precalculus

[fading monolith]

And argue why tan(π/2) goes inf on the plot

[solar olive]

wouldnt that require u to know that tan(x) = sin(x)/cos(x) ?

[pastel fossil]

I have that as one of my answers, that sin and cos make tan

[fading monolith]

I assume that he knows already

[solar olive]

Oh, then yeah

since tan(x) = sin(x)/cos(x), tan(x) is undefined if cos(x) = 0

u could show that the points where cos(x) = 0 makes tan(x) have asymptotes

[pastel fossil]

Also, since they are all translations of each other, I can say that they are all related in a way

I got a lot of differences like the range, domain, cycle length etc

[solar olive]

eh, i would just say that cos and sin are a translation of eachother

"cos" is just a shifted sine wave and vice versa

they have the same shape

tan is the outlier

[viscid thistle]

11:25

[pastel fossil]

Thanks for the help

[solar olive]

Your welcome

[pure lily]

equate both of them to as powers of 2
4^x+5 = 2^ 2. (x+5) for example

do it for both and then calculate the x

if you have a problem with a specific one you can point it out

[hollow stump]

https://en.m.wikipedia.org/wiki/Lanchester's_laws hi, could anyone help me out? I want to know what math topics I should study to learn about the lanchester laws for a project

[grim valve]

Yo Wass up

[mental sonnet]

hello @grim valve

[gaunt carbon]

seems like all you need is a basic understanding of derivatives in calculus and an idea of what hyperbolic trig functions are

[amber ridge]

hey does anyone know how they broke 2ζω into ζω - ζω?

[twilit eagle]

because the original fraction had a minus at the front

so they break up into being both negative

also laplace in #precalc 💀

[amber ridge]

yea mb I thought I would put it here since it was an algebra thing lol

[proper creek]

needing help with carious questions, iv someone wants to vc and help for a bit i can pay 5 for every 30 min

[twilit eagle]

no that's fair lol

[hallow lodge]

what can i search on yt for problems with Normal, sample distribution connected with the z scores and percentile

[raw comet]

does anyone here have ap precalculus

[polar hemlock]

thats a thing now?? 😭

[raw comet]

yes bro

it is actually the worst thing ever

[polar hemlock]

oh

i just had pre calc honors 2 years ago

[raw comet]

i tried taking ap statistics but they refused to switch my class

[polar hemlock]

thats

depressing

what was the reasoning

[raw comet]

my pre reqs are gonna suck so bas

bad*

[raw comet]

for switching? it was just not my type, it still isn't. im literally here in the server to try to grasp pre calc better so i can have a chance at passing the AP test lol

[polar hemlock]

no i mean for ur counselor not being able to switch u

well thats good at least youre trying

[raw comet]

oh it was simply that she didn't want to

[thin roost]

Is 36.36(11.07)^x roughly approximate to 40(11.07)^(x-1)

the problem was find the value of k if the graph was exponential

X = 1, 5, 13
f(x) = 40, k, 135

my teacher got the second graph using a geometric sequence

i got the first graph using algebra methods

which is more correct

[shadow summit]

Do you know how to solve for common ratio of a geometric sequence?

Because with some modifications to it you can easily find the common ratio for this series.

[thin roost]

yeah

i didnt really make the geometric sequence correlation until after i saw her solution though

[shadow summit]

Fair enough.

[thin roost]

are both solutions correct? when i graphed mine it seemed to be pretty correct

my mindset was that

ab = 40
ab(b^12) = 135

and i solved for a and b

to get the exponential function

[shadow summit]

Here’s a pro tip, there are infinitely many solutions, so it really depends on how you are taught it in class.

[thin roost]

Got it

Thank you 🙏

[stiff oar]

can someone walk me through how to do this step by step

what I tried so far:
moving (4x+2)/(2x-3) over to the other side
then getting common denominators and solve the problem normally
I got to (-18x+7)/((x+1)(2x-3)) and I don't know what the next step is

[summer ruin]

what can you say about numerator and denominator of a fraction if you know that the fraction is negative?

[willow bear]

you mean that you got to (-18x+7)/((x+1)(2x+3)) <= 0, yes?

[solar river]

T-T

[obsidian monolith]

Removed the studying! role from you.

[tulip flicker]

Hello Guys!

good night

i am starting precalculus today

and i have a question: How i can apply my studies?

just in exercises?

[hexed apex]

I need to describe the transformations of f(-4x-2)-3. But I got it wrong can I get help

[fading monolith]

You mean like real life applications

?

[tulip flicker]

yeah

i am a programmer

i dont study much for think on a project yet

[fading monolith]

Precalculus is too basic to apply to programs in my opinion. Maybe calculus help you but for programming is better linear algebra

[willow skiff]

You need to replace x with x - 2

So -4(x - 2) becomes -4x + 8

The best way to avoid this problem is to do the shift first, and then the compression

Alternatively -4(x + 1/2) does give you -4x - 2

[dry fractal]

lim x-> infinity of f(g(x)) is equal to f(lim x-> infinity of g(x))?
I am getting conflicting information

[lean tundra]

For composite function $\lim_{x\rightarrow +\infty} (f\circ g)(x)$ $\lim_{x \rightarrow +\infty} g(x)=a$ then $\lim_{x\rightarrow a} f(x)=b$ then $\lim_{x\rightarrow +\infty} (f\circ g)(x)=b$

It's basically what you showed

[obsidian monolith]

Birby

[lean tundra]

Note that a and b can represent either reals or+infinity or -infinity

[dry fractal]

Thank you!

just making sure, same logic if the initial lim is to any value right?

[lean tundra]

yep

[spiral sonnet]

I'm not quite sure what I've done wrong here, the graph seems identical except between 2-5 on the x axis

And obviously that the graph is going down on the correct one at 5 😂

Oh wait, I f igured it out I think

Yep. I didn't think that -4 was a very smooshed parabola, once I fixed that it worked out when looking back for a

[fossil badge]

Hello fellow AP precalc people

[lavish heath]

Can someone find the equation for this graph?

[shadow summit]

its a quartic, so do foot form of quartic

[lavish heath]

I guessed it. Idk how

[shadow summit]

looks good.

[willow bear]

you need to scale it so that the y intercept is -48 tho.

[rotund oyster]

i need help graphing 😭😭😭🙏

are we all on the same unit?

[shadow summit]

Wdym?

[rotund oyster]

are we learning the same unit?

[shadow summit]

No

[rotund oyster]

aw

[shadow summit]

I’m still able to help if you are still confused with the question.

[rotund oyster]

YES PLEASE KIND SIR 🙏🙏🙏

[shadow summit]

Hold on, I’m gonna get it up on my computer

[rotund oyster]

ok ok

[shadow summit]

we have the function $g(x)=2\cos^{-1}(-(x+1))$ and we want to graph it.

[obsidian monolith]

TheLord26

[shadow summit]

first consider what the 2 does, it affects the amplitude of the function.

[rotund oyster]

yuh

[shadow summit]

and we know that it has a y domain of [0, pi] initially, right?

[rotund oyster]

?

[shadow summit]

the height of the function.

i meant pi

[rotund oyster]

oh it has a range of 0,2pi

[shadow summit]

*pi, sorry mb

[rotund oyster]

wait wah

[shadow summit]

if you look at the graph of the original function it only goes up to pi.

