precalculus

[valid violet]

choose N = M^2 + 1

then sqrt(N) = sqrt(M^2+1)>sqrt(M^2)=M

and then the most important part of the proof

$\blacksquare$

[obsidian monolith]

gfauxpas:

[sturdy haven]

So what would l and m be

[valid violet]

?

[sturdy haven]

L and M

[valid violet]

don't know what L is

[sturdy haven]

Because there was no epsilon delta

[valid violet]

there's M and N

oh you're thinking of the ML inequality maybe

that's different

[wise bison]

My text mentioned under the section on partial fractions that substituting certain values of x to "clear out" one or more of the variables does not work. Can someone give me an example of when this is the case?

[wise bison]

Never mind, I think I see how now. Seems to be the case if there is a repeated prime quadratic factor in the denominator.

[warm crescent]

$\blacksquare$

[obsidian monolith]

Raftaar:

[fading token]

[dense oar]

[rigid beacon]

[bold isle]

My book did this to show that f(x) = 1 - sqrt(1-x^2) is continuous in [-1,1] But it just plugged the -1 and 1 in order to claim that f(x) is continuous in -1 and 1. My problem is that I don't get how just plugging the values directly into the limit means that it is continuous for -1, I mean how do I know the function is continuous to the left of 1 and to the right of -1 because just plugging in the values just means that that specifically point is defined in f(x)

[willow bear]

asserting that these limits are equal to the function's values at the corresponding points only shows continuity at -1 and at 1

[bold isle]

oh okay that's what I thought

[prisma prairie]

how do i find out if this is symmetric about the y,x or origin?

[rigid sun]

well

it's impossible to be symmetric over the x axis

or it wouldn't be a regular function, it would be an implicit function

if you arrive at the same solution by substituing x for -x, then it is even

if you arrive at a negative version of the original solution by plugging in -x, then it is over the origin

[prisma prairie]

how do we check for odd?

so i guess its not symmetric over the y axis either ?

[rigid sun]

it can't be symetric over both

that wouldn't make sense

[prisma prairie]

what i did nt say that

or did i?

[rigid sun]

"it wouldn't be symmetric over the y axis either"

yes

it cannot be both over the y and the origin

[prisma prairie]

oh

so it is symmetric over the origin?

[rigid sun]

idk

why don't you plug in

-x

[prisma prairie]

i did

[rigid sun]

and find out

[prisma prairie]

wait

dont you have to plug in -y and -x

not just -x

[rigid sun]

no

[prisma prairie]

oh

[rigid sun]

when you plugged in -x what did you get?

[prisma prairie]

=tex \frac{-x}{\sqrt{x^2+5}}

$\frac{-x}{\sqrt{x^2+5}}$

[obsidian monolith]

The-Elite:

[rigid sun]

ok

is that = -1*y?

[prisma prairie]

no cause i only made the x -?

[rigid sun]

what

[prisma prairie]

im confused

[rigid sun]

is that -y?

[prisma prairie]

idk

[rigid sun]

what do you mean

[prisma prairie]

what are we finding right now?

[rigid sun]

what is y

this is y

[prisma prairie]

yeah

[rigid sun]

is this

[prisma prairie]

yea

[rigid sun]

is this -y?

[prisma prairie]

i think

[rigid sun]

you think?

[prisma prairie]

lol

[prisma prairie]

yeah it is -y

[rigid sun]

ok

you plugged in -x and got -y

therefore its odd

[prisma prairie]

true

can we check for symmetric over origin now?

not sure what to do

[rigid sun]

that is symmetric over the origin

[prisma prairie]

wait how did you know

[rigid sun]

odd=symmetric over the origin

bruh

[prisma prairie]

lool

ohho k

didnt know that

[rigid sun]

remember

this means that all x, where x is postive, have a -x that =-y

essentially, when you cross the y axis, everything flips upside down or rightside up

that is over the origin

[prisma prairie]

i see

[rigid sun]

,w plot -x^3=y

[obsidian monolith]

Display results online and refine query

Requested by maleb1964#6739

[rigid sun]

have you ever heard how sin(x) is an odd function?

same reason

sin(-x)=-sin(x)

[prisma prairie]

yeah i have

if i have this would i just plug 0 into the top function?

then whatever i get from f(0) is the value of c?

[viscid thistle]

@prisma prairie try pluggin 0 into the top function

[prisma prairie]

yeah i did

[viscid thistle]

and what'd you get?

[prisma prairie]

=tex \frac{12}{7x^2}-\frac{4sin\left(3x\right)}{7x^3}

$\frac{12}{7x^2}-\frac{4sin\left(3x\right)}{7x^3}$

[obsidian monolith]

The-Elite:

[prisma prairie]

that ^

[viscid thistle]

typo left side, yes, what happens when x=0?

[rigid sun]

stop immediately

[viscid thistle]

err

not typo

oops

reduction.

my b

[prisma prairie]

what

[viscid thistle]

Either way

What happens when x=0?

what happens when you substitute x=0?

[prisma prairie]

it turns to that function i just sent

[viscid thistle]

when x=0, you get that function?

