precalculus

[gloomy bane]

one sec

lemme try one of my physical ones

nope

gives 1

[jagged glade]

Anyways.
S12 = a(1 - 0.006^12)/(1 - 0.006) this is good.
We can plug numbers in and make the equation better.

S12= 345.64*(1+20%)

[gloomy bane]

how did you reach that, other than replacing a with the total?

[jagged glade]

They have to be total +20% at then end of the year right?

[gloomy bane]

yes

[jagged glade]

The S12 which is one year after all investments, it will be total +20%

So that it can cover any price rises within the year.

[gloomy bane]

makes sense

[jagged glade]

S12= 345.64*(1+20%) = a(1 - 0.006^12)/(1 - 0.006)

[gloomy bane]

got it

[jagged glade]

[gloomy bane]

although

I'm supposed to calculate the amount invested monthly

[jagged glade]

Which is a.

[gloomy bane]

ok let me attempt

I think I'm making a mistake somewhere

after working the equation I received a = 412.279392

[jagged glade]

Hmm....

The problem is the r.

It's going on a rate of 1+0.006

So the r=1.006

[gloomy bane]

hmm I see

it was 0.6% so I input it as 0.006

[jagged glade]

That's the problem :D
It's increasing by 0.006 so it's 1.006 😂

[gloomy bane]

makes sense lol

how would I adjust the equation then

just replace the old r with new?

[jagged glade]

Yep

[gloomy bane]

worked it

$33.44

[jagged glade]

,calc 345.641.2/(1-1.006^12)(-0.006)

[obsidian monolith]

Result:

33.438170370925

[jagged glade]

Nice!

[gloomy bane]

tysm

it all feels so straightforward now

[jagged glade]

You're good

[raw fjord]

Help 👉👈

[jagged glade]

[raw fjord]

Transforming an equation of a parabola to standard vertex form by completing the square

There's a lot of instruction but I don't what's the way to answer it

[jagged glade]

Okay... So do you have any example we can work on?

[raw fjord]

-x^2-2x+y+7 =0
the answer should be: (x + 1)2 = y + 8

[raw fjord]

I am confused on what to do, the instruction said transform this to standard vertex form

the standard vertex form I know is the y = ....

[jagged glade]

Oh I see what you mean.

[raw fjord]

not the complete the square one

[jagged glade]

The standard vertex form I know is y=a(x-h)²+k

[raw fjord]

exactly

Should I send the verbatim instruction here?

[jagged glade]

But in case your teacher want it in
-a(x-h)²=y-k, then just change it to that form.

You can send it here :)

[raw fjord]

Here's the answer keys if needed.

[jagged glade]

So.... They like it in (something)²=x or y + number.

[raw fjord]

That's my point where I'm kinda confused or completely confused

[jagged glade]

I understand your confusion...

That's what I am feeling right now

Anyways, if they want it that way, make it that way

[raw fjord]

Thanks for the advice!

[jagged glade]

You're welcome!

[raw fjord]

I'm stuck at this one particular task for damned days and all it needed is just to divide the goddamn equation to -1. Oh Lord.

[void cedar]

hello i need help

[viscid thistle]

Do you know wavy curve method?

[void cedar]

no

i do npt

[viscid thistle]

Well then learn it

https://brilliant.org/wiki/wavy-curve-method/

[viscid thistle]

can someone explain how you would solve this problem "A box with a lid is to be created from a 52 cm by 33 cm piece of cardboard by cutting x by x squares from the four corners of the cardboard, and at the centre of the two sides, as shown in the diagram. Determine the function that represents the volume of the box in terms of x, and state the restrictions on x. If the box is to have a volume of 1904 cm3, determine the side length of the squares that need to be cut. "

https://cdn.discordapp.com/attachments/689902007758487818/783532939656036402/capture_1.JPG

i got a solutions for x= 8 cm and x= (155+-root12601)/12

but i wanted to ask how you would know which one to select so how do you determine restrictions

[sour hemlock]

what's the height of the box?

[viscid thistle]

isnt that what we have to solve for?

@sour hemlock

[sour hemlock]

"Determine the function that represents the volume of the box in terms of x"

need to know the height

to get the expression

also please post the original text of the question

the description you posted seems to be missing some stuff

[viscid thistle]

that is the original text

[sour hemlock]

so, you have a flat piece of cardboard

and you want to make a box out of it

[viscid thistle]

i think thats what they are asking for

I did 1904 = ((52-3x)/2)(33-2x)(x)

and I got my solutions

but how do I get the restrictions for this/

?

[sour hemlock]

ah ok

I think I understand now

you cut those holes

and then lift up the cut portion

to make a box

so the height is x

[viscid thistle]

im pretty sure

[sour hemlock]

so volume should be x(52 - 3x)(33 - 2x)

[viscid thistle]

why would it be 52-2x

[sour hemlock]

oh sorry

52 - 3x

corrected

[viscid thistle]

btw why is it 3x and 2x for 33 and 52

[sour hemlock]

for restrictions on x, I think it should be < 33

look at the diagram, on the side of 52 you have three cuts of x

and on the side of 33, you have two cuts for x

[viscid thistle]

ah okok

when my teacher solved it he did 52-3x/2

why did he do that

why is divided by two

[sour hemlock]

it is wrong what your teacher did

the length is 52 - 3x

you can see the base of the box in the middle

[viscid thistle]

dosent it say that you have to cut the two centers as well

[sour hemlock]

oh ok

I see it now

so one half is actually the lid

ok ok

yes

then it should be (52 - 3x)/2

[viscid thistle]

ah so its divided by 2 because of the centers

@sour hemlock

[sour hemlock]

yes

[viscid thistle]

okok thanks

how do I solve this 2 and 1 are roots of the function f(x)=2x^3+px^2-qx-2. Find the values of p and q and the other root

[sour hemlock]

well you could check with a calculator

[austere verge]

Yes they are complementary

tan(x) = cot(90 - x)

Hence the prefix "co" in front of tangent

Cotangent

Same goes for sine and cosine

sin(x) = cos(90 - x)

The co stands for complementary

Meaning angles that add up to 90

And think of it like this

In a right angled triangle

There are 180 degrees in total and 90 is allocated to the right angle, leaving the two other angles adding up to 90

You see sin(x) and cos(90 - x) relating to the same sides

From a visual perspective

@bitter basin

Also sec(x) = cosec(90 - x)

You're welcome

[viscid thistle]

👀

[jolly gust]

Hi does anyone know when sin x = 2

[gaunt mason]

Never. sinus takes its maximum at 1.

