precalculus

[pliant locust]

That only happens if you let both sides be exponents to e

[lilac storm]

log(1/4)=(t/27) x log(1/2)?

[pliant locust]

👍

and then t is equal to ? (in terms of logs, dont compute it)

[lilac storm]

log(1/4) / log(1/2)

which gives you 2

[pliant locust]

You forgot the $\frac{1}{27}$

[obsidian monolith]

rts

[lilac storm]

yes and then 2 x 27 = t which is 54

[pliant locust]

👍

[lilac storm]

and i just had a question so what is the point of finding k or there is no point because the answer key shows the answer for k and then this answer that we just found?

[pliant locust]

I have no idea what your k in this case is

[lilac storm]

yeah np, i was just asking to make sure.

[pliant locust]

If something is halved after T years you can express it as a function in terms of time t in the following:

$$\left(\frac{1}{2}\right)^\frac{t}{T}$$

[obsidian monolith]

rts

[pliant locust]

I havent looked in physics formula sheets in a long while, this is just what i came up with myself

pretty straightforward tbh

[lilac storm]

yes

Ok, thank you so much for your time and help, i really appreciate it

[wide void]

I know this might be really simple, but we got these two questions as an extra challenge at the end of a homework and i dont even know where to start

[ornate lintel]

your profile pic is so funny

[wide void]

Thank you🥸

[pliant locust]

Use pythagorean theorem to get the length of the hypotenuse then tell me what cos(x) is (x is theta here)

[wide void]

Hyp would be 26 for the first one but then do i use the secant to find the cosine?

[pliant locust]

Recall that cosine is just adjacent divided by hypotenuse

And then just use half-angle identity to compute sin(x/2)

[wide void]

Sqrt of 1-10/26/2?

Very very confused

[pliant locust]

Again, recall that cosine is adjacent divided by hypotenuse

[hoary rune]

<@&286206848099549185>

<@&268886789983436800>

Any help?

[stuck lark]

this isn't what the mod ping is for

[hoary rune]

Ok

[stuck lark]

and please wait 15min before pinging helpers

[hoary rune]

Could you help tho

[stuck lark]

i'm not available rn to stay and help. wait for an available helper, and it doesn't hurt to also post what you've tried to solve the problem

[hoary rune]

<@&286206848099549185>

[hoary rune]

I did

Lol are you gonna help or just be one of those discord mods

<@&286206848099549185>

[sick steppe]

Patience is key

[smoky pagoda]

Christ, a triple helper ping

For a question you could likely look up

As rokabe said, post your work so we can guide you through instead of giving you the answer

[obtuse locust]

Is k -0.253

[winter quartz]

Some people just aren't interested in following the rules 🤷

[winter quartz]

Walk us through what you did

[obtuse locust]

This is what I wrote

[winter quartz]

Okay. Walk me through it

[obtuse locust]

I subbed in the variables

Then subtracted to isolate k

Use log for the exponent

Do u think k is 0.253

[obtuse locust]

<@&286206848099549185>

[sick steppe]

Based on what you have written, k isnt .253

[obtuse locust]

actually I forgot to put e next to 32

And is it right based on the problem

[sick steppe]

what?

the base isnt 32e

$17=32e^{-25k}\to \frac{17}{32}=e^{-25k}$

[obsidian monolith]

moshill1

[obtuse locust]

Oh thank u

But I got the same thing for k

is the final answer 36 when the time is 110

[obtuse locust]

i got the same answer

[sick steppe]

Ok well what you wrote wasn't right

,w solve 17=32e^(-25x) over the real numbers

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by moshill1#9771

[sick steppe]

which is approx .0253

[obtuse locust]

when the time is 110 do you get 36

[sick steppe]

your question had no time

[obtuse locust]

what does that mean

[sick steppe]

also im not going to check an answer, use a calculator/wolfram to check

[obtuse locust]

ok

[subtle mist]

Josh starts at the point $A=(0,0)$ and is heading towards the point $P=(12,3)$ and he uses 3 hours to reach this point. Tommy is located at $B=(0,9)$. One hour after Josh starts to walk towards $P$, then Tommy starts to head towards point $P$ as well. I need to find a parametrization for the path to Tommy such that he and Josh ends up at $P$ at the same time.

[obsidian monolith]

please request a new nickname

[subtle mist]

I found where Josh is after one hour, which is just $\mathbf{x}=(0,0)+t(12,3) = (0,0)+1/3(12,3) = (4,1)$

[obsidian monolith]

please request a new nickname

[subtle mist]

but I'm not sure how to go on about making sure that they end up there at the same time

[obtuse locust]

Is this right

[main cedar]

yeah its correct

[grand barn]

idk if this belongs in pre calc but does anybody know how I can find the function of a building which has the shape of a paraboloid using the information that the maximal height is 36m , the width is 72m, the length is 140m

and that the maximal height touches the y axis

if we only had the two dimensions of height and width

this would be the shape

but I need 3 dimensions does anybody have a clue

[pliant locust]

Well, solid of revolution comes in mind. If you let the parabola in 2D have 140 meter in width and the height remaining 36 on y-axis a 180 degree revolution around the x-axis would give you a shape with a width of 72 meters.

A full spin will give you an ellipsoid, but i guess if you cut it in half you get a similar structure, hence a revolution of 180 degrees

[grand barn]

ok how does that help me find the function

im missing it

[pliant locust]

True, solid of revolution would just describe the volume of it.
What exactly do you mean by a function that describes the 3d of it?

You want like a funciton that would draw the building in geogebra?

[grand barn]

yeah

I think it is f(x,y) = (x^2/-36) + (y^2/- ((70^2/36)))+36

[pliant locust]

Pretty close actually

but yeah, this isnt precalc btw

[glossy geyser]

is the derivative of a functions in calculus or precalculus?

[blissful ridge]

Calculus

[glossy geyser]

wait is it the same as increasing and decreasing functions?

but thats the general name

[echo wagon]

No, different concepts but related in that the derivative can tell you whether a function is increasing or decreasing

[glossy geyser]

so they are both in calculus thank you, i have been looking for resources about them in precalculus and i was wondering why i didnt find anything

[echo wagon]

You should know what an increasing function is before precalc. Like when you learn about linear and quadratic functions.

[glossy geyser]

wait i am so confused so what field are increasing functions and reference functions are in?

[echo wagon]

Idk what a reference function is

You should learn about increasing functions almost immediately when you start learning about different functions. Idk when that happens or what that is called where you are.

