precalculus

[harsh smelt]

@glad pasture

[glad pasture]

i was not sure what RHS and LHS meant

[harsh smelt]

right hand side

and left hand side

basically LHS is what is to the left from <= (>= or = )and RHS is what from the right

[glad pasture]

I don't think it would be clear which side those equations would be on

[harsh smelt]

ok let me rewrite it

[obsidian monolith]

Commander Vimes:

[harsh smelt]

now assume a is nonnegative and b is negative

we have two cases

when a+b is >= 0

then we have a-|b| inside modulus

i mean for example |2+(-1)|=2-|1|=1

clear?

[glad pasture]

yea

[harsh smelt]

and on the right it would be still a+|b|

so we have for case a+b >= 0
a-|b|<=a+|b|

is it clear that this inequality will hold always?

[glad pasture]

yea i think i get it

[harsh smelt]

can you consider case a+b <0 analogously?

after you did it you have |a+b| <= |a|+|b| proved

and we will be able to proceed to main problem

[glad pasture]

ok thanks for the help

[harsh smelt]

but we have not proved that |a|-|b| <= |a-b| yet

[glad pasture]

i just wrote the explanation that if b is larger than a then |a-b| would be larger since |a|-|b| would be negative and |a-b| would always be positive

[harsh smelt]

it is wrong

can you write down complete proof for |a+b| <= |a|+|b| first?

like combine all what i ve shown

and then we will be able to prove |a|-|b| <= |a-b|

(i mean yes you can do proof by cases for |a|-|b| <- |a-b| but bruh)

[umbral sky]

can someone help me with my hw?

i have this so far but im confused for the second function

https://imgur.com/a/NUNBmn7

[uncut mulch]

your third piece is incorrect too

are you able to draw the line y=2x+1 normally?

[umbral sky]

y=2x+1 should go through (0,1) correct?

[uncut mulch]

yes

[umbral sky]

I see, whats wrong with the 3rd function?

[uncut mulch]

what you have there is f(x) = 2 for x>0 (instead of x>2)

[umbral sky]

hmm i see, x is supposed to be greater than 2

would this be better

https://imgur.com/0oxCK6w

dont mind the red dots. it's an open circle too.

the -2<x<2, is that the same as x > -2 and x < 2

[uncut mulch]

not sure on that system but are you using the 3rd line option when drawing?

[umbral sky]

the 2nd one

I dont think i have the 2nd function correct? https://imgur.com/mSBqkkJ

[uncut mulch]

no its not

you only need two points construct a line,
it is most convenient here to consider the boundaries,
i.e. looking at the value of 2x+1 when x=-2 and when x=2

and use the appropriate open/closed circles

[umbral sky]

hmm but the line doesnt go through -2 or 2?

[uncut mulch]

wdym

[umbral sky]

ohh nvm thanks i got it

could you also help me with a few more questions that im stuck on?

the -1 exponent, what does that do

https://imgur.com/a/TaiHJGZ

[uncut mulch]

denotes inverse

[hollow stirrup]

guys

help?

to find the limit of 2 i have to find the limit from 2- and 2+

but the values when i calculate the y

makes me confused

[uncut mulch]

why is the limit from the left and right of 1 relevant here?

you can factorise the denominator then cancel and/or apply certain limit laws/properties

[hollow stirrup]

sorry i meant to mention 2

not 1

how?

[uncut mulch]

x^2 - 4 is a difference of 2 squares

[hollow stirrup]

aha

[ornate wolf]

what should i do since it say no real solution, i can't use inverse method?

[viscid thistle]

think about intersection of lines

[hexed otter]

Hello who is only

Hello who is online here

I have some question about ellipses

[viscid thistle]

Currently almost 5000 people

[harsh smelt]

@viscid thistle and one ape

sorry

[viscid thistle]

An extraordinarily intelligent ape

[hexed otter]

So like

There is this question

Ellipse equation

The given are, vertices: (5,0) (-5,0)
Pass through (2,4)
Write it in standard form

I got confused on the the pass through part please help thanks!

[tardy ridge]

if it passes through that point it'll satisfy the equation

[obsidian monolith]

Albot1288:

Albot1288:
Compile Error! Click the reaction for details. (You may edit your message)

[viscid thistle]

?

[ember crane]

I would like to ask a question. May I?

[vital wedge]

yeah go ahead?

[ember crane]

am I do this right? the question asks yo find the limit

[brisk fiber]

hello

what unit would this be in?

[novel cargo]

meters

t is in seconds

[vital wedge]

@ember crane yeah your answer is good, i normally wouldn't have done it by dividing through by x^3 but that works!

[ember crane]

great! thnks~ @vital wedge

[brisk fiber]

@novel cargo

its asking for instant velocity

that cant be in meters

[novel cargo]

you showed us the function of the position

[brisk fiber]

sorry

[novel cargo]

that's definitly in meters

[brisk fiber]

my bad

[novel cargo]

the rest of the Q isn't there

np

[brisk fiber]

i didnt post the whole question

[novel cargo]

velocity is in m/s

[brisk fiber]

well i get that the distance is 97.408

and the time is 7

so wouldnt the answer be 13.91542 m/s

but im getting it wrong

@novel cargo

[novel cargo]

please show the whole question, where do you get 7 s?

[novel cargo]

you must have learned derivatives, r?

[vital wedge]

wouldn't this be in #calculus?

[novel cargo]

yeah

true

@brisk fiber

[brisk fiber]

yeah

so take the derivative of the function?

[novel cargo]

yes

velocity is the first derivative of the position function

[mighty onyx]

For each angle below, determine the quadrant in which the terminal side of the angle is found and find the corresponding reference angle $\overline{\theta}$. You can enter $\pi$ as 'pi' in your answers. Do not use any space. You can round your answers to four decimal places.