[viscid thistle]

Yeah the range is [-1,1] so it never reaches 2

[jolly gust]

Ok thanks

[austere verge]

It can if x is complex

We would use z to represent that tho

[jolly gust]

If sin x = 1/2, how do I prove that cos x = √(3) / 2

[uncut mulch]

you can't

that isn't true in all cases

[lucid hinge]

@jolly gust

[hallow bison]

How does -cosx/sinx-1 become cosx/1-sinx ?

[deep umbra]

i need help on how to graph polar curves

like theta=5π/6​

[austere verge]

There are formulas, for example to find sin(15deg), namely that you can do
sin(45 - 30), both of which are on the unit circle.

[austere verge]

By multiplying the top and bottom by -1

@viscid thistle You can google them

sin(x + y) and similar formulas

You can expand them

Also if y = x, then there are double angle formulas for sin(2x) and similarly for others

That's ok

[clever yacht]

how to solve this? derivative question

i know how to solve it normally,but using definition of derivative is hard

[grave gull]

yes lengthy

f'(x) = lim h ->0 [f(x+h)-f(x)]/h

I think

then use trig identities to simplify

@clever yacht

wait... what am I doing

[clever yacht]

i already tried that,its too complicated

[grave gull]

so you are not allowed to use l'hospital rule ?

[clever yacht]

we can

[grave gull]

wait you wont need that

cos 0 is 1

sec 0 is also 1

so answer is sin 1 :0

[clever yacht]

we need to use definition of derivative

[grave gull]

what's that ?

[grave gull]

u mean this ?

[clever yacht]

yes

but it toocomplicated for me

[grave gull]

that's for finding derivative, not the limit itself

I'm confused

sorry man

[fringe narwhal]

A certain celestial body is orbiting around a star somewhere in space. An astronomer tries to graph out the elliptical orbit of the celestial body using dimensional analysis and scaling, In his drawing, the celestial body is in the origin, and he finds out that the equation of the elliptical orbit is 4x^2+9y^2-256x+342y+k=0 for some value of k. If his findings are true, find the location of the star (the center of the ellipse) in the cartesian plane.

Based on this problem, is the answer that symbolab gives correct? https://www.symbolab.com/solver/step-by-step/solve for center%2C 4x^{2}%2B9y^{2}-256x%2B342y%2Bk%3D0
(trying to study it rn, the book we were given gives poor explanations and examples)

[solar bloom]

does anyone know the answer to this question, I have to find the "tensions" of each one, because there's 2 angles I need to solve and only 1 value given, which is the weight so I don't get how to solve it

[solar bloom]

<@&286206848099549185>

(oh yeah it's 1:43 AM right now 😭)

[grave gull]

2 equations

Tl sin theta1 + Tr sin theta 2 = mg

and Tl cos theta1 = Tr cos theta 2

find Tl and Tr

@solar bloom hope u didn't sleep lol

[opal totem]

Anyone help, 😫
using derivatives, is this method can be used for exponential and logarithmic function? I'm bit confused which method I use first.

[viscid thistle]

hint: || e^(lnx) = x ||

[opal totem]

y = d/dx e^ln(2x/x+3)

[trim hemlock]

You should start simplifying first lime hobosas said

And then just use some basics rules to differentiate that

[robust salmon]

you can use the quotient rule

your final answer should be 3/(x+3)^2

[earnest coral]

can someone help me?

[velvet blade]

the segment intervals go from 0/n to n/n where's k can go from 0 to n. How can k go from 0 to n when (k+1)/n can be (n+1)/n?

[viscid thistle]

probably a typo

k = 0,..., n-1

[velvet blade]

Ok, but..

they're using this equation for it

give me a sec

[velvet blade]

<@&286206848099549185>

[viscid thistle]

im not sure where you are stuck

[regal linden]

are you asking why that formula holds?

it should just be induction

[viscid thistle]

no

[velvet blade]

Ok, so k goes from 0 to n, where each segment goes from k/n to (k+1)/n

But if k = n, then it's outside the cone right?

[viscid thistle]

this

[velvet blade]

The summation goes $\sum_{k=0}^{n}$

[obsidian monolith]

Researcher in Pre-algebra

[viscid thistle]

yes

[velvet blade]

not n-1

[viscid thistle]

there is no +1 here though

those are different things

[velvet blade]

Okay so it was a typo in the end

[viscid thistle]

yeah just dont look at the interval thing

it's just confusing

[velvet blade]

Man I wish someone updated this book

Like I read somewhere it says "you can use a computing machine" instead of calculator

like for some logarithms I think

Anyways, thanks man

[viscid thistle]

ahah a bit old

[velvet blade]

yeah XD

[rich grotto]

Hi

How do I type in foundations

[viscid thistle]

Type ,iam adv in #bots to get access to the advanced section

[viscid thistle]

Yes.

[short sorrel]

@frozen sparrow please don't phantom ping.

[blissful ridge]

Wait, what's phantom ping??

Is it the same thing as ghost ping?

[viscid thistle]

I suppose?

[viscid thistle]

hii this channel is for calculus like limits and derivative right ?

[eager atlas]

can someone help me find an equation of an exponential function that goes through (1,2) and (3,18)

yes

[viscid thistle]

perfect

[astral mantle]

no this channel is for precalc

#calculus is for calculus

this is mostly trig and simple function related stuff

@eager atlas still need help?

[eager atlas]

yes

[astral mantle]

alright

what have you tried?

whats the form of an exponential equation

if you write both points in the form then you can solve one for a and substitute to find b

[eager atlas]

ohhh. i think that’s all i needed to hear thank you.

[viscid thistle]

does someone have a paper with all the derivative for the exponential ? like all d/dx

[astral mantle]

derivative of e^x = e^x thats all you need

[viscid thistle]

but with ln

nvm

[austere verge]

d/dx (ln(x)) = 1/x

And generally

d/dx (ln(f(x)) = f'(x)/f(x)

@viscid thistle

When integrating, you need to slap on a +C and also an absolute value around the ln(|x|)

I mean when integrating something that looks like f'(x)/f(x)

[stuck lark]

you mean ln|f|

[austere verge]

Yes sorry

[viscid thistle]

@austere verge thank you sm !!! 💓

[austere verge]

That's ok

[idle moat]

how do you get x = 5pi/6, 7pi/6?

