precalculus

[lavish edge]

yes

its given in the fig

[vapid plaza]

Well (x,0) doesn’t always lie on the circle. (r,0) does

[lavish edge]

no noo

i know that cos0 is 1 and sin 0 is 0

[vapid plaza]

But in the special case where x = r,
(x,0) is (r,0) and therefore lies on the circle.

[lavish edge]

ok but

look look

in the figure

the terminal side

as in figure

theta is given as theta say 30

alright

so how can you have theta = 0 for x in the same figure

[vapid plaza]

Theta is 30 in the figure, but the figure represents one only possible point on the circle.

[lavish edge]

so if theta is given as 0

30

can we use the same variable theta for 0

[vapid plaza]

If we were to consider all the possible points on the circle, we would take theta as many different values, including 0.

The figure only shows what happens when theta = 30!!!
when theta = 0, we would get a different figure

[lavish edge]

NO i am not saying that

[full pagoda]

i have not the slightest clue what you're trying to ask

[vapid plaza]

Bruh

[lavish edge]

just look at the initial side x. If you see its coordinates youll get x,0 which is the same as xcos theta and xsin theta when theta is 0 but in the figure theta is already defined and is a positive angle

[vapid plaza]

Yeh but in the figure (x,0) is not on the circle we were talking about anyways!!!

[lavish edge]

so how cn we take theta as 0 for x and theta = some positive angle for another side

if theta is already mentioned to be some value can like 30 for one question can we take theta as some other value for the SAME question?

[vapid plaza]

What question?

Wdym question??

[lavish edge]

A question?

a sentence worded or expressed so as to elicit information.

[vapid plaza]

There was a question? You never mentioned there was a question.

[lavish edge]

look at the figure

[vapid plaza]

There is no “question” in the figure.

The question is because you’re confused

💀

[lavish edge]

yeah i am

thats why im asking

[vapid plaza]

Then what question are you talking abt

[lavish edge]

does x form a 0 degree angle

with the base

or not

[vapid plaza]

Yeh so what?

[lavish edge]

whatever im talking about is from the figure

[vapid plaza]

In the figure, “x form a 0 degree angle with the base”

But why is that a problem?

[lavish edge]

its not

if it does form 0 degrees with the base

does that mean theta is the theta in the figure or 0

[full pagoda]

theta is an arbitrary angle measurement

[vapid plaza]

The figure is trying to show that (x,y) lies on the circle.
So of course it would display theta as the corresponding theta for (x,y)

[lavish edge]

and for x,0 whats the corresponding theta

[vapid plaza]

0

[lavish edge]

so therer 2 thetas right

[hushed sphinx]

If theta=0 then there isn't really any triangle, just a single line from the origin out to (1,0).

There are infinitely many possible thetas.

[vapid plaza]

But only 1 case, and only 1 theta is the focus of the figure!!!

Besides, (x,0) doesn’t lie on the circle, (r,0) does

[lavish edge]

yes

but (r,0) is the same as rcos0 and rsin0

[vapid plaza]

Yes

So r lies on the circle

[lavish edge]

does that mean theta is 0

[vapid plaza]

Theta is not just “some value” for an entire circle

[hushed sphinx]

Yes, theta is whatever you give as input to the trig functions.

[lavish edge]

look what im trying to say is some point is given P(x,y) with some theta as the focus. If somehow theta is 30 we will get x as root 3/2 and y as 1/2 for a unit circle. Theta is defined to be 30. But if the side x lies on the circle for the SAME example, it will still have the points r,0 which is the same as rcos theta and r sin theta. Now if theta is defined as 30 degrees how come we still get r,0 which occurs when theta is 0

[hushed sphinx]

Theta is defined to be 30.
No it isn't.

[full pagoda]

you just said theta is two different angles

[vapid plaza]

There is no problem!!!!! If theta is 30, we get a point that lies on the circle, if theta is 0, we get another point that lies on the circle.

[hushed sphinx]

Just because you can choose to let theta be 30 doesn't mean theta must always be set to 30.

[lavish edge]

im not saying theta is 30 and 0

if theta is 30

we will still get the base as r,0

[full pagoda]

"the base"?

[lavish edge]

yes

initial side

[full pagoda]

if theta is 30, you're not looking at the coordinates of the initial side

[vapid plaza]

Hmm Vedanth what are your hobbies?

[lavish edge]

physics and maths studies

[vapid plaza]

Anything else?

I was trying to say do sth else then return to this “problem”.

[lavish edge]

No

i understand everything you guys are saying

but it doesnt answer my question cause i dont know how to explain my question in text

[vapid plaza]

:/

[lavish edge]

read this again

[hushed sphinx]

It still falsely claims, "theta is defined to be 30".

[lavish edge]

theta is defined to be 30

[hushed sphinx]

No it is not.

[lavish edge]

as per question

[full pagoda]

what question???

[lavish edge]

a sentence worded or expressed so as to elicit information.

[hushed sphinx]

The question you still haven't shown us?

[lavish edge]

theres no question

[full pagoda]

???

[hushed sphinx]

Please post a picture of the text that defines theta to be 30.

[lavish edge]

read the damn thing.

[vapid plaza]

“I have discovered a truly marvellous question which this text is too narrow to contain ”

[lavish edge]

i created a scendario where theta is 30

[hushed sphinx]

The damn thing claims that theta is defined to be 30, which is not true, It is false. it is not the case.

It is not even an EX-fact.

[lavish edge]

If somehow theta is 30 we will get x as root 3/2 and y as 1/2 for a unit circle. Theta is defined to be 30.

first i said if somehow theta is 30

[full pagoda]

bro

[lavish edge]

then i claimed theta is 30

[lavish edge]

dont bro m

me

[hushed sphinx]

No, theta is not defined to be 30, no matter how often you repeat that false claim.

[full pagoda]

[lavish edge]

ok

theta is defined to be 30 as per question

[full pagoda]

lol

[hushed sphinx]

You repeat your false claim again. That does not make it more true this time either.

[lavish edge]

its true

[hushed sphinx]

You lie.

[lavish edge]

truth is relative

[hushed sphinx]

It there was a definition of theta as 30, it would have been a simple matter for you to post a scan, photo, or screenshot of that definition.

[lavish edge]

did i ever say

theta us defined to be 30

is

[full pagoda]

yes

like 40 times

[lavish edge]

no lemme finish

[hushed sphinx]

You have not provided that. I confidently conclude that you have not in fact seen such a definition in whatever it is you're quoting from.

[lavish edge]

did i ever say theta is defined to be 30 for every scenario

[hushed sphinx]

You keep saying that theta is definied to be 30 in the thing you're talking about.

[lavish edge]

yes

[hushed sphinx]

From which you have still not shown the definition you keep claiming exists.