[obsidian monolith]

Albot1288:

[mighty onyx]

$\theta$ = 3 in quadrant I but what does $\overline{\theta}$ mean is this case?

[brisk fiber]

@novel cargo

[obsidian monolith]

Albot1288:

[brisk fiber]

i got 19

ad thats wrong too

[novel cargo]

what is the derivative you calced?

[mighty onyx]

oh In thought yall were done sorry

[novel cargo]

np

[brisk fiber]

aw damn

put a plus where a minus should have been

nevermind

it worked

thanks

[novel cargo]

hahah

I make so many similar mistakes

dw

that's the reference angle @mighty onyx

[mighty onyx]

how would I find what the angle is?

[novel cargo]

gimme a sec

[viscid thistle]

in radiant?

[viscid thistle]

$3\approx \pi$

[obsidian monolith]

HoboSas:

[novel cargo]

so, if it's 3 rad, it's in the second Q

also, the ref angle is pi - 3

look at the left most pic

@mighty onyx

[mighty onyx]

it says that it's in 1st quadrant

[viscid thistle]

3 degrees then

[novel cargo]

yep

ref angle is 3

Q one ref angle = the angle itself

[mighty onyx]

it says it's wrong

[novel cargo]

show

[mighty onyx]

Don't worry about the others I use 6 as a place holder

[novel cargo]

@viscid thistle

look at this please

[viscid thistle]

maybe the other angle is the supplementary?

idk just guessing

[novel cargo]

looks funny to me

@mighty onyx how did you reason about the correct quadrants?

[mighty onyx]

I thought of it like a graph and picked first quadrant because it was positive but then when it got to theta = 16 idk why it's in quad 4

and theta = -8 had to be in 2-4 because it had negative

I could be wrong as just got lucky but idk

[novel cargo]

have you checked other values in the quadrant input boxes?

I mean, do they show incorrect for any input?

or have you guessed all of them correct?

[mighty onyx]

The first one for quadrant 4 work I haven't checked the others. My teacher could have put it in wrong because he sometimes does that

[novel cargo]

look, if this is in rads (which I think it is because you are talking about pi in the question), then theta = 3 is in quadrant 2

this means that this question is broken, more or less

[mighty onyx]

let me try it as if it is in quadrant 2

yep he put it in wrong and pi-3 =0.1416 works

[novel cargo]

to work out the rest of the ref angles, look at the picture I sent you earlier

also, you can get extra credit by showing off that you found the bug

[mighty onyx]

yeah I should also right it down in my work book

thanks

[novel cargo]

np

[terse ravine]

@novel cargo This question is weird.

[novel cargo]

do you know about shifting?

[terse ravine]

I have infosheet on Transformation rules for functions.

That someone posted in here.

(x-5) is right c units and -2 is down d units.

[novel cargo]

nice

you almost answered the question

take the know point (3, -3) right and down by the specified vals

what do you get?

[terse ravine]

(8, -5)

[novel cargo]

yeah, this the only option IMO

I'm not sure if this would be always true tho

I mean provably true for all and any function

because the Q says "which ordered pair must be on the graph"

[dusk phoenix]

hey guys

can someone help me with this question

[novel cargo]

what's the problem @dusk phoenix ?

[viscid thistle]

Hey I'm having a bit of trouble with this problem

i keep ending up with something that I can't factor

and I don't know if I'm just doing it wrong or what

[echo wagon]

Show your work

[viscid thistle]

huh ok 1 sec ill write it down

like

subtract both sides by 1
take the second term, -8, divide it by 2 and then square it, you end up with 16
add that to both sides

now you have 4x^2-8x+16=15

i tried factoring the left hand i cannot

[echo wagon]

Your problem is the 4 in front of x^2, before you complete the the square you must factor out 4. Or because you have an equation, divide through by 4

[viscid thistle]

Yea, I did

then you get x^2-2x+4 which you just cant factor

[echo wagon]

You have to get rid of the 4 before you decide what to + and -

[viscid thistle]

oh

uh

so i would just divide the initial equation by like

4

and i would have x^2-2x+1/4

right?

[echo wagon]

Yed

Yes

Then complete the square as you were doing

[viscid thistle]

hm i just

yea i dont think this is working

i got x^2-2x+5/4=1

i cant factor the left hand side again

is there some trick that im missing

Okay so

Idk what kinda math this is

But I have a very specific problem

I have this equation

And it is supposed to Sequantially output digits of Pi

But

It outputs them as decimals, and I need outputs as integers

0 -> 3.1333333333333333
1 -> 0.12942612942612944
2 -> 0.04222052457346574
3 -> 0.02075533661740559
4 -> 0.012313749155854425
5 -> 0.008145077498205814
6 -> 0.005784671552865616
7 -> 0.004319630382142555
8 -> 0.0033482289236764964
9 -> 0.0026712052667351854
10 -> 0.002180579380761778
11 -> 0.001813703745553055
12 -> 0.0015322017344245025
13 -> 0.0013114985592049856
14 -> 0.0011352659599557589
15 -> 9.923110450969705E-4
16 -> 8.747532268239622E-4
17 -> 7.769139035694725E-4
18 -> 6.946173203983882E-4
19 -> 6.247373535173935E-4

This is the output

But

[novel cargo]

is this precalculus?

[viscid thistle]

I think so

Im not sure tho

This is for a CS project

Anyway

I need a Sequantial function that outputs digits of Pi accurately and as integers

For example

1st iteration = 3
2nd = 1

3rd = 4

Ect

[novel cargo]

you want a new equation or a hacky programming workaround is OK?

