precalculus

[slate scroll]

yeah I'm sure your technique would work

and your solution makes more sense intuitively

I just used some extra stuff that comes when you learn this topic

but It is kinda hard to understand why the coefficients of x,y and z are always in a direction normal to the plane

[novel cargo]

Oh, didn't know that

so in a plane equation, the coeffs of x, y, z, are always perpendicular to the plane?

[slate scroll]

yes

[novel cargo]

we could maybe prove that with some trigonometry?

[slate scroll]

thats why cross product is useful, as it generates a vector perpendicular to the two vectors used

[novel cargo]

I don't know exactly what the corss product is

I know matrix dot multiplication

dot product

[slate scroll]

yeah that is sometimes called scalar multiplication as it generates a scalar

cross product is often called vector multiplication and it generates a vector

cross product only really works in 3 dimensions

you basically just follow an algorithm to calculate the cross product

[novel cargo]

is all this stuff precalc really?

[slate scroll]

i dont really know what precalc is

im not in the american system

[novel cargo]

do you know calculus?

[slate scroll]

yeah

[novel cargo]

precalc is all that comes before it

I assume

[slate scroll]

our classes are much more integrated than yours

like we do all this stuff together

[novel cargo]

ahh, I see

less modular

more a to z

[slate scroll]

like I learnt a little bit of calculus, like derivatives

then we did complex numbers and vectors

then we did integrals

[novel cargo]

in high school?

[slate scroll]

then some proofs

then more advanced integrals and differential equations

yeah high school

in Australia

[novel cargo]

diff equations ins high school?

that's gangsta

[slate scroll]

yeah quite simple ones

[novel cargo]

I like this approach

[slate scroll]

separable ones were the most advanced it got

[novel cargo]

circliing subjects and approching the core gradually

[slate scroll]

yeah it is good, because we get to revise older stuff while learning new stuff

[novel cargo]

gives your brain time to develop intuition

and really grasp concepts

[slate scroll]

like we are starting vector calculus now which means that I can revise both vectors and calculus while learning new stuff

[novel cargo]

yeah, that's real cool

hey, man

ping wiz earlier in the chat

like above

edit your message and ping

[slate scroll]

ah ok

what level are you up to?

[novel cargo]

I'm early calc

limit, diff

derivative

[slate scroll]

yeah I have seen that you guys focus a lot on limits

I think your system does that better

[novel cargo]

limits are really fundamental

[slate scroll]

it introduces calculus much much better

[novel cargo]

true

[slate scroll]

we pretty much dont even look at limits

[novel cargo]

nooo

even in uni?

[slate scroll]

im not in uni

so idk

[novel cargo]

nah, I think they'll teach you rigorous limits in uni

[slate scroll]

but like we look at how limits work to generate the first principles formula but not much more than that

[novel cargo]

that's ok for high school

[slate scroll]

then we use that formula for about 2 weeks

then go straight into shortcut rules

but I feel like a lot of kids don't actually understand how calculus works fundamentally

they just apply the rules and move on

[novel cargo]

that's how life actually works lmao

for most ppl

applying the rules and moving on

[slate scroll]

yup, then as soon as they get a question that is slightly out of the box they have no idea what to do

[novel cargo]

but where do I plug in this number?

no, yeah, a little bit to the left... down... an epsilon more to the left...

yeah, right there!

put it in

[slate scroll]

haha

thats what it be like

and there is another problem with our system

well sorta our system and sorta my school

I've never been taught matrices or sequence and series

completely missed those topics

because in eighth grade, smart kids were put into and extension class

I was really bad a maths back then

so I didnt make it

However in my last two years I've had to do some classes that involve proofs on summations and matrices

and like calculating the cross product for example uses matrices

I dont know how it works, but I just follow the formula like one of "those people"

[novel cargo]

take up the initiative and teach yourself

[slate scroll]

yeah I have just started it

[novel cargo]

I imagine you're quite capable of that

[slate scroll]

using khan

[novel cargo]

nice

there are tons of resources out there

[slate scroll]

but I should have been formally taught

[novel cargo]

many great men didn't have formal education

[slate scroll]

and In some ways learning integration before summations kinda works

I can see that integrals are just continuous versions of a summation

[novel cargo]

where you consider the limit when n approaches infinity

[slate scroll]

yeah

and not just using integers within the summation

or specific discrete values

but every real number attached to the particular function

[novel cargo]

after developing the intuition for tangent lines from the secant line in the build up for derivatives

the idea of the area under the graph divided into increasingly thinner rectangles feels intuitive

[slate scroll]

yep

and then you just use a summation that is continuous to add them all up

[ornate wolf]

therefore c=-2 @ornate wolf
@slate scroll thank you'~

1 + 2a + 3b = c
@novel cargo sorry,you plugged in the value 1,2,3 given by the question (point 1,2,3) right? how does that link up?

[novel cargo]

I'm not sure I understand your question correctly, but I assume you mean how I can plug those values into the plane equation.

when the point (1, 2, 3) lies on a plane that is define by the equation x + ay + bz = c

it means that a, b, and c are constants such that if you plug in (1, 2, 3) for x, y, z in the plane equation, the equation holds (is true)

therefore 1 + a * 2 + b * 3 = c is true

@ornate wolf

[maiden pelican]

Is there a formula for a sum of infinite arithmetic progression?

[willow bear]

yeah, it's +∞ or -∞ depending on the sign of the common difference

[cedar cipher]

I don't think any infinite Ap converges

[somber yew]

Or it could be 0+0+0+...

[cedar cipher]

Sorry,forgot about that

[willow bear]

$\sum_{n=0}^{\infty} (a + nd) = \begin{cases} +\infty & d > 0 \ -\infty & d < 0 \ +\infty & d=0, a>0 \ -\infty & d=0, a<0 \ 0 & a=d=0 \end{cases}$

[obsidian monolith]

Ann:

[willow bear]

@maiden pelican is this what you were looking for

[maiden pelican]

Ohh yes

[novel cargo]

oh man, I learn stuff by visiting this channel

thanks Ann

Also OP for raising the question

[willow bear]

this was the "fuck off" answer tbh like

i really hope you aren't going to just memorize that

[novel cargo]

no, I'm not good at memorizing at all

went through it and it was like something I needed

it made logical sense

[willow bear]

i was talking to exynouz

[novel cargo]

I'm sorry

[maiden pelican]

Nah, I forgot that aps almost always diverge haha

i really hope you aren't going to just memorize that
@willow bear

I mean that thing isn't that hard to memorize, you can easily derive it

[willow bear]

if you derive it you don't need to memorize

[maiden pelican]

Ohh you meant it that way

[rapid falcon]

what does -theta look like compared to theta?

