help-38

[sacred sapphire]

just rotate QLRM by 90 degrees

[open fulcrum]

what does that do

[sacred sapphire]

makes it easier to see

because SM is the long side of the PLMS, its horizontal, but LM is the long side of LMRQ and its vertical

by rotating LMRQ 90 degrees, LM will be horizontal

then use side ratios

[open fulcrum]

when did you learn side ratios?

[sacred sapphire]

?

[open fulcrum]

i dont get what u mean by side ratios

[sacred sapphire]

similar shapes

corresponding sides are proportional

[open fulcrum]

oh right

i got 2, says incorrect

since PLMS~QLRM and QLRM has side lengths of 1 and x-1, the proportion seems to be 1/1 = x-1/1, which is 1 = x-1 and x=2

[sacred sapphire]

which sides are you using

[open fulcrum]

RM/SR = RQ/ML

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[celest gazelle]

Answer key says t=5.5 years

[digital bison]

you can't divide by 1.0044 like that

you have to use logarthms

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@celest gazelle Has your question been resolved?

[celest gazelle]

So how do I solve for T then

[celest gazelle]

20000=150000(1+0.053/12)12t

T=?

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@celest gazelle Has your question been resolved?

[celest gazelle]

It was not but I will try again later

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[sudden stream]

Let's say I have a number 10000 and I need to find the range 5% above and below that.

Would something like this work if lower range is a, upper range is b

[sudden stream]

hold up typing it out now hit send too early

a = 10000 - (10000 * (5/100))
b = 10000 + (10000 * (5/100))

[crisp forge]

sure

[sudden stream]

Okay sweet

[flint sluice]

a = 10000 * 1.05 b = 10000 * 0.95

[sudden stream]

oooh

but that would be b I think

[crisp forge]

i guess thats a better way of doing it

[flint sluice]

yeah

[sudden stream]

so essentially what you havbe but swapped around

[flint sluice]

yup

[sudden stream]

a is upper b is lower

[flint sluice]

it just takes the entire 10000 and adds that 5% range

for 1.05

[sudden stream]

Yeah I see that's smart

Okay thank you very much guys!

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[languid cargo]

are these correct?

[languid cargo]

the rectangular prism question and area of a rectangle

wait erasijg the W variable that doesnt exist

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[languid cargo]

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[barren patrol]

can someone help me with this please

[trim jolt]

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

[blissful bison]

Generally, there are two methods for such an inequality: by the definition of the absolute value or using the quadratic inequality

[barren patrol]

can you help me iwth the definition of absolute value definition because i havent learned quadratic inequality yet

[blissful bison]

i give you a hin deifnion:

$|x| = \left{ \begin{array}{cl}
x & : \ x \geq 0 \
-x & : \ x < 0
\end{array} \right.$

[solid kiln]

Joanna Angel

[blissful bison]

this definition shows how to leave out the absolute value symbol

and if you d like to use it in yoru case, you can notice that you have different absolute values on both sides of your inequality

so you have to apply to th left side and to the right side

[blissful bison]

can you see it ?

$|a| = \left{ \begin{array}{cl}
a & : \ a \geq 0 \
-a & : \ a < 0
\end{array} \right.$

[solid kiln]

Joanna Angel

[blissful bison]

and

$|2a-3| = \left{ \begin{array}{cl}
2a-3 & : \ 2a-3 \geq 0 \
-\left( 2a-3 \right) & : \ 2a-3 < 0
\end{array} \right.$

[solid kiln]

Joanna Angel

[blissful bison]

and then you need to connect them

[timber grotto]

Hint: \par\begin{itemize}
\item $2a-3\geq 0$ implies $a\geq 0$.
\item $a<0$ implies $2a-3<0$.
\end{itemize}

[solid kiln]

luke1337

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[wraith hinge]

Yooooooooooooooooooooo

[wraith hinge]

I am so confused

I need to find the length of this curve

$r = 2(1+\sin(\theta))$

[solid kiln]

we

[wraith hinge]

I thought about doing $2\int \sqrt{(1+\sin(\theta))^2 + (\cos(\theta))^2} d\theta$

[solid kiln]

we

[wraith hinge]

but it becomes

$2 \int \sqrt{1+ 2 \sin(\theta) + \sin^2(\theta) + cos^2(\theta)} dx$

[solid kiln]

we

[wraith hinge]

which is $2 \int \sqrt{2} \sqrt{1+\sin(\theta)}$
an elliptic integral i can not integrate

[solid kiln]

we

[wraith hinge]

maybe i need to keep the 2 inside

nah

still dont work

[burnt mulch]

You can integrate 1+sin x

https://math.stackexchange.com/questions/1901857/how-to-evaluate-the-integral-int-sqrt1-sinx-dx

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[wraith hinge]

wtf

TY

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[wraith hinge]

how do i do d

using this

[turbid socket]

^ i actually also have a questions about this, specifically,

D and F,

for D, the (e^x - 1) - the one goes to the (n+1)!

for F, how does the Zs turn to x^2n -1

[wraith hinge]

<@&286206848099549185>

D:

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[wraith hinge]

@atomic socket

[atomic socket]

can u post the laws

or idk what it calls

the series

@turbid socket

[atomic socket]

k

here u can actually take out 8/7x

and find the series for e^x - 1

[wraith hinge]

and then j have it start at 1?

to get rid of the -1 part

[atomic socket]

so e^x - 1 is the same one as e^x but it starts from n = 1

ye

easy? right?