[glossy geyser]

we follow the french system thats why im confused

reference functions are the one thats are known like quadratic and square root and stuff like that im not sure my self i was looking for resources on them

[astral apex]

reference function is probably another name for what's sometimes called "parent function"

Like standard things that have horizontal and vertical translations happen to them

[mild swan]

^^

[mild swan]

probably

[stark fulcrum]

can someone pls help with this?

[willow bear]

whats troubling you?

@stark fulcrum

[stark fulcrum]

how to make the function

[willow bear]

make what function

[cunning spear]

look at the requirements

[stark fulcrum]

yes but how can i make c'(1) equal them added together?

[cunning spear]

basically make any qudratic or higher degree function of ur choice

[willow bear]

pnksociety

you need to write down the formula for c(x) first

and then take its derivative

and then evaluate said derivative at x=1

[stark fulcrum]

but how do i know it equals this

[willow bear]

that's the definition of c(x)

c(x) is g(f(x))+h(x)

[stark fulcrum]

OH okay

would that be the chain rule ?

or product rule

[willow bear]

what do you think?

these are functions being composed, not multiplied.

[stark fulcrum]

so chain rule?

[willow bear]

yes.

[stark fulcrum]

oh alright, thanks

[fossil verge]

guys i’m lost

[viscid thistle]

could someone tell me if the professor made a mistake or I'm high

[willow bear]

this looks ripped out of context

[mystic basin]

im scared

[main cedar]

f(x) > 0, when the y value is >0, i.e. the areas where the graph goes above the x-axis

[sinful gulch]

Are there any integers whose squared difference equal 4?

I've highlighted the solutions I am struggling to find

[past meadow]

the only integer solutions for a^2-b^2=4 are (a,b)=(-2,0) and (a,b)=(2,0)

you can prove this by factorising both sides

[past meadow]

not sure of the physics/whatever that is but thats the answer to this lol

[round bane]

Do people say cosine-i-sine instead of just cis?

[past meadow]

i say e^(itheta) personally

[round bane]

I've heard it as cis for a while but then my precalc teacher this year keeps saying cosine-i-sine lmao

[past meadow]

but i've only ever heard people say cis when they're pronouncing it lol

[round bane]

Yeah ikr!! It was getting on my nerves haha

It took him sooo long to say cis every single time bc he kept saying cosine-i-sine

[runic flower]

$\lim_{x\to \infty} e^{-x}=0$

[obsidian monolith]

canol

[runic flower]

how to prove this using taylor series?

[willow bear]

why do you need to prove it using taylor series?

[runic flower]

was just trying to proof it by definition

[willow bear]

you can prove $\lim_{x \to +\infty} e^x = +\infty$ instead

might be easier

[obsidian monolith]

Ann

[runic flower]

well would it imply the other one then?

[steel tulip]

$e^{-x} = \frac{1}{e^x}$

[obsidian monolith]

Tiessie

[runic flower]

hmm right

and so how to prove this then?

[willow bear]

you could maybe prove that e^x > x

(for all x)

[runic flower]

oh yes that would work i think

and is there a way to directly prove it using taylor series of e^(-x)?

[willow bear]

meh

i guess not?

if you try to examine the taylor polynomials of e^-x as x goes to +∞ then youll alternately get them blowing up to +∞ or -∞

which tells you basically nothing

[runic flower]

right

its kind of surprising that it goes to 0 then

i mean just looking at the expansion of e^(-x)

and even more so that it goes to 0 faster than any polynomial

[willow bear]

i mean... the taylor series only really tells you local information normally

it just so happens that e^x is one of those really nice functions whose taylor series converges everywhere

[runic flower]

hmm

[viscid thistle]

hi guys

can anyone help me out with a couple of problems?

I can find the sum of these easily

but I dont know how to tell if their values do not exist

<@&286206848099549185>

[lilac pier]

@viscid thistle For the first, you can separate the sum into sum(k^2) - 5 sum(k) + 7 sum(1)

you should know the formulas for sum(k^2), sum(k) and sum(1)

[viscid thistle]

yup

did that

[lilac pier]

okay

so you're done?

[viscid thistle]

in hindisght yes, i shouldve come back here and said so

I kinda had a brain fart when i asked the question

and then it resolved itself and im like

[lilac pier]

ok np

[viscid thistle]

oh shit?!

yeah my b once again

[neon venture]

hey can I ask a precalc question here?

or is this channel being used im confused lol

[stable yacht]

You can ask a question here. It’s the “questions” channels that are more for 1-on-1 help so your question doesn’t get buried if the topic channels are busy.

[neon venture]

@stable yacht alright thx i asked in #help-8 🙂

[lilac storm]

this is what i have done so far but now i'm stuck so if anyone could help me out?

i think i made a mistake that's why i am getting that t^-1+1/-1+1 which is 1/0

i don't know what to do so if anyone is there that can please help me out?

[brave snow]

Well for one i don’t think splitting the fractions that way in line 2 is allowed

[lilac storm]

oh ok

so then how would i write out line 2?

[cedar pawn]

@lilac storm

Numerator is the derivative of denumerator

f'(x) =3t^2 {f(x)=t^3-3}

g(x)=t^2

So the only difference is the 3

So just put the 3 before the integral and u get 1/3integral of (1/t)

[cedar pawn]

f(x) =t

The formula is integral of f'(x)/f(x) = ln|f(x)|

[edgy blaze]

https://dontasktoask.com/

[lilac storm]

sorry, i wasn't here, can you explain where are you getting all of this, i'm confused?

[lilac storm]

are you there?

[cedar pawn]

From my textbook

Whenever the numerator is the derivative of the denumerator u can use that cheat

[lilac storm]

but it isn't because the derivative of the denominator is 3t^2 and the numerator is t^2?

[cedar pawn]

Yeqh5

Only difference is 3

Put it before the integral u get 1/3

[lilac storm]

i don't get what do you mean by put it before the integral?

[cedar pawn]

1/3integralofsomething

,w integral of x^2/(x^3-3)

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Z = Z²+c#2374

[cedar pawn]

^

[lilac storm]

sorry can you tell me where are we getting the 1/3 from?

[cedar pawn]

From the 3 of 3t^2

3t^2dt=du => t^2dt=du/3

[lilac storm]

ok

[harsh cipher]

Hello

Can someone help me with this question

[opal kettle]

ig ask question first

[opal kettle]

well at least thatll help me know if i can solve it to begin w/ kek

oh no word problem xd

[harsh cipher]

haha

c^2 = a^2 + b^2

I think a = 80 because it tells us that the aircraft is 20km closer to station b.

