help-0

[formal dust]

R, mb.

[carmine reef]

ah ye

that's right

you can replace 1-P(R) with P(L) + P(W) if you want

since P(L) + P(R) + P(W) = 1

doesn't matter either one is fine

[formal dust]

i'm saying that P(R) is (1- (b+1/6))(c/6)

[carmine reef]

i dont think it would be a product like that

[carmine reef]

the benefit of this is P(W) is the easiest to calculate probably

P(R) and P(L) probably won't be expressible as a single function

it will be a matter of whether c > b+1 or not

[formal dust]

let ℓ simply mean not rolling within the success range and ℜ simply mean rolling within the reroll range

R = ℓ ∩ ℜ

So P(R) = P(ℓ ∩ ℜ) = P(ℓ|ℜ)⋅P(ℜ) = P(ℓ|ℜ)⋅c/6

[carmine reef]

you can't multiply here because the events aren't independent

oh

l given R

then its ok

[formal dust]

hmm yea that conditional probability seems impossible to simplify

[carmine reef]

ye

P(L) can be expressed using the min function probably

[formal dust]

since the two events that it relates are not determinately related

[carmine reef]

7-c is the largest number to get rerolled

so 6-c is the smallest number not to get rerolled

and 5-b is the smallest number that's still a success

so P(L) = min(6-c, 5-b)/6

[formal dust]

wouldn't that be 6-b?

An individual roll is called a “success” if it falls within the range {6, .., 6-b}.

or you mean smallest number that's not a success

[carmine reef]

oh i mean largest for both

so 6-c is the largest number not to get rerolled
and 5-b is the largest number that's still a success

[formal dust]

5-b is not a success

I understand the logic if you meant smallest that's not a win/roll for both, because you're finding the probability of getting the range of the die that is neither a win nor a reroll

your equation makes perfect sense to me, just not your current explanation of it

P(W) = (b+1)/6

P(L) = min(6-c, 5-b)/6

P(F) = P(L)/(P(L) + P(W))

P(F) = [min(6-c, 5-b)/6]/[(min(6-c, 5-b)/6) + (b+1)/6]

answer: P(at least one success with a die) = 1 - [[min(6-c, 5-b)/6]/[(min(6-c, 5-b)/6) + ((b+1)/6)]]^a

[lone heart]

@formal dust Has your question been resolved?

[carmine reef]

not a success ye

Largest number that's not a success ok

I'm not sure I understand the product in the denominator

[formal dust]

i mean an addition

ahhh

i make about as many trivial errors as i do breakthroughs :p

wait no

i make significantly more trivial errors

case in point

[carmine reef]

Yeah I do a lot too

But yeah this looks right

[formal dust]

@carmine reef well i get this for some random values

but b=2 should guarantee that the probability of success for an individual coinflip is 1/2

c being greater than 0 should ensure that the probability for any success is greater than 1/2

[carmine reef]

1/2 is right

if b=2 then c=0,1,2,3 are all the same

1,2,3 means failure without reroll

You're only rerolling when you've already succeeded

[formal dust]

@carmine reef all makes sense

thank you so much for your time

i gotta remember that if c ≥ b + 1, then you only reroll sucesses

[carmine reef]

less than or equal

but yes

you're welcome

[formal dust]

yeah thats right

[formal dust]

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[orchid sparrow]

Can someone help me to solve this

[vague mirage]

Translate please

[orchid sparrow]

pro means "for"
výsledek means "solution"

[vague mirage]

Oh

Ok

First find f'(x)

$\frac{d}{dx}(√x)+\frac{d}{dx}(e^x)+\frac{d}{dx}(3)$

[ocean seal]

Arnab Pal

[vague mirage]

First solve this

And then you have to find the derivative of those terms again'

You could just write

$\frac{d²}{dx²}(√x)+\frac{d²}{dx²}(e^x)$

[ocean seal]

Arnab Pal

[vague mirage]

And you know the derivative of 3 is 0 cause it's a constant

[vague mirage]

Do you need help in this?

[orchid sparrow]

I think I do not know how to make the second f''(x)

[vague mirage]

You have to find f'(x) or the derivative of f(x)

And derivative it against

[orchid sparrow]

Alright I will try it, thank you for your time

[vague mirage]

Again*

[vague mirage]

Yeah welcome 🫂

[lone heart]

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[hard garden]

what is the difference between a vector span that is homomorphism and one that is isomorphism?

[vale wigeon]

that seems to be a little nonsensical

are you sure no terminology is being mangled here?

[hard garden]

let me check

[vale wigeon]

"a vector span that is homomorphism" sounds like nonsense.

[hard garden]

oh sorry, i meant linear transformation

[vale wigeon]

where was it that you were checking, exactly?

"homomorphism" is a word that, in the context of linear algebra, bears no different meaning than simply "linear map"

an isomorphism is a linear map which has an inverse, or equivalently a linear map which is both injective and surjective.

[hard garden]

interesting. In fact i dont understand what these terms mean in the context of linear algebra. i was looking in the textbook but it is filled only with endless definitions and proofs

[vale wigeon]

these terms
which terms?

[hard garden]

homomorphism and isomorphism

in my notes we say that a linear transformation from V to W is isomorphism if it is homomorphism and a bijection. but the thing is that i cant wrap my head around this definition. i dont understand it

[vale wigeon]

for linear algebra "homomorphism" is just fancy speak for "linear map"

bijection means injective-and-surjective

[hard garden]

ok, i understand. one thing is bugging me. then, what is the difference between linear transformation and a linear operator?

[vale wigeon]

"operator" is sometimes used for maps from a vector space to itself

but this usage is neither universal nor mandatory

if you say "a linear map T : V -> V" this is in no way wrong

[hard garden]

oh ok. So linear operator is just another way of saying linear transformation

[vale wigeon]

no, they are not quite synonyms

when people say "linear operator" they typically mean that its domain and codomain are the same space

[hard garden]

ahh make sense

so when people say "linear operator" this is like working and staying in the same domain?

[vale wigeon]

kind of

[hard garden]

got it

thank you!

[lone heart]

@hard garden Has your question been resolved?

[lone heart]

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[dim root]

guys

[dim root]

lim f(x) = lim f(1-h) = lim 1+(1-h)^2
x → 1− h →0 h →0

how did you get from the 2nd one to the third one

[hollow sparrow]

does f(x) = 1 + x² ?

[kindred anchor]

Need more context

[dim root]

yes

[pliant cedar]

$\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)=\lim_{h\to0}1+\left(1-h\right)^2$

[ocean seal]

SilverSoldier

[dim root]

f(x) = 1+x^2

[pliant cedar]

is that what u have

[dim root]

yup

?

[kindred anchor]

Just substitue back in

[dim root]

what are you typing?

[kindred anchor]

f(x)=1+x²

f(1-h)=1+(1-h)²

[dim root]

oh

OH

how did you get 1+(1-h)^2

how did the 1 come in

[kindred anchor]

From here ^

[dim root]

ohhh

so (1-h)^2 is basically the limit of x^2?

[kindred anchor]

No

Not in this case at least.

[dim root]

?

why

[hollow sparrow]

wherever you see x change it to 1-h

[kindred anchor]

^

[dim root]

k

thx

a ton

appreciate it

[ocean seal]

Mehdi_Moulati

[dim root]

and if x tend to 1+ the h will tend to 0-

?

[hollow sparrow]

yes

[dim root]

okay

again, thx a ton

[kindred anchor]

.close if you are done with your doubts!

[dim root]

yes will do

[hollow sparrow]

np

[dim root]

.close

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[steady basin]

bit confused on why we have a small minus sign on top of the 1 in the limit

and confused on why the answer is 2

[hollow sparrow]

1/sqrt(1-x) not defined when x = 1

[steady basin]

ye i get that

but why the minus sign

on top of the 1

[hybrid heron]

It means approaching 1 from increasing side right

Like from behind

[steady basin]

so 0.9999 for example?

[hybrid heron]

Yeah

I mean that

[steady basin]

ok

and why is answer 2 then?

[hybrid heron]

I think this works this way

[hollow sparrow]

$1-1^- = 0^+ = 0$

[ocean seal]

Mehdi_Moulati

[steady basin]

oh so as u sqrt 1 - a value that gets closer and closer to 1 from behind

it tends to 0

?

[hollow sparrow]

yeah

0^+ is a very small number 0.000000000...1

[hybrid heron]

Because maybe when k = 1, 2√0 = 0

[hybrid heron]

Oh that means the 1- sign is = 0

?

[hollow sparrow]

$\lim_{k \to 1^-}-2\sqrt{1-k} + 2 = -2\sqrt{1-1^-} + 2 = -2\sqrt{0^+} + 2 = 2$

[ocean seal]

Mehdi_Moulati

[steady basin]

yh this makes sense

very logical

cheers all

[hollow sparrow]

k is tend to 1^- let's say 0.99999..9
1 - 0.99...9 = 0.00...01 = 0^+
the sqrt of 0.0000...1 is tend to 0 so we say that it's equal to 0

[hollow sparrow]

$1^- = 1 - \frac{1}{n}$ when n tend to $+\inf$

[steady basin]

i see

.close

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[worthy sparrow]

.reopen

[queen gull]

Probleem with a problem partial integratienota

[queen gull]

Integration

[limpid spade]

Assuming your previous works are correct, a way to solve the last integral is to complete the square so that the integrandum can be in the form of 1/(cx+-a)²+-b², which can be transformed to 1/u²+1, which is arctan(u)

[queen gull]

Right under is that correct of nog

Not

[limpid spade]

Can u put it into (cx+-a)²+-b² form, quite not sure what u did there

(cx+-a)²+-b² = 2x²+2x+1

[queen gull]

Can you mark whats fault

Please

Whats the b

Please help me out tomorrow exam

[limpid spade]

(cx+-a)²=c²x²+-2acx+a², right?

where do u see a x² in 2x²+2x+1?

[queen gull]

Yes

I really dont knop

Know

Im stupid you know, waking up like someone in the forest

[limpid spade]

does 1 have x² in it?

[queen gull]

Yes

I think it

[limpid spade]

how

where do you see x² in 1

[queen gull]

Ow ow i see it know

You are my savior

I really appreciate your help

[limpid spade]

so which term has x² in it?

[queen gull]

Im solving it nog

Is this correct

<@&286206848099549185>

[limpid spade]

c should be 1/sqrt(2)

$(\sqrt{2x}+\frac{1}{\sqrt2})^{2}+\frac{1}{2}$

[queen gull]

Why

can you send with $

No

[ocean seal]

Fucktalogist

[queen gull]

C has to be 1/2

[limpid spade]

2ac=2, knowing c=sqrt(2) => a*sqrt(2)=1 => a=1/sqrt(2)

my a is your c

[queen gull]

why?

[limpid spade]

why not

[queen gull]

is c = 1/2

[limpid spade]

no

[queen gull]

because wolfram said yes

[limpid spade]

its 1/sqrt(2)

wait

[queen gull]

my b is 1/sqrt(2)

[limpid spade]

^ use these letters

c=sqrt(2), a=1/sqrt(2) and b²=1/2

[queen gull]

my b is your a

[limpid spade]

yes

[queen gull]

and my a is sqrt(2)x

[limpid spade]

is your c my b?

[queen gull]

no

[limpid spade]

huh

[queen gull]

another excercises everything ok

but this excercise it breaks my head

can you write it on a paper and send me the solution so I can see where my fault is please

I thank you from the buttom of my heart

[limpid spade]

sorry amma gonna continue studyting

[lone heart]

@queen gull Has your question been resolved?

[late mango]

Take what's inside the square as the u then you derive it.
du = sqrt(2)dx implies that dx 1/sqrt(2) du

You get 1/sqrt(2) * integral(1/(u² + 1/2))

Multiply the whole thing by 2 and write the final result in the form of arctan(something) + c

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[lone heart]

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[alpine sable]

help

[alpine sable]

i want to prove that 4 divide 7^2n +3

my friend did this

7^2n+3 ≡ 0 [4]
(7^2)n
7^2 ≡ 1 [4]
(7^2)^n ≡ 1^n {4]
7^2n + 3≡ 1+3[4]
7^2n + 3≡ 0[4]

and i did this

can you guys help us

Are you familiar with proof by induction?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

no

[alpine sable]

can you just tell us wich answer is true and why

[alpine sable]

or provide us with correction

Ok your proof checks out provided you can show explicitly 7^2 is congruent to 1 (mod4)

Your friend has shown that this is equivalent to (49)^m is congruent to 1 (mod 4) but hasn’t proven this is true

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

my friend said how to prove it's true

[hollow sparrow]

\begin{align*}
(7^{2n} + 3 )\mod 4
&= ( (49)^n + 3 )\mod 4 \
&= ((49 \mod 4)^n + 3 )\mod 4\
&= (1^n + 3) \mod 4 \
&= 4 \mod 4
&= 0
\end{align*}

[ocean seal]

Mehdi_Moulati

[alpine sable]

@hollow sparrowcan you do it with congruence

[alpine sable]

Write 49 = 48 + 1 = 4 *12 + 1 is congruent to 1 (mod 4) and then we have know if x congruent to y (mod n) then x^n is congruent to y^n (mod n) this 49 is congruent to 1^n (mod 4) which is 1 (mod 4)

.close

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[alpine sable]

How do i solve this

x-y=1
a²x-y=a

[alpine sable]

I think it has no solution but I dont know how to check

[hard patio]

well it does havea solution

[echo socket]

Would recommend subtracting the equations

[echo socket]

Generally it may depend on the value of a

Let's see

Here we get that x(a^2 - 1) = a - 1

If a = 1, x could be any number
if a = -1, there is no solution for x
Otherwise x = (a - 1)/(a^2 - 1) = 1/(a + 1)

[alpine sable]

I see

Well thank you

.close

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[open roost]

how would u get the horizontal asymptote for this?

[celest phoenix]

do you mean at what x would it be?

[open roost]

na like u know when u graph it

there is a horizontal and vertical asymptote?

[celest phoenix]

yeah

[open roost]

how would u find the horizontal?

asymptote?

or would there be no way??

[celest phoenix]

well you could see the derivatives

up until the n-th derivative gets towards infinity it is a vertical asymptote

and if the derivative tends toward zero at that point it is a horizontal asymptote

[autumn vessel]

try to see limits in infty, -infty and -5

[open roost]

yea but there is an intercept at 0

[celest phoenix]

this is the graph for a high order derivative of x/x+5

[celest phoenix]

so

[open roost]

wait how did u get that graph?

this is wut i got

[outer lark]

a good way to find the singularity

[tacit arch]

You just take the limit as x goes to positive and negative infinity for horizonal

[outer lark]

is to see where the function is undefined

😸

also

a very very useful property

[tacit arch]

And find where the denominator is zero for vertical asymptotes

[outer lark]

A horizontal asymptote for f(x) produces a vertical asymptote for f-1(x).

you can swap and change those around if you want both the horizontal and vertical asymtotes of functions

as opposed to just the vertical one

[ocean seal]

riemann

[open roost]

wait so to find a horizontal asy for f(x) u would just find vertical asym it for f(x)-1?

[outer lark]

yes

[open roost]

oh k

[outer lark]

although it wouldnt be undefined, it would be 0

I think...

might have to double check

[celest phoenix]

yeah 1/0 = inf your right

[outer lark]

lets double check

,w plot x/(x+5)

,w plot x/(x+5) and (5x)/(x-1) for -2<x<2

[tacit arch]

,w inverse function x/(x+5)

[ocean seal]

Results provided by WolframAlpha

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

Requested by riemann#2529

[outer lark]

oop forgot to write the negative

still how to I fix the picture not showing everything :((

there

it does it fact work 😎

[tacit arch]

Desmos > wolf for plotting. Too bad there's no desmos bot though

[lone heart]

@open roost Has your question been resolved?

[open roost]

k thx

[lone heart]

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[indigo thorn]

can someone explain me this step by step

[alpine sable]

I dont know sorry

[indigo thorn]

.close

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[lone heart]

@trail wadi Has your question been resolved?

[trail wadi]

<@&286206848099549185>

[hollow sparrow]

there is 12 different ways

[trail wadi]

there is 12 different ways if we start at 1?

can you clarify what you mean?

[alpine sable]

Note you only need to look at the cases where the knight starts at 1,3 or 5 and then use symmetry for the remaining squares

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[rose sigil]

that sounds confused

ok you could read it like this:
for all x in A such that 0 < |x-c| < delta_1, we have |f(x) - L_1| < epsilon/2

this doesn't stand alone

yea, this is one thing

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[rose sigil]

yep ^-^

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[potent ruin]

i’d like to know if the highlighted step is allowed, i’m guessing no…

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[potent ruin]

sorry, my earlier message had a mistake in it

[marsh rapids]

No it isn't the same x at all

It's even worse when you then say that d is just du(x)

Because if you look at the first and last step of this line you basically just turned dx/du(x) into x

Which is most likely incorrect

[potent ruin]

true

thank you

[lone heart]

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[gloomy citrus]

help

[gloomy citrus]

8A

i used the formula a-b/2 (f(a)+