precalculus

[willow bear]

but yw

[willow bear]

w/o modifications synthetic division only rly works when your divisor is linear

while long division works for any degree

[harsh rampart]

https://i.imgur.com/a7ZCobv.png this means the derivative of the function equals to 0? since its a horizontal tangent so gradient is zero?

[willow bear]

yes

[harsh rampart]

i find the values of x for 3x^2 -2x -2 = 0, then i plug it into the original function and those will be my coordinates? 2 points right?

[willow bear]

yes

[velvet blade]

when doing long division, should I stop when the divisor has higher degree than the remainder?

[blissful ridge]

Yes

[velvet blade]

thanks

[quaint mason]

can someone walkthrough on how to do these 2 problems

[sick steppe]

isolate for the trig, then think of CAST rule

[quaint mason]

Cast rule?

whats that?

[sick steppe]

cast rule tells you where trig is positive in the quadrants

[gaunt mason]

Is there a table, or calculator operation for asec (that is sec^-1) ?

[sick steppe]

sec = 1/cos

[quaint mason]

i use a TI84 so i dont have sec^-1

[gaunt mason]

We don't use sec in east europe

[quaint mason]

so would i have to get sec(theta) on one side by dividing the 3 over to the other side?

for the first problem

[sick steppe]

no, the 3 is in the argument

[quaint mason]

oh

[sick steppe]

$1.4150 = \sec{3\theta} \implies 3\theta \in Q1,3$

[gaunt mason]

You need to convert sec to a function that has an inverse. So you can apply the inverse on the left side.

[obsidian monolith]

moshill1:

[quaint mason]

whats that E sign?

[sick steppe]

In

3 theta is in quadrant 1 and 3

1.4150 = sec(3theta) means 1/1.145 = cos(3theta)

agree?

[quaint mason]

yes

[sick steppe]

ok so we want values for when cos(something) is positive right?

since 1/1.415 > 0

[quaint mason]

yeah

[sick steppe]

ok so by CAST rule, cos is positive in quad 1 and 4 (not 3, my bad)

you can also get that from the graph of cos

[quaint mason]

which is in q1 and q4

[sick steppe]

quadrant

[quaint mason]

q3 is tan

[sick steppe]

yeah

Ok so from taking cos^-1 of both sides, what do you get?

[quaint mason]

45?

[sick steppe]

question 1 is in radians

[quaint mason]

oh oops

[sick steppe]

but yeah around 45 degrees

[quaint mason]

.79

[sick steppe]

Yeah, this is an intermediate step, so it's best to keep a few more digits in the decimal

0.78595 is good enough

[quaint mason]

and then im assumin its askin for in the 4th quadrant which would be the next value? so im assumin at 315 degrees, so find the radians of 315 degrees?

[sick steppe]

Yeah OR

[quaint mason]

OR?

[sick steppe]

go 2pi - the principle angle

the principle angle being the value you get directly off calculator displays (typically)

so in this case, $2\pi - 0.78595 \approx 5.49723$

[obsidian monolith]

moshill1:

[quaint mason]

ah bcuz 2pi is just 360 degrees in radians form

[sick steppe]

ye

so you have .78595 and 5.49723, what's the next step?

[quaint mason]

graph it?

[sick steppe]

nope

you dont need to graph

[quaint mason]

oh

[sick steppe]

$1/1.145 = \sec{3\theta} \implies 3\theta \approx 0.78595, 5.49723$

[obsidian monolith]

moshill1:

[sick steppe]

what do you do to get theta?

[quaint mason]

are u able to divide by 3 now? since its no longer in the argument

[sick steppe]

yep

the 2 decimals we got were for sec^-1(1.145)

so that leaves 3theta

[quaint mason]

so we divide 3 into both values?

[sick steppe]

yeah

then round to 2dp like the question asks

[quaint mason]

oh its wrong

[sick steppe]

umm

[quaint mason]

did u get .26 and 1.83

[sick steppe]

yeah

what did you put in the box?

[quaint mason]

.26, 1.83

[sick steppe]

When you type in the box, do you get access to inverse trig pre-written out?

like a button that says [cos^-1] ?

[quaint mason]

nope

hm its says we r missin like

4 more values

[sick steppe]

Hmm not sure then sorry

[quaint mason]

all gud, tyvm though

appreciate it

[viscid thistle]

I have an inequality which is already factored

(x-3)(x+1)(x-1) < 0

I need to portray all the possible values for X in a numberline or something, but is there a way to know which sides to shade without plugging in numbers?

I would also like to understand it without the tricks

[north island]

you could draw the curve

[viscid thistle]

I mean

[north island]

just sorta scribble it out quickly

[viscid thistle]

There's another method that it has a lot more work, and it doesn't involve plugging in numbers between the intervals to check which ones to shade

@viscid thistle

If you are curious ab it let me know

is plugging in numbers the easiest way other than sketching?

cuz i dont know what the graph looks like

[north island]

it's a cubic and you know all the roots

[viscid thistle]

i dont understand it in my head
what does it mean when its bigger than 0? what exactly is bigger than 0? the Y values?

[north island]

yes

set y = (x-3)(x+1)(x-1) and draw it out

and then you want the bits where the curve goes below the x-axis

[viscid thistle]

Ah ok, thank you

Does this mean, if I plot the 3 roots in the numberline, then there won't be 2 shaded regions that touch?

between them

[north island]

yes

[viscid thistle]

ok that helps a lot

[sand gate]

What does this mean

[sick steppe]

write the lhs in function notation

f(x) = x+7 for example

[paper wigeon]

C=f(n)?

[sand gate]

What is lhs

[sick steppe]

left hand side

[sand gate]

Can I please get more description I’m confused

[sick steppe]

do you know what function notation is?

[sand gate]

Not Really

[sick steppe]

Ok so if i have something like y=x+7, i can plug in x values to get y values, right?

if i plug in x=1, i get y=8

[sand gate]

Yeah

[sick steppe]

so the value(s) of y depends on what I put in for x, agree?