[opal kettle]

im confused on how this navigation system works xd

[harsh cipher]

Radio transmitter situated in a straight line.

Center (0,0)

Center to station B is 100km

[opal kettle]

oh nvm understood it

so points A and B on a plane 200 units apart

find the loci of pts P=(x,y) such that PB+20=PA

which define a hyperbola

[harsh cipher]

hyperbola

(x-h)^2/(a^2) - (y-k)^2/(b^2) = 1

need to find value of a and b using pythagoras

[opal kettle]

oh i dont have the formula memorized but

u could also just use

[opal kettle]

^

let A = (-100,0) and B=(100,0)

and then use the distance formula

[harsh cipher]

Yes

We don't know the coordinates of the plane except that it's 20km closer to station B

or aircraft

[opal kettle]

@harsh cipher

so I just let P=(x,y) be the location of the plane

and they said PB+20=PA

so then i used the pythagorean theorem /distance formula and simplified the expression

[harsh cipher]

Center to the vertice is the value of a. Center to the foci (100,0) is the value of c.

[opal kettle]

or x^2 - y^2/(30sqrt11)^2 = 10^2 if u want it in that form

[harsh cipher]

I'm not sure why you used the distance formula here.

[opal kettle]

oh thats cz i dont have the a/b/c stuff memorized oops xd

[opal kettle]

oh ok

[harsh cipher]

In that image the question was different.

[opal kettle]

c = half the distance between the foci

so c = 200/2 = 100

[harsh cipher]

It said the difference of the distance from boat to A and B as 386km

[opal kettle]

the center, we could choose so i chose the origin to simplify things

[harsh cipher]

In the question I asked, the aircraft is 20km closer to station B

[opal kettle]

so (h,k) = (0,0)

now a^2+b^2=10000

[harsh cipher]

100-20= 80

a = 80

[opal kettle]

wait where r u getting a = 80

[harsh cipher]

100-20?

[opal kettle]

why?

[harsh cipher]

the plane is 20km closer to station b

[opal kettle]

oh shoot did i mess up somewhere?

[harsh cipher]

or I did

[opal kettle]

the plane is 20 closer to station b

which means

20 = (100+a) - (100-a) => a = 10

try drawing a diagram; it might help

[harsh cipher]

You're right

🙂

@opal kettle Thanks for helping

[tough trail]

answer is C

all i did was substitued numbers, the easiest numbers to use is 1 and 5 since they are inclusive.
u see where it says -2X - 2 for all of X which is equal to 1 or less, substitute 1 since the rule applies to it and then see if it correlates with the graph or not, if it does test another number such as 5 to double check. @rain monolith

[past meadow]

square x and rewrite one of the cos^2's in terms of sin^2 so you can use pythag identity

[rustic glacier]

i need helpp

[lilac storm]

can anyone please help me with this question?

[still depot]

u-substitution

Set u to t^3-3 find derivative cancel t^2 solve u terms and then substitute back in

[lilac storm]

so i have u=t^3-3 and du/dt=3t^2, so what next then?

[sick steppe]

do the sub...

$u=t^3-3\to \frac{1}{3}\dd{u}=t^2\dd{t}$

[obsidian monolith]

moshill1

[lilac storm]

oh ok

sorry but i am stuck again so i have this so far

what do i like do?

[still depot]

So you have du/dt = 3t^2

Isolate dt

Substitute dt for du/3t^2

[lilac storm]

so du/3t^2

[still depot]

Yes

[lilac storm]

sorry so i don't get the substitution part

what do i sub in into what?

[still depot]

So u have t^2 / u as of now

Substitute dt for du/3t^2

So you have integral of t^2 / u dt

Make that now (t^2 / u) * (du/3t^2)

[lilac storm]

sorry so where did you get the t^2/u?

so i have du/3t^2=dt

so what now?

[lilac storm]

are you there?

[still depot]

Uh yeah

You susbsituted u for t^3-3

[lilac storm]

so i have du/3t^2=dt

[still depot]

Thats ur original part

[lilac storm]

so what do i do now?

[still depot]

cancel

if u have t^2/u dt and u want to get rid of dt just substitute du/3t^2 so you just have t^2/u * du/3t^2

[lilac storm]

sorry so i don't get where do we get this t^2/u from?

[still depot]

ur original prolem

problem

[lilac storm]

oh sorry my bad, being stupid

[still depot]

np

[lilac storm]

so then you have du/3u, right?

are you there?

[sick steppe]

Are you able to do work by yourself @lilac storm ?

You've gotten help, why do you need the answer spoon-fed?

[lilac storm]

sorry about that

so i'm still confused, so i have du/3u so when will i be writing everything in integral form, right now i'm just substituting and simplifying so i just need help that is the du/3u correct so what will be my next step?

[sick steppe]

if everything's in terms of u (and was correctly substituted), integrate...

if you have t's and u's in the same integrand, dont integrate until it's all 1 or the other

[still depot]

Yah i think u should watch this video first

https://youtu.be/IAh00vU3FSY

[eternal stone]

sqrt(25x) = 5(sqrt(x)) is this legal?

[sick steppe]

yes

[eternal stone]

bless

[sick steppe]

by exponent laws

[eternal stone]

the part underlined by the pen

hopefully this is all valid

(x^2 + 1)^(3/2) = sqrt((x^2 + 1)^3))

is this also valid?>

nm wolfram says it is

[jolly raven]

Ye bc that's definition of fractional exponent

[lilac storm]

can anyone please help me start off part a?

[lilac storm]

anyone there to help?

[sick steppe]

a's just integrate then use the initial conditions

[lilac storm]

if you don't mind, can you help me a bit with the integrating for this eq?

[sick steppe]

u-sub

[unborn nimbus]

hello does anyone know how to solve this? do i have to draw the graphs first or...

[tough trail]

yea draw graphs just sketch it nothing fancy plot few points and u will get the answer straight away

[dusk oriole]

Can someone help me find the sin equation of the indicated period?

[stark fulcrum]

can someone please help with this?

[eternal stone]

to be clear ln(1/sin(x)-1/tan(x)) is the same as ln|cscx + cotx| right?

[eternal stone]

what have you tried?

[viscid thistle]

not really.

first off, you pulled out a plus sign out of nowhere.