[somber yew]

Ummm, when denoted by a vector it will be rotated by 180 degrees? I don't know.

[ornate wolf]

therefore 1 + a * 2 + b * 3 = c is true
@novel cargo thank you

[rapid falcon]

ok. thanks

[somber yew]

I'm not sure if I completely understood what you asked, I guess someone more qualified could step in.

[novel cargo]

yw @ornate wolf

[novel cargo]

first column and second column

[viscid thistle]

Look at g(x)

[novel cargo]

(f circle g)(x) = f(g(x))

[terse ravine]

the first two are obviously 10.

I'm confused about the last 4.

[viscid thistle]

(f circle g)(x) = f(g(x))

[terse ravine]

what is f(g(x))?

[novel cargo]

f(g(-7))

since g(-7) = -7

f(-7)

= -1

[terse ravine]

I still don't get it

[lime bolt]

What do u not get about that

[terse ravine]

(f circle g) (x) = -1?

[novel cargo]

have your read the intermediate messages?

if so, what you think about them?

(f circle g) (x) = -1?
@terse ravine is the last part. You must have gone confused somewhere inbetween

[terse ravine]

My apologies I don't understand this at all.

[willow bear]

is the concept of "plugging the output of one function into the input of another" foreign to you?

[terse ravine]

The table feels foreign, plugging the output of one function into the input of another isn't.

the f circle g is messing with me too

[willow bear]

just pretend it's f(g(x))

[terse ravine]

Doesn't look right

[somber yew]

Look at it this way: (f o g)(x) = f(g(x)). For example, (f o g)(-7)=f(g(-7))=f(-7)=-1.

Does this make sense?

[terse ravine]

Does (g o f)(x) =g(f(x)). (g o f)(-1)=g(f(-1))=f(-1)=-7?

@somber yew The layout is clearer for me to understand, but I don't know where you got the -1 from.

[somber yew]

Well, for x=-7, what value of f(x) do you have?

[terse ravine]

-1

@somber yew Is the table correct now?

[somber yew]

Let's see

Yes, it's correct :)

[terse ravine]

@somber yew If I want to find the slope of (-9,10) and (10,-8) I use m=y2-y1/x2-x1 which equals -18? This seems wrong.

[somber yew]

Let's see. By the way, there's nothing wrong with the slope being -18(it can be any real number for that matter)

$$m=\frac{-8-10}{10-(-9)}=-\frac{18}{19}$$

[obsidian monolith]

TedNotKaczynski:

[terse ravine]

Ok I re calculated and got -18/19

[somber yew]

Cool :)

Hmm? Well, the function is piecewise. For instance, -2<0, so you'll plug it into the expression 7x+7. Similarly, 20>0, so you're supposed to plug it in the expression 7x+14

A piecewise function like this one tells you which definition is to be used when dealing with a given element of the domain.

[terse ravine]

hmm

[somber yew]

You could try these values and tell me the answer

[terse ravine]

7(-2)+7 = -7 and 7(20)+14 = 154

[somber yew]

I meant the ones in your problem, but this is correct too 😅

[terse ravine]

Ok, working on it.

f(-1)=7(-1)+7 = 0

f(0)=7(0)+14=14

f(2)=7(2)+14=28

@somber yew Is this the answer?

[lime bolt]

yea

[terse ravine]

Consider the function P(x) = (x+8)^2(x+1)

The y-intercept is the point y = (x+8)^2(x+1)

To find the y intercepts, set x=0 and solve for y

y = (0+8)^2(0+1) = 65?

The x-intercept is/are the points: 0=(x+8)^2(x+1)

(x+8)^2=0 or (x+1)=0?

x=-8^2 or x= -1 ?

@somber yew Is this correct?

or x= -64 or x= -1 ?

[lime bolt]

lol is 64*1=65

also is (a+b)^2=a^2 +b^2?

these are 2 blunders which u made

[terse ravine]

Oh I start off with (a+b)^2=a^2 +b^2

[lime bolt]

what?

[terse ravine]

y = (0+8)^2(0+1) = 64

The y-intercept is the point: 64

[lime bolt]

yea there also is an issue with the x-intercept

[terse ravine]

To find the x intercepts I set x = 0 and solve for x

0=(x+8)^2(x+1)

Then (a+b)^2=a^2 +b^2?

[lime bolt]

lol no it isnt

im saying that was the issue which u made

so that will be zero when either (x+8) or (x+1) =0

[terse ravine]

what happened to the square?

it just disappears?

[lime bolt]

because a^2=0 if and only if a=0

as 0 doesnt have an inverse

[terse ravine]

Ok so, x-intertcept is/are the point(s): (x+8)(x+1)=0

(x+8)=0 or (x+1)=0

[lime bolt]

yea

[terse ravine]

x= -8, x= -1

correct?

[lime bolt]

yes

[terse ravine]

or is it 0, -8, -1?

or just -8, -1

[lime bolt]

if u plug in x=0 that doesnt make y=0

[terse ravine]

yes

[bleak lance]

Why is the domain of f(x)=sqrt(4x-28 [7,infinity)?

[terse ravine]

so 0, -8, -1?

[bleak lance]

I know that 7 would make the sqrt 0

But why including 7?

How do you know when to include a number and when not to?

[lime bolt]

so 0, -8, -1?
i just told u why it cant be x=0

u need sqrt(n) to be nonegative @bleak lance

and n>=7 is nonegative

[terse ravine]

Ok, I think this is correct?

No this is all wrong...