[sand gate]

Agreed

[sick steppe]

Ok so what people did was they made this fact more apparent by introducing function notation

instead of y, we write f(x), f for function

[sand gate]

Ok

[sick steppe]

so we can now say f(x) = x+7, cause x is what makes things happen

people adapted this idea and you could have, for example, C(n)

which might mean Cost function

and the cost depends on the number of items, n

[sand gate]

Yes

[sick steppe]

so a from your homework: C = 20n+8

what do you think the function notation could be?

[sand gate]

f(c) = n +8?

[sick steppe]

Close

n is what you want in the brackets since that's your input

[sand gate]

ok so f(n) = c + 8?

[sick steppe]

Nope, the function is still 20n+8

[paper wigeon]

if the equation uses a variable then it depends on that variable, so lets say the variable is x, that means the equation depends on x

because it depends on x the function notation would be f(x)

[sand gate]

Ok thank you for the help

[paper wigeon]

gl

[viscid thistle]

There's another method that it has a lot more work, and it doesn't involve plugging in numbers between the intervals to check which ones to shade
@viscid thistle what is the method called?

There's no name to it i believe

It's just another method that involves playing with negative or positive intervals

Divide into cases, etc.

Ah

Is it making a table, and then seeing which regions are positive/negative?

Not exactly

For example, by doing the method i'm mentioning, it'd start by $$(x-3)(x²-1)<0$$ dividing into cases, the only way for a product to be negative is if one of both factors is negative and the other positive, $a\cdot b<0$ is only possible if\ $\begin{cases} a>0 \ b<0 \end{cases} \ \begin{cases} a<0\ b>0\end{cases}$, \ hence we'll do it here $$\begin{cases} x-3>0 \ x²-1<0 \end{cases} $$ $$ \begin{cases} x-3<0\ x²-1>0\end{cases}$$ and by solving for each, finding the intersection and then the union you get to the same result

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

ah I get it. very nice, thanks!

[serene heath]

just sketch the cubic and save yourself hella effort

[pastel pecan]

Does anyone know how to figure out which half angle formula to use?

[serene heath]

45-30=15

[viscid thistle]

just sketch the cubic and save yourself hella effort
@serene heath we already mentioned it, but they wanted to see if there was any other method, so i went to write that one

[sick steppe]

Cases made me bomb a test cause I was stubborn and forced myself to answer it with them

(I had the option to not use cases, i just chose not to)

[pastel pecan]

How did they go from sinx to cosx??

[sick steppe]

what do you mean?

they didnt

@pastel pecan

[pastel pecan]

well they had (cos^2/sin^2)sin

[sick steppe]

yeah

[pastel pecan]

Then it went to (cos/sin)cos

[sick steppe]

yeah

canceled the sine and pulled a cosine out of the fraction

[pastel pecan]

owwww

shoot that makes sense now

thanks

[sick steppe]

ye

[pastel pecan]

didnt know that was allowed

but that makes sense now

[sick steppe]

you can pull stuff out of fractions if they're multiplied in the numerator

$\frac{ab}{c} = a\frac{b}{c}$

[obsidian monolith]

moshill1:

[umbral current]

how on earth do i do this

they didn't give me an interval

[sick steppe]

Well you know it has at least 1 R root since it's a cubic

and it says to use a graphing software

[wintry yacht]

Yow, is this free?

Ok. Later

[sick steppe]

this isnt a help channel, so yes it's free

[wintry yacht]

Ok

How do I find a function with vertex and that passes through a point P?

[sick steppe]

depends on the info given

[wintry yacht]

With vertex (1,-2) and point passes through P(4,16)

[sick steppe]

is it a quadratic?

[wintry yacht]

Yeah I'm a bit confused

[sick steppe]

do you know vertex form?

[north island]

y = |6(x-1)| - 2

[sick steppe]

1. that's wrong
2. dont just give answers

[north island]

in my defence

it was funny

[sick steppe]

do you know vertex form?
@wintry yacht

[wintry yacht]

The vertex form? Uhmm, the square of the half coefficient of x?

[sick steppe]

y=a(x-h)^2 + k

the thing you get after you complete the square on a quadratic is called vertex form

[wintry yacht]

Oh yeah, the points after you subtract and add the half coefficient

[sick steppe]

yeah

$y = a(x-h)^2 +k$ has vertex (h,k)

[obsidian monolith]

moshill1:

[wintry yacht]

I then just plug the values?

[sick steppe]

yeah, then you need to find a

[wintry yacht]

Ok. Thanks!

[sick steppe]

np

[chilly apex]

can somebody please help me with physics 12 ?

<@&286206848099549185>

[wintry yacht]

@sick steppe

[sick steppe]

hmm?

[wintry yacht]

Is there a way to find the cubic equation roots by the standard algebraic evaluation? e.i, $ax^(3)+bx^(2)+cx=-d$?

Evaluate*

[north island]

cubic formula

[wintry yacht]

cubic formula
@north island I mean. Yeah, but no?

[north island]

no? rip

howso

[sick steppe]

if you want to write out cubic formula, go ahead @north island

[wintry yacht]

Just simple as the quadratic equation

[sick steppe]

If it's factorable, that's it

[north island]

oh jesus what the hell happened to that image

[sick steppe]

like i said

[wintry yacht]

[north island]

no why is it dark

[sick steppe]

cause the pic you used is dark?

[north island]

it was not

[viscid thistle]

I'm guessing it had transparent bg

[north island]

ohhh

wild

[wintry yacht]

So it just the standard cubic formula

[fleet yew]

You should know that by heart

[wintry yacht]

What about approximation? For the zeros of cubic equation?

Ahh.. this is dead

[viscid thistle]

if y = 8cos(xpi)+12, how do i find the equation of the second derivative in terms of y?

[willow bear]

have you found y'' in terms of x thus far?

$y = 8\cos(\pi x) + 12$ should give $y'' = -8\pi^2 \cos(\pi x)$

[obsidian monolith]

Ann:

[viscid thistle]

I need help factoring .

