is

you can plug the height in as 0

and s(t) is the height

[hardy zenith]

wait is this undergrade math?

undergrad

isnt undergrad after college

[calm belfry]

uh im in college

[hardy zenith]

oh

damn

just asking cause i did that last year

[hardy zenith]

so u can plug in s(t) as 0

so u get the equation

0=64+96t-16t^2

and you can solve that equation

by factoring

do u get it

[devout snow]

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[jaunty gazelle]

does anyone understand why i got this wrong?

[devout snow]

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[full sand]

Let B lie in the sigma-algebra generated by some set A. Show that B lies in the sigma-algebra generated by an at most countable subset of A. Billingsley, Probability and Measure, anniv. ed., Ex. 2.8

[full sand]

I'm not used to working with smallest structures generated by a set where there's no easy internal form in terms of the generating set like with groups.

definition of smallest sigma algebra being the intersection of sigma algebras (closure under union, complement, and containing omega) that contains the generating set

I'm thinking that I should consider the smallest sigma-algebra generated by subsets of A that contain B, then show that it's countable by contradiction.

So I suppose that I may consider an uncountable generating set, and then show that this always results in a sigma algebra that contains excess stuff; that is, some subset of this uncountable set generates a strictly smaller sigma algebra. Thus it cannot be the smallest sigma algebra generated from a subset of A containing B.

Actually I should see if there is a smallest sigma algebra generated by subsets of A containing B.

[untold saddle]

.reopen

.reopen

.reopen

.reopen

.reopen

[full sand]

wrong channel bro #help-12 @untold saddle

[full sand]

Ah, indeed, any nonempty intersection of sigma fields is itself a sigma field

I can take the intersection of sigma fields generated by subsets of A that contain B.

But does this mean that some subset of A generates thes intersection sigma field?

I suppose we can: take the intersection of A and this intersection sigma field as our new generator.

Call that intersection sigma algebra N so that our generator was A intersect N

Then sigma(A intersect N) is a subset of N

To have equality we need to show that B is an element of sigma(A union I)

[restive river]

yes

[full sand]

Is it possible to construct this subset or do we do it by contradiction?

[full sand]

Consider a "finest" chain of subsets of A, with the empty set at one endpoint and A itself at the other end. We can partition the sets in this chain, with one group being sets generate s-algebras not containing B, the other being sets that generate s-algebras containing B. Consider the intersection of the sets that generate s-algebras containing B; it is still not clear to me that this is again in the same partition.

<@&286206848099549185>

I give up I'm going to peek at the hints

I'm still stuck
the hint is
where calA_(B_n) denotes a subset of A that generates B_n (that we hope we can find a set having this property that is countable)

I'm still stuck

[devout snow]

@full sand Has your question been resolved?

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@full sand Has your question been resolved?

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[full sand]

.reopen

[devout snow]

✅

[devout snow]

@full sand Has your question been resolved?

[full sand]

I peeked at someone's answer on the internet (they were wrong for another problem) but their appriach fir thes problem seems fruitful

Define by G the subset of sigma(A) that lie in a sigma algebra generated by a countable subclass of A. Show that G is a sigma algebra, so we have equality.

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[onyx scroll]

Hi, if we have a ceiling function, for all x in ]1;e] ln(x) = 1. If we have a flooring function then for evry x in [1;e[ ln(x)=0. Correct ?

[onyx rune]

that doesnt make sense

do you mean ln(ceiling(x)) = 1 for all x in [1;e]?

if yes, thats not even true for any x

or do you mean ceiling(ln(x)) = 1

[onyx rune]

if you mean that then, ceiling(ln(x)) = 0 (for x = 1) and celinig(ln(x)) = 1 (for x in (1;e])

so floor(ln(x)) = 0 for x in [1;e) and floor(ln(x)) = 1 for x = e

[onyx rune]

so thats mostly correct ig

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[hardy cloud]

how do yall do fractions

[hardy cloud]

er

real

[hexed holly]

what do you mean specifically?

[hardy cloud]

uhh

lemme show

[hexed holly]

mk

[hardy cloud]

like these

[tawdry wing]

you mean in latex?

$\frac{a}{b}$

[woven radish]

proofAd

[hardy cloud]

.

i mean uh

44/104

4

[tawdry wing]

oh you mean simplifying?