[sick steppe]

Are there options...?

[bitter basin]

oh

[sick steppe]

Cause sin30 != A

since A is not defined

3

[bitter basin]

o

[sick steppe]

yeah...

[uncut mulch]

what's wrong with 6 or everything except 6?

so yeh, 3 was wrong

and its so fkn annyoing to check

why don't they just tell you to enter the numerical value

[sick steppe]

looks like only 3 and 6 are wrong in the ss

[bitter basin]

ikr

[uncut mulch]

so i don't have to keep cross referencing the stupid letter

[sick steppe]

Easier to type a letter than math pepega

[bitter basin]

yea

this right?

[hallow bison]

What's the half angle identity for csc(x/2)?

[sick steppe]

1/sin(x/2) @hallow bison

so whatever the half angle for sine is, flip it

[hallow bison]

so csc(x/2) = 1/sin(x/2) = 1/sqrt((1-cos(x))/2)?

[sick steppe]

Imma go out on a limb.. they dont want cscx=cscx as an answer

[hallow bison]

1st one

[sick steppe]

if that's sin(x/2) is, then yes

[sick steppe]

@ this

[hallow bison]

ye the bottom part of the last one is identity for sin(x/2)

[sick steppe]

read

i responed to dad

@bitter basin I doubt the answer to 1 is K, since that's trivial

[bitter basin]

oh

[sick steppe]

1 and 5 are wrong

[bitter basin]

ok

[sick steppe]

$\csc{\frac{x}{2}} = \pm \sqrt{\frac{2}{1-\cos{x}}}$

[obsidian monolith]

moshill1

[sick steppe]

@hallow bison

[uncut mulch]

not cot(5)

[uncut mulch]

cot and tan are multiplicative inverses

[bitter basin]

so

ik

[uncut mulch]

tan(90° - B) = cot(B)
and you can get the value of that using tan(B) = 5

and the mentioned property

what do you get when applying the mentioned property?

no

tan(B) isn't the same thing as cot(B)

[bitter basin]

soo

[uncut mulch]

no

tan(B) * cot(B) = ?

numerically correct answer to their question but they don't want decimals

[lucid hinge]

u can try drawing a triangle with the values u are given

[uncut mulch]

so simply 1/5

[round quest]

k

[bitter basin]

ok

[ripe adder]

Julia's soybean field is 3 meters longer than it is wide. To increase her production,
she plans to increase both the length and the width by 2 meters. If the new field is 46
square meters larger than the old field, then what are the dimensions of the old field?

I know that: L=W+3, and (L+2)(w+2) = x + 46

how do I find 'x' aka the original size of the field?

[astral mantle]

x is equal to the area of the original field

[ripe adder]

x = (lw)

I just got it

[astral mantle]

yes

im glad

[ripe adder]

thank you though

[austere verge]

I don't understand how they factorised this

I think the solution is missing a negative sign in front too

But

Idk

How would I factor that thing?

Do I use poly division?

Bc there are fractions there.

I have only used poly division for whole number coefficients.

Wait nvm

I got it

They just pulled out the fraction

[keen turret]

Please

[worthy seal]

Hey guys struggling a bit on this

Magnitude of V minus the Magnitude of W, I'm reading that correctly, right?

[willow bear]

yes

[worthy seal]

You see I thought it was saying the absolute value of

and my answers were way off xD

This discord is so nice

[viscid thistle]

@worthy seal

[worthy seal]

Sup

[viscid thistle]

What grade you in

nvm

[worthy seal]

Pre-calc 1st semester of college, didnt get into calc 1

Needed it anyway, not that good at math

Calc 1 ima need to gear it up though

[viscid thistle]

ph

oh

hm

[velvet blade]

hey so I resumed studying again

I'm doing the exercise in summations

having very hard time deriving it.

so you have height as $\frac{kh}{n} ; k = 0,1....n$

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

height of a cylinder means value of x. It's confusing

then, volume of every cylinder of radius kh/n = $\pi(\frac{kr}{\frac{h}{n}})^2 \frac{kh}{n}$

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

this doesn't be a multiple of sum $\sum_{k=1}^{n} k^2$

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

And I don't get h in the numerator for $\frac{1}{3} \pi r^2 h$

[obsidian monolith]

Researcher in Pre-algebra

[viscid thistle]

gimme a sec

which exercise are you doing exactly?

[velvet blade]

16.2

[viscid thistle]

but which exercise in 16.2

16.2 is a section

[velvet blade]

page 410

first question

page 407, there's the explanation below the picture

[viscid thistle]

determinants start at pg 401

the chapter you're talking about ends at 399

maybe you're using pg numbers provided by the pdf viewer

let me see

oh ok

[velvet blade]

I'm using pdf numbers

yeah

[viscid thistle]

wait the first exercise is already solved in the chapter isn't it

[velvet blade]

No, he gave a hint

he solved for a cone of radius and height 1

not radius r and height h

[viscid thistle]

oh right

[velvet blade]

did you get it?

[viscid thistle]

not really, I don't understand how to generalise it to cylinders of an arbitaray height and length

sorry researcher

[velvet blade]

it's ok

thanks for attempting

Can I ping helpers now?

[viscid thistle]

probably

[velvet blade]

<@&286206848099549185>

[viscid thistle]

I only read the part on induction and geometric series

[velvet blade]

The exercise also has derivation for a sphere

[viscid thistle]

hmm

seems interesting

I'll read up this part tomorrow actually

I'm just too sleepy to focus rn

[velvet blade]

Oh it's okay

[brazen tundra]

hey i need help solving this problem, i have no clue on how to t started with finding radius or area with just an equation of a graph

get*

<@&286206848099549185>

[arctic fractal]

The equation of the circle will be

(X-r)^2 + (y-r)^2 = r^2

Because if u think avout it

The circle has been moved r units to the right and r units up

And then sub in the line equation into this formula

Once uve simplified everything

Use b^4-4ac = 0

And workout the value of r

@brazen tundra

[vocal tree]

how would you find a function that satisfies f(2x) = 1/2 f(x) that goes through a point (a, b)

[brazen tundra]

yeah i understand thanks IC

[static locust]

@vocal tree Have you already looked at a concrete combination of values?