Come on. Pics or it didn't happen.

[vapid plaza]

You say that theta is defined to be 30 for this scenario, but then screams in terror when another point with a different theta also lies on the circle.

[lavish edge]

no

is theta 30 or 0

[hushed sphinx]

You keep claiming that you have a definition that says it is 30.

If it is 30, it is not 0.

[lavish edge]

no

just listen to me

if we have an example

where it says theta is 30 degrees

[vapid plaza]

(This is based on the figure btw
Figure just looks like 30, doesn’t even say so)

[hushed sphinx]

Please show the place where it says so.

[lavish edge]

thanks for the great help

[hushed sphinx]

The figure looks like a fair bit less than 30, but never mind. The author might have intended it to be 30 when they wrote the definition that Vedanth refuses to show.

[lavish edge]

its from a google image

i-

[vapid plaza]

Go do something else

[lavish edge]

can a question have two different thetas'

These are the most important rules. For more info see info.
‌
If you see anyone violating these, feel free to ping <@&268886789983436800>.
‌
GUIDELINES:
● Respect topics of the channels: make sure that you are asking in the right place.
● Avoid Directly Messaging people for help without their consent. In general it is bad etiquette to DM people you do not know.
● #help channels are for homework type questions/problems. Topical channels prioritise open-ended discussion about particular topics.
● Do not spam your questions in the wrong channel or across multiple channels. Be patient and wait for someone to answer.
● When asking a question, show what you tried to solve the problem. We are not here to do your homework for you, but we are here to help you learn how to do your homework. See ❓how-to-get-help for more info on how to ask questions.
● Respect that other people might be at a different stage in their education than you, what is obvious to you might not be obvious to them.
● Be rational and exercise critical thinking. If someone makes a mistake in their reasoning, politely point it out.
‌
SOCIAL CHANNEL GUIDELINES:
● Keep it civil in the chill channel. Treat it as a channel for friendly banter with your classmates or possibly your teacher/professor/advisor.
● If there is a debate, be polite and civil. Try to back up your arguments with objective facts, and avoid fallacies. When arguing you should seek to find out the truth, instead of trying to be right. There is nothing wrong with changing your opinion based on presented information and admitting that you were wrong.
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mniip — 06/05/2021
(Revision as of June 4, 2021)

[slender river]

oops

[lavish edge]

<@&268886789983436800>

[slender river]

you accidentally pinged us

or

twice

what

[stuck lark]

???

[lavish edge]

yes

Respect that other people might be at a different stage in their education than you, what is obvious to you might not be obvious to them.

[full pagoda]

if you are looking at the set of coordinates that lie on the unit circle when theta = 30 degrees, you are not simultaneously looking at a different set of coordinates that relates to a different theta

[lavish edge]

this rule has been broken

[full pagoda]

we've told you time and time again that what you're saying makes zero sense yet you keep saying it over and over again and expecting things to change

definition of insanity

[lavish edge]

Be rational and exercise critical thinking. If someone makes a mistake in their reasoning, politely point it out.

can a question have two thetas

[full pagoda]

entirely depends on the context of the question...

[short sorrel]

i dont see anyone being rude or untoward

youre asking a vague question thats hard to interpret with imprecise wording and unclear intent/clarification

people are asking you for an explanation

you are being dodgy

almost makes me think this is for a test or something

[hushed sphinx]

(He's moved to #geometry-and-trigonometry now, btw).

[short sorrel]

i was responding to the mod pings

[dapper brook]

How come I’m learning trig while I’m in pre-cal

[hushed sphinx]

Why wouldn't you? It's pretty common to teach trigonometry before calculus.

[viscid thistle]

ik this is a lot to ask but could someone plz offer a step by step by step explanation for how each step works. I’m very lost here; the only step I understand is get is when we distribute the square of the absolute value of the difference of Ba_i and Cb_i

the book I’m using rn is skimming complex numbers and assumes the reader already knows about em ;/

[low slate]

Bit of a multi step problem, but i dont really understand how to find the answer and if someone can help me out it would be appreciated

[full pagoda]

have you drawn pictures yet?

[fast monolith]

I don't know what the first step is, if possible, can someone explain how to solve this?

[full pagoda]

try getting everything in terms of sin and cos

see what you can do from there

[low slate]

i have but im confused on where the angle goes

[full pagoda]

the ladder makes an angle with itself and the ground

[low slate]

correct

[full pagoda]

so where the ladder touches the ground

[low slate]

so the angle on the top would be 15

right?

[full pagoda]

i mean yeah but that's not needed

well i guess

you could

but

lol

[low slate]

and the hypotunese is 24

[full pagoda]

right

[low slate]

but then how do u go from there

cuz i only have angles and a side

[full pagoda]

can i see the diagram you've drawn?

[low slate]

yea sure gimme 1 sec

[full pagoda]

you can set up a trig ratio with a missing variable using a right triangle, an angle, and a side

[full pagoda]

looks good

so using that, you need to find both other sides

you can use two different trig functions to set up an equation for either side

[low slate]

ok

how to i figure out the trig functions tho

thats the main part im confused on

[full pagoda]

well what ratio does sin represent?

[lofty yarrow]