[hardy cloud]

yes

no letters

jus numbers

[tawdry wing]

you can divide the top and the bottom by the HCF of 44 and 104

or prime factorize both and cancel out the numbers that occur at the top and bottom

for 44/104 specifically, it would be:

$$\frac{44}{104}$$
$$\frac{11 \cdot 2 \cdot 2}{13 \cdot 2 \cdot 2 \cdot 2}$$

[woven radish]

proofAd

[tawdry wing]

so cancel out 2 twos at the top with 2 twos at the bottom

and you get $\frac{11}{13 \cdot 2} = \frac{11}{26}$

[hardy cloud]

aa

[woven radish]

proofAd

[tawdry wing]

so thats the simplified answer

[hardy cloud]

hm

thanksi

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[restive river]

what kind of function can be used to make a line that connects the red dots?

[solar goblet]

vaguely similar to lambert W function

[lost laurel]

You could even make a quadratic , or linear function IMO

[restive river]

my friend told me i could use a quadratic

[solar goblet]

maybe a really wide parabola might work?

[restive river]

and i could also use cos or sin

[restive river]

how does a lambert w function look like

how do i write it

[solar goblet]

you have to rotate it and turn into parametric equation though

[solar goblet]

pretty sure its not writable in desmos anyway

[lost laurel]

as the only condition is that the curve has to pass via the two red dots, there are infinitely many such functions

a linear function is probably the easiest to construct here

[restive river]

i don't want to use ellipse or seperate functions to do it

i know i could make it work

with a linear function

and an ellipse

[lost laurel]

what's the original question?

[restive river]

but ive done that a lot, and want to use something else

[restive river]

i just want a function that would fit there

[solar goblet]

would logarithm do the trick?

[restive river]

i dont know

[solar goblet]

you have to rotate it though, like i said

because that curve is not a function as per vertical line test

[restive river]

rotate what

a parabola

[solar goblet]

https://math.stackexchange.com/questions/17246/is-there-a-way-to-rotate-the-graph-of-a-function

this might help

[signal violet]

W is multibranched, a concept you'll learn more about if you get to complex analysis

that graph only shows the real part of W, in practice W is usually expressed as a multibranched complex-valued function

[solar goblet]

interesting, though i saw some people use W to solve real log function?

there seems more to this than meets the eye

[signal violet]

when i first saw this questoin my first intuition was "smooth" but i honestly don't know what is being asked here, still don't

[restive river]

i will check it out now, thanks!

[solar goblet]

he is asking which equation fits the curve connected by the 2 red dots

[solar goblet]

as shown here

desmos art i believe?

[onyx scroll]

Why are you talking about x=1 if it's not in the interval ?

[signal violet]

with the graph lines provided, there's no "function" that does so if we assume a standard x-y axis

[solar goblet]

yea, it isnt a function, a parametric equation is possible though

[restive river]

yes, unfortunately

[signal violet]

we cano bvoiusly parameterize a curve (x,y) = f_x(t), f_y(t)

there are many such curves, though

[solar goblet]

he can either make a wide parabola or some exponential curve

[restive river]

some people also said exponential could be possible with some manipulation

[solar goblet]

then finetune it

[restive river]

but i couldnt put that into work

[signal violet]

yes, but we shoudl take that discussion to a different forum if you wish to continue it

[solar goblet]

no thanks, for now i cant quite grasp complex math

[restive river]

I think im going to use a normal parabola

x^2

but how do i know rotate the function

i think expanding this a bit and rotating it would resolve the issue effectively

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[unique sequoia]

given a line AD with I as the midpoint
draw another line segment Ix perpendicular to AD at I
draw a circle (O) which touches AD at A, also intersects with Ix at B, C (B is between I and C)

question: line segment DB intersects (O) at D': prove that D' is symmetrical with D with respect to AC

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<@&286206848099549185>

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[sinful imp]

sorunun orjinal hali bu mu. çeviri kullandığımda ne yazdığınız anlaşılmıyor

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[lyric belfry]

equation of C

for b(ii), i was trying to create the equation that passes through the centre of C

and use the equation of C and the striaght line to solve

so i get 2 points which touches 2 tangent line

so ill know the equation of L1 and L2

but why cant it be the way to solve

the solution says i have to create y=3/4x + C, where C is a constant

and use that equation to solve

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@lyric belfry Has your question been resolved?

