precalculus

[verbal leaf]

Adding two negatives together is the same thing as adding two positives together and multiplying by -1

Unless I’m wrong and that’s not the reasoning behind that

[fading token]

Yes

[rigid beacon]

Well kinda but really you should see it as factoring

Being able to look at one step to another and realize it's factoring is important

[grizzled totem]

I've completed all of a) and b), I'm not sure how to answer c).

I don't know what y = (tanØ)x with Ø = 120° is

[patent beacon]

You'd need to know tan(120) from a unit circle

You'd get y = -√3 x

[grizzled totem]

Cheers

I've just attempted writing down the composite functions of voh

V o H V(H(x)] = V[2x-1] = 2x-1 +3 = 2x + 2

V says = x²

and I didn't do anything with the ²

because I'm not sure if I have to

[patent beacon]

Like you said,
v(x) = x² + 3.

v takes the input, squares it, then adds 3.

In this case, your input is 2x - 1

[grizzled totem]

2x² - 1?

[patent beacon]

v(2x - 1) = (2x - 1)² + 3

From v(x) = x² + 3

v() = ² + 3

[grizzled totem]

V o H V(H(x)] = V[2x-1] = (2x-1)² +3 = 2x² + 2

I'll go back

[patent beacon]

Now we have (2x - 1)² + 3
In order to evaluate (2x - 1)², we should FOIL:
(2x - 1)(2x - 1)

[grizzled totem]

I'll recalculate, i think I got this

4x² −4x + 1

4x² −4x + 1 + 3 =

4x² −4x + 4

Would that be the final answer for V o H?

[patent whale]

https://www.symbolab.com/solver/step-by-step/\frac{1}{x}-\frac{2}{3x%2B3}>1

why <-1,0> is out of the domain? I thought that only x =/= -1 and x =/= 0

[uncut mulch]

those are the solutions to the given inequality
(not the domain of y = 1/x - 2/(3x+3))

[patent whale]

uhh how do i calculate these solutions or do something so i know what to give as an answer because on paper i only get x1, x2 and that x=/= -1 and x=/= 0

[uncut mulch]

combine the 3 terms into a single fraction and analyse signs

I would imagine that symbolab showed the work for it

wdym by x1, x2?

[patent whale]

(-1+sqrt(10))/3 is x1 and (sqrt(10)-1)/3 is x2

[desert socket]

I thought i knew drt problems, but this one is messing me up

all the other prior examples were simple i guess

[serene heath]

what did you try

[desert socket]

well basically i got 4.4(50+60) = 484

might be completely off in my approach though

[willow bear]

why did you do that calculation? what is it supposed to represent?

[desert socket]

Sorry I’ll respond in a few.

[desert socket]

Okay I took a break and re looked at the problem.

The format when we have the time is t=D/r so I set it up as D/50 + D/60 = 4.4

Then I get D = 120 after simplifying.

[grizzled totem]

Sorry for cutting you mate

I want to know if my final answer is correct

V o H V(H(x)] = V[2x-1] = (2x-1)² +3 = 4x² −4x + 1 + 3 = 4x² −4x + 4

VoH only

[clever inlet]

Looks fine

[grizzled totem]

cheers, i'll attempt VoU now

[grizzled totem]

My attempt for V∘U

(V∘U)(x) = V(U(x)) = V(x/2 - 1) = ((x/2-1))² + 3 = 1/4(x-2)² + 3

[harsh cipher]

question

[clever inlet]

🤔

[harsh cipher]

On the larger grid. I don't understand why the graph's minimum is at -2 when the maximum is at 1

what I'm asking is....because of the v expansion of 2

the minimum should be at down at -3?

look at that graph...its not even from the working axis

but then he plotted the coordinates correctly on the smaller grid?

[autumn parcel]

how to solve for the next number in the series: 75, 69, 63, 67, 51, [ ? ]? can someone help

[valid violet]

Next number is 0

[clever inlet]

[valid violet]

The numbers satisfy the polynomial p(x)= (2 x^5)/15 - (29 x^4)/12 + (41 x^3)/3 - (331 x^2)/12 + (51 x)/5 + 75

[clever inlet]

Lel

[valid violet]

Starting at x=0 that is

[clever inlet]

n or x 🤔

[valid violet]

Right

There's no way to know the next number without context is my point

You can make a polynomial for any next number you choose

[autumn parcel]

choices are 44,45,46,50,51

[clever inlet]

Any

[valid violet]

If you want the next number to be 44

(17 x^5)/45 - (1007 x^4)/180 + (1264 x^3)/45 - (9673 x^2)/180 + (373 x)/15 + 75

If 45

(23 x^5)/60 - (17 x^4)/3 + (341 x^3)/12 - (163 x^2)/3 + (126 x)/5 + 75

any number

You need context to have an answer

oeis.org has many important sequences and you can look up if your terms are some existing sequence that's been important in some context

your sequence isnt in it

[autumn parcel]

I see. Thanks a lot

[viscid thistle]

given f(3) = 4, f'(3) = -1. f'6=5, g(3)=6, and g'(3) = 2. Find : (f * g)'(3)

[clever inlet]

Do you know the product rule?

[viscid thistle]

I don't, I'm actually kinda lost

[clever inlet]

(fg)' = f'g + g'f

And then evaluated at the point of interest

[viscid thistle]

i think i understand

so it's asking to find f*g'3, and I'm not sure how to attempt this

[clever inlet]

What do you get from apply the product rule?

[winter isle]

find the exact value of sin(u+v) given that sin u = 7/25 and cos v = 4/5, 90<u<180 and 270<v<360

[lilac tide]

what kind of trig are you using? (sum formula?)

[marble crown]

So, I'm currently in need of some things in real life that represents y=Sec theta. If anyone is able to give me some ideas it would be nice.

[trim fable]

question so

for x is an element of the real x != 0 how would u write that in absolute notation

would it just be |x| != 0 ?

[bronze canyon]

x != 0 would be enough

Or say “x is an element the reals (excluding 0)”

[viscid thistle]

How do I solve sinx+4cscx+5=0

[clever inlet]

what have you tried?