⠄⠄⠄⢰⣧⣼⣯⠄⣸⣠⣶⣶⣦⣾⠄⠄⠄⠄⡀⠄⢀⣿⣿⠄⠄⠄⢸⡇⠄⠄
⠄⠄⠄⣾⣿⠿⠿⠶⠿⢿⣿⣿⣿⣿⣦⣤⣄⢀⡅⢠⣾⣛⡉⠄⠄⠄⠸⢀⣿⠄
⠄⠄⢀⡋⣡⣴⣶⣶⡀⠄⠄⠙⢿⣿⣿⣿⣿⣿⣴⣿⣿⣿⢃⣤⣄⣀⣥⣿⣿⠄
⠄⠄⢸⣇⠻⣿⣿⣿⣧⣀⢀⣠⡌⢻⣿⣿⣿⣿⣿⣿⣿⣿⣿⠿⠿⠿⣿⣿⣿⠄
⠄⢀⢸⣿⣷⣤⣤⣤⣬⣙⣛⢿⣿⣿⣿⣿⣿⣿⡿⣿⣿⡍⠄⠄⢀⣤⣄⠉⠋⣰
⠄⣼⣖⣿⣿⣿⣿⣿⣿⣿⣿⣿⢿⣿⣿⣿⣿⣿⢇⣿⣿⡷⠶⠶⢿⣿⣿⠇⢀⣤
⠘⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣽⣿⣿⣿⡇⣿⣿⣿⣿⣿⣿⣷⣶⣥⣴⣿⡗
⢀⠈⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡟⠄
⢸⣿⣦⣌⣛⣻⣿⣿⣧⠙⠛⠛⡭⠅⠒⠦⠭⣭⡻⣿⣿⣿⣿⣿⣿⣿⣿⡿⠃⠄
⠘⣿⣿⣿⣿⣿⣿⣿⣿⡆⠄⠄⠄⠄⠄⠄⠄⠄⠹⠈⢋⣽⣿⣿⣿⣿⣵⣾⠃⠄
⠄⠘⣿⣿⣿⣿⣿⣿⣿⣿⠄⣴⣿⣶⣄⠄⣴⣶⠄⢀⣾⣿⣿⣿⣿⣿⣿⠃⠄⠄
⠄⠄⠈⠻⣿⣿⣿⣿⣿⣿⡄⢻⣿⣿⣿⠄⣿⣿⡀⣾⣿⣿⣿⣿⣛⠛⠁⠄⠄⠄
⠄⠄⠄⠄⠈⠛⢿⣿⣿⣿⠁⠞⢿⣿⣿⡄⢿⣿⡇⣸⣿⣿⠿⠛⠁⠄⠄⠄⠄⠄
⠄⠄⠄⠄⠄⠄⠄⠉⠻⣿⣿⣾⣦⡙⠻⣷⣾⣿⠃⠿⠋⠁⠄⠄⠄⠄⠄⢀⣠⣴
⣿⣿⣿⣶⣶⣮⣥⣒⠲⢮⣝⡿⣿⣿⡆⣿⡿⠃⠄⠄⠄⠄⠄⠄⠄⣠⣴⣿⣿⣿

niceeee

[viscid thistle]

Here you go

Let me know if you want something explained in more detail

[viscid thistle]

https://math.stackexchange.com/questions/446148/cauchy-schwarz-for-complex-numbers
also you can prove a stronger version of this with cauchy-schwarz

[low slate]

opposite/hypotonuse

do u have to use sin ?

like how do u know which one to use

[full pagoda]

depends on which sides you have relative to your angle

in your triangle, you have an angle and the hypotenuse

and you're looking for both sides, so to find the first side, you can use either sin or cos

since both utilize the hypotenuse that you know, the angle that you know, and an unknown side

so one equation, one variable

solvable

[sick steppe]

🤨

[vivid void]

@viscid thistle could you like

not post this in 3 channels

[viscid thistle]

for calculus questions dm

[mellow tapir]

hey guys my friend has this question and i don't know how to solve it either so can someone help us

[hushed sphinx]

Hmm, there are five "I don't know yet"s and 96 is 3·2^5.

That indicates there's probably a progression where each successive declaration of ignorance doubles the numbers.

The guess that would fit best with that hypothesis would be ||oops, I see it's a still open contest||.

[mellow tapir]

nah the contest is over but that was a cool observation

[tribal lodge]

Can someone help explain something

[mellow tapir]

yeah i can help

[tribal lodge]

Can you vc???

[mellow tapir]

nah

[tribal lodge]

🥲

[hushed sphinx]

If you state what it is you need explained, perhaps someone who knows will reply.

[tribal lodge]

I’m slow that’s why I need to be in a vc

[viscid thistle]

i am not sure how to find the real zeroes

[hushed sphinx]

Numerically.

[viscid thistle]

can you clarify?

[hushed sphinx]

Basically systematic trial and error until you know a short enough interval where the function changes sign that it approximates the root well enough for your purposes.

[cerulean flower]

hey guys, when theres a slant asymptote of a rational function, how do you know when and where it gets crossed by the function?

[jolly raven]

i dont think functions can cross slant asymptotes but i might be wrong

if they were to be passed, it would probably be in a central part of the function bounded by vertical asymptotes where the function is not continuous all the way to both infinities

[cerulean flower]

i know that slant asymptotes can be crossed, and you are right

but i wanted to know how to find the exact point

of where the function crosses the slant asymp

[hushed sphinx]

An example of a crossing would be f(x) = x - x/(1+x²)

I don't think there's a particular procedure for that, though. Subtract the equation for the asymptote, and look for roots of the numerator of the resulting rational function, is the best general thing you can do.

[cerulean flower]

i know that for finding the slant asymptote, you have to divide the numerator and the denominator, and whatever remainder you get determines if and where it crosses the slant asymp

wait a second

i answered my own question

thanks for clearing some things up guys

so if anyone wants to know ig, if the remainder you get when dividing the num and denom of a rational function is a constant, then the slant asymptote doesnt get crossed. if it has a variable (ex: 2x-1), then solve for the variable (x=1/2) and plug the answer into the og rational function to get the y value of the coordinate. (not so sure ab the last part but its what i remember)

[grave tartan]

I have to do this without calc, how would i go about doing this

[willow bear]

you can construct a piecewise-linear function that does this

[grave tartan]

see that came to mind

except we havent covered piece wise

[willow bear]

[grave tartan]

[willow bear]

no calc and no piecewise functions...

i mean like

you COULD do it all exclusively with absolute values i guess

[grave tartan]

im like stumped

[willow bear]

at least tell me you have access to that

[grave tartan]

wouldn't it also be piecewise tho

[willow bear]

the piecewiseness would be hidden

[grave tartan]

no we do I just don't know how to combine like

[willow bear]

you said "we haven't covered piecewise functions"

[grave tartan]

absolute values with different vertexes

[grave tartan]

yeah

[viscid thistle]

i dont think the question is asking for a function using conditionals, more likely a polynomial

[willow bear]

a polynomial would require calculus to properly construct

and we are told we aren't allowed calculus

[grave tartan]

^^

I tried using like different bell curve shaped graphs but they don't work nice

[viscid thistle]

or maybe not, (1,1) (2,2) (3,3) are "good values", probably we can do something

[willow bear]

you'd need the polynomial to have at least degree 4

[grave tartan]

which means 5 unknown variables

[grave tartan]

and these give 3 equations if plug into f(x)=ax^4+...+e

so its not nice

unless I just give myself like (0,0) and (5, 5)

but its a cancerous system of equations

[willow bear]

this ban on piecewise functions is super annoying ngl

[grave tartan]

IKKK

i am just going to assume the worksheet was meant for after and she either assigned it too early or changed the flow of the curriculum this year

[viscid thistle]

wait, how many degrees a function with 3 relative maximums has?

[grave tartan]

but either way wanted piece wise

[viscid thistle]

?

