precalculus

[round geyser]

Yea

[sullen gull]

Log4(x/y)=1

[round geyser]

Then it’s
log_4 (x/y) = 1

[sullen gull]

Yea

You're getting the hang of it, good

[round geyser]

Then it’s (x/y)=4

[sullen gull]

Now 1 is just log of base 4 of 4

Yes

[fiery spear]

yes yes there's your proportion

[sullen gull]

Good

[fiery spear]

now top equation similar

[round geyser]

Great

[sullen gull]

X=4y

Then just use substitution

I correct myself

[round geyser]

4y^(log_4 y) + y^(log_4 x) = 4

First one

[sullen gull]

Yea

[round geyser]

4y^(log_4 y) it’s just y

[sullen gull]

Yess

[round geyser]

As the rule
a^log_a c = c

[sullen gull]

Yup

[round geyser]

y+y^(log 4 x)=4

Look we can sub it in again

x=4y

[sullen gull]

Yes

[round geyser]

y+y^(log_4 4y)=4

Can we cancel

[fiery spear]

what do you want to cancel

[round geyser]

log_4 4y

[fiery spear]

i want to go back to 4y^(log_4 y) it’s just y

[round geyser]

Yeah

y+y=4

And 2y=4 so y=2

[fiery spear]

4 y^(log_4 y) is what we start with right

what rule applies to y^(log_4 y)

[round geyser]

power rule

[fiery spear]

show the rule?

[round geyser]

Wait

That doesn’t look right

[fiery spear]

yeah you need a different rule

you have a^(log_b c)

[round geyser]

change base

[fiery spear]

we know a = e^p for some p

yes

[round geyser]

Make it

y^(log y/log 4)

[fiery spear]

a^(log_b c) =
e^p^(log_b c) =
e^(log_b c)^p =
e^(log c / log b)^p =
e^(log c)^(1 / log b)^p =
c^(1 / log b)^p = 
c^(1 / log b)^(log a) =
c^(log a / log b) =
c^(log_b a)

[round geyser]

1.0?

[fiery spear]

programming habit

[round geyser]

So

What’s the p,c, b, and a

[fiery spear]

p is log a as defined but i ultimately proved the power rule you need

a^(log_b c) = c^(log_b a)

[round geyser]

Let’s do 4y^(log_4 y)

4y=a

[fiery spear]

a = 4y b = 4 c = y

[round geyser]

So it’s y

[fiery spear]

y^(log_4 4y)

[round geyser]

Question where’s the e from

[fiery spear]

i define a = e^p i.e. log a = p

[round geyser]

y^(log_4 4y) =
4y^(log_4 y)

[fiery spear]

anyway you have y^(log_4 4y) + y^(log_4 4y) = 4 now yw

[round geyser]

Ok

Since these two are the same

Then it’s
y^log_y a = a

[fiery spear]

y^(log_4 4y) + y^(log_4 4y) = 4
y^(log_4 4y) = 2

not sure where you went from there

[round geyser]

Oh

Where’s the 2 from

[fiery spear]

half of 4

z + z = 4

[round geyser]

Okay

[fiery spear]

therefore

z = 2

[round geyser]

Oh

x=2 now we can plug it back into
x/y = 4

y=2

[fiery spear]

i think you're jumping to conclusions

we're here

y^(log_4 4y) + y^(log_4 4y) = 4
y^(log_4 4y) = 2

since z + z = 4  =>  z = 2  for
z = y^(log_4 4y)

so log_4 4y = log_y 2

[round geyser]

Can we set proportion

[fiery spear]

why not just expand the log_bs

[round geyser]

Oh

[fiery spear]

log_b a = log a / log b

[round geyser]

Change base

[fiery spear]

yeah normally just change base to or from e

[round geyser]

Then log 4y/log 4 = log 2/log y

[fiery spear]

ye now proportion again

[round geyser]

Now can we just add or we multiply inside the logs

Is it like
log(y(4y))

[fiery spear]

(log 4y)(log y) = (log 2)(log 4)

product of log is power of interior

a log b = log b^a

[round geyser]

Oh

But it looked like it’s not

[fiery spear]

hmm

[round geyser]

Let’s expand (log 4y)

[fiery spear]

log 4 + log y

[round geyser]

Into (log 4 + log y)(log y)=(log 2)(log 4)

We can’t cancel

But can we divide

[fiery spear]

yeah let's go back here

log 4y/log 4 = log 2/log y

1 + log y / log 4 = log 2/log y

easier

[round geyser]

Okay

[sullen gull]

Btw are partial derivatives calc 2?

Or 3

[fiery spear]

usually intro'd in 1 and used in 3

[round geyser]

Can we split into (log 2 + log 2)

But that won’t do anything

[fiery spear]

fyi here we are assuming y != 1, so could miss that solution

[fiery spear]

collect by log y

[round geyser]

Collect ?

[fiery spear]

like collecting coefficients by terms

here i'll make simpler

[round geyser]

Ok

[fiery spear]

1 + log y / log 4 = log 2 / log y right?

1 + z / log 4 = log 2 / z

[round geyser]

Yes

[fiery spear]

solve z (it = log y)

[round geyser]

Then it get

z^2+z=log 4*log 2

[fiery spear]

that's more like it

quadratic now

[round geyser]

Can we subtract log 4*log2

[fiery spear]

ye that's ur c in quadratic equation

solve quadratic and then exponential and get y

[round geyser]

Can I complete the square

[fiery spear]

objective is just to solve

use standard QE

[round geyser]

Ok

[fiery spear]

as usual

[round geyser]

Ok

Is this right

-1 - 4(log 4)(log 2)

That’s the discriminate part

[fiery spear]

(-1 - sqrt(1 - 4(log 4)(log 2))) / 2
(-1 + sqrt(1 - 4(log 4)(log 2))) / 2

i got both of those

[round geyser]

Me too

[fiery spear]

so exp each of those

gives each y

and x is ez from there

what funny numbers

[round geyser]

Log can’t have negatives

So - part can cancel

[fiery spear]

i checked and 1 is not a solution so we didn't miss one by assuming on y

that's the thing you have to care for with equations like this. taking assumptions causes you to miss solutions

[round geyser]

2 log y = log y^2

[round geyser]

It’s the negative

[fiery spear]

oh both of those are solutions

but at one point we assumed y != 1

and that could be invalid but was not

[round geyser]

Ok

Take two cases

Now I’m on log y^2=-1 + sqrt(-1-4(log 4)(log 2))

[fiery spear]

i got these numbers

[round geyser]

Could we add one

This is funky

[fiery spear]

just log y = ...

so y = e^...