[prisma prairie]

oh wait

0/0

[rigid sun]

0/0 does not equal 0

it means

[prisma prairie]

yeah i didnt mean to say that

[rigid sun]

do some more work

[prisma prairie]

so i deleted it

[viscid thistle]

yeah, the function isn't defined at 0

so you can't plug in the values and figure out what it should be

however, in your context

you can find the limit as the function approaches 0 from the negative and positive side

if the limits are equal, that is what C should be

[prisma prairie]

oh ok

i seee

thx all

[summer sierra]

How to solve this question?

[viscid thistle]

@summer sierra what have you tried?

[summer sierra]

I was thinking of using one of the sum and differences formulas but I’m not sure if that would work

[uncut mulch]

what do you get when you use it?

[summer sierra]

The wrong answer cuz u just realized u have to use the double angle identities

I*

[viscid thistle]

the hint itself says "use the double-angle formulas"

I'd suggest using that

[uncut mulch]

the angle sum identities should work fine

they probably meant that instead of double angle

[viscid thistle]

they should result in the same thing

[uncut mulch]

what was your working out the first time you tried applying it?

@summer sierra

[summer sierra]

I'm not entirely sure if I combining the terms correctly so thats probably where I went wrong

[uncut mulch]

which identity would apply here?
cos(a)cos(b) + sin(a)sin(b) = ?

[summer sierra]

thats what i used

but i do not know what to do with the one

[uncut mulch]

well what did you get when you used it?

[summer sierra]

1.000

I mean 0.100

yea so im not doing it right

[uncut mulch]

huh?
cos(a)cos(b) + sin(a)sin(b) \neq 0.1 ?

when applying the identity, you should get the
cos of something.
what was the thing that you got?

[summer sierra]

cosine 6x?

[uncut mulch]

no

cos(a)cos(b) + sin(a)sin(b) = cos(fill this in)

[summer sierra]

cos a-b

[uncut mulch]

yes

[summer sierra]

thats the idenitity

[uncut mulch]

cos(a-b)

and how would you apply it to your question

[summer sierra]

but I dont know what to do with that

i inputted like 3-2

for like the insides

but im sure thats not what im supposed to do

[uncut mulch]

there are x in those terms

not 3-2, but 3x-2x

[summer sierra]

yea that was what i was thinking at first

but

[uncut mulch]

=cos(3x-2x) = cos(x)

[summer sierra]

yes

[uncut mulch]

now you need to solve
cos(x) = 1

(and the solution to that isn't -pi)

[summer sierra]

0

[uncut mulch]

yes. x=0 would be your solution

[summer sierra]

Thank you so much I really appreciate it

[wise bison]

Hey. Suppose I have a matrix and applied a row operation to it, can I write that the old matrix is equal to the new one using the = sign? If not, how would I connect each line in writing?

[stuck lark]

row ops can be written as left-multiplication by an elementary matrix

[willow bear]

no, you can't say they are EQUAL

[stuck lark]

the result of a row op on A can be written as EA where E is the result of applying the same row op on the identity I

[willow bear]

two matrices are only EQUAL when their corresponding entries are also equal

[wise bison]

So what symbol may I use then?

Can I just point an arrow?

Also, RokettoJanpu, I'm not really sure what you mean. Just started. Sorry. 😰

[stuck lark]

if B is the resulting matrix of a row op on some matrix A, then B=EA...

where E is the result of applying the same row op on the identity matrix I

any unfamiliar words in there?

[wise bison]

Identity matrices, and the multiplication indicated by EA? The only thing I've learnt is representing a linear system as an augmented matrix and am starting to learn Gaussian elimination.

I'm just trying to figure out how to present my work for the practice problems.

[stuck lark]

an arrow showing work flow is fine

just know that row ops generally do not make the old and new matrix equivalent

[wise bison]

OK. Thanks! I wasn't thinking they'd be equivalent, hence why I asked if I could use the = sign. Didn't think I could.

[stuck lark]

no prob

[wise bison]

Is this presentation OK?

[willow bear]

no, it's verbose and the brackets delimiting your matrices overlap between lines

[wise bison]

Oh. By verbose, do you mean I should remove the intermediate steps where I multiply and add?

[willow bear]

yes, and instead you should write the row operation above the arrow

[wise bison]

Ah! OK. Thank you!

Sorry, never did this type of thing before.

[wise bison]

Is this better?

[willow bear]

somewhat

[wise bison]

If you don't mind me asking, how else am I supposed to improve it?

[willow bear]

your notation for row operations is unclear

[wise bison]

Is there any standardised notation? It's not given in my book. They give all the intermediate steps like the first one I sent.

Oh! Am I supposed to write -2R₁ + R₂ → R₂ and -R₂ → R₂?

Ah. I'm still confused as to which side of the arrow is the side to be replaced and which side is the side we are replacing it with.

[plush matrix]

I need to check my answer

anyone willing to solve a property of log question?

3^(x+5)=4^(-3x+2)

[uncut mulch]

what did you get?