[grave tartan]

yeah 4

but I think I am allowed to have more

just has to have a minimum degree of 4

[viscid thistle]

this way of solving ignores the fact that the maximums are on the line y = x

[viscid thistle]

the author of the problem wouldnt give the three maximums on y =x for no reason

[grave tartan]

true but at this point I don't want the ingenius solution

[grave tartan]

nasty nasty

[willow bear]

just add a +b term at the end or something to shift it vertically

if necessary

[vivid void]

dear god

[viscid thistle]

maybe rotating a sine or cossine function?

idk if this is possible

[vivid void]

you know what

a sine function would actually work

you just need a period of 1

[viscid thistle]

but the function needs to go upwards

[vivid void]

i’m aware

[viscid thistle]

sine functions dont go above 1

[vivid void]

that’s why you give it a variable coefficient

[willow bear]

doing that will screw up the location of the maximum points

and pinning those down will take calculus

so that is a no go

[grave tartan]

tried that too

[viscid thistle]

something like this

[viscid thistle]

the bad part is that this isnt a function

[grave tartan]

fortunately $f\left(x\right)=-2.5\left|x\right|-1.5\left|x-1\right|+3\left|x-1.5\right|-4\left|x-2\right|+5\left|x-2.5\right|-6\left|x-3\right|+10.5$ is

[obsidian monolith]

vrang

[grave tartan]

was also thinking about it

would x^2 - x*ceil(x-1) work

do you need points on either side to be considered a max

[peak moss]

Sorry to interrupt, but I'm stuck with this question. I've tried to factor the (sin2x•cos2x)² • cos 2x, but the answer that I got is too different compared to the scheme. I'm a bit stuck, and in need of help 🤲🏻

[opaque idol]

sub sin^2(2x) = 1- cos^2(2x)?

then let u = 2x or something?

or u = sin 2x

idk

try it and see if it works

[peak moss]

I’ll try in a bit

Thenk youuue

[vivid void]

precalculus

[viscid thistle]

wdym quantum that looks like precalc to me

antiderivatives are like algebra or somthin

[vivid void]

ok let me just post a nonhomogeneous ode here then

[vapid plaza]

no homo

[viscid thistle]

the minimum degree of a polynomial that satisfies the conditions turns out to be 7

[willow bear]

oh yeah you're right

you need two minimum points between the maxima

so that's 5 turning points

...why degree 7 tho

[viscid thistle]

so i know this wasn't what the question was asking but you can find the minimum polynomial that satisfies this with some calc

https://www.desmos.com/calculator/neunoxvobc

[willow bear]

it is impossible to do it with 6?

[viscid thistle]

yeah it is impossible with 6

i worked it out

,rotate

[viscid thistle]

,rotate

[viscid thistle]

original problem seems impossible without using either piecewise functions or calculus

[viscid thistle]

ok it is possible to do it with 6, just looked back through my work and saw that i found the degree of the polynomial incorrectly at the end

everything else still checks out though

[viscid thistle]

minimum degree should have 6 boxed in here instead of 7

[peak creek]

Hello

Why does "Out of these, only x=1 is within the interval (0,3)"?

The mean value theoreme does not apply to the limit of the interval?

[willow bear]

the mean value theorem asserts the existence of a point in the open interval (a,b)

[gloomy gulch]

I need to create a function with all these requirements. Cannot to it. Need help please

[vivid void]

it would help to write it as this

$\frac{f(5)-f(1)}{5-1} =5,\ \frac{f(10)-f(1)}{10-1} =-20 \ \frac{f(9)-f(2)}{9-2} =0,\ f’(2) = -10,\ f’(3) = 0$

[obsidian monolith]

quantum

[vivid void]

notice what the third one implies

i would assume that this is a uh

maybe third degree polynomial

not sure honestly

actually no

assume it’s in the form of ax^2+bx+c

since you would have three equations in three variables and two equations in two variables then

@gloomy gulch

this is just a system of equations in disguise

seems kinda fun honestly

[gloomy gulch]

@vivid void Hey. I just came sorry

[vivid void]

it’s fine

i didn’t expect you to respond immediately

[gloomy gulch]

Lol. I’m trying to solve this that’s why I came in quickly

[vivid void]

yeah considering that you have the two derivatives this shouldn’t take too long

[gloomy gulch]

Well, it is to me 😂

The thing is, I was able to make a function that meats the first requirement

But then, when I try to meet the second, my first requirement is lost

I’m doing trial and error

[vivid void]

you shouldn’t do that

[vivid void]

.

[gloomy gulch]

That actually made things easier now that I know how my function would look

[gloomy gulch]

I don’t understand what’s here

Nvm

I have two functions. One that matches the first requirement and one that matches the second. Maybe I should combine them?

[vivid void]

note that this guess of it being a quadratic could be wrong

if it is wrong we can try something else so it’s fine

[gloomy gulch]

Okay

Hmmm this is weird

[vivid void]

$f(5)-f(1) = 20 \ f(10)-f(1) = -180 \ f(9)-f(2) = 0 \ f’(2) = -10 \ f’(3) = 0$

[obsidian monolith]

quantum

[gloomy gulch]

Nvm nvm. Those numbers are supposed to be my numerators?

For example, the first requirement. My denominator is 4 so we need 20 in the numerator so when we divide, we get 5 which is the rate of change

Idk I’m confused

[vivid void]

yeah i just multiplied both sides by the denominator

[gloomy gulch]

Ok that helps

[gloomy gulch]

Well if its in the form of this, I’ve noticed that the c value doesn’t have an effect on my rate of change

Well if I spend 20 hours straight of trial and error I would be able to get it lol but that’s not the efficient way

Hopefully we don’t need to include cos, and sin in the function

[vivid void]

@gloomy gulch i’ll just write it out and see if my guess is wrong

[gloomy gulch]

Alright

[vivid void]

$25a+5b+c-(a+b+c) = 20 \ 100a+10b+c-(a+b+c) = -180 \ 81a+9b+c-(4a+2b+c) = 0 \ 4a+b = -10 \ 6a+b = 0$

[obsidian monolith]

quantum

[vivid void]

$-4a-b = 10 \ 6a+b = 0$

[obsidian monolith]

quantum

[vivid void]

@gloomy gulch this might actually be a quartic lol

or something

[gloomy gulch]

Lol please no

[vivid void]

maybe even a quintic

remember that it’s really just a big system

$24a+4b = 20 \ 99a+9b = -180 \ 77a+7b = 0 \ 4a+b = -10 \ 6a+b = 0$

[obsidian monolith]

quantum

[vivid void]

this is what i get after simplifying

i actually got a = 5, b = -30 for this

but i’m pretty sure that’s wrong

yeah that’s definitely wrong

this would be a uh

hmm

quintic yeah

i am not writing all that out on latex lol

a quintic has 5 coefficients excluding the constant at the end

and notice that none of these have that constant term

[gloomy gulch]

Hmmmm

[vivid void]

i’m pretty sure this if in the form of $ax^5+bx^4+cx^3+dx^2+ex+f$

[obsidian monolith]

quantum

[vivid void]

so many letters i needed to recite the alphabet for the last one

not even a joke

[gloomy gulch]

😂

I think your going too far tho

[vivid void]

no i’m seriously not

this isn’t a joke

[tight compass]

just use a_i instead.