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@lyric belfry Has your question been resolved?

[lyric belfry]

whatsoever

.close

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[restive river]

So I'm trying to do this problem and all you have to do is derive the function correct? Because whenever I do it and sketch v(t)=-100t+300 it doesn't work

[near trout]

that should be right?

weird

[restive river]

It's a straight vertical line too correct?

[near trout]

no

it should be sorta diagonal

it is a straight line though

[restive river]

oki thats prob where im messing up

I'm using symbolab and is doing the vertical line thing

oki I figured that one out, I was being a little dense

so this uses (x2-x1) / (y2 -y1) and 20 is x2 correct? and what is x1 10?

[near trout]

you could use x_1 = 10, x_2 = 20 or x_1 = 10, x_2 = 30 or x_1 = 15, x_2 = 25 or anything

[restive river]

oki just realized t =20 meaning it in the denominator. making (3.5 - x1) / (20 - y1)

[near trout]

i think the middle one is best

not sure though

idk how you're meant to be precise here

[restive river]

yeah I love how they make these problems

would you say x_2 = 30 is v=3.75 on the graph

[near trout]

maybe a little higher idk

[restive river]

so it would be (3.75 - 2.5)/(30 - 10)

[near trout]

it goes like 0, 2.5, 3.5, ??,

so difference from 4 is like

4, 1.5, 0.5, ??

which isn't a nice pattern

so you have to just guess yeah idk

[restive river]

oki oki

[near trout]

i'd go for like 3.8 or 3.85 or something idk

[restive river]

oki, ill try it

It's not working, every answer I'm getting with the numbers I'm using is coming out to 0.07 and it saying that's wrong

[near trout]

scammed

try 0.06

[restive river]

just did (3.85-2.5) / (30 - 10) = 0.065

oki

[near trout]

they might want 3.75

it looks higher than 3.75 to me but yeah

[restive river]

that worked

lolol

[restive river]

facts

Estimating numbers to use for an equation from a graph is the worst

Oki, I'm on this now. Idk if I'm just flat out missing something but I have no clue how to start this

[near trout]

do you know the formula for the volume of a cube given the side length

[restive river]

not off the top of the head atm

[near trout]

like

if you have a cube that's 2x2x2

what's the volume

[restive river]

Oh lol

L X W X H

[near trout]

sure

but it's a cube so all of the sides are equal

[restive river]

oki 10 X 10 X 10

[near trout]

sure

so if the side length is s

what is the volume

[restive river]

1000?

[near trout]

no

i mean

the volume when s = 10 is 1000

but if you don't know s

what is the volume just in terms of s

in algebra

[restive river]

hmm I'm drawing a blank

[near trout]

so it's just going to be

s * s * s

so V = s^3

[restive river]

Oh, I missed cubing it

[near trout]

anyway you want the rate of change of V when s is 10

so first you want to find the rate of change of V wrt. s

and then you want to let s = 10

[restive river]

Sorry If I'm missing something but is that not just 1000?

[near trout]

no

find the rate of change of V wrt. s just generally first

if V = s^3

[restive river]

Does it change things if we're looking for Instantaneous Rate of Change?

[restive river]

V(s) = s * s * s?

[near trout]

yes

so what's the rate of change of V wrt. s

or the instantaneous rate of change

whatever you want to call it

[restive river]

V(10) = 10 * 10 * 10
V(10) = 1000

I feel like I'm so lost

omg, V'(s) = 3s^2

We've never done that so I didn't realize

[near trout]

yeah there we go

then you want V'(10)

[restive river]

Yea, got 300

I'm working on a problem similar to that one atm root to the power of 7 sqr root x to the respect x when x = 3

have 1/7 * (1/(3)^6/7) so far

here's the problem

as I said I got to 1/7 * (1/(3)^6/7) but idk If I'm calculating things right

[devout snow]

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[restive river]

.close

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[vernal grail]

but a neighborhood of x0 is just an interval containing a set of the form x0 - δ , x0 + δ. But for example [1,2) is not around 1 because if I take x0-δ I go outside the interval? For every δ > 0

[near trout]

yes

[vernal grail]

thx

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[oblique spire]

Looking to get some help with this question

[faint hearth]

!status

[devout snow]

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

[oblique spire]

This is my current working out and would just like it going over

[faint hearth]

I think the 4th line is incorrect

Howd you get k=340?