[wraith hinge]

Ye, thanks!

[atomic socket]

ur welcome

@turbid socket cool?

[turbid socket]

yesss

[atomic socket]

🙂

[turbid socket]

thank you bro

[steel moon]

alright i'm starting to be afraid of math exams on the future

[wraith hinge]

its like i know wut u mean bro

[turbid socket]

who invited bro

[steel moon]

sorry

no one was responding and got bored

[turbid socket]

,rotate

[turbid socket]

n was orginally n=0 but i said n=1 to cancel it out

[atomic socket]

x ^ n-1

not x^n

x^n/x --> x^n-1

and dont forget the sigima notation

[wraith hinge]

wait im gonna try that problem ou trq

[turbid socket]

ighty

[turbid socket]

this is the answer

[atomic socket]

?

[steel moon]

the hell i got @d for

[turbid socket]

thats the answer for D

[atomic socket]

it actually the same bro

[wraith hinge]

bros imagining stuff

[atomic socket]

it sdtarts from n =0

its the same answers

[turbid socket]

OH sorry

im slow

[steel moon]

i had a notification for here

meh

[turbid socket]

ok got it

[atomic socket]

🙂

[turbid socket]

can i do one more problem

[atomic socket]

@steel moon open ur own channel pls

[steel moon]

I did

[atomic socket]

no worries if @wraith hinge accept

[steel moon]

trying to understand this one because i got a similar issue

[atomic socket]

since he opened the channel

[atomic socket]

or u can open ur own @turbid socket

[turbid socket]

i think they have the same study guide as me

^ kitefly

[atomic socket]

can u post the series again haha sorry im too lazy to search

[wraith hinge]

waiiit

give me ooone sec

[atomic socket]

k

[atomic socket]

lets wait 1 min for kite

[turbid socket]

oki

[wraith hinge]

i dont understand how they got n+1

[atomic socket]

what did u get?

@turbid socket if we take the x out then we get arctan(3x)

just subsitute instead of x do x--> 3x

[turbid socket]

thats what i did so far

[atomic socket]

amazing

and all g?

[atomic socket]

ye thats right

since u starts the n from 1

[wraith hinge]

Why does it look different from the answer key then?

[atomic socket]

in the answer key they start from n = 0

[wraith hinge]

ohhh so if subtract 1 there i j need to add 1 wherever theres an n?

[atomic socket]

ye

[turbid socket]

almost, i got 3^2n+1 but after you add x its supposed to be x^2n+2
which is what i dont get

[atomic socket]

u dont add c

x*

u multiply by x

[turbid socket]

okk

i got it

@wraith hinge

[wraith hinge]

got it!

[atomic socket]

good guys?

if needed something else lemme know

[wraith hinge]

Actually

[wraith hinge]

Alr, thanks! Im prolly gonna have a lot more questions lmaoo

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[obsidian hornet]

can we go over this question?

so what I did was found x=t^2/2, y=t, z=t^3/6

and ds/dt = (t,1,1/2t^2)

oh

so i got this,

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[remote creek]

can someone explain how to do this?

[remote creek]

so far i did this but i dont know how to continue

[hallow spruce]

ooh this is a fun one

my first hint to you is to rewrite the denominator of the original fraction as x-0

see if that makes anything jump out at you

[remote creek]

i dont think i get it, i rewrite it as x-0 but does that change something?

[hallow spruce]

well in the numerator you have some function of x minus some constant

in the denominator you have x minus a constant (let's say c2 to make them distinct)

you're taking the limit as x approaches c2 (which is 0)

[remote creek]

i still dont think i understand... 😭

[hallow spruce]

$\lim_{x\to c_2} \frac{f(x)-c_1}{x-c_2}$

[solid kiln]

Steakanator

[hallow spruce]

this should look almost identical to, but not exactly like, something you've (hopefully) been shown before

[remote creek]

oh i think ive seen it before, is it something like x approaches a

the derivative?

[hallow spruce]

well a is already being used in the question but it does tend to be the more used letter

[remote creek]

definition?

[hallow spruce]

yes, the definition of the derivative

now in my expression, what must c_1 be replaced with to match the definition?

[remote creek]

f(c2)?

[hallow spruce]

indeed

and what is c_1 for this problem in particular?

[remote creek]

1

[hallow spruce]

so you know f(x) requires you to solve for two parameters a and b; you know f(0) = 1 and f'(0) = 2

get solving

[remote creek]

okk

is b = 1/2

and a = 4

okk i got it verified, thanks a lot

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[wraith hinge]

is tjis correct

is the process correct i mean

btw fast integrations arnt allowed here, you have to use reimann sum method

<@&286206848099549185>

i'm not sure its safe to say $\sum_{i=1}^{n} |2i - n|$ is = to $2\left( \sum_{i=0}^{\frac{n-1}{2}} (n-2i) \right) - n$ as if they were equal in every case

ik numerically its true but idk if thats enough

[solid kiln]

Retribution | ᵇᵃˢᵉᵈ

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@wraith hinge Has your question been resolved?