[vivid void]

are you triggered by me using e as a coefficient

i was about to say something about that lol

i don’t like it either

[gloomy gulch]

No no, it’s weird, but no worries

[tight compass]

Imagine you need to 26 + 1 degree polynomial.

[vivid void]

i would just not do the assignment at that point

[gloomy gulch]

I wish I can’t 😂

[vivid void]

augmented matrix time 😎

[gloomy gulch]

If I decide to surrender I’ll just create a function that meets one requirement. That’s what I’m capable of 😅

[vivid void]

well you have all you need now

just gotta power through i guess

[gloomy gulch]

Yup. I’ll try my best

Thanks anyways

[vivid void]

no problem

oh and if you couldn’t tell

[gloomy gulch]

I hope no one falls in this type of question 😂

[vivid void]

do the last three equations first

well

at least the last one and the third one

since they both equal 0

so you don’t have to worry about any constants there

[gloomy gulch]

Yes yes

Well the question is out of 4 marks. If I manage to get at least 3 (which I think I can). There’s at least 2.5/4

[vivid void]

although i’m not sure how you would get the constant term at the very end

hmm

[mild root]

hi, could some EXPLAIN to me how I could solve this?

#3

[full dock]

divide all 3 vectors into component form

then

find the resultant vector's magnitude and direction by taking the sum of the y and x components

[viscid thistle]

f(t)=5t
f(t)=-20t
f(t)=0
f(t)=-10t
f(t)=0

[vivid void]

cringe

[low slate]

what does 6 1/2 represent?

is that the radius or the circumfrence?

[hushed sphinx]

My first guess would be diameter.

[velvet jungle]

It's typically diameter e.g.
https://www.lowes.com/pd/SKIL-12-Volt-5-1-2-in-Brushless-Cordless-Circular-Saw-with-Brake-and-Punch-Steel-Shoe-Charger-Included/5001375947

[low slate]

seems simple but i feel stuck

[vivid void]

same honestly

wait

i assume that the information given means that 108 = 2x+2y

so just solve for y @low slate

[low slate]

ok ill try

[graceful fiber]

y=(x^3-9x)/(x^2+2x-15) graph

i know that it looks like a slanted hyperbola but i cant figure out how to get the second asymptote

[gloomy gulch]

@vivid void hey, remember me from yesterday? The guy who asked for help in finding average rate of change. Well guess what? It turns out the question only asks to sketch a graph that meets those requirements and not build a function with numbers. I think that’s suppose to make it easier 😂

[vivid void]

@gloomy gulch bruh

did you solve it though? lol

[gloomy gulch]

I’m working on it lol

After yesterday I just quit

Good practice tho

[vivid void]

yeah if it wasn’t such a big system it would be fun

[gloomy gulch]

Graph will still have big numbers in it

It’ll be a sinusoidal behaviour too

Just a quick question, what do you recommend should my scale go by?

Because there will be some big numbers, I don’t want to have a graph that looks gigantic @vivid void

[vivid void]

repost the question and i’ll help in a minute

[gloomy gulch]

Just a quick question, what do you recommend should my scale go by? Because there will be some big numbers like -180, I don’t want to keep extending my graph for ever

[vivid void]

@gloomy gulch no i meant the question you sent yesterday

[gloomy gulch]

My bad

I’m thinking a scale of 5?

[vivid void]

yeah that’s a good idea

because the height will have to be at least like 90 because of the second condition

[gloomy gulch]

Yes

I’ve calculated and the max hight would be 200

😬

That’s a lot. My teacher would hate my graph lol

[graceful fiber]

y=(x^3-9x)/(x^2+2x-15) graph finding the asymptotes

[dim lodge]

hey can anyone help me with precalculus

[willow bear]

http://dontasktoask.com

[dim lodge]

its delta math work 😭

ive been at this for like 4 hours now

but theres a penalty so if i get it wrong it makes me go back

[willow bear]

well if you don't show us the problem(s) then we can't help you now can we

[dim lodge]

alr ur making me feel like a dumbass now

lmfao

but im trynna figure out how to send pics cuz im on pc

[uncut mulch]

snapshot tool

win+shift+s

[willow bear]

or Prt Sc

should be a key on your keyboard

[willow bear]

okay

have you made any progress on this or are you stuck not knowing how to begin?

[dim lodge]

i know how to begin but its just i make some sort of mistake somewhere

so heres what i do

instad of cosx i just substitute x

like 4x^2 - 3 = 0

then sqrt 3 / 2

[willow bear]

then sqrt 3 / 2

then what?

okay first off

maybe you shouldn't use the same letter for two different things

let's call cos(x) something like t instead to avoid confusion

so you have the equation 4t^2 - 3 = 0

[dim lodge]

ok ok

[willow bear]

what did you do afterward?

[dim lodge]

afterwards i do cos^-1 (sqrt3/2)

[willow bear]

no

what did you do immediately after writing down the equation 4t^2 - 3 = 0?

[dim lodge]

oh

solved for t

[willow bear]

show your work in solving for t.

i have a suspicion you're making a mistake somewhere in that part, and i want to ensure it doesn't happen.

[dim lodge]

uhm

add 3

to 0

then divide by 4

so 3/4

then square root

[willow bear]

i said show not tell

[dim lodge]

ohhh

[willow bear]

do you have it written out on paper somewhere

it needs to be written out on paper in order to diagnose potential errors

[viscid thistle]

would this be 8x?

[vivid void]

no

do you know how to find an inverse function

[tight compass]

Forgot about using the technique for finding the inverse of f, think about what f is doing to x. It scales it by 5 then added to it by 3. So what do you think the inverse should do?

[viscid thistle]

figured it out sorta

[viscid thistle]

(x-3)/5

[tight compass]

Yep.

[viscid thistle]

how about this?

[tight compass]

You can split it to 1 + 2/x

[viscid thistle]

ok

[tight compass]

So apply what we did above to this.

[viscid thistle]

ohh

okay 1 minute

-2 then times x?

so -x?

nope

[tight compass]

Think about what f is doing to x. Clearly we first take x to 1/x.

[viscid thistle]

im going to be honest. im really confused rn lol

maybe im tired but i do not understand

[tight compass]

We take x to 1/x then multiplies by 2 then adds 1

So x-1 then (x-1)/2 then 2/(x-1), for x in the domain of inverse f. Do you see it?

Generally to find the inverse of f change f(x) to y to make it easier, then switch the variables y and x then solve for y.