[near trout]

it's 540

just bad handwriting

[faint hearth]

Ah, sorry

Looks good then

[oblique spire]

sorry my writing is awfuly

awful*

[near trout]

that's radians

[faint hearth]

Yes i know

,w arctan(3/4) radians to degrees

[near trout]

yeah looks good

first page is a bit messy but i've been there

[faint hearth]

We hope you solved for x correctly

After finding k

I didn't check that

[oblique spire]

overall I think I have a good understadning the bit I was unsure about was when to do the converstion from cot to tan

[faint hearth]

You can do it anytime

Before the k substitution or after

It wouldn't make a difference

[oblique spire]

okay sounds good

[oblique spire]

I calculated the conversion to be (X)+90/3 and I used a calculator to perfomr the sum will double check again they are correct

thanks all.

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[cunning roost]

can someone explain this?

[misty crest]

which part

[misty crest]

ok so i’m assuming ur confused by the 4(x+h)

[cunning roost]

Yeahh im confused how he plugged it in i

[misty crest]

we know that f(x) =1-4x correct

[cunning roost]

Yes

[misty crest]

now our input is x+h as opposed to simply x

so the function essentially tells u to subtract 4 times the input from 1

or 1-4(input)

[cunning roost]

f(x+h)

[misty crest]

now our input is x+h instead of x

so we do 1-4(x+h)

if it was f($+6) it would be 1-4($+6)

[woven radish]

knief

[misty crest]

then just distribute

[misty crest]

oops

sorry i meant just any random thing

where i wrote $

[cunning roost]

I kind of dont understand why we subtract 4 times ?

[misty crest]

imagine i wrote &

[misty crest]

wdym when they distributed

1-4x-4h

this

or 1-4(x+h)

[cunning roost]

so thats their way of plugging in sqrt(1-4x) into f(x+h)

[misty crest]

or yea sorry the function is sqrt

but yea the x is just the input

wherever u see an x u replace it with whatever the input is

so in this case the input is x+h

so we replace the x with x+h

[cunning roost]

wait ima paste the image again and try to comprehend it

[misty crest]

if it was f(😃) it would be sqrt(1-4(😃))

[cunning roost]

ohh im starting to get it

OHH

OMG

[misty crest]

yes good

[cunning roost]

they plugged it into THIS x

[misty crest]

it makes sense now

[misty crest]

yes this is f(x)

[cunning roost]

ohhhhhhhhhhhhh

that helps so much

[misty crest]

it’s a function of x

whereas the other is a function of x+h

[misty crest]

yes exactly

because that’s the function

[cunning roost]

so when it says -f(x)

[misty crest]

just subtract f(x)

it’s essentially the slope formula

[cunning roost]

ohh

[misty crest]

if ur aware of that

[cunning roost]

y2-y1/x2-x1

[misty crest]

(y2-y1)/x2-x1

yea

notice

h is the change in x

or x2-x1

[cunning roost]

oh okay

[misty crest]

we’re taking the limit as h approaches zero because we want a tiny nudge in the x value

[cunning roost]

like the distance

[misty crest]

we want to go a little to the right of the x value for f(x)

[cunning roost]

oh ok

[misty crest]

also notice

f(x+h) is y2

because since h is a tiny change in x

the input will be a little more than x

[cunning roost]

OMG

[misty crest]

so it’s essentially the function evaluated a little to the right of it

[cunning roost]

ur right

[misty crest]

or y2

i’d recommend watching an animation

3blue1brown probably has one

or just look up derivative visualized

because all the derivative is is the slope at a single point essentially

that’s y we take the limit as h goes to zero

we want it practically to be the same point

[cunning roost]

okay im gonna do that

I watched a nancy pi vidfeo on it and she gave a good visualizatin

[misty crest]

she doesn’t have the animations tho

the animations r better in my opinion

[cunning roost]

oh ur right

ill look for one

[misty crest]

3blue1brown should have it in his essence of calculus video

series

[cunning roost]

oh but she showed it moving down

[misty crest]

he does the same for integration

[cunning roost]

is that what it is?