@viscid thistle

[viscid thistle]

Not even programming

[novel cargo]

algorithm

[viscid thistle]

Im trying to design an ALU that uses an equation to calculate Pi in binarh Sequantially

I need a new equation

Binary*

[terse ravine]

Hmm

[novel cargo]

divide by 2

what are the angles in the interval 0 <= theta < 2pi for wich sin = -sqrt(2)/2

@terse ravine

[viscid thistle]

So uh

Is there a formula that fits the criteria

?

[novel cargo]

Im trying to design an ALU that uses an equation to calculate Pi in binarh Sequantially
@viscid thistle this probably the wrong channel for this. Ask maybe in #discrete-math or the more advanced channels

[viscid thistle]

Okay

[terse ravine]

5pi/4, 7pi/4

@novel cargo what do I divide by 2?

[novel cargo]

2 sin (theta) = -sqrt(2)

sin (theta) = -sqrt(2)/2 | divide by 2

which you basically got right

also, 7pi/4

[obsidian monolith]

JakeHillman12:

[viscid thistle]

any help would be greatly appreciated 🙂

A hint would do nicely, how would I start solving this equation?

[blissful ridge]

Can I see the original question?

Looks like simple factorisation to me

[viscid thistle]

this is the question

would you multiply the top and bottom by something, I'm confused

[blissful ridge]

$\frac{x^{-2}-1}{x+1}=\frac{\frac{1}{x^2}-1}{x+1}$

[obsidian monolith]

The Godfather:

[blissful ridge]

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

[obsidian monolith]

The Godfather:

[blissful ridge]

Now factorise the numerator and cancel out whatever you can

[viscid thistle]

Ah! Ok, thanks! 🙂

wait @blissful ridge (sorry for the ping!), how would you factorize the numerator??

[blissful ridge]

$a^2-b^2=(a-b)(a+b)$

[obsidian monolith]

The Godfather:

[viscid thistle]

$$\lim_{x\to 0} 4x \cot(\pi x)$$

[obsidian monolith]

JakeHillman12:

[viscid thistle]

need help on this one as well

for friend

would greatly appreciate some hint!

[hallow thunder]

@viscid thistle maybe this will help: $$\cot(x) = \frac{1}{\tan(x)}$$

[obsidian monolith]

jedben2:

[viscid thistle]

yeah but what is tan(0)

[hallow thunder]

0

[viscid thistle]

but 1/0?

Oh wait! so then it would end up being 0 then

[hallow thunder]

no

[viscid thistle]

1/tan(pi * x)

x is 0

[hallow thunder]

what happens if you use this in the actual limit?

[viscid thistle]

wait it wouldnt, would it?

use what in the actual limit?

[hallow thunder]

$$\lim\limits_{x\to 0} 4x\cot(\pi x) = \lim\limits_{x\to 0} \frac{4x}{\tan(\pi x)}$$

[obsidian monolith]

jedben2:

[hallow thunder]

what next? @viscid thistle

[viscid thistle]

is tan(pi*x) some sort of identity

[hallow thunder]

no

have you heard of L'hopital's Rule?

[viscid thistle]

no?

[hallow thunder]

that would help here

[viscid thistle]

is there a way to solve this algebraically

[hallow thunder]

you could use a taylor series

but l'hopital's rule is easier

in my opinion

@viscid thistle this goes over it: https://en.wikipedia.org/wiki/L'Hôpital's_rule

[viscid thistle]

Oh ok!

[hallow thunder]

you can use this here because plugging in 0 gives us a $\frac{0}{0}$ indeterminate form

[obsidian monolith]

jedben2:

[hallow thunder]

@viscid thistle are you able to evaluate the limit now?

[viscid thistle]

Don't use L'Hopitals...

$$\lim_{x\to 0} \frac{4x}{\tan(\pi x)}=\lim_{x\to 0}\frac{\pi x}{\tan(\pi x)} \frac{4}{\pi}$$

[obsidian monolith]

HoboSas:

[viscid thistle]

tan(f(x))/f(x) is a well know limit

[misty ocean]

If cos(t) = 0.7 for a constant t, explain why cos(t + 6π) must also equal 0.7.

i dont get it

[viscid thistle]

$\cos(x+2\pi n)=\cos(x) \forall n \in \mathbb{Z} $

[tardy ridge]

yes

n E Z lol

why'd you write x

[viscid thistle]

Fuc

[obsidian monolith]

HoboSas:

[viscid thistle]

It's late

[viscid thistle]

Hey, this may not be suitable for pre calculus, but can anyone please help me find a design and equation for this question? It the "wrap up problem" part

[terse ravine]

[rigid beacon]

uh

looks good to me?? @terse ravine

check in desmos

looks fine

[ember crane]

from step 3 to step 4 is confusing, that is mathematically possible?

someone explain step 3 to step 4 please

[simple edge]

Split the fraction into two simpler limits. If you multiply the expressions in step 4 you get step 3

[mossy tiger]

I'm in precalc right now and I just have a question so I don't mess up in the future. I am trying to find the domain of a rational function.

How do I solve for -x^2-11x-30

there's a -x^2 so do i factor -1 out?

[onyx wharf]

you can factor -1 out sure

[bleak lance]

How do I clear fractions in an equation?

In a y=mx+b and just in general?

[livid turret]

hi

trying to find vertex of this parabola and i get (2,8) but calculator says its -1,9

[uncut mulch]

show work

[livid turret]

i made mistake bc i did the 2 in 2a as an exponent

[marsh idol]

I think the limit is r but Im not sure

[willow bear]

as r approaches 0

the value of the limit cannot involve r

@marsh idol can you show your work for this problem

[marsh idol]

Im not exactly sure how to do this, need someone to explain

[willow bear]

i would begin by finding the coords of the intersection point of your two circles

or maybe even before that, you could write down the equations of the circles

to then find the intersection point from those

[marsh idol]

yes, i can do that

[willow bear]

you know the general form for the equation of a circle right?

yeah well do that

[marsh idol]

yes

and then?