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[cunning inlet]

There are 13 people and i want to make a group filled with 7 members
2 of the 13 people doesn't want to work with each other
Is the answer 11C6 x 2?

[shrewd ridge]

no, that would force one of them to work

[cunning inlet]

oh wait thats right

11C6 * 2 + 10C7?

[trim lichen]

better to use * rather than x as a multiplication symbol

also why 10C7?

[cunning inlet]

mb

[cunning inlet]

all the cases where theyre not on the team

[cunning inlet]

this one means the only cases im getting are the cases where they work on a group and 10c7 are the cases where they dont work in the group

can it go that way

[quiet fractal]

Are you looking for the number of possible groups?

[cunning inlet]

yes

[trim lichen]

who are "they"?

also btw what are your pronouns

[cunning inlet]

you can refer to me as he dw

[quiet fractal]

Are you familiar with the choose function

[trim lichen]

we have a dedicated he/him role

[cunning inlet]

the 2 person that doesnt want to work together

[shrewd ridge]

what you said makes sense, just it;s not 10

[cunning inlet]

yes n!/n! (r-n)! right?

[cunning inlet]

i misclick and idk how to change roles

[trim lichen]

yeah but aside from them there are 11 people not 10

[cunning inlet]

do explain

oh wait yeah it was 13 not 12 mb

[trim lichen]

"channels and roles" at the very top of the channel list

[cunning inlet]

oh right, thank you for that

[cunning inlet]

ok so its 11C6 * 2 + 11C7

[shrewd ridge]

i like that

[cunning inlet]

thank you so much for helping

[cunning inlet]

I'm done with the question, thank you for your response too

.close

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[coral abyss]

lots of questions
contour maps are stumping me
have I indicated a saddle point at A correctly for the second part? I read somewhere contour lines cross each other at saddle points

my reasoning is that the gradient of f will always be perpendicular to the constraint at saddle pts or maxima/minima since gradient points towards maxima/minima

[trim jolt]

@coral abyss Has your question been resolved?

[cunning flicker]

You need tangency between a countour line of f and the constraint g=0

For the perpendicular gradient condition to be met

And to have a saddle point, you'd want to have that such tangent countour line crosses the constraint at the tangency point and is not always inside or always outside

@coral abyss

Your sketch lacks tangency, also, you don't need countour lines of f to intersect

[coral abyss]

idk how to sketch tangency at the minimum/maximum

[cunning flicker]

For minimum maximum you did almost correctly, in the sense that that is a minimum maximum point for f but the gradient is 0 there, so it depends if they would accept the 0 vector to be perpendicular or or not.
To be sure though, you could instead draw a contour line which is tangent but is always outside (or inside) of the constraint g=0 in that case, instead

[coral abyss]

I am so confused lol

how can I do this at a minimum/maximum

[cunning flicker]

Imagine two tangent circumferences

Make a drawing and send it

[cunning flicker]

Good, now add random contour lines to complete the plot

[cunning flicker]

Only black lines, and you can also draw larger ones. Black lines are contour lines of f, red line is g(x,y)=0

[coral abyss]

this work?

[cunning flicker]

You can overlap larger ones with the red line

Start again with the first pic, draw some other black lines inside the tangent one, some other outside which overlap the red line

[coral abyss]

any conditions on the overlap?

?

[cunning flicker]

They should still be all concentric though

[coral abyss]

?

[cunning flicker]

Oh, now this works. Do you see why at the tangent point we have a minimum/maximum of f on the constraint?

[coral abyss]

no lol

[cunning flicker]

Well, on each contour line f has the same value c (different for each line), and the line divides the plane in two regions, one where f>c, one where f<c. Now, in light of this, consider the tangent line

[coral abyss]

ooooo I see

it can't help but be a min/max over g

[cunning flicker]

Yep, well done

And that is also why we need the tangent line to cross g=0 to have a saddle point

[coral abyss]

actually, why does the req. point in b have to be a saddle point at all?

[cunning flicker]

Yes, sorry, i was abusing the term. A point which is neither a max nor a min of f restricted to the constraint

Which is also a tangency point (what is said to be a critical point of f on the constraint)

Now, i don't know if a critical point on a constraint which is neither a local max nor min is named a "saddle point on the constraint". Similar to how a critical point of f which is not a local extremum is called a saddle point

Anyway, sorry for wandering off

[coral abyss]

wouldn't anything tangent be a min/max?

[cunning flicker]

Eh, no since we can make a tangent line which also "cuts" g=0 at the tangengy point

[coral abyss]

how can a tangent line cut at the tangency point

[cunning flicker]

Think now, for instance, as an analigy, of a cubic profile, which is tangent to the x-axis but crosses it

[coral abyss]

I have no idea what you mean lol

[cunning flicker]

That is, draw the plot of the function f(x)=x^3

,w plot x^3

[solid kiln]

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Requested by Mat#9616

[cunning flicker]

Can you see what I mean by crossing the x-axis at the tangency point?

Before, instead we were making something similar to this
,w plot x^2

,w plot x^2

[solid kiln]

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Requested by Mat#9616

[cunning flicker]

Consider the x-axis as the constraint g(x,y)=0 in these examples

And as the blue lines of the plots as the tangent contour line of f

[coral abyss]

umm

I don't see how this would translate to our contour map

we're essentially trying to make a map

such that we're not dividing into two parts w f>c and f<c right?

at the tangency point?