[rotund oyster]

uh huh yeah

[shadow summit]

so the new function will have twice the amplitude

meaning it will go from 0 to 2pi.

now, as for the difficult part of this question

you need to determine how the range is affected (these things will help you graph it)

i would recommend expanding the inside

,,2\cos^{-1}(-x-1)

[obsidian monolith]

TheLord26

[rotund oyster]

ok im lost here

[shadow summit]

the two makes the graph more elongated (red is original amplitude, blue is new)
dont mind my shit drawing which is completely wrong for the x-axis

[rotund oyster]

oh

[shadow summit]

now we need to figure out how it is affected along the x axis

firstly, identify that the minus sign in front of the x means the function will be flipped.

[rotund oyster]

ok ok

[shadow summit]

now we idetify that the -1 inside the function moves the graph left 1 unit.

[rotund oyster]

yeah

so it’s kinda like that?

[shadow summit]

nice

that is correct

just make sure you make it more clear

it should be much more obvious that it goes up to 2pi

[rotund oyster]

makes more sense now tho

[shadow summit]

thats good

i will say inverse trig functions are pretty difficult

[rotund oyster]

i thought the 2 changed the x value

[shadow summit]

do you know how to graph regular trig functions?

[rotund oyster]

yeah thats light

[shadow summit]

$a\cos(bx-c)$

[rotund oyster]

a is amp

b is period

c is phase shift

[shadow summit]

if you can understand how each of those variables affect the cos function, then understanding how they affect inverse trig functions becomes much easier

[obsidian monolith]

TheLord26

[shadow summit]

**made a mistake, should be -c

[rotund oyster]

wait why?

[shadow summit]

its just easier to understand with a minus sign

because if c is positive, it moves the function in the positive x direction.

in the other form it would have to be negative to move the functino in the positive x direction

[rotund oyster]

?

[shadow summit]

try graphing it on desmos!

[rotund oyster]

i thought -c meant right c and +c meant left c

for cos

[shadow summit]

yeah it does

[rotund oyster]

oh

[shadow summit]

can you see how the green graph is moved left even though c=1, the minus sign just corrects that.

[rotund oyster]

yeah

[shadow summit]

same with inverse trig

[rotund oyster]

wait for the inverse of tan, a changes the horizontal asymptotes right?

[shadow summit]

it looks like a regular tan graph, but flipped on its side

[rotund oyster]

this what i got so far

[shadow summit]

good

[rotund oyster]

the 2 still confuses me

[shadow summit]

you just have to double the amplitude

make sure to keep the "center"

[rotund oyster]

erm... what center

[shadow summit]

do you see where it passes through (0,0) in the original function?

[rotund oyster]

yeah

[shadow summit]

i like to think of that as the center of the function (at least for tan and inverse tan)

[rotund oyster]

oh

so the 2 changes the range right?

[shadow summit]

no

the amplitude

this is the graph with only the amplitude affected (red is tan^-1(x), blue is 2tan^-1(x)

[rotund oyster]

oh

OHHHHHH

i get it now

[shadow summit]

thank god my art skills are good enough to help people understand math.

[rotund oyster]

real

[shadow summit]

good job

[rotund oyster]

that’s all for tn thank god 🙏🙏🙏

i have no more graphing

[shadow summit]

nice

graphing definitely isnt my favourite part of mathematics.

[rotund oyster]

fr

[rotund oyster]

wait these have to be between pi and negative pi right?

(ignore reflection)

(not done with ca2 9)

[river drift]

not quite. arcsine outputs values between -π/2 and π/2, and arccos outputs values between 0 and π

[rotund oyster]

i turned it in already 💀

[rotund oyster]

noted

[river drift]

which basically just comes from considering the smallest interval containing 0 that outputs all possible values of sine and cosine respectively

[rain scroll]

whats after pre-calc?

i wanna get ahead

[fading monolith]

Calculus ig, limits, sequences, series, continuity, derivatives,Taylor, Riemann integration

[honest hill]

Guys please explain,i dont understand calculus,i understand pythagoras theorem,area of triangle,square,square roots,negative and imaginary numbers,exponents,but not this.

[rancid forum]

I find this approach to solving radical equations a bit frustrating.

Considering we algebraically work our way to the statement x = 4/3 which is false, we must have performed some invalid logic, right? What is this invalid logic and could we have avoided it in the first place?

[summer ruin]

squaring both sides

the original equation only makes sense if x >= 1 and x >= 3

[unique oasis]

yup. Just need to note the restriction in the radicand

[rancid forum]

I understand x>=1, but why x >= 3?

[summer ruin]

if you hope for sqrt(3x) - 3 to be equal to another square root, you better ensure that sqrt(3x) - 3 >= 0

[unique oasis]

Can't have negatives under a square root

[summer ruin]

no, that's not why there's a restriction to x >= 3

a square root of something can't be equal to a negative number

which is why x = 4/3 didn't satisfy the original equation, even though it's both positive and bigger than 1

it failed to meet x >= 3

[rancid forum]

ahhhh i got it

thx for the help guys that made sense, but i'll need to study this for sure 🤓

[rancid forum]

tl;dr: i just need to keep in mind that a^2 = b^2 does not imply a = b

[rotund oyster]

precalc: ap stats or calc ab

[winter plover]

how would you split the function |x^2-4|/x-2 into its peice wise function and domains

[fading monolith]

Remember the definition of abs value

[uneven sequoia]

hi guys, am I starting this right?

[viscid thistle]

Bruh can’t you write properly or what

[uneven sequoia]

lol srry

is getting the y value the correct first step for each of them or no

oh I realised it's y=mc+c nvm

[rugged crater]

if i'm reading these questions right you are trying to find an equation parralel to the other equation right?

so for parallel on the first one you can just change the y intercept

you can just change the y intercept because it is already in y=ax+b form and all you have to do is change b

the rest of them it is the same, just make it so there is just y on each side first

[zealous mango]

Can someone pls pls pls help me with bearings Im so confused 😦 Heres a question: The bearing from the Pine Knob fire tower to the colt station fire tower is N65°E and the two towers are 30 kilometers apart. A fire spotted by rangers in each tower has a bearing of N80°E from Pine Knob and S70°E from colt station. Find the distance of the fire from each tower. I know how to solve them I just am so confused on how to draw them.

[uneven snow]

I need help with inverse trig domain restrictions aswell as graphing all the sinusoidal functions

[tired sorrel]

I’m going into precalc starting this year in August. What should I know about precalc before going into the class.

[shadow summit]

Just know north is 0 degrees, and north is “upwards” when you draw.

And you go clockwise when talking about degrees.

[willow bear]

the usual convention when drawing diagrams for these is that north is up

bearings in the format you gave them are read as "this many degrees from <north/south> to <east/west>", but typically one also sees bearings as a single number from 000 to 360

and that is the angle as measured clockwise from north

does this address your confusion?

[trim cedar]

yo guys quick question, so in limits, like it says x --> c^-, what does the negative mean there, sometimes its a postive and im not sure what the difference is between them

i think its just the direction on the graph but someone please clarify 🙏 ^

nvm no need to answer i figured it out

[lucid trench]

oh you use the same textbook as me! dope

[proven void]

,tex $$\lim_{{x \to +\infty}} \frac{2x^2}{x^2 + 1}$$

[obsidian monolith]

レナト (renato , ping if reply)

[proven void]

guys how do I calculate the limit of this ?

[river drift]

dividing each term by the highest power of x is a good start

[proven void]

thanks

that did the trick

[minor nexus]

I need help solving rational equations. Can anyone help me with that?

[chrome ether]

give a specific example

[willow bear]

we can't really help you much if you don't show us examples of what you are struggling with.

[willow bear]

ok, and can you show all your progress on this problem?

if you have made absolutely zero progress, as in not even a failed attempt, then say "I have no idea where to even begin with this problem."

[chrome ether]

oh dear what happened here 😭 your first step is incorrect

so usually the first thing we want to do with these rational equations

is get rid of the fractions

is there something we can multiply both sides by to get rid of the fractions?