[eternal stone]

oops

im mostly concerned with the 1/sin = csc

and 1/tan = cot

[viscid thistle]

and while one has the absolute value, and the other one does not, both aren't equal when the argument of the first is negative, ie where 1/sin(x)-1/tan(x)<0

but yeah, if you were to fix that, $\\ln(\frac{1}{\sin(x)}-\frac{1}{\tan(x)})=\ln(\csc(x)-\cot(x))$ is totally correct.

[obsidian monolith]

Al𝟛dium

[lilac storm]

so i have solved this question but the answer in the back of my textbook says S(t) = 2.3e^0.234t + 0.6, but the answer that i am getting is 2.3e^0.234t + C, so how do i get the 0.6, if anyone there that can please help me out? like how do i solve for C?

[lost bane]

how do i do this

[lilac storm]

anyone there to help?

[unborn nimbus]

does anyone know how to solve this? its going to be on my exam later but i've never learned how to do these type of questions

[faint nest]

is the range of arctan (-pi/2, pi/2) or is that with brackets??

[mild swan]

parentheses

[mild swan]

If you don't know about logs, I'd look them up on mathisfun.com

or other online sources

[unborn nimbus]

ill do that thank you!!

[mild swan]

❤️

🙂

@unborn nimbus https://www.mathsisfun.com/algebra/logarithms.html

[unborn nimbus]

ah okay for that one i've watched a video about already

but i think thats all i know

[viscid thistle]

Just use Photomath and you can scan the problem and it tells you the answer and also shows you the work to get the answer

I mean sure it’s cheating but it’s a good way to learn by reading the ways to get the answer given

[unborn nimbus]

hahaha okayy!! thank you sooo much

[civic sierra]

@unborn nimbus if you want to I can go over a couple of those with you

[unborn nimbus]

i've actually used photomath, should i send you my working to make sure?

[civic sierra]

sure

[unborn nimbus]

okay should i send a pic here?

[civic sierra]

yeah

[civic sierra]

yep

that's the way to do them

[unborn nimbus]

okay, thank you so much!!

[stark fulcrum]

oh I ended up solving it

[stark fulcrum]

I think it’s wrong but it’s ok

sorry it’s all over the place I kept running out of space LOL

[viscid thistle]

anyone know this?

[willow bear]

im sure a lot of people do

but this looks like a test

[viscid thistle]

no lmao

[willow bear]

1 point

why are there point values for questions if its not a test

[viscid thistle]

we're just supposed to read a lesson in our textbook

and the answers are in it

[willow bear]

great then why not read it

[viscid thistle]

but i cant find this

[willow bear]

then think about it

[viscid thistle]

ok

[blissful ridge]

Ask mods, they generally handle these kind of requests

[echo wagon]

Pinging the mods for your own banning

[stuck lark]

👢

[slender raptor]

Finding this matrices questions a bit hard

[willow bear]

direct calculation?

,w simplify (sin(a) + cos(b))(sin(a) - sin(b)) - (cos(b)+cos(a))(cos(b) - cos(a))

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ann#0413

[willow bear]

this won't simplify, will it

was that your issue? @slender raptor

[slender raptor]

Yup, that's it

I tried various trig formulas and none of it worked

[willow bear]

yeah looks like it ain't your fault

the question is fucked

[slender raptor]

We can't apply properties of matrices to solve it?

I'm pretty bad with the properties so I haven't been able to figure out if it would simplify that way

Thank you btw!! @willow bear

[willow bear]

well it just isn't constant

i had WA simplify it right there

[raven tusk]

anyone able to help me with a question real quick?

[lilac storm]

so i have solved the question but the answer in the back of my textbook for this question is S(t) = 2.3e^0.234t + 0.6 and the answer I have is 2.3e^0.234t + C, so my question is how do i find C, if anyone there that can please help me out?

[viscid thistle]

https://prnt.sc/10umcjf

[sudden bluff]

help? pls

[bold lintel]

You need to set the problem =0 and then solve for x

Plugging in your answer will tell you if you have one solution, 2 or none

[lilac storm]

so i have solved the question but the answer in the back of my textbook for this question is S(t) = 2.3e^0.234t + 0.6 and the answer i have is 2.3e^0.234t + C, so my question is how do i find C, anyone there that can please help me out?

[sick steppe]

you're given a boundary condition

S(0)=2.9

[lilac storm]

sorry isn't S(0)=2.3 + C and if you rearrange then C=-2.3?

[sick steppe]

,w integrate 0.528434e^(0.234t) dt

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by moshill1#9771

[sick steppe]

not sure how you got -2.3 from 2.9-2.3

[lilac storm]

sorry where are you getting the 2.9 from?

[sick steppe]

the boundary condition

S(0)=2.9

[lilac storm]

sorry but if you don't mind, can you tell me what is the boundary condition?

[sick steppe]

S(0)=2.9

that's the boundary condition

at t=0, S is 2.9billion

[lilac storm]

no, the concept or like the definition of boundary condition, what exactly are you calculating which might help me understand how we get 2.9?

[sick steppe]

Im reading the question

"The expenditure on media spending on ads in 2012 stood at $2.9 billion"

[lilac storm]

i'm so sorry, thank you so much for helping out, i really appreciate it

so i just want to make sure so whatever so far i have written in my solution which i have also posted above, underneath i just simply write now that S(0)=2.9 so 2.9-2.3 which gives 0.6 which is the value of C, right?

[sick steppe]

yes

[lilac storm]

ok, thanks a lot once again.

[hexed depot]

hello kind fellows

I NEED HELP!!!!!!

[small trail]

SHE NEED HELP IMMEDIATELY

[hexed depot]

https://prnt.sc/10v9qpp

may someone help me walk thru this Plz

[sick steppe]

what are you stuck on?

[hexed depot]

im confused what does it mean by the most complicated side

math is not my good subject but im trying to learn

[sick steppe]

well which side looks more complicated?

[hexed depot]

um the cos squared ect

[sick steppe]

something with 3 trig ratios or something with just 1?

[hexed depot]

left side

[sick steppe]

yes

so typically you want to start with the messier side of an identity and simplify it, cause we can just simplify stuff better

[hexed depot]

okay so im just simplifying it?

[sick steppe]

simplifying/molding it into the right side

[hexed depot]

YAS OKI

[unborn blade]

How do I do question B?