[cunning flicker]

The point is, after seeing the plot of x^3 above, would you be able to make a correct drawing if the constraint g(x,y)=0 was just a line?

[coral abyss]

um I don't see how that prevents it from being a maximum/minimum tbh

crossing over

[cunning flicker]

Complete first something like this in the desired manner to have a crossing tangent line

[coral abyss]

how is the crossing line tangent here tho

[cunning flicker]

The blue line is tangent to the x-axis at (x,y)=(0,0) and crosses it there

[coral abyss]

doesn't a tangent only intersect at one point around the point of intersection

"touches" the curve or sth

[cunning flicker]

Yes, by crossing I mean that on the left the blue line is under the line given by the x-axis while on the right it is above

That is, the x-axis doesn't lie, near the tangency point, in only one of the two regions in which the blue line splits the plane

Instead, it is in both

[cunning flicker]

In this case on the contrary, the blue line is always above

[coral abyss]

hmm

[cunning flicker]

Perfect

[coral abyss]

wait can this even be the map of a continuous and differentiable f

idk

[cunning flicker]

Yes, why wouldn't it?

[coral abyss]

yeah got thrown off for a sec

how does the Lagrange multiplier method avoid providing cases like part b as the answer

far as I can see gradient would also be proportional for the two curves at A in case b

[cunning flicker]

Indeed, that's an acute observation. For local extrema on a contraint, being solutions to the lagrange multiplier system is only a necessary condition

So, that is a way to provide a set of candidate points to be local extrema, a narrowing down of the problem of reserching them, if you want

But then we need to analize each solution and discard cases alike the last drawing you sent

[coral abyss]

how would we discard cases?

[cunning flicker]

It depends on the specific situation normally, there are some shorcuts

For instance, now you know that drawing a contour plot is a way to identify points to be discarded

[coral abyss]

so the lagrange multiplier system basically gives us the critical points?

and it's upto us to check for extraineity?

[cunning flicker]

Yes, precisely

[coral abyss]

ah

trying to learn how to use the lagrange system

does this look alright?

[glad valley]

lambda 2z = 2

not quite

z=1/lambda

y/2x=lambda

and x/2y=lambda

$z=\frac{1}{\lambda}$

[solid kiln]

naokye

[glad valley]

and

$\lambda = \frac{y}{2x}$

and

[solid kiln]

naokye

[glad valley]

$\lambda = \frac{x}{2y}$

[solid kiln]

naokye

[glad valley]

$\frac{x}{2y}=\frac{y}{2x}$

[solid kiln]

naokye

[glad valley]

$x^2=y^2$

[solid kiln]

naokye

[glad valley]

uh fuck i cbb solving this sorry

[coral abyss]

sorry what?

I can't figure out how to solve this further

got to here. not sure how to find the minimum and maximum values

[cunning flicker]

It is 2λ(2xλ)=x, hence 4λ^2x=x

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[novel stump]

the small formula at the top is the small photo and why is this proof wrong can someone explain me why does zeta work like this?

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[novel stump]

help

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@novel stump Has your question been resolved?

[novel stump]

<@&286206848099549185>

[wraith hinge]

No idea sir

I am still doing high school maths but it seems like ramanujan paradox

Is that true

?

As I see inf=-1/12

[novel stump]

ramanujan found out that zeta(-1) is -1/12

[wraith hinge]

Yeah good luck finding some helper

[novel stump]

but zeta function says that all numbers from 1 to inf summation 1/n^s is the formula

if u put -1 to "s" 1/n^s value becomes n value

[wraith hinge]

Chill chill bro

I know nothing about this

Sorry

[novel stump]

;-;

Im also highschooler

gl bro

[wraith hinge]

Thanks same to you

[trim jolt]

@novel stump Has your question been resolved?

[novel stump]

<@&286206848099549185> pls anyone

[wraith hinge]

Alr I'll help bro

(I'm also high schooler 💀)

I think the problem is that your applying infinite directly

But you need the lim

It's like when you divide by zero in a lim equation, you can't just replace all variables with 0 it will be an error. Instead you need to do some shenanigans

I believe it is called Dirichlet Regularization in this case

@novel stump

[novel stump]

but wolfram alpha can do it without limits when it is a condition of infinity

[novel stump]

yeah wolfram alpha says both inf and -1/12 at the same time because of Dirichlet Regularization

[wraith hinge]

https://youtu.be/sD0NjbwqlYw?si=vWATXnG5fkDN-0PS

[novel stump]

.close

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[loud egret]

How to solve this

[jovial shell]

Try taking log both sides

Since x^2-8x+15 cannot be 0 (or else result would never be 1) you can move it

[loud egret]

This sum needs to solved with polynomials

[jovial shell]

Oh

[loud egret]

they havent taught me log yet

[jovial shell]

Well we know that a^0 is 1 when a is not zero

[loud egret]

yes

[jovial shell]

So put the power as equal to zero

As the base can't be 0

[jovial saddle]

you can get 1 from this equation either the exponent is zero or the base is 1 or -1

[jovial shell]

Yes

[jovial saddle]

but x-2 can't be 0

[jovial shell]

X can't be 2

[jovial saddle]

because its in denominator

yeah

[loud egret]

we have to find the number of x intercepts. The answer is 4

but i dont how to approach this sum as the question equates to 1

[jovial saddle]

even by solvin this you will get 6 answers but i gues you will have to reject 2 of them because they are irrational or smth

[jovial saddle]

that's what we explained

it can only be zero when the exponent is 0 or when the base is 1

so equate it accordingly

[loud egret]

im not getting the idea here.