[willow bear]

once you have the coords of the intersection point you can write down the equation of the line

all in terms of r of course

and then find its x-intercept

also in terms of r

and once you have that, you can then take the limit as r -> 0

[viscid thistle]

Is the limit 0?

[willow bear]

no

[marsh idol]

ok thanks, I'll get back to you once ive done these

[willow bear]

anyway the problem really isn't that hard, it just requires a little bit of calculation but not much more than that

that doesnt sound right at all

no

[viscid thistle]

made a mistake

(r^2/2, r(1-r^2/4)^1/2)

[willow bear]

that sounds more like it now

great you spoiled it

[viscid thistle]

After applying l'hopital

[willow bear]

...bruh

[boreal geyser]

||Uh the limit is 0 though right||@willow bear

[willow bear]

no it's not

[boreal geyser]

Lmao I can't do algebra mb

[viscid thistle]

Reee.exe has failed and stopped working

And apparently so has my brain

[proven marten]

Oh, hello @boreal geyser

[boreal geyser]

Hey bby

[stiff oracle]

Can someone explain to me how log9(cos2x+2) is equal to log3√(cos2x+2)

[lime bolt]

log_a^2(b^2)=log(b^2)/log(a^2)=2log(b)/2log(a)=log_a(b)

[strong ermine]

what are roots of unity and what are they used for

i have hw but we didnt get that far into notes

so i am a little stuck

[lime bolt]

the n^th roots of unity are the n solutions to x^n=1

[strong ermine]

um can you give an example?

its hard for me to understand cuz its a little abstract for me

[lime bolt]

well they are all exp(2kiπ/n), from k =0,1,..n-1

[pure delta]

I am not sure if this belongs here.. but:
"For every x> 0 we consider the quadrilateral Q with vertices A (x, 0), B (0,1), C (-x, 0) and D (0,1). When the centers of the sides of this quadrilateral connect, we obtain a rectangle R. Calculate algebraically the limit of the ratio of the circumference of these two figures as x approaches 0"

I hope someone could help me out!

[tardy ridge]

D=0,-1

?

ratio of circumference?

well one perimeter is 4sqrt(x^2+1), the other one is 2x+2

[tough shell]

What does that hole mean?

[tough shell]

Bro we have the same hw?

At least help me bro

[viscid thistle]

how do you do something like this?

I forgot the method

[rigid sun]

pythagoerean theorem

[viscid thistle]

huh

[rigid sun]

what is csc(x)=?

[viscid thistle]

1/sin

[rigid sun]

ok

[viscid thistle]

so sin theta is 4/7

[rigid sun]

good

what is cos?

[viscid thistle]

x/r

adj/hyp

im so confused

@rigid sun are u trolling?

I doubt he’s trolling

[rigid sun]

no

how can you find cosine from sin?

[warm nebula]

you can find the value of tan if you know the values of sin cos and tan using pytahgoreous therom

[rigid sun]

look at your formula sheet

or try to think of one

which one can you use

there's quite a few

how about this one

sin^2+cos^2=1

do you have that one?

[viscid thistle]

yea

[rigid sun]

ok

[viscid thistle]

Now use it

[rigid sun]

and solve for cos

[viscid thistle]

^^^

At which point you will have acquired tan(theta)

Make sense @viscid thistle ?

what does the restriction do

0<theta<pi/2

theta is an angle between 0 and 90 degrees (first quadrant)

That’s all it says

i got tan theta = 4/sqrt(33)

is that the final answer?

or do i do soemthing with that

angle

i rougly recall having to find multiple answers to a set

[warm nebula]

it should only be that 1 step. the restriction makes tan positive if you mean that as a second step

[viscid thistle]

so tan can be negative here because its +/ when u root both sides

+/-

?

i used to know this stuff

haven't done it in a year

and kinda freaking out

[warm nebula]

the first quardrent is (+) for all values how could it be negative

you dont need to calculate the root just make sure the value is true to the condition

[limber heath]

Can someone explain this to me?

[viscid thistle]

yes

you multiply the negative by
FOILing

(-) * 4 = -4 and (-) * -r = +r

so you get -4 + r

FOIL = First Outside Inside Last

its the distributive property when multiplying with variables

does that make sense (-) * (-) = (+) and (-) * (+) = (-)

thats the minus-plus rule

brackets mean multiplication

[limber heath]

I'm getting -4 - r because (-) * (4) = -4 and (-) * (-r) = +r so it's (-4) - (+r) = -4 - r

[viscid thistle]

its (-) * 4 and (-) * (-r) not (-) * (-) and then -*r)

you don't seperate - and r

-r is one componenet

[blissful kayak]

When you just see a floating negative sign, just think of it as -1**r*

[limber heath]

I'm lost

[blissful kayak]

And when you see something like the expression you were given at first (-(4-r)), you can just switch the terms inside the parenthesis and that will cancel out with the negative out front

[viscid thistle]

valk ur over complicating it

[blissful kayak]

So you can rewrite -(4-r) as (r-4)

[viscid thistle]

-4 is - * (+4)

[blissful kayak]

Yes

[viscid thistle]

so when you multiply another - it cancels out the negative and leaves you with +

I hope that clarifies

[blissful kayak]

Just like any negative number squared yields a positive number

[limber heath]

I believe I figured it out now. There is a invisible 1 before the brackets. -1 * (4 - r) = -4 + r

[blissful kayak]

Exactly!

[limber heath]

and something like this would be 5 − (4 − r) = 5 - 4 + r right?

[blissful kayak]

Mhm

[limber heath]

okay that's confusing because it's acting as both subtraction and multiplication but ill just roll with it

[blissful kayak]

But that's because that 5 out front isn't being multiplied by anything else

[limber heath]

thanks

[blissful kayak]

Ye!