[round geyser]

So the the e

[sullen gull]

Can somebody help me solve this problem:

[round geyser]

Okay

[sullen gull]

Int(0 to 2) (arctanx)/1+(1+x)²dx

[round geyser]

Honestly that’s the hardest one

Also I never seen these questions

[fiery spear]

unfortunately there was a mistake

but it can be found, just collect all the steps

[round geyser]

These two

It took me almost 1 hr

[sweet flower]

i think ure supposed to know the integral of arctan

lemme search it up rq

oh right it's just parts

wait nvm im stuck asw 😭

[topaz maple]

what question you working on?

[acoustic apex]

What is up with that dubious bracketing

[topaz maple]

Is that church hill in your pfp

[acoustic apex]

no

[tawny viper]

is trigonomentry precalc?

[gusty merlin]

i failed my precalc test

[vast rivet]

Damn.

[main peak]

guys is the area of this slope this integral: (if i seem dumb just know that i have not reached grade 9 bruh.)
2
∫ x^2 dx
0

[sullen gull]

Yes.

No you don't seem dumb

But also depends which slope u talking about

The moment it becomes dy shit changes

[main peak]

bru wth is dy

ok nvm

[jaunty rock]

theres no calculus in GCSE bro

[sullen gull]

Bro glitches for a moment

Glitched*

[main peak]

wait what

[sullen gull]

Forget it dude lol

This guy probs just enjoys calc

And so do I

I'm in class 10

But I learnt calc aswell

[light oxide]

additional maths?

[main peak]

yes i like learning high-level topics earlier

i want to join kumon but some people say it's hell for kids

[vapid coral]

they're american

[rigid tusk]

If u want I can send some good materials for learning them

[round geyser]

Wow

[main peak]

YES YES YES YES YES I DO TYSM

bru what am I doing

[round geyser]

Question for u

How come pulling Leading coefficient works

On both 8 and 9

[raw hill]

divide both the numerator and denominator by $x^2$ and notice how everything goes to zero except for 6 and -3

[obsidian monolith]

Civil Service Pigeon

[raw hill]

you can generalise this logic quite easily as well to any rational limit tending to infinity/-infinity

[round geyser]

Like 8 is x^2/-5x^2

Cancel x^2

Got me -1/5

9 is 6x^2/-3x^2 is just -2

[round geyser]

I assume pulling leading coefficient works

[raw hill]

My point was that
$$\frac{x^2+3x-4}{1-5x^2}=\frac{1+\frac{3}{x}-\frac{4}{x^2}}{\frac{1}{x^2}-5}$$
and notice how $\frac{3}{x}, -\frac{4}{x^2}, \frac{1}{x^2} \to 0$.

[obsidian monolith]

Civil Service Pigeon

[round geyser]

It’s better than this

[raw hill]

I mean that's basically where it comes from

it's just that you don't rlly have to write it out

[round geyser]

Aka complex fractions

[raw hill]

cause it's the same every time

like I said, you can generalise this to any rational limit that tends to inf/-inf

[round geyser]

My teacher taught me on pre-calculus that we can pull out like terms

And it’s the only time we are allowed

[raw hill]

alright

[round geyser]

Good

I don’t need to do the complicated way

[round geyser]

3 ways to solve limits, direct sub, rewrite expressions, and trial of error

Am I right

[raw hill]

eh depends on what you've been exposed to

everything has levels to it

like you have l'hopitals, polar, series expansion, etc as well

[round geyser]

Oh I’m on pre-calc

[round geyser]

No, I didn’t see it or learned it

[round geyser]

U know what’s trial of error right

[polar phoenix]

in absolute values y= |-2x+2| does it stay the same or change

i have no clue on how to graph it

[round geyser]

Ok

Do u know what’s y=-2x+2 look like right

Taking absolute value means you are reflecting the negative part up into the positive

[polar phoenix]

this is my small idea on how it looks like

im so cooked

[round geyser]

No it doesn’t look like that

Start with 2

(0,2)

Rise/run

Make -2 into -2/1

Down 2 left 1

[polar phoenix]

-2x/1

was my vertex correct

[round geyser]

You should have (1,0) graphed

Then connect a line down

Since it’s a negative slope

[polar phoenix]

so vertex is 0,2 and goes up -2/1

[velvet meadow]

ily

[round geyser]

You know what I mean right

[round geyser]

Yes

[polar phoenix]

anything below the negative becomes the opposite above?

[round geyser]

And take the negative and shift it up

Up 2 right 1

Then your vertex is (0,1)

[polar phoenix]

how did it become 0,1

i understand everything else except the vertex

[round geyser]

Do you see (0,1)?

It’s the tippy bottom

[polar phoenix]

oh i see what u did there

thank you

[round geyser]

Yw

[polar phoenix]

did i have the correct idea the red is the y=(x+2)squared -2 while the green one is the absolute version

hear me out, so its basically joe basic, but since its asking for the absolute value of that, negative 2 becomes positive 2 and -1 just becomes positive 1

[round geyser]

Yes

I did these before

[polar phoenix]

do we also mark where they intersect or is it not required

[round geyser]

Mmm

Depends

Just in case do it

When I did these

I mark the steps

By using different colors

First one is the parent graph

[polar phoenix]

was there a case where it becomes -2 is 1/2 im pretty sure im thinking of something else but i remember my teacher showing sum like that

in the same lesson of this

[round geyser]

1: y=(x+2)^2
2: y=(x+2)^2 -2

The last step

Graph it’s absolute value

I recommend these three use three different colors

Like choose your 3 favorite color pencil

[polar phoenix]

oh ok i'll do the same then when handing this in, just to avoid confusion

[round geyser]

For example
1: y=(x+2)^2 (blue)
2: y=(x+2)^2 -2 (orange)
3: y=|(x+2)^2 -2| (green)

You get it

If on the quiz your classroom doesn’t have color pencil

Mark the equation you used

And also use dotted lines

In my case I get to use color pencils

[polar phoenix]

luckily we use pens so im fine

just gotta get the graphing correct lol

[round geyser]

Ok

[polar phoenix]

so i did it algebraically, as for graphically do i need to change it to a different form?