[willow bear]

can you explain what you tried to do in the first step? not only is it incorrect, i for one can't make heads or tails of it at all.

anyway there's multiple ways to solve this problem. i see at least 2, slightly different but largely equivalent both logically and in terms of difficulty.

maybe i even see 3.

[minor nexus]

I think that this was copied from a tutor. Let me see if I can find the transcript.

[willow bear]

do you understand what the tutor has done in the first step?

or would you like us to explain

[minor nexus]

Can you explain?

is it distributing?

[willow bear]

no, not at that step.

in the first step we multiply both sides by 7(x-6).

do you understand
(a) what that means, and
(b) why we do it?

[minor nexus]

the 5(x-6) looks like cross multiplying? But idk after that.

[willow bear]

no

ok, so full disclosure: i myself can't follow your tutor as to what they did in their second step.

but let me try to explain and write out more fully what they did in the first.

im writing it out on paper right now jsyk

"move" bad

[solar olive]

what

[willow bear]

you should not speak of "moving" things between sides.

that way lies confusion and misunderstanding.

[solar olive]

Ah right, i meant subtract 4/(x-6) from both side

[willow bear]

@minor nexus here is the tutor's first step, written out in more detail. does this make sense to you? Y/N/R

[willow bear]

ok yeah. that's possible, actually. but rn i am explaining to op what her tutor did.

[solar olive]

Oh

[vestal bison]

can someone explain me this question:

If A=[[3,2],[-2,-1]] write A^2 in the form pA + qI where p and q are scalars.

[solar olive]

have u tried calculating what A^2 is

[willow bear]

#linear-algebra ?

this channel is a bit busy rn

[vestal bison]

yes [[5,4],[4,3]]

alr

im sorry i sent the wrong one

[solar olive]

it should be -4, -3

[minor nexus]

Okay so I have to find the least common multiple then.

[solar olive]

not 4, 3

anyways move to #linear-algebra

[willow bear]

you don't "have to" do anything.

[minor nexus]

Okay I think I will work on these tomorrow. I have to go to bed. Thanks for helping me and I think I'm starting to get these now.

[vestal bison]

Can someone answer the question below? I found an answer that has nothing to do with the correct answer.

Find the values of x for which the matrix [[x,2,9],[3,1,2],[-1,0,x]] is singular.

So since this is singular i've concluded that det(A) ( assume that the matrix name is A) = 0

Therefore, I had to use b^2-4ac formula to get the answer but the answer is an integer not an irrational number.

[summer ruin]

the answer is an integer

[willow bear]

show your work

we have to rule out arithmetic errors

also since when was it known that x had to be irrational?

[vestal bison]

i found an irrational thats why i wrote that

to find det(a) i did :

x(x-2)-2(3x+2)+9(1) = 0

probably this is the part where im incorrect but i did find denominators of 6 other matrices and i've had problem finding this one only

[summer ruin]

it's x*x instead of x(x-2)

[vestal bison]

why

[summer ruin]

the determinant of ((1,2),(0,x)) is x, not x-2

[vestal bison]

wait i've been using a really different way though

[willow bear]

show your work

[vestal bison]

how do u find the determinant of 3*3 matrices usually?

okay hold on i need to send a picture

[summer ruin]

there are tons of ways

[vestal bison]

could you at least explain me one of them? Clearly I've been doing the wrong one since the beginning

Here’s my work

I used this page to compelte it

Complete*

[summer ruin]

well this is not any different to what I was talking about

[summer ruin]

you computed first 2x2 determinant incorrectly

[willow bear]

2*0 is 0 not 2.

[vestal bison]

ohh okay

[willow bear]

and that arithmetic mistake screwed you over.

[vestal bison]

its due to calculations

alr

i knew something was wrong but couldnt find where did i do wrong calculation

is this type of solution great? or do i need to search for more types of "finding the determinant of 3x3 matrices" solutions

im preparing for university this is why it matters to me

[summer ruin]

as long as you get the correct answer, the choice doesn't matter

[vestal bison]

alr thank u

[proven void]

,, \lim_{x \to +\infty} (\sqrt{x^2 - 6x - 40} - x)

[obsidian monolith]

レナト (renato , ping if reply)

[proven void]

please send help

[willow bear]

multiply and divide by the conjugate

[proven void]

,align
\lim_{x \to +\infty} \left(\sqrt{x^2 - 6x - 40} - x\right) \times \frac{\sqrt{x^2 - 6x - 40} + x}{\sqrt{x^2 - 6x - 40} + x}

[obsidian monolith]

レナト (renato , ping if reply)

[willow bear]

indeed.

[proven void]

xdd what now

[willow bear]

what do you usually do when you do stuff with conjugate multiplication like this

[proven void]

,align
\lim_{x \to +\infty} (\sqrt{x^2 - 6x - 40} - x) \
&= \lim_{x \to +\infty} \left(\sqrt{x^2 - 6x - 40} - x\right) \times \frac{\sqrt{x^2 - 6x - 40} + x}{\sqrt{x^2 - 6x - 40} + x} \
&= \lim_{x \to +\infty} \frac{(\sqrt{x^2 - 6x - 40} - x)(\sqrt{x^2 - 6x - 40} + x)}{\sqrt{x^2 - 6x - 40} + x} \
&= \lim_{x \to +\infty} \frac{x^2 - 6x - 40 - x^2}{\sqrt{x^2 - 6x - 40} + x} \
&= \lim_{x \to +\infty} \frac{-6x - 40}{\sqrt{x^2 - 6x - 40} + x}

[obsidian monolith]

レナト (renato , ping if reply)

[proven void]

what now?

xdd

[fading monolith]

Factor x from the numerator and denominator

[proven void]

thanks, that did the trick

[proven void]

im so stupid sorry

[cosmic citrus]

Anybody enjoy math and would like to help me with precalc

[proven tulip]

Is brief applied calculus by berresford rockett a good intro to calculus?

[thorn creek]

it's probably ok if you want to understand calculus as a financial or business major

[proven tulip]

What about just in general would it make learning regular calculus easier?

[thorn creek]

i don't know if it would have that effect. the book is intended for people who want an intuitive understanding of basic calculus sufficient to satisfy the needs of people in nontechnical disciplines. such approaches do not serve the needs of engineering or sciences students and are actively problematic for those who intend to study higher mathematics

[rugged galleon]

,,( \left(\frac{3 + \sqrt{33}}{4}, \frac{3 + \sqrt{33}}{4}\right) ) and ( \left(\frac{3 - \sqrt{33}}{4}, \frac{3 - \sqrt{33}}{4}\right) ).

[obsidian monolith]

emss

[rugged galleon]

oh period

[glad geyser]

Find the equation of the the exponential function whose graph will pass through the points (-1, -27) and (1,3) with y = 5 as a horizontal asymptote.

[wise belfry]

hey gang

shoot its sideways

cant figure out 28 :(

[river drift]

,rotate

[wise belfry]

thats handy

[river drift]

have you made a diagram?

[wise belfry]

naur

i dont understand the golf terms used

[river drift]

"straight" means the course doesn't bend. 400 yards is the playing distance from the starting point to the hole. par 4 means you'd expect a decent golfer to get from the start to the hole in 4 swings

[wise belfry]

oh

[river drift]

so basically it means that he started 400 yards away (straight-line distance) from the hole, then he hit it 280 yards right, then 170 yards to the hole

[wise belfry]

mkay

20.73 degrees

how does that sound

[river drift]

seems reasonable

[wise belfry]

tyty

[rotund oyster]

i’m confused on why sin and inverse sin end up having the same answer

like isnt sin(-1/2) also 7pi/6 and 11pi/6?

why is inverse sin the same thing?