[rigid sun]

can't be bothered to work with this many fractions

[plush matrix]

kek

nah I think I got it

x=-0,5174

[uncut mulch]

did the question ask for a decimal approximation?
otherwise you should leave the answer exact

[trim fable]

i need help ;-;

with that

[plush matrix]

ya it asks to round up to 4 decimal places

[serene heath]

@trim fable try multiplying left fraction by 1-tanx

Top n bottom

[trim fable]

oh

but the cot?

do i multiply that too

[serene heath]

Wdym

[trim fable]

like

(1+cotx)(1-tanx)

so

(1+cotx-tanx-tancotx)

@serene heath i have a question is there a trick to make trig identities easier coz

it makes my brain go

🤯

[serene heath]

Uhh

Yea but tanccotx is just 1

So it's just cotx-tanx

[ivory echo]

Your best bet is probably to derive them imo.

At least that's how I remember identities.

[trim fable]

what do u mean

are u in uni? @serene heath 😮

also are u french 😮

[serene heath]

Na 7th grade

[trim fable]

whattttt

and how do u know trig???

[serene heath]

Did u figure out ur question

[trim fable]

no lol

[serene heath]

Ok so

[trim fable]

coz

i went to #9

lol

my question is like

[serene heath]

$\frac{(1+\tan(x))(1-\tan(x))}{\cot(x)-\tan(x)}$

[obsidian monolith]

the one n only:

[trim fable]

how do u plan everything in ur brain for trig coz

[serene heath]

This is what we have rn

[ivory echo]

I'll actually butt out while you're explaining lol.

[trim fable]

?

[serene heath]

U with me?

[ivory echo]

Was gonna explain what I meant by "deriving the identities"

[trim fable]

ik what u mean by it

but my question was how lol

for identities like that question

[serene heath]

Practise

Nothing more to it

[trim fable]

:c

[ivory echo]

Most trig identities come from just "playing" with it.

[trim fable]

:c

ok

ill play with them i guess 😦

I love mathhh but im getting bad at it ;c

i think

[ivory echo]

Just google "Proof of (blank)"

[trim fable]

or am I still good but not great

:c

[ivory echo]

Understanding why something works is much more effective at learning how, why, and where to use it.

[trim fable]

yess

wait so

[serene heath]

So u gave up on that q?

lol

[trim fable]

@serene heath can i work with the left side

[serene heath]

Yea

[trim fable]

ok

[serene heath]

Either side

[trim fable]

ok

im gonna attempt

[serene heath]

That's what we were doing all this wth

[trim fable]

once more

oh

[serene heath]

What have u been doing lol

[trim fable]

nothing

really

[serene heath]

Ok try what I suggested then

[trim fable]

ok

can u do me a favour

can u help me through the homework questions i dont get

coz i have a test on this and other stuff

[serene heath]

Just ask

[trim fable]

ok

🙂

yayyy

[serene heath]

Sm1 will.hel eventually

Probably

[trim fable]

yeahhhh

ok

[ivory echo]

Don't ask to ask.

[trim fable]

so

uh

rn im at

1-tan^2x / 1+cotx-tanx-tanxcotx so im left with cotxtanx?

right?

[serene heath]

Uhh

Dont expand the top

Leave it as it is

[trim fable]

oh..

huh

[serene heath]

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

From here

[trim fable]

ok

[serene heath]

Divide top n bottom by tanx

[trim fable]

huh

ok

hmm

im :c

confused

[serene heath]

With what

[trim fable]

life?

lol

[serene heath]

What do u get after dividing

[trim fable]

uhh

1+tanx/cotx

i hope

[serene heath]

Uh

No?

[trim fable]

lol

[obsidian monolith]

the one n only:

[trim fable]

huh

[serene heath]

After dividing top n bottom by tanx

Or equivalently multiplying by cotx

[trim fable]

wow umm

[serene heath]

Now u can do some simplification

[trim fable]

;-;

[serene heath]

By factoring the bottom

[trim fable]

;-;

ok

[serene heath]

So u got it?

[trim fable]

umm

ill see

lol

[ivory echo]

The bottom is a difference of squares. Factor it similar to how you might factor x^2 - 1

[trim fable]

maybe

ill moveon

[ivory echo]

Idk how to do the Latex so. ¯_(ツ)_/¯

[trim fable]

to the other homework and do

identities last

idk

[serene heath]

All u had to do was factorise and cancel tho lol

[ivory echo]

^

@serene heath Btw, just my opinion, but cross multiplying probs would've been a better idea at the start.

No need to convert trig identities other than multiplication identities.

[serene heath]

Cross multiplying what

[ivory echo]

https://cdn.discordapp.com/attachments/363224154469826562/650461070825095178/unknown.png

Might've been easier for him to wrap his head around.

[serene heath]

Ur assuming what u wanna prove tho

Bad habit

[ivory echo]

I take it you mean by modifying both sides?

[serene heath]

Yea

It's better to pick one side and turn it into the other

[ivory echo]

'Cause I'd say it's equally valid imo, it's one of the 3 ways to prove something.

[trim fable]

her*

[ivory echo]

My bad.