[tulip olive]

u didn't settle out the "|"?

[polar phoenix]

im not sure if ik what you mean

[tulip olive]

coz its should be -x² + 3x + 15 = 25

you reversed the symbol as we want to eliminate the modelus sign

[polar phoenix]

oh wait

i see know

[tulip olive]

@polar phoenix i got X1 = -2 and X2 = 5

[polar phoenix]

is this for the 2nd part

cus i think thats where i went wrong

[tulip olive]

wait ah

[polar phoenix]

wait actually

is this not correct? if your solving absolute value u gotta do 2 versions of it one negative and one positive? im prolly mixing the terms up or whatever u call em

hence x²-3x-15=-25 and x²-3x-15=25

im still not sure where you got the -x² though, im prolly missing something or i cant see it

[tulip olive]

the modelus function as to make negative become positive

@polar phoenix

[polar phoenix]

wait, i didnt know it changes like that

[tulip olive]

blamed the modelus XD

so ur graph should be like this

[polar phoenix]

god damn i think i just did my whole paper wrong then

take 2 for example

is this not correct though? i checked the answer key for this and it was correct

if its the same as your way, it would be x+3=11

[tulip olive]

so the answer is correct?

[polar phoenix]

for the one we did im not sure, i dont have the answer key but im sure yours is correct

[polar phoenix]

this was the answer key for this

[tulip olive]

oh i see

yeah its correct (No need to redo)

[polar phoenix]

alr, thank you

[round geyser]

Set it as plus or minus cases

[round geyser]

Also this works where you pull out like terms

8/3

U know how I got this

[round geyser]

Can some one give me an example on this “The Assessment - You will ONLY NEED TO KNOW HOW TO GRAPH LOG FUNCTIONS and also as well as knowing how to Expand and Condense Logs using the Laws of Logs for the Assessment you will have 2 BONUS questions - 1 Expanding and 1 Condensing that you will be able to do.”

[echo sinew]

Probably "simplify log_10(4) + log_10(25)"?

I don't know, this seems something you should ask your teacher directly

[round geyser]

Okay

[round geyser]

log (4 x 25) =2

[echo sinew]

Yup

[frosty dust]

L'hospital

[echo sinew]

No need at all!!

Please, tell me you never used L'H with such questions...

[frosty dust]

Ik

Only x terms matter

So you can remove the other part

Cancel x's

[echo sinew]

Oh lol I didn't realise you were joking 😁

[frosty dust]

Done

[echo sinew]

Because unfortunately I'm pretty sure there are people that would use it there too

[frosty dust]

How did u do it then?

Only X's terms matter

[round geyser]

Yea

That’s what I did pulling out the leading term

[frosty dust]

Yes

[round geyser]

And one and done

[frosty dust]

That's what I'm sayin

Take leading terms

Cancel x

Done

[round geyser]

Yep

That’s also on my unit in Pre-Calc

If we have drive it

Thanks to L’H rule

[obsidian solar]

you are right

[delicate grail]

For the first verification here can someone explain why its not -2(x/2) and then its equal to -x i dont what i messed up on

[raw hill]

Recall that the domain of the inverse is the range of the original function. Since the range of the original function is all non-positive reals, the domain of the inverse is all non-positive reals. \ \

Thus, $x$ is non-positive. As a result, we have that
$$\sqrt{x^2}=|x|=-x.$$

where the last part comes from the piecewise representation of the absolute value function:
$$|x|=\begin{cases} x, & x>0 \ -x, & x \leq 0 \end{cases}$$

[obsidian monolith]

Civil Service Pigeon

[raw hill]

,w graph |x|

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[delicate grail]

oh i see it now thanks

[raw hill]

[round geyser]

I can tell u the domain since sqrt(x) is x>0

[round geyser]

What about this

Two cases

[raw hill]

Me when equality for triangle inequality holds so I can |5x+2|=25

[round geyser]

I got this

[hollow widget]

What can you cancel?

[round geyser]

25^0

And (120/5!)

Since 5!=120

[hollow widget]

Yea you can cancel a lot of stuff and then have an equivalent expression and solve that for x

[round geyser]

I’m I remember correctly it’s sqrt(8x)

[visual plaza]

can someone plx explain

for horizontal compression and stretch

and its relation to this?

because i assumed you would divide x by 2 or smth because we tend to do the opposite to x

sorry if this is a dumb question im half asleeo rn im not even joking 😭

[ripe scarab]

whats the range of f(x)=sinx-3?

[serene dagger]

-4 <= f(x) <= -2

[round geyser]

Wait so

On exponential their Range is the constant like 2^x and y=0

Then Logs Domain is the one inside

[ripe scarab]

pretty sure

[quiet stirrup]

that questions phrased horribly😭

[wanton ermine]

for points, you can translate them literally, so for example the root (1, 0) goes to (2, 0) then to (5, 0)

oh wait, I bet they mean reflected across the y-axis

the roots already lie on the x-axis by definition, so they already sit on the mirror line and get reflected to themselves

anyways, the point is that 'horizontal stretch by 2' -> (1, 0) goes to (2, 0) ✅

but then if you have the equation y = x - 1 (which the point (1, 0) satisfies)

if you want (2, 0) to satisfy this equation, you have to first get back to the original point, so you divide x by 2
only then does y = x - 1, for y = 0 and x = 2/2

so if you put that together using mathematics, you end up having y = x/2 - 1

[rare crystal]

how is using L'hopital a problem

let people do what they want

[little dragon]

I have a question, I’m in 8th grade, what are the things I should learn before calculus

[round geyser]

One word, factoring

[little dragon]

Be more specific 😭

[round geyser]

Anyway like difference of squares/ cubes and sum of cubes

[little dragon]

Do you have any worksheets you could recommend

[little dragon]

?