The trick is -- if you want to call it like that -- to find one function that satisfies the equation and then to adjust that function to go through (a,b).

An easy way to find a function that satisfies the equation is to take a look at a concrete combination (a,b), to draw the graph and then to ask yourself "Do I know a function that looks similiar?".

However I do not want to help you too much. If you need another tip, ping me.

Hint: Do not use a combination (a,b) with a = 0. Once you solved the puzzle you will see why. ;)

[vocal tree]

1/x seems to satisfy this, though I don't intuitively see why

is there a rigorous way to solve this sort of equation

[astral mountain]

if f(x) has limits infinity at infinity, does that mean it's derivatives always have the same?

I'd think yes, but maybe their are special cases where it isn't the case?

[stuck lark]

take a sec, you can think of plenty of functions that don't follow this

[astral mountain]

goniometric functions?

i mean if it goes to infinity its slope must be the same ain't it

hmm

ok nvm lol

big derp

[static locust]

@vocal tree Well, I just solved this question by experimenting around until I found the solution. There may be an algorithm that solves equations like this, but there is none I know of.

There are other solutions to this equation, but they mostly are pretty weird -- aside from the very easy f(x) = 0.

One way to look at the problem intuitively ist to take a look at the following set:

Ok, scrap that. Just pick a point (a,b) and take a look at all the other points that have to be part of the function if (a,b) is part of the function.

So ..., (a / 4, b * 4), (a / 2, b * 2), (a, b), (a * 2, b /2), (a * 4, b / 4), ...

You will see that if you multiply a by some factor y, you multiply b by 1/y. So you need a function in which you can multiply x with some factor y and this produces the same result as if you multiplied f(x) with 1 / y. The only "simple" function I can think of that has this property is 1/x.

[velvet blade]

$\sum_{k=1}^{n} \pi[\sqrt{\frac{k(r^2 - x^2)}{n}}]^2 (\frac{x}{n})$

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

is this right?

[willow bear]

what's x @velvet blade

are you sure you didn't mean to have only r^2 - (kr/n)^2 under the root

[velvet blade]

x is value of x on the graph

the curve is on an x,y plane right?

y gives the radius of each small cyinder?

[velvet blade]

yeah I just did it the same way I did the previous problem

[willow bear]

the cylinder at x = kr/n has radius sqrt(r^2 - (kr/n)^2) and thickness r/n

so what you should have is $\sum_{k=1}^n \pi \left( \sqrt{r^2 - (\frac{kr}{n})^2} \right)^2 \cdot \frac{r}{n}$

[obsidian monolith]

Ann

[velvet blade]

ok so I'll get $\pi \frac{r^3}{n} \sum_{k=1}^{n} [1-(\frac{k^2}{n^2})]$?

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

@willow bear you there?

[willow bear]

i am

assuming you didnt fuck up anywhere this should be correct yes

[velvet blade]

Is there a summation formula for $\sum_{k=1}^{n} [1-(\frac{k^2}{n^2})]$?

[obsidian monolith]

Researcher in Pre-algebra

[viscid thistle]

yes

split into sum 1 - sum k^2/n^2

[willow bear]

split

i would advise against using "divide"

[viscid thistle]

agreed

[velvet blade]

okayy

there's a formula for summation of k^2, but not k^2/n^2 right?

[viscid thistle]

you can bring the 1/n^2 out of the sum

[velvet blade]

oh okay, forgot that

$\pi r ^3 - \frac{1}{6} (1+(\frac{1}{n})(2n + 1)$

[obsidian monolith]

Researcher in Pre-algebra

[viscid thistle]

what happens when n is very big

[mystic umbra]

when n is very big 1/n = 0

therefor 0(2n+1)=0

@viscid thistle

[viscid thistle]

yes I know

I was asking to Researcher

[willow bear]

when n is very big 1/n = 0

[viscid thistle]

I’m confused on there the minus 2 came from. I understand everything else

[viscid thistle]

you have the form of 2 factors equal to 0, $ab=0$ one way the product of both factors is 0, is by a=0, because if $a=0$, we get $b\times 0=0$ which is true and therefore we want to analyse $a=0$ and same thing occurs when the other product is 0, $b=0$, we get $a\times 0=0$ which is true, so we want to analyse as well $b=0$, which is exactly what they did. They solved for $$3t-1=0$$ $$t=\frac13$$ and they solved for the other case $$t+2=0$$ $$t=-2$$

[obsidian monolith]

Al𝟛dium

[viscid thistle]

i'm kind of surprised that you understood the t=1/3 part but not the t=-2 though

@viscid thistle

No I meant the 3sec^2theta + 5sectheta -2 =0 @viscid thistle

Sorry for not being clear enough

oh

[viscid thistle]

ok let me latex it

overdetailed and this time unnecessarily overdetailed tex at your service.

[obsidian monolith]

Al𝟛dium

[viscid thistle]

@viscid thistle

Ohh that makes sense thanks

[gilded brook]

Just a quick question for Sequences

How does "a" get included would a just be the same as a,sub1?

for a_2 it would be 3/a_3 - 1

But then how would I find a?
Or would I just do what a_1=4 and then multiply by each sub?

[willow bear]

you can use underscores for subscripts

also what do you mean by "find a"

$a$ without a subscript doesn't refer to any of the numbers $a_1, a_2, a_3, ...$

[obsidian monolith]

Ann

[willow bear]

in fact it doesn't refer to anything at all as of now. you haven't told me how it's defined.

[gilded brook]

Sorry I think I'm asking this weirdly, but to get a purely number answer would a_1 =4 mean a_3=16?

[willow bear]

what do you mean by "a purely number answer"

what are you asked for? a_4?

a_3?

[gilded brook]

if a_1 =4 then what would a_2 = or a_3 is what I'm trying to ask

Is it based on multiplication or addition?

[willow bear]

what

a_1 = 4
a_2 = 3/4 - 1 = -1/4
a_3 = 3/(-1/4) - 1 = -13

subscripts don't denote an arithmetic operation like superscripts do.