[zealous surge]

convert r=2sin3θ to rectangular form can someone please help me in this

[winter plover]

the two concepts are a bit different let me explain, inverse sin or inverse of any trig function is typically used to find angles but a trig function like sin(x) is used to find the trig ratio of the angle so in summary inverse is used to find angle and trig function themselves are for finding trig ratios

[rotund oyster]

OHHH

that makes sense

ty

[empty trellis]

guys i have smth that im so confused abt

it says 1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6. I get why n(n+1)/2 is in there, but why (2n+1)/3?

can somebody tell me whether i find this myself or is this a rule to memorize

[willow bear]

I get why n(n+1)/2 is in there
sus

[empty trellis]

what...

[willow bear]

not all formulas for the sum of a progression from 1 to n involve n(n+1)/2

[empty trellis]

yes but this specific one does

[willow bear]

anyway, if you know how to manipulate summations with big sigma you can in fact calculate this sum

with some trickery

[empty trellis]

oh...

ok

have u heard of this rule before?

[willow bear]

as in me specifically?

yes, i have. what's it to you?

[empty trellis]

i recently started studying Calc and i didnt rlly know abt this formula until yesterday. turns out it's just a rule that i have to memorize myself.

[willow bear]

and what exactly does my knowledge of it have to do with that

[grim valve]

Wass up

I have questions have when I do x=2a/b, and like xb =2a

When I sub to b/a/2 re written to b/2a then b/xb which becomes x=b/xb , x²=b/b, which is 1 so x is equals to √1

M I doing right

[tight sleet]

b/(a/2) equals 2b/a

like 1/(1/2) equals 2

x^2=(2a/b)*(2b/a)=4 so x=2,-2

It's in real positive number, so I think it'll be a=b >> x=2

[grim valve]

Can u explain

[tight sleet]

oh yes

[grim valve]

Holy typo

[tight sleet]

aa

[grim valve]

It's like fractions do first or something like that is it like that

[grim valve]

Can i get reply pls

Hands trembled thats y typo

[tight sleet]

Are you saying that a form like 1/(1/a) is hard to understand?

[grim valve]

Yes

Unfortunately

Is it like fractions do first or something

[tight sleet]

make it easy form
1/(1/a)=b

then multiples 1/a each term

then it will makes 1=b/a

multiples a

a=b

[grim valve]

What Abt like xb/2 =a

2b/a, then x =2b/xb/2

Will it be 2b/xb*1/2like this then x=1

[tight sleet]

thats be x=2a/b

multiples 2 and 1/b

[tight sleet]

right

[grim valve]

The answer is x =1?

[tight sleet]

which question?

[grim valve]

This one

[tight sleet]

x=2

In my country most members like this

I'm not used to the explanation cause high school student

sry

[grim valve]

Good try though

Thank u for helping

I really appreciate

[tight sleet]

If there's a fraction in the denominator, you'd better multiply that fraction on both sides to make it a little easier

I'll cheer for you!!

[grim valve]

Bud it gives two different answers

I really appreciate u keep it up

2b/1 * 2/xb is different and 2b/xb * 1/2 is different

[tight sleet]

It could be because of the sign
If b/(a/2) is written as b/a/2, you can think of it as (b/a)/2=b/2a

[grim valve]

It's actually 2b/a, ur correct

[tight sleet]

😄

[grim valve]

Ur knowledge is correct

I have a misunderstanding

So let it be

Maybe others who know better can help but ur correct I don't want u to get misunderstanding

There is some misunderstanding on my side

Ur very helpful and good keep it up

[empty trellis]

guys can somebody give me the full worked out solution for \lim _{x\to \infty }\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)

if the ai doesnt simplify it, i'll send an image

of the problem

[willow bear]

!nogpt

[tender quest]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[willow bear]

!noans

[tender quest]

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

[proven void]

,, \lim_{x \to \pm\infty} \left( \sqrt{x^2 - 2x + 3} - x \right)

[obsidian monolith]

レナト (renato , ping if reply)

[proven void]

I have a question?

why is the left side of this limit infinity but the right side -1?

,w limit of x to -inf for sqrt(x^2 -2x + 3) - x

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by xbz24

[willow bear]

$\sqrt{a^2} = |a|$

[obsidian monolith]

Ann

[willow bear]

sqrt(x^2 - 2x + 3) behaves approximately like -x, not like x, as x goes to negative infinity.

[proven void]

ohh

I think I am starting to understand

,align
\lim_{x \to \pm\infty} (\sqrt{x^2 - 2x +3} - x) \
&= \lim_{x \to \pm\infty} \left(\sqrt{x^2 - 2x +3} - x\right) \times \frac{\sqrt{x^2 - 2x +3} + x}{\sqrt{x^2 - 2x +3} + x} \
&= \lim_{x \to \pm\infty} \frac{(\sqrt{x^2 - 2x +3} - x)(\sqrt{x^2 - 2x +3} + x)}{\sqrt{x^2 - 2x +3} + x} \
&= \lim_{x \to \pm\infty} \frac{x^2 - 2x + 3 - x^2}{\sqrt{x^2 - 2x +3} + x} \
&= \lim_{x \to \pm\infty} \frac{-2x+3}{\sqrt{x^2 - 2x +3} + x}

[obsidian monolith]

レナト (renato , ping if reply)

[proven void]

so after I factorize x from numerator an denominator here

and I factor the x^2 out of the sqrt

then it goes out as |x| not x

[willow bear]

yes but since you've got x approaching +∞

|x| is x

[proven void]

what about when x is approaching -inf

[proven void]

,align
\lim_{x \to \pm\infty} (\sqrt{x^2 - 2x +3} - x) \
&= \lim_{x \to \pm\infty} \left(\sqrt{x^2 - 2x +3} - x\right) \times \frac{\sqrt{x^2 - 2x +3} + x}{\sqrt{x^2 - 2x +3} + x} \
&= \lim_{x \to \pm\infty} \frac{(\sqrt{x^2 - 2x +3} - x)(\sqrt{x^2 - 2x +3} + x)}{\sqrt{x^2 - 2x +3} + x} \
&= \lim_{x \to \pm\infty} \frac{x^2 - 2x + 3 - x^2}{\sqrt{x^2 - 2x +3} + x} \
&= \lim_{x \to \pm\infty} \frac{-2x+3}{\sqrt{x^2 - 2x +3} + x}

[obsidian monolith]

レナト (renato , ping if reply)

[proven void]

I made a mistake here, now its corrected

,w limit of x to +inf for sqrt(x^2 -2x + 3) - x

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by xbz24

[proven void]

I see the point when +inf, that is -1
since I can get the x^2 out of the sqrt

but what about when the limit is x to -inf

[proven void]

,w limit of x to -inf for sqrt(x^2 -2x + 3) - x

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by xbz24

[bitter rose]

not exactly; $\sqrt{i^2}=i\ne |i|=1$

[obsidian monolith]

Treidex

[broken swan]

can someone please explain question 3

[river drift]

you can think of 0.11111... = 0.1 + 0.01 + 0.001 + ...

[willow bear]

i was obviously talking about real x. your pedantry is unhelpful.

[lime cloak]

first off

root i^2 cannot exist

it can only in the complex plane

not in the real plane

are we talking in general or wrt to the complex plane?