[trim fable]

its fine 😛

[ivory echo]

So much for Rule 16 xD

[plush matrix]

guys

how do I find the domain of (Sqrt(x+4))/x^2-5

[viscid thistle]

where does this 3 go...

https://i.imgur.com/PGj90Dy.png

in the last step

nevermind i know

[obsidian monolith]

JY1853:

[viscid thistle]

ty

i did it something similar i just did

3^1 * 3^(-1/2) = 3^(1/2)

[viscid thistle]

can somebody tell me what happened in the b lue circle?

https://i.imgur.com/37keAcq.png

nevermind... i just firgur out.. add 4 both sides

[rigid sun]

im glad you firgured it out

[viscid thistle]

well i wanted to ask waht happen, buit when i asked myself out loud i thats how i knowed

i say how did 4 happen to both sides, then i say oh, both sides

[shrewd tendon]

Could anyone help me solve #50

I’m pretty lost

[craggy olive]

i didnt even know matrices were in precalc

[summer monolith]

How do I go about a quadratic equation when I get a negative discriminant?

Thks

[earnest barn]

I used the binomial expansion formula to obtain a formula in terms of variables called k and r for the powers of x (where k ranges between 0 and 5 for the first bracket, and r ranges between 0 and 6 for the second bracket)

That formula was:
26 - 6k - 2r = ...

And that will give me the values of k (where k ranges between 0 and 5) and r (where r is between 0 and 6) for which I get the desired power of x

This worked for the coefficient x^-12. I used k = 5 and r = 4 to get 5C5 * 6C4 = 15 (exploiting the binomial coeffiicients)

For the coefficient of x^2, I've got that the pairs of k and r which give me 2 are

k = 2, r = 6
k = 3, r = 2
k = 4, r = 0

From this I got the equation

5C4 * 6C0 + 5C2 + 6C6 - 5C3 * 6C2

In order to get the coefficient of the x^2 term

The reason why there is a minus sign is because in my binomial expansion I had a term (-1)^k and since k = 3 that one term there will be negative

But this isn't giving me the right answer

Why might it not be giving me the right answer?

Here's the binomial expansion formula that I used to get the formula for the exponents

215 is meant to be the answer, not -135

[earnest barn]

Oh wait

I think I made an error typing it into my calculator

Ugh

Yeah

That was the problem

Sorry, don't need help anymore, typed in a '2' instead of a '3'

[odd helm]

Do I need a calculator to solve this?

Because I’m not sure I can’t figure it out

[rigid sun]

Not possible

the adjacent cannot be larger than the hypotenuse

[willow bear]

undefined*

[karmic wave]

at least when working in the reals...

[ripe dust]

I don't understand how it went from x to x/2

[rigid beacon]

@ripe dust (x/2)^2 = x^2/4 right?

also (1/2)^2 = 1/4

that should lead you on the right path

[ripe dust]

I mean

I did that

inside I have 64 - x^2

which is... what is in the picture since 4*16 is 64...

[rigid beacon]

yea

[ripe dust]

i just don't understand the x/2 outside..

[rigid beacon]

they factored out a 1/4 from the inside right?

[ripe dust]

yeah

[rigid beacon]

well more specifically

they factored out a sqrt[1/4]

[ripe dust]

oh ok thanks

[rigid beacon]

that's how 16 became 64

they multiplied 16 by 4/4

[karmic wave]

typo? 4/4 is 1 tho

[rigid beacon]

what typo?

[karmic wave]

"multiplied 16 by 4/4"

[karmic wave]

in other words "multiplied 16 by 1" unless I'm misunderstanding something

[rigid beacon]

well yea multiplied it by 1

but more specifically you need a 4 in the denominator

so like

4/4

so you get 4 in the denominator

[karmic wave]

maybe I should go to bed lol

[ripe dust]

Wait are you still here I'm a bit confused @rigid beacon

[rigid beacon]

yea what's up

[ripe dust]

so like

thats what I did but the answer is actually 0.5xetc etc

but like

shit

Can you re-explain again how it become 0.5

I removed the denominators 4 because they were both on 4

oh shit wait

Oooooooooooooooooooooooooooooo

I didn't think about factoring it,

I just removed the denominators since they were the same lmfao

goddamn it

[rigid beacon]

there ya go

[karmic wave]

moment_of_clarity.jpg

[odd helm]

When I did this I got y as my answer

I did inversef(x)= x since it is (y,x) and then f(x)= y

[rigid beacon]

Let's say f(a) = b

f^-1(b) = a

f(a) = b

so f(f^-1(b)) = b

f(f^-1(x)) = x

[odd helm]

But wouldn’t b=y though because it’s f(a)= b so b is the output right? And y values are the output when it’s a function that isn’t an inverse right?

[rigid beacon]

you're assuming that because f(a) = b

f(b) = a

which isn't true

[odd helm]

Hm so like I’m thinking that since it is (y,x) when we do inverse if we plug in a y we will output an x. And then we plug that x into f(x) and we output a y

I think I understand now

Thank you

[rigid beacon]

yea

I mean that's also part of the definition of an inverse function

f^-1(x) is the inverse of f(x)

where f(f^-1(x)) = x

[odd helm]

Yeah that makes sense

[unborn python]

I literally have never understood domains

can someone help me

[valid violet]

of a function?

[unborn python]

yeah

i never know how to find the function

like this

[viscid thistle]

@unborn python what do you have problems with?

(aside from cropping)

[unborn python]

how does this become -

when it says to add them

[viscid thistle]

how does what become? @unborn python

[unborn python]

with the + and -

[nocturne hollow]

hi for this problem do i add the vectors then find the magnitude of the sum? https://i.imgur.com/yByDnJN.png
or is that the wrong order

[native timber]

yes

[nocturne hollow]

ok thank u

[viscid thistle]

@unborn python the what

put the problem here

[unborn python]

ok

i solved it actually

[viscid thistle]

@warped willow that does indeed look like a problem

[plush matrix]

can anyone try and graph this parabola using Vertex and Focus?