[jovial saddle]

1^a=?

a^0=?

where a is a constant

this is the only way you can get one

[loud egret]

yeah... how can i apply it to the question

we need to find the number of roots

[jovial saddle]

In
case 1 : the exponent should be zero
case 2 : base should be 1

from the above logic

do you get it now?

[loud egret]

(eqn)^0 = 1

[jovial saddle]

yes

and 1^eqn = 1

[loud egret]

im trying it

[jovial saddle]

ok

[loud egret]

but if i do this how will i get the number of roots?

[jovial saddle]

once you get the roots you can just count them to get the number 🙃

[loud egret]

but im not getting the roots here, its becoming a more complex equation, btw this sum in a polynomial sum.. i have no idea about logs

Can you solve it on paper and show me

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[feral ocean]

This confuses me. The b part

[feral ocean]

Upon equating the velocity function to 0, I get a negative time an imaginary number

Did I go wrong?

[lone river]

That is one of the answers, you just got the wrong one for this question :P

You are getting negative time and an imaginary number because that is what would happen if ... I think if time was flowing backwards? it is the 'mathematical' answer that is impossible in real life

[feral ocean]

So I didn’t go wrong

[lone river]

yeah it should be a mathematically correct answer, i have not looked too close at your work but you can get that as an answer with projectile problems

This is a projectile motion question without or with gravity?

[feral ocean]

Without

[lone river]

just checking 👍 let me look 🤔

[feral ocean]

Okay

[lone river]

ok I see what you meant

i went through your math and don't see a problem with a, i followed along and got the same number

for b the velocity would by definition be 0 of an at rest object...?

so it may be a trick question?

it doesn't ask the time, and it starts with the object moving and accelerating, so the only time when it is at rest has to be an imaginary value

but the only value for velocity you will ever get at rest is 0

(unless this is a special relativity question)

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@feral ocean Has your question been resolved?

[feral ocean]

Ohh I understand

[lone river]

yeah that's not a math question that's a 'definition of being at rest' question :P

[feral ocean]

How do I solve this though?

[lone river]

solve for what? the velocity for b?

it doesn't ask for time, or anything

the answer I would put down for question b would be 'by definition, the velocity is zero at rest'

[feral ocean]

No it’s a trigonometry question

Sorry it didn’t send

[lone river]

ah i think you're supposed to only ask one question at a time

just do .close and then send it immediately after

[feral ocean]

.close

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[lone river]

looking at this, I would use trig rules

do you know how to get secant from sine, and how to get tangent from secant?

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[halcyon quiver]

What i have so far is it right? and i need help on how to figure out the units

[halcyon quiver]

@mystic latch Could you check mesaages or request and try to help me

With a couple of other geometry questions cause im on limited time today

[wraith hinge]

tf is 2tan^2theta

[proven hound]

Ig it's tan²theta

Most likely a printing error

Or even tan(2theta)

[wraith hinge]

assuming its a printing error

simple trigonometry

[feral ocean]

I don’t know anything about trigonometry

Or rather I do to some@extent

Extent

[feral ocean]

Ohh

Thanks

[halcyon quiver]

anyoen can help with this??

[ornate oasis]

You're trying to map figure I into figure J
so the translation isn't down, it's up, the rotation part seems correct though

[halcyon quiver]

Ok

[ornate oasis]

as for the number, we'll hafta count the squares

[halcyon quiver]

Half count like the distance? or the unit squares that are apart of the figure

[ornate oasis]

you just count the number of squares that figure i moved up before rotating

you can pick a point between them

since it's rotated though

try to imagine how it would look like before rotating figure j

[halcyon quiver]

Alright

[ornate oasis]

the top left point in figure j is the same as the bottom right point in figure i

before j was rotated, that top left point in figure j would've at (7,5) I believe

visually it's not easy to tell at first, but with enough practice this types of problems get easier

[halcyon quiver]

do u mean like count the squares from up here before it rotates over? Sorry if im not making sense new topic form e

[halcyon quiver]

Yea

[ornate oasis]

that's alright, I can probably sketch something to help you understand since it's hard to explain with words, jimmy a moment if you have time

[halcyon quiver]

Ok no problem

[ornate oasis]

When it was moved up it went 13 squares up

It's easier to pick a point and count how many squares it went up

I picked the bottom right point

@halcyon quiver

[halcyon quiver]

Oh and thats how u find the units?