[lime cloak]

you can use the method to derive it into its fractional form

from there things get very easy

it is actually a CBSE question from India for Grade 9

[lime cloak]

also If I wasnt clear I was talking about the 4th question

[grim valve]

Ik it's the first answer but like how do I write the steps

[viscid thistle]

simple

as they say one of the two numbers need to be negative

so case 1 if $x+2<0$

[obsidian monolith]

EinPest

[grim valve]

Why is it isn't x=-2 or x=5

[viscid thistle]

in that case the polynomial will become equal to 0

but it should be less than 0 (negative)

[grim valve]

Ok

So we test by ourself to find out

[viscid thistle]

u can make two case, one in which x+2 is negative and one in whcih x-5 is negative

[grim valve]

Shouldn't it be x<-2 or x<5

[viscid thistle]

yeah

[rotund oyster]

YOOOO CHAT LETS GO

#blessed

[viscid thistle]

great

[grim valve]

I mean like

Like if it was equals to 0 the it will be x=-2 or x=5

[viscid thistle]

yeah

[grim valve]

Why in the above case like x must be -2<x<5

[viscid thistle]

and hold up answer wont be opyion 1

[viscid thistle]

its wrong, that will make the polynomial positive or 0

you will get $x<-2$ or $x<5$

[grim valve]

No

[viscid thistle]

theree interesection will be for $x<-2$

[grim valve]

Why it's and, and not or

[viscid thistle]

which is the answer

[obsidian monolith]

EinPest

[viscid thistle]

it will be or

[grim valve]

I don't think so

Ur good helper

But I don't understand

[viscid thistle]

it cant be both tho

like lets say x=-3

now evaluate the polynomial

[grim valve]

It will become positive

[viscid thistle]

yeah

and it shoul be negative

so two cases arise

[grim valve]

The range of x should be between larger than -2 but smaller than 5

[viscid thistle]

yeah u are right

i made a blunder

[grim valve]

I show u chatgot

[viscid thistle]

100% correct

[grim valve]

The inequality (x+2)(x−5)<0 is a "strict" inequality, meaning it's asking for when the expression is less than zero, not less than or equal to zero.

If we plug in x = -2 or x = 5, the expression becomes zero, not less than zero. Therefore, x = -2 and x = 5 are not included in the solution.

If the inequality was (x+2)(x−5)≤0, then x = -2 and x = 5 would be included in the solution.

[viscid thistle]

yeah thats right

answer will be infact option1 for this

[grim valve]

So I have to manually plug in the numbers to find

[viscid thistle]

yup

[grim valve]

Sadge

Thanks mate

[broken swan]

can you like show me how to do that??

[lime cloak]

https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-numbers-operations/cc-8th-repeating-decimals/v/coverting-repeating-decimals-to-fractions-1

go here dude

there's a whole course

it is pretty intuitive

[warm vault]

Yo, is Permutations and Combinations part of this category?

[willow bear]

yeah you can post P&C questions here why not

they also fit in #discrete-math

[warm vault]

Oh ok cool

[willow bear]

if you have one then just post it

& all your progress etc.

[warm vault]

Oh okay

Im actually stuck on a problem that my friend and I are baffled about

[willow bear]

ok but show us the problem

[warm vault]

I will just give one sec

I’ve been going back and forth with my classmate on how letter d is plus and not times

[willow bear]

do you mean that 4!+3! is the answer as claimed by you, or as claimed by them?

cause to me it looks BS

[warm vault]

It was claimed by one of my other friends, but I sided with him.

[willow bear]

do you have reasoning to support this answer

[warm vault]

Well theres 4 boys, so all the boys are one unit, same with the girls. So I thought we add because theyre independent from each other

[willow bear]

surely their independence would indicate multiplication?

addition would mean you choose an arrangement for one xor the other.

but you need to choose arrangements for both,

and also decide if you're going for BBBBGGG or GGGBBBB

[warm vault]

Well I thought we can choose either way because multiplication is inverse so it's the same in any order

[willow bear]

multiplication is inverse

what

what do you mean when you say "multiplication is inverse"?

[warm vault]

3 x 4 is the same as 4 x 3 basically

[willow bear]

oh, you meant multiplication is commutative.

[warm vault]

My bad

[willow bear]

"inverse" is the wrong word.

anyway

my point is that (d) will be 4! * 3! ** * 2**.

the 2 is for the two ways you could arrange the two groups (boys and girls) in the row.

either boys first or girls first.

[warm vault]

Ohh makes sense

[warm vault]

I guess it ties with dis

[willow bear]

of course it does.

i repeated the same point but more explicitly.

[warm vault]

Okay, my friend is on discord too

I will negotiate this with my friend.

[willow bear]

negotiate??

this is not a trade agreement

what is there to negotiate?

[warm vault]

Sorry about my vocabulary. I meant I would "discuss" with my friend about this.

[grim valve]

I don't think this is precalculus

Better u upgrade on roles to get access to undergraduate math

[solar olive]

oh

whats undergraduate math ? i dont use that term in my country

[fading monolith]

When you are at uni

[willow bear]

undergrad is typically reserved for when you are in a bachelor's degree program

anything higher like a masters is graduate

[proven void]

,,\lim_{x \to -3^-} \frac{2x+1}{x+3}

[obsidian monolith]

レナト (renato , ping if reply)

[proven void]

guys how does this work?

why is it infinity as x approaches left side of -3

,w limit x to (-3) for (2x + 1)/(x+3)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by xbz24

[uncut mulch]

consider whether x+3 approaches 0 from the negative or positive side as x→-3^-

[willow bear]

do you mean this in a "what's going on?" way or in a "i think it should be sth else" way

[proven void]

I am not saying wolfram is wrong

im just curious why is it +inf for x to -3^-

[proven void]

why

I understand x+3 approaches 0 from the -3^+ case

[willow bear]

but does it approach 0 from above or from below?

that's important

because that is how you tell if your one sided limit is +∞ or -∞

also take note 2x+1 approaches -5, a negative number.

[proven void]

yeah I understood the right limit one

I am having trouble with the left side of the limit I think

[uncut mulch]

same principle applies

to understand the right limit properly you would pretty much have had to gone through the same thing

[proven void]

mmm

[willow bear]

can you spell out your reasoning IN FULL for why $\lim_{x \to -3^+} \frac{2x+1}{x+3} = -\infty$?

[obsidian monolith]

Ann

[proven void]

I just replaced -3 for x

= [2(-3) + 1] / [-3+3]
= -5/0 = -inf ?

xd

,w -5/0

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by xbz24

[proven void]

xd

maybe I am not following

[willow bear]

well, your reasoning is lacking.

[uncut mulch]

division by 0 is undefined

[willow bear]

2x+1 approaches -5, a negative number. that much is okay.

x+3, meanwhile, approaches 0 FROM ABOVE -- meaning x+3 is POSITIVE while doing so -- and thus 1/(x+3) approaches +∞.

[proven void]

what

x +3 approaches 0. from above since -3^+. meaning x + 3 is positive while doing so?

[willow bear]

yes

[uncut mulch]

consider what you get if you plug in values above -3 into x+3

or even look at the graph of y=x+3

[proven void]

,w plot y = x+3

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by xbz24

[solar olive]

OHh

Ty

[warm vault]

Yo I think the answer to c is also 144

Correct me if I'm wrong, but since you can just arrange the girls (3!) and then add the rest (4!), the answer is just the same

[willow bear]

no, you forgot to take into account the 2 possible arrangements of genders

boys on the left vs. girls on the left

[warm vault]

So does this mean it's the same as d? Multiply by 2?

[willow bear]

oh.

oh god, sorry.

i thought you were talking about d again.

no, c isn't 144.

for c, the best way to think about it is like this:
there are 3! ways to arrange the girls among themselves,
and then for the purposes of seating in a row, we've got 5 people to seat.

the 4 boys, plus the girls considered as a single unit (because they are together).

this gives us 3! * 5!.