[somber yew]

All but the last one are correct

[terse ravine]

I didn't get the right y-intercept and x-intercepts

[lime bolt]

yea z^2 is always nonegative

[terse ravine]

hmm

[somber yew]

[lime bolt]

so if u multiply it by negative it is negative

[somber yew]

Wait

[lime bolt]

lol that isnt a point

u need to include both x and y coordinate in a point

[somber yew]

Yepp, state it like (0,64)

[terse ravine]

y-intercept=(0,64)

x-intecept=(-8,0),(-1,0)

x>infinity, y>infinity

[bleak lance]

So why is the answer to f(x)=5x/sqrt(x-7 (7,infinity)?

7 is a nonnegative number.

[terse ravine]

x>-infinity,y>-infinity

[bleak lance]

So is 0.

[lime bolt]

no, f(7) is

[bleak lance]

How?

[lime bolt]

which [7,infinty) is the domain

[bleak lance]

That's incorrect.

[lime bolt]

have u tried plugging it in

[bleak lance]

The answer to that was actually (7,infinity).

[lime bolt]

oh wait this is a different question from before

i thought this was the same one from earlier

[bleak lance]

Ik why

Because think about it

[lime bolt]

i also know why

[bleak lance]

This rational function can't evaluate to undefined.

[lime bolt]

obviously u cant dvide by 0

in the future, try and be clearer by adding a closing bracket to each starting one

this is specifically what im talking about

ok

[terse ravine]

When the question asks for points I give them point(s).

When the question asks for intercepts I give them intercept(s).

[lime bolt]

what?

[terse ravine]

The wording caught me off guard.

[lime bolt]

not sure what your trying to say, but yea a point in the cartesion plane has both and x and y ordinate which u need to state

[terse ravine]

@somber yew Is this correct?

[somber yew]

Let me see

You're supposed to give equations, not numbers

[terse ravine]

hmm

[somber yew]

The second one is correct because coincidentally the equation of line over that interval is y=2

[terse ravine]

How do I solve top and bottom?

f(x)=mx+b?

[somber yew]

Yes

Exactly

[earnest coral]

question, can the absolute min/max can be a local min/max as well? Or can they only be different??

[terse ravine]

I'm not sure if this is correct.

[tardy ridge]

desmos.com

[viscid thistle]

hmm

[terse ravine]

2x+7 looks right

-1x+2 looks right

[viscid thistle]

Im going through introductory precalc atm

and I cant really figure out this question

does anyone know how to do it?

[blissful kayak]

Well

[viscid thistle]

heres a similar question's explanation for reference

[blissful kayak]

First you need to find the center of the two circles

[viscid thistle]

alrighty, does that mean i just throw it into standard form?

[blissful kayak]

Yuh

And then you find the slope with their centers

[viscid thistle]

y1-y2/x2-x1 ?

[blissful kayak]

Ye

Back to the basics

[dawn coral]

Hello, I am trying to learn integration (antiderivatives) by myself and I am struggling with a particular point. Could someone help me understand? Thank you ^^

[viscid thistle]

@blissful kayak different question similar topic

apparently this is incorrect

not sure what im doing wrong

[blissful kayak]

Well

Try graphing it

I can't see it

Oh I see

It's the radius

[viscid thistle]

ah should it be 7^2?

[blissful kayak]

9^2

[viscid thistle]

where'd ya get 9?

[blissful kayak]

It's shifted 9 units parallel to the y-axis

That's by definition the x-coordinate

(Sorry if I'm acting like a know-it-all lol)

[viscid thistle]

thanks!

[blissful kayak]

Yay!

[viscid thistle]

[blissful kayak]

Of course!

[viscid thistle]

yep it was 81

phew

[tame wedge]

yo if I proved that an infinite sum doesnt converge, is it true that the sum diverges instead?

[bitter basin]

yo

[stuck lark]

@tame wedge it's a binary thing, either converges or it doesn't ie diverges

[blissful ridge]

Do you know how to write h(g(x))

[bitter basin]

i got it

now

yeah i plugged it in

is this right

[blissful ridge]

Looks alright to me

[bitter basin]

ok

i dont know this

[blissful ridge]

Do you know how floor function works?

[bitter basin]

is it tihs

[blissful ridge]

I mean do you know the idea behind it

[bitter basin]

we didnt learn that yet

[blissful ridge]

Then how are you expected to solve it?

[bitter basin]

idk

we never went over it

lol

maybe the graphs

[blissful ridge]

Go over floor function first, then try this question

[bitter basin]

ok

i just dont understand the graphs

[blissful ridge]

Well, how could you if you don't know the floor function

Yes, what does it do?

[harsh smelt]

up or down

floor(x) gives greatest integers <= x

@bitter basin why you think it is first

[uncut mulch]

the floor function isn't the same as the round function

floor(1) is 1

[terse ravine]

@uncut mulch here to ans a question?

[viscid thistle]

Hello, is this channel free?

[blazing raven]

Pascal’s triangle rocks. Did you know you can get a cool triangle if you subtract instead of add?

1
-1 1
1 -2 1
-1 3 -3 1
1 -4 6 -4 1

Happy to help Serena. What is up?

[viscid thistle]

I'm having trouble with this question

The whole thing is confusing to me

[blazing raven]

Cool. So it says the illumination at any point is the sum of the illumination of the two lights.

Without worrying about the rest of the problem how would you write that out?

We’re going to pick apart the sentences one by one.

[viscid thistle]

I'm not sure

I tried doing it but I couldn't get it

Hello?

[viscid thistle]

Hey, I've gotten an answer to part (a). I've formed an expression

Is this correct? If it is, can someone please help me with parts (b) and (c)?

[whole thicket]

@viscid thistle from what level?

[viscid thistle]

What do you mean? My grade?

@whole thicket

[whole thicket]

@viscid thistle yes

[viscid thistle]

Grade 12 calculus

@whole thicket

[odd abyss]

can the cube root of -24 be simplified

cube root of 8 * cube root of 3?

im going with that

[blissful ridge]

Don't forget about the negative sign

[odd abyss]

so -2 times the cube root of 3

?