[viscid thistle]

ok 1 second

need to read it all lol

where did you get the f from originally?

[tight compass]

What do you mean?

[viscid thistle]

in a message before you said f-1 first or something similar i think

you edited it though so it doesnt matter

one minute

thanks for your help so far but i think ill try tomorrow.

its currently 3 am lol and i think im too tired to understand

[tight compass]

Go to sleep and rest on it.

Wake up, read your notes and watch videos explaining it.

[viscid thistle]

@tight compass a friend explained it too me and made me understand it better

it was 2/x-3

[tight compass]

Nope it 2/(x-1).

[viscid thistle]

oh yeah sorry mistype

figured out the other few questions after too lol

[dawn sun]

Hello is there someone here who could help me please?

[stuck lark]

[limpid dune]

Hello

[desert badger]

Anyone help

[full pagoda]

we can't see any of the information given

[full pagoda]

,rotate -90

[full pagoda]

where'd you get 56 degrees from?

[desert badger]

Idk

I try to solve it

[full pagoda]

how exactly did you do that

i say that because it's close

i'd like to know where you went wrong

[desert badger]

I did

180-29-95

[full pagoda]

where'd 95 come from?

[desert badger]

B

[full pagoda]

b is not 95

3 degrees more than vertical

what angle would represent vertical?

[desert badger]

Not sure

[full pagoda]

what is angle C in this?

[desert badger]

c

[full pagoda]

what is the value of angle C

[desert badger]

90

[full pagoda]

right

and that's vertical

straight line from the ground up

so what's 3 degrees more than vertical?

[desert badger]

93

[full pagoda]

👍

[desert badger]

K

How I find height

[full pagoda]

using your new equation

solve for h

[desert badger]

@full pagoda Ik how to plug it in

[full pagoda]

ok

[desert badger]

@full pagoda I can’t get the answer tho

Idk how to use the calculator

[full pagoda]

that might be a problem

you just uhh

type

[desert badger]

I can give u the number can u try it

[full pagoda]

you can use the bot here

,w sin(50 degrees)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by a disappointing son#0540

[desert badger]

44/sin 58 (sin 29

That

[full pagoda]

your syntax is not the best

[desert badger]

Wym

[full pagoda]

this is not typed properly

[desert badger]

Can u help me plug that in

[full pagoda]

are you meaning $\frac{44}{\sin{58}}\sin{29}$

[obsidian monolith]

a disappointing son

[desert badger]

Yes

[full pagoda]

,w (44/sin(58 degrees))sin(29 degrees)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by a disappointing son#0540

[desert badger]

Ty

Got it right

@full pagoda r u good with stats

[snow ravine]

ayo why is the derivative of sin = cos

[uncut mulch]

you can try differentiating by first principles

[hushed sphinx]

How is that PREcalculus?

[uncut mulch]

its a fine line. difference quotients are pre-calc and then involving limits in the last steps takes it up to calc

[obsidian skiff]

What’s precalculus

[sick steppe]

foundations for calculus

[viscid thistle]

any help

[willow bear]

@viscid thistle do you still need help with this?

[viscid thistle]

yes

[willow bear]

how familiar are you with basic trigonometry?

[viscid thistle]

umm not really fimilar teacher literally did one example

[willow bear]

so your teacher did not even explain to you the words "sine", "cosine" and "tangent", and did not teach you the SOH-CAH-TOA mnemonic to remember which is which?

[viscid thistle]

oh yeah he tought the sine, cos, tan but did not really touch on soh, cah and toa

[willow bear]

...

either you have a shitty teacher or you just didn't pay enough attention in that one particular class.

do the words "hypotenuse", "adjacent" and "opposite" sound familiar at all to you or are they completely alien

[viscid thistle]

i know what they are

[willow bear]

okay but you don't know the definitions sin(θ) = opp/hyp, cos(θ) = adj/hyp and tan(θ) = opp/adj? is that what's happening?

[viscid thistle]

Yes but is it the same thing for this

[willow bear]

what do you mean by "the same thing"

[viscid thistle]

Do I use them to find x

Bc I think the degree is throwing me off

[willow bear]

you are expected to use one of them to find the value of x, yes.

you are also supposed to know what the value of sin(30°) is.

[viscid thistle]

Is it 0.5

[full dock]

1/2 yes

[viscid thistle]

So is it relative to we’re the degree is

[willow bear]

???

[viscid thistle]

nvm i got it

[viscid thistle]

heyy

how do i type "sin x cos x = sin 2x" in calculator

[willow bear]

what calculator?

[viscid thistle]

casio fx-991ex

[willow bear]

read the manual for the casio fx-991ex

[viscid thistle]

and also for the number 6

[viscid thistle]

in the menu/setup?

[willow bear]

wait, you're not asked to solve these equations using your calculator, are you?

[viscid thistle]

no, i mean he said we can use our calculators so

either we solve these equations manually or by using calculator

[timber obsidian]

tan (x)/tan(x) = 1

true for all values where tan (x) != 0

[timber obsidian]

2sin(x) cos(x) = sin(2x)

[timber obsidian]

hence it’s not true for any value of x

except x = 0

or values where sin(2x) = 0

[timber obsidian]

x sin (30) = 7
x/2 = 7
x = 14

[viscid thistle]

what number is this

[timber obsidian]

y = (x+2) / x
yx = x+2
x(y-1) = 2
x = 2/(y-1)
therefore inverse of f(x) = 2/(x-1)

[timber obsidian]

infinitely many values of tan unless given bounds / a domain

[viscid thistle]

wait, where did 2sin(x) come from?

[viscid thistle]

i mean what number is this equation, is it for number 6?

[viscid thistle]

i got it already but ty

[vivid void]

why did you reply to a 3 day old message?

[timber obsidian]

let cos x = y ( for ease of typing )
4y^2 = 3
y^2 = 3/4
cos ( x) = root(3)/2
x = pi/ 6
x = 11 pi/6

[timber obsidian]

nws matey.

[timber obsidian]

idk i was just looking at them now

[timber obsidian]

6

[timber obsidian]

it’s an identity which you can prove by using two right angles or e^ix conversions. i’ll show you

[viscid thistle]

oohh but where are cos and sin

[timber obsidian]

sin(x)/ cos (x) = tan(x)

[viscid thistle]

do i need to change the given from "sin x cos x" to "2sin x cos x?"

or is it the rule or something to make it like that?

im sorry for asking too much :( i kinda dont get it

[timber obsidian]

sin(x)cos(x) = (sin(2x))/2

[vivid void]

yes because that’s what makes them equal

[timber obsidian]

you can look at the derivations if you like

there is one using triangles and another using euler’s fomula

[viscid thistle]

what about the euler's formula

[viscid thistle]

is this the euler's formula?