Makes more sense

[ornate oasis]

yep

[halcyon quiver]

I knew it was soemting like that but wasnt sure

So my rotation is good

[ornate oasis]

It's very easy once you understand it

[halcyon quiver]

And the 180 clockwise

[halcyon quiver]

Yea

[ornate oasis]

yep that's right

[halcyon quiver]

Do you mind if i add you if i ever need any other suggestions or tips or Na

[ornate oasis]

clockwise or counterclockwise, 180º still looks the same

[ornate oasis]

yeah oc

my dms always open

[halcyon quiver]

Yea

Idk why i thought it was different Lol

[halcyon quiver]

Thanks

[ornate oasis]

it's aight, I overthink a lotta things like that lol

[halcyon quiver]

Lol

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[ornate timber]

if theres an increase of 16% over 5 years, what yearly percent increase is required to fulfill that agreement

[ornate timber]

whats the formula

to calculate this

[cedar flower]

it must be some increase, applied every year, that joins up to make 16%

so for the initial amount x, per year it must be n
and (x*n)*n)*n*n*n = 1.16

[ornate timber]

do

so

x^5

or what

nvm i found the formula

.close

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[tribal fractal]

how do i know if my estimation using the euler method is an under or over estimate

[tribal fractal]

when the second derivative is negative?

anyone?

<@&286206848099549185>

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[tribal fractal]

anyonee?

<@&286206848099549185>

.close

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[tribal fractal]

.reopen

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[tribal fractal]

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[wraith hinge]

WTF

[wraith hinge]

What is the method for solving $\frac{1}{\sin^2(\theta)} d\theta$

[solid kiln]

we

[wraith hinge]

I know how to do the left fraction with u sub

but the right!!!!!!!!!!!!!

[vagrant marsh]

csc^2(x) integrates nicelg

[wraith hinge]

wtf is a csc

tan(x) + c

[vagrant marsh]

no

[wraith hinge]

how tho

[vagrant marsh]

csc is cosecant

[vagrant marsh]

I suggest looking up trig derivatives

[wraith hinge]

uuuuh

i refuse

[vagrant marsh]

Itll help you figure out the integral of csc^2(x)

[wraith hinge]

it has not

I have looked it up many times

still have no idea how to integrate this

[vagrant marsh]

,w d/dx (cot x)

[solid kiln]

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[wraith hinge]

The step by step guide is

[vagrant marsh]

Indeed

[wraith hinge]

Yeah

but how

Just saying the integral of x is y isn't really helpful

[vagrant marsh]

Idk what to tell you except the derivative of x is y tho

cot = cos/sin

You can use quotient rule

Its kinda one of those things you need to know

[wraith hinge]

so

do i just reverse quotient rule?

and solve

uuuuh

[vagrant marsh]

Yeah

[wraith hinge]

$sin(x)f'(x) - cos(x)f(x) = 1$

[solid kiln]

we

[wraith hinge]

$sin(x) f'(x) = 1 + cos(x)f(x)$

[solid kiln]

we

[wraith hinge]

uuuuuh

[vagrant marsh]

Im telling you, you just gotta know derivatives of the 6 trig funcs

[wraith hinge]

Not happening

If it could i would had remembered by now

$f'(x) = \frac{1}{sin(x)} + tan(x)f(x)$

[solid kiln]

we

[wraith hinge]

$f'(x) = \frac{cos(x) +sin(x)cos(x)}{sin(x)cos(x)}f(x)$

[solid kiln]

we

[wraith hinge]

$\frac{f'(x)}{f(x)} = \frac{cos(x) +sin(x)cos(x)}{sin(x)cos(x)}$

[solid kiln]

we

[wraith hinge]

Is this solved?

$$f(x) = sin(x)cos(x)$$
$$f'(x) = cos(x) + sin(x)cos(x)$$

[solid kiln]

we

[fair bison]

Brandon do you know the integral of sec^2 x dx?

[wraith hinge]

tan(x) + c

[fair bison]

Right, so now just using the co-angle identity

[wraith hinge]

only because i was recently given a hint

[fair bison]

csc^2 x is just sec^2 (pi/2 - x)

Set u = pi/2 - x

Have a party

[wraith hinge]

but i will forget the integral of sec^2(x)

[fair bison]

Keep using it then

And you'll remember

[wraith hinge]

wait so

is $\frac{1}{cosh^2(x)} dx = \tanh(x) + C$

[solid kiln]

we

[wraith hinge]

,w 1/(cosh^2(x)) dx

[solid kiln]

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[wraith hinge]

oh

i need to use the exponential forms

ty ty

.close

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[wraith hinge]

,w 1/(cos^3(x)) dx

[solid kiln]

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[clever ferry]

Hello. Very late , just need re assurance on this question im about to send.

[clever ferry]

Is the answer 51/416

51.416

[cedar flower]

what is the perimeter in words
what is each part of that perimeter equal to

[clever ferry]

well what i did is

180/360 = 0.5

0.5 x pi

x 20

• 2-

+20*

if you can confirm that my answer of 51.416 that would be great

[cedar flower]

so yeah the straight part is 20, because it's just the diameter, check

[clever ferry]

my bad

wait

[edgy cedar]

It is the right answer.

[clever ferry]

yeah 20 is diameter right?

[cedar flower]

the circumference of a circle is pi*d, check
and you've got half a circle so that's 1/2*pi*d

check

[clever ferry]

yes

[cedar flower]

sounds good to me

[clever ferry]

but

if we are basing it of pi

no

i mean like

if pi = 3.14

in this case

it would be

0.5 x 3.14 =1.57

1.57x20 + 20 = 51.4

[cedar flower]

that...is what I got, yeah

[clever ferry]

alright

[lyric nymph]

the circumference of a circle is 2πr where r is the radius so that of the semi circle is πr and r=diameter/2=20/2=10

[edgy cedar]

You got the right answer but idk what you're really doing with your calculations.