[round geyser]

Let me see

[little dragon]

Ok thanks 🙏

[round geyser]

Here

[little dragon]

Ty

[round geyser]

Easy or not

[little dragon]

I’ll do it tomorrow at school, bc I had taekwondo training so I’m tired

[sullen gull]

I saw that u didn't get the answer for this question @spiral shuttle

It's 4

You need to use a simple rule called l'hopital's rule

(I am assuming u know what derivatives are)

So when

lim f(x)/g(x) =0/0
x->a
Which is the inderminant form, and cannot be determined,

We differentiate f(x) and g(x)

And then plug in the value a

Into the equation

So by differentiating cot²x -3 we get 2cotxcsc²x

And for cscx-2, we get cscxcotx

So by plugging in π/6,

We get 8√3/2√3

Which is really just 4.

[round geyser]

Ok

[wooden bison]

Teach me your ways in math

[warped cipher]

!da2a

[tender quest]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[hard kernel]

is a b in precalc bad i needa lock in second semester

[brave portal]

hmmm, what's with this, i can help if needed

[round geyser]

That worksheet is for

[brave portal]

for?

[round geyser]

Him

[grand yoke]

I could help too, the baseline stuff to do is to find the GCF of all terms (if applicable), then if its a four term polynomial you'd group it into two, and for others apply special patterns if there are any

and then you'd find the factors of the starting and constant terms that sum up to the middle term

[round geyser]

Okay great

[round geyser]

Ik it’s gonna be among Difference of squares/cubes and sum of cubes

[dry verge]

Is there an easier way of factoring the second denominator? It seems pretty complex for no calculator problem.

I don't think 64, 48, and 9 share a common denominator outside of 1, so I can't factor anything out before factoring it again.

[uncut mulch]

since 64=8^2, and 9= 3^2

you could consider whether you actually have a perfect square trinomial

also based on the type of question you'd expect more cancellation to happen
there's currently an excess of the factor (8n-3)
in the denominator
so it's likely that there'd also be another factor of that present

[dry verge]

Oh! So it's most likely something like (8n-3)^2... which seems to be correct, as 8*3 = 24.

[uncut mulch]

yes

not always guaranteed to be like this but it's a good place to start

[dry verge]

Thank you! I got this as the result. Does this look correct?

[uncut mulch]

Yes

I'd also recommend putting a dot between the 1 and the n in
21n

When writing your work

you'll start using
ln (natural log)
later on and what you have could be misread

[fiery spear]

it's also pretty common to parenthesize coefficients

[uncut mulch]

not common at all

[fiery spear]

you are wrong

e.g. when writing products of several integers (12)(15)(3)n

i have seen this from purdue to harvard

[uncut mulch]

but that has little to do with them being coefficients of the variable

and I would argue that the () on the 3 are unnecessary here

and I was referring to cases like (21)n not being common

[fiery spear]

it's just another notation that audiences will recognize

the dot enjoys a lot of the same benefits

https://tutorial.math.lamar.edu/classes/alg/factoring.aspx here's an example of an author preferring parens

[uncut mulch]

Situations where there are () around coefficients would be stuff like
x^2 + (m+n)x + mn

[fiery spear]

this author makes some similar choices to that

https://www.cliffsnotes.com/study-guides/calculus/precalculus/polynomial-and-rational-functions/factoring-polynomials

[uncut mulch]

there aren't any () on coefficients in any of that

[fiery spear]

(2) is parenthesized several times
there i danced like good monkey

[uncut mulch]

yes, but not as a coefficient of a variable
and also completely unnecessary

for the cube

not denying that ()() is common notation to imply product,
just that it's not common to parenthesis coefficients like (21)n

[fiery spear]

unless i'm missing a lot of literature i think the most common way to write the numerator in question would have been 21 (3n + 8) n

leaving the zero roots at the end, that is

[uncut mulch]

21n(3n+8) would be the most common

having () in the middle looks ugly

[fiery spear]

i think a lot of authors avoid putting the zero roots in that position for the very same reason the conversation began

the ambiguity between 1 and l is not exactly rare knowledge among typesetters

[uncut mulch]

when typing math, they should be using something like latex or math type

$$1, l ,\ln$$

[obsidian monolith]

ραμOmeganato5

[fiery spear]

in type it's usually fairly clear yes

[uncut mulch]

$21n, 2\ln$

[obsidian monolith]

ραμOmeganato5

[fiery spear]

isn't that a dread

the 1 and l are so close and the n just gives it away

[uncut mulch]

typesetted properly there is no issue
and easily resolved with a dot or small gap when writing

[fiery spear]

i've seen other authors parenthesize all factors except the leading coefficient

i've also often seen those authors write factors in root order

e.g. 21 (x + 3)(x + 1)^2(x)(x - 10)

[uncut mulch]

still valid, certain authors have their own preference

[fiery spear]

ultimately notation offers a lot of ways to cope with any particular collision or organization problem

[uncut mulch]

But I was only taking issue with specifically how you worded what you said at the very start

[fiery spear]

i'd stand by my statement it's common to parenthesize coefficients

i can count probably eight or nine students i interacted with during undergrad who resorted to it consistently

i've also pointed out where professional authors resort to it

[uncut mulch]

3x, 3 being the coefficient of x
It is not common for that 3 to have () like (3)x

is what I interpret from the phrase
coefficient having parentheses

the examples you've shown are not that

[fiery spear]

when the author consistently writes the constant coefficient apart from the varying coefficients you'll see things like 3(x)

[uncut mulch]

still not coefficient having ()

[fiery spear]

x is a varying coefficient in the product

however you're right that parenthesizing the leading coefficient is far less common then any other

it's more common to write 21(14)(2) than (21)(14)(2)

[uncut mulch]

x would be the variable
and the numerical value being multiplied to it being considered as the coefficient

[fiery spear]

3 (x + 3) (x - 5) is a product with three coefficients, two of which are non-constant i.e. varying

[uncut mulch]

that's the point I'm making, not really anything else

just taking issue with that

just take out leading

Perhaps this is a language barrier issue?