[viscid thistle]

I know 1/sin is csc

But like I’m confused on how to do it without fucking up the whole equation

[willow bear]

would you be able to solve the equation $u + \frac{4}{u} + 5 = 0$?

[obsidian monolith]

Ann:

[viscid thistle]

Yea multiply the whole equation by u

OH SHIT

But like

I think I’d need to use the quadratic formula

[willow bear]

ok quadratic formula it is then

is that an issue?

[viscid thistle]

Yea

Wait it’d be no solution

Cause discrimination

[willow bear]

[patent beacon]

"Cause discrimination" is the best reason I've ever heard for anything ever

[viscid thistle]

Discriminant

That’s what I mean XDD

[willow bear]

well when you multiply through by u

what do you get

[viscid thistle]

U^2=4=5u

Oh wait no there an answer

Nvm

[willow bear]

U^2=4=5u

how did an equality become a double equality

[viscid thistle]

Cause discrimination

[willow bear]

...

how about we get back on track here

$u + \frac{4}{u} + 5 = 0$

[obsidian monolith]

Ann:

[willow bear]

you multiply both sides of this equality by u. what do you get upon doing that?

[viscid thistle]

U^2

=5u

=4

Bruh the fucking plus and equal won’t work

Fuck me

[willow bear]

surely pressing the = button while holding down shift ought to produce a plus sign?

[grizzled totem]

V o H V(H(x)] = V[2x-1] = (2x-1)² +3 = 4x² −4x + 1 + 3 = 4x² −4x + 4

VoH Only

Would it be correct to leave my final answer for VoH as "(2x-1)² +3"

[fluid shore]

I mean, I guess it's okay to leave it like that. I'm just a little perplexed about your choice to change the notation given in the question to something different.

Like, in the question, h(x) = 2x - 1. You changed that to f(x), for no good reason :/

You should also state the domain and range of your composite function. @grizzled totem

[grizzled totem]

so it won't matter to leave my answer as either 4x² −4x + 4

or (2x-1)² +3

because apparently you have to foil the (2x-1)² + 3

but here in the solution and answer, it doesn't do so.

so I'm not sure which one is the correct answer

[fluid shore]

I assure you, (2x-1)^2 + 3 is not a villain. There's no need to foil them.

[grizzled totem]

Ah, someone earlier said I would need to foil them

[fluid shore]

On a more serious note, yes, you may leave it in the expanded form. You may also leave it like how it is. If your question asks you to leave it in a particular form, then leave it in that form.

[grizzled totem]

My attempt for x = 2

(2(2+1)/2-1) = 6

Is this correct

[stuck lark]

sure b(2)=6

[leaden stratus]

How would you verify this identity?

[willow bear]

do some simplification on the lhs

like writing tan as sin/cos and sec as 1/cos

[leaden stratus]

Yeah, I did that @willow bear

But then, Idk what to do.

[willow bear]

can you show me what you ended up with

[leaden stratus]

@willow bear I don't know if doing common denominator

Was a right thing to do

[willow bear]

that was fine

i see you've got $\frac{1 - \cos^3(\alpha) - \sin^2(\alpha) - \cos^2(\alpha) + \cos(\alpha)}{\cos(\alpha)}$ in there

[obsidian monolith]

Ann:

[willow bear]

can you explain how you went from that to $\frac{1 - \cos^4(\alpha) - \sin^2(\alpha)}{\cos(\alpha)}$

[obsidian monolith]

Ann:

[leaden stratus]

I've wrongly summed the exponents

palmface

But... anyways what to do in the first TeX image?

[willow bear]

well you see that $-\sin^2(\alpha) - \cos^2(\alpha)$ there

[obsidian monolith]

Ann:

[willow bear]

maybe something can be done to that

[leaden stratus]

🤔 Pythagorean identities?

[willow bear]

.....whatever you choose to call it

but it's just -1

so surely your numerator becomes $\cos(\alpha) - \cos^3(\alpha)$

[obsidian monolith]

Ann:

[leaden stratus]

And group it as $ \cos(\alpha) (1-\cos^2(\alpha))?$

[obsidian monolith]

TheUndergroundSet:

[willow bear]

why not

[leaden stratus]

Well, at least I verified the identity. Thanks

[warm crescent]

if a and b are complex cube root of unity show that $a^4+b^4+a^{-1}b^{-1}=0$

[obsidian monolith]

Raftaar:

[valid violet]

@warm crescent if x is a cube root of unity then x⁴ simplifies

[warm crescent]

I have googled the problem

And the people there assume a=w b=w^2

w is complex cube root of unith

Unitu

Unity*

[valid violet]

If a is a cube root of unity what's a³?

[warm crescent]

1

[valid violet]

Yes, so a⁴+b⁴simplifies

Makes me think there's a trick they want you to use by leaving it in terms of ⁴ though

Uh i dont think it's true unless you impose that a and b are distinct

[warm crescent]

So what should i do

It doesn' simplifies

[valid violet]

a⁴=a³a¹=1a¹=a

[warm crescent]

I know but still

a+b+1/ab

[valid violet]

If a=b, you have 2a=-1/a² or 2a³=-1 i.e. 2=-1 which is false

[warm crescent]

[valid violet]

So maybe it's true but only if a and b are distinct

[valid violet]

In which case you might as well assume a=w and b=w² likewhat you found googling

So you have to assume the book is assuming distinct

Otherwise it's false

If they're distinct and nonreal, one is the square of the other

And the equation is symmetric in a and b. That means if you switch a with b and b with a the equation stays the same

So that's why your googling gave you that you assume a=w and b=w²

Because either a=b² or b=a²

Actually aren't both true lol

Either way

w+w²+w-³=0

[sharp mulch]

hi can somebody help me solve this inequality

[valid violet]

Chat is in use

Do you see how we got this? w+w²+w-³=0

@warm crescent

[fleet yew]

@sharp mulch divide both sides by 500

[fleet yew]

@glossy yew no

This is not the right channel lol

Go elsewhere

[meager hemlock]

what is this? https://cdn.discordapp.com/attachments/319703641458081794/669419879652130856/unknown.png

https://cdn.discordapp.com/attachments/319703641458081794/669420008287371284/unknown.png

[stuck lark]

what're you confused on

[meager hemlock]

i dont know how to evaluate this equation

i only just learned this, im reading the notes here

but the notes do a shit job of explaining it so idk what im supposed to do

why does it equal 8

[stuck lark]

are you evaluating integrals purely by looking at graphs

[meager hemlock]

no

the drawing there is just to help explain the question

[willow bear]

the area under the graph is a rectangle

do you know how to find the area of a rectangle

[meager hemlock]

uh yeah

[willow bear]

so?