6B***

i got A correct

but i keep getting 0/0 for B

but the correct answer is 4

[mild swan]

@unborn blade send your work

[unborn blade]

whats that

[mild swan]

@unborn blade you can factor the numerator

sry that's a derivative thing

[unborn blade]

i tried factoring but didnt work

wait or do you mean before i adding the -5

ok yeah that would factor

so i guess my mistake was adding -5

[mild swan]

3h^2 - 2h - 1

factors

[mild swan]

nope, this is good

$3h^2 - 2h - 1 = (3h+1)(h-1)$

[obsidian monolith]

Shen

[unborn blade]

uh oh i made a typo when calculating the discriminant

i multiplied by 1 instead of -1

so i assumed it didnt factor

tyty

[mild swan]

👍

[unborn blade]

@mild swan if you're still there can you confirm whether I did 6c correctly?

[ornate lintel]

$$Tan(\theta)>1$$ in right triangle?

[obsidian monolith]

lebedevhツ

[fallow slate]

woah how did you do that

tan is greater than 1 when sin is greater than cos

[ornate lintel]

is this affirmation true?

[mild swan]

@unborn blade looks good to me

[unborn blade]

thank youu

[viscid thistle]

hello

anyone on?

[lilac storm]

can anyone please help me with this question?

[sick steppe]

integrate, apply the boundary condition, then solve for the debt in 2012

[sick steppe]

Can't read that

,w integrate 0.070251t^2-0.51548t+2.1667

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by moshill1#9771

[sick steppe]

looks right

[lilac storm]

yes, that is what i got when integrating the eq

[sick steppe]

Ok so apply the boundary condition

[lilac storm]

and we know that R(0)= 10.025

[sick steppe]

Not R, whatever you call the integral of R

$S(x)=\int R(x)\dd{x}$ (for example)

[obsidian monolith]

moshill1

[sick steppe]

then S(0) = what?

[lilac storm]

yes

S(0)=10.025

[sick steppe]

yes

so that makes C=?

[lilac storm]

so now do i rearrange the eq that i just found and solve for C?

[sick steppe]

Yep, but hopefully you'll see you dont need much re-arranging

[lilac storm]

yes i can factor it and so on

[sick steppe]

No

all the terms except C have t, and you're plugging t=0

so anything with a t disappears, leaving you with S(0)=C

[lilac storm]

oh, yes

[sick steppe]

so what's C=?

[lilac storm]

0?

[sick steppe]

No

[lilac storm]

C=10.025

[sick steppe]

S(0)=10.025
S(0)=C

C=S(0)=10.025

[lilac storm]

ok so i just want to make sure but the answer in the back of my textbook, the answer for this question says 16.067

[sick steppe]

yeah, have you found the debt in 2012?

[lilac storm]

yes

[lilac storm]

are you there?

[sick steppe]

well if you found the debt in 2012, then you're done

[lilac storm]

so i don't get why my answer is so different from the one in the back of the book?

[sick steppe]

double check your math

and the back of the book might be wrong

[lilac storm]

so i am double checking my math now and just want to make sure that this this 2012 year value which is 4, i am subbing it into the original eq, right?

[sick steppe]

yes you find S(4)

[lilac storm]

oh, i found the answer

ok, thank you so much for your time and help once again.

[solemn flame]

Please excuse the numbers around this equation. I want to learn how to solve this on Gaussian Elimination but I don’t understand it.

[willow bear]

have you done Gaussian elimination before?

[mystic basin]

can someone help me in #help-4 its precalc related ;)

[willow bear]

.-.

[mystic basin]

sorry

[willow bear]

@solemn flame ?

[solemn flame]

@willow bear yes

[willow bear]

is that a "yes i'm here" or "yes i have done GE before"

[solemn flame]

Not for a long time tho

[willow bear]

ah

[solemn flame]

Yes I’ve done GE before

Sorry

[willow bear]

i see you've written out the matrix for your system below

$\left[\begin{array}{ccc|c} 2 & -1 & 2 & 10 \ 1 & -2 & 1 & 8 \ 3 & -1 & 2 & 11 \end{array}\right]$

[obsidian monolith]

Ann

[willow bear]

so you know the three types of elementary row operations, right? (row addition, row swapping, row scaling)

[solemn flame]

Yes

[willow bear]

okay

[solemn flame]

I swapped R1 and R2

[willow bear]

great, that's what i was gonna suggest

$\left[\begin{array}{ccc|c} 1 & -2 & 1 & 8 \ 2 & -1 & 2 & 10 \ 3 & -1 & 2 & 11 \end{array}\right]$

[obsidian monolith]

Ann

[willow bear]

we now have this

did you do anything afterwards?

[solemn flame]

I timed row 1 by -1

[willow bear]

mkay

[solemn flame]

Then added R2

[willow bear]

you... added row 2 to row 1?

[solemn flame]

Yes

[willow bear]

hm

a bit unorthodox, but legal nonetheless.

there's a shortcut here that just became apparent to me, by the way.

two shortcuts, even.

[solemn flame]

Really

[willow bear]

subtracting row 2 from row 3 (in the matrix i wrote out above) will give [1 0 0 | 1]

and subtracting 2 times row 1 from row 2 will give [0 3 0 | -6]

which gives you x=1 and y=-2 almost for free

[solemn flame]

Can U explain that first part

Please

[willow bear]

the part about subtracting row 2 from row 3?

[solemn flame]

Nvm

I completely dismissed that you were subtracting Nvm

[willow bear]

do i need to continue?

[solemn flame]

No I got it

[willow bear]

ok

[solemn flame]

Thank youuuu

[severe sandal]

[slender raptor]

is f: I --> R defined as f(x)=x^3 a one-to-one function?

[willow bear]

what is I?

[harsh smelt]

no need in particular info on I tho

x^3 is one-to-one on the whole R

[willow bear]

how do we know $I \subseteq \bR$ tho

[obsidian monolith]

Ann

[willow bear]

if ${1, e^{2i\pi/3} } \subseteq I$ then $f$ won't be injective

[obsidian monolith]

Ann

[harsh smelt]

true

#precalculus tho

[willow bear]

sometimes people ask questions about complex numbers here.

[slender raptor]

Ahh this topic came under precalc for me that's why I asked it here

[slender raptor]

Isn't I imaginary numbers?

[viscid thistle]

no that's C

[sick steppe]

i is the imaginary unit, C is the set of complex numbers, I usually denotes a general interval

[viscid thistle]

if i send math problems here, could i get help ?

[willow bear]

yes, and in fact that's exactly what you need to do to get help here

can't help you if we don't know what to help with, if you catch my drift

[viscid thistle]

yeah i got that haha

i'll be needing help soon, cheers

[viscid thistle]

how do i do this problem?