V:(0,0)

F:(-3,2)

[warped willow]

i got how to do my problem

[lethal oracle]

Question is as asking to solve for X

[viscid thistle]

@lethal oracle what are you struggling with?

[lethal oracle]

Well I don’t really know how to get started tbh

[viscid thistle]

well, A + ? = B?

[lethal oracle]

Well A+3X=B in this case

[viscid thistle]

correct, but ignore that for now.

A + ? = B?

[lethal oracle]

A+0=B?

No sorry

I’m not sure

[viscid thistle]

So, think of elementwise addition for now.

row -> row

so, first row

1 + ? = 4?

[lethal oracle]

3

[viscid thistle]

ok, and 3 + ? = 2?

[lethal oracle]

-1

[viscid thistle]

sweet!

so you know A + [3, -1] = [4, 2]

Now, the question is looking for 1/3 of [3, -1], because it multiplies it by 3

so what is 1/3 * [3, -1] ?

[lethal oracle]

[1,-1/3]

[viscid thistle]

yeah! now, double check that

what does [1, 3] + 3[1, -1/3] equal?

[lethal oracle]

[4,2]

[viscid thistle]

yup!

so X = [1, -1/3]

[lethal oracle]

Thanks so much dude

[viscid thistle]

np, gl.

[inner knot]

hey, so when are you supposed to use ln instead of log? i thought that it was ln only if it involved e in the problem, but i have a problem for homework and the book uses ln but theres no e in it and im confused, i have a test tomorrow and ive got to know why lol

2^(1-x) = 3^(2x+5) is the problem, and solving for x

[chilly lotus]

Hey, I need help with radicals to exponent
Can anyone do radicals to exponents?

Solve for y, given 5y−8 = 5y−9:

changed subject

[rigid beacon]

That has no solution

[chilly lotus]

Exactly

Undefined

[rigid beacon]

So like

There's no radicals and exponents anywhere

[chilly lotus]

Oh yeah, I am past that

Spamakin, how old are you and where did you go to school?

[fleet yew]

@inner knot the use of ln is mostly arbitary

[inner knot]

define arbitary

im very big retard

[fleet yew]

you could just as easily solve it with the base 10 log, or the base 2 log, or any other log

it's just used for convenience

[inner knot]

so if you use ln you get the same answer as if you used log

?

[fleet yew]

yeah

[inner knot]

wtf

maybe i put it in calculator wrong

[fleet yew]

https://www.symbolab.com/solver/algebra-calculator/2^{\left(1-x\right)} %3D 3^{\left(2x%2B5\right)}

[inner knot]

oh damn

well then i just did it completely wrong

thanks man

[rich kite]

Hey does anyone know how to prove this identity?: 2tanx/cosx = 1/1-sinx - 1/1+sinx

[viscid thistle]

@rich kite what have you tried?

[rich kite]

I tried changing the 1's on the right side to sin^2x + cos^2x but it got too complicated

[uncut mulch]

try adding the two fractions on the right together

[rich kite]

ok

ok ty I got it

[distant basin]

If I have this equation, can I simplify the exponents between the parentheses first (by adding them), or do I have to multiply each one by the exponent of the parantheses?

Probably a silly question but I just want to make sure lol

[stuck lark]

either's fine

[distant basin]

Thanks!

[fleet yew]

What is the difference between principal root and real valued root

Wolfram always asks that for anything more than a square root

Like cube roots and 4th roots etc

[native timber]

A real valued root is a root which is real, a principal root is solely positive roots

For example, $\sqrt[4]{16} = 2$ is the principal root, but the real valued roots would be $\pm2$

[obsidian monolith]

Jiramide:

[astral mountain]

so how do you know when u need what?

[junior roost]

what in gods name

[warped willow]

draw it on a cord plane

turn it into a triangle and solve with terms x y

[strange adder]

hey can someone help me with this question

i got the answer 30sin(6pi(x))

not sure if thats right or wrong

[viscid thistle]

@strange adder negative distance?

[ripe dust]

babymod @native timber

[trim fable]

i have a question for

is there any double angle formula for sec^2x?

like is there any identity?

[viscid thistle]

@trim fable did you google it?

[trim fable]

google what

[viscid thistle]

"sec^2 identity"

[trim fable]

no

hmm

[viscid thistle]

try it

[trim fable]

ok

[native timber]

@ripe dust buby mode

[trim fable]

tan2 x + 1 = sec2 x

OH

WHAT

U CAN DO THAT?

who knew

hmm

wow thank u

lol

hmm wait

but its 2sec^2x

[viscid thistle]

@trim fable

2a+2b = 2(a+b)

2(sec^2(x)-tan^2(x))

[trim fable]

hmm

oh also

im stuck on a different one

i asked this here before but kinda like

skipped it coz i found it confusing but ye i get it but im stuck on one part

what i did so far is

cosx/cosx + sinx/cosx

sinx/sinx + cosx/sinx

=cosx+sinx/cosx

sinx+cosx/sinx

=cosx+sinx * sinx

---

cosx sinx+cosx

which got me to

sinxcosx+sin^2x

cosxsinx+cos^2x

so like sin/cos would be tan and cos/sin would be cot but like

ye..

idk where to do from here tho im close?