[gilded brook]

So how do you determine what you put in for a_n? in general

Like how do you go from a_1 =4 to a_2=3/4-1?

[strong ermine]

quick question: is $\theta=\frac{2\pi}{5}$ in the 2nd quadrant of a unit circle

[obsidian monolith]

e^{\pi i}+1 = 0

[willow bear]

no it's in the first

[strong ermine]

ok

[willow bear]

@gilded brook you don't

[drowsy helm]

could someone tell me if I did the conversion from cartesian to polar correctly pls?

[willow bear]

i went from $a_1 = 4$ to $a_2 = \frac{3}{4} - 1$ by using the recurrence relation $$a_{n+1}=\frac{3}{a_n}-1$$ with $n=1$ (giving $a_2 = \frac{3}{a_1} - 1$)

@drowsy helm seems ok

[obsidian monolith]

Ann

[willow bear]

@gilded brook is this your first exposure to sequences as a mathematical concept

[gilded brook]

Yes

[drowsy helm]

aight well I hope it is lol

[willow bear]

i see

the number at the bottom is simply the position of your term within the sequence

like

[gilded brook]

So I'm just trying to figure out how it all works. Usually I watch videos on the subject but no video I could find had the a actually in it. only n

[willow bear]

a_5 is just "the fifth term in the sequence a_n"

"a", if you insist, is the name of the entire sequence. but that's rare to hear in practice.

there are many kinds of number sequences out there and many ways to define them

what you gave there is a recursive definition

also known as a recurrence relation

the gist is that each number in the sequence is defined as a function of the number before it.

[gilded brook]

Ah alright.
Oh I think it just snapped into my head. Do you just take

Yeah that's what I was gonna just say

So for a_4 it would be 3/-13-1

[willow bear]

yes

[gilded brook]

Alright I get it now thanks a bunch

[quaint mason]

for the first one is the problem just saying that from the origin, the end of one vertex to the other for the x axis is 35 and the y axis being 17.8?

[woven verge]

im so dumb can someone help im at 8n^3-3n^2-72n+27=0

and i dont know how to go any further bruh

[neon mural]

what's the question lmao

[woven verge]

solve for n

[neon mural]

hmmm

hmmm

wait

n^2(8n-3)-9(8n-3)=0

(8n-3)(n+3)(n-3)=0

$n=\pm3,~n=\frac{3}{8}$

[obsidian monolith]

lofi hip hop radio

[neon mural]

@woven verge

[woven verge]

how do u know how to factor like that

like is there a rule or smth when theres a degre of 3

[sick steppe]

Looks like they common factored the 8n-3

[neon mural]

yeah

8n-3 was a common factor

so basically between 8n^3-3n^2, you can take out n^2, and in -72n+27, you can take out 9

and by doing that you get (n^2-9)(8n-3)

as n^2-9 is a difference of squares

you can further factorise that

[quaint mason]

can someone confirm w my clarification question?

[woven verge]

how do i solve for R2

I got R2 = -(R1R3RT/R3RT+R1Rt=R1R3)

but thats wrong

[viscid thistle]

Guys how do you solve questions you weren’t taught? My teacher gave me a speed time graph bs from physics. If I didn’t take physics there would be no way for me to solve it. Please help.

[astral mantle]

There is probably a way for you to solve it

just because the problem has concepts from physics doesnt mean you need to know physics to solve it

[viscid thistle]

But how am I supposed to solve these types of questions if I haven’t been taught them beforehand. Everyone says practice practice, but it’s not working.

What theory book should I read anything helps people, I’m super hopeless.

My final precalc was 75 😦

[astral mantle]

I dont know if you have been taught them or not or how you view the questions

i cant really give any solid advice for ya there

all i can say is thoroughly read the question and dont overthink you can most likely do questions you are given

[viscid thistle]

I appreciate it bro, but it was an integration question that wasn’t in any of my textbooks.

[quaint mason]

for the first one is the problem just saying that from the origin, the end of one vertex to the other for the x axis is 35 and the y axis being 17.8?

[astral mantle]

what?

[quaint mason]

oml sorry

the pic hasnt loaded

earlier i asked for confirmation but didnt get response so im still stuck kinda

[viscid thistle]

@ me if someone solved this

[viscid thistle]

How can I find the distance of conjugate axis by knowing only the foci and vertex length?

The question:
If the focal distance of a hyperbola is 10 units and half of its transverse axis is 8 units, what is the distance from the center of the hyperbola to one of the endpoints of the conjugate axis?

Found the answer

[worthy seal]

are these answers actually different in any way?

[blissful ridge]

They are essentially the same

[willow bear]

Simplify your answer. Type a...

this got cut off

can you show what it says in full?

[worthy seal]

Says that on every question so I didnt even bother adding it

sec

[willow bear]

i think there might've been a format requirement that you didn't meet

[worthy seal]

​(Simplify your answer. Type an exact​ answer, using radicals as needed. Type your answer in the form ai​ + bj. Use integers or fractions for any numbers in the​ expression.)

[willow bear]

ah

[willow bear]

got it

Type your answer in the form ai​ + bj.

thats the problem

you didnt do that

thats why yours was rejected

because of the wrong form

even though it had the right value

[worthy seal]

Ah I see

Well that's fucky

Rip, thanks

Would this be a proper way of writing it too then?

[willow bear]

proper in the sense of accepted by the homework system?

idk if it's lenient enough but i wouldnt bet on it

[harsh moon]

Does anyone know why these two formulas aren't the same? Why is the sign inbetween the two terms different?

[willow bear]

expand the right-hand sides and you'll see.

[harsh moon]

Like when do I know to use one vs the other

[willow bear]

it does not matter.

you could also have used the first formula with x and y swapped to get sin(y) cos(x) = 1/2 [sin(y+x) + sin(y-x)]

since sin is odd, sin(y-x) is the same as -sin(x-y)

[harsh moon]

I saw that they're only equal if the angles are the same, but it still feels weird

O.k.

[leaden stratus]

Excuse me, can I re-write this denominator as 4(x-1/2)^2?

Because doing it so, makes it impossible to simplify it with the numerator

I've semplified the numerator as (x+5)(2x-1). Ruffini rule.