[willow bear]

try to think about building your word letter by letter

[viscid thistle]

so can I put b any where?

like in the last letter of the four word letter

[willow bear]

there are 4 possible places for you to put the B

[viscid thistle]

is there some kinda formula

[willow bear]

no, just think about the multiplication principle

after you place the B, there are 3 slots left to fill with a letter each

the first slot can be filled by any of the 5 remaining letters

afterward, the second slot can be filled by any of the 4 letters that remain

[viscid thistle]

5 times 4 times 3

times 2

right?

[willow bear]

no

don't forget to multiply by 4 for the possible positions of the B.

and the 2 should go.

[viscid thistle]

so 5 times 4 times 3 times 2 times 4?

why shouldnt the two go?

nvm got the answer thanks

[willow bear]

the 2 should go.

[viscid thistle]

so answer is 480?

5 4324

[willow bear]

no! the 2 should go!

[viscid thistle]

oh lol

okay then

[willow bear]

at no point are you making a choice with 2 options in this problem!

[viscid thistle]

how about this one?

[echo wagon]

Hold up

That is definitely a test

[willow bear]

oh yeah, good point

it is

[echo wagon]

There are even instructions to prevent cheating

Lol

[willow bear]

there are point values

[viscid thistle]

its a practice test

for the real thing

[echo wagon]

Can you take a picture of whole page?

These instructions are very sus

[uncut mulch]

unmute mic? proctored?

[willow bear]

sus

[viscid thistle]

can u pls help me out ?

[willow bear]

from #❓how-to-get-help

6. Do not ask for help on tests. Any violation of this will lead to appropriate action being taken at mod discretion.

<@&268886789983436800>

[uncut mulch]

if you can alleviate out concerns that this is an assessed test

[obtuse locust]

Do u think I picked the right answer

[sick steppe]

Assuming convention, no

[obtuse locust]

The one right to it?

[sick steppe]

I mean you tell me

[echo wagon]

Is 16 a1 or a0 for you?

Normally sequences start at a1

But you could be starting at 0

[obtuse locust]

it doesn’t say but probably a1

[echo wagon]

Then clearly the one you chose doesn't work

Because 16+2(1) = 18 not 16

[obtuse locust]

Oh then it’s the one next to it

[slender river]

were they banned yet or no

looks like it

[haughty pendant]

just gonna put this here

#help-2 message

i'm stuck on this problem

[misty imp]

If theres an angle theta where csc(theta) = -35/24 and tan(theta) > 0, what are the ratios of: sin(theta) = , cot(theta) =, and cos(theta+pi) =

can someone help me out on this one

[lilac storm]

i just want to ask that have i done this question correctly?

and for f, so i have the area which is 17.5 found in a so do i just simply write in f that as we increase n, the area becomes more accurate and to the actual exact value as you can see from each graph respectively

and here's part b as well

anyone there to help?

[lilac storm]

is anyone there to help?

[fading token]

looks good

[lilac storm]

thank you so much

[willow bear]

derivative

$f'$ is the derivative of $f$

[obsidian monolith]

Ann

[willow bear]

leibniz notation

[upper kelp]

You don't need to go back to square one. It's just a different notation.
The $f'$ notation is the one most commonly used for single-variable functions, whereas the $\frac{d}{dx}$ notation is required in case you have functions with multiple variables.
As an example, if you had the function:
$f(x, y) = 3xy + y^2$
then, when you differentiate, you have to specify which variable you're differentiating by. For example, if you differentiate by x, you get:
$\frac{df}{dx} (x_0, y_0) = 3 y_0$
and if you differentiate by y, you get:
$\frac{df}{dy} (x_0, y_0) = 3 x_0 + 2 y_0^2$.
(The point $(x_0, y_0)$ is the point at which you measure the derivative of the function.)

If it makes more sense, as an example, in Leibniz notation, the sum rule can be written as:
If $h(x) = f(x) + g(x)$ [another way to write it would be $h = f + g$], then: $h'(x) = f'(x) + g(x)$.
In the $\frac{d}{dx}$ notation, the rule would be: $\frac{d}{dx} h(x) = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)$.

[obsidian monolith]

RoiKadmon

[vapid mica]

hello, i was wondering if someone could check what i did wrong with this problem:

https://www.calculator.net/triangle-calculator.html?vc=&vx=&vy=27&va=&vz=8&vb=65&angleunits=d&x=88&y=15

this gives me something completely different and im not sure why

any help is appreciated.

[uncut mulch]

there is ambiguity when applying the sine rule

where possible, apply the cosine rule instead to avoid situations like this

[stable yacht]

The question asks: “For which values of a and b does this function have a limit at x = 0?”

I think it’s the wording, but I don’t really know how to go about it.

Do I just set the limits approaching zero from the left and right equal to each other?

[copper vigil]

set the upper and lower functions equal to 5

[stable yacht]

Is that because at x = 0, f(x) = 5?

[upper kite]

Every question channel is being used, and no one has answered here for more than an hour now, so I'm going to go ahead and post my question here :

In the following image, the problem basically asks what's the probability of at least 2 people having the same birthday in a room of 30 people.
So what he's doing is getting the probability of no one having the same birthday, and well substracting it from 1 to get the answer we need

But I'm confused about this part, why is he using the permutation formula instead of the combination formula here ? It's not like it matters in which order the birthdays are, so I'm confused

[upper kite]

<@&286206848099549185> please

[rare galleon]

Can you state the difference between permutation formula and combination formula?

[upper kite]

From what I've learned, the permutation formula is :
nPr = n! / (n-r)!
whereas the combination formula is :
nCr = n! / ( r!(n-r)! )

[winter quartz]

Qualitatively: Permutations care about order, Combinations do not.

[upper kite]

Exactly, which is why I'm confused as to why permutation is used here
It doesn't matter in which order we're taking people's birthday here, as long as they're all different

[winter quartz]

So it's hard to tell based on your screenshot, but are you sure he is using the permutation formula? Perhaps the r! Got cancelled out.

[upper kite]

Do you want me to send you the link of the video instead ?

[winter quartz]

Or perhaps he is using permutation and making it a combination later for the final solution

[upper kite]

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:prob-comb/x9e81a4f98389efdf:prob-combinatorics-precalc/v/birthday-probability-problem?modal=1
That's the video if you prefer that, because it's hard to explain from screenshots

But at the end, to get the probability of no one having the same birthday, he's doing : Permutation / 365^30

[winter quartz]

I agree, and that is simply because that is how you mathematically express $365363\dots*335$

[obsidian monolith]

dackid

[winter quartz]

$365364363*\dots335=\frac{365\dots335334*\dots1}{334\dots*1}=\frac{(365)!}{334!}$

[obsidian monolith]

dackid

[winter quartz]

We weren't using the permutation or combination formula, we were simplifying our term here.