[wise bison]

Well, from (cos x + sin x)/cos x × sin x/(sin x + cos x), notice that we can cancel out the common (sin x + cos x) and be left with tan x.

Is there some way for us to show that the right side is also equivalent to this?

[trim fable]

well

we work with one side ;-;

to prove that its equal to the other

[wise bison]

Oh. In that case, multiply the simplified tan x by (cot x - 1)/(cot x - 1) and try to simplify it, perhaps?

[trim fable]

oh

[odd helm]

I’m not sure why it isn’t -35cos because instead of starting at the max we are starting at the min

I’m thinking this is a typo but if not could someone explain why?

[viscid thistle]

@odd helm why what isn't?

[viscid thistle]

please help

what is the area of the triangle ?

@viscid thistle what level of math are you allowed?

pre calc 11

i got an answer of my own with a scientific calc but i think it's wrong

ok

show your work

y1 = (x-2)^2
y2=|x-4|

2nd calc to find the intercepts

P = (3,1)
Q= (5,1)

Area = 1

to find the y intercept I used Tan

with the length on the x-axis

I think maybe I should used pythagoras to find the one side first

lemme work through it myself real quick

or perhaps law of sines

I solved it a different way and got 1 as well

your intercepts are correct

omfg thank god

it was on my final exam

lemme double check, but I think you're right

ahahaa

i was so concerned

how about this question...

double checked, you should be good.

sorry

bottom was x first I think

so it was $\frac{x+y}{x^{-2}y^{-2}}$?

[obsidian monolith]

Sun:

[viscid thistle]

I think answer is either X^2Y^2 / X-Y or X+Y

oh y

yeah

ye

sweet, considering $\frac{A+B}{C} = \frac{A}{C} + \frac{B}{C}$, you can do it that way

[obsidian monolith]

Sun:

[viscid thistle]

answer should be $x^3y^2 + y^3x^2$

[obsidian monolith]

Sun:

[viscid thistle]

no

answer is in a fraction

I can never use the bot when I need to

[uncut mulch]

comma w

space

[viscid thistle]

,w simplify (x+y)/(x^(-2)y^(-2))

[obsidian monolith]

Display results online and refine query

Requested by Sun#3606

[viscid thistle]

ah, there it is. love ya, @obsidian monolith

no

i told it to you wrong

one sec

,w simplify (x+y) / (x^(-2) - y^(-2)

[obsidian monolith]

Display results online and refine query

Requested by damanug#2689

[viscid thistle]

it doesn't solve the next step though..

,w simplify (x+y) / (y^2 - x^2)

, w simplify (x+y) / (y^2 - x^2) / (x^2*y^2)

brackets are a fickle mistress

[obsidian monolith]

Display results online and refine query

Requested by damanug#2689

[viscid thistle]

,w (x^2y^2) / (y-x)

[obsidian monolith]

Display results online and refine query

Requested by damanug#2689

[viscid thistle]

i dont think wolfram does simplifying fraction over fractions

to lowest terms

[uncut mulch]

depends on what terms you have

did you find out what your expression was supposed to be?

[viscid thistle]

(x+y) / (x^-2) - (y^-2)

that was the question - simplify in lowest terms

[uncut mulch]

so this one:

[viscid thistle]

yes

[uncut mulch]

wolfram's result looked quite simple

what's the issue with that?

[viscid thistle]

i think that's what I got

i factored out an x+y

i am stressed out because i dont know if i did well but seeing as we got the same area for the triangle question and this one i feel much better

the one i did the worst on I think was expanding a polynomial.

, w -(x+2)^2 (x)(x+2)^2

[obsidian monolith]

Display results online and refine query

Requested by damanug#2689

[uncut mulch]

was it actually written like that in your test?

[viscid thistle]

they gave me a picture of a graph

then I had to write the equation then expand

[uncut mulch]

strange that you have (x+2)^2 twice

[viscid thistle]

it was -(x+2)^2 x(x-2)^2

[uncut mulch]

was one of them supposed to be (x-2)?

[viscid thistle]

all they gave me was the graph i figured out the function but i fudging wasn't able to expand it to polynomial form.. DERRRP

did i write the function wrong ?

[uncut mulch]

was it cut off past x=-2?

[viscid thistle]

yeah

[uncut mulch]

the first pic you had was correct assuming it's monic
or if you were asked for a possible function

[viscid thistle]

the function is what i came up with

[uncut mulch]

just (x+2) without the square. since there's no indication of even multiplicity at x=-2

[viscid thistle]

oh no

i thought i was going to get at least 1 mark for getting the function right.. damnit

yep you're right

damnit

hm when i put in both are squared i get symmetry between the min and max on each side of the y axis

like the +y value is the same as the -y value in terms of displacement from the x axis

but if i get rid of the square like you said, then the y positive absolute value is way more than the -y value

but i think your way makes more sense

i guess ill see when it gets marked tomorrow

feel like i failed myself anyways, ill probably pass but still

[uncut mulch]

its a bit bad that they cut it off at a potential critical point

[viscid thistle]

true

to illustrate what I was talking about

on the test question the picture of the graph looked more like the one on the left

where the mins and maxs LOOKED like they were the same absolute value

@uncut mulch

[uncut mulch]

ic

[pseudo sonnet]

can someone tell me what formula was used kn the middle problem?

in*

the problem with A1 and A2

[stuck lark]

A_1, A_2 are triangles

[pseudo sonnet]

yeah i know

i know of heron’s formula and also A=1/2bh

is there another triangle area formula im missing?