X^3 - 2x^2 - 5x + 6

How do I recognize first that factoring by grouping isn't an option, and once I recognize this, what method do I use to factor?

(or how do I find the roots)

[north island]

doing it really easily doesn't seem like it'll work bc the coefficients are like 1, -2, -5, 6

if they were like 1, 2, 1, 2 you could see that you could get that by (1, 2)(1, 0, 1)

and similarly if they were like 1, 3, 2, 6, you could pretty easily see that you could get that from (1, 3)(1, 0, 2)

but there aren't easy multiples

to factor something like that you can apply factor theorem

(ie. try random numbers and see if they work)

[viscid thistle]

ruffini works too

i'm not sure if that's how you call it in english

synthetic division is what it's called (I think)

I call it ruffini method too

the guy in utube said "rational zero theorem" is this a thing

[willow bear]

ah yes, "the" guy. the only person on YouTube who can be described as a guy.

[viscid thistle]

yea that guy

[willow bear]

that was sarcastic. i have no idea who you are talking about.

but there is a theorem called the rational root theorem.

[viscid thistle]

is there a difference between rational root theorem vs ruffini method

wdym Ann dont you know that guy?

[north island]

i've never heard of it either, but google does have results for rational zero theorem

[viscid thistle]

yeah cringe

[north island]

seems like it's basically just the factor theorem

?

[viscid thistle]

it's not about one random guy but the guy

how do I know something cannot be factored by grouping without trying it first?

is there a way

[north island]

you just stare at it and look for easy multiples

and then go 'meh, looks hard'

[viscid thistle]

alrighty

[viscid thistle]

is this always true?

can i trust it

[astral mantle]

its more like at most the same amount of real distinct roots as its order

[willow bear]

the "exactly as many roots as its degree" version is true in the complex numbers but you need to count them with multiplicity

[viscid thistle]

alright

[hollow magnet]

Think this goes here

Trying to differentiate y=ln3x-e^(-2x)

i have got ln3x(2e^(-2x))-e^(-2x)(1/x)

although solution is 1/x + 2e^(-2x)

Not sure where im going wrong, Let me know if you need working

[viscid thistle]

yes, post your working if possible

it'll be easier to spot the mistake

[willow bear]

already answered in calculus

er

no, not there

in alpha

[viscid thistle]

oh ok, then @hollow magnet don't multipost

[fossil crown]

From

#❓how-to-get-help

Rule 3

Stick to one channel and don't post the same question in multiple channels. Please don't ask for help in other channels if no one is responding in the one you have posted your question in.

[hollow magnet]

oh ok, then @hollow magnet don't multipost
@viscid thistle Sorry, Wasnt sure where to post

[viscid thistle]

I have this inequality, and I need to find roots and shade the numberline
why should I not multiply both sides by (x-1) to get rid of the denominator ?

because this is an inequality, not an equation

you'd have to consider the signs of x-1

you can't just blindly multiply by x-1 both sides like in an eqn

[lapis sphinx]

If x was any number less than 1, the denominator would be negative, hence you’d have to flip from a < to a >

Like al3dium said

If x was any number greater than 1, the denominator would be positive, hence the < would stay the same

It wouldn’t work either if x was 1

I think that’s why

I may be wrong

And plus the other side is 0, so the numerator values are important too

[viscid thistle]

alright

[leaden stratus]

Hi, the exercise asks to deduct the following limits by studying these plots

[sick steppe]

It's just asking about the y values as you get close to x = a

[leaden stratus]

Wait

Wdym

[sour hemlock]

for example, what is the value of f(x) at x = -4

[leaden stratus]

-4

[sour hemlock]

?

no, look at the graph again

the y axis is f(x)

[leaden stratus]

Ah, it's 0

[sour hemlock]

there you go, use the same idea for the rest

[leaden stratus]

Ahhh it's easy then

[patent beacon]

There's a bit of subtlety going on with the word "lim". It won't matter until you approach x = 4

[leaden stratus]

But why for x =4, f(x) has a white dot? Maybe it's excluded

There's a bit of subtlety going on with the word "lim". It won't matter until you approach x = 4
@patent beacon can you explain?

[patent beacon]

Yes. Note specifically that f(4) ≠ 6

[sick steppe]

the white dot is hollow

[patent beacon]

Still, the limit as x→4 = 6

Because a limit ignores what happens AT a point, and only considers what's happening AROUND it.

[leaden stratus]

Ah ok

[leaden stratus]

So...

Lim x -> - 4 f(x) = 0
Lim x -> 4 f(x) = 6
Lim x-> 0 f(x) = 4
Lim x-> 6 f(x) = 4

Right?

[serene heath]

ye

[lunar bone]

Construct a polynomial function with the following properties: fifth degree, 2 is a zero of multiplicity 3, −2 is the only other zero, leading coefficient is 2

sad

Hawkes keeps telling me my answers are wrong when every source i check and my teacher say I am right and I am very confused

[patent beacon]

A naive answer may be
y = 2x³(x + 2)

Problem is, that's not 5th degree

[north island]

nah, 2 is a root of multiplicity 3

it would have to be 2(x+2)^2(x-2)^3

[patent beacon]

Oh oops misread

[ancient rampart]

im kinda stuck on this

[patent beacon]

Yeah, the solution is to make the other zero multiplicity 2

[ancient rampart]

i started doing some stuff and i keep getting stuck

[patent beacon]

That's an identity I assume? Make sure you know what the problem is asking, haha

[ancient rampart]

yes i need to prove it

or verify it

[lunar bone]

The answer worked thank you very much

[ancient rampart]

should i try multiplying by conjugate on the left side

[patent beacon]

Change both sides to sin/cos is usually a good first step. In this case, the left is multiple terms and should be combined into one fraction

[ancient rampart]

ok

ok im at 1/sinx - 1/sin^2x -1

so since i know the identity is true that means - 1/sin^2x -1 has to cancel out somehow

hm

i think i did something wrong

oh it would be 1/sinx + 1/sin^2x -1

[patent beacon]

I'll turn the left onto sin/cos:

[ancient rampart]

oh ik what i did wrong

cosx * cotx * sinx = cos^2x

[patent beacon]

$\frac{cos^2(x)/sin(x)}{1 - sin(x)} - 1$

[ancient rampart]

i wrote it as cos^2x/sin^2x

[patent beacon]

Good then

[ancient rampart]

ok i got it now

ty :P

btw is it worth taking time to learn latex formatting

[patent beacon]

https://www.codecogs.com/latex/eqneditor.php?latex=D

Using that you can learn the basics pretty easy. Normally the server supports it but the bot seems to be dead atm

[ancient rampart]

is it useful

[patent lance]

Hey i was having some trouble with this

I can get the basic equation but i have trouble determining what a is or how to deal with the y int

[willow bear]

what's a

[patent lance]

I am not sure

do i just plug in one of the xint points

to find

probably gimme sec

didnt work or it did and im dumb

[deft sparrow]

can someone help

you would multiply both things by numerator right

but then youre left with 9x^2+2 over (4x+3)*numerator

idk how to find limit from that

[sour hemlock]

divide both numerator and denominator by x

the x would go inside the square root in the numerator

[deft sparrow]

you mean divide them seperately?

or the whole fraction divided by x

[north island]

don't do any of that

lim as x goes to inf of sqrt(9x^2 + 2) is more or less lim as x goes to inf of sqrt(9x^2)

you can ignore the constants when there's something going to infinity like that that's just way bigger

[sour hemlock]

yes divide them separately

you cannot divide the whole fraction by x, that would change the expression which is not allowed

[deft sparrow]

@sour hemlock but do I have to get rid of the square root?

[sour hemlock]

what?

have you solved these kinds of questions before?

[deft sparrow]

not in a while tbh

I'd usually get rid of the square root tho

[sour hemlock]

you are not allowed to get rid of anything

you are allowed to move things around

to get an equivalent expression on which it is easier to apply the limit

[vapid mica]

tan^2(x) - tan^2(x) times sin^2(x) = sin^2(x)

can someone help me prove this

[sour hemlock]

wait what?

tan(2x) - tan(2x)?

[vapid mica]

idk how to use the math type thing

no no

wait

14

[sour hemlock]

convert tan to sin/cos and just do fraction addition to get your answer

you will see the pattern emerge

[vapid mica]

got this so far

[sour hemlock]

take sin^2 out

[vapid mica]

out?

[sour hemlock]

factor it out

from numerator

[vapid mica]

this?

[sour hemlock]

sin^2(1 - sin^2)

do this

do you see the pattern

[vapid mica]

no what???

[sour hemlock]

sin^2 + cos^2 = 1

do you see it now?

I guess you should review your basic trig identities

[vapid mica]

cud u like write it out?

im still confused

[viscid thistle]

\begin{align} LHS&= \frac{\sin²(\theta)-\sin⁴(\theta)}{\cos²(\theta)} \ &= \frac{\sin²(\theta)[1-\sin²(\theta)]}{\cos²(\theta)} \ &= \frac{\sin²(\theta)[\overbrace{1-\sin²(\theta)}^{\cos²(\theta)}]}{\cos²(\theta)} \ &= \frac{\sin²(\theta)\cancel{\cos²(\theta)}}{\cancel{\cos²(\theta)}} \ &=\sin²(\theta) \end{align}

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Written out would be like this

And yeah review your trig identities

@vapid mica

[vapid mica]

wait how does -sin^4(x) = cos^2(x)?

[viscid thistle]

What? I factored out sin²(theta)

This is algebra

[vapid mica]

OHHHHHHHHHHHHHHHHH

sin^2(x) times sin^2(x) right?

and then one cancels out

and leaves sin^2(x)

no 1-sin^2(x)

[viscid thistle]

Uh yes sin²(theta) times sin²(theta)=sin⁴(theta), so we factor sin²(theta) out, and we get sin²(theta)(1-sin²(theta))

[vapid mica]

im lost on the factoring part man

[viscid thistle]

Oh well

What's the step that confuses you

[vapid mica]

the factoring

thing

[viscid thistle]

(they are numbered)

[vapid mica]

1 to 2

everything else makes sense

substitution

wait wait

[viscid thistle]

Do you understand how to factor with variables, eg x²+x

[vapid mica]

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

[viscid thistle]

You'd factor out an x, leaving us with x(x+1)

[vapid mica]

wait leeme try this out

yeh yeh

[viscid thistle]

Yes

[vapid mica]

i got that

[viscid thistle]

Okay

[vapid mica]

and then wait lemme do it out

i still dont get it man

😔

[viscid thistle]

You don't get the factorisation?

[vapid mica]

its just that one part

yeh

[viscid thistle]

Okay

Do you understand how to factor with variables, eg x²+x

[vapid mica]

the -sin^4(x) to the identity

[viscid thistle]

Would you be able to factor out polynomials like that

[vapid mica]

yeh no that makes sense

[viscid thistle]

Oh

[vapid mica]

x(x+1)

how did u go from the -sin^4(x) to 1-sin^2(x)

[viscid thistle]

Yeah i mean, if you are able to factor out x²+x, the same logic applies to here

Would you be able to factor out x²-x⁴

[vapid mica]

yeh

x^2(1-x^2)

[viscid thistle]

Yes

Exactly what happened on our case

[vapid mica]

🤦‍♂️

im so dumb man

[viscid thistle]

The same thing

Nah don't worry

[vapid mica]

right?

and then substitution and what not

[viscid thistle]

Exactly

@deft sparrow that doesn't belong here, it'd be more appropiate at #calculus if it's free, if not, go to another #question free channel

[deft sparrow]

oke

[vapid mica]

idk man i cant even factor

[viscid thistle]

?

Are you confused about it again?

[vapid mica]

no no i was just joking lmoa

[viscid thistle]

I mean, you can just make a substitution t=sin(theta), and it'd be t²-t⁴ on the numerator, which is t²(1-t²) and then subbing back sin²(theta)(1-sin²(theta))

[vapid mica]

lmao

[viscid thistle]

Oh lol

It'll stick to your head when encountering more problems

It's about practicing more

[vapid mica]

wait aledium

i need help on 17 too

just scroll up and look at the pdf

can someone else help?