And it just so happens to be a permutation

Ah, they don't have the 335 in the problem.

Either way the logic is the same.

So $365*\dots*336=\frac{365!}{(365-30)!}$

[obsidian monolith]

dackid

[upper kite]

This is a weird one, I hope this isn't a common occurence.
But so far I've been able to work through every exercice using one of those two formulas. This one struck me as a combination one so that's how I interpreted it
It's still a bit fuzzy to me but I think that's just me, I'm gonna try to work through this another time and see if I get it

[winter quartz]

Don't focus on the formulas.

Focus on how he solved the problem. The logic is sound

[sick steppe]

Birthday Paradox?

[winter quartz]

This is the probability that out of 30 people, at least 2 people share the same birthday

[upper kite]

The thing is the logic he went though only works for this problem, and maybe a handful of other ones, what I'm trying to do is find a pattern to generalize one technique that works on a lot of problems (which I'm aware all problems are different, but there's always this one pattern you can figure out once you understand it correctly)

[winter quartz]

Combinatorics is a bit more involved than you are giving it credit for

If it was that simple, there wouldn't be college classes on the subject

[misty marten]

Hi, I have a question for trigonometry I don’t understand how to start this.

[upper kite]

I guess you're right, I'm just upset I guess that unlike everything I've learned so far, there's no magic formula that can help me breathe through every exercice

[winter quartz]

Do we agree that $\sec(\theta)=\frac{1}{\cos(\theta)}$?

[obsidian monolith]

dackid

[winter quartz]

If it worked like that, I'd have my degree years ago 😆

[misty marten]

Yes! It’s primary reciprocal

[winter quartz]

Okay, so we can rearrange the question

$\sec^{2}(\theta)=2$ is the same as $\frac{1}{\cos^2(\theta)}=2$

[obsidian monolith]

dackid

[winter quartz]

But that can be reexpressed as $\frac{1}{2}=\cos^2(\theta)$

[obsidian monolith]

dackid

[misty marten]

Ahh ok.

[winter quartz]

And squarerooting gives us two equations

$\frac{1}{2}=\cos(\theta)$ and $-\frac{1}{2}=\cos(\theta)$

[obsidian monolith]

dackid

[winter quartz]

I think you can solve from here.

Note, there should be four solutions for your final answer

[misty marten]

Ok I’ll try it out.

Wait 4 solutions?? How please explain.

[sick steppe]

2 from the positive 1/2, 2 from the -1/2

since cos is positive in 2 quadrants, and negative in the other 2 by CAST

[misty marten]

Ok thank you. See what you guys are talking about now.

Don’t laugh but am I getting somewhere? Lol

[sick steppe]

double check your exact values

[misty marten]

Should I not do the cos inverse?

[sick steppe]

I mean you can, but it's not 30 deg

[misty marten]

Oops I did the wrong sign. It’s 60 degrees

Ok thank you

Now I have my final answer as 300 degrees

[winter quartz]

60° and 300° are two of your answers

What about the other two?

[misty marten]

Oh for the second set I got 120 and 240 degrees

[winter quartz]

Yep!

You have all 4 solutions

[misty marten]

Yesss!!!😀

Thank you guys for helping me!! 💯💯

[winter quartz]

You betcha :)

[grand barn]

how many solutions can a biquadtrical equation have ? including complex solutions

[willow bear]

biquadratic?

do you mean specifically ax^4 + bx^2 + c = 0 or do you mean general 4th degree (quartic)

in either case, up to 4 solutions.

[fossil ravine]

Why does y^2 - x^2 = 2y^2 - π
draw a circle

the left side is a difference of two squares

-y^2 - x^2 = -π

y^2 + x^2 = π

Explanation please?

[echo wagon]

Rewrite in the standard form

y^2 + x^2 = π

[fossil ravine]

Ok

I wrote that bit

[echo wagon]

Well, how do you know if something is a circle? Because to me, being able to write it in that form makes it a circle centered at the origin with radius sqrt(π)

[fossil ravine]

Ok so that's how we get 1.77 as radius
Thanks @Lunasong#7197

[lilac storm]

can anyone please help me start off this question?

[uncut mulch]

consider the general equation for exponential growth
and what the variables represent

[lilac storm]

so i don't get how do you solve for k when k is not even in the eq, like i have solved a question similar to this one before but in that one you have k with the t but over here you don't even have k so how would you solve for it?

[uncut mulch]

consider the general equation for exponential growth

[lilac storm]

are you talking about the Q(t) = 500e^0.02t?

[uncut mulch]

no

the general equation i'm referring to involves stuff like k and t

[lilac storm]

oh, that is Q(t) = Qoe^-kt

[uncut mulch]

parentheses

[lilac storm]

yeah

[uncut mulch]

and _ for subscript

[lilac storm]

that's the intial value, right?

[uncut mulch]

and then compare that to the equation you're given

also it seems you've written decay instead of growth

$Q(t) = Q_0 e^{kt}$

[obsidian monolith]

ℝamonov

[lilac storm]

oh i think i get it, so i sub in 500e^0.02t in for Q(t) in the Q(t)=Qoe^-kt eq, right?

[uncut mulch]

no - in the exponent

[lilac storm]

sorry, where?

so i sub in the eq in k?

[uncut mulch]

no - (in the exponent)

compare:
$Q(t) = Q_0 e^{kt}$ with $Q(t) = 500e^{0.02t}$

[obsidian monolith]

ℝamonov

[uncut mulch]

Q_0 = ?
k = ?

[lilac storm]

Q(0) = 500

but i already have 500

[uncut mulch]

now compare the stuff in the exponent

[lilac storm]

do you mean kt=0.02t?

[uncut mulch]

yes

[lilac storm]

so i have k=0.02

[uncut mulch]

yes

[lilac storm]

so for b, do i use the Q(t) = Qoe^-kt formula?

because i have k now

[uncut mulch]

whut

Q(0) = 500
but i already have 500

[lilac storm]

oh, yeah sorry

[uncut mulch]

also you already have the equation. explicitly knowing the value of k is irrelevant

[lilac storm]

oh, ok

so for c, i just sub in the values in the eq that is given, right?