[stuck lark]

that's only for right triangles. for non rights: https://www.mathwarehouse.com/geometry/triangles/area/images/SAS-formula/thumb1.jpg

[pseudo sonnet]

ohh

thanks man

[stuck lark]

no prob

[summer monolith]

Is this legal? To put a common multiplier behind the brackets for the 1st two summands neglecting the third one?

[willow bear]

why wouldn't this be legal

[summer monolith]

No idea. I imagined it would not for some reason.

Until i saw this in a textbook solution that is

[wise bison]

So I have this problem over here:

Describe a viewing rectangle that shows a complete graph of each polar equation while minimising unused portions of the screen: r = 4/(5 + 5 sin θ)
The supposed answer key answer is [-4, 4, 1] by [-10, 0.4, 1]. Not sure how this answer is reached, since these choices seem completely arbitrary. Normally the answer key places "Answers may vary.", but not for this one, which I find odd.

I suppose I can choose various values for y and plug it into a graphing calculator to see which is most appropriate, but I'm entirely clueless how they chose those values for x.

[rigid sun]

Unused portion of the screen?

Curves are infinitesimally thin

and have no area

[wise bison]

Beats me. Looks like a poor question paired with an equally poor choice of answer.

[valid violet]

I guess they converted to rectangular and optimized x and y

[willow bear]

$\log(x^2)$ is defined for $x < 0$ but $2\log(x)$ is not

[obsidian monolith]

Ann:

[whole socket]

how would i convert log base 3 of X to a base of 27?

[blazing raven]

oh

[willow bear]

using the change of base formula

[blazing raven]

yar (what she said)

then log base 27 of 3 is 1/3. so 3*(log_(27) (X))

You can also use the exponent method.

$$ K = \log_{3}(X) $$

$$ 3^K = X $$

$$ 27^{K/3} = X $$

$$ \log_{27}(X) = \frac{K}{3} $$

$$ 3*\log_{27}(X) = K $$

[obsidian monolith]

DumbRanterHehe:

[steel venture]

might be the wrong place for this (its an easy question either way) but
if i have a graph of y=1/x I can linearize it by plotting y vs 1/x
by that logic i should be able to linearize a graph of y=a*b/(a-x) by plotting y vs 1/(a-x) correct?

[valid violet]

looks right

[steel venture]

and slope should just be a*b with a y-intercept of 0 right

[harsh cipher]

Hello

Question: I don't understand how the equation went from P=Po*5.278)

P=Po(5.278) to P= 5.278Po. Then after 60 days (5.278Po)/(2Po)

[errant timber]

its in spanish but its finding the values, dont know how to do this

[dusky wave]

Oh ok

So I’m

Doing partial fraction decomposition and

I got to here right

And I’m trying to find B

Can I multiply 27 on both sides or no?

[wise bison]

Why not?

[obsidian frost]

as long as you do it for both sides i dont see the problem

[wise bison]

Yeah.

[obsidian frost]

are you trying to simplify the equation?

[dusky wave]

Yes

I got to here

I found A and C already

[steel venture]

@harsh cipher
at 60 days the population is Po(2^(60/25))
at 25 days the population is Po(2)

[harsh cipher]

hi

[steel venture]

then the population at day 60 is 2.64 times greater than at day 25 by doing Po(2^(60/25))/Po(2)

which equates to Po(2^35/25)

2^(35/25)=2.64

[harsh cipher]

and that is doubling time formula

[steel venture]

doubling time is Px=Po(2^(x/n))

x is the day at which you find the population

n is the number of days it takes to double

[harsh cipher]

okay

[steel venture]

and Po is the original pop, Px is pop at day x

[harsh cipher]

ty much appreciated.

[steel venture]

@errant timber you know C has to be equal to 1 as log(1) = 1 for any base, and the graph goes through 0,0 which means log(A*0 + C) = 0 => log(0+C) = log(1)

you know that logb(1.5A + 1) = -1 and logb((-3/8)A +1) = 1

therefore logb(1.5A+1)=-logb((-3/8)A+1)

1.5A + 1 = 1/((-3/8)A + 1)

(1.5A + 1)((-3/8)A + 1) = 1

(-9/16)A^2 + (9/8)A = 0

A = 0, A=2

and from there logb(2(3/2)+1) = logb(4) = -1
and logb(2(-3/8) + 1) = logb(1/4) = 1

as A cannot be zero because the function would just be logb(1) = 0 for all real b (i think)

and then 1/4 = B and 1/b = 4

hope that helped

[errant timber]

i see now, i mustve made a mistake in my algebra then

thank you, appreciated

[viscid thistle]

f(x) has the following zeros -2, i, -i; find f(x)

not sure how to do this

[steel venture]

(x-i)(x+i)(x+2)

[hollow horizon]

how do i solve this

[steel venture]

(x^2+1)(x+2)

[viscid thistle]

maximo, the human calculator.