[meager hemlock]

how does the equation tell me that theres a rectangle im supposed to find the area of

[stuck lark]

because integrals are usually introduced to calc students as area under graphs of functions

[willow bear]

you know that integrals can be represented as areas under graphs right

[meager hemlock]

not well

[stuck lark]

now you know

[meager hemlock]

ok?? well i still dont know how i would go about solving this without the answer already being there

[stuck lark]

$\int_1^34\dd x$ is the area under the graph of $f(x)=4$ from $x=1$ to $x=3$

[obsidian monolith]

RokettoJanpu:

[meager hemlock]

so for this kind of equation the numbers on the top and bottom of the first symbol is the sides of the area im measuring and the value before dx is the height?

so 2dx would be 4 with the same limits

[stuck lark]

we call the top & bottom numbers on that integral sign the limits of integration

visually, they correspond to vertical lines at x=1, x=3 bounding the area

[meager hemlock]

so is the 2dx thing i said accurate

[stuck lark]

$\int_1^32\dd x=4$

[obsidian monolith]

RokettoJanpu:

[stuck lark]

visually, that's the area of a rectangle of width 2 & height 2, so area of 4

[meager hemlock]

alright thanks

[stuck lark]

you're welcome

[urban moat]

anyone knoww?

[serene heath]

@urban moat sum to product

[rigid kestrel]

how does sin²0 work

How do i solve

[willow bear]

...do you know what the notation $\sin^2(x)$ refers to?

[obsidian monolith]

Ann:

[willow bear]

@rigid kestrel

[stuck lark]

answer ann's question

[rigid kestrel]

No

[stuck lark]

$\sin^2(x)$ is another way to write $(\sin(x))^2$

[obsidian monolith]

RokettoJanpu:

[rigid kestrel]

So

What would happen to the ²

[stuck lark]

wdym

[rigid beacon]

@rigid kestrel what's sin(0)

[rigid kestrel]

0?

[rigid beacon]

Yes

What's 0^2

[rigid kestrel]

0

[rigid beacon]

So then what's sin²(0)

[rigid kestrel]

0

[rigid beacon]

There you go

Take it step by step

[rigid kestrel]

So the ² can be after sin

I thought i needed to square the sin but how lmao

Ty anyways

[rigid beacon]

You did square the sine

Order of operations tho

[rigid kestrel]

now that i think about it

[grizzled totem]

my answer for x = 0 is: 2/-1

is my answer correct

[stuck lark]

2/(-1) can be simplified to what?

[grizzled totem]

-2?

is that the final answer?

[stuck lark]

closer. "write down the 3 values of b(x), for x=0,... using ordered pair representation"

[grizzled totem]

it wants 3 values for x = 0?

I thought it meant the 3 values 0, 2 and 3

[stuck lark]

that's not what i said

you have to actually write the ordered pair (x, b(x)) for x=0. repeat for x=2,3

[viral imp]

can someone explain me how do I get from this form

$\lim_{x\to\infty} n\left(\log(1+\frac{2}{n})\right)$

[obsidian monolith]

Umma.Gumma:

[viral imp]

to this one

$\lim_{x\to\infty} \frac{\log(1+\frac{2}{n})}{\frac{1}{n}}$

[obsidian monolith]

Umma.Gumma:

[viral imp]

I would usually raise the log argument to the nth power

[stuck lark]

$\frac1{\frac1n}=\frac1{\frac1n}\times\frac nn=\frac n1=n$

[obsidian monolith]

RokettoJanpu:

[viral imp]

clear thank you

[stuck lark]

np

[vast rose]

My teacher told me precalc is harder than calc is this true?

[patent beacon]

I think I know what you're referring to. Algebra tends to be the hardest part of calculus for many students, as it's pretty intense, and most find they don't know the properties they've been taught

However I would say that calc is the more intensive class compared to precalc

[valid violet]

I found precalc harder than calc because of the trig

Before i got gud

[pale bison]

i still suck at trig

[fading bloom]

what the fuck

limit as x approaches infinity of something with Ns

beautiful

[clever inlet]

🤔

[restive mauve]

[viscid thistle]

Hey i have a question about trig? so my problem is Sin(Cos-1(-√2/2))

and i have no clue what the hell to do

[clever inlet]

Draw a triangle or something

[viscid thistle]

hieroglyphics make more sense then that

what would i do with the trianlge

trianlge

trianlg

TRIANGLE

[clever inlet]

Actually

-sqrt(2)/2 is a convenient value

[viscid thistle]

no clue what that means either

[clever inlet]

Do you know what we can do when we have cos inverse of a negative value?

[viscid thistle]

its not cos negative its CSC sorry

[clever inlet]

🤔

[viscid thistle]

Cosecant

Sin(Csc(-√2/2))

[clever inlet]

There's like

Nothing meaningful you can do

Other than utilise odd function properties

[viscid thistle]

im fucked

[clever inlet]

Do you have a picture of your question?

[clever inlet]

That's cos inverse.....

[viscid thistle]

what

but isnt cos inverse just CSC?

[clever inlet]

No....

It's confusing notation, yes

But they are different

[stuck lark]

Don’t mix a function’s inverse (though cos only has an inverse for a certain restriction on its domain) with its reciprocal

[viscid thistle]

oh

sorry

[clever inlet]

It's ok

Do you know how you can work with negative arguments for cos inverse?

[viscid thistle]

nope

[clever inlet]

Have you seen

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

[viscid thistle]

nope

[clever inlet]

Hmm

[viscid thistle]

i dont know what that means

like at all

[fading bloom]

jesus fucking christ. just say arccos

also if i'm not mistaken since cos(x) = cos(-x), arccos(x)=arccos(-x)

between -1 and 1 you should be fine for reals

[clever inlet]

🤔

[fading bloom]

it's a nice 0, just like my exam grades

[viscid thistle]

i solved the other one

[urban moat]

anyone know why (3,5) is a point?

[clever inlet]

Well

What's f(3)?

What about g(3)?

@urban moat

[urban moat]

ohh wait, f(3)=1 and g(3)=5? so 1*3 and then 1 * 5? @clever inlet

[clever inlet]

Yes

Wait

It's simpler

(fg)(3) = f(3)g(3) = 1 * 5 = 5

[urban moat]

gotchaaa

[viscid thistle]

in my precal class i have a problem where i have the hypotenuse of a right triangle given and i know its a 45 45 90 triangle. How do i calculate the other two legs if the hypotenuse is 8?

[clever inlet]

Well they have to be the same length

[uncut mulch]

multiple ways to go about this

[viscid thistle]

i know its a 45 45 90 because of the line thats drawn for each of the legs to show they're =

[clever inlet]

Ok

Well

It's a right triangle

Certain theorem?

[viscid thistle]

yes, i dont remember which one i got to use theres so many i have to remember like finding altitude and stuff i forgot what i got to use

[uncut mulch]

the theorem that relates the 3 sides of a right angled triangle

starting with P

alternatively, you could also use trig

[viscid thistle]

okay. i mixed it up with cutting the triangle in half but thats for a 30 60 90 triangle

I am having problems with this : evaluate log 26 1/8. 26 is base and 1/8 is X btw

[uncut mulch]

mixed what up?

so your triangle is actually a 30°,60°,90°? (not 45°,45°,90)

[viscid thistle]

Okay so can some one help me on my question if u want

[uncut mulch]

what have you tried?
are you expected to give a decimal approximation?

[viscid thistle]

It says to round to the thousandth my bad forgot to mention that lmao

[uncut mulch]

try looking up the change of base law

[viscid thistle]

Kind of confused on the change of base law is the base always 10 when you're converting them

[uncut mulch]

the new base doesn't matter as long as its a legitimate base

but if you wanted to approximate it using a calc,
you would generally choose ln or log_10

[viscid thistle]

So what do you do after putting it in the format of change of base law

Do you just divide

The 2logs

[uncut mulch]

yes

[runic cradle]

hey im kinda stuck on this problem. Our unit has mostly been on vectors and the law of sines/cosines if that helps at all. Anyone have any ideas on how to do this? (please ping me if you respond)

[viscid thistle]

I got my answer as approximately -0.638 rounded to the thousandsth of course

[viscid thistle]

Having problems with this problem : 4^(x+3) = 5^x

[clever inlet]

What have you tried?

[viscid thistle]

I not sure where to start in this instance

[clever inlet]

Well we have some exponential equation

[viscid thistle]

Would you start by transferring it to log

[clever inlet]

We can simplify a bit by taking the log of both sides can't you?

Yes

[viscid thistle]

So how do you transfer it I thought I knew how but I think I'm wrong

Is there a particular equation or principal

[clever inlet]

What do you get when taking the log of both sides?

[viscid thistle]

I'm not sure as I think I'm doing it wrong. I got the left side as log ,4 (5^x) which doesn't seem right to me at all

I was using the equation log b x equals y is equal to b ^ y equal to x

[clever inlet]

That's fine for the right side

You can just apply the log function to both sides

Giving you

log(4^(x+3)) = log(5^x)

[viscid thistle]

So I transferred the log 5 ^ x over?

[clever inlet]

What log laws can you apply?

[viscid thistle]

Wait what sorry confused

[clever inlet]

What can you rewrite log(5^x) as?

[viscid thistle]

Wait now I'm not sure. To refresh how do you transfer a log to a expontial

I I was using the assumption that log subscript b x equal to y is equal to b ^ y equal to x

But this does not work out

Atleat I don't think so

least

[clever inlet]

That is correct

But it's not particularly useful here

[viscid thistle]

So do u have any equations I can use

[clever inlet]

Log laws

[viscid thistle]

I looked up a log laws on Google. I got an equation for exponents. So would this equation be good (x^m)^n = x^mn which is equal to log(a^n) =n-log(a)

@clever inlet

[clever inlet]

nlog(a) yes

[viscid thistle]

Oh okay

Oh that's a multiplication symbol my bad lmao let me work It out now

[slow roost]

How come the f(-2) is -4 for this problem and not -2? Is there a rule? My teacher taught this in 5 minutes lolllll

[heady jewel]

see the first conditions inequality and then the seconds'

x<=-2

and x>-2

so you take the first one

2*(-2)=-4

[slow roost]

So I only do the ones with the >= sign or <= sign?

It can’t be > < cause those aren’t technically equal to?

[heady jewel]

yea

if i understand you correctly

a<b means a can be any value less than be

[slow roost]

So then

[heady jewel]

not equal

[heady jewel]

$a\leq b$ means a can be less than b but also equal to b

[slow roost]

So then for f(3) it’s 4

[obsidian monolith]

Lionel:

[slow roost]

Bc of that

[heady jewel]

yes

[slow roost]

Not f(3)=2 since the 2 is at an open circle

So technically

It’s not a correct output

Since it isn’t even equal to it

[heady jewel]

yea

[slow roost]

Okay thank you!!

[heady jewel]

np

[slow roost]

@heady jewel what about this

What would i do for f(4)

My key says 2 but (4,2) is the x>0 one

Is there something about if the equal to problems don’t have the following input to look at other ones w the input?

[heady jewel]

you have to check if x>0

iirc 4 is >0

so you get the principal square root

f(4)=4^(0.5)=2

dont create a method for what to evaluate based on the signs

[willow bear]

^(0.5)
ew

[smoky needle]

hello, just wanted to ask how they solved this problem? the problem is

: y=1-2sqrt. of x+3

[grizzled totem]

Is this correct?

[rigid beacon]

Your magnitude formula is wrong

For vector ai + bj +ck

The magnitude is not Sqrt[a-b-c]

It's Sqrt[a^2 + b^2 + c^2]

You should have been thrown off when you got the magnitude being Sqrt[-1]

Your part a looks right @grizzled totem

[willow bear]

$\hat{c} = \frac{(\hat{\imath}, -\hat{\jmath}, -\hat{k})}{\sqrt{\hat{\imath}, -\hat{\jmath}, -\hat{k}}}$ is simply nonsensical

[obsidian monolith]

Ann:

[rigid beacon]

Also yes that

Your notation needs work

[willow bear]

also the screenshot and the paper seem to have little if anything at all to do with each other

[rigid beacon]

$ \hat{A} = \frac{\hat{A}}{||\hat{a}||} $

[obsidian monolith]

Spamakin🎷:

[rigid beacon]

That formula you wrote is also bad

$ \hat{A} = \frac{A}{||A||} $

[obsidian monolith]

Spamakin🎷:

[rigid beacon]

That's the one you want. It's again just notation

And @willow bear I think the screenshot he sent was just an example problem he was given and then what he did was another similar problem

[grizzled totem]

@willow bear I'm so sorry, that's the incorrect screenshot

[willow bear]

my other point still stands

[obsidian monolith]

Illimar:

[oak arch]

It's $\frac{1}{2\sqrt{x}y}$

[obsidian monolith]

Illimar:

[grizzled totem]

https://cdn.discordapp.com/attachments/533043064965693460/669974669725663232/unknown.png

B = {1,4,9} C = {2,3,4}

Can someone check if my answer to 1. C) BxC is correct or not

B x C = { (1,2), (1,3), (1,4), (4,2), (4,3), (4,4), (9,2), (9,3), (9,4)}

[serene heath]

what is N?

[grizzled totem]

N just mean the natural numbers in the universe which is 1,2,3...,10

[serene heath]

looks good

[grizzled totem]

Cheers

[grizzled totem]

For (¡¡) is my attempt correct

[stuck lark]

👍🏾

[finite iris]

someone asked me a problem, and I have no clue what the scenario is trying to convey

what is this situation lmao

[serene heath]

wow julie is tall

its asking you to use similar triangles

wording is meh but yea

[finite iris]

could you draw it or something xD I don't understand the wording

if I had a picture I could do the trig or similar triangles or whatever no problem

or if you want don't to draw it, could you reword the problem or something?

I don't understand what it means by "hold a square, edges line up with top and bottom of the flagpole"

[serene heath]

i tried my best

but this is the basic idea

[runic tapir]

👏 10/10 drawing

[serene heath]

if u look at it from the point of view of julie the edges 'line up'

h is height of pole

[finite iris]

uhhh

I don't actually understand how you got that from the description..
and I appreciate the drawing x-x but it's umm.. kinda difficult to understand

what do you mean by the edges "line up", I guess this is my main confusion

Is this the scenario?

[serene heath]

yea, that's actually a way better diagram

[finite iris]

... why was it worded that way Dx

this scenario/diagram is easy af

but the wording is just a middle finger

thanks btw

[serene heath]

yea it's a shitty question ngl

and np

[viscid thistle]

Shit I need help with identities

Idefk my question that’s the thing

That whole area is gray for me :/

Is there a good lesson for identities here

I mean like half and double identities

<@&286206848099549185> do u guys have toolkits

Like papers that just explain a topic

[runic tapir]

Don't ping helpers until waiting at least 15 minutes

[viscid thistle]

Oh oops

I’ll wait then

[patent beacon]

What do you mean "toolkits"? We can answer a question you may have @viscid thistle

[viscid thistle]

A tool kits like a lesson on a paper

Just one page that explains everything u need to know

I made one for graphing here

[patent beacon]

I don't think we have anything Google doesn't

[viscid thistle]

I’m not asking if u have anything exactly like it

[patent beacon]

But search Google, if you haven't yet

[viscid thistle]

I just mean like short summaries of a lesson

I’ll check google

[grizzled totem]

Are you trying to figure out what the topic is called for your work

[viscid thistle]

Nah I just need to learn what identities are

And toolkits always help

[grizzled totem]

Is that for revision?

[viscid thistle]

Yeah

[patent beacon]

Something like this?

[viscid thistle]

Nah just like a mini lesson or like a summary of a lesson about it, if that makes sense. Something that shows examples of when to use identities and stuff like that... kinda like a notes page but really organized and broken down

I know it’s too much to ask for an entire lesson up in here so that’s why I was hoping there would be like server shared slideshow lessons or something here

[runic cradle]

I've been stuck on this one for a while. Could any of you help me? Thank you in advance! (please ping me if you respond)

[viscid thistle]

@runic cradle I assume you've sketched the triangle

[runic cradle]

I have. Would you like a picture or are you just ensuring?

[viscid thistle]

are you given any values for DEF and def?

@runic cradle

if D is a 90 degree angle, it's pretty trivial

cuz

$d = \sqrt{e^2 + f^2 - 2ef \cos D} \ = \sqrt{e^2 + f^2 - 2ef \cos \frac{\pi}{2}} \ = \sqrt{e^2 + f^2 - 2ef \cdot 0} \ = \sqrt{e^2 + f^2}$

[obsidian monolith]

Carter:

[viscid thistle]

which you already know from the "distance formula" (pythagorean theorem)

[runic cradle]

No we aren’t given any values sadly

https://media.discordapp.net/attachments/387449305885048852/670096393662234644/unknown.png

This is the rest of the problem but I’m not sure if it really helps

[viscid thistle]

i never learned my basic geometry because i always decompose everything into right triangles

i could derive the solution that way with some work but the answer will be clunky and your teacher will probably not enjoy reading it

to the man with a trigonometry, every problem is a right triangle

[runic cradle]

I mean the right triangle assumption is the best thing I have going for me at the moment so I’ll take that gladly as things stand

[viscid thistle]

ok i will try it using decomposition

[thin granite]

I’m confusioned

[willow bear]

what is confusing you here

[thin granite]

Okay so I need to find root of the function first?

[willow bear]

no?

the whole point is that you don't need to find any roots

[thin granite]

What I do

[willow bear]

the point is that you're meant to see if the intermediate value theorem applies to this function on this interval

[thin granite]

[willow bear]

what

[viscid thistle]

@thin granite do you see why that's the point

[thin granite]

for all value of x from 1 to 2 if the fucntion is continuous there's a place that the root must exist

[viscid thistle]

try getting rid of the = zero 😉

and then working with what you have

like, just delete it

$x^4 + x - 3 \ (1,2)$

[obsidian monolith]

Carter:

[viscid thistle]

what can you do with that info?

to find any roots

if you know it's continuous over that domain

@thin granite

[thin granite]

plug chug

[viscid thistle]

what are you plugging?

and, then, what are you chugging?

[thin granite]

f(2)-f(1)?

[viscid thistle]

why that

[thin granite]

i'm blind rn i'm clueless help me

[uncut mulch]

can you state the IVT?

[thin granite]

ok

[viscid thistle]

i'm helpin' ya

won't do ya no good if i just tell ya the answa

[thin granite]

i understand that for all of value from [a,b] in a continuous function, there's always a place where x exist b/t a,b

idk if i word it right or not

[nova wedge]

Sorry if this question doesn't fall in this category (I think it does).

What is the name of the rule for this thing. I have tried looking it up on google and I cannot find anything.

k = ln(c)

e^k = c

[viscid thistle]

nope

@thin granite

well

yeah

but that's not IVT

that's just the definition of continuous functions

IVT deals with the actual values of functions, not their parameters

so x is not what you're actually thinking about here

if f(x) is continuous on x=[a,b]

then that means

any value between f(a) and f(b)

[thin granite]

i'm following...

[viscid thistle]

MUST be

the result

of some f(x) where x is in [a, b]

does that make sense?

cuz u gotta go in a continuous path from f(a) to f(b)

when ur x is going from a to b

therefore

ur gonna go to all of the INTERMEDIATE VALUES

on ur way

from f(a) to f(b)

its really quite simple

it just means "you cant jump around if its continuous"

[thin granite]

ok lemme process that

[viscid thistle]

imagine a really complicated function

squiggly line

doin all kinds of wacky waves

then pick 2 points on the function

those 2 points have x values and y values

well

if u travel from the left point

to the right point

the y values between the left and right points

must ALL eventually be reached along ur travels

[thin granite]

okay

so it's the f(x)

[viscid thistle]

is that obvious to you

[thin granite]

i'm dumb

lol

[viscid thistle]

😛

so try again

[thin granite]

i was thinking in terms of x

[viscid thistle]

we are all dumb until we are not

[thin granite]

well now that you explain that it's F(x) not x i understand

[viscid thistle]

well, what's the solution!

[thin granite]

must be btw F(1) and f(2)

[viscid thistle]

what must be between them?

[thin granite]

value of f(x) where x is btw 1,2

[viscid thistle]

yes, that's true

so how, from that, can you arrive at a solution?

[willow bear]

don't mix f and F

[viscid thistle]

(that factoid does not help you, btw)

also this

[thin granite]

sorry ann yes that was my mistake

[willow bear]

also "btw" doesn't mean "between"

[viscid thistle]

you will see in calculus why not to mix f and F

[willow bear]

math is case sensitive that's why

[thin granite]

yeah F is like anti right?

[viscid thistle]

yep

[willow bear]

it doesn't matter

[thin granite]

k

[willow bear]

no, F by itself has no meaning until you assign it a meaning

it just doesn't need to mean the same thing as lowercase f

[viscid thistle]

so whats the solution @thin granite

[thin granite]

yes it is guaranteed there is a root in [1,2]

[viscid thistle]

why?

[willow bear]

a root in [1,2]

[viscid thistle]

you must convince me

for i remain unconvinced

[runic cradle]

@viscid thistle thank you :) ping me if/when you do because I don't want to have you waste all that time just for me to not see it

[viscid thistle]

your problem is unsimple with the proper tools i would have to spend a few sheets of paper to do it my way

@runic cradle

without the*

[runic cradle]

which tools would those be?

[viscid thistle]

basic geometric laws

[thin granite]

ok so i found that f(1) is negative and f(2) is positive which means there's a point where the function will cross the x-axis which by definition the root?

@viscid thistle

[viscid thistle]

absolutely correct

[thin granite]

[viscid thistle]

thus it is guaranteed

,

[thin granite]

you are a lovely person

[viscid thistle]

yes?

that

there is a root!!

[thin granite]

thank you so much

[viscid thistle]

np

[thin granite]

idk how i passed smh

[runic cradle]

@viscid thistle If you're willing to do that I'd be very grateful but I don't want you to bother if it'd be too much extraneous effort

[viscid thistle]

Unfortunately i cannot today baby boy

Rest easy

[dusty wasp]

can any1 help me with this please

[clever inlet]

what have you tried

[grizzled totem]

Attempt for VoU

a) (V∘U)(x) = V(U(x)) = V(x/2 - 1) = ((x/2-1))² + 3 = 1/4(x-2)² + 3

When I expand (x-2)^2 I then get (x-2)(x-2), when I solve that, I then get X^2 - 4x + 4

I'm stuck at putting the 1/4 + 3 with X^2 - 4x + 4

[willow bear]

wdym

you have $\frac14(x^2 - 4x + 4) + 3$

[obsidian monolith]

Ann:

[willow bear]

what's giving you trouble simplifying that?

@grizzled totem

[grizzled totem]

Simplifying that gets me my final answer to the question?

[willow bear]

yes

[grizzled totem]

1/4x^2−x+4

Is that correct

[willow bear]

seems so.

[grizzled totem]

Q 6) VoH
a) V o H V(H(x)] = V[2x-1] = (2x-1)² +3, (2x-1)(2x-1) = 4x^2 -4x + 1 = 4x^2 - 4x + 1 + 3 = 4x² −4x + 4
a) (V∘U)(x) = V(U(x)) = V(x/2 - 1) = ((x/2-1))² + 3 = 1/4(x-2)² + 3. (X-2)(×-2) = (x^2-4x+4) = 1/4(x^2-4x+4)+3 =
1/4x^2−x+4

b)VoH -1 place = (2(-1)-1)² + 3 = 12
VoH 0 place = (2(0)-1)² + 3 = 4
VoH 2 place = (2(2)-1)² + 3 = 12

Is there anything wrong with my full attempt

[willow bear]

your notation needs work

if this was in an exam and i was grading it you'd lose a lot of points because your notation is sloppy

if you want, i can give you a detailed list of exactly what parts of it i would mark as such

[grizzled totem]

Ok

Yes

Out of 10 Marks btw

[willow bear]

you would receive 2 marks each for (a) and (b) on account of having the correct answer for both

do note however that this may not match your markscheme and is merely my own subjective grading

[grizzled totem]

For the equal signs, shall I use an arrow instead

How shall I explain how I get from one to the other

[willow bear]

would you like to see how i would've laid out the same work as you did

[grizzled totem]

Yes please

[willow bear]

optionally put an arrow at the beginning of line 9 or delete the arrow at the beginning of line 4 for consistency

[grizzled totem]

Ah, I think my answers for Q6 B) are incorrect

[rigid beacon]

If my teacher took off points for notation like that (still in HS btw)

People would have alot lower grades

[willow bear]

no, your answers for Q6(b) were correct

what you just wrote however

is bullshit

$v(h(x)) = 4x^2 - 4x + 4 \ v(h(-1)) \neq 4-1^2 - 4-1 + 4$

[obsidian monolith]

Ann:

[grizzled totem]

So is there no need to change my answers for 6 B)?

[willow bear]

there is no need to change your answers for Q6(b) besides the notational/phrasing issue i pointed out.

[grizzled totem]

I'm not sure why I don't need to change it though

[rigid beacon]

You plugged in the numbers wrong what

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

[willow bear]

that's what i said, spamakin.

[rigid beacon]

The way you said it sounded like you were talking about notation not the answer my bad

[grizzled totem]

Because for my 6 B), I've done (2(-1)-1)^2 , shouldn't it be 4-1^2-4-1+4 ? Like the answer of VoH

[willow bear]

enchant, what i am saying is that the work you sent earlier (i.e. the work i dissected) is correct

[grizzled totem]

But the work now is incorrect

So must be adjusted

[willow bear]

if you wanted to use the same formula as you obtained in the final answer, you would do 4**(-1)^2 - 4(-1)** + 4.

[grizzled totem]

Ah

4(-1)^2-4(-1)+4

[willow bear]

your work had correct answers but bad notation.

[grizzled totem]

= 12 still gets 12

[willow bear]

yes

[grizzled totem]

4(0)^2-4(0)+4 = 4

[rigid beacon]

Yup

[willow bear]

there is nothing wrong with using (2x-1)^2 + 3 instead of 4x^2 - 4x + 4. these expressions are the same thing.

[grizzled totem]

4(2)^2-4(2)+4 = 12

I'm just worried about losing marks

[willow bear]

i suppose that'll depend on your graders in particular

[rigid beacon]

If you get the right answer the right way what do you have to worry about

[willow bear]

i would not dock points for using one formula over the other

but that of course is just me

[rigid beacon]

I don't think any sensible teacher would

[grizzled totem]

Ann, if someone submitted this paper(the one you did), would you not question any working out methods such as actually seeing the user do notations such as adding numbers, would you be fine with it as long as you understand the answers

For example, lines of working out on the (2x-1)(2x-1), the user showing us how to actually multiply these brackets to get to the next point

[rigid beacon]

Nah

If they move from one line to the next

And it's correct

Then it assumed they know how to do it

[grizzled totem]

Perfect

[rigid beacon]

Just like how if I had this

=2 + 3 + 7
=12

I don't need to show how I added up the 3 numbers

(basic example but you get the idea hopefully)

[grizzled totem]

Is Ann here

According to my friend, he said that the answers I've given for VoH, 4x^2 -4x + 1 can be simplified

[uncut mulch]

you could express
4x^2 - 4x + 4 as 4(x^2 -x + 1) if you want

it isn't necessarily simpler, but it will make calculations for 6b) less tedious

[viscid thistle]

Will this work?

[fluid shore]

[viscid thistle]

Lmfao

[fluid shore]

Nice meme

[viscid thistle]

Not even a meme

[willow bear]

uh

what is that signature

[viscid thistle]

Ofc if i change my name at the end

Will it be realistic

[fluid shore]

Yea that name at the end

[willow bear]

at the very least, it's a bit risqué

[fluid shore]

Is what makes it a meme

Okay also, you need to separate things out into paragraphs. That looks ugly as fuck

And spacing is very important at the beginning of paragraphs

[viscid thistle]

Should I complete a math problem for her too

Lmfao Jkjk

[fluid shore]

Also, your sentences are not entirely grammatically correct

Depending on how dumb your teacher is, that shouldn't be too much of a problem

[viscid thistle]

Will it say sent from my iPhone

[willow bear]

Also, your sentences are not entirely grammatically correct
which sentences do you consider grammatically incorrect

[viscid thistle]

^

Ok I broke it down into paragraphs

Indented

But I’m concerned about how it’s going to say sent from my iPhone

Why tf they do that

It’s not even in the email so I can’t backspace it out

Is it just going to get thrown into the email once I send it

[uncut mulch]

you should be able to remove it in options

[willow bear]

But I’m concerned about how it’s going to say sent from my iPhone
why would that be a concern, exactly

ok it'll say it's sent from your iphone so what

[viscid thistle]

Unprofessionalism

[uncut mulch]

how so?
ideally you want to use w/e method to reach them asap

[viscid thistle]

Ok

I’m sending it

Fuck it

It hurts to send it

I get a sharp pain in my heart

I think it’s called fear

[grizzled totem]

Is this correct for the value x = 2

Or have I understood the question incorrectly

[willow bear]

yes b(2) = 2(2+1)/(2-1) = 6

[obsidian monolith]

WhipStreak23:

WhipStreak23:

WhipStreak23:

[grizzled totem]

for value x = 3, i get a fraction

4/1

4/1 = 4?

[stuck lark]

yes

[grizzled totem]

Do I have to explain that

in my coursework

why it = 4

sweet

Is there anything else that I'm missing out on, with that question?