[uncut mulch]

yes\

[lilac storm]

ok, thank you so much for your help, i really appreciate it

[mystic basin]

theres a question on my preclalc homework

"If a graph is vertically stretched or shrunk, explain why points on the x-axis remain fixed." my answer would be like

"The y is modified because it is being stretched or shrunk, so the x remains the same" is this a valid answer?

[uncut mulch]

no

[mystic basin]

hm

why

[primal quail]

Can anyone explain this to me

Why is d/dx of lny=1/y dy/dx

I am confused about that part

Help is greatly appreciated

[onyx wharf]

chain rule basically

$\frac{\dd}{\dd x} f(g(x)) = f’(g(x)) \cdot g’(x)$

[obsidian monolith]

rjsp

[lilac storm]

can anyone please help me with this question?

[dim rampart]

for (a): $S=\frac{1}{2}*f(4) + \frac{1}{2}*f(4.5)$

[obsidian monolith]

veryhappyperson

[lilac storm]

sorry, if you could just explain to me where are you getting the 1/2?

and also the f(4.5)?

[dim rampart]

The interval [4,5] has the "length" of 1. I want to put to rectangles in there so I split it into 2 parts, each of them with a width of 1/2

[lilac storm]

i get f(4) because 4 is given but where is the 4.5 coming from?

[dim rampart]

I just split the interval up into n parts

The middle is 4.5

[lilac storm]

oh, ok

[sick steppe]

$\Delta x = \frac{b-a}{n}$ for n rectangles on the interval [a,b]

[obsidian monolith]

moshill1

[lilac storm]

so for (b), would it be S=1/5 x f(4) + 1/5 x f(4.5)?

[sick steppe]

no

since that's 2 rectangles.. not 5

and * for multiplication

[dim rampart]

You would have $S= \frac{1}{5}f(4)+\frac{1}{5}f(4.2)+\frac{1}{5}f(4.4)+\frac{1}{5}f(4.6)+\frac{1}{5}f(4.8)$

[obsidian monolith]

veryhappyperson

[lilac storm]

oh

[sick steppe]

$\Delta x \sum_{i=0}^{n-1}f(a+i\Delta x)$

[obsidian monolith]

moshill1

[lilac storm]

can anyone please help me with this question?

[sick steppe]

find the inflection point, then find the slope at that point and the point itself

[lilac storm]

so i have y' = e^-2 - e^-2 (x)

[sick steppe]

how did a constant show up in your derivative?

$f'(x)=e^{-x}(1)+x(-e^{-x})$

[obsidian monolith]

moshill1

[lilac storm]

yeah so i have that but i then subbed in 2 for the x's

[sick steppe]

,w solve d/dx (e^(-x)-xe^(-x)) = 0

right ok

so you need the slope at x=2 and the point at x=2

[lilac storm]

so i have f'(x) = e^-2 - e^-2 (x)

[sick steppe]

poor notation

$f'(2)=e^{-2}-2e^{-2}=\frac{-1}{e^2}$

[obsidian monolith]

moshill1

[lilac storm]

yes, ok

so that is the m value right?

[sick steppe]

yes

[lilac storm]

so i have y = -1/e^2x + b, so i sub in 0 for y and 2 for x to solve for b, right?

[vast beacon]

uhh brainfart

why are these equal

[willow bear]

why would they not be

a^n * b^n = (ab)^n

[vast beacon]

oh i was looking at it like a^4 * b^4n and thought since the powers arent the same

silly me

ty

[steep anchor]

@vast beacon

        [-1 * x^4]^n =
            (-x^4)^n```

[vast beacon]

yup thx

[steep anchor]

what does this mean?
{𝑥 ∈ ℜ / 𝑥 ∉ ℜ− ∨ 𝑥 > 6}
i think it mean X has to be 0 or +, and/or greater than 6
am i right?

[stark helm]

Definition of the 6 trig functions for reference

[lilac storm]

so i have almost solved all of the question but i am stuck on a step so i have y-2/e^2 = -1/e^2 (x-2)

so my question is, the thing that i need help with is that how do i get from that to the y=mx+b form?

anyone there that can please help me out?

anyone there to help?

would it just be y = -1/e^2(x-2) + 2/e^2?

[viscid thistle]

Did they give any (x,y) values?

[lilac storm]

no

it says in the question at its inflection point, that is why it is a lot more tougher

[dim rampart]

you have to calculate the inflection point by taking the second derivative and setting it 0, have you done that?

wait

[lilac storm]

yes

[dim rampart]

why is calculus precalculus? xD

[lilac storm]

yes, i got x=2

[dim rampart]

then plug the x value of the inflection point into the first derivative then you have your slope

for y=mx + c

[lilac storm]

yes i got that

[normal night]

: D

tf happened to this server

[lilac storm]

the slope is -1/e^2

[normal night]

damn

now we are 57651

the hell

🤡

Calculus is hard

[dim rampart]

now you put that into f(x) to get a y value

so you have two points and a slope

[viscid thistle]

I got y=x

[lilac storm]

so i put the slope value into the original eq?

[dim rampart]

no, you put the x coordinate of your inflection point into f(x)

[normal night]

💀

[lilac storm]

so i have solved to get the points which i got (2, 2/e^2)

[normal night]

😭

[viscid thistle]

No you have to find the derivatives of the given equation first

[normal night]

GUYS GRADE 11 MATH IS HARD WHY

😭

[dim rampart]

no it isn't.

just focus xD

[viscid thistle]

Like I did it in two ways and got two different answers lol

[normal night]

its the teacher

[lilac storm]

ok, so i'll tell you from the beginning, i took the first and second derivatives, then set the second derivative equal to 0 and got x=2

[viscid thistle]

X+ye^2=4, that’s what I got

[dim rampart]

Now let me summarise: You need to find the inflection point. Then you write the coordinates of that point down. Then you calculate the slope at that point. Then you use this

Then you have the equation of the tangent line

[lilac storm]

then i did f(2) which gave me the y point so i have (2, 2/e^2)

oh i think i get it

or no

[dim rampart]

(2,0.27) is correct

now use your knowledge from like pre algebra and use the formula I sent you

[lilac storm]

so what i'm getting is y = -1/e^2(x-2) + 2/e^2, is this correct?

[viscid thistle]

Like for the derivative?

[lilac storm]

no, as the final answer in y=mx+b form

[viscid thistle]

Umm I got a different one

[lilac storm]

oh

[viscid thistle]

I got x+ye^2=4