[clever ferry]

also pure is my method elegible for all type of perimeter questions

[lyric nymph]

so 31.4+20=51.4 your answer is correct

[gray rapids]

where does the addition come from? i feel like im going crazy here

[edgy cedar]

I don't really get your method.

Like what is happening?

[clever ferry]

wdym

[lyric nymph]

basically he is adding the diameter which is considered a part of the perimeter

[clever ferry]

angle/360 x pi x d + d

thats the method

[lyric nymph]

thats what he did there

[edgy cedar]

Ah okay.

[clever ferry]

also can y ou check if this asnwer is correct 6.2831

[edgy cedar]

That would work for any angle yes.

[clever ferry]

alr thats good

[edgy cedar]

You should show your work.

[clever ferry]

mb

12 x pi = 12pi

12pi x 60/360

i think i did it wrong

supposed to be 24

give me a sec

[edgy cedar]

angle/360 x pi x d + d
Here d is not 12.

yeah

[clever ferry]

this was for perimeter not arc lenth

length*

[edgy cedar]

Sure but that just means you don't add the d.

[clever ferry]

pure

[edgy cedar]

?

[clever ferry]

wait so u dont add the d in the perimeter?

or u do

[edgy cedar]

No for the arclength it should be angle/360 x pi x d and that's it because you don't need to add the two radii

[clever ferry]

tell me ur answer

[edgy cedar]

tell me ur answer

[clever ferry]

i have one but i wanna confirm

so if i can get a graduates answer it would be great

[edgy cedar]

Well tell me your answer and I'll tell you if it's right or not.

[clever ferry]

nHail

got an answer?

[glossy root]

if pure answers first, you might just steal it, which is not allowed.
there is no reason for pure to steal your answer, so you should answer first.

[clever ferry]

Mb

12.6 ( rounded to the nearest decimal is this correct )

[edgy cedar]

Yes, that's correct.

[clever ferry]

W

WW

anywyas

one mroe inquiry

volume of cylinder formula?

[edgy cedar]

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[viral kernel]

i keep getting this as my answer (for sin)

i split 17pi/12 to (3pi/4)+(2pi/3) so i could use the formula
sin(u+v)=sin(u)cos(v)+cos(u)sin(v)

and used the sum and difference formulas

sin(3pi/4)cos(2pi/3)+cos(3pi/4)sin(2pi/3)

sin(3pi/4)= sqrt(2)/2
cos(2pi/3)=-1/2

cos(3pi/4)=-sqrt(2)/2
sin(2pi/3)=sqrt(3)/2

sin(3pi/4)cos(2pi/3)=-sqrt(2)/4

cos(3pi/4)sin(2pi/3)=-sqrt(6)/4

(-sqrt(2)/4)+(-sqrt(6)/4)= -(sqrt(2)-sqrt(6))/4

dont get what i did wrong

bruh i had the minus wrong

nvm

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[viral kernel]

.reopen

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[viral kernel]

wait is this

more correct than

[fathom ivy]

seems identical and unnesting radicals is somewhat tedious

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@viral kernel Has your question been resolved?

[viral kernel]

ic thanks

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[wanton mortar]

can someone help me solve this one: 2*(sinx)^2 + 5*sinx = 3, x>=-180 and x<180

[wanton mortar]

I get the solution of 30 and 150 degrees

but is it more than that?

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[calm adder]

assume sinx as x

and solve the quadratic

[wanton mortar]

I did

and got 30 and 150 degrees

are there any more solutions ?

[calm adder]

can i try ?

[wanton mortar]

yes

srry it is a bit hard to read it

is it more solutions or not?

[calm adder]

2

sol

[trim jolt]

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[supple locust]

Can someone help me how to find int x and int y with this equation

[vapid lynx]

,rotate

[vapid lynx]

what is "int x" and "int y"

this is a circle

well

its a long circle

whatever they are called

ellipse?

[trail ingot]

prolly x intercepts and y intercepts

[vapid lynx]

makes sense

if thats the case set the respective variable to 0 then solve for the other

@supple locust

[supple locust]

X intercept and y intercept

[supple locust]

That’s what I did but I’m having trouble solving the equation as a whole

I get stuck

[vapid lynx]

show your work

[vapid lynx]

move the 64 and 81 to the other side, then divide by 9

also after we finish this process remind me to say the shortcut way to do this problem

[supple locust]

Like this?

[vapid lynx]

well thats not the right answer, lemme double check the work

hey you broke the rules!

[supple locust]

?

[vapid lynx]

$(x-3)^2 \neq x^2-3^2$

[vapid lynx]

line 2

[supple locust]

Why not?

Isn’t it exponent rule?

[vapid lynx]

$(x-3)^2 = (x-3)(x-3)$

[supple locust]

OHHH

[vapid lynx]

exponents cant distribute over addition or subtraction

[supple locust]

That makes sense

[vapid lynx]

anyways, you should re-do the work and let me know what you get

[supple locust]

Okay let me do it rq

From here what do I do with -81x

[vapid lynx]

,rotate

[vapid lynx]

well this is a quadratic formula, are you familiar with how to solve one?