<@&286206848099549185>

[daring dew]

Multiply LHS by 1 - sin theta both in the numerator and denominator

@vapid mica

[vapid mica]

?

this is what i have so far

@daring dew

[umbral current]

prove

[sick steppe]

Factor denominator

[vapid mica]

yeh yeh i got i made the denominator (a-b)^2 form and then cancelled it out

[hard hedge]

can someone help me with this problem please?

[uncut mulch]

where are you stuck

[hard hedge]

I’m not sure where to start

should I turn cot into tan?

[uncut mulch]

you could express it in terms of tan if you think it'll help

[hard hedge]

cos is x right?

[uncut mulch]

you're gonna have to be more clear

[hard hedge]

like on a unit circle the x is cos and y is sin?

For the radians

[uncut mulch]

cos(theta) will give you the x-coordinate on the unit circle
sin(theta) will give you the y-coordinate

[hard hedge]

Did I do it right?

[uncut mulch]

,rotate

[uncut mulch]

no

you didn't write the $\cot( \red{\frac{\pi}{3}})$. also how are you getting from: \
$\frac12\cdot \frac{2}{\sqrt{3}}$ to $\frac{2}{2\sqrt{6}}$

[viscid thistle]

are anti-derivates and integrals the same thing?

[obsidian monolith]

ramonov:

[hard hedge]

Am I on the right track?

[uncut mulch]

still some algebra issues there

how are you getting from $\frac{2}{2\sqrt{3}}$ to $\sqrt{3}$

[obsidian monolith]

ramonov:

[hard hedge]

?

[uncut mulch]

that can be simplified further

[hard hedge]

-sqrt3/6?

[uncut mulch]

yes

[hard hedge]

okay thank you

[urban urchin]

could someone tell me what the rg in the second dotpoint is referring to?

[uncut mulch]

it might be range of g

[urban urchin]

i thought so, but in the question it says the range is R?

[uncut mulch]

says the codomain is R

[urban urchin]

oh ok thank you

[supple sigil]

hey

7. a carpenter is buying supplies for a job. the carpenter needs 4 sheets of oak paneling and 2 sheets of shower tile board. the carpenter pays 99.62 for these supplies. for the next job the carpenter buys 12 sheets of oak paneling and 6 sheets of shower tile board and pays 298.86. he also spends 139.69 on 1 sheet of shower tile board and 8 sheets of oak paneling. how much does each item cost individually? please help

[jagged glade]

Hello @supple sigil

[supple sigil]

hi

[jagged glade]

So what will your initial approach be

[supple sigil]

i want to solve it

and i have been trying for 1 hr

i am abt to give up...

[jagged glade]

Oh...

Can we see your work?

For me, I will first
Let the cost of 1 sheet of oak paneling be x.
Let the cost of 1 sheet of shower tile board be y.

[viscid thistle]

@supple sigil has been answered, see #prealg-and-algebra

Could someone maybe point me in the direction to where I can find some info on how to do this? Please and Thankyou.

[willow bear]

make a diagram & it might become a bit clearer to you

if you're familiar with trigonometry on the unit circle

[viscid thistle]

when you see trigonometry, you should think triangles @viscid thistle

tri
triangles
trigonometry

I see in my head already a nice grid with the x and y axis, and there is the point at the bottom left quadrant

connect that bad boy from origin (0,0) to it, then go into the y axis to form a right angle triangle

hint: make sure you end up to the left and to the bottom

also hint: radians

[viscid thistle]

Thanks for the help

https://tenor.com/view/dancing-cow-soh-cah-toa-funny-meme-gif-19135335

remember

@viscid thistle

[viscid thistle]

what do I have to multiply this by in order to get rid of the cube root?

[willow bear]

$x^2 + x\sqrt[3]{3} + \sqrt[3]{9}$

[obsidian monolith]

Ann:

[sick steppe]

@viscid thistle it's a factor of a difference of cubes

[viscid thistle]

can you show me

[sick steppe]

Factor $x^3 - 3$

[obsidian monolith]

moshill1:

[sick steppe]

$x^3 - 3 = (x-3^{1/3})(x^2 +x \cdot 3^{1/3} + (3^{1/3})^2)$

[obsidian monolith]

moshill1:

[random cloud]

I'm just wondering

When calculating the slant asymptote why doesn't the remainder matter?

[patent beacon]

Something like
3/(x + 1)
Will vanish (go to zero) as x gets very large

Remainder terms tend to look like this

[random cloud]

oh

[violet knoll]

Anyone know how I would write the domain and rage for this?

(infinity, -4)U(-4,2)U(2,infinity)??

[willow bear]

thats the domain yeah

[violet knoll]

YEah and I have no Idea what the range would look like

[blissful ridge]

Do you know what range is??

[ancient rampart]

how would i do this

sin is reciprocal of csc right

so just -1/3?

[willow bear]

yes

[last tinsel]

i just don't know how to find it with 2 arbitrary points

<@&286206848099549185>

[north island]

could try using the conic section stuff, should get simultaneous equations? idk

[last tinsel]

ok

[viscid thistle]

how would i turn this into an expression with integers

[sour hemlock]

what?

I did not understand

expression with integers?

[viscid thistle]

uhhm

like

i want to get rid of the square root

[sour hemlock]

y = x - 1 + sqrt(3)

[viscid thistle]

i got (x-1)^2 = sqrt(3)

or i mean

[sour hemlock]

I guess you could carry sqrt(3) to the other side and everything else on the opposite side

[viscid thistle]

(x-1)^2 = 3

[sour hemlock]

and just square it

woah

what are you doing?

you have an algebraic expression with one unknown variable, how did you get an equation with one unknown variable?

y = x - 1 + sqrt(3)

[viscid thistle]

oh

[sour hemlock]

manipulate this to get what you want

[viscid thistle]

sorry I wrote the wrong sign

I got it now

[last solar]

do I just add 360 or subtract it?