[misty crest]

yea that’s the secant line

but the two points become infinitely close

which gives the derivative

[cunning roost]

the limit is approaching 0 because ........

wait why does it want to approach 0

[misty crest]

yea watch the essence of calculus series or atleast the first or second video

[misty crest]

because we want the change in x to be essentially zero so it’s practically the same point

[cunning roost]

ohhhhh

[misty crest]

to get a good approximation for the exact slope of the curve at a single point

[cunning roost]

I seee

[misty crest]

i’ll see if i can find an animation

[cunning roost]

https://youtube.com/shorts/8Q5HsVrcu6g?si=S3LeoX-u-W0A8yTK

is this one good?

I dont understand how that line is equal to the slope 🤔

[misty crest]

here

sorry for the audio

again think of the f(x+h)

the h is a horizontal shift

as it gets smaller and the smaller the shift gets smaller and smaller

like let’s say h=0.001

and f(x)=x^2

[cunning roost]

Ohhh that makes more sense

[misty crest]

if we say the slope at 1

we can approximate it by taking a point really close to 1 and doing the slope formula

so with ya shift of 0.001

we’d have f(1+0.001) or f(1.001)

-f(1)

/(1.001-1)

[cunning roost]

I dont see how y changed by 0.83 if its going from 2 to 3?

[misty crest]

well that was just a random function

it doesn’t really matter

[cunning roost]

oh ok

[misty crest]

it’s some random function

you’ll learn much easier methods for finding the derivative

but what ur doing now is the long tedious way to make sure u understand the concepts

[cunning roost]

oh ok i see

[cunning roost]

which part is shifting

the tangent line?

[misty crest]

the x value

[cunning roost]

oh

[misty crest]

when u learn transformation of functions you’ll learn it

r u familiar with that

[cunning roost]

yea i am

[misty crest]

like f(x-a)

[cunning roost]

yea

[misty crest]

shifts the graph a units to the right

[cunning roost]

(x-5) shifts right 5

[misty crest]

yea thinks of h as the same thing

and h units is near zero

because we’re taking the limit as h goes to zero

so the function near the point we’re interested in finding the slope at

if f(x+h)

[cunning roost]

ooh ok

so the secant line becomes the tangent line

[misty crest]

yes

[cunning roost]

in order to find the slope accurately

[misty crest]

mhm

[cunning roost]

ok tysm

[misty crest]

ur welcome

[cunning roost]

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[vivid estuary]

is the website wrong

[vivid estuary]

tried it without +C also

[flint prawn]

Unless im braindead rn wouldnt you multiply the 17?

[misty crest]

cosh

[misty crest]

no u divide

it’s integration

u put cos

[flint prawn]

I missed that part

[vivid estuary]

oh mb I had cosh before

changed it because I thought they made an error and had cos instead of cosh

[flint prawn]

Then its not negative

[vivid estuary]

hyperbolic functions for cos and sin don’t have a negative when integrating

so the negative would stay

[misty crest]

do they want it in terms of e^x?

[vivid estuary]

hmm

I could try that

[flint prawn]

Ima just not say anything anymore cause idk what hyperbolic functions are qre and I can’t function rn

[misty crest]

x^2-y^2

,w integrate sinh(17x)

[woven radish]

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Requested by c.knief

[misty crest]

yea and it should be negative

so ur right

[vivid estuary]

yeah I kind of want to click on view answer

to see what they have

but it won’t give me my points

[misty crest]

imagine they have 1/17

[flint prawn]

What’s symbolic notation?

[misty crest]

multiplied by it

oh maybe

they want the division symbol

but it says fractions

it’s says a variable is wrong

is there a cosh command

instead of writing c o s h

or maybe it’s the +C

[vivid estuary]

yeah I tried the cosh command

and taking away the+C

still was wrong

[misty crest]

did u do +C for the other ones

or is there a command for that too

[vivid estuary]

this one worked fine

and it seems like my classmates had the same answer as me and there’s worked too

oh

nvm

it was the plus c

though I removed it before

my classmate said he reformatted it to -cosh(17x)/17 also

[misty crest]

yea do the 1/17

[vivid estuary]

with the negative right by the cos

[misty crest]

instead of the fraction

oh

hmm it should be +C tho

that’s weird

[vivid estuary]

I got it right now

[misty crest]

it doesn’t indicate to do so

[vivid estuary]

It was the +C that was counting it wrong

I guess they didn’t want it

[misty crest]

but u should

[devout snow]

@vivid estuary Has your question been resolved?