[fiery spear]

no, consider this

It is common for berries to grow on bushes.

Some berries never grow on bushes.

It is common for coefficients to be parenthesized.
Some coefficients are never parenthesized.

The principle here is that the ones which defy the common trait are not the prevailing cases.

[uncut mulch]

not sure I get the point ur trying to make

[fiery spear]

The existence of certain coefficients which are never parenthesized does not abrogate the commonality of parenthesizing coefficients.

[uncut mulch]

I don't think your definition of coefficient matches mine

[fiery spear]

right because i use the definition in differential equations

where constant coefficient has a substantially different meaning from coefficient

So, very precisely, the constant coefficients are the coefficients we're often discussing in a polynomial.

You get carried away when you say things like "linear ODEs can be solved by a matrix"

because the ones with varying coefficients can't in general

linear ODEs with constant coefficients can be solved by a matrix

[echo sinew]

I believe you have another idea of coefficient

The coefficients in 3x + 4x² are the 3 and the 4, do you agree?

[fiery spear]

I agree those are coefficients but I do not agree those are all of the coefficients.

I agree those are all of the constant coefficients.

[uncut mulch]

if you really want to get into pedantry
do you agree that it's not common for the constant coefficients to have parentheses here

[fiery spear]

No because it's actually common for constant coefficients to have parentheses, in particular when they are not leading coefficients.

I'd argue that parenthesizing leading coefficients is right on the margin of common.

It's arguably common and arguably rare.

[echo sinew]

So you're saying that **(3)**x + **(4)**x² is common?

[uncut mulch]

So you're saying it's common to see
$$(3)x + 4x^2$$

[obsidian monolith]

ραμOmeganato5

[fiery spear]

Those are leading coefficients in their products and what I've just said about those is:

Parenthesizing leading coefficients is right on the margin of common. It's arguably common and arguably rare.

The ones which defy the common trait are not the prevailing cases.

Leading coefficients are not the prevailing cases of coefficients.

[echo sinew]

Mmh I'm not getting your doubt honestly

[echo sinew]

Huh? Wdym???

[fiery spear]

You're saying it's rare to parenthesize leading coefficients.
I'm not disagreeing, but it remains common to parenthesize coefficients.

[uncut mulch]

can you find 2 examples in the wild where the constant coefficients in a polynomial of a single variable in expanded form have ()

[fiery spear]

https://tutorial.math.lamar.edu/classes/alg/factoring.aspx here's an example of an author preferring parens
https://www.cliffsnotes.com/study-guides/calculus/precalculus/polynomial-and-rational-functions/factoring-polynomials

[uncut mulch]

being rare by definition means it isn't common

[fiery spear]

I would fully agree with that.

[uncut mulch]

and as mentioned those aren't examples of this

[fiery spear]

These are not rare.

You can encounter these frequently in literature and study.

It is about as common as the dot.

[uncut mulch]

yes, but not what we're arguing against

[fiery spear]

both of those have constant coefficients in a polynomial of a single variable in expanded form with ()

[uncut mulch]

Not around the constant coefficient of a variable

[fiery spear]

both of these have polynomials of single variables in expanded form with a constant coefficient in ()

[uncut mulch]

First example isn't in expanded form

[fiery spear]

these just aren't the only examples i'll find haha

you can meet real humans who do this

i don't need to stake my reputation on that

you don't have to search high or low

[uncut mulch]

If it wasn't clear, for the third time
My main issue is with you stating that it is common to have stuff like $$(3)n$$, I'm not arguing anything against have stuff with multiple adjacent ()

[obsidian monolith]

ραμOmeganato5

[fiery spear]

and i'm not going to turn guns and say dots are uncommon either

because that would be an intellectually dishonest debate even if i was advocating devil

[fiery spear]

I've already said this is arguably rare.

I just also gave credit where it's due and said it's arguably common.

That's because my observations have told me that it's marginal.

[uncut mulch]

Wdym by arguably common

[fiery spear]

Like you can argue it's a nap or a rest.

There's a sorites paradox built into the idea.

How many cases before the illness is 'common'?

[uncut mulch]

I'm going to bail. Because you're saying it's both rare and arguably common.

[fiery spear]

There's nothing inconsistent about this.

It's possible for opponents to have good cases.

[uncut mulch]

I've yet to see a single case

[fiery spear]

You can make a good case for its rarity but I also think a good case can be made that it's not.

[uncut mulch]

of that situation

[fiery spear]

I don't have to show you one.

But I also showed you this case

[uncut mulch]

And I said that is not the issue here

[fiery spear]

If you are unwilling to survey the matter then you don't really know.

If you survey it, you will find it is on the margin of common.
I don't need to stake my reputation on that.
It's just what's out there.

[uncut mulch]

absence of evidence not evidence of absence yada yada,
Regardless from what I've seen, u have nothing to justify your claim of it being arguably common

[fiery spear]

I'm not here for a win.

If the truth is different from what I think, I'd like to know.

But I'm primarily here to promote information that meets my standards of evidence.

It's up to you to determine what meets yours.

However, if yours is met by my say so then you really need better.

[uncut mulch]

My claim that it isn't common,
You're claiming that it is both rare and arguably common but have no evidence to show that

[fiery spear]

I'm not claiming it's rare.

I'm claiming the argument could be made that it's rare.

I'm also claiming the argument could be made that it's common.

That's because the frequency I can observe is on the margin of what could be considered common.

[fiery spear]

How precisely would you like me to report my survey?

Would it matter?