[onyx lotus]

im super tweaking rn can someome walk me thru how its 3/2

[warm vault]

Wait so I think the answer to f is just a - e then, correct me if I’m wrong

[chrome ether]

not quite

[light flume]

can someone help me

Elimate parameter for the following set

x = 2sint
y = 3sin^2t
and identify range

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by 317evade

[feral tulip]

IN WHAT ORDER SHOULD I APPLY VERTICAL/HORIZONTAL SHIFTS, STRETCHES, AND SHRINKS?

order of graph transformations

[chrome ether]

depends on the equation

NO NEED FOR THE ALL CAPS BTW

generally work from the inside out

so f(2(x-3))

means first you horizontal shift by 3

and then horizontal squeeze by factor of 2

[fleet arch]

im pre hihgschool but im tryna learn calc to prepare me, any tips?

(pls @ me if u have response)

[burnt wolf]

You still have Algebra I, Geometry, Algebra II, and Pre-calc to go through....if you're in a pre-High School grade

[fleet arch]

yeah

but i wanna learn it now

cause i wanna learn other math stuff

but i need to know calc for it

so i am turning to the internet

is their like a step by step explination i can find?

[thorn creek]

first tell me what a function is

if you can't explain what a function is, you're not ready for calculus

[fleet arch]

it hasnt been taught in school yet, so i might be completely off base here. F is the function symbol, f(x) = x>2 means that we take x and square it. f(p) = 2p x hv means that we take p and apply the part of the equation after the = sign. so in this case 2p x hv. i think

idk man, i havent read much abiout it

[river drift]

could you post a picture of this?

[fleet arch]

me?

[river drift]

yes, a picture of the place you're seeing this notation

[fleet arch]

oh dw it wasnt a question, i didnt see it anywhere

i was asked to explain waht a function is

and i tried

but im in 8th grade

so i prob missed a few things

[uncut jay]

something im missing here?

[burnt wolf]

But you can’t just skip steps. Wherever you’re planning on learning Calculus from, you can probably learn the first four classes from as well?

[fleet arch]

?

[burnt wolf]

You can’t just jump from math 8 to Calculus

[willow bear]

it is like trying to run the marathon when you are barely able to crawl

[fleet arch]

ik

which is why im tryna find something that can help me go from crawling to running

so far that seems to be khan academy

[chrome ether]

don't rush it

i could've learned calculus as early as eighth grade if i wanted to

but i deferred it to sophomore year of high school just so i could let deeper understanding of other "precalculus" topics really sink in

[thorn creek]

sadly, if this is the best you can do you're probably not ready for calculus. spend some time learning algebra first

there's no shame in not knowing a thing, but you really shouldn't try to learn "too far ahead". math builds cumulatively on previously learned math and if you try to skip ahead too far you'll just be wasting time trying to understand stuff you don't have the groundwork to understand yet

[chrome ether]

^ don't go ego chasing, learning calculus early is really nothing special in the long run

and if you rush into it too early you WILL struggle unnecessarily

[solar olive]

listen to them
source : i tried to do the exact thing, it was awful

[uncut mulch]

attempting to do calculus without the pre-requisite foundations makes the solving process take 100+ times longer

(not an exaggeration)

[solar olive]

(been there before)

(trust me, do NOT )

[thorn creek]

i tried to teach myself calculus in 10th grade. i could memorize the rules for derivatives, but i did not understand them and basically just learned nothing

to be fair, this probably did make it easier to learn calculus when i actually took it in 11th grade

[crude umbra]

do not just jump from yr 8 math to calculus just cus u wana learn smth else

unless ur actually crazy good at math and can recall everything from ur textbook and the textbook above without struggling

i tried to do what u did when i was in yr 8 too

didnt understand anything cus i had no fundamentals

i only started learning in yr 9 but that was only learning techniques and when or how to use them

i didnt understand how they worked or why

or what they even did

and i only learnt that in the second half of the year cus my teacher told me how important it was to actually understand why how and where those things came from

and still learning calc at yr 8 is worthless and wastes ur time

what

u still have like

2 years

until u actually need to understand anything in calc

why not spend those 2 years polishing up the fundamentals for precalc and calc

the only reason i learnt it was cus my teacher saw i was interested in it so he gave me extra work for me to look at if i was bored and had finished all my work

id already finished the whole book for yr9 and yr10

so he let me do it

[crude umbra]

also im not trying to hate on u

but based off ur definition of a function

im gonna assume u havent done much research/learning abt math by urself outside of the content in school

if ur gonna learn calc at such an early age its gonna involve a bunch of extra learning

[burnt wolf]

I did (and am currently doing) the same thing actually

[gaunt glen]

Hello. tmr i have a math comp for famat in precalc and these are the standards tested..
Does anyone have any helpful resources

1. Demonstrate an understanding of the theory of functions.
   • find domains; ranges; an specific values of functions in functional notation.
   • given two functions perform the algebra of functions including composition of functions.
   • determine if a given function is:
   a. symmetric (with respect to the axes and/or origin.
   b. periodic
   c. monotonic
   d. bounded
   e. continuous
   • identify and graph polynomial and rational functions and determine asymptotes.
   • define and use parametric forms of functions and convert from parametric to Cartesian form.
   • given a function; determine the inverse and state whether or not the inverse is a function.

2. Demonstrate an understanding of connection between circular and trigonometric functions and theiinverses.
   • evaluate circular and trigonometric expressions involving any of the six functions and
   their inverses.
   • given the equation for a circular (trigonometric) function; identify and/or sketch the graph
   and the graph of its inverse relation and state the domain and range of the original
   • function and its associated inverse function.
   • identify its equation when given a graph of any of the six circular functions.
   • state the period; amplitude; phase shift; and vertical shift of a circular function and/or
   graph of the function.
3. Demonstrate an understanding of the trigonometric identities.
   • prove that a given trigonometric equation is an identity by applying the Pythagorean
   relation and reciprocal identities.
   • prove that an appropriate trigonometric equation is an identity when given the sum and difference formulas
   for the cosine; sine; and tangent.
   • prove that an appropriate trigonometric equation is an identity when given the double order formulas for sine;
   cosine; and tangent.
   • prove that an appropriate trigonometric equation is an identity when given the half-angle formulas for sine;
   cosine; and tangent.
   • evaluate circular and trigonometric expressions involving any of the six functions and
   their inverses.
   • given the equation for a circular (trigonometric) function; identify and/or sketch the graph
   and the graph of its inverse relation and state the domain and range of the original
   • function and its associated inverse function.
   • identify its equation when given a graph of any of the six circular functions.
   • state the period; amplitude; phase shift; and vertical shift of a circular function and/or
   graph of the function.

4. Demonstrate the ability to apply trigonometry to problem solving situations.
   • solve a right triangle given two sides; or a side and an acute angle.
   • use the appropriate trigonometric function(s) to solve problems involving right or oblique triangles.
   • apply the Law of Sines.
   • apply the Law of Cosines.
   • find the area of an oblique triangle.
   • estimate the solution to a problem involving a right or oblique triangle.
   • in the SSA case determine whether 0; 1; or 2 triangles exist and determine the
   • triangles (if they exist)
5. Demonstrate the ability to solve a variety of trigonometric (circular) equations.
   • find the general solutions to a trigonometric equation
   • find particular solutions to a trigonometric equation within a given domain.
   • solve equations involving inverse of circular/trigonometric functions.