[blissful ridge]

Looks alright to me

[odd abyss]

ok so if I had cube root of -24 * x^12 * y^23

I would simplify to -2 x^4 y^7 * cube root of 3 y^2

[viscid thistle]

Yep

[shadow plaza]

@viscid thistle do you have a question?

[viscid thistle]

@shadow plaza yeah I do

How do I figure out parts b and c to this question?

[shadow plaza]

Okay so for b part you just gotta find the point of maxima of the function you obtained in part a

And for c, you need to find point of minima

You do know how to do this, right?

[viscid thistle]

I forgot

It's been a very long time since I've learned that

[shadow plaza]

Okay so you differentiate the function with respect to x, and put it equal to 0. In this case, you'll obtain at least two values of x.

Furthermore, find the second derivative of the function and have the obtained values of x satisfy it. For which ever value of x, if you get f"(x)<0, then x is the point of maxima and if f"(x)>0, then minima

[viscid thistle]

So from the two x values, one will be the maximalist and the other will be the minima?

Also, why would I obtain two x values?

[shadow plaza]

Okay so since the question is directly asking the distance of maximum and minimum illumination, i am assuming that such distances must exist. Therefore you must obtain at least two values of x.

So from the two x values, one will be the maximalist and the other will be the minima?
@viscid thistle yepp

[viscid thistle]

Okay thank you!

And I just take the derivative from my equation in part a?

Also, I hope I'm not taking up your time. But would you mind helping me with part d as well?

[shadow plaza]

Ok so first of all i need to know what answers you got for b and c

[viscid thistle]

I can't take the derivative of this function

[shadow plaza]

Ok so let me get this straight. You don't know how to differentiate and yet you're trying to solve this question?

[cyan glen]

Why is b) incorrect?

[viscid thistle]

take 0<a<1

[cyan glen]

Ohhhhhh

I was only thinking about integers

Thanks

[frosty spire]

Is this true: Let {a_i} be set of positive integers such that log a_i satisfy no non zero linear relations with integer coefficients. Then log a_i satisfy no non zero polynomial relations with integer coefficients.

[willow bear]

i... think you might be in the wrong channel? but also what even is your question

[frosty spire]

Is the statement true? My friend is teaching pre calculus class. This question came up in his class.

[willow bear]

oh did it

ok wait let me try to digest this

so youre saying that for any choice of integer coefficients $c_i$ not all zero we have $\sum_{i=1}^n c_i \log(a_i) \neq 0$

[obsidian monolith]

Ann:

[willow bear]

and you're asking whether or not that implies $$P(\log(a_1), \log(a_2), \dots, \log(a_n)) \neq 0$$ for every nonzero integer polynomial $P$?

[obsidian monolith]

Ann:

[frosty spire]

Yes. Exactly.

[willow bear]

hm

say, what context did this even come up in?

[frosty spire]

My friend's students complained that there were no hard questions about logs. So I wondered if that was really true. Thus this question.

[willow bear]

this looks like a case of over-generalization to me bc there's like... nothing to grab onto

[frosty spire]

Well, can you find a counter example?

[willow bear]

can't think of one off the top of my head

[frosty spire]

Actually, I suspect that it is true. But I have no idea how to prove something like that. Never studied transcendence.

[cyan glen]

This is the first time I’ve ever heard about precal students complaining that the questions were too easy

[frosty spire]

I searched Wikipedia. Turns out the exact question I asked is in Wikipedia. Sorry for wasting your time.

@cyan glen My friend has lots of smart students in his pre calculus class. He wants to challenge them so he asked me if there are any hard questions about logs. I guess the answer is that it's really really really hard to find any hard algebraic questions about logs.

[cyan glen]

In my precal class no one dared to ask for more challenging questions lol. There was one super smart kid, but he wouldn’t speak to anyone so he never asked for more questions anyway

[frosty spire]

My friend teaches at a school that is 97% minority. He says the students are better than anywhere else he has taught. Where do you teach?

[cyan glen]

Oh I don’t teach, I just finished highschool last year

[frosty spire]

I see. You mentioned your precal class--thought it was YOUR class. My misunderstanding.

[cyan glen]

Oh sorry I meant my class as in the one I took

[winged meteor]

Hey yall hows it going. I am trying to complete my summer homework for entering into Pre Calc, but there is this quadratic inequality problem that I forgot how to do.

I need to find 2x^4-19x^3+57x^2-64x+20<0 in the real number system.

I know that I need to find the turning points, maximums and minimums, and x intercepts but I for got how to do those. Can anyone help me or point me in the right direction?

[willow bear]

this... is a quartic inequality, not quadratic

also wait

are you asked to graph y = 2x^4-19x^3+57x^2-64x+20 or are you asked ONLY to solve 2x^4-19x^3+57x^2-64x+20<0

[winged meteor]

only to solve

[willow bear]

ok then what's this talk about turning points and maxima and minima

where's THAT coming from

[winged meteor]

I think from what I remember I use those values as boundaries for testing.

[willow bear]

what values.

[winged meteor]

the intercepts and like the turning points

[willow bear]

you don't give two shits about the turning points

the x-intercepts yes those are important

[winged meteor]

ok

[willow bear]

i.e. the roots of your polynomial

THAT is something you care about

a lot

[winged meteor]

oh ok

so I would just factor 2x^4-19x^3+57x^2-64x+20=0 to find the roots

[willow bear]

yes

[viscid thistle]

There’s something funny about “just factoring” that

[winged meteor]

Yeh

this is quite hard

[willow bear]

throw RRT at it and it becomes piss easy

[winged meteor]

what does rrt mean

oh rational root theorem

[cyan glen]

So it wants you to find when the function is below the x axis

So just imagine generally what the function looks like on a graph and then use that and the x-intercepts to find the intervals of x where the function is below the axis

So the highest degree term is positive, so the arms of the function will be pointed up

[viscid thistle]

Can anyone help me with this question

[blissful ridge]

Do you the formula for sum of an AP

[terse ravine]

I'm having trouble finding the correct first ans.

mx+b=-2/5+(-7/2)?

[tardy ridge]

which part do you have trouble?

what part do you think is not right

[terse ravine]

I typed it in desmos and it doesn't look right

[tardy ridge]

can you take a screenshot and send it here

[terse ravine]

m=-0.4 what does b=?