I don't think it should.

I don't even think you should accept a comprehensive and professional survey report from me.

[uncut mulch]

I don't care anymore because I think you're just here to waste time

[fiery spear]

I think you should survey this in your own professional conduct.

I think you should be able to report what you have seen with as little bias as possible.

[round geyser]

Can someone derive sin(90-x)=cos(x)

[drifting oyster]

Flat earthers argue the same way ifndr argues

🙏

[drifting oyster]

$\sin{(a-b)}=\sin{(a)}\cos{(b)}-\cos{(a)}\sin{(b)}$

[obsidian monolith]

Enterlessguy

[drifting oyster]

Then take it from there

[round geyser]

Sin cancels out

[round geyser]

Can we use SOHCAHTOA

[drifting oyster]

Indeed sin(pi/2)=1

[drifting oyster]

use what

[round geyser]

Triangle trig

[drifting oyster]

I'm not sure what you're talking about

like describing a triangle with angle 90-x?

[spice crown]

what are you talking about sin(100)=1

[drifting oyster]

what

[spice crown]

lol

$2\pi^r=360^{\text{o}}=400^{gr}$

[obsidian monolith]

Tangent

[drifting oyster]

Errr

ok 💀

[drifting oyster]

But yes I think you can describe a right triangle such that one angle is x and the other one naturally 90-x

And then derive the $\sin{(90-x)}$ from that

[obsidian monolith]

Enterlessguy

[drifting oyster]

If thats what you mean, sure.

[round geyser]

So that’s why it’s cos

[drifting oyster]

Yeah sure...

[somber hound]

currently in precalc to build a good foundation engineering degree after not being in school/doing math for a while
there's a lot of factoring...

might be algebra though or a mixed course

[polar phoenix]

can someone take a look and tell me where i made a mistake on c? answer is supposed to be 113.3 degrees

i used cosine to find the missing side first then used sine law to find the missing angle

[round geyser]

Use laws of cosine

[uncut mulch]

you didn't apply cos law correctly,
if using the sides 21 and 30 like that, the angle would've been the one in between

also you shouldn't have the sqrt like that in an attempt to save time

there is a way to apply the cos law using 21,30, and 40° though

$c^2 = a^2 + b^2 - 2ab \cos(C)$

[obsidian monolith]

ραμOmeganato5

[uncut mulch]

I'd recommend you identify the angle you're going to use
(In this case 40°)
and then based on that, the opposite side which will be 30,
then the adjacent sides
and rearrange and solve from there

[wanton ermine]

for the cosine rule, you need the angle in between the two sides

and that's definitely not 40

so you need to take a different approach, since you don't know side MP (opposite angle x) either

there's one angle-side pair you haven't labelled

sine rule on that, and then you don't need another cosine rule after

[frosty folio]

Hey is anyone online able to help with a couple of problems?

[rough leaf]

yeah sure i can try

[tender quest]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[dreamy cobalt]

Just a tip as a calculus student, when you start pre calc , its really really important to start understanding more than memorising

For example understanding graphs and function notation will help you visualise and understand calculus (and pre calculus) alot

Finding out why certain things in maths are the way they are helps broaden understanding greatly

[fallow lagoon]

can i get help on number 10 iput logbase5x into desmos and all the answer choices and c is the one that matched the expression in the question

how can i do it by hand though?

[sullen gull]

It's option c

Do you know the formula for changing of base

@fallow lagoon

[fallow lagoon]

unfortunately no

oh wait

[sullen gull]

Well I'll enlighten u ig

[fallow lagoon]

i just saw it on my notes

💔

[sullen gull]

Bruh

💀🙏

[fallow lagoon]

it’s that right

[sullen gull]

Yes.

[fallow lagoon]

sorry for wasting your time😓

[sullen gull]

Np lol

[fallow lagoon]

i will ask more useful questions next time

[round geyser]

I am also doing that

But not the change base

It’s the addition, quotient, power rule(log), and radical

[fallow lagoon]

i have three questions i don’t quite get and it’s not in my notes

for 12 i substitute w and z into log form

but idk what to do with -3

i know you divide logs if you subtract

i think 13 might be B because the 4 is inside parentheses and the horizontal dilation is like the opposite of what it says

so changes to 1/4

[drifting oyster]

unfortunately my computer screen doesn't turn sideways and I don't know the command

💀

[fallow lagoon]

sorry about that

[drifting oyster]

for b try using these following properties $\$
1)$a\log{b}=\log{(b^{a})}$ $\$
2)$\log{a}-\log{b}=\log{(\frac{a}{b})}$

[obsidian monolith]

Enterlessguy

[drifting oyster]

assume equal bases for all of them

[fallow lagoon]

so the -3 becomes the exponent of logbase6 (z)?

[drifting oyster]

it becomes the exponent of the argument inside of log

aka z

[fallow lagoon]

hmm

[drifting oyster]

(so z^3)

[fallow lagoon]

so it’s b

[drifting oyster]

yep

[fallow lagoon]

since negative

wow thanks

you seem quite smart

can you help me with 13 and 14

[drifting oyster]

so for 13, notice that $g(x)=f(4x)=\log{(4x)}$

[obsidian monolith]

Enterlessguy

[drifting oyster]

so which statement would then follow?