[steel venture]

x^3+2x^2+x+2

@viscid thistle

yessir @viscid thistle

[viscid thistle]

uh

Wow

I want that skill

im not entirely sure what you did

magic

[steel venture]

so you know the zeroes are i,-i, and -2

[viscid thistle]

yes

[steel venture]

that means that the function can be written as 0=(x+2)(x-i)(x+i)

because if you plug any of those numbers

the function is zero

[viscid thistle]

right

[steel venture]

then you do (x+i)(x-i)

which is x^2+ix-ix-i^2

which is x^2+1

then you do (x^2+1)(x+2)

which is x^3+2x^2+x+2

did i lose you

[viscid thistle]

no not at all

[steel venture]

ok

[viscid thistle]

i got it now thank you

just looked weird on screen but wrote it down makes sense

[steel venture]

yeah i messed it up a lil up top but the second explanation is right i think

@hollow horizon you can multiply everything by sinx yielding sin^2x + 2sinxcosx = 1/2 sinx

or wait no you cant

[hollow horizon]

yea i tried that lol

[steel venture]

sin^2x +sin2x -1/2sinx = 0

sinx=u

yup nvm lol

@hollow horizon square both sides? 1+sin2x = 1/4

or 1+2sin2x = 1/4 i believe

[hollow horizon]

yeaa but thats a gay solution

i hate doign solutions like thaaaaaaaaaaaaaaaaaaaat

but ig its the only thing to do

[icy spruce]

can someone help me with this problem i have no idea what to do

[wise bison]

This is a parabola which is shifted down by 2 units. For y = x²/4, we can rewrite this as x² = 4y. Comparing this with the equation of a parabola, x² = 4py, we see that p = 1. The directrix has the equation y = -p, so in this case, y = -1. We shift the whole graph down by 2 units, giving us y = -3 as the directrix.

[icy spruce]

How'd you get p to =1?

[wise bison]

The standard equation for a parabola is x² = 4py, yes?

[icy spruce]

yes

[wise bison]

Ignoring the -2 for now (that's a vertical shift), we have the equation y = 1/4 × x², which simplifies to x² = 4y.

We can compare 4y to 4py to deduce that p = 1.

Because p = 1 is the only value of p that makes 4y equal to 4py.

[icy spruce]

oh ok i understand now

Thank you very much

[timid dock]

anyone on

how do i do sin(arcsin(-1.6))

wait

[willow bear]

uhh

[timid dock]

what about cancelation laws

[willow bear]

no

no

no

[timid dock]

^

[willow bear]

no

[timid dock]

^

[willow bear]

no

[timid dock]

^

[willow bear]

no

no

[timid dock]

^

[willow bear]

arcsin(-1.6) is undefined

[timid dock]

so sin(undefined) = undefined

[willow bear]

what else could it be

[timid dock]

ok

thank you

for helping me again

one more question

why does arccos(cos(3.8)) only get 1 value

[willow bear]

what do you mean by "only get one value"

[steel venture]

it should give you 2 values if im not wrong

[timid dock]

how many answers should there be

[steel venture]

infinitely many

but 2 values is standard

for example, the cos of 3.8 is in the third quadrant

[timid dock]

mhm

[steel venture]

instead lets use the cos(5pi/4)

[timid dock]

what about the entire "being in the domain crisis"

[steel venture]

the what in the what

oh

cos and sin can take all real numbers as their inputs

however arcsin and arccos only take numbers in the [1,-1] range

[timid dock]

ok lemme try this problem

arccos(cos(3.8)) = x
arccos(-0.78) = x
x = 0.7

[hexed bolt]

Could I get help for this question please?

[steel venture]

koala can you go to #help-4

kenja

i don't think that's right

[timid dock]

me too

i might be terribly confused

help me

[steel venture]

okk

so the cos was right

the arccos however was not

or well the cos was almost right

[timid dock]

idk

tried estimating

[steel venture]

its -.79

can u use a calculator by any chance?

[timid dock]

no

not allowed for my class this unit

[willow bear]

y'all fucking nuts
arccos is single-valued

[timid dock]

aha

[steel venture]

ann no

fight me

whatever

kenja

the answer will be 3.8

thats what theyre looking for

however 3.8+pi will work too

and so will 3.8+2pi

[willow bear]

3.8+pi won't work

[steel venture]

oh oops

yup

it wont

[willow bear]

but the thing is

[steel venture]

its 2npi

[willow bear]

arccos

is

single

valued

it

always

returns

a

single

value

between

[steel venture]

within 0 and pi

[willow bear]

zero

and

[steel venture]

yeah

[willow bear]

pi

[steel venture]

but

yeah

ur right

[willow bear]

no buts

[timid dock]

so it cant be 3.8

[willow bear]

it cannot be 3.8

it's 2π - 3.8

[timid dock]

why do we subtract from 2pi

[willow bear]

cos(2π - x) = cos(x), and 3.8 is in (π, 2π)

[timid dock]

y does it seem like i'm not taking in any information from class

[willow bear]

🤷‍♀️