[tardy ridge]

well b is not right

where did you get -7/2

[terse ravine]

The line crosses at -3.5 so I just turned -3.5 into a fraction which is wrong ik.

[tardy ridge]

actually the slope isn't either

the coordinates of the ponits are -6,0.5 and -1,3

you should be able to get an equation from that

[mighty onyx]

The positive angle between 0 and $2\pi$ that is coterminal with the angle $\frac{48\pi}7$ is

[obsidian monolith]

Albot1288:

[mighty onyx]

I don't know how to use a Unit Circle

oh sorry i'll let yall finish

[patent beacon]

@mighty onyx
"Coterminal" means "the same angle"
Note that π and 3π look the same on the unit circle, we call these "coterminal".

Any two angles that have a difference of 2π are coterminal

[terse ravine]

@patent beacon How to do I calculate mx+b with -2.5 down and 5 across?

[patent beacon]

@terse ravine
You're trying to calculate the slope with a rise of -2.5
And a run of 5

Slope = rise/run

[terse ravine]

=-0.5?

[patent beacon]

Yup! That's the slope of that section

[terse ravine]

How do I find b?

it looks like it crosses the y at -3.5

[patent beacon]

So pick some point on your line. That can be x and y. You already know m. So, you can calculate b in y = mx + b

But yes it will cross at -3.5, you can catch that with your slope

I'll pick (-1, -3) as a point on my line. Then:
y = mx + b
-3 = (-0.5)(-1) + b
-3 = 0.5 + b
-3.5 = b

[terse ravine]

Ok so, -0.5=m and -3.5 = b

So the answer is -0.5x+(-3.5)?

That looks correct to me.

[viscid thistle]

@terse ravine seems fine to me

[terse ravine]

Now I need to figure out how to shift x^2 to the right 1 unit.

[viscid thistle]

This'll help you

If you understand the difference between for example f(x)+c and f(x+c)

Can someone please tell me how to find the maxima and minima of this function?

@viscid thistle not the appropiate channel

Don't multipost too

[somber yew]

The question's already being discussed in #calculus

[viscid thistle]

Oh sorry

@terse ravine did the pic help you fully?

[viscid thistle]

Good job

[mighty onyx]

how do you convert degrees to radians

I have to covert 7pi degrees to radians and idk how to even start it

couldn't you just put f(x-1)?

[terse ravine]

@viscid thistle How do I type the ans in correctly?

[viscid thistle]

@terse ravine i mean, f(x-1) works but i think you need to actually put the formula of g

Like g(x)=(x-1)²

[terse ravine]

g(x)= is already there

[mighty onyx]

then just type (x-1)^2

[terse ravine]

ok it accepted that.

[mighty onyx]

gg

now how would I transfer 7pi degrees into radians?

would I need to turn 7pi into a number and then convert it with a formula?

[rigid sun]

$radians= degrees* \frac{\pi}{180}$

[obsidian monolith]

maleb1964:

[viscid thistle]

You just need to multiply by $\frac{π{\text{ rad}}}{180^{\circ}}$ as said for the units to cancel

Can someone in this channel please help me? I don't think anyone is able to in the other channel

[mighty onyx]

thx

[obsidian monolith]

Al𝟛dium:

[mighty onyx]

(7pi^2)/180 is the answer

[viscid thistle]

Yeah

Okay

How do I find the maxim and minima of this function? This is the factored solution

[tardy ridge]

set f'(x)=0

[viscid thistle]

Not here.

Why not? This is grade 12 calculus

[rigid sun]

e

[viscid thistle]

Because this channel is precalculus and not calculus?

Oh okay

[viscid thistle]

Like

Why you using desmos

[terse ravine]

I don't know tbh.

f(x)-d down 4 units and f(x+c) left 5 units correct?

[viscid thistle]

@terse ravine yes

So literally f(x+c)-d

[mortal vector]

Okay so I'm going through these exercises and the third one confuses me a lot

I'm meant to find a function g by transforming a given function f

But how can I find g if f isn't defined?

[terse ravine]

I think I made some errors sec.

[terse ravine]

@viscid thistle 4\sqrt{x+5}-4

let me try that ans

Got it.

[viscid thistle]

Yes

Good job

[viscid thistle]

@mortal vector what do you mean if f isn't defined?

if you want to write it with more function notation, you can put f(x) = 2^x for 1)

then f(x-3) = 2^(x-3) would be your function shifted to the right 3 units

carry on with the next 2 steps, then define g to be whatever the expression is after those 2 steps

[terse ravine]

Oh, the line crosses at -5

What did I do wrong hmm

@somber yew Did I make a mistake here?

m=-4/4=-1, b= -5

-1x+(-5) isn't correct either.

[bleak lance]

Ik that this maybe a bit too much to ask from someone here. But may someone please help me study for an exam?

My exam is in 4 days.

[uncut mulch]

the slope of the rightmost piece is positive

[terse ravine]

because its going left to right correct?

So, m=4/4 = 1 and b= -5

1x+(-5) is answer correct @uncut mulch ?

[uncut mulch]

that would be acceptable
but just simplify it to x-5

so many unnecessary 1s, parentheses and signs

[terse ravine]

ok

Now I need to find the solution to the equation sinx−1 =0 on 0 ≤ x < 2π *(I'll work on this tmr).

[tardy ridge]

so sinx=1

that would be 2pik+pi/2 or k E Integers

and you can use that with the original restricition

[terse ravine]

@uncut mulch Is this correct?

[uncut mulch]

yes

[terse ravine]

Goodnight 🙂

[sinful pecan]

im asked for an x-intercept coordinate in a question but there is none

is this possible?

[rigid sun]

let me see

@sinful pecan

[sinful pecan]

also is this even possible to graph on this chart?

[rigid sun]

your numbers are scuffed

$y=-x^2+2x+8$

[obsidian monolith]

maleb1964:

[willow bear]

this thing does have x intercepts, what are you talking about?