[limber heath]

what are you taking right now?

[viscid thistle]

what does it mean to solve by inspection?

am i meant to just magically think of the value?

[blissful kayak]

Oh what the fuck?

I've never heard of that before...

[viscid thistle]

i mean 2 is the answer but how do u show steps

for that crap

or what do u write

cos(pi) = -1

[blissful kayak]

My friend came up with something

I'm not completely sure how he got it, but he says it's also 2

"-1≤19-5x²≤1 and then find the possible set for x
-2≤x≤2 we get
And put in values 2 and -2 so you get the answers.
The answers are x=-2 or 2"

He basically composed the cosine into its set of values for x

[tardy ridge]

you solve it by inspection because |cosx|<=1

there aren't many integers to guess.

theren't

[viscid thistle]

how on earth do you do this?

its the exact value

[blissful kayak]

theren't
@tardy ridge Good language my freind

*friend

how on earth do you do this?
@viscid thistle Oh these are pretty fun!

So basically when you have composite inverse Trigonometric functions, it's best to construct a right triangle using the value inside the inverse function

And since it's arccosine/inverse cosine, it's the adj/hyp

And that's enough to construct a right triangle with a leg of 9, a hypotenuse of 10, and another leg of √19 (Pythagorean Theorem)

And then what we do from there is we just find the tangent of this triangle we constructed, so opp/adj

And we get that to be √19/9

[quick mirage]

tan x = sin x / cos x

sin(cos^-1(x)) = sqrt(1-x^2)

cos(cos^-1(x)) = x

[viscid thistle]

what happens with the inverse?

[quick mirage]

might as well do this step by step

tan(cos^-1(9/10))

= sin(cos^-1(9/10)) / cos(cos^-1(9/10))

= sqrt(1 - (9/10)^2) / (9/10)

= sqrt(1 - 81/100) / (9/10)

= sqrt(19/100) / (9/10)

= sqrt(19) / 10 * (10 / 9)

= sqrt(19) / 9

[blissful kayak]

:3

[quick mirage]

i think this is what it is lol

[blissful kayak]

That's my result also

[quick mirage]

; )

whew

[feral monolith]

I did (a) but dont know how to do (b)

[blissful kayak]

No Trig?

[feral monolith]

no

[blissful kayak]

Awh...

[elder junco]

@feral monolith what have you tried?

This is a nice problem here.

[feral monolith]

I only learned how to do A so I dont really know what to do for B

[quick mirage]

pythagorean theorem seems to be all you need here

[elder junco]

Yeah

[quick mirage]

length = x, height = sqrt(x)

[feral monolith]

ohh

[quick mirage]

hypoteneuse = sqrt(length ^2 + height^2) = sqrt(x^2 + sqrt(x)^2) ) = sqrt(x^2 + x)

[feral monolith]

i thought i would need to use the distance formula to get measures

[quick mirage]

perimeter = length + height + hypoteneuse (a + b + c)

= x + sqrt(x) + sqrt(x^2 + x)

[feral monolith]

Thanks!

[quick mirage]

no problemo

[fiery wren]

is there anything else im missing

[quick mirage]

@fiery wren i believe the function is in lowest terms

[fiery wren]

dude

i JUST submitted my quiz

and that was the only thing

i got wrong

LOL thanks anyway

[quick mirage]

oof

srry man i was too late D:

[fiery wren]

noo its ok

thanks buddy

[viscid thistle]

I can't seem to get this question to work using y=ax^2+bx+c

it doesn't return a quadratic

i get a=0

how is it done

c = 5
4 = a +b +c
3 = 4a +2b +c

a+b = -1
4a+2b = -2

2a+2b = -2
4a+2b = -2

-2a = 0
a = 0

b = -1

but that makes no sense cus it needs to be a quadratic

anyone knows?

[tardy ridge]

then it is not a quadratic.

[viscid thistle]

but the question says it is

and when i use that equation it doesn't intersect with the first point

[tardy ridge]

f(x)=-x+5

f(0)=5

f(1)=4

f(2)=3

[viscid thistle]

how do i find a quadratic that passes?

its not asking for linear

am i missing something?

[tardy ridge]

then the question is probably wrong.

[lethal oracle]

Alright I’ll help

[tardy ridge]

linear eq. could be considered quadratic for some people.

[viscid thistle]

ooo

ur right

@tardy ridge

why though?

[tardy ridge]

0x^2+bx+c?

[viscid thistle]

why do some ppl consider linear quadratic

[tardy ridge]

I don't know

[lethal oracle]

Ya you’re right about it being linear

[tardy ridge]

the question probably just messed up.

[viscid thistle]

oh lol

this is looked at by pHD in math

idk man

[tardy ridge]

I don't blame him for messing up if it is a mess up because imagine having a PHD and writing quadratic questions.

[viscid thistle]

tru

if u root (y-9)^2 is it (y-9)?

[lethal oracle]

No

[viscid thistle]

sheiit

x^2 + y^2 -18y +75 = 0

and then how would u write that in acceptable form if there are 2 variables of y with different powers?

[lethal oracle]

Wait sorry you’re right

[viscid thistle]

oh okay

[lethal oracle]

I misunderstood you

[viscid thistle]

y = -x^2- sqrt(6)-9

is that correct?

it doesn't look right

[lethal oracle]

one sec

it should be y=-sqrt(6-x^2) +9

you can verify via graphig

[lethal oracle]

correct

[viscid thistle]

from y = sqrt(6) -x +9

how did u get that?

oh nvm

i see

ur right

can u show steps?

@lethal oracle

im having trouble getting to it

[lethal oracle]

ya give me a sec

[viscid thistle]

I figured it out

no worries

its what u did right?