I think I have to do both with Ruffini or both with the rule a(x-x1)(x-x2)

[cobalt swallow]

use any method hook or crook, provided it makes sense to find the limit

[viscid thistle]

i mean yeah, 4(x-1/2)^2 and (2x-1)^2 are basically the same thing, you can check out by expanding

[leaden stratus]

How? @viscid thistle

[viscid thistle]

how are they equal you mean? like \ $4(x-\frac12)^2=4(x^2-x+\frac14)=4x^2-4x+1$ while $\ (2x-1)^2=4x^2-2(2x)+1=4x^2-4x+1$

[obsidian monolith]

Al𝟛dium

[leaden stratus]

The problem is: how do I simplify both numerator and denominator?

[viscid thistle]

by factoring as you just did, but we want the form (2x-1)^2 and not the form 4(x-1/2)^2, both forms are equivalent, but we want the first one to cancel with the numerator

[leaden stratus]

Yes

[viscid thistle]

so you didn't factor incorrectly, it was just another form of what we wanted

[leaden stratus]

But how can I cancel with numerator?

[viscid thistle]

$\lim_{x\to \frac{1}{2}^+}\frac{2x^2+9x-5}{4x^2-4x+1}=\lim_{x\to \frac{1}{2}^+}\frac{(x+5)(2x-1)}{(2x-1)^2}\ =\lim_{x\to \frac{1}{2}^ +}\frac{(x+5)\cancel{(2x-1)}}{(2x-1)^{\cancel{2}} }=\lim_{x\to \frac{1}{2}^+}\frac{x+5}{2x-1}$

hold up

[leaden stratus]

Ok

[obsidian monolith]

Al𝟛dium

[viscid thistle]

this is all what we talked about

nothing new

[leaden stratus]

Ah ok, just factorised as (2x-1)^2

I'll write like that

[viscid thistle]

yeah, remember what i said here

[viscid thistle]

https://cdn.discordapp.com/attachments/504653099793645568/784999841796849664/Capture.JPG

can someone help me

[sick steppe]

undo the 2nd set of transformations / apply the opposite transformations

[viscid thistle]

could you explain a bit more

@sick steppe

[sick steppe]

ok so first you undo the 6 units down translation

so you translate it up 6 units

[viscid thistle]

okay so it becomes f(x) = -4(4(x+2))^2 +1

now what do I do

[sick steppe]

you undo the horizontal compression by 1/2

[viscid thistle]

how do i do that

do i multiply 4 by 1/2

[sick steppe]

what do you do to a function that horizontally compresses by a factor of 1/2?

[viscid thistle]

you multiply the function by 2

[sick steppe]

yeah, so to undo that, you divide by 2

or multiply by 1/2 like you said

[viscid thistle]

okay so its now f(x) = -4/(2(x+3))^2 +1

now what should i do

[sick steppe]

im not doing the entire question for you, I did the 1st 2 for you

Apply the logic i did

[viscid thistle]

okok

if i solve it can you just identify if its correct

@sick steppe

[sick steppe]

sure

[viscid thistle]

okok thanks

i got f(x) = 5(2(x+3))^2 +1

is that right?

@sick steppe

[sick steppe]

looks good

[viscid thistle]

so its right?

[sick steppe]

yes

[viscid thistle]

okok thank you

[quaint mason]

can someone help me with this problem?

[viscid thistle]

im confused by what this question is asking

[bronze star]

change of z for each change in t

[viscid thistle]

yeah i got delta z / t

but how do i represent that as a derivative

[bronze star]

well it’s dz/dt
Or z’(t), depending on your notation

[viscid thistle]

LOOL

WAS IT THAT EASY

THANKS

[bronze star]

No problem

[viscid thistle]

whats the difference between the derivative and the gradient function

cause our teacher told us dx/dy is another way of representing the gradient function

[bronze star]

I believe the gradient is just another word for derivative
Except gradient is more commonly used in 3 dimensions and higher

[viscid thistle]

so its the same thing but used in different contexts?

like how perpendicular, normal and orthogonal is used in different contexts?

[bronze star]

Yeah, pretty much

[sick steppe]

Derivative is just more specific gradient function, as in it tells you slope anywhere whereas gradient is between 2 points only

[solar blaze]

for the half angle formulas when proving trig identities

how do we decide when to put a positive or negative before the square root?

[quaint mason]

unit circles or CAST

[solar blaze]

but if youre solving with x

how do we apply cast

[quaint mason]

wym solving for x

[solar blaze]

like im trying to solve this: "Prove (sin2x/1+cos2x) x (cosx/1+cosx) =tanx/2"

and on the right side it has tanx/2

so that could be (sinx/2)/(cosx/2)

and then i would sub in these

but how do i know whether it should be positive or negative

[quaint mason]

sorry tryna visualize it erm

wouldnt it depend on which quadrant its in

for example since tangent is positive in Q3, both sin and cos would be negative

but then if u take other quadrants

for example Q4. cos is positive there

but sin is negative, then tang is just negative

not sure if thats what ur asking

[viscid thistle]

I always thought that matrices computation is studied during the multivarible calculus level "formally" but why how come matrices and complex numbers are in the same category "precalculus"?

Just 2 different fields, idk

[stuck lark]

matrix arithmetic & algebra on C aren't anything exciting

[winter spire]

functions and graphs is it calculus or algebra? i dont live in us/uk etc

[viscid thistle]

You aren't annoying at all and are you still wondering?
@leaden stratus

[quaint mason]

how would i find the coordinate using that?

[copper vigil]

@bitter basin not the most ethical tip

but are you allowed to use a calculator?

so everything by hand?

in that case i would recommend to always factor out constants first

in fact don't even do that. divide each sum by 3/2

you just need to find the odd one out, not actually evaluate the sum

yeah

0 - 1 = ?

i just told you

no

first term is not 3/2

would you agree

that 0 - 1 = -1

we're just looking at the first sum. that's the odd one out.

3/2 is a constant

that doesn't change as i changes

so you factor out the constant

anyways, the first term is not 3/2. what is (1/4)^(-1) ?

dude no offense but how are you doing sums without knowing basic exponents

yes

(1/4)^-1 = 4

3/2 * 4 = ?

and what is the first term in each sum supposed to be?