[rigid sun]

just check your numbers

make sure you're including your signs when you're adding them

and then it will make more sense

like 5 for example

-5^2+2*5+8 /=/ 43

[willow bear]

bruh

did you really just write /=/ for "not equal"

[rigid sun]

langauge of the gods

anyway yeah that's just one of the values that's wrong

just do the whole thing over again

you didn't even get that far anyway

[uncut mulch]

triple check your order of operations

$-x^2 \not\equiv (-x)^2$

[obsidian monolith]

ramonov:

[rigid sun]

see thats so much worse to type

efficiency 100

[sinful pecan]

are all of them wrong

or only some of them

[uncut mulch]

redo all of them

[sinful pecan]

alright cool

ty for catching that

[sinful pecan]

so the first 3 are

-7

0

5

right?

[uncut mulch]

yes

[sinful pecan]

x | y
-3 | -7
-2 | 0
-1 | 5
0 | 8
1 | 9
2 | 12
3 | 5
4 | 0
5 | -7

is this correct?

[willow bear]

is this meant to be a table for y = -x^2 + 2x + 8?

[sinful pecan]

yes

[willow bear]

how did you get y=12 for x=2

all the other entries are correct but this one sticks out like a sore thumb

[sinful pecan]

-2^2 + 2(2) + 8

-4 + 4 + 8

= 12

[somber yew]

-4+4+8=8

[sinful pecan]

damn

thanks

how did i miss that

[rare zephyr]

Before the invention of calculators, how did people find out the angles of numbers not obtained from special angles

Like how did they find out the answer for sin^-1 0.73621938721 or etc

[fading token]

Well have you ever wondered how a calculator itself does it?

[uncut mulch]

magic

[rare zephyr]

Well have you ever wondered how a calculator itself does it?
@fading token i mean someone has to know the solutions to program the solutions into the calculator right

[somber yew]

I don't think solutions are programmed into calculators. It's the algorithm which is programmed.

The algorithms/methods to estimate the values of trigonometric functions have been around well before the advent of calculators. I guess in historic times, geometric measurements were an important source, and later they were more formalised(I don't know when power series, etc. came into being, but I'm almost certain they arrived before calculators came into being).

[rigid sun]

no

there are just cats inside that are really smart

[somber yew]

catculators

[blissful ridge]

[chilly ruin]

i think this is precalculus?
i just need some help with my calculator

this is meant to help

me to solve this

yes that is variable x on the fraction

but i have no idea which numbers to put where to actually get a solution

[ornate wolf]

i've got the euqation

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

However I'm stuck at what to do next

how do i compare th parts?

[blissful ridge]

Group all the real and imaginery parts together respectively

[ornate wolf]

Group all the real and imaginery parts together respectively
@blissful ridge should i expand it first?>

[blissful ridge]

Yeah

[ornate wolf]

so i've got 9x^3 + 9xy^2 + 2x +9iyx^2 + 9iy^3 - iy

[blissful ridge]

What did you do?

Why am I seeing cubic terms

[ornate wolf]

i multiplied 9(x+iy) by x^2+y^2

to remove the denom in the second part

[blissful ridge]

Okay

Why did you remove the denom?

[ornate wolf]

to make everything on the same "level"?

sorry im so confused

[blissful ridge]

You cannot just remove denominator

[ornate wolf]

when you said yes when i asked if i expand, which one were you referring to?

[blissful ridge]

Just put the denominator back

You'll be fine

[ornate wolf]

okay

9x+9iy + (2x-2iy) / x^2 +y^2

like that ?

You'll be fine
@blissful ridge what do i do ?

[blissful ridge]

What you did above was fine,
Just put the denominator back

[ornate wolf]

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

yeah i dont see it

i have no clue what im suppoosed to do, tried to expand etc

[uncut mulch]

set Im of that to 0

[ornate wolf]

set Im of that to 0
@uncut mulch so all iy=0?

[uncut mulch]

no

[ornate wolf]

how to set to 0? I thought if Im is 0, it means iy = 0?

but question says Im is not 0

[uncut mulch]

Im(z) isn't 0

9z + 2/z is real
Im(9z + 2/z ) = 0

[ornate wolf]

yeah that's the part where im very confused

we substitute x+iy into Z, right?

[uncut mulch]

expand it out, separate you real and imaginary terms

[ornate wolf]

we sub x+iy into Z right?

and x+iy the Im cannot be zero

[uncut mulch]

so far that's fine

[ornate wolf]

however 9z + 2/z is real? how can it since it's made up of the x+iy?

[uncut mulch]

why don't you think it can be real?

Im(z) = Im(x + iy) = y

it's y that can't be 0

lets take a very simple case

,w i + 1/i

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by ramonov#0921

[uncut mulch]

similar setup. resulting in something real

expand it out, separate you real and imaginary terms

[cyan glen]

Apparently the answer to #2 is d. Isn’t it c?

Cuz the parabola is y=(x+1)^2-2. And when you expand that you get c

[rigid sun]

where is number two?

[sturdy haven]

Hello and plz give tips for matrix algebra unit

[cyan glen]

@rigid sun it’s on a separate page, but am I right?

[viscid thistle]

could anyone help me with this question

[proven marten]

could anyone help me with this question
@viscid thistle What have you tried

[viscid thistle]

well i've tried setting up an equation with two variables where one is for the first amount invested into sharings, and the other one for the investment into the bonds

but Im not really sure how to set up for question a

[proven marten]

If Karim invested $x

How much would be in bonds?

[viscid thistle]

(x + (1,5x/100) +1000) *(1,025)^n

thats how I set it up for the bonds one

[proven marten]

I was going to say $(x+1000)

[viscid thistle]

but I think i have to account for the simple interest as well

[proven marten]

Thats why x is the savings?

[viscid thistle]

yeah

the first equation is the savings + $1000

[proven marten]

Yes

Whats the shares

[viscid thistle]

you would multiply that amount by the compound interest in a multiplier form

but im not sure how to set it up for a, because i dont really get the wording

[proven marten]

Honestly Im pretty sure IB hasnt asked any questions like these

but

You have to get the value of the shares, savings and bonds seperatly

at least thats how I did it in my notebook

then the loss would mean -0.01

times shares

[viscid thistle]

so would I use different variables?

or try use the same one?