So I've found out each of the 3 values

[willow bear]

you haven't written down the ordered pairs as instructed

[grizzled totem]

Yes, that's what I don't understand

Ordered Pair Representations

[stuck lark]

you have to write 3 pairs of the form (x, b(x)) for x=0,2,3

[grizzled totem]

and "Explain when B is a function"

Is it alright to ask for an example of Ordered Pair Representation in a question and answer scenario

[willow bear]

honestly? idk what is meant by "explain when b is a function" either

[grizzled totem]

Throughout this coursework, it's been asking me questions like "Explain whether this is possible or not", maybe it literally wants me to answer why B is a function

[willow bear]

i mean

b is a function

[grizzled totem]

What about Ordered Pair Representations, can I get an example of an Ordered Pair Representation via a Scenario and the answer to the scenario using ordered pair representation

(x, b(x)) for x=0,2,3

would that need any working out?

or can I just state it like you've written?

[grizzled totem]

So would I have to give my answers like this

(0,2)

(2,6)

(3,4)

[willow bear]

yes

(0,2), (2,6), (3,4)

[grizzled totem]

So my answer

Would look like this for x = 2

[willow bear]

no your answer would be (2,6)

[grizzled totem]

= (2,6)

[willow bear]

the = is inappropriate

just write

[grizzled totem]

But I would need that part in aswell right

For marks

Working out

[willow bear]

Answer: (0,2), (2,6), (3,4).

[grizzled totem]

With working out right

[willow bear]

once you've worked out b(0), b(2) and b(3), just write the above

that should suffice

[grizzled totem]

Ah

Ah it was -2

[grizzled totem]

I've completed 7 a)

And I've had my work checked

Can someone give me feedback on b), c) and d) please

[cerulean solar]

hey

whats the standard form of an equation?

like if the slop is undefined

what would be the standard form of that equation

[valid violet]

@cerulean solar Ax + By + C = 0

if you're talking about a line

or Ax + By = C maybe

[obsidian monolith]

WhipStreak23:

[valid violet]

that's if there's a slope

they were asking for a form that works even if there is no slope

[serene heath]

undefined slop?

like, a vertical line?

[languid trout]

guys

if i have a function f(x)

and another function g(x)

does fg(x) count as a new function

and what if f(x) has the domain x>1

does fg(x) also have the same domain as f(x)

[undone drift]

Do you mean F(g(x))?

[languid trout]

yeah

[undone drift]

Yeah so the domain of f(x) will be the range of g(x)

[languid trout]

g(x) is not the inverse tho

[astral mountain]

How does one call points like cusps in general?

(im trying to find a list of all of them)

[deft briar]

quick question, is the domain of x+2/x+1: x != 1 or -1 < x < 1

[uncut mulch]

are non-differentiable points what you're looking for?

$\frac{x+2}{x+1}$?

[obsidian monolith]

ramonov:

[uncut mulch]

neither option you listed is the correct domain

[deft briar]

oh

i dont really know about differentials

[uncut mulch]

that wasn't for you

[deft briar]

oh, oops

[uncut mulch]

what's the definition of the domain and what method would you use to determine it for
y = (x+2)/(x+1)

[viscid thistle]

Photo de Cαstilhο

Any thoughts about this system of equation? I've tried everything I know, but I couldn't solve it

[uncut mulch]

show us what you tried

[fleet yew]

What does that italian mean

[hexed sparrow]

note that $\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$

[obsidian monolith]

ville:

[willow bear]

@fleet yew "solve the system" i'm pretty sure

[viscid thistle]

@uncut mulch I tried to solve it by substitution

@willow bear yes, that's it. That's Portuguese

[fleet yew]

Looks Italian

[viscid thistle]

Portuguese and Italian have a lot in common

[uncut mulch]

what were you substituting?

[viscid thistle]

I tried to isolate one of the variables. The second one looked easier to do that

[fleet yew]

its a quadratic lol. xy=12, x+y=7. Second equation is redundant

[uncut mulch]

be more specific
you substuted:
__ = ____
into equation __

[viscid thistle]

(X=12/Y) into the second one

[uncut mulch]

not quite redundant amd, you can only get:
xy = 12a, x+y = 7a
from equation 1
eq2 confirms that a=1

you mean x=12/y into the first one?

[viscid thistle]

it seems that wasn't the smartest way to solve it lol

[uncut mulch]

calculations are around the same

$\frac{y}{12} + \frac{1}{y} =\frac{7}{12}$

[obsidian monolith]

ramonov:

[viscid thistle]

oh, I see

when I put the x=12/y into the first one it's like: (1/(12/y)+1/y)=7/12

but you divided 1/(12/y), that's why it became y/12

I feel like the dumbest person now lol

[uncut mulch]

can you do the rest now?

[viscid thistle]

yes, I'm pretty sure of it. Thank you so much!

Also, I'm looking for getting better at algebra. Do you have any tips?

[uncut mulch]

practice

[earnest barn]

How can I understand the way in which terms in polynomials affect the shape of the graph?

Ie

If I take a graph of the form

ax^3 + bx^2 + cx + d = 0

What would be different about another cubic graph but with, say, the bx^2 term missing, so ax^3 + cx + d = 0?

Ie what I'm asking if

For a graph of a polynomial

How do terms lower than the highest order in the polynomial affect the resulting graph?

What I've noticed by messing around in a graphing calculator is that if I start with x^3 + x^2 + x = 0, which has one solution, and then plot x^3 + x^2 = 0, the latter has two solutions whereas the former has one solution

Whereas x^3 + x = 0 has only one solution, in contrast to x^3 + x^2 = 0

Is there a way to formalise what impact the lower-order terms have?

[fleet yew]

$a(x-x_1)(x-x_2)(x-x_3)$

[earnest barn]

Ahh good idea

[fleet yew]

If you multiply it out you get a system of equations for coefficients b, c, and d

[earnest barn]

I was asking the question the wrong way round wasn't I - starting with the wrong representation of a polynomial

[obsidian monolith]

AMD:

[harsh cipher]

Hi question