3/2

[bitter basin]

o ye

[copper vigil]

so there you go

that's why a is wrong

[bitter basin]

oh

thx

[fleet yew]

@worn violet ...

[worn violet]

nvm got it

[fleet yew]

If you're texting on discord why cant you use an online calculator

[worn violet]

i used 2 different sites but i got diifferent answers but i got it wrong so dont matter

[fleet yew]

Its probably a units issue

Did you check deg or rad

[worn violet]

yes idk what i did wrong

but dont matter

[fleet yew]

Always check trig units

Always

It saves lives

[worn violet]

yeah

@fleet yew can u see what im doing wrong

[cobalt stream]

Sin(theta) = opposite / hypotenuse

[upbeat moss]

https://gyazo.com/c1b1c8e3f12da23e284778d103f1422b

i did the same question yesterday but with different numbers and got everything correct

today, im all wrong??

[uncut mulch]

you seem to be taking the definite integrals of the definite integrals

[upbeat moss]

yea. I placed 19 and -4 within the brackets

[uncut mulch]

f(x) and g(x) aren't 19 and -4 (respectively)

[upbeat moss]

Oh, I though it was...
so im supposed to find f(x) and g(x)

[uncut mulch]

nope

you're supposed to apply linear properties of the integral

eg

applying linearity for a):
$$= \underbrace{\int_2^7 f(x) \dd{x}}{19} + \underbrace{\int_2^7 g(x) \dd{x}}{-4}$$

[obsidian monolith]

ℝamonov

[uncut mulch]

you are not required to find f(x) or g(x),
there isn't sufficient information to get them anyway and it isn't required to complete the question

[upbeat moss]

WOW

ok I understand

[viscid thistle]

Hello

[strong ermine]

hi

[robust nest]

Need help with finding the second triangle

btw the drawed os are supposed to be degrees

idk what to do from here

[willow bear]

what's 'the second triangle'

[robust nest]

https://www.youtube.com/watch?v=ijGxnVb9b9c&ab_channel=Mathispower4u this type of stuff

[willow bear]

also this is wildly not to scale

[robust nest]

ye ik

im using paint lol

cant change triangle shape

[willow bear]

you can use the line tool thrice

anyway whatever

[robust nest]

here 1 sec

nono

ill fix it

[willow bear]

also you have given no indication of whats given and what you calculated

so like

i have zero clue what your problem ac is

[robust nest]

one sec

nvm i got it

I have one question tho

for the video I sent

[willow bear]

mmh?

[robust nest]

This is kinda a dumb question, but isnt the way they drew the triangle actually ACB and not ABC?

I thought ABC would have B at the top angle

[willow bear]

no

the order in which you name the vertices of a triangle is immaterial

[robust nest]

Ok so its not important

hmm

[viscid thistle]

I would say that functions basics is precalculus

Graphing occurs everywhere so far

[rigid finch]

I think this is the right spot

anyone know how I prove that (1,0) (0,-1) and (-1,-2) are part of the solution set?
I tried plugging them in, but it doesnt work for the inequality x -y > 1
for example
(0, -1)

Y <= 3x + 1

-1 <= 3(0) + 1

-1 <= 1
so (0,-1) works for the inequality Y <= 3x + 1 for instance
but then you do x -y > 1
x - y > 1
(0) - (-1) > 1
-1 > 1 is not true

so I am confused on how I would prove that its actually a part of the solution set

even if I convert x - y > 1 into slope intercept form (addx to both sides) it doesnt work

y > 1 + x
-1 > 1 + (0)
-1 > 1
is -1 greater than 1? no it isnt

[velvet jungle]

You're already showing it that those points are NOT part of the solution set.

[rigid finch]

dang what did I do wrong

[velvet jungle]

?_? you're doing it right

[rigid finch]

how would I find the correct solution set

thats what I thought was correct

[velvet jungle]

Yes

Basically (0,-1) is not in the solution set

[rigid finch]

it is not?

[velvet jungle]

But (0,-1.05) is part of the solution set

They are not

Because they break the x-y>1 inequality.

[rigid finch]

oh thats why its a dashe dlnie rip

[velvet jungle]

Yes

[rigid finch]

do you know how I find precise answers

like whole numbers I don't know how to get em

[velvet jungle]

You might want to show that (-1,-2) is on both inequalities

So basically they form a vertex

Now (-1,-2) is not part of the solution set

uhh you shouldn't only want whole numbers

[rigid finch]

oh then idk, what random 3 points should I select lol

[velvet jungle]

Anything deep into the black and red zone

But not only-black or only-red

Ok so (-1,-2) is not part of the solution set

[rigid finch]

but (-1,-3) is

[velvet jungle]

But (-2,-5) is also

Yes (-1,-3) is

Basically anything on the black line below, but not (-1,-2) itself, is also part of the solution set

[rigid finch]

very nice and third point for example (-3,-8)

[velvet jungle]

Because you have just an inequality, and not a strict inequality

Yup yup

[rigid finch]

thanks for your help Shattered that makes a lot more sense now

[velvet jungle]

You should

Find a way

[rigid finch]

when it says "Solve the system of inequalities" it just means graph, right?

[velvet jungle]

To write out this set

When it says solve, you need to write out the solution set

[rigid finch]

how would I find that

[velvet jungle]

Ok hmm

[rigid finch]

isnt the first one y <= 3x + 1 basically already the solution set

[velvet jungle]

I'm also a little unsure of the notation actually

no, you need the intersection

But writing the intersection doesn't mean much, you need to really describe the solution set well

[rigid finch]

wouldn't shading/graphing indicate that?

[velvet jungle]

Well yes it does

I think shading is fine for your level

But maybe you want to define a case-based function

[rigid finch]

👍 I don't think we have learned how to actually solve them without graphing/shading

[velvet jungle]

Ah I see

[rigid finch]

but if I could learn how to it would be dope

[velvet jungle]

Yes then just shade the both-red-black region

But dash line the red-one

Do a open circle on their intersection

And bold-line the black line

That should be the correct notation

[rigid finch]

thanks Shattered, so on (1,0) (0,-1) and (-1,-2) I put a open circle (to show they are not included) -- dash red line to show whatever is on that point isnt included

but why bold-line the black line?

[velvet jungle]

Because they are included

[rigid finch]

to show that it is not dashed?