[proven marten]

Use the same one

[viscid thistle]

how did you know it was IB?

[proven marten]

I mean Ive seen those questions in the oxford book

Also

[viscid thistle]

ohhh

[proven marten]

We share a mutual server called the IBO

[viscid thistle]

so the 1% loss is only regarding the compounded interest?

and the added $75 is for both the investments into both shares and bonds?

[proven marten]

yes

wait

well no, because when you are doing investment

you calculate the interest acquired

not the total

so keeping that in mind

all the investments together will add to $75

If that makes sense

[viscid thistle]

so it my answer would be savings + shares = total + 75?

oh i meant

savings + bonds + shares

[proven marten]

question a is only asking for what x is

[viscid thistle]

yep, alright thank you so much

[proven marten]

Yeah no worries

[daring yarrow]

@cyan glen i also think it's c, i just plugged the numbers into the vertex x-coordinate formula and with d it doesnt work out, with c it does

[plain root]

yo

does anyone know why this does this

this is adding the -4 and -4

-3*

and making it go down -7

but im trying to move it to the left 4 and down 3

nvm

[rigid sun]

lol

[rigid beacon]

@plain root put the -4 under the square root

[plain root]

yea i got it

[rigid beacon]

what you've written is -sqrt(-x) - 4 -3

aight

[plain root]

i just messed up typing it

i hate graphing tho

this is my night mare

[plain root]

could someone help me solve this

[tardy ridge]

does it want you to do g(x)=((-1/2)^(1/3)*(x+2))

[hearty bronze]

could someone help with these problems

[tardy ridge]

1. substitute.

2. substitute

(ax+b)/(cx+d)=a/b=a/c=a/d test them all

and remember the restrictions

(The restrictions are insuring it is a rational function and not just a function with a hole in it)

(if ad!=bc, it would mean that the bottom fraction is not a factor of the top fraction)

[terse ravine]

@tardy ridge How to find the solution to the equation sinx−1 =0 on 0 ≤ x < 2π ? I see that sin -1 is 3pi/2 on the unit circle.

[tardy ridge]

sinx=1 has a solution of x=pi/2 in those restrictions

[plain root]

@tardy ridge im not really sure whats its asking

function notation

[tardy ridge]

g(x)=f(((-1/2)^(1/3)*(x+2))) if that is what it's asking

you can check with your calculator to see if sinpi/2=1

[terse ravine]

sin pi/s = 0.02740

I'm probably not understanding the question correctly.

[tardy ridge]

you are in degree mode

change it to radians

what??

[terse ravine]

Ok in radian mode sin(pi)/2 = 0

I still don't understand what its asking me.

[tardy ridge]

sin(pi/2)

it asking what x satisfies sinx=1

[hearty bronze]

@tardy ridge i kept trying with problem 3 but im still stuck

do you have any tips to help me figure it out?

[uncut mulch]

it's asking you to solve for x (in that interval)

Ok so sin(1)-1=0 is what they asking for?
no

[terse ravine]

From 0 to 2pi is 360 degrees.

I see sin-1 is 3pi/2 and opposite of that is sin1 which is pi/2

[uncut mulch]

you aren't using and/or reading the functions properly

@hearty bronze find inverse of g first

[terse ravine]

sinx = 1 = pi/2

Is that the y?

[uncut mulch]

"1 = pi/2" makes no sense

[terse ravine]

Sin 1 in Q1 =pi/2

[uncut mulch]

you aren't using and/or reading the functions properly

[terse ravine]

Sin-1 = 3pi/2

[hearty bronze]

while trying to find the inverse, im getting stuck at x(cy-d)=ay+b

im trying to isolate y

[uncut mulch]

distribute first and regroup

[hearty bronze]

so then its xcy-dx=ay+b

[uncut mulch]

sin(pi/2) = 1
sin(3pi/2) = -1
sin-1 makes no sense

[terse ravine]

ok

[hearty bronze]

and wdym by regroup?

[uncut mulch]

group your terms with y

factor, divide and you'll have what you need

[terse ravine]

@uncut mulch What do I need to do to find the solution?

[hearty bronze]

so i got xcy-ay=b+dx

[uncut mulch]

unit circle and/or know your trig functions
you sorta seem to know a bit about using the circle and/or may be poorly representing it.

[hearty bronze]

so far i now have y=(b+xd)/(xc-a)

[uncut mulch]

and then determine what won't be in the domain of that (from looking at the denominator)

[hearty bronze]

can you explain what you mean by that?

[uncut mulch]

when won't it be defined?

what can't you divide by?

[hearty bronze]

the question tells me to find weather its a/b, a/c, a/d, or none

which is what im stuck on

[uncut mulch]

answer my questions

sin (pi/2)=1 = 0,1 (cos,sin)
just ** sin (pi/2)=1 **

[terse ravine]

@uncut mulch I added one to both sides and now the equation is sinx+1, thus x=pi/2

[uncut mulch]

which = pi/2

please stop using = signs inappropriately

what you're writing implies that sinx = 1 = pi/2
clearly write x = pi/2

ugh.

x = pi/2 is the solution to sin(x) = 1 for 0<=x<=2pi

@hearty bronze what can't you divide by?

[hearty bronze]

thats what im working on

but its just confusing for me

[uncut mulch]

don't overthink it

dividing by this numerical value won't get you anything meaningful

and some might consider to be illegal

looking at something a lot simpler, what would be the domain of 1/x?

[hearty bronze]

wouldnt it be negative infinity to 0 and 0 to infinity

[uncut mulch]

(-inf,0)U(0,inf)
and how did you reach that conclusion?

[hearty bronze]

i graphed it on my calculator and saw that the line on the x axis was getting smaller but never hit 0

[uncut mulch]

eeep.

division by 0?

[hearty bronze]

on the graph

[uncut mulch]

can you divide by 0?

[hearty bronze]

no

i cant

[uncut mulch]

hence 0 is excluded from the domain of 1/x

(-inf,0)U(0,inf) represents the set of reals excluding 0

and use a similar idea to determine what can't be in the domain of g^(-1)

[hearty bronze]

can you explain a bit more? im still stuck on finding what the domain cannot be

[uncut mulch]

do you know the definition of domain?