And thats the answer?

[sour hemlock]

stop spamming

[velvet blade]

I got the whole $(a^m)^n = a^(mn) thing. Should I just write down the property and call it a proof?

[viscid thistle]

try to write the proof

it's good exercise

[velvet blade]

Yeah I did

[topaz flint]

A triangle ABC lies within a parabola y = 9x -x^2. Vertex C lies on the parabola and the base of triangle AB is on the x-axis. Calculate the maximum possible area of the triangle and justify your answer.

I got 13564.5 m^2, can someone verify

[gentle vigil]

@topaz flint is vertex C in the first quadrant?

[topaz flint]

no

Area of triangle = 1/2*BH

[gentle vigil]

Doesn’t that make the maximum area unbounded?

[topaz flint]

so Area = 0.5(9x - x^2)(9x-x^2)

not sure

[gentle vigil]

When you say triangle ABC is within the parabola, what do you mean?

[topaz flint]

i will give u a rough diagram

[gentle vigil]

Sure, thank you

[topaz flint]

j

@gentle vigil

[gentle vigil]

So is this a right angle triangle?

[topaz flint]

ye

i forgot to mentionj

[gentle vigil]

Okay I’ll try it out

[topaz flint]

k

[harsh rampart]

quick question, is it possible to differntiate this using power rule? https://i.imgur.com/bReDw9g.png

[gentle vigil]

@topaz flint I’m getting 19.65 approximately

[topaz flint]

how did u solve

[gentle vigil]

@harsh rampart yes

@topaz flint

[topaz flint]

oh i see

i made the wrong equation for the area

anyways thanks.

[gentle vigil]

You’re welcome

[ancient rampart]

could some1 help with 29 and 31

[gentle vigil]

What have you done so far? @ancient rampart

[ancient rampart]

i made sin(x+y) to sinxcosy + cosxsiny

but idk what to do with the cos stuff

[gentle vigil]

Can you use trigonometric identity or Pythagoras theorem to find cos?

[ancient rampart]

uh

idk

all i have is sin2x +cos2x =1

[gentle vigil]

Yes, exactly

You know sinx. Using that identity, can you find cosx?

[ancient rampart]

oh you mean plug in 0.8

:O

[gentle vigil]

Yes

[ancient rampart]

ok thanks

could u help me with 2 more

[gentle vigil]

Sure

[ancient rampart]

that i dont know how to do

25 and 27

[gentle vigil]

Do you know values of sin(pi/4) and cos(pi/4)?

[ancient rampart]

root2/2

[gentle vigil]

Yes. Can you find cosx using the same method as before?

[ancient rampart]

why do i need to find cosx

[gentle vigil]

Because you will need it when you use formula for sin(x+y)

[ancient rampart]

why do i need to use sin(x+y)

[uncut mulch]

for these questions it is implied that they want exact values

[ancient rampart]

yes

i dont understand what i should start by doing

[uncut mulch]

and applying compound angle identities would be the recommended approach

sin(a+b)= sin(a)cos(b) + cos(a)sin(b)

[ancient rampart]

wait so are we saying sin(pi/4+x) is sin(y+x)

[uncut mulch]

sin(pi/4 + x)

is in the form sin(a+b) or whatever variables you want to use

[ancient rampart]

oh

[uncut mulch]

and you can use compound angle identites to expand that out

[ancient rampart]

does it matter which value is which

pi/4 and x

[uncut mulch]

no

[ancient rampart]

even tho we are given value for sinx?

[uncut mulch]

sin(a+b) = sin(b+a)
order here doesn't matter

[ancient rampart]

o

[uncut mulch]

cos(x) can be determined from pythagoras (also make sure you have the correct sign from unit circle, astc and/or otherwise)

[ancient rampart]

ok ty:D

[velvet blade]

Hey is the answer just $C = 5/(e^4)$?

[obsidian monolith]

lazypawtato:

[velvet blade]

<@&286206848099549185>

[sick steppe]

yes

[velvet blade]

thanks

[lapis sphinx]

$test$

[obsidian monolith]

North:

[upper furnace]

bruhhhh

i need help.com

<@&286206848099549185>

[viscid thistle]

Have you considered even posting the q?

Have you even considered reading #❓how-to-get-help

If you had, you'd prolly have gotten help already

[foggy lake]

can someone explain why the answer is (sin^2(x)+1)/cos^3(x)

[sour hemlock]

you can take the derivative and see for yourself

[foggy lake]

is the derivative of that sec(x)tan(x)?

[jolly gust]

Hey we just started logs and I dont get this question

10^2log6

[gentle vigil]

Can you write that with brackets?

[jolly gust]

10^(2log6)

[old nimbus]

Can I get help with integration?

[steel venture]

sure just go to one of the question channels

integration isnt really precalc

[rapid lance]

@jolly gust do you know that a*log(b) = logb^a?

and that a^(log_a(b)) = b

[jolly gust]

Yeah

[ripe adder]

Im stuck on a systems of equations word problem

I understand that (x - 0.077x) + (y + 0.022y) = 9378.50

but dont know how to solve from here

[patent beacon]

@ripe adder
That's a good equation. However you have two variables so you need two equations

The other is
x + y = 10000

[ripe adder]

how can i find the solution?@patent beacon

im sorry im very new to systems of equatios

[patent beacon]

With that second equation, you could write
y = 10000 - x

Then sub that into the first. Solve for x.

[ripe adder]

ok so

what do you mean by sub into that

in this case @patent beacon

[patent beacon]

That is, replace every y with (10000 - x)

Which you can do, since they are the same

[ripe adder]

so

[patent beacon]

"sub" is a common lingo for "substitute"

[ripe adder]

(10000 - x) = 10000 - x?

or wait

i see

in the original equation

(x - 0.077x) + ((10000 - x) + 0.022(10000 - x)) = 9378.50

[patent beacon]

Haha yeah you can't put the info from the second equation into the second equation or else you get that

[ripe adder]

like that?