[lethal oracle]

You add a negative since it’s the bottom half

[viscid thistle]

ooo

that makes sense

do you know what this is

[viscid thistle]

Is that abs value

[flint spruce]

I think they just want you to define absolute value

[ornate wolf]

Given two planes:

7x+8y+5z= 0,
7x+8y+5z= d,

Suppose the distance in between the two planes is 12 and d> 12, find d

This means two planes are parallel since their normal are parallael

what should i proceed with?

[viscid thistle]

Do these look right?

[willow bear]

f(g(x)) is ok, g(f(x)) isn't

[viscid thistle]

E

The answer is E

[terse ravine]

I thought it would take 2pi but it doesn't like it

[viscid thistle]

That's correct but there's an easier angle

[hidden willow]

just did math homework and im not very confident, can someone check it and maybe help if i got any wrong, not asking to cheat

[viscid thistle]

Sure just post

[hidden willow]

Not too sure about how to do that one

i know the x is replaced with -5/4

[blissful kayak]

Yes

[agile owl]

,w (x-2)^{4}

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by wolfram#9936

[uncut mulch]

not quite.
roots

[bitter basin]

i got it

[plain root]

anyone here good at pre calc and would we willing to do a voice call

to help me w my homework

[brisk ruin]

is this correct
@bitter basin Useful hint: complex roots always appear in conjugate pairs. So if 1+i is a root 1-i is also a root

[willow bear]

complex roots always appear in conjugate pairs.
this only applies to polynomials with real coefficients

which is the case here, but it's worth pointing out

[brisk ruin]

Ah of course. Thanks for clarifying 👍

[left fable]

Yep and also, it the polynomial is of odd degree, there is at least a real root.

[patent junco]

Working with Parametric Equations. Question is asking if there is a difference in the equation if b was 1 more than a instead of vice versa. I've ended with both:
2cos(t) and 2cos(-t). How are these different? I am drawing a blank atm

y(t)=2con((2-1)t) and y(t)=2con((1-2)t)

[willow bear]

cos(-t) happens to be the same as cos(t)

i'm not sure what your question is tho like

can you show the exact wording

[patent junco]

Writing a paper, ofc. So it's #5.

[willow bear]

wait ok so like

so point 5 is to examine the case where b = a+1?

so $x(t) = a\cos((2a+1)t), y(t) = a \cos(-t)$

[obsidian monolith]

Ann:

[patent junco]

I have just been assuming it's if we switched around a and b, where we use a=1 and b=2, instead of vice versa

but his wording is always weird

[viscid thistle]

How do you do this?

[viscid thistle]

@viscid thistle still need help?

[viscid thistle]

yes please

@viscid thistle

[limber heath]

^ help

nvm

im just dumb

[viscid thistle]

f(x) = [[x]] is the "greatest integer" function. The range of f(x) = x - [[x]] is:

I believe it is 0, but this is a multiple choice quiz and 0 isn't one of the options.

@viscid thistle have you tried anything?

Also you need a or b?

@viscid thistle quiz?

No its online homework

I believe it is 0, but this is a multiple choice quiz and 0 isn't one of the options.

My bad. It online homework. May of said quiz because I'm not used to hw being multiple choice.

[gritty stratus]

Hi does anyone know how to do this

I’m supposed to verify the identity

[viscid thistle]

Have you tried anything?

[strong ermine]

can someone help in #help-1

[viscid thistle]

@gritty stratus

[gritty stratus]

Hi no not yet I’m not sure where to begin

[viscid thistle]

Okay

Start by the defn of cot

Basically express cot² in terms of cos² and sin²

[gritty stratus]

👍

[viscid thistle]

Post what you get

[gritty stratus]

Cos^2x/sin^2x

[viscid thistle]

,rccw

[viscid thistle]

We don't want to express cos in terms of sin

So you should have $\frac{\cos²(x)}{\sin²(x)}-\cos²(x)$

[obsidian monolith]

Al𝟛dium:

[gritty stratus]

👍

[tardy ridge]

it should be -1 anyways.

[viscid thistle]

Now proceed with multiplying on the numerator and denominator by sin²(x) on the -cos²(x) to have the same denominator

Post what you get

[gritty stratus]

Is this what u mean?

[past meadow]

,rccw

[viscid thistle]

No

Only multiply the numerator AND denominator by sin²(x) on the -cos²(x) only

[gritty stratus]

Ohh

[viscid thistle]

It is also known as common denominator

[viscid thistle]

You know you can multiply the numerator and denominator by the same thing because it's basically like multiplying by 1, because they cancel

Okay good job

Now basically $\frac{\cos²(x)-\cos²(x)\sin²(x)}{\sin²(x)}$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Can you try to continue from here?

If not, tag me and we can continue

You are basically 2 steps from getting to cot²(x)cos²(x)

[gritty stratus]

Yay I solved it tyty

[viscid thistle]

Yw!

@viscid thistle both

i dont know how to do inverse

[tardy ridge]

for f inverse of 5 it's asking when the graph is hitting y=5

for the second one, it's asking when the graph is hitting y=f(11)

[viscid thistle]

what are the steps

[tardy ridge]

5=f(x)

[viscid thistle]

do u do e^ of both sides?

and then what is b asking

[tardy ridge]

it's asking when the graph is hitting y=f(11)

[viscid thistle]

im still confused

what is when the graph is hitting mean

[tardy ridge]

f(x)=f(11)

when the line of the graph is

ok I will show you picture

first one

[viscid thistle]

its asking for the y value?

[tardy ridge]

the definition of f inverse, first, f only has an inverse if f(x)=f(k) if and only if x=k

second f(x)=y, then f inverse of y= x

[viscid thistle]

what is f(k)

[tardy ridge]

f inverse of some number is asking what x value makes it output that number
f inverse of 5 is asking what value of x makes f(x)=5

k is some other value.

