help-28

[torn jolt]

how do I find the divergence of polar coordinates?

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[cursive jacinth]

hey! i'm a bit stuck on where to start and i was wondering if someone could help me figure out this question? thank you!

[cursive jacinth]

sum of parellel sides divided by 2 x height?

so it would be x+2+x+4 over 2 x x-1?

(x+3) (x-1)?

would u multiply them out to make x^2+5x-3?

32?

[rough tinsel]

a simple equation

( x + 4 + x + 2) ( x - 1) / 2

[cursive jacinth]

Oh wait i think i get it now, thank you for helping

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[minor wadi]

b

i only manage to 1/sin(theta-pie/2)

[vapid barn]

Cofunction identity

[minor wadi]

whats that

[vapid barn]

sin(90-x) = cosx

[minor wadi]

nvm i understand that

ik but idk how continue

[vapid barn]

But you have to factor out the negative from the sin

[minor wadi]

how

[vapid barn]

Sin(-x) is -sinx

[minor wadi]

so -sin(90+x)?

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@minor wadi Has your question been resolved?

[gritty rose]

No. How did you get a +x from what EndTimes said?

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[gritty rose]

.close

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[runic blaze]

I don’t know how to word the transformations

[runic blaze]

Woukd the first question just be “moves up by 4 units”

?

[dusk inlet]

Yes.

[runic blaze]

Oh

What about the second question

How would i explain the 2 before the parenthesis?

because the x-3 would be moves 3 to the right

and 8 units down

wait so i would have to expand the x - 3^2?

because if im factoring out 2 would i have to expand

[dusk inlet]

disregard that deleted statement.

[runic blaze]

alright

wait i think i got it

correct me if im wrongb

but would i flip the parabola uopside down and move it all down by 2 units

[dusk inlet]

Yes but what happens first?

[runic blaze]

not sure, when graphing the 3 points would i multiply all of them by 2?

is that what comes first

or negative 2

sorry

[dusk inlet]

Focus on (x-3).

[runic blaze]

just move that 3 units to the right?

[dusk inlet]

Yep.

[runic blaze]

gotcha

[dusk inlet]

Then what you said up top happens next.

[runic blaze]

okay so it would be opening downards, 3 units to the right, and 10 units down?

[dusk inlet]

Here.

[runic blaze]

yeah but do i just add -2 and -8

or is it separate

[dusk inlet]

It will translate to the right by 3 units then it would open downwards and scaled by 2.

The final one would do what?

[runic blaze]

move 8 units down

[dusk inlet]

Yep.

So where do you think the vertex is?

[runic blaze]

3,-8

[dusk inlet]

Yep.

[runic blaze]

alr i understand now

ty

[dusk inlet]

np.

Remember to close channel when done. “.close”.

[runic blaze]

.close

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[jagged vigil]

can someone give me a crash course on scientific notation

[jagged vigil]

ah nvm--

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[pliant jungle]

i need help doing problems d. and e.

[pliant jungle]

i am stuck because i don't know which strategy to use

[kind jay]

e root test

[pliant jungle]

i am not familiar with that test

lol

[kind jay]

look it up

[pliant jungle]

k

[opaque helm]

For d) you'd want to use Riemann criteria

And polynomial approximation of cos at 0

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[jagged vigil]

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

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[brittle hinge]

How do you do d

Fill in 170,000 for c but what do you do after with the rest of the formula

[gritty rose]

,rotate

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[brittle hinge]

<@&286206848099549185>

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@brittle hinge Has your question been resolved?

[neat dirge]

You solve for p

[brittle hinge]

How?

Just guessing numbers till its close to 170k?

[neat dirge]

Nah

You have to isolate it using algebra

Getting it in the form p= some stuff

[brittle hinge]

Oh yh

But whats the some stuff 💀

[neat dirge]

That’s the final answer

Since p is equal to it

@brittle hinge are you getting it or do you wanna go step by step?

[brittle hinge]

If you could step by step would clear it up moee

[neat dirge]

Kk

I’m on mobile so I’m gonna hand write it all out

So far you have this down?

@brittle hinge ?

[brittle hinge]

Tysm

[brittle hinge]

Yes

[neat dirge]

@brittle hinge This form makes it hard to isolate for p, since we can’t separate the denominator easily (without partial fractions). So the next step is to multiply by (100-p).

Since it’s multiplied to both sides it keeps the balance of the equation.

[brittle hinge]

Ooh ok

Then you expand the bracket ?

[neat dirge]

Yup

[brittle hinge]

Oh thats easier than i thought

Tysm!

[neat dirge]

No problem

Tell me your final answer just to check

[brittle hinge]

17,000,000 = 250,000p

Then devide by 250,000

Which is 68

[neat dirge]

https://tenor.com/view/cowboy-emoji-stick-man-yee-haw-gif-14108336

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[dull copper]

how do you solve this? ive tried using the law of sines but it doesnt say to round and you cant really put it in radical form

[vivid coral]

What type of triangle is this?

[dull copper]

a right one?

[vivid coral]

yes good

[dull copper]

everything in the screenshot is all the info we get

O

30-60-90

[vivid coral]

:)

[dull copper]

is that what you were going for

[vivid coral]

yeah thats 1 way

another way is basic right angle trig

both work

[dull copper]

law of sines?

[hot herald]

is a bit excessive

[vivid coral]

no just normal right triangle trig

what is sin(60) in terms of this diagram

and what is the value of sin(60)

set these equal done

for 30-60-90 write each side as variables for this way and then substitute to find x,y

[dull copper]

its what we're learning in math, so i just assumed

[vivid coral]

always try the easiest option first

dont do extra work

[dull copper]

wait

so then you can just use 30-60-90 rules

divide 12 by sqrt(3) to get length X

but how would you express that radically

[hot herald]

simplify the fraction

simplify
$$\frac{12}{\sqrt{3}}$$

[glossy valve]

ℝamonov

[dull copper]

multiply by the conjugate?

but then that just gets $$\frac{-12sqrt(3)}{-3}$$

[glossy valve]

MMMMM

[acoustic pebble]

Uh

Don't do that

Like

Use

[dull copper]

oke

[acoustic pebble]

$\frac{\sqrt{3}× \sqrt{3} × 4}{\sqrt{3}}$

[glossy valve]

बाशुदेव

[acoustic pebble]

Now cancel out

Then

Ur left with

4×sqrt(3)

[dull copper]

how did you get 4?

[acoustic pebble]

Liatwj

Listen

[dull copper]

if you multiply 12/sqrt(3) by sqrt(3)

[acoustic pebble]

Sqrt(3) × sqrt(3) = 3 right?

[dull copper]

yea

[acoustic pebble]

I hope u know this

Sqrt(3) × sqrt(3) = 3 then we can right 3 × 4 as it results into 12

3×4=12

[dull copper]

yea

[acoustic pebble]

Lik eg: we can right 4 as 2+2

So

U got me

?

[dull copper]

yea

[acoustic pebble]

If not then

Let me explaine it in a better way

So

$\frac{\sqrt{3} × \sqrt{3} × 4}{\sqrt{3}} = \frac{12}{\sqrt{3}}$

[glossy valve]

बाशुदेव

[little helm]

why not just multiply by √3 and rationalise the denominator tho? its much simpler

how did you get -12?

[acoustic pebble]

He conjugated

[little helm]

yeah ik

-12?

[acoustic pebble]

Ye

I'm too confused

Rn

🗿

[dull copper]

i did it wrong?

[little helm]

no it should just be $\frac{12\sqrt{3}}{3} = 4√3$

[glossy valve]

kinglacto

[little helm]

yep

[dull copper]

well

that explains it

[dull copper]

so when simplifying that kinda stuff, dont multiply by its conjugate?

[acoustic pebble]

Nah

Dude

U need to

[little helm]

honestly i dont know what conjugate means sorry lol

[acoustic pebble]

But u did it wrong

[dull copper]

sqrt(3) ->conjugate-> == -sqrt(3)

quick explaination

[acoustic pebble]

💀

Let me explain it

[little helm]

its alright

i looked it up

[dull copper]

so if x is 4sqrt(3)

y is just 8sqrt(3)

cause its double x

[little helm]

i have already learnt it, but never was taught the terminology

[acoustic pebble]

U can just use law of sins

[little helm]

yep so ig your conjugate should be √3

[dull copper]

tried

needs to be either a whole number or be able to be simplified radically

[acoustic pebble]

Oo

So

[dull copper]

law of sines gets y as 12.85640646

[acoustic pebble]

Sin(60)/12=sin(30)/x

[dull copper]

6.92820323

[acoustic pebble]

Ye

[little helm]

why dont you just use simple trig?
12/y = sin(60°)
12/x = tan(60°)

[acoustic pebble]

$x \approx 6.9$

[glossy valve]

बाशुदेव

[acoustic pebble]

Now just use

[dull copper]

he told us not to round

[acoustic pebble]

Ok

So

[dull copper]

hence radicals

[acoustic pebble]

Ok ok

[dull copper]

welp

very much thanks for the help

much appreciated

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[tribal path]

I know that it is made up of the zero vector and all possible eigenvectors

but I don't understand why its equal to null

[torn jolt]

help

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[smoky lintel]

Find the solution to the recurrence relation $a_n=a_{n−1}+6a_{n−2}$ with initial terms $a_0=2$ and $a_1=1$

[glossy valve]

M00NLIG7

[smoky lintel]

So i have found the first few terms in the sequence but i do not know what to do from there

[zenith kernel]

Find the root of x^2-x-6=0 first

So -2 and 3

So it is of the form u(-2)^n+v(3)^n

Then use the first two terms to solve u and v

[smoky lintel]

where did you get x^2-x-6=0 from

[zenith kernel]

I didn’t learn it in English

Seems called finite difference method

But it doesn’t matter

Just sequences satisfying $\sum_{0 \leq k \leq r}c_{k}a_{n+k}$ for any n form a linear space

[glossy valve]

Cogwheels of the mind

[zenith kernel]

For the corresponding polynomial $p(λ)=\sum_{0 \leq k \leq r}c_{k}λ^{k}$ the basis is $(… x^{n},nx^{n},…,n^{j-1}x^{n},…)$ for a j-multiple root x of $p(λ)$

[glossy valve]

Cogwheels of the mind

[zenith kernel]

We assume that $c_{r}=1$,This can be proved using Jordan form just like the very similar ODE case: $\begin{pmatrix}a_{n+1}\…\a_{n+r}\end{pmatrix}=\begin{pmatrix}0&1&0&…&0\0&0&1&…&0\0&0&0&…&0\…&…&…&……&…\-c_{0}&-c_{1}&-c_{2}&……&-c_{r-1}\end{pmatrix} \begin{pmatrix}a_{n}\…\a_{n+r-1}\end{pmatrix}$
just like ode case find the Jordan form of that matrix then you can prove that the basis is for the form as I said.

[glossy valve]

Cogwheels of the mind

[smoky lintel]

Honestly this seems far above anything i have come close to in the class i am taking

[zenith kernel]

If you don’t have time to prove it just remember this result

Okay then remember the result

Give you two examples: like if a(n+2)=a(n+1)+a(n), Fibonacci, λ^2-λ-1=0 has two roots (1+/-sqrt(5))/2, so a(n)=u((1+sqrt(5))/2)^n+v((1-sqrt(5))/2)^n

Another example:

a(n+2)=-6a(n+1)-9a(n), so λ^2+6λ+9=0 has a 2-multiple root -3 so a(n)=(u+vn)(-3)^n

Just remember the result, hope those two examples and your question itself can help you memorize it, but still the best way is to prove it yourself

[smoky lintel]

Ok well thank you for the help i will try my best

[zenith kernel]

Np

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@sterile fossil Has your question been resolved?

[sterile fossil]

<@&286206848099549185>

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@sterile fossil Has your question been resolved?

[glass cave]

let the number be (\overline{ABC}), so reversed it is (\overline{CBA}).

[glossy valve]

ricey

[glass cave]

for it to be a 3-digit number, (1 \leq A + C \leq 9) (without considering the carrying of (B)---that is, if (B < 5))

see if you can work from this

[glossy valve]

ricey

[sterile fossil]

ohhh okayyy

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[prime fern]

i just want to make sure i'm on the entering the right info into the graphing calculator. What would I enter for a0, a2, a3, etc when putting it into the graphing calculator on problem 2?

[prime fern]

here's the photo so it can be seen in discord

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[prime fern]

<@&286206848099549185>

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[torn jolt]

help

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[torn jolt]

<@&286206848099549185>

[torn jolt]

should it be 40x x

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[reef oriole]

I already know the answer because of my graphics calculator but i'm very lost as to how to get there

i tried doing simultaneous equations but yeah i'm not good haha

my a value differs to the answer

[reef oriole]

i multiplied by 4

oop can i not do that

oh wait

I'm so stupid

i got it

i think

I subtracted 14 instead of adding it 💀

i will keep this open until i finish

okay i've gotten it anyway ty

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[silk solstice]

is that correct ?
if not where is the mistake

(minus sign at the end *)

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[gritty rose]

,w derivative (x^2+a^2)^(-3/2)

[glossy valve]

Results provided by WolframAlpha

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

Requested by riemann#2529

[gritty rose]

multiply that by (a^2 / 2) and that should be your answer

[gritty rose]

it looks like you're trying to use quotient rule. you don't need to. but if you insist, you simplified powers incorrectly at the beginning of the second line.

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[old rapids]

Prove the identity 1/cosθ - cosθ/1+sinθ = tanθ

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Please don't occupy multiple help channels.

[old rapids]

1-cos²(θ) +sin (θ) / (1+sin(θ)) cos(θ)

sin²θ + cos²θ-cos²(θ) +sin (θ) / ( sin²θ + cos²θ+sin(θ)) cos(θ)

this is where i got

[shy valley]

I thought darkside already helped you

did it not work?

[old rapids]

he said he was going to be back in 10

half an hour passed

and its 10 pm i need to finish this or i am a dead man

this is the first of 10

questions

[shy valley]

first turn the two fractions into one

[shy valley]

you're overcomplicating it here

[old rapids]

ok

[shy valley]

then write everything on the top side in terms of sin

[old rapids]

it is already one fraction

[shy valley]

yeah but you have 1-cos^2+sin

on the top

1-cos^2=sin^2

sub that in and simplify

[old rapids]

sin'2(x)+sin(x)

on the top

[shy valley]

take sin out as a factor

[atomic blade]

$\frac{1}{\cos{x}} - \frac{\cos{x}}{1 + \sin{x}} = \tan{x}$

Oh

That's great

[old rapids]

nice one

i understand it

hahahaha

[shy valley]

I wish I could use latex

[old rapids]

math on text is something else to understand man

next level stuff

[atomic blade]

This

[old rapids]

?

[glossy valve]

Umbraleviathan

[atomic blade]

Is that correct

For what you're trying to prove?

[old rapids]

i dont think so

it's an identity

we need to show that left side is equal to the right

[atomic blade]

Can you send a picture then

[old rapids]

first one

[atomic blade]

I just used x instead of θ cause I don't wanna switch between English and Greek keyboards lol

Well, the goal of tan is to get sin/cos, correct?

So I would make the left side into one whole fraction

@old rapids I can prove it

But first, to make shit easier for latex:

a = sin(θ)
b = cos(θ)

First, make the left side into an entire fraction

[old rapids]

yes

you move tan to the left

or sin over cos

since tan = sin/cos

[atomic blade]

Wait no

[atomic blade]

That's just for conceptualization; don't touch the right side

Just make the left side a big fraction

[old rapids]

ok boss

cross multiply?

nevermind

one sec

Is it this?

[atomic blade]

Yeah

[atomic blade]

Notice something you can do?

[old rapids]

hmmm

open the brackets

[atomic blade]

Well no

[atomic blade]

Look at the numerator

Notice any identity you can use?

[old rapids]

sinx +1?

you use an identity there?

[full forum]

@old rapids Has your question been resolved?

[atomic blade]

what is $1 - \cos^2{x}$

[glossy valve]

Umbraleviathan

[old rapids]

sin2x?

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[atomic blade]

I'll claim this for you

[atomic blade]

sin^2 (x)

But yes

[atomic blade]

THIS IS THE QUESTION, I JUST RESERVED IT FOR HIM

It'll be sin^2 (x)

You then can factor sin(x) in the numerator

[gritty rose]

you can use .reopen if their question was the last one that was closed

bot is bad sometimes

[atomic blade]

Oh thanks

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@atomic blade Has your question been resolved?

[atomic blade]

I hate this

I think the guy went to sleep

@old rapids lmk when you get back online; for now, I'll close it

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[pure quail]

Hello

[pure quail]

Is this as easy as it sounds?

[gritty rose]

why do you think it's easy

[pure quail]

Because it's a possible function

Possibly

and -4v sounds like we're moving to the negatives or adding/subtracting by a negative

NEvermind it was correct

Thanks for your time

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[south shale]

a cylindrical pail of icecream has a diamtere of 10in and a height of 12in. You are going to scoop sphere shaped scoops of icecream with a diameter of 1.5in. How many full scoops of ice cream could this pail hold?

[south shale]

i got 554.1176

is that right

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[peak sigil]

hello! my problem is - a collection of 77 quarters and dimes is worth $12.50 how many quarters are there? i cant figure out the two problems i need. i used 25q +10d =12.50 and q+d =77 but the answer wasnt right after i checked it

[fathom saddle]

0.25q + 0.10d = 12.50

[peak sigil]

wait that makes so much more sense i forgot about that part

ty !

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[winter hedge]

aye i remember these problems

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[kindred ermine]

is there a quick way to compute 2^(51) mod 103?

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@kindred ermine Has your question been resolved?

[fresh cypress]

Hello

Try using the fact that 2^103 - 2 is a multiple of 103 :D

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[ruby helm]

does anyone know why it isn't d?

the answer is c

But I thought you take the area 36pi subtract from the intersection, which is approximately 18sqrt(3)

[vapid barn]

Are you familiar with polar integration?

If not thats totally ok

But the techniques of calculus make this problem trivial

Like really trivial

@ruby helm

[ruby helm]

no I don't know calculus

unfortunately, is it necessary to solve this problem?

[vapid barn]

Are you familiar with polar coordinates at all?

[ruby helm]

uhh, barely

I only know that it's in the form (radius, theta)

is anyone able to help?

<@&286206848099549185>

[vapid barn]

The only other way I can think of doing this is the dumb one

Can you use a calculator here?

[full forum]

@ruby helm Has your question been resolved?

[ruby helm]

no it's not calculator allowed

How were you gonna solve it with polar coordinates?

[soft pier]

ok so you have two identical circles with radius 6. they pass through each others center. that means you have 3 segments of length 6 in total. this equals to 18. now what im thinking is complementary counting, where you find the area of a rectangle around the whole thing and subtract out the unshaded circle as well as the corners around the two circles.

[ruby helm]

how do you find the area of the corners?

[soft pier]

imagine a circle internally tangent to all four sides of a square. the square has side length 12 (the diameter of the circle). using this we can find the area of the square, which is 144. next, we subtract the area of the circle, 36π, which leaves us with the area of the 4 corners, 144-36π. to find the area of, say, 1 corner, divide it by 4. this gives us 36-9π.

The only problem with this method is that the area of the middle portion isnt really solvable maybe

i think i have an idea though, let me work it out

[ruby helm]

Ok, I'm just a bit confused as where you get the square from, shoulnd't it be a rectangle?

[soft pier]

well just one circle should have a square

but the two circles should have a rectangle, yes

[ruby helm]

but are u accounting for the overlapping area if you're breaking it into two circles?

[soft pier]

no unfortunately, thats what I'm trying to solve

but i have an idea

okay so now we can make a Rectangle that contains a twelfth of both circles. this is confusing but i can try to demonstrate

so 1/12th of the circle is just 36π/12 = 3π, and we have two of those areas so that is 6π. Now we find the area of the rectangle, which is 6 * 1 (the rectangle has vertices (-3,5); (3,5); (3,6); (-3,6) )

wait actually is it 1/12 lemme check my math

yeah it is 1/12

[ruby helm]

how is it 1/12?

[soft pier]

wait no its 1/24 i believe

so 36/24 * π = 3/2 * π

[ruby helm]

how is it 1/24?

I'm confused where you got this number from

[soft pier]

we take 1/2 of the circle, and because the diameter is 12 units, we take 1/12 of the circle because its just nice

wait no im just being really stupid right now

[ruby helm]

sorry I'm still confused

[soft pier]

you should be, what i did was completely incorrect lol

I was trying to divide the circle up like you would a square or a rectangle

part where the circles intersect the square are not evenly 1/24th of a circle

[ruby helm]

oh ok

so do you know a way to solve the question?

[soft pier]

may I ask where you are getting the problem from? Just so I have more context about what methods are used to solve it

[ruby helm]

I got this from a random math competition website

[torn karma]

really bad drawing, hope it helps

[ruby helm]

it only goes up to trig

[torn karma]

if you haven't noticed already you have equilateral triangular symmetry

kind of

[soft pier]

yeah this guy seems like he is a better guy to ask

[ruby helm]

yes

you can construct two equilaterals

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[past talon]

Can I get help on these two questions?

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@past talon Has your question been resolved?

[past talon]

no

<@&286206848099549185> Can someone help me get started on 3 and 4?

[grizzled lance]

So we have $\log_6{16} = a$ and $\log_{12}{27} = b$

[glossy valve]

azeem321

[grizzled lance]

We need to express $a$ in terms of b

[glossy valve]

azeem321

[grizzled lance]

Have you learned change of base?

log change of base

[past talon]

so log 16/log6

=a

liike this?

[grizzled lance]

not sure why u converted them all to what i assume is base 10 logs

[past talon]

yeah was trying something

can you let me know the right direction?

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[dusty vector]

am i tripping?

have i forgotten how to do 8th grade math?

WAIT

WHAT

ok the area of triangle i get but area of circle what

pi r squared

.cloes

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[shadow holly]

This my question A family of polynomials has equation f(x) = a(x-6)^2(x-9)^2(x+9)^2. What is the value of a for the family member with a y-intercept of 2?

[shadow holly]

would I set f(x) = 2 and x = 0 and solve?

<@&286206848099549185>

[hot herald]

#❓how-to-get-help

[shadow holly]

I literally read it. wht did i do wrong..

[hot herald]

multiposting
multipinging

[shadow holly]

i've waited so long and yet havent got any help.

[hot herald]

despite breaking the rules, I provided you with some and then you continued to break more rules

wait patiently

would I set f(x) = 2 and x = 0 and solve?
in simpler terms, set f(0) = 2

in a single step

[shadow holly]

Thank you.

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[frigid oak]

Hi, what is that symbol in the curly bracket?

[frigid oak]

The Pk (symbol) x

[sweet ingot]

$\sqsupset$

[glossy valve]

ScapeProf

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[frigid oak]

I see, thank you @sweet ingot

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[young storm]

Why is C true? Since it's a rhombus and not a square shouldn't it not be?

[light sonnet]

Is that saying both B and C are answers?

[young storm]

Yes

[hoary summit]

Draw the line in part C

Or one of them

[hoary summit]

Should be obvious enough

[young storm]

That wouldn't work

Right?

I think the answer key is just wrong

[hoary summit]

You could just reflect the points in the line and draw the resulting rhombus

It would make it pretty clear

(But yeah pretty sure you’re right)

[young storm]

Yeah alr

,close

thx

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[tacit shale]

Hello!

[tacit shale]

Ok so it’s a word problem with Intergers

I’m stuck on problem 2 as I do not know how to find the net change of somthing

[tacit shale]

.close

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[tacit shale]

Hey so I’m working on this math escape room homework

[tacit shale]

And I’m extremely confused how to find the answer to this one

Shawna reduced the size of a
rectangle to a height of 2 in What
is the new width if it was originally
24 in wide and 12 in tall

How can I figure out the new width without the volume

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[karmic narwhal]

i dont know how to begin?

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[karmic narwhal]

.close

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[viral zenith]

Hello

[viral zenith]

I could please receive some assistance with this question

I’ve done the question but I need someone to double check my calculator work

My friend and I have the same working out but different answers

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@viral zenith Has your question been resolved?

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@viral zenith Has your question been resolved?

[stray abyss]

got the same answers 🙂

[viral zenith]

Thank you!

[viral zenith]

Hey if you're still available in the next five minutes, could I get you to look over a question you helped me with the other day? The circuit one haha

[stray abyss]

lol i'll try

[viral zenith]

Okay what do you think of this

[stray abyss]

can't say for sure about the transient state, got the same for the steady state tho

[viral zenith]

What did you get for the transient state?

I made a few algebraic mistakes

So I deleted the old pics and uploaded new ones

[stray abyss]

nono i meant we haven't learnt that yet so can't say

[viral zenith]

My new transient state is e^-2t(cost-3sint)

Oh okay

[stray abyss]

hold on

[viral zenith]

I was just told transcient state is anything attached to e^-t

[stray abyss]

oh

i'll give it a shot

[viral zenith]

I'm just checking my algebra haha

All good

I think I got it

This is my TC

[stray abyss]

yeah ig looks good

[viral zenith]

Yay I finally did a maths question correctly

[stray abyss]

lol

[viral zenith]

Thanks! 🙂

[stray abyss]

np

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[full forum]

Please don't occupy multiple help channels.

[kind jay]

what u need help with

[balmy ice]

solving the whole equation

[kind jay]

don't

open an other channel

if ur occupying one

close one of your channels

[balmy ice]

the person who was helping me left

[kind jay]

doesn't matter

[balmy ice]

ok

i wont

[kind jay]

close one channel

.close

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[thorny cape]

I need to find the derivative

[thorny current]

split it into x/3 - 1/3 and then differentiate

you'll get the answer as 1/3

[thorny cape]

no thats not right

when I split into that I get the wrong answer

[thorny current]

whats the answer in your book?

[thorny cape]

this is an assignment

[thorny current]

its 1/3

a fraction can be split at anytime cus its using a common denominator of 3

x/3 - 1/3 = x-1/3

[torn jolt]

You can start by splitting them up (x/3 - 1/3)

Then rewrite them in a better format.

[thorny cape]

(x+h)-1 / 3 (x-1) / 3

[torn jolt]

So for example, x/3 you can rewrite as 1/3x

And then differentiate normally.

[thorny cape]

x remains?

[torn jolt]

Do you know how to differentiate?

[thorny cape]

yes

this is what I did so far

[torn jolt]

Ohh you're doing it from first principles?

[thorny cape]

yes?

[torn jolt]

If the question doesn't specify that you have to do that, there's a much easier way

[thorny cape]

this is what the question wants me to do

[torn jolt]

Okay, from first principles:

[torn jolt]

You can simplify this further: (x+h-1-x-1)/3

Because both the fractions share common denominators

[thorny current]

just add up the top fraction you get h/3 , and thats being divided by h

then u use the rule h/3 / h = h/3 * 1/h = 1/3

[torn jolt]

Try not to just tell people the answers @thorny current

[thorny current]

alr

[torn jolt]

$(x+h-1-x-1)/3$

[glossy valve]

Glitch

[torn jolt]

$\frac{x+h-1-x-1}{3}$

[glossy valve]

Glitch

[thorny cape]

so x-1 and x-1 cancelled out

[thorny current]

yep there's a slight error in the second equation glitch wrote but they cancel out

the last -1 was meant to be a +1 cus the - sign distributes to x-1

[torn jolt]

Sorry my mistake, I'm getting used to using Latex 😳

[thorny current]

latex was a pain for me, i had to use it for one of my math papers thank god its over xD

[torn jolt]

Haha that sounds awful

@thorny cape What are you left with after everything is cancelled out?

[thorny cape]

h/3/h

yeah

[thorny cape]

yo

[thorny current]

divide the fraction h/3 by h

[thorny cape]

I get 3

h has visible 1 right?

so 1/3 ?

[thorny current]

yep

[torn jolt]

Sorry, I was doing something - yes that should equal 1/3^^

[thorny cape]

why is h below 3 gone?

[thorny current]

division rules

so h/3 is one fraction and you're dividing it by h again

and you can rewrite division in the form of multiplication with the inverse of the value

which is 1/h

it'll cancel it out since theres a numerator h and denominator h

[thorny cape]

do you know about this too?

[thorny current]

yep, but do you need to solve it using fundamental theorem of calculus?

[thorny cape]

same with before

[thorny current]

ok just do the same thing f(x+h) - f(x) / (h)

make sure to use f(x+h) and not just add h

[thorny cape]

okay

so

[2(x^2+h)^2-1] - (2x^2-1) / h

[thorny current]

yes exactly!

thats the numerator btw, you need to divide that by h

[thorny cape]

I forgot

[thorny current]

so first expand , simplify and then factor out h, and you'll get your answer

[thorny cape]

2x^2 +2xh^2 -1 -2x^2 + 1 ??

[thorny current]

nope

[2(x^2+2xh+h^2) -1] -[ (2x^2-1)]/ h

[thorny cape]

why did my h doubled?

[thorny current]

okay remember it was 2(x+h)^2 ?

recall the formula for (a+b)^2

which is a^2 + b^2 + 2ab

[thorny cape]

ohh I see

how do I solve that?

Can I @ you btw?

[thorny current]

@thorny cape just expand your equation carefully and you'll be able to get rid of terms

yeah sure thing

i might go to bed in a little bit tho its already pretty late over here

[thorny cape]

same btw

2xh+h^2/h

[thorny current]

nope but its close

you forgot to multiply everything by 2 on the numerator

thats the step you missed out on

[thorny cape]

4xh+h^2/h

[thorny current]

yep

[thorny cape]

how about this

[thorny current]

2h^2 btw

then you notice how 4xh and h^2 have h in common?

[thorny cape]

my final answer would be 4x

[thorny cape]

cause?

[thorny current]

yes

[thorny current]

4x(h) + 2(h)^2

[thorny cape]

4xh+2h^2/h]

[thorny current]

you have to factor it out to cancel with the denominator

[thorny cape]

is equal to?

[thorny current]

h(4x+2h)/ h = 4x + 2h = 4x

since h tends to 0

[thorny cape]

Im confused

[thorny current]

okay so you understand this part right

[thorny cape]

yes

whats after tht?

[thorny current]

h is a common term of both the things on the numerator right

[thorny cape]

yeah

[thorny current]

cus 4x * h + h^2 ?

u take h out

as a common factor

h(4x + 2h)

make sense?

[thorny current]

try multiplying this you'll see its the same as 4xh + 2h^2

you basically factor out the h since the limit is h tends to 0

if you dont cancel out the h you end up with a 0 over 0 situation

which is undefined

[thorny cape]

isn't it 4x+2h/h ??

[thorny current]

nope

4xh + 2h^2/h

which reduces to 4x + 2h

[thorny cape]

I mean after that

[thorny current]

thats it

[thorny cape]

whats after h(4x+2h)

[thorny current]

oh

you cancel with the denominator

cus its h(4x+2h) / h

[thorny cape]

(4x+2h) remains

then

[thorny current]

yes

lim h -> 0

so 4x + 2(0)

[thorny cape]

is my answer not 4x only?

@thorny current

[thorny current]

@thorny cape it is 4x

thats the right answer

[thorny cape]

why is there 4x+2 ???

[thorny current]

2 times 0

cus its 2h

h tends to 0

so you can just say h = 0

2*0 = 0

[thorny cape]

really?

[thorny current]

yep

[thorny cape]

okay thanks

[thorny current]

np

do .close so you close the channel

[thorny cape]

.close

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[silk solstice]

how go get the real part and the imaginary

[tall depot]

do you know what polar coordinates are?

[silk solstice]

yeah i wrote i^i in the polar form

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[proud shore]

how do i figure this out? without drawing a graph..

[atomic blade]

plug in the x values, solve for y

and see which ones correspond

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.reopen

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✅

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[torn jolt]

when it comes to dedekind cuts why is it that they are irrational if the larger set doesn't have a smallest element? e.g smaller set: (-infinite, 4) bigger set: (4, infinite). 4 isn't irrational? first paragraph of wiki page for context:
https://en.wikipedia.org/wiki/Dedekind_cut

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[torn jolt]

<@&286206848099549185>

[fair spindle]

hm?

[torn jolt]

^ need help understanding how the set of irrational numbers can be constructed from dedekind cuts

an example would be nice

[fair spindle]

a im not fimiliar with this, but i'll try

[torn jolt]

ight

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[torn jolt]

I am wondering how to get the initial distribution vector from this ? This is an example post from my teacher but this class ends in 1 day and I can’t ask him what it means now because it’s too late : (

[torn jolt]

I have tried so many methods for the past few hours and I’m having trouble figuring out how he got .8 and .2 as an answer

This is for discrete math btw

<@&286206848099549185>

Anyone 🥺

<@&286206848099549185>

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[torn jolt]

<@&286206848099549185>

If no one understands it’s okay I just was wondering how the initial distribution vector answer became .8 and .2 because I don’t know how to calculate that

But I truly appreciate any help ! : )

[spare jungle]

80% still maintained, 20% not

[torn jolt]

How do you get that answer🥺

From the graph

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@torn jolt Has your question been resolved?

[torn jolt]

: (

[gritty rose]

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[grand zephyr]

Does anyone know how I would do this

[grand zephyr]

i think maybe it something with first differenes or common ratio

[tender stump]

I guess I would take log on both sides
log(y)=xlog(b)
which is technically the equation of a straight line
where log(b) is the gradient

[candid island]

you'd look at the y coordinate where x = 1

[grand zephyr]

that may be correct btu we havent taught was log was

[candid island]

play around with desmos and see if you can find a pattern

[grand zephyr]

ye i think comomn ratios

cause thats what he showed

common ratio of y values of the points

and the ratio betweenthe consecutive y values is the same

its exponentaianl

[candid island]

so have you tried it

[grand zephyr]

my teacher made a chart and ye it works but

i wanted to check with others

if im right

because im not 100 percent sure

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[torn jolt]

I need help with finding the area of a circle

[torn jolt]

I wasnt there that day when we just started on area of circles

[topaz notch]

Tell me what you know

[torn jolt]

Is it okay if i just show what i was told to write down today? The problem is also part of my homework

My substitute teacher just told us to write it down

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[torn jolt]

how do i solve 11-13?

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[torn jolt]

use law of cosines to find one of the angles

after that finding the height should be straightforward using the pythagoras thm.

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[gritty rose]

Show what you've tried

[unreal hearth]

Can't get started on it

[gritty rose]

You have a half angle on the right side, so use those identities for all the terms on the left side

[unreal hearth]

I don't really understand

[gritty rose]

What part

Do you know trig identities?

[unreal hearth]

Can you give me an example please?

I might know it but don't know what it's called

[gritty rose]

https://www.onlinemathlearning.com/image-files/double-angle-formulas.png

Use those on the left hand side

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[hollow raptor]

are there any ways to find $a^ab^bc^c$ with vieta's formulas?

[glossy valve]

Darwn

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