(note that the vertical shift is $\log{4}$ and not 4 because of $\log{ax}=\log{a}+\log{x}$)

[obsidian monolith]

Enterlessguy

[drifting oyster]

and for 14 its similar, namely $k(x)=h(8x)$

[obsidian monolith]

Enterlessguy

[fallow lagoon]

if it’s inside the parentheses isnt it a horizontal transformation

[drifting oyster]

because again, $\log{ax}=\log{(a)}+\log{(x)}$

[obsidian monolith]

Enterlessguy

[round geyser]

Think of power rule

[fallow lagoon]

so you’re saying g(x) is log(4) + log(x)

[drifting oyster]

precisely

[round geyser]

Consider it as (f(w))-(3*f(z))

[drifting oyster]

or rather g(x)=log(4)+f(x) or just f(4x)

[round geyser]

What’s question is that sorry for disruption

[drifting oyster]

13 and 14

[fallow lagoon]

and 4 inside the parenthesis is the dilation

[drifting oyster]

yes

the coefficient is the factor of dilation

[fallow lagoon]

is horizontal dilation the reciprocal

1 over whatever the number is

[drifting oyster]

yes the factor would be 1/4

because for example, you want to solve for the zero of the graph

[round geyser]

Number 12 is C am I right

[drifting oyster]

the zero of g(x) would lie at $\log{(4x})=0 \iff 4x=1 \iff x=\frac{1}{4}$

[obsidian monolith]

Enterlessguy

[drifting oyster]

so you can see our dilation factor is 1/4

[fallow lagoon]

yes

that makes more sense

thank you

but on 14

[drifting oyster]

B

[fallow lagoon]

the 8 is on the inside again

[drifting oyster]

you forgot the negative

[round geyser]

Yea

[fallow lagoon]

but 1/8 isn’t an answer choice

[round geyser]

Yea you are doing power rule then the quotient rule

[drifting oyster]

no you just missed the negative

you could do either way

but either way z^3 ends up in the denominator

[round geyser]

My teacher states that

[drifting oyster]

right so

for 14 its basically the reverse argument

[fallow lagoon]

so k(x) has dilation of 1/8

[drifting oyster]

instead of saying that g (the funcvtion with the coefficient in the argument) is a dilation of factor 1/k, its saying that f is a dilation with factor k

[fallow lagoon]

ohhhh

[drifting oyster]

k(x) has dilation 1/k relative to h yes but h has dilation k relative to h

[fallow lagoon]

yes

makes sense

thank you

[round geyser]

The positive number (log_6 w) goes to the numerator and the negatives (-log_6 z^3) goes to the denominator to make it into log (w/z^3)

[drifting oyster]

yes, the answer is hence B not C

[round geyser]

Yes

[drifting oyster]

either order is correct either way z^3 goes to the denominator

[round geyser]

Yes

Like 2 (log AB) can be log (A^2B^2) or log (AB)^2

[polar phoenix]

i can cancel the (x+6)

but for the (X+1) and (1+x) can you cancel those?

[raw hill]

yes b/c addition is commutative (the order in which you add things doesn't matter), so they're the same

[polar phoenix]

so the order doesnt really matter? as long as they have the same signs

[raw hill]

order does not matter for addition or multiplication

[polar phoenix]

i see alr ty

[raw hill]

[arctic dragon]

any help pls?

[serene dagger]

Looking at the shape of the graph, it is a graph of an exponential function.

[arctic dragon]

yeah but i cant get the function for it can you help?

[serene dagger]

yes

[arctic dragon]

you got it?

its a online practice and its pretty vague and annoying to do

[polar phoenix]

when stating npvs: for ex 3(a-4)

a cannot be 4 right, but what about the 3 do i also put 0 as an npv?

[arctic dragon]

um

no?

idk

whats going on bruh

can someone just tell me the function pls

[serene dagger]

As the graph approaches x=1, the general form is y=a*log(1-x)+b

[arctic dragon]

yes

[serene dagger]

also, since it passes through the two points (0,2) and (-15, 4), we can solve the system of equations.

y=a*log(c)(1-x)+b and it passes through the three points (0,2), (-3, 3), (-15, 4)

[arctic dragon]

yeah i solved it and tried and its still wrong

brooooo

you want to try putting a specific function?

this website is stupid man

[serene dagger]

I'll calculate it

[arctic dragon]

okok

[serene dagger]

b=2

[arctic dragon]

i have tried like 10 function

[arctic dragon]

yep

i think so

atleast

you got it? @serene dagger

[serene dagger]

c=2^(2a)

so let a=1, then c=4

thus y=log(4)(1-x) + 2

[serene dagger]

how about you?

[arctic dragon]

its wrong

i just entered it

[serene dagger]

why?

it's right

[serene dagger]

Check it out.

[arctic dragon]

i gave up

thanks tho

[serene dagger]

Don't give up

[arctic dragon]

its probably bugging it happened before

i will ask my teacher and i will let you know

@serene dagger

[round geyser]

Use y=log(-x-h)+k

Domain is inside the function

[sullen gull]

Log52.631(-2x+1.9)

Very specific

Imo

@arctic dragon

[arctic dragon]

ty!

@sullen gull

[polar phoenix]

can check if i did this right, since im not sure which is which

since its in the quadrant where only cos is positive would my sin and tan still be negative?

or would it be positive since my opposite is also (-9)

ik its d btw i just need clarification about that part

[warm panther]

What book do yall recommend for self-learning precalculus?

[vast rivet]

Most syllabuses in year 11+ have precalc

[sullen gull]

Np 😄

[sullen gull]

It will be negative.

Cuz the y is negative

[stiff cypress]

I cant even understand step 1

What are the roots

Like

What

😭

[obsidian monolith]

Civil Service Pigeon

[raw hill]

[https://math.libretexts.org/Bookshelves/Precalculus/Precalculus_1e_(OpenStax)/08%3A_Further_Applications_of_Trigonometry/8.05%3A_Polar_Form_of_Complex_Numbers]

<@&268886789983436800>

[stiff cypress]

Do i just look at everything

[raw hill]

the part about finding roots is what's directly relevant here

you may need to backread to some extent for context though

[stiff cypress]

I'll find myself some time to read allat

[round geyser]

I thought we can use a^3+b^3

And just square it

Like we can make a^5

(a^2)^3-(1^2)^3

[raw hill]

$(a^b)^c=a^{bc} \implies (a^2)^3=a^{2 \cdot 3}=a^{6}$

[obsidian monolith]

Civil Service Pigeon

[raw hill]

5 is prime

[round geyser]

Oh

Is there a sum of the 5th power

[raw hill]

what

[round geyser]

a^5+b^5=?