[supple locust]

Haven’t done one in a long time

[vapid lynx]

also to be clear, if your unit is on shapes then we are not doing this the intended way

if ellipse rings a bell?

[supple locust]

No😭

[vapid lynx]

okay

then i guess we have to do it this way

[supple locust]

My unit is on functions and this specific problem is how to find x and y intercepts

[vapid lynx]

,tex .quadratic formula

[supple locust]

I genuinely don’t know if im doing this right rn

[vapid lynx]

I have to go, so im going to just do spoiler steps for the answers:

Once you solve your current equation: ||you will get a negative number under the square root - this means there are no (real) solutions - which also means there are no intercepts for that||
Hint/once you finish the x=0 version ||you will get a negative number under the square root again||
answer to the problem: ||there are no x or y intercepts||

Way to do this faster: ||this is the equation of an ellipse centered at (3,4) with x-radius (wrong terminology) of 2, and y-radius of 3. This means that it never hits either axis. I will spoiler attach an image.||
SPOILER image from previous post:

[supple locust]

Thank you sm 🤍

.close

[trim jolt]

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[supple locust]

.reopen

[trim jolt]

✅

[supple locust]

I can translate everything but I need help figuring this out

Example #7: Identify the characteristics of a function. Use the drawing of the function f to answer the questions.

Determine f(3)
Determine f (-3)
Obtain all the x for f(x) = 0
Obtain all the x for f(x) = 2
Determine the x intercept(s)
Determine the y intercept(s)
Determine the domain of f
Determine the range of f

[trim jolt]

@supple locust Has your question been resolved?

[trim jolt]

@supple locust Has your question been resolved?

[trim jolt]

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[novel jay]

in a time crunch situation, how do i figure out which one it is?

[novel jay]

the 1 slopes, 0 slopes, and undefined slopes are the same afaik

[low gazelle]

What is the question

[novel jay]

Select the differential equation that matches the given slope field.

nothing really important

[low gazelle]

What are the choices

[novel jay]

the other two are irrelvent ones that can be instantly ruled out

my main issue is with the squareds

when the 1, 0, undef are all the same

and there's no -1

[low gazelle]

The choice u selected allows the slopes to be negative.

When x < 0 and y > -2

The square forces the slopes to all be positive

[novel jay]

oh so when its squared i have to find when its positive and negative

[low gazelle]

It seems you noticed that all the slopes are positive. You can rule out the option you chose because it is possible to get negative values

[low gazelle]

When its squared it wont ever be positive

[novel jay]

oh ok

.close

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• Show your work, and if possible, explain where you are stuck.
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[obtuse brook]

Can someone pls tell me how to do this

[trim jolt]

Please don't occupy multiple help channels.

[obtuse brook]

Closed my other one

[past river]

!status

[trim jolt]

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

[obtuse brook]

Im on 1

[past river]

Alright, well, they say that f(x) is linear, which is a big hint. What do linear functions usually look like?

[obtuse brook]

y = mx+b?

[past river]

yup

do you see how we can use that here?

[obtuse brook]

Uhhh not really

Well a little bit

But I dont know where

[past river]

So, we just stated that f(x) = mx + b. They don't give you much info about f(x), but they do give you some about 1/f(x)

[obtuse brook]

Yeah it passes through (1,4) and (3, 1/7)

[past river]

Alright, well, if f(x) = mx+b, 1/f(x) = ?

[obtuse brook]

1/mx+b

[past river]

yeah! (careful with the brackets though), you have 2 unknowns, 2 points

you can build a system of equations and solve from here :)

[obtuse brook]

How do I do that

[past river]

Do you agree that the points they give you are of the form (x,y)?

[obtuse brook]

yes

[past river]

wait a second, have you done this kind of stuff in class/by yourself before? Finding the equation of a function from points and such?

[obtuse brook]

Uhhhhhhh

A tiny bit but I didnt really understand it much

[past river]

I see. Well, the core concept is to plug in both points they give you into two seperate versions of the function ( in our case, 1/(mx+b) ), and using some tricks to isolate one of the variables we are looking for. I'll write it down on paper and send you a screenshot, it might be easier to understand

[obtuse brook]

kk

[past river]

,rotate cw

[past river]

Alright! If something's not clear let me know

[obtuse brook]

Thanks

What do I do from here though

[past river]

There are a couple things you can do, but I think the easiest here would be to isolate a variable in the right equation and subbing it into the other

if you notice the right equation is of the form 1/something = 1/something, which makes it pretty easy to work with

[obtuse brook]

Are we just trying to find m?

oh wait

theres m and b

[past river]

yeah exactly, which is why we need 2 equations

[obtuse brook]

so what do we isolate

[past river]

doesn't matter which you pick

you'll end up with the same result in the end

[obtuse brook]

so can isolate b

Doesnt that become 4 - 1/m(1) = b

or 4/1 - m(1) = b

@past river

[past river]

let me see

i think you messed up a bit in a division

it should be 1/4 - m

[obtuse brook]

Oh

So 1/4 - m = b?

[past river]

yup

[obtuse brook]

What about the other one

[past river]

the point of doing it this way is you have a way to express b only with m, and so, you can use that definition for b inside the other equation, and be left with an equation that only contains m

[obtuse brook]

Is m = to 27/8