[patent beacon]

You did that fast. Yeah exactly like that

[ripe adder]

nice

sorry the word problem -> actual problem had me confused for a minute

so it would be 8500$ at 7.7% loss

and 1500$ with 2.2% gain

thanks @patent beacon

I understand now

[patent beacon]

Np, feel free to ask if you need anything else!

[viscid thistle]

Hey guys, can anyone help me with this question:
3sin(2x+20°) = 3cos(2x+k)
I thought all about it and it just isnt clicking, I looked at my friends work and I just don't understand it. The answer key says that it is 70° but I don't know why.

I tried graphing 3sin(2(x+10)), and figured I would just make (k/2) = where the maximum x coordinate is but that doesn't seem to work (or more likely I did something wrong)

[patent beacon]

Oh it's an identity for all x haha

[viscid thistle]

lol yeah

i was aboutt to mention that

when u sent the original message

[patent beacon]

'member that sin(x) = cos(x - 90)

Now, x can be anything. So why not just replace x with 2x + 20? Seems to fit

That gives the identity
sin(2x + 20) = cos(2x - 70)

[viscid thistle]

wait where did the 70 come from?

oh wait nvm

i see

OOooooo

[patent beacon]

Then multiply both sides by 3 as that changes nothing

[viscid thistle]

ooooh jeez that is smart

i didnt think of that

[patent beacon]

Now you will!

[viscid thistle]

thank you so much that was so helpful!

[patent beacon]

Np. Feel free to ask if you need anything else

[viscid thistle]

thank you

is that a way i can find the solutions to any of those types of questions?

provided the amplitude and sinusoidal axis are the same

[patent beacon]

I mean if it looks similar to an identity you already know, then odds are it is just that identity in disguise

I just did a little algebraic manipulation to show that your question and sin(x) = cos(x - 90) are pretty much saying the same thing

I can't say this will always work but if you think it might, it's worth trying.

[viscid thistle]

yeah okay

I mean if it is just sin(x) = cos(x+90) in disguise, i can just use that method right

[patent beacon]

Oops, I messed up the identity didn't I? Haha

But yes

[viscid thistle]

no u didnt mess up the identity

wait what about -cos(x) = sin(x-k)? what r those identities

[patent beacon]

Note that -cos(x) = cos(x - 180)

[viscid thistle]

ooh so -cos(x) = sin(x-270)?

or wait, would it be -cos(x) = sin(x-90)

because cos(x) = sin(x+90) right
therefore cos(x-180) = sin(x-180+90) = sin (x-90)
therefore -cos(x)=sin(x-90)

[willow bear]

Give the exact answer, not a decimal approximation

[gives decimal approximation anyway]

[pastel pecan]

yeah Ik

I just didnt know what to put

[willow bear]

how did you get this decimal anyway

[pastel pecan]

I plugged it into the calculator using the sum or difference identity for tan(a+b)

[willow bear]

plugged what into the calculator

[pastel pecan]

tan(4/3)+tan(5/12)

1-tan(4/3)tan(5/12)

thats over each other

[willow bear]

tan(4/3)

are you sure you want to take the tangent of four-thirds of a radian?

[pastel pecan]

hmm thats a good point

Should I convert it into degrees

[willow bear]

you do not need the values of alpha and beta themselves

you should have written that tan(β) = 5/12

and then you should have written: $$\tan(α + β) = \frac{\tan(α) + \tan(β)}{1 - \tan(α)\tan(β)} = \frac{\frac{4}{3} + \frac{5}{12}}{1 - \frac{4}{3} \times \frac{5}{12}}$$ and simplified carefully

[obsidian monolith]

Ann:

[pastel pecan]

Ok this makes a lot more sense now

Thank you very much

63/16

[grand belfry]

Hey guys, maybe I am missing a key concept, I understand why we're getting the bottom part of the function like it is, but why do we have the left and right parts?

[willow bear]

what do you mean

the function y = (3x^2 - 12)/(x^2 - 9) is defined for any x not equal to ±3, and this includes x < -3 as well as x > 3

[grand belfry]

@willow bear Yeah I analyzed it but I am not sure why we have this part

Sorry if I am unclear but after writing everything down, I understand only why we drew the bottom part

[willow bear]

do you think the parts you circled just... shouldn't be there at all?

[grand belfry]

From my understanding yes

I cannot figure out why they're there

[willow bear]

so you think the function should not be defined at all for x > 3 or for x < -3?

[grand belfry]

Couldnt it be drawn like this then?

as well or

[willow bear]

why would it be like that?

that would mean the function's value is less than 3 in those regions

do you think the value of (3x^2 - 12)/(x^2 - 9) is less than 3 when, say, x = 10?

[grand belfry]

Oh, okay, so that's the reasoning behind it

Thank you

[viscid thistle]

how would i go about finding the roots and the y-intercept value of the function: π = q^3 - q^2 - 18

[willow bear]

what's y

[viscid thistle]

i'm not sure, as the problem doesn't state it

[willow bear]

can you show the entire problem

[viscid thistle]

Find the roots (if any) of the following profit function. Give the economic interpretation of the root(s) and the y-intercept value. π = q^3 - q^2 - 18

[willow bear]

i'm not sure this graph corresponds to your equation

[viscid thistle]

Which information is missing from the initial problem

[uncut mulch]

is pi supposed to be the constant pi or is the symbol being used to represent something else?

[viscid thistle]

It is being used to represent the profit function

pi is used to denote economic profit.

[uncut mulch]

then I guess you could try factorising q^3 - q^2 -18 with the aid of rational root theorem or otherwise

if you wanted to plot it,
y = x^3 - x^2 - 18

[viscid thistle]

alright

[amber linden]

$f(x)$

[opaque olive]

how do i like show theres no derivative at x=0

[uncut mulch]

piecewise definition of the abs value

[opaque olive]

yeah i know that

y = x for x>= 0

y = -x for x < 0

[uncut mulch]

then piecewise definition for tis deriviative

[opaque olive]

y = 1 for x >= 0

y = -1 for x < 0

[uncut mulch]

then consider the limits from the left and right of 0

[opaque olive]

well zero

[uncut mulch]

i mean for the derivative

[opaque olive]

ahhhhhh

limit doesnt exist

thus no derivative at zero

how did i not realise that :/

[opaque olive]

unsure about this question

[velvet blade]

how do I write the equation of this parabola in desmos?