[viscid thistle]

can you explain it

[toxic cradle]

any resources for pre calc?

[viscid thistle]

aops

[tight compass]

Probably khan academy, Stewart’s book

[broken pelican]

correct me if im wrong but
for example, the given is 30°

if the question is "find the SMALLEST POSITIVE co-terminal angles", then the answer is 390°

if the question is "find the SMALLEST NEGATIVE co-terminal angles", then the answer is 330°

is it correct? idrk my answer on the second part

[queen ermine]

the wording on the second question is annoying but I’m assuming it’s asking for the closest co-terminal to 30° and if it is you’re answer is correct.

[broken pelican]

thank you

and uhhh sorry bout that

[timber obsidian]

it would take a while plus my ipad is not with me, just try expanding the brackets out and you’ll see it yourself. or you could simply accept the identities. you can still try expand them. do you know e^ix = cos(x) + isin(x) right? that’s how you get the first line. substitute these values in.

[timber obsidian]

agreeably one of the best resources

[shy helm]

Anyone know how they would go about solving this? thanks

[sick steppe]

you have a quadratic in x

so you want the quadratic to have no roots in R.

[hushed sphinx]

You also want the leading coefficient to be negative, such that the parabola points downwards.

[potent raft]

question about #23 on this review, do I cancel the LNs leaving me with x = (y - 4)^5 or am I meant to do something else

[hushed sphinx]

Do you mean y = (x-4)^5?

[potent raft]

It's finding the inverse so I have to swap the x and y first

[hushed sphinx]

Hmm, it's strange to ask about the "given function" when the formula then doesn't define a function but just a relation between two variables ...

[viscid thistle]

aightt i'll try to learn this. thanks for helping!!

[timber obsidian]

** y should also be always be negative,

as in the thing should be <0

[tender quarry]

can anyone help me with parabolas in a call?

[viscid thistle]

if possible could we hop in a call so i can get a full understanding

[uncut mulch]

convert <LKM to radians and apply the formula for arclength

[viscid thistle]

so 20/360= 1/18

[vivid void]

yes, what do you think you’d do next

[viscid thistle]

umm Substitute

[uncut mulch]

write down the formula you intend to use

[viscid thistle]

is there away to do it in the server

[uncut mulch]

take a snapshot of a formula online / draw in paint / type it out in text properly with sufficient and appropriate notation

[uncut mulch]

bad notation at the end

it looks like you crossed out r at the end
and your final line isn't an equation telling you exactly what r is

sure, 36/3 = 12. but so what?

[viscid thistle]

is the answer 12

[uncut mulch]

simplify your equation properly and it'll tell you

you'll either reach something that tells your
r = 12 or r = something else

[viscid thistle]

r=12

[uncut mulch]

yes

[radiant zodiac]

how

idk how to mess with that tan3x

imma isolate it

[mellow tapir]

hey guys this is on my friend's homework im not really sure what would be the answer

[limpid nest]

the last 2 terms simplify to 1, by the identity

so tan3x = 1

sry, -1

3x = y

so y = 45

so, therefore x = 15

15 degrees

[compact tulip]

do u know expression of 1+tan^2x=sec^2x

[compact tulip]

correct vsg

[timber obsidian]

-1x + 6

[timber obsidian]

sec2 = tan2 +1
and tan(x+ 2x) = (tan addition formula)

[tepid cloak]

any desmos experts here knows how to shade this shape?

ill share the link https://www.desmos.com/calculator/nnsphpejqd

[marsh cipher]

@tepid cloak use inequalities

[mellow tapir]

Ty

[fickle robin]

this is the best I could get with the shading, hope it helps https://www.desmos.com/calculator/g4oo7dg1y1

[keen eagle]

Can someone explain me this question

Like I dont understand what we are suposed to do

[tight compass]

I remember you, I told you before to go to desmos and play around with different transformations of functions to understand how to solve it. Also read your notes/book on it to understand it.

[hushed sphinx]

There are three operations you're being asked to do in sequence. Can you do the first one?

[keen eagle]

I did I dont understand it

[keen eagle]

No

[keen eagle]

Like I went to demos and played around with it and still dont understand it

like I did with 2^x

then 2^x + 2

[tight compass]

Read this. https://www.mathsisfun.com/sets/function-transformations.html

[keen eagle]

Thank you

[proper dome]

Anyone else doing precal 1 and 2 in one semester?

[tepid cloak]

help how do I shade the unshaded area (the inside of the rectangle) in this figure?

https://www.desmos.com/calculator/sqjv5hrbus

[uncut mulch]

consider first splitting the region into two parts

and isolate y for each circular arc

[tepid cloak]

how

[willow bear]

wait are you required to do it as an integral

can't you just subtract the areas of both quarter circles from that of the rectangle

@uncut mulch @tepid cloak

[uncut mulch]

they wanted to shade it in desmos

not calculate the area from what i interpreted

like

[willow bear]

oh shade

my bad

right so this region is defined by four inequalities
$$\begin{cases} \left(x-16\right)^{2}+\left(y-8\right)^{2} \geq 1 \ \left(x-19\right)^{2}+\left(y-12\right)^{2} \geq 9 \ 16 \leq x \leq 19 \ 8 \leq y \leq 12 \end{cases}$$

[obsidian monolith]

Ann

[willow bear]

or six inequalities i guess

min( (x-16)^2 + (y-8)^2 - 1, (x-19)^2 + (y-12)^2 - 9, x-16, 19-x, y-8, 12-y) ≥ 0

@tepid cloak sorry for the double ping but

[tepid cloak]

it showed me this

[willow bear]

?

did you input the ineq as i typed it

[tepid cloak]

uhh

im not really good at desmos

[uncut mulch]

type
min( (x-16)^2 + (y-8)^2 - 1, (x-19)^2 + (y-12)^2 - 9, x-16, 19-x, y-8, 12-y) ≥ 0

[tepid cloak]

oh

it worked omg thanks!

[tepid cloak]

what if I tried to flip it

like this?

https://www.desmos.com/calculator/qrsf3zkium

like this

[willow bear]

swap out the equation of the bigger circle

[tepid cloak]

oh I got it

thanks

[timber yarrow]

Hey guys does anyone know how to do no. 3?

[astral prism]

did you distribute the negative?