[raw hill]

[https://math.stackexchange.com/questions/4952780/factorization-of-an-bn]

[round geyser]

Okay

So we have
(a+b)(a^4-a^3b+a^2b^2+a+b^3+b^4)

[raw hill]

,w factor a^5+b^5

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[ripe scarab]

Bruh can anyone help with domain range and graph of functions

[round geyser]

Sure

[normal kayak]

Please teach me calculus

Im in year 8

Helpp

[echo sinew]

We can't teach you here 😅
It's too much to be covered in a chat

I suggest you pick a textbook or an online course and start from there

Btw, ask in #calculus or #book-recommendations or also in #study-discussion

[grim pulsar]

Does anyone know how to complete za square when x isn't 1?
I have a math exam coming up that i need to study for but i cant remember how to do that.

[round geyser]

Sure

I’ll take care of you

Lui

[grim pulsar]

o sry i didnt see that

[round geyser]

It’s okay

[grim pulsar]

but, you can help me with za square?

i wanna find an example

Solve: 2x^2 - 6x + 3 = 0 by completing the square

[grim pulsar]

i found an example

[round geyser]

Sure

Okay

First divide all the b and c term by the a

a=2, b= -6, and c=3

[grim pulsar]

it would be x^2 -3x + 3/2 = 0?

wait i also divide the a right?

[grim pulsar]

i mostly dont know what to do with the fraction

[round geyser]

Okay

Move 3/2 to another side

[grim pulsar]

alright

[grim pulsar]

done

[round geyser]

What do u get now

[grim pulsar]

x^2 - 3x = 3/2

[astral apex]

should be -3/2 on the right side

[grim pulsar]

o yeah!
mb

[grim pulsar]

I fixed it back

[round geyser]

Use this formula (b/2)^2

And add

Like

x^2 - 3x+ ? = -3/2 + ?

Usually I would leave a blank

[round geyser]

(-3/2)^2

What do u get

[round geyser]

?

[grim pulsar]

i had added the - to the 3/2

rn i have
x^2 -3x + 9/4 = -3/2 + 9/4

[astral apex]

that's right

now you can factor the left side as a square

and combine the fractions on the right side

[grim pulsar]

wait that confused me😭 wha?

do yo mean like multiplying till the denominators are equal? or something else i cant remember what it means

[round geyser]

Okay so

[round geyser]

Combine -3/2 and 9/4

-3/2 x 2

[round geyser]

So do you know where I did I get (x-3/2)^2

[grim pulsar]

rn i have x^2 -3x + 9/4 = 3/4

[round geyser]

Okay

[round geyser]

It looked like a^2+2ab+b^2

b^2=9/4

Sqrt it to get b=

[grim pulsar]

so i have x^2 - 6 +9/4 = 3/4
I'm thinking i should move 3/4 to the other side to make
x^2 - 6 + 3/4?

unless its all wrong...

[grim pulsar]

.

[round geyser]

Wait

So

[round geyser]

Trick to do this is take the square root

Now you have (x-3/2)^2 = 3/4

Sqrt them

Now it’s x-3/2=+/-sqrt(3)/2

Add 3

To get x=(3+/-sqrt(3))/2

[grim pulsar]

what does the sqrt stand for?

i dont recall learning that

[astral apex]

sqrt is square root

it's not normally used in school classes but it's a common shorthand online

[grim pulsar]

o

[astral apex]

the 3/4 is good, but where did the -3x on the left side go? why did it become -6?

[grim pulsar]

i thought it would be multiplied by 2 for some reason

[astral apex]

let's take it from here

at this stage the three terms on the left side can be factored

[grim pulsar]

alright

[astral apex]

into something that looks like (x - number)^2

and that number follows a pattern (if you're doing the process properly)

it is always b/2

where b comes from the previous work

in your case b = -3

it's what's in front of x in the last line

[grim pulsar]

i reread that a bit then i wrote it down

so i think i understood that

[astral apex]

do you think you see what the left side factors into?

[grim pulsar]

yeah (x-3/2)^2

[astral apex]

yup

[grim pulsar]

alr, about it tho i hae a question , the (x-number) part, it always has to be a negative right?

[astral apex]

yeah

[grim pulsar]

Alright

I get it now! :]

Thank you

[round geyser]

Yw

[compact moat]

I mean this says precalc but like can I ask Calc 1 , 2 ,3 questions I think thats what it corresponds to in Uk

[rough leaf]

there’s a channel for calc

[compact moat]

Is it under the pre univerity package thing

[rough leaf]

early uni

[viscid thistle]

Hello how can one master high school calculus(continuity and differentiability, application of derivatives, definite and indefinite integration, application of integrals, differential equations) within 3 days

[echo sinew]

It depends a lot on what you already know

And whether you have the prerequisites

[round geyser]

I thought of using a^b=c

And this one is log_a c=b

How would u convert this

[atomic flare]

Ask yourself 9 to the power of what equals $3\sqrt[4]{27}$

[obsidian monolith]

USS-Enterprise

[atomic flare]

Of course perhaps simplify that number first

rewrite as 3^(something)

And then that must equal to 9^(of that), or 3^(2*(of that))

[round geyser]

9^x=3 x 3^3/4

We know 9 can be written into 3^2

3^2x=3^7/4

[atomic flare]

yes

[round geyser]

According to
a^x=a^n
a as a base cancels

2x=7/4
multiply by 1/2

And get x=7/8

[atomic flare]

uh

It doesn't "cancel" algebraically like a factor

If that's what you meant

[round geyser]

Yea

Same base can cancel

And I just has a test

On these

With graphing

I think that test was easy

[atomic flare]

they are pretty nice, yes

[round geyser]

Ik on the quiz on exponential equation and graphing it I got an 100

[atomic flare]

nice

me too

I think that was the only test I ever got a 100 on

[round geyser]

The first 100 was the Unit 4 of Rational expressions

Then these

[atomic flare]

This seems more confusing than it should be

4B.2

[round geyser]

That has fence problem

The cut out problem

Line close to a coordinate

[valid topaz]

hope and pray

all i have 4 u

[polar phoenix]

when solving for d do i do this