[viscid thistle]

what do i do from here?

or did i do something wrong

@tardy ridge

I can see that x=4 from just looking at it

but how do u do it though

like steps instead of just examining

[umbral cloud]

Hey guys

[viscid thistle]

hi

[umbral cloud]

E(x^2+1)=E(2x)

How to find the x?

[viscid thistle]

what is E

is this ur question?

x^2 + 1 = 2x

first step is to arrange it into ax^2 +bx +c form

then here I see it's a perfect square

[umbral cloud]

No

Hahahhahaha

[viscid thistle]

lol

[umbral cloud]

Not that the question

[viscid thistle]

what's e

[umbral cloud]

Wait XD

I know it in French dude.. Sorry..
But... Here is a hint
If x € [1;2[ then E(x) = 1

😂 I hope that it was simple as you wrote

[tardy ridge]

x=4 by inspection

[viscid thistle]

how to do it without inspection?

actually lets go to 2

f^-1(f(11))

you set y =12 + ln(8)

or did i understand wrong?

[tardy ridge]

it's asking f(x)=11 what the value of x is

but there is an easier way.

[viscid thistle]

wait so its the same as a?

[tardy ridge]

f^-1(f(x))

what is a

no

[viscid thistle]

f^-1(5)

i made f(x) = 5

then solved

wait

right?

so ur saying i want to now find what make f(x) = 11

[tardy ridge]

yeah sorry I meant f(x)=f(11) instead of that

[viscid thistle]

oh so i make it 12 + ln(8) = 1 + x + ln(x-3)

[tardy ridge]

you can get the answer directly from f(x)=f(11)

[viscid thistle]

ah yes

cus they cancel out

but im still solving by inspection

is there a better way for when the questions get harder?

[tardy ridge]

https://courseware.cemc.uwaterloo.ca/8

it'll be much easier if you take this course here.

[umbral cloud]

No dude not like that...

[tardy ridge]

I'm saying this will help him understand the material.

[viscid thistle]

bruh how u gonna do my uni like that

[umbral cloud]

Look...
Here is an exemple...
let x=0.6
the x^2+1=1.36
And 2x=1.2
Then E(x^2+1)=E(2x)=1

[viscid thistle]

f^-1(f(x) is a cancellation equation

[tardy ridge]

uni doesn't teach pre calc.

[viscid thistle]

bruh i know that

[tardy ridge]

yes that's right, only if f(x) is 1:1 correspodence.

[viscid thistle]

yea

are u in waterloo?

[tardy ridge]

yes

[viscid thistle]

what year

[tardy ridge]

can't tell you.

[viscid thistle]

im first year and cus we had corona before midterm i slacked off heavy in math

so i didnt really learn the content

[umbral cloud]

Well XD...
I'm not French but...
It's like I'm studying the last year in highschool

[viscid thistle]

im tryna put in more work now to catch up honestly

so i can do well

[umbral cloud]

Thx bruh

[tardy ridge]

muka who are you talking to?

[umbral cloud]

😂 I thought he was talking to me...
I've read only the last message XD

Excuse me

[spring thunder]

@umbral cloud have you solved it ?

well ✝️

[umbral cloud]

I have a mistake but idk where X")

[viscid thistle]

@umbral cloud

I dont understand where I went wrong

I got a 3/6 are my answers right

[umbral cloud]

1/ E
2/ D
Idk the answer of the last question cause I have not studied the exponential function yet...
Sorry

[viscid thistle]

I have a super difficult question

[unborn mist]

dont think you can get a closed form solution so id suggest doing it numerically

[shy salmon]

I have a super difficult question
@viscid thistle answer is pi

and its the only answer because the function is inversible

[viscid thistle]

Don't give away answers. The point of this server is to make them to understand it.

@viscid thistle do you still need help anyways?

[blissful kayak]

Graph it...

[gritty stratus]

I’m not supposed to I think

Is this right?

[blissful kayak]

That looks right to me

[gritty stratus]

Ok tyy

[blissful kayak]

Ye!

[uncut mulch]

,rotate

[uncut mulch]

nope

you can't apply the period post like that

[gritty stratus]

O

How am I supposed to solve it?

[uncut mulch]

$\cos\br{-3\theta + \frac{\pi}{2}} = -0.16 \
-3\theta + \frac{\pi}{2} = 2k\pi \pm \arccos(-0.16)$

[obsidian monolith]

ramonov:

[uncut mulch]

where k is an integer

[gritty stratus]

Ooo

[uncut mulch]

solve for theta and generate the solutions in the interval using appropriate values of k

[gritty stratus]

Okok thank

[severe bison]

info:

[past meadow]

,w 4^4+256

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Sneaky#0928

[past meadow]

crack

(i.e. the solutions aren't real)

[strong ermine]

i am retard

this is what happends when u are up for 21 hours

@past meadow

[olive quartz]

@gritty stratus if you are still on that here is what u need to do

256 can be expressed as 4^4

x^4 + 4^4

You won’t be able to do this with reals cause you’ll need a sum of squares here

[gritty stratus]

Oooh

Ok thank u

[viscid thistle]

hey

[agile moat]

Just transpose 256

$x^4=-256$

[obsidian monolith]

abcdefg:

[agile moat]

$\sqrt[4]{-256}=x$

[obsidian monolith]

abcdefg:

[agile moat]

$x=(2+2i)\sqrt{2}$

[obsidian monolith]

abcdefg:

[willow bear]

congratulations you missed three of the four solutions this equation has in C @agile moat

[agile moat]

+/-

[opaque olive]

quick question. ive got $\frac{dy}{dx}=\frac{-(8x+16y+24)}{16x+2y}$ the numerator lets me find out the stationary points, what does the denominator mean again?