[velvet blade]

<@&286206848099549185>

[steel venture]

x = 1-y^2

[velvet blade]

thanks

[velvet blade]

I didn't understand how to map this

[jolly gust]

I dont get what to do when a log is in the exponent

Like 10^(2log6)

[viscid thistle]

You may want to transform it to 10^(log(6²)) so that you can use a law

$10^{\log_{10}(a)}=a$

[obsidian monolith]

Al𝟛dium:

[velvet blade]

<@&286206848099549185>

[round quest]

How many multiples of 7 are in between 29 and 361

[idle moat]

is this right?

<@&286206848099549185>

[uncut mulch]

yes

[velvet blade]

<@&286206848099549185>

[ripe adder]

I am stuck on a math word problem about apples and bannanas

It wants me to convert it to a systems of equations

I currently know that (x*(100c+4f)) + (y*(125c+2f)) = 3875c+86f

however, i have been using substitution to do most of these problems, and dont see how it can be applied here

[sour hemlock]

number of apples = A
number of bananas = B

now write down the two equations

show it to us and we can guide you if there are still errors

[ripe adder]

ah I see

one sec

so X = 100c+4g?

because X=A?

[sour hemlock]

best you use the variables A and B

[ripe adder]

so X=A

[sour hemlock]

as indicated in the question

[ripe adder]

Y = B

[sour hemlock]

okay

let's walk through this

[ripe adder]

X+Y=3875c+56f

[sour hemlock]

first let's talk about the calories

we need a total of 3875 calories

if one apple has 100 calories and one banana has 125 calories

how many apples and bananas we need to get 3875 calories

let A be the number of apples we need and B be the number of bananas we need

are you able to write the equation for calories?

[ripe adder]

25 A = 3B

[sour hemlock]

????

what

[ripe adder]

35A =3B

fuck

bot tgat

[sour hemlock]

I think you are confused

[ripe adder]

35a+3b

[sour hemlock]

let's try this again

let A be the number of apples

[ripe adder]

35 apples + 3 bannanas

[sour hemlock]

and one apple gives 100 calories

[ripe adder]

= 3875 calories

[sour hemlock]

so how many calories would A apples give us?

[ripe adder]

A * 100 calories

[sour hemlock]

good, let's do the same for bananas

if one banana gives 125 calories

how many calories does B bananas give us?

[ripe adder]

B*125

so

[sour hemlock]

good

can you now tell me the total calories that A apples and B bananas would give us?

[ripe adder]

A(100c+4f)+B(125c+2f) = 3875c+86f

[sour hemlock]

please answer my question

[ripe adder]

A(100)+b(125)

[sour hemlock]

why are you multiplying B with 100?

[ripe adder]

125*

[sour hemlock]

please write it again

[ripe adder]

its edited

[sour hemlock]

ok good

so, this total calories is actually given to us in the question

can you read the question and tell me what it is

[ripe adder]

ah

so

A(100)+B(125)=3875

and

[sour hemlock]

good

please do the same for fibre

[ripe adder]

a(4)+b(2) = 86

[sour hemlock]

yes

solve for A and B

[ripe adder]

@sour hemlock

A=10

B=23

it says my answers are incorrect

[sour hemlock]

I guess check your calculation for errors then

[ripe adder]

10(100)+23(125)=3875

thats true

and it was found through systems

[round quest]

how to do this

[bold meadow]

Arithmetic Progression

[round quest]

i used the formula

and keep getting 15625

[tough thicket]

Is that incorrect

[round quest]

yes

either its incorrect or the answer key is messed up

[tough thicket]

Oh god, I wish I remembered more from earlier classes

Okay do you know the ratio @round quest

I’m guessing that would be the point of error if anything

[round quest]

-5

[tough thicket]

And what is the formula you’re using

[round quest]

a1r^n-1

[tough thicket]

I would use the bot if I knew how to

But it would be set up so that you’re finding a(7) = a*r^(7)-1

[round quest]

yes

a=-1

r=-5

so 5 ^6

[tough thicket]

-5 you mean

The first a represents the first term of the sequence

[sleek ruin]

Anyone know how to do this?

[round quest]

a is -1 right

and the common ratio is -5

so -5 x -1 is 5

[sour hemlock]

@sleek ruin there should be only one value for theta between 0 and 180

so 143.1 is wrong

53.1 is the only answer

[tough thicket]

@round quest order of operations my friend

Do your power first

[sleek ruin]

@sour hemlock I tried that. It came back wrong as well.

[round quest]

so 78125

[sour hemlock]

what the heck?

I mean you can even verify this from the graph of sin

there is only one value of theta for 0.8

[sleek ruin]

Is there something wrong with the system or something lol

[tough thicket]

Axle is that correct

[round quest]

yes

[velvet blade]

How did he come to a conclusion that this is a rectangle?

How do I map it?

[tough thicket]

You’re welcome @round quest lol

[round quest]

z

[velvet blade]

?

should I use another channel?

[tough thicket]

I’m not positive as to that being pre-algebra

But I think that it’s important to note the part where it says (r, theta) being a point

[velvet blade]

how do I draw the image?

Is it just busy work? Can I skip it if they're not numericals?

[velvet blade]

<@&286206848099549185>

[bronze sandal]

can someone rewrite this in terms of n? ive been trying to use the sum of an arithmetic series, but the resulting equation wont match my empirical data

[proud raven]

support incoming

[surreal crest]

based lang reader

[velvet blade]

Hey thanks a lot for coming

[proud raven]

this, yea?

[surreal crest]

@bronze sandal first thing is u can factor out 1/2

[velvet blade]

I'm very confused with problems in mapping

this, yea?
@proud raven yes