[devout snow]

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[heavy sail]

how do i do this

idk the rule for f(x)

[devout snow]

@heavy sail Has your question been resolved?

[heavy sail]

<@&286206848099549185>

[tame palm]

Estimate the gradient line.

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[outer spade]

Which number is greater: ( 5 - 4 \sin \frac{3\pi}{14} ) or ( 4 \cos \frac{\pi}{7} - 5 \sin \frac{\pi}{14} )?

[woven radish]

kitten

[outer spade]

Helppp

Pls

[near trout]

what have you tried

[devout snow]

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[thorn lagoon]

The number 2022 is the product of exactly three primes. Find their sum.

[dawn iris]

!status

[devout snow]

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

[thorn lagoon]

What I have so far is that I know one of the numbers has to be 11

1

[dawn iris]

i dont think 11 divides 2022

[thorn lagoon]

I mean 22

[dawn iris]

22 doesnt divide 2022 either

[thorn lagoon]

I think 2 could be one of them

Wait nevermind I think I'm gonna make a factor tree and go from there

I got the answer its

3, 337, and 2 thanks for the help

.close

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[light pecan]

my last line of working is this

$\frac{\theta}{2}$

[woven radish]

Kap

[light pecan]

$ \frac{sin(\theta * 2)}{4}$

[solar goblet]

$\frac{sin(\theta * 2)}{4}$

[woven radish]

FungusDesu

[light pecan]

$\frac{\theta}{2} - \frac{sin(2\theta)}{4}$

[woven radish]

Kap

[light pecan]

evaluated between 0 and 1/2

where x = sin(theta)

how do I undo the substitution

of sin(2 theta)

in this scenario

would love some help with undoing the substitution

<@&286206848099549185>

[solar goblet]

this is the result in terms of theta after the integration correct?

[light pecan]

yes

this is the result after integrating

[solar goblet]

$x = \sin\theta \newline \theta = ?$

[woven radish]

FungusDesu

[light pecan]

sin^-1(x)

i know how to do the first part

[solar goblet]

correct

[light pecan]

which is that on 2

but its the second term that i cant do

$\frac{\sin(2\sin^{-1}(x))}{4}$

[solar goblet]

^{-1}

[light pecan]

you know hwat i mean mate

i dont know how to simplify that

bc the answers have only x

[lusty sapphire]

@solar goblet use $\sin$, not $sin$

[woven radish]

SWR

[solar goblet]

yeah i was using that

[light pecan]

why doesnt that work

[lusty sapphire]

Use {-1}, not (-1)

[light pecan]

ok

[woven radish]

Kap

[light pecan]

there we go

so

how do you simplify that

down to only ahving x's

[lusty sapphire]

double angle formula is your friend

[light pecan]

omg

forgot about that

so i got this

$\frac{\sin(\sin^{-1}(x)) * \cos(sin^{-1}(x))}{2}$

[woven radish]

Kap

[light pecan]

thats what i got

so sin(sin^-1(x)) = x right?

[solar goblet]

yes

the latter is actually a semicircle

[light pecan]

how do you know that

[solar goblet]

,w plot cos(sin^-1x)

[woven radish]

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Requested by fungusdesu

[light pecan]

i know it is

but how do you konw

[solar goblet]

honestly, i dont know either

i know because i played around in desmos and found out

[light pecan]

fair

[solar goblet]

https://math.stackexchange.com/questions/1824176/why-is-sin-arccosx-a-semicircle-with-radius-1

[light pecan]

im reading that now

[solar goblet]

this article explains it pretty well

[light pecan]

haha

thanks mate

[solar goblet]

by the way if we switch the sin to cos and cos^-1 to sin^-1, its still a semicircle

fun fact

same way of proving

[light pecan]

ik

[solar goblet]

anyway. i digress

[solar goblet]

using that information, surely you can simplify this