[velvet jungle]

Yeah

[rigid finch]

👍

thanks for your help Shattered I now understand

[neon mural]

uh

yeah

but idk how much i can help thi

[marsh idol]

I could do upto C but not sure what to do for D

Anyone help?

[strange arch]

@marsh idol

For a) just integrate the upper curve - lower curve

Same thing for b)

[sick steppe]

^

[strange arch]

Oh lol

For D u have to find the length of a vertical strip as a function and integrate to find volume

As height is not constant

Like integral of length_of_vertical_line_at_x.height(x).dx

The length of the vertical strip is given be subtracting the lower function from upper

[viscid thistle]

I'm not able to understand this problem

how do I prove this

I think it has something to do with ${n\choose k}=\frac{n(n-1)\dots(n-k+1)}{k!}$

[obsidian monolith]

Alexander Grothendieck

[viscid thistle]

I don't have any leads tho

[blissful ridge]

Yes,

We start select k elements in any arbitrary order
i.e n.(n-1).(n-2)...........(n-k+1)

Now we know that for every group of k element is counted k! Times

So we adjust for that

n(n-1).......(n-k+1)/k!

And this must be an integer because we are selecting elements of a set.

Now n(n-1)......(n-k+1)/k!=n!/(k!(n-k)!)=nCk

[viscid thistle]

n(n-1)...(n-k+1)=n!/(k!(n-k)!)???

[blissful ridge]

Missed divided by k!

[viscid thistle]

ah

yeah

[blissful ridge]

There you go

[hazy grove]

anyone wanna teach me how to graph some functions by hand? there are two, a logarithmic function and a cubic one. if anyone would like to, I would really appreciate it 🙂

[sick steppe]

@hazy grove SIAP: shape, intercept(s), asymptotes (if applicable), point for scale (if needed)

[hazy grove]

my teacher says i have to graph it by transforming the parent

[sick steppe]

ok, so apply the transformations

[pastel cloud]

happy studying every1

🎄📖

[viscid thistle]

^^

[drowsy karma]

So I had a math quiz today

I got a question wrong, idk what the correct answer was

The question was to find an equation

Vertex = -2,4

Point = 0,0

I got y = -2(x+2)^2 + 4

[pastel cloud]

@drowsy karma where did you -2 come from in front of the brackets?

it should be y = a(x+2)^2 + 4

then plug in the pont (0,0) and solve for a

[drowsy karma]

Isn't-2 = a? @pastel cloud

Oh I'm dumb

its 4a

it's-1

rip

[pastel cloud]

nice

[drowsy karma]

big brain

[pastel cloud]

simple mistake

you caught it easily

[gloomy mortar]

These are the last problems in my practice assignment before the upcoming test. We’ve gone through so many different things in such a short period of time that I’m so confused. Could someone please help? I literally have absolutely no idea how to do any of this, and yes I’ve looked at stuff before asking here

[astral mantle]

for 13 you only need to find the radii and the center with what you are given and place them in the standard form equation

[gloomy mortar]

Talk to me like I don’t at all understand because I don’t, that’s why I’m putting this here

[astral mantle]

do you know the equation for the standard form of an ellipse

[gloomy mortar]

No

[astral mantle]

(h,k) is the center, a is the horizontal radius and b is the vertical radius

you can find the distance between the vertices and the foci for the radii

and for the center you can find where the two radii intersect

[gloomy mortar]

I don’t know what the foci are and stuff

When I tell you I don’t understand this, I literally seriously don’t understand this at all

@astral mantle

[astral mantle]

why do you have an assigment on it if you dont know it

did you just not pay attention

you can just google what foci are

[gloomy mortar]

I did pay attention, but I told you we’ve gone over so many things, I don’t understand anymore

My brain has reached capacity

Because in such a short time there was so much to take in

@astral mantle

[astral mantle]

ok sorry

I dont have much time to teach a lesson about ellipses sorry

[gloomy mortar]

Well can you do something with it

Like teach me something to help me

[pastel cloud]

@gloomy mortar hi

If you give me a second I can find a formula for you

[storm spoke]

HII CAN SOMEONE HELP ME

https://cdn.discordapp.com/attachments/785617075740999690/785619482885226516/unknown.png

<@&286206848099549185> do i ping this

[sick steppe]

ping after 15 mins

[storm spoke]

im sorry

but can you please help 😭

[strong ermine]

I have an exam on this tmrw and i dont remember any of this lol

[drowsy spear]

Hello so this is an assignment for my class. I got quarantined and missed irl classes iam lost and really need some help here 😔

[strong ermine]

i cant read any of that lol

[drowsy spear]

Blurry?

[gloomy mortar]

@pastel cloud Did you get anything

[drowsy spear]

<@&286206848099549185> ...

Would the period be 13 months?

[gaunt mason]

The period is the distance (interval) over which the function (graph) repeats

@drowsy spear

The function (graph) repeats itself. It is obvious, but you can also transcribe it in your notebook and extend it to the left and right to make it even more obvious

[drowsy spear]

Hmm yes, well I usually refer to the midline to see where the graph repeats

[gaunt mason]

The midline is irrelevant for this.

[drowsy spear]

😶 I see

I mean isint it just every year it repeats right?

[gaunt mason]

If you look at the graph, that is correct.

[simple helm]

I’m so lost as well

[gaunt mason]

Also check this @drowsy spear
This might represent the hearthbeat which is a periodic (repeating) phenomena.
It is slightly more disguised but it is still periodic (repeating), and the period (duration of one cycle) is kind of obvious when you draw it like this.

[simple helm]

Also learning about sin and cosine graphs

[gaunt mason]

@simple helm What is confusing you?

[drowsy spear]

Yes its is

[gaunt mason]

@drowsy spear The yearly temperature is logically repeating every 1 year, or 12 months.

But they can make it differently, sneaky, in the graph, to repeat every 13 months.

They havent done that though.

The temperature every January is the same (very similar), every February also...

[drowsy spear]

Its just my teacher used the midline as a reference where the function kept repeating thats why I mentioned it

[gaunt mason]

@drowsy spear I dont see the midline have any relevance. You kinda need to visually decide the repeating distance

[simple helm]

How do you find the asymptotes ?

[gaunt mason]

This is not cosine

I think this is cosecant