[obsidian monolith]

Yes:

[opaque olive]

points where its parallel to the y axis was it?

[shadow plaza]

Those would be the points where denominator tends to zero

points where its parallel to the y axis was it?
@opaque olive i hope by "it", you mean tangents, right?

[spring thunder]

@umbral cloud déso je dormais à 2h du mat

t'y es presque, le problème c'est quand tu ajoutes les deux inégalités

le raisonnement ne marche que dans un sens

si je te dis qu'il y a deux réels a,b tels que 1<a<2 et 3<b<4 alors oui on aura 4<a+b<6

mais si je te dis que a+b est entre 4 et 6 c'est pas forcément le cas que a est entre 1 et 2 et b entre 3 et 4

c'est ça le truc, oui si x est solution alors x est dans (0,2) mais tous les x dans (0,2) ne sont pas forcément solution

ça te permet de réduire grandement là où tu dois regarder

tu passes de checker tous les réels à checker dans (0,2)

maintenant il suffit de regarder quels nombres dans (0,2) sont effectivement solution

et ça c'est assez facile tu peux raisonner par cas

[opaque olive]

@opaque olive i hope by "it", you mean tangents, right?
@shadow plaza lol yep xD

[viscid thistle]

could anyone help explain how to do this question

[scenic slate]

I need help on understanding how to do this

[lime bolt]

so basically u are looking for the term with degree -7, so when multiplied by x^7 is constant term. looking purely at the exponents, you can solve simultaneous equations for a+b=7, and -3a+4b=-7, yielding b=2. now u can just look at the binomial coefficient and solve for k.

(the 4 and -3 here come from the x^4 and x^-3 in the binomial expansion, and u are only interested in the term i mentioned, and obviously a and b are the exponents in the binomial thing)

[scenic slate]

im sorry I don't understand why the -3a+4b=-7 instead of 7

[lime bolt]

because u want the exponent to be -7

[scenic slate]

so the -3a refers to the x that is a denominator and the 4b refers to the x^4

[lime bolt]

yea that is where the -3 and 4 came from

the a and b are terms in the binomial expansion

well not terms but exponents from the expansion

[scenic slate]

so the a in -3a is 5 and the b in 4b should be 2

so it would be 7C5

[lime bolt]

what, k isnt 7C5

but now we have an equation where we can dervie k

[scenic slate]

ohhh I finally solved it out

THANK YOU!!!!

wait no nvm

[lime bolt]

[scenic slate]

I got to this point: 84 x X^8 x K^5 = X^8 x 168

[lime bolt]

(k/x^3)^5=k^5/x^15

so u get 168=x^7((1/x^7)k^5/4)*7C2=k^5/4*7C2

[scenic slate]

ohhh I tried doing it in multiplication so the it would be K x X^-3

[lime bolt]

it would still be x^-15 which is the same thing

but yea the whole point is so that the overall effect is 1/x^7 so the x's cancel and leave constant term

[scenic slate]

why did you do 1/x^7 though?

like wouldn't you have to multiply/divide all "terms"

and why did you divide the "K" by 4?

[lime bolt]

multiplication is commutative

[scenic slate]

if you multiplied the X^7 wouldnt you have to do the same with 168?

[lime bolt]

bruh have u read the question

it is saying the term that is constant equals 168 not some random one, and we are muliplying our whole equation by x^7, so u obvioulsy need the one with x^-7

i have already said this

[scenic slate]

ohh so when we multiplied the X by their coefficients they got cancelled out and therefore it is K^5/2^2= K^5/4?

[lime bolt]

i dont know what u mean by that

by yea if u are saying is (n/2)*y=(y/2)n then yes

[scenic slate]

when we multiplied -3 by 5 and 4 by 2 so that it would be X^-7 and therefore cancel the X^7 which is outside of that bracket

so the remaining is K^5 x 2^-2

[lime bolt]

well yea obviuolsly that is the point in the question

[scenic slate]

so K^5 x 0.25=168

THANK YOU

[lime bolt]

not quite

u also need to multiply by 7C2

in full,7C2(x^4/2)^2*(k/x^3)^5

[scenic slate]

thats what the a and the b represent in the -3a and 4b

b=2 and A=5

[lime bolt]

yea the exponents in that binomial thing

[scenic slate]

YES THANK YOU

BLESSSSS

[opaque olive]

What does constant term mean

[lime bolt]

x^0 coefficient

[opaque olive]

Oh i see

[scenic slate]

hold on I need help again

(X^4/2)^2 so does that mean the numerator is multiplied so it would be 2X^4/2^2

[lime bolt]

bruh

[scenic slate]

yo im just here for help

ohh do the exponents(powers) multiply

so it would be X^8

[viscid thistle]

does anyone know how to solve this question?

my answer is always bigger than the actual answer

Maybe I just dont get what this means but what is this?

Like what is this question asking

[viscid thistle]

@viscid thistle are u sure ur supposed to put brackets

I see brackets outside the question

[willow bear]

@viscid thistle why did you put a second pair of parentheses there lol

[viscid thistle]

Oh pffft I’m dum

[wicked hearth]

Can you cancel the x’s?

[gritty stratus]

How do I get from the first part to second part

[tardy ridge]

,r

,rwwc

,help

[obsidian monolith]

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[tardy ridge]

,rotate

[tardy ridge]

multiply by conjugate

@gritty stratus

[strong ermine]

does this look right?

[gritty stratus]

@tardy ridge ok thank u

[viscid thistle]

This would be helpful

[strong ermine]

any help is appreciated

[viscid thistle]

@viscid thistle yeah, it is (-6+/- *sqrt(36-12))/(-6)

because of y=0, hence -3x^2+6x-1=0