[hearty bronze]

yes

[uncut mulch]

in your words what is it?

[hearty bronze]

the domain is what all the x-values are

[uncut mulch]

poor description

[hearty bronze]

then what would be a better description of it?

[uncut mulch]

domain of a function would be the set of inputs where the function is defined
would be one way

looking at, 1/x it should be clear that it's not defined when the denominator is 0 and won't be in the domain

[terse ravine]

A function f is the rule that assigns to each element x in a set D exactly one element, called f(x), in a set E. We usually consider functions for which the sets D and E are sets of real numbers. The set D is called the Domain of a function. the number f(x) is the value of f at x and is read f of x. The range of f is the set of all possible values of f(x) as x varies throughout the domain.

[hearty bronze]

i get that

but im still stuck on how to find which of the answers wouldnt be the domain

[uncut mulch]

similarly determine the value of x when the denominator of (b+xd)/(xc-a) is 0
the identify what won't be part of the domain

[terse ravine]

The Domain is the set of all input values that have associated y-values.

[viscid thistle]

ah im also dealing with domain here

I was literally going to ask what this is

Im currently loosing braincells by the second because I dont know how to get that answer

[uncut mulch]

factorise denominator, determine when the function is undefined

[viscid thistle]

I dont really understand how to go from that fraction to a set of ordered pairs

dont understand what its even trying to get me to do or what the U means

[uncut mulch]

they're not ordered pairs

in context, its interval notation

[viscid thistle]

this is going over my head ngl

[uncut mulch]

what they have there represents the set of all real numbers excluding -2 and 2

U is symbol for union

[viscid thistle]

oh ok

how come its written as -∞,-2) and not (∞,-2)

[uncut mulch]

there's a distinction between -inf and (+)inf

[viscid thistle]

hmm

maybe i should just look up videos on this

cause the way this is wrriten is completely missing me

I cant interpret what this says

[terse ravine]

goodnight friends 🙂

[uncut mulch]

smaller value is also on the left

[viscid thistle]

gn varn

[uncut mulch]

look up something like interval set notation

[viscid thistle]

I see, so in the case of (2,∞) two is greater than ∞?

[uncut mulch]

no

[viscid thistle]

then why is 2 on the left

[uncut mulch]

in interval notation (2,inf) represents the set of values greater than 2

between 2 and inf (excluding 2)

[viscid thistle]

(2,inf) represents everything greater than two

then (-inf,2) represents everything lesser than 2?

[uncut mulch]

parentheses are needed

set of values less than 2
() indicate exclusion, brackets for inclusion, reverse brackets may be used for exlusion

[viscid thistle]

how would I represent everything lesser than 2

[uncut mulch]

** (-inf,2)**

[viscid thistle]

I see

U(-2,2)U

would mean

[uncut mulch]

as written the U there would be inappropriate

[viscid thistle]

(concerning the screenshot)

(-inf,x)U(-2,2)U(x,+inf)

where x can be -2 or 2

x is both -2 and 2

?

[uncut mulch]

poor wording

[viscid thistle]

Im drawing a parallel to numberlines with this

am I on the right mindset?

these inf signs just represent the arrows

for integer number lines right?

[uncut mulch]

number lines are continuous

[uncut mulch]

(-inf,2) would be represented by the first line

[hearty bronze]

@uncut mulch ive been trying to figure it out but i still cant seem to get it

[uncut mulch]

for what value of $x$ will the denominator of $\frac{b+xd}{xc-a}$ be 0?

[obsidian monolith]

ramonov:

[uncut mulch]

(resulting in division by 0, and should hence cannot be part of the domain)

don't overthink it

[hearty bronze]

the denominator would have to be 0 but a, b, c, and d have to be potitive

ad cannot equal bc

so that means that a b c d are all different numbers

[uncut mulch]

don't overthink it

[jaunty spruce]

can i get help with this?

[uncut mulch]

is it not clear to find when the denominator is 0, you would equate the denominator to 0?

[viscid thistle]

(used the letter U instead of the symbol U)

[hearty bronze]

so i find weather a/b or a/c or a/d makes b+xd/xc-a = 0

[uncut mulch]

your inverse: $\frac{b+xd}{\red{xc-a}}$ is undefined when the denominator: $\red{xc-a}$ is 0. \ To determine the value of $x$ when that happens (the value that cannot be part of the domain), all you need to do is solve $\red{xc-a}=0$ for $x$

[obsidian monolith]

ramonov:

[hearty bronze]

so then itd be a/c

[uncut mulch]

yes

[willow bear]

grumble grumble parentheses

[glad pasture]

could someone help me im not sure how you would prove this ?

[glad pasture]

<@&286206848099549185>

[harsh smelt]

@glad pasturehave you already proved original triangle ineq?

that is

[obsidian monolith]

Commander Vimes:

[glad pasture]

well its asking me to prove that equation

[harsh smelt]

it is asking to prove |a|-|b| <= |a-b|

but have you proved one above?

[glad pasture]

no

my teacher just told me to do that one

[harsh smelt]

prove first it

by cases

i mean if you want we may take it as granted

but proof is really simple

[glad pasture]

how do you prove an equation exactly ?

[harsh smelt]

consider cases

when a and b are both nonnegative it is obvious

when a and b are both negative it also reduces to that

since || destroys -

remains cases when a is nonnegative and b is negative

you will have to consider two cases a+b <= 0 and a+b >=0

[glad pasture]

so they way you prove it is by words and there is no mathematical way to prove it ?

[harsh smelt]

wdym

i just expressed in words mathematical thoughts

that reasonins is in words does not make it math

[glad pasture]

i just thought there was a specific way you had to prove it with math like how you could prove a answer is true by plugging it in the equation

[obsidian monolith]

Commander Vimes:

Commander Vimes:

[harsh smelt]

can you consider these cases by your own?

[glad pasture]

don't think so

[harsh smelt]

bot slow

[obsidian monolith]

Commander Vimes:

[glad pasture]

a is larger or equal to b ?

[harsh smelt]

yes

and b is negative

[obsidian monolith]

Commander Vimes: