help-23

[fleet gale]

*x²

Yeah

[weak spire]

Right

[fleet gale]

And {sqrt(1+x³) + sqrt(1-x³)} is not 0

[weak spire]

No matter what I do inside the parenthesis, the answer will still be 0 right?

x² = 0

0 multiplied by anything is 0

[fleet gale]

And x² can be cancelled be x³ on numerator

Numerator is 2x³ ryt

[weak spire]

Yeah

Then it will be 2x

[fleet gale]

Yeah so 0

[weak spire]

Yea

Ig makes sense

Denom is 2 right (although it doesn't really matter)

[fleet gale]

Yeah

[fleet gale]

0/0

Ok let me try

[weak spire]

ight

[fleet gale]

Is the x under root?

[weak spire]

Yep

[fleet gale]

Then the denominator is not 0

[weak spire]

Actually I think it should be outside the root, only that would make the denom 0

Inside it doesn't make sense

[fleet gale]

$3sqrt(2sqrt(2)) - 4$

[fleet gale]

Yeah that makes sense

Ok we will take it out side

[weak spire]

Aight

I got 8/5

Used lhop

[fleet gale]

Oh I won't use L hospital for this u can mid term the denominator here

4√2 - √2

[weak spire]

Yep

[fleet gale]

And for x⁴ - 4 = x⁴-(√2)⁴

[fleet gale]

Then (x-√2)(x+√2)(x²+2)

U can cancel the x-√2 factor

[weak spire]

x² + 4√2 x - √2 x - 8
= x(x + 4√2) - √2(x+ 4√2)
= (x+4√2)(x-√2)

[fleet gale]

It is 8/5

[weak spire]

Right

?

[fleet gale]

Yeah

[weak spire]

Aight

Makes sense

8√2/5√2, that gets us 8/5

Using l hop is no fun haha

[fleet gale]

Ryt 😂

[weak spire]

Fr

[fleet gale]

I use it when we have log 0 or something

[weak spire]

Log0?

[fleet gale]

I meant ∞/∞ form and numerator or denominator is log 0

Log 0 = -∞

Very annoying

[weak spire]

x³ - x² - 2x² + 2x - 2x +2
x²(x-1) -2x (x-1) -2(x-1)
(x-1)(x²-2x-2)

[storm herald]

could anone help me on #help-19 its a basic question but i dont understand 😭

[weak spire]

Got the denominator

What about the numerator

💀

[fleet gale]

Hmm this one may be long if L hospital is not used

[weak spire]

It will take me 10 sec if I use lhop fr

[fleet gale]

Dividing by (x-1) 💀

[weak spire]

BRO

Lets di it

Do

[fleet gale]

Haha

Ok

[fleet gale]

Bro is it statistics

[storm herald]

yeh 💀

[fleet gale]

Shit i forgot most of the formulas and stuff i need to revise it

[storm herald]

its a question on linear interpolation

[fleet gale]

Hmm

[storm herald]

i just did another linear interpolation question and i got it correct

[weak spire]

x⁷ - x⁶ + x⁶ - x⁵ - x⁵ + x⁴ - x⁴ - x³ + x³ - x² + x² - x + x + 1

x⁶(x-1) + x⁵(x-1) - x⁴(x-1) - x³(x-1) + x²(x-1) + x(x-1) - 1(x-1)

LMAOOO

[storm herald]

just that question is weirddd

[weak spire]

@fleet gale check if this is correct

(x-1)(x⁶ + x⁵ - x⁴ - x³ + x² + x -1)

I think I messed up

[fleet gale]

Hmm

Nah it's correct

[weak spire]

YOOOO

Lessgo

That shit made me sweat fr

[weak spire]

I got similar problems, can u help?

[storm herald]

now wanan try out #help-19 🙃 ??

[weak spire]

I'm not done 😭

[storm herald]

😭

[fleet gale]

Bro i don't know linear interpolation

I know standard variation

[storm herald]

ohh ok np

[weak spire]

@fleet gale ?

[fleet gale]

Yeah i just checked the LHL and RHL they are equal

If u don't want to do it with L hospital u can substitute x = a±h

Lim x -> a will become h-> 0

Then we can do binomial approximation

Wanna try?

[weak spire]

Hey can you explain this briefly? Im sorry

[weak spire]

Sure

[fleet gale]

,rotate

[flat frigate]

Couldn't find an attached image in the last 10 messages.

Oct 2, 2023, 7:03 AM

[fleet gale]

See this

,rotate

[fleet gale]

See this

[weak spire]

Yup makes sense

[fleet gale]

So separate ((a+2)+h)

Where h is very small so u know how to do binomial approximation ryt

[weak spire]

5/2(a+2)?

[fleet gale]

$((a+2)^\frac{5}{2})(1+\frac{h}{a+2})^/frac{5}{2}

$((a+2)^\frac{5}{2})(1+\frac{h}{a+2})^/frac{5}{2}$

[flat frigate]

Pinball Wizard

[fleet gale]

F

[weak spire]

5/2 right

I get it fr

[fleet gale]

Well something like this

[fleet gale]

Now h/(a+2) is very small

[weak spire]

Yes

[fleet gale]

So $1+/frac{5h}{2(a+2)}

So $1+/frac{5h}{2(a+2)}$

[flat frigate]

Pinball Wizard

[fleet gale]

I am so bad at using this

[weak spire]

1+{5h/2(a+2)}

Haha no worries

[fleet gale]

1 +

[fleet gale]

This

1 + 5h/2(a+2)

[weak spire]

Yea

[fleet gale]

And u have ((a+2)^5/2)(1+5h/2(a+2)) - (a+2)^5/2

This is numerator

And denominator is just h

[weak spire]

Yeapp

What should I do with the num

[fleet gale]

,rotate

[fleet gale]

As u can see everything falls in place

[fleet gale]

@weak spire u understand this ryt

[weak spire]

I still don't understand one part, how did u find that {1+5h/2(a+2)}part

[fleet gale]

Binomial approximation

If it is (1+x)ⁿ and x is very small u can do
(1 +nx)

So u need to make this in a form of 1+x before using binomial approximation

@weak spire ?

[weak spire]

Idk

[fleet gale]

Hmm let's start from the beginning

[weak spire]

(a+h+2)^5/2

How did u make that into that

[fleet gale]

Yeah only this term

Take out a+2

[weak spire]

(a+2)^5/2 + h^5/2

?

[fleet gale]

No

[weak spire]

I'm dumb de

Fr

[fleet gale]

Divide and multiply the factor with (a + 2)^5/2

And when u take the divided
(a+2)^5/2 term inside the power of 5/2 it will become only (a+2)

[weak spire]

Can u write it in copy

And send it

[fleet gale]

Yeah wait

Now?

aⁿbⁿ = (ab)ⁿ

Exponents rule

[weak spire]

AHHHH

Yep

[fleet gale]

Ok

[fleet gale]

Then this

U get (1+h/(a+2))^5/2 ryt

[weak spire]

Yes

[fleet gale]

Which is in the form of (1 + x)ⁿ so u can directly approximate it now

So (1 + 5h/2(a+2))

Just drop the power to the second term

Now did u understand it?

[weak spire]

Yep

Next?

[weak spire]

This

[fleet gale]

Yeah

[weak spire]

The positive and negative get eliminated

[fleet gale]

Yeah

At the end the h on denominator and numerator also gets cancelled

[weak spire]

5/2(a+2)^3/2

Righr

Right?

[fleet gale]

Yeah

U can check by L hospital it's correct

[weak spire]

Ok

This was quite the hard way

[fleet gale]

Yeah but it will enhance ur skill

[weak spire]

Yep

What's ur age?

[fleet gale]

18

[weak spire]

Ohh dem

Big bro 😃

[fleet gale]

Oh ur?

[weak spire]

15

[fleet gale]

Oh dem

[weak spire]

Wait I got another last

[fleet gale]

I got calculas at age 17😂

For the first time that is

[weak spire]

Ahhh niceeee

[fleet gale]

Ok

[weak spire]

This looks similar to the definition of derivative

Anything related to that?

[fleet gale]

Lol use the thing i just taught u

[weak spire]

Lemme try fr

[fleet gale]

Here x is the constant

[weak spire]

I sent u friend req

I'll go watch anime now

[fleet gale]

Ok

[weak spire]

[fleet gale]

U got the ans ryt?

[weak spire]

Aight lemme just find it and go fr

[fleet gale]

U got it ryt here u need to deal with (x + h)^1/2

And h is already small

[fleet gale]

just divide and multiply √x like this

[weak spire]

Ok

[fleet gale]

I mean this

[weak spire]

Hey

[fleet gale]

?

[weak spire]

Wait

[fleet gale]

U got it ryt

[weak spire]

@fleet gale hey

I got

h/√2x

💀

[fleet gale]

By h ryt

There is a h in the denominator

[weak spire]

Ohh tuck

I forgot

[fleet gale]

Here

[weak spire]

HAHAHA YEP

[fleet gale]

Lol

[weak spire]

1/√2x

?

[fleet gale]

Yeah

Nice

[weak spire]

I'd prefer to write it as (2x)^-1/2

😭

[fleet gale]

2^(-1/2)

💀

U mean (2√x)^-1

?

[weak spire]

wait i got

{h(2x)^-1/2}

(2x)^-1/2

-1/2

MB

[fleet gale]

Wait what?

Wait u got 1/√(2x)

[weak spire]

Yea

[fleet gale]

,rotate

[fleet gale]

How did u get √2 tho

[weak spire]

Is the exponent

Is only of x

And not 2x?

[fleet gale]

(x + h)^1/2 here

Give me ur solution

[weak spire]

Wait

Shouldn't it be 1/2-1?

Instead of 1-1/2

[fleet gale]

Yeah but 2x is not the constant here here constant is x

There only 2^-1 (x)^1-1/2

[weak spire]

Ahh right

Looool right

[fleet gale]

Ok go watch ur anime

Cya

[weak spire]

💀

[fleet gale]

Don't forget to close this

[weak spire]

Yup

I'll dm u

If i

Have doubts

.close

[safe radish]

Channel closed

Closed by @weak spire

Use .reopen if this was a mistake.

[fleet gale]

Ok

[safe radish]

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[urban eagle]

How do I do 16

[reef estuary]

split it up into 4x/sqrt(1-x^2) and 5/sqrt(1-x^2)

for the first one u = x^2 is quite simple

second one is a standard integral

[urban eagle]

Ohh

[safe radish]

@urban eagle Has your question been resolved?

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[pulsar pecan]

Need help

[pulsar pecan]

Leave c be a positive real number. Show with induction after n the following statement:

ignore the first sentence

[tiny wraith]

Can you assume that ln(a * b) = ln(a) + ln(b)?

for positive real numbers a and b

[pulsar pecan]

sure

[tiny wraith]

Have you done the base case?

[pulsar pecan]

yes

base case is ln(c^0) = 0

[tiny wraith]

Yes

Now you need to prove that if it's true for some n = k, then it's also true for n = k + 1

So suppose ln(c^k) = k * ln(c), what can you say about ln(c^(k+1))?

[pulsar pecan]

mhm

[pulsar pecan]

(k+1)*ln(c)

[tiny wraith]

Yes, so the identity holds for n = k +1 and you are done

[pulsar pecan]

explain what you mean with identity holds

[tiny wraith]

Did you do this step with algebra?

[pulsar pecan]

yeah

because ln(c^a) = a*ln(c)

[tiny wraith]

that's what we are trying to prove, so you can't use that

[pulsar pecan]

well I know that its (k+1)*ln(c)

but idk how to proceed. not sure you mean with identity holds

[tiny wraith]

The identity is what you are trying to prove, so ln(c^k) = kln(c)

[tiny wraith]

You need to prove using other rules that this is (k + 1) * ln(c)

*and using the inductive hypothesis

[pulsar pecan]

hmm

not sure

[pulsar pecan]

I need to prove that k+1 * ln(c) = ln(c^k)?

or what

[tiny wraith]

No, that if ln(c^k) = k * ln(c), then ln(c^(k+1)) = (k + 1) * ln(c)

[pulsar pecan]

well i've already proven that

i mean

its quite obvious

[tiny wraith]

What did you do then?

[pulsar pecan]

well we just use the same rule for k as k+1

[tiny wraith]

Do you know how induction works?

[pulsar pecan]

Barely

Im learning it today

[tiny wraith]

If you want to prove something for every n, then first prove it for n = 0 and then prove that if it is true for some specific k then it is also true for k + 1. That way if n = 0 is true, then n = 1 also works since 0 + 1 = 1, and if 1 works, 1 + 1 also works, and 2 makes 2 + 1 = 3 work etc...

[pulsar pecan]

Right

so we have a case with ln(c^k+1)

and we can put (k+1) outside

because thats the rule for ln

[tiny wraith]

That is the exact rule you have to prove so it doesn't work

[pulsar pecan]

mhm

[tiny wraith]

You have only proven it works for 0

[tiny wraith]

This is what you need to prove

Note that it is given that the rule does work for k, but you have to prove it works for k + 1

[pulsar pecan]

mhm

[tiny wraith]

Try writing c^(k + 1) as c^k * c

[pulsar pecan]

ln(c^k*c)

mhm

.close

[safe radish]

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@wind siren Has your question been resolved?

[safe radish]

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[fickle trail]

This demonstrates F(b) - F(a) for definite integrals with x=0 as the reference point

But what about curves that are both positive and negative on the x axis?

[lean otter]

What is the question?

May I help? :>

[fickle trail]

We are doing b - a
(0 to b) - (0 to a)

Does that make sense? But what if the graph starts at -2 and ends at 10

[hasty wagon]

yes it does ~

[lean otter]

Can you please give me all the details of what you are doing and place the questions?

[hasty wagon]

since the bounds would be reversed

[fickle trail]

Is this guy trolling?

[lean otter]

Yes

I can help. Sorry for joking LOL.

[hasty wagon]

so, it will be F(a) as in negative area

from 0 to a

e.g.

$\int_0^{-2}$

[flat frigate]

Biscuity

[fickle trail]

https://tenor.com/view/mods-discord-polo-g-gif-19016309

[lean otter]

yall talking about notation?

haha

[fickle trail]

Oh so I still use x=0 but I take absolute value from the left?

[hasty wagon]

hmmm

[fickle trail]

Or I start at the negative x value rather than x=0

I will graph it

[hasty wagon]

because according to your notations
we have [0,b] and [0,a]

so, [0,a] will be like reversed

and note that

$\int_0^{-2}f=-\int_{-2}^0f$

[flat frigate]

Biscuity

[fickle trail]

Assuming a = -2

[hasty wagon]

yea

[fickle trail]

Or do we have to take are from left and right with this

Using x=0

[hasty wagon]

if it's this case

it's werird

a is negative on yhe first, and positive on the third

[fickle trail]

Yeah

We want absolute area

[hasty wagon]

well

since you were using different a's, I can't tell anything about that

-2 was like a relative point just like x=0 back there

you can use -1, 1, 3, etc.

and still get the same result

[safe radish]

@fickle trail Has your question been resolved?

[safe radish]

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[wind abyss]

this is vertical shrink?

[cunning shard]

absolute value

[wind abyss]

since the number is <1

Oh

[quasi bison]

it's a vertical flip followed by a stretch by a factor of 2

shrinking means numbers between -1 and 1

[wind abyss]

ohhhh Ok

thnx

.close

[safe radish]

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[cunning shard]

.i read what is that vertical line 🥹

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[tidal island]

-3+√ x-1=-x+10

[cosmic grove]

hmm ?

[frigid locust]

solve for x i think?

[tidal island]

the radical is x-1

i got this 0=x^2-25x+170

[lean otter]

shwo your work

[frigid locust]

if u got this , just use quadratic

[tidal island]

the quadratic give me error

[cosmic grove]

,w x² -25x +170

[flat frigate]

Results provided by WolframAlpha

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

Requested by herels

[cosmic grove]

there is no solutions in R

[tidal island]

yea, but i got the solution and it has

[safe radish]

Show your work, and if possible, explain where you are stuck.

[tidal island]

just dont know where did i do wrong

[cosmic grove]

then you did a mistake

[worthy hemlock]

Then show your work

[tidal island]

√ x-1=-x+13

x-1=(-x+13)^2

x-1=x^2-26x+169

Oh wait

[worthy hemlock]

You went from -26x in your work to -25x here

[cosmic grove]

for x>=1, we have
√(x-1) = 13-x (x<=13)
x-1 = 13² - 26x + x²
x² -27x +13² +1 = 0

[tidal island]

just find it

well, that's it tysm

.close

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[rotund wolf]

Consider two infinite sets F and J. The question wants us to create a function f(x) that maps values from F to unique values that appear in J

[rotund wolf]

for the first part, they want us to create a function that maps all the prime numbers to the composite numbers

would this just be something like f(x) = 2x

[safe radish]

@rotund wolf Has your question been resolved?

[rotund wolf]

<@&286206848099549185>

[bold sedge]

What about 2 it’s prime

And wouldnt be unique right?

[rotund wolf]

nah like the input of f(x) would be 2

giving 4

which is a unique composite no.

the x values are the element of the set of infinite primes numbers

[bold sedge]

unique in what way

[rotund wolf]

like f(3) shouldn't be equal to f(4)

or something like that

every single input should give a different output

[bold sedge]

Ah then yeah

That should work

Does it want you to prove it

[rotund wolf]

nah it just wants a function

but if you wanted to prove it

how would go about that

[bold sedge]

Well you could define the function is linear

And then show that every element multiplied by 2 is a composite

[rotund wolf]

you wouldn't go term by term right

cause there's infinite terms

[bold sedge]

I meant via induction

[rotund wolf]

oh wow

[bold sedge]

n + 1 + ….. 1_n

Yeah you could prove it by showing the elements applied to the function never decrease and that the slope is not exponential

[rotund wolf]

okay this is kinda going over my head

so im not confuse myself more

[bold sedge]

Fair enough

[rotund wolf]

I have one last question, would there theoretically be infinite amounts of functions

because, wouldn't f(x) = 3x work also?

[bold sedge]

Yes

[rotund wolf]

ah cool

[bold sedge]

Any f(x) = Ax function should work

Since A can only be even or odd

Assuming A ∈ ℕ

[rotund wolf]

yeah cause they give multiples

which can't be prime

[bold sedge]

Oh my bad

I meant A ∈ ℕ | A ≠ 1

[rotund wolf]

because you want to actually generate something lol

[bold sedge]

Ye

[rotund wolf]

nice nice

thanks for all the help

[bold sedge]

Np

[rotund wolf]

.close

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[tidal island]

Let x and y be the measurements of the sides of a rectangle inscribed in a circle of diameter 2cm.

[safe radish]

Please don't occupy multiple help channels.

[tidal island]

a)Check that the area of the rectangle, as a function of x, is given by the expression

[lean otter]

#help-18

[icy lance]

dont advertise

[tidal island]

<@&286206848099549185>

.close

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[frigid spruce]

a,b,c are in hp then 1/a,1/b,1/c are in ap, common difference of this sequence would be 1/b-1/a= a-b/ab, so can we say that common differnece of the hp sequence would be ab/a-b

[frigid spruce]

.close

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[north stag]

A Level Pure Maths
I have no clue what this is even asking me, ive dont know where to begin and my teacher has never shown me this before

[lean otter]

i think the question is for which values c the equation doesn't have a real root

try taking everything on one side of the equality then calculate the discriminant

[north stag]

how can i calculate the discriminant for that

[lean otter]

wdym

[north stag]

ive been given the solution but i just cant follow it

like i dont get why any of it is done

[lean otter]

if the discriminant is positive then the equation has a real solution

[normal moss]

2* it's a quadratic

[normal moss]

Do you know when a quadratic equation doesn't have real roots?

[north stag]

When it is less than 0

[normal moss]

When what is less than 0

[north stag]

the discriminant

[normal moss]

Yes.

We've got $\frac{3}{x} + 5 = -2x + c$

[flat frigate]

USS-Enterprise

[normal moss]

We gotta get rid of fractions, so we multiply by x

We get $3 + 5x = -2x^2 + cx$

[flat frigate]

USS-Enterprise

[normal moss]

Rewriting, $2x^2 + (5-c)x + 3 = 0$

[flat frigate]

USS-Enterprise

[normal moss]

Now just solve when D < 0

$a = 2, b = (5-c), c = 3$

[flat frigate]

USS-Enterprise

[north stag]

thats an annoying B to have

[normal moss]

There to mess with you

🙃

[north stag]

i love maths

ill try and go from here, thank you

[normal moss]

No problem 🙂

Not sure why they equaled D to 0, because when it's 0 there's 1 (double) real solution nevermind I get it

[north stag]

what happens to the X after the (5 - c)

do we just leave that

wait nevermind sorry dumb question

lol that went right over my head

[normal moss]

Quadratic equation is in the form of $ax^2 + bx + c = 0$

[flat frigate]

USS-Enterprise

[normal moss]

Same reason you leave x^2 at a

😄

[north stag]

yeah

do i expand the B after i put it through the equation

ive left myself with c^2 - 10x + 1 = 0 and i feel like that leads nowhere

im assuming this is a rearrange

ok i rearragned to it but i dont get why it is a +/- instead of just a -

ok i got informed by another friend on why, thanks for the help

.close

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[fickle trail]

am I overthinking this?

it's literally just copy and pasting the entries into each box?

oh but it's first row = x
second row = y
third row = z?

[quasi bison]

no, the variables correspond to columns.

but yes other than that it is just copy paste.

[fickle trail]

so
-1 2 3 7
for the first row

[quasi bison]

-1, 2, 3 for the first row of the matrix on the LHS, and 7 for the first entry of the RHS.

but yes

[fickle trail]

OK

ty

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[potent moth]

Can someone help ☠️☠️

[lean otter]

Count the lines

[potent moth]

what lines

[lean otter]

there are 20 40 lines on the graph

[potent moth]

thers only 1 line or am i trippin

[lean otter]

Grid lines

[potent moth]

the U shape oner

ohhh

[lean otter]

You have lines coming off the x-axis

and lines coming off the y-axis

[potent moth]

ok now wat

[lean otter]

$c(2)$ means $x = 2$

[flat frigate]

please request a new nickname

[potent moth]

wtf

[lean otter]

so you would count 2 lines to the right

[potent moth]

ohhhhh

[lean otter]

then you would go up or down untill you intercept the curve

at that point you should intercept another line that comes off the y-axis

count how far up or down that line is

that will be your answer

[potent moth]

wait

stay here

im gonan couint it

so i count -10,10,1

[lean otter]

Ignore that part of the question

[potent moth]

ok

where do i start counting from 💀

[lean otter]

$c(2)$ means $x = 2$

[flat frigate]

please request a new nickname

[potent moth]

ohhhh

ok i get it

ok

so i count from 2 until i hit the liune

[lean otter]

no?

[potent moth]

tf

[lean otter]

do you know how to draw the line x = 2?

[potent moth]

hollon lemme se

isnt it just a vetical line

[lean otter]

yes

You would go (up or down) that line, counting the number of horizontal lines you cross untill you reach the curve

[potent moth]

so i go down 3

[lean otter]

Correct

If you go down 3

that means

y = ?

Fill in the ?

[potent moth]

3

[lean otter]

no

[potent moth]

omg

[lean otter]

y = -3

[potent moth]

-3

yeaaa

[lean otter]

because its down

[potent moth]

i forgoe

so 1 would be -3

[lean otter]

he forgor 💀

Yeah,

you solve all of those problems the exact same way

[potent moth]

i got a 6/50 on my test

[lean otter]

skill issue

[warm oyster]

outstanding !

[potent moth]

wait what about 3

what do i do there

[lean otter]

i did not see 3

Do the same thing to fin the c(x) s

fill in their spots with the y values you find

then evaluate the expression

[potent moth]

but theres expontents

[warm oyster]

is there a derivative version of the super hexagon

[lean otter]

you know how to use exponents right?

[potent moth]

ithink

[lean otter]

so what's the problem?

[potent moth]

so the asnwer i just do the exponent

[lean otter]

no

[potent moth]

whe i get the answer

[lean otter]

It's a math expression

you evaluate the entire expression

[potent moth]

my head is hurting

[lean otter]

c(x) = y

[potent moth]

ok

[lean otter]

so

[potent moth]

so it would be

fopr 2

for 3

[lean otter]

$(y_1)^4 * (y_2)^2 = ?$

[flat frigate]

please request a new nickname

[potent moth]

ohhhhhh

wait let me go in mathwya

[lean otter]

????

[potent moth]

for calculator

[warm oyster]

anyways

is there a derivative version of the super hexagon

it would save braincells

[lean otter]

why not

[warm oyster]

OH MA GAAAD

show me it

i need it

[potent moth]

the answer is

2916

wait npo

[lean otter]

what are you even talking about boss

[warm oyster]

so you know the super hexagon

[lean otter]

My advise is to forget about the super hexagon

It's handicapping you

[warm oyster]

oh ....

[potent moth]

is the answer for 3 2916

[warm oyster]

but it works really good

[lean otter]

what is c(2)

what is c(-1)

[potent moth]

-3

6

c(2)=-3

c(-1)=6

[lean otter]

You are correct

Good job

[potent moth]

yippee

thank u uuu

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[icy tide]

https://cdn.discordapp.com/attachments/905285721886187580/1158488845705289898/image.png?ex=651c6e50&is=651b1cd0&hm=4e0b81e672ac56155ac2f21874260deef52aa841c03bc72b147cfb618f05c4ff&

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@icy tide Has your question been resolved?

[icy tide]

can anyone helop

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@icy tide Has your question been resolved?

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[keen furnace]

I’m stuck on finding out the value of the slope and the values of B and C.

[safe radish]

@keen furnace Has your question been resolved?

[keen furnace]

<@&286206848099549185>

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@keen furnace Has your question been resolved?

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[vocal cliff]

How would I find the domain of f(x) on [0, 2]

[lean thorn]

it depends on what f(x) is

[vocal cliff]

average rate of change

sorry, thanks

[lean thorn]

the formula for average rate of change for interval $[a, b]$ is $\frac{f(b) - f(a)}{b-a}$

[flat frigate]

MellowDramaLlama

[lean thorn]

oh looks like the bot is down

[vocal cliff]

but if I do that, don't I get one?

[lean thorn]

depends on the funtion

[vocal cliff]

it said the answer was 5

Idk

[lean thorn]

what is f(x)?

[vocal cliff]

just f(x) on [0, 2]

[lean thorn]

do you have a graph

[vocal cliff]

nope

[lean thorn]

that doesn't make sense

[vocal cliff]

it's a written question

[lean thorn]

can you post a screenshot of your question?

f(x) can be anything

[vocal cliff]

give me a moment

[lean thorn]

,rccw

[lean thorn]

so this is part c, what are parts a and b? Do they define f(x) previously?

[vocal cliff]

unrelated

no, they don't define

oh wait

I think I got it

thanks!

[lean thorn]

lol, was it defined above?

[vocal cliff]

yeah lmao there was a table, it just didn't look like it applied

that's my bad

[lean thorn]

gotcha

no worries

[vocal cliff]

thanks again

[lean thorn]

sounds like you got it from here

[vocal cliff]

yeah lol

.close

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[delicate trellis]

is this correct?

[safe radish]

@delicate trellis Has your question been resolved?

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[low orbit]

is this simplification incorrect?

[low orbit]

shouldnt it be negative at the bottom?

[static flame]

Yeah

[low orbit]

cool, thanks

+close

.close

[static flame]

it’s .close

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[compact bough]

how do i do this

[static flame]

try doing polynomial long division with $x^4 - cx^3 + 7x -6$ and $x - 2$, keeping $c$ throughout as a placeholder

[flat frigate]

FirstRuns

[static flame]

Then you might see

[safe radish]

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[pulsar sorrel]

don’t know where to start

[pulsar sorrel]

bottom one

[static flame]

What’s the angle of a straight line

[pulsar sorrel]

0?

[static flame]

it’s 180° lol

[pulsar sorrel]

oh yea whoops

so where do i go from there ?

[static flame]

So you have two angles that split a straight line

so their sum has to equal 180°

[shadow epoch]

Same thing you did with the 90 degree angle

[pulsar sorrel]

add them set them equal to 180

[static flame]

yeah

[shadow epoch]

Yup

[pulsar sorrel]

thanks again you guys better than my teacher

[shadow epoch]

Lmk what u get as the answer

[pulsar sorrel]

x = 10

[static flame]

Yeah

[pulsar sorrel]

for x i just set then equals to each other and for y i plug in the x right

[shadow epoch]

Yup

[pulsar sorrel]

then i set them equals to each other for the y?

[shadow epoch]

Well once you find what x is you can find out the measure of the angles

And all of the angles will add up to be 360 degrees

[pulsar sorrel]

and because they’re vertical both angles are the same right

[worthy hemlock]

You don't need this

Once you find x, using the concept of vertical angles

Then you can apply supplementary angles concept

[worthy hemlock]

x i just set then equals to each other
Yes

Then you can apply supplementary angles concept

[pulsar sorrel]

140 + 8y = 180?

y=5 is what i got

[worthy hemlock]

What did you get for x?

[pulsar sorrel]

14

[worthy hemlock]

Good

[worthy hemlock]

How did you get this equation?

[pulsar sorrel]

i added the y equation twice after i plugged it x

70 + 4y

[worthy hemlock]

Why twice?

[worthy hemlock]

.

[pulsar sorrel]

i thought that i would need to add it twice to get 180

[worthy hemlock]

The 5x + 4y is on a straight line with the other two angles

[worthy hemlock]

No, because if you did that, that's assuming you have two of 5x + 4y, which you don't

[pulsar sorrel]

but isn’t it a vertical angle?

[worthy hemlock]

With what?

[pulsar sorrel]

ooh

i see

[worthy hemlock]

Vertical angles is opposite angles are equal

Does 5x + 4y have an opposite angle that is given?

[pulsar sorrel]

no

i don’t why i thought that

if you don’t mind could you explain the supplementary angles concept

because i don’t know what you mean by it

[worthy hemlock]

So supplementary angles add to 180

Or a straight line

You have 5x + 4y, it is supplement with 5x - 20 or with 3x + 8

[pulsar sorrel]

i can use either one?

[worthy hemlock]

Yes

[pulsar sorrel]

so add then set equal to 180

[worthy hemlock]

Yes

[pulsar sorrel]

alright i think i just need to go over the vocabulary of the types of angles

Thanks

[worthy hemlock]

Did you find the value of y?

[pulsar sorrel]

15

[worthy hemlock]

Yes

[pulsar sorrel]

how to i find angle 2?

is it 43?

or 47

[worthy hemlock]

Vertical angle concept

[pulsar sorrel]

so the same at the angle on the left

[worthy hemlock]

Yes

[pulsar sorrel]

what is complement and supplement?

[worthy hemlock]

I told you supplement already

Complement/complementary angles add to 90

[worthy hemlock]

If you don't know any terms, you should try googling it

[pulsar sorrel]

alright

thanks for the help i understand most of it now.

[safe radish]

@pulsar sorrel Has your question been resolved?

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[barren ledge]

How do I go about his question

I did M=n/v

and assumed the mol ratio was 1:1

but that didn't work

[safe radish]

@barren ledge Has your question been resolved?

[safe radish]

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[quasi bison]

why should it be

[slate thorn]

doesnt R^n notate there should be n collumns?

[quasi bison]

no

R^n just means the dimension of the space is n

[slate thorn]

why is that information useful then?

[quasi bison]

there's nothing illegal about considering a set of vectors which has more vectors than the dimension of the space

your matrix will be 3 by 4

tho tbh you would not really do it much differently if it said that the a_i live in R^100

[slate thorn]

oh so is R the rank?

[hard crest]

R is the set of real numbers

[slate thorn]

would it be 100 by 4?

[quasi bison]

yes in that case A would be 100 by 4

[slate thorn]

ty

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[paper kernel]

could anyone help with a question pls

[paper kernel]

its a question that i need to use further trig

[stuck karma]

You can write ADB in terms of r and theta.

And BC too.

[paper kernel]

how?

[stuck karma]

theta is in radians right?

So from the circular sector arc formula r*theta

And BC is OB * tan(theta)

Hence, ADB and BC can be written in terms of r and theta.

The sum is L, divide ADB+BC gives L/r

[paper kernel]

how dod u get this part?

[stuck karma]

Oh

Let me draw it on paper.

[paper kernel]

ty

[stuck karma]

Oops, I forgot the negative sign.

Sorry.

[paper kernel]

where should the negative sign go?

[stuck karma]

BC ×-tan(theta) + r × theta

BC = r

So we get r(theta - tan(theta))

and this equals to L

[paper kernel]

shouldnt this

be BO= BC/-tan theta

[stuck karma]

Ah wait.

Okay, here we go.

[paper kernel]

how would we get the answer tho

[stuck karma]

'ere.

[paper kernel]

ty

could you help me with partc b and c too?
]

[stuck karma]

Sure.

[paper kernel]

ty

im just plain confused on these two qeustions

[paper kernel]

thank you

.close

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[rough flower]

how would i find the exact perimeter of this sector?

[dapper venture]

what's the circumference of a circle of radius 10?

[rough flower]

20pi?

[dapper venture]

now the sector is a semicircle

what would be the length

[rough flower]

idk soz

[dapper venture]

it's half a circle

[rough flower]

ooo wait i got it

180/360 ?

[dapper venture]

so the length of the arc would be?

[rough flower]

is it 10pi?

[dapper venture]

also remember to add the diameter for perimeter

[rough flower]

10pi + 20

[dapper venture]

ccorrect

[rough flower]

ooh thx sm : D

[dapper venture]

no problem

[rough flower]

.close

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[paper kernel]

help pls

[paper kernel]

i need help with part a and b using furter trig

[hasty wagon]

what have you tried?

[paper kernel]

i used s= r theta to find arc length

but it was uselss so not im just stuck

ive tried to find other angles and other sides with no luclk too

[hasty wagon]

i see, did you find the radius of the smaller Quadrant?

[paper kernel]

yes

well dct

[hasty wagon]

let the radius of the smaller Quadrant be r, then length of PQ=√2 • r

oh wait

lemme try again

Let radius of the big Quadrant be R
Let radius of the small Quadrant be r
then
PB=R-r
using the right angled triangle PBC, we know that
tan(angle BCP)=(R-r)/r

similarly for triangle QDC, we have
tan(angle DCQ)=(R-r)/R

then angle PCQ=π/2-angleDCQ-angleBCP

[paper kernel]

sorry im confused

[hasty wagon]

lemme try to draw it out

[paper kernel]

ok

[paper kernel]

ok ty

how would we solve for theta tho

if we have two unknowns

[hasty wagon]

you can find r here

hint: Pythagorean

[paper kernel]

ah

1-r aquared +1 squared=1?

but that would give us 0

which wouldnt be possible

[hasty wagon]

consider the diagonal of the square

[paper kernel]

diagonal of square is 1.41

[hasty wagon]

yea

[hasty wagon]

and it equals to?

[paper kernel]

uh

[hasty wagon]

red dotted line + blue dotted line

[paper kernel]

oh

how did i not see that

omg ty

.close

[safe radish]

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[serene quest]

help

[serene quest]

Good morning or good afternoon, I need you to help me solve this exercise step by step applying derivatives