[timber yarrow]

I used sin(A-B)

[tepid cloak]

does anyone know how to shade the red line on this figure?

https://www.desmos.com/calculator/j0xpxwhovv

[trim burrow]

this is kinda unorthodox and long but it does the job (obv you can fix the colors)

https://www.desmos.com/calculator/cdznd06j5r

[marsh cipher]

Convert arctan to arcsin

[glass relic]

hi, can someone help me to resolve this but using properties of natural log

If $2000 is invested at an interest rate
of 3.5% per year, compounded continuously, find the value
of the investment after the given number of years.
(a) 2 years
(b) 4 years
(c) 12 years

[willow bear]

@glass relic what do you mean

did you receive explicit instruction to do this "using properties of natural log" under threat of punishment?

[hushed sphinx]

Does "compounded continuously" mean that after one year $100 has grown to $100·e^0.035, or that we select a rate of continuous exponential growth that will make $100 reach exactly $103.50 after a year has passed?

[willow bear]

i would think the former

[hushed sphinx]

Hmm, that agrees with what Google seems to be telling me.

[viscid thistle]

Are you doing grade 10 FMP?

[hushed sphinx]

I'm not -- just musing after reading Andrusita's question.

[teal plinth]

is anyone around

need a little help

[viscid thistle]

same

what help you need

[teal plinth]

how do you find b. I haven't done precalc in a long time

[viscid thistle]

bro this is the same lesson

i’m struggling with

that’s quadratic functions right?

[teal plinth]

yep

actually nvm i found the answer to b

[viscid thistle]

do you understand that lesson?

that’s like the lesson i’m struggling with and i have a test tomorrow

[teal plinth]

not well enough to help you

[viscid thistle]

fair enough

[teal plinth]

https://www.youtube.com/channel/UCQv3dpUXUWvDFQarHrS5P9A

checkout this channel. he explains stuff really well.

just type the type of problem and search if he has (mostly likely does)

thats how I get thru my math classes

[viscid thistle]

oh ik that guy

aight man thanks

[teal plinth]

np

[vagrant lintel]

Hi, anyone can help me finding this?

[vivid void]

it’s a linear function

so just find the slope

and then find what the added constant is

[vagrant lintel]

It's not lineal.

[vivid void]

yeah it is

we have two points

wait oh gosh

,w plot (3.3,8.9),(0.38,40)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by quantum#6320

[vivid void]

[rustic forge]

maybe a family of circles is needed? Isn't there more context behind the question?? otherwise there's a hell lot of graphs passing through these two points out there

[willow bear]

the person left

[uncut mulch]

supposedly there's an image that comes with the question

[stuck lark]

@strong ermine dont shitpost here

[strong ermine]

Ok

Precalc meme

[hollow vapor]

PLUS CCCC

[lean cove]

Can anyone help me really fast? I don't remember what the thing on the right of the 8 is for

also it's not a quiz for my class just got an example off khan academy

[uncut mulch]

to indicate which sides you're approaching from

[lean cove]

So - would come from the left?

[uncut mulch]

yes

[vivid void]

$x\to 8^-$ means you’re approaching 8 from values that are less than 8

[obsidian monolith]

quantum

[vivid void]

so the left

[lean cove]

Oh ty that makes sense

[deep pewter]

HELP

[viscid thistle]

<@&286206848099549185>

[heady mango]

Do you know the equation for the area of sector?

[viscid thistle]

to late already got it

[steel fern]

I have an old calclulator that my dad used in college. A casio 7000-GA is that something i could now days ?

if not waht should i buy calculator wise for precalc

[uncut mulch]

pretty much any scientific calc is good enough for precalc

[steel fern]

this calc is a graphing calcualtor

[uncut mulch]

a graphing calc can do what a scientific calc can and more

[tepid cloak]

how do u read this again

[willow bear]

this looks like desmos notation for a restriction

[tepid cloak]

yeah

[runic ravine]

can someone help? the topic is pascal triangle

<@&286206848099549185>

[tight compass]

You can either write out the pascal triangle up to 5 and 9 to figure it out or just compute it explicitly using combination formula.

[fiery creek]

$\frac{5!}{2!(5-2)!}$

[obsidian monolith]

mniip's foe

[fiery creek]

do you know how to do factorials?

[crystal stream]

isnt the formula dA/dr = dA/dt * dr/dt

[fiery saffron]

dA/dt=dA/dr•dr/dt is the correct chain rule

[hoary rose]

I dont get wat I did wrong

the steps look correct yet for some reason I got the wrong answer

3h^2 - 22h + 7
3h^2 - 21h - h + 7
(3h^2 - 21h) + (-h+7)
3h(h - 7) - 1(h - 7)
(3h - 7)(h - 7)(h - 7)

I dont know what order the factored out numbers for 22h should be in

so I dont know how to factor it correctly

[heady mango]

you did everything right until after the last line. how would you factor this:

$3h \cdot a -1 \cdot a$

[obsidian monolith]

Taltost

[cold kestrel]

lmao

[viscid thistle]

huh

hmmmmm secx?

No

3sec(x/pi) or smth

[viscid thistle]

pix?

Yeah?

[amber badger]

Yoo can someone help me understand this

Im lost

[hushed sphinx]

Do you have a place to look up trigonometric product formulas?

(at least you presumably have Google).

[amber badger]

i dont think so

this was given as a quiz on paper in class and your allowed one cheat sheet

[hushed sphinx]

If this kind of questions are on the quiz, the trigonometric product/addition identities should definitely be on your cheat sheet.

[amber badger]

ok

are they these

ohhh i get it

you just have to find which ones correspond

my bad

[hushed sphinx]

Yes, that's the ones.

There should be ones that take sums to products too, for the second part.

[short pilot]

can someone explain me how i find the nht term

[sick steppe]

what have you tried/thought/noticed?

[short pilot]

i tried to find the d like others excersice

but my book do something and doesnt show a formula or something

[sick steppe]

Geometric Sequence/Progression is the key term

[warm nebula]

@frank acorn expand it using foil

[tight compass]

Expand it using distributive property which says a(b+c) = ab + ac.

Apply it to this (a-b)^2 or just (a+b)^2 then you have a nice quicker way to always expand expressions in that form.

[meager cairn]

2 Cosec(πx) + 2

[viscid thistle]

Expand this using (a-b)² = a² + b² - ab . There you go :-
(2x-3)²
=> (2x)²+(3)²-2×2x×3
=> 4x²+9-12x

[serene pewter]

The range of this function would be [3,0) right?

[vivid void]

no

[serene pewter]

what the heck

What am I doing wrong?

[vivid void]

what’s the largest y value

[serene pewter]

3?

oh wait

nvm

hahaha

[vivid void]

[viscid thistle]

Wth are h&k?

[hushed sphinx]

Hopefully that is explained on one of the preceding pages; otherwise you're sunk.

[viscid thistle]

?

[hushed sphinx]

There's no standard concept of "parameters h and k" for a function. It must refer to something defined specifically in that book/course/module/test/whatever.

[hollow idol]

I think question 2 explains what h and k are