6. Demonstrate an understanding of conic sections and loci.
   • given the description of a locus determine the equation of the locus.
   • given the equation of a line determine slope and y-intercept; and graph the line.
   • given the equation of a circle determine the center and radius; and graph it.
   • given the equation of a parabola determine vertex; focus; and directrix; and graph it.
   • given equation of an ellipse in standard form; determine the center; foci; and vertices; graph it.
   • given the equation of a hyperbola in standard form; determine the foci; vertices; and asymptotes; and graph it.
   • determine new equations resulting from translation or rotation of axes.
   • identify the graph of any second degree equation.
   • express a quadratic equation in general form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 and use B2 - 4AC to
   distinguish conics.
   • recognize degenerate and imaginary cases.

7. Demonstrate an understanding of the relationship between exponential and logarithmic
   functions and their application to problem situations.
   • evaluate expressions involving rational exponents.
   • sketch the graphs of exponential functions and logarithmic functions of different bases.
   • solve equations involving exponential functions and logarithmic functions.
   • solve real-world problems involving exponential functions and logarithmic functions.
   • simplify expressions using the relationships between logarithms and exponents.
   • express the number e and the expression 'e to the x' as infinite series
8. Demonstrate the ability to solve problems using concepts from matrix algebra.
   • apply determinants to solve systems of equations.
   • invert a square matrix

9. Demonstrate the ability to solve problems using vectors.
   • find a vector in standard position equal to a given vector.
   • determine magnitude and direction of vectors.
   • identify perpendicular and parallel vectors.
   • determine the measure of the angle between two vectors.
   • resolve a vector into component vectors.
   • add and subtract vectors and multiply a vector by a scalar.
   • find the dot product of two vectors.
   • use vectors to solve real world problems.
10. Demonstrate an understanding of polynomial and rational functions; their parametric
    equations and their graphs.
    • given a polynomial function determine intercepts and sketch the graph.
    • given an equation of rational function determine intercepts and asymptotes and sketch the graph.
    • given a set of parametric equations sketch the graph.
11. Demonstrate an understanding of graphs in the polar coordinate system and their relation to
    the Cartesian coordinate system.
    • graph points in the polar coordinate system.
    • convert between polar coordinates and Cartesian coordinates.
    • express complex numbers in polar or trigonometric form.
    • convert equations in polar form to Cartesian form.
    • convert equations in Cartesian form to polar form.
    • graph polar equations and identify specific types (roses; limacons; spirals; and conics)
    • use de Moivre's theorem to find powers and roots of complex numbers.

if anyone has any resources and tips, please share 🙏

[chrome ether]

do past papers if they’re available

those are generally the best way to get a feel for the contest

[fleet arch]

the gist ive gotten from all the messages (thanks for replying everyone!) is that i should learn some fundamentals and foundational stuff before jumping ahead to calculus. So my question is what kind of fundamentals should i work out, how can i find how to learn them? thanks!

[random yoke]

Khan academy is good

I self-studied calculus after a year or so of doing all the pre calc on there

[spiral sonnet]

Okay so I'm stupid, I am currently looking for a slanted asymptote but when I do

6x^2 - 30  
----------
-2x-4

with polynomial long division, I get -3x-6
Yet the answer is -3x+6
what am I doing wrong.

[chrome ether]

show us your work

sign errors are kinda easy to make with long division

[hybrid epoch]

,rotate

[chrome ether]

you wrote the thing you’re dividing incorrectly

it’s 6x^2+0x-30

remember that there’s that hidden 0x term you have to include

otherwise it would be like saying dividing 103 by 13 is equivalent to dividing 13 by 13

obviously a ridiculous statement

run the division again with that correction and that should get you the correct answer this time

[spiral sonnet]

Alright I'll try it real quick ty

[chrome ether]

np

[uncut mulch]

that isn't the issue

its helpful to keep stuff aligned (and highly recommended) so you don't mistakenly combine unlike terms
but it isn't essential

[chrome ether]

oh d’oh

[uncut mulch]

the issue is that you added 12x instead of subtracting

[spiral sonnet]

Well this became more of a logic failure is what I am realizing. I was used to the -'s subtracting into positives and forgot by nature this is a subtraction thing

[chrome ether]

yea

[spiral sonnet]

yeah I just realized this, ty

0x did help me visualize it. I just wasn't naturally taught to use it so Ididn't think about it

[uncut mulch]

(writing the 0s for absent terms is essential if you're doing synthetic division though)

[unreal lantern]

does anyone know how to derive this? im not sure if i have to use pre calculus rules as i have to find stuff like whys the maximum points y is bigger than 0

or do i just derive a^2 x as multiplicative?

[summer ruin]

a^2 is constant

[unreal lantern]

OH

thank u i understand now

so the derivative should just be x^2-a^2

[uncut mulch]

yes

[unreal lantern]

got it

[willow bear]

btw English is stupid

and the verb "to take the derivative" is NOT "to derive"

it is "to differentiate"

[worthy bridge]

would be asking members in here for tutoring in my server go against the rules?

[willow bear]

might want to ask <@&268886789983436800> about that

[short sorrel]

we typically do not allow you to use this server to advertise, no

whether that be advertising other servers or advertising "recruitment"

[worthy bridge]

okay thank you

just checking

dont wanna be banned lol, id fail my classes

[short sorrel]

if you need a personal long-term tutor, youll probably have better luck asking around your school/community anyway

though youd probably have to pay

[analog flower]

and we tend to prefer that interactions stay on the server since we've had some people get harassed or taken advantage of (mostly monetarily) in DMs

[worthy bridge]

makes sense

[short sorrel]

fyi feel free to ask questions like this in modmail in the future

not saying its inappropriate here, just pointing out that thatll probably get you a faster response

[worthy bridge]

sorry whats modmail?

a channel in here?

[short sorrel]

@tall sequoia

can just DM it

[worthy bridge]

oh i see

okay il do that in the future

[candid brook]

would you say precalc is more similar to the algabraeic type of math or geometric?

or is it completely unique and it’s own separate way of thinking

[summer ruin]

you should ask people who do algebraic geometry

[fading monolith]

Spooky stuff

[signal sigil]

wdym whily

[willow bear]

a poorly written and poorly spelled problem that's what

[nocturne jackal]

Using ideas from precalc i was able to graph the mandelbrot set in desmos what do yall think

40 iterations

[old zephyr]

thats amazing

[normal kite]

hi guys i am making my math flashcards about functions and logic quantifiers but i don't know if i made this one correctly

is it correct?

[warped cipher]

amazing

[hallow tide]

very nice indeed

always a pleasure to see Desmos used for abstract? endeavors like this

[hallow tide]

i do not think

see:

where “X” refers to the domain of the function

[willow bear]

40 iterations and it didn't choke?

god damn

[shadow summit]

you have to account for x1=x2

[shadow summit]

bro desmos breaks when i have a taylor series with more than 3 terms

[willow bear]

don't call me bro.

please edit that word out of your msg.

[uncut jay]

why is midline not equal to 3?

[river drift]

it may want the equation of the midline

[uncut jay]

how is this wrong?

this is the graph btw

[willow bear]

uhh

feels like some fuckup w the system

your midline and eqn are correct as drawn...

[river drift]

it's probably something like "the system wants y= at the start"

[willow bear]

it marks the midline as wrong too

[viscid thistle]

Hi, just yes or no please, no explanation. Does this identity exist?

[fading monolith]

Yes, you can figure out where it comes

[uneven flame]

No, this is not good. You can't go willy-nilly throwing quantifiers all over the place. You need to put them together at the start. It's the rule of first-order logic. Quantify stuff, get some names and then do the logic. There are some texts which define functions as sets and then builds on from that, but that's not happening here. Get the statement in English, first (or, your language if it's not English). Think about, what you want to say. You need to state that if $x \neq y$ then $f(x) \neq f(y)$. You must use a quantifier for object you introduce here, so you $\forall x \forall y$, now that you've declared the stuff you need, you have the logical statement, so you must write it: $x \neq y \Rightarrow f(x) \neq f(y)$. Now you've got it and you just put it together: $\forall x \forall y (x \neq y \rightarrow f(x) \neq f(y))$. This is typical definition, but often the contrapositive is used in practice. The contrapositive of an implication is logically equivalent to the implication, so there's no trouble here.

[obsidian monolith]

Saurus

[normal kite]

I just tried to replicate the order I saw in an ebook that I found

[uneven flame]

Alright, I suppose just do what the book shows if that's how you're learning it. If it's not for formal logic then do it so that works for you. This one is talking about sets, though, so it is using the sér membership symbol. As long as you can read those statements, understand them and do problem, then it will do the job. There's nothing wrong with writing stuff in words, though.

[normal kite]

OOH i thought that it was required to write math without using words in university

idk why lol

thanks for the explanation!

[uneven flame]

No, only in logic. What you really need to know is, more than first-order logic, the semantics and all of that, is propositional logic. You need to be able to read a statement, a theorem for example, and know exactly what is going on. Knowing when different logical statements are equivalently is valuable, but you can get it from propositional logic. For fun, you can look at modal logic, and then they start doing logic on lattices, but to do the rest of maths and computer stuff, propositional is good to go.

[uneven flame]

Sure, no problem, if it helps!

[nocturne jackal]

It was pretty laggy but desmos's implicit grapher is pretty good

[waxen flame]

idk why i cant solve this geometic series question

im using the formula Sn= (a(1-r^n))/1-r

[fading monolith]

Geometric is for |r|<1

That one is a power series

[willow bear]

incorrect

this is a finite GP so r can be anything save for 1

[willow bear]

show your work, you're probably screwing up the arithmetic

and/or misidentifying n

@fading monolith what's the defn of a "power series" according to you?

[fading monolith]

Right

[fading monolith]

Sum a_n x^n

[fading monolith]

Probably not working bc its alternating

[fading monolith]

Is that wrong?

[willow bear]

ok but like this is a sum of raw numbers, there's no x. and if you put x=-3 anyway, you will find that all the coefficients are the same. so it is still a geometric series.

[willow bear]

alternating just means the common ratio is negative.

[fading monolith]

Maybe have the form (-1)^n(2)3^n

[fading monolith]

You can put a_n=2(-1)^n and x=3

But anyway

[willow bear]

you are overcomplicating it.

[fading monolith]

Then how would you proceed?

[fading monolith]

Tried this and is not working with r=-3 and a=2

@waxen flame i get it now, the formula works if your sum starts with n=1 and starts with n=0 if you take 2(-3)^n. So the formula if you start with n=0 is a(1-r^n+1)/1-r that way it will give you a positive answer and notice that 1458=2*3^6

[red meadow]

can somebody give me some calculus questions pls

[willow bear]

show YOUR work.

[red meadow]

yup

[fading monolith]

Calculated the sum by hand and doesnt coincide with the formula lol. Is just he must be using this formula but there n=7

[willow bear]

i said show your work

not tell

showing your work means presenting a screenshot or picture of the actual calculations that you did on paper

[fading monolith]

Used a calculator(? I mean, the series is 2(-3)^n so

[willow bear]

ok then show your exact calculator input.

[fading monolith]

Its just is 1094

[willow bear]

like you realize there is a reason why im so anal about this right

[fading monolith]

And using the formula of S_n it gave you a negative answer

[willow bear]

with a calculator, there is always, ALWAYS the possibility of input error.

[fading monolith]

Then I realize that the formula he had is asuming the series starts with n=1

Used wolfram to check it too

[willow bear]

first you don't comply with my (apparently simple) request to show your work, then you show something that contradicts your previous claim that something "wasn't working" or the answers "did not coincide".

[red meadow]

am i wrong

[fading monolith]

Im not in home rn, how can I show you my work?

[red meadow]

no i think i was correct

[fading monolith]

@willow bear look I just figured out the general term that is 2(-3)^n smth that we coincide Ig. Then calculate using wolfram and gave me 1094. Used the formula he showed and doesnt work bc it gave me a negative value. So I search the formula of geometric series bc maybe I was using it wrong, then notice that the sum is took from k=0 to n-1 so I got it the n=7 so it will have more sense. Then used wolfram to check that everything was ok. Thats all

If you dont like my work cant do anymore, Im not in home cant write on a paper rn

[willow bear]

sorry i am just frustrated right now

[fading monolith]

Looks good

[fading monolith]

Dont worry, Im not the best explaning myself

[red meadow]

are there more questions?

[fading monolith]

For now nope

[onyx lotus]

is this function continuous at 4?

[summer ruin]

check the definition

[onyx lotus]

i dont think it is since it appears to be a jump discontinuity

[summer ruin]

then it's not

[novel mica]

No
lim x~4- gives the values 5 and the RHL is 0

[torn atlas]

How can I solve b and c?

[summer ruin]

-f = (-1) * f
a+b = a - (-b)

[chrome ether]

b, c should just be solving for y no?

[summer ruin]

no, they would not be in the form af(b(x-h)) if you only solve for y

[stone crystal]

!help

[tender quest]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[willow bear]

!status @stone crystal

[tender quest]

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

[stone crystal]

1

im cooked

@willow bear

[willow bear]

ok so do you know what exponential decay is

[stone crystal]

yea

[willow bear]

ok right.

they want you to write an exponential equation in the form y = a * b^x for this data.

b will of course have to lie betweeen 0 and 1, bc the function will be decreasing.

[stone crystal]

oh ok

it all makes sense now

but how do i get x or the years

[willow bear]

wym

[stone crystal]

like 9 years

[willow bear]

you're gonna have

an exponential equation in the form y = a * b^x

a and b are numbers for you to figure out, but x and y are x and y

here x stands for time in years and y for price in dollars

[stone crystal]

alr

[willow bear]

afterward, it is a matter of finding the value of y at x=9 ...

[stone crystal]

alr thx

[night surge]

why are related rates questions so hard

im solving thomas' calculus

and right upto iimplcit differentiation everything is easy

and then related rates questions which come next seem like a whole different level

[willow bear]

imo the hardest part is translating them from words into symbols

the calculus side of things almost always involves the chain rule in some way but otherwise is not anything more complicated than stuff you've seen before

[night surge]

yea exacty

wriiting the correct equation or even finding the equation is real hard

[willow bear]

do you have an example we could work thru

btw what is related rates talk doing here and not in #calculus

derivative shit is #calculus for sure

[willow bear]

hmm let's see

[night surge]

ah i didnt have that channel on my channel list because i chose pre-universty

[willow bear]

$P(t) = (10 \sin(kt), 10 \cos(kt))$ is immediate-ish.

[obsidian monolith]

Ann

[night surge]

were u able to solve it

my first attempt on the question was futile

ig i'll try it later

[willow bear]

i didn't really bother

[summer rivet]

shortest path from x to y

(you can do it by brute force or using other methods)

[random yoke]

bruh

google dijkstra's algorithm

[willow bear]

ok and what does this wonderful graph have to do with us?

[summer rivet]

maths

I don't know

[willow bear]

so you don't know why you are posting it here.

[summer rivet]

no I do

you can do it if you wnat

[white rapids]

X, B, G, K, M, Y

Just a guess

[summer rivet]

XAEDJMY

13,150 vs 11,450

Not a bad guess

It's only about 1 node off

[white rapids]

A

nice

[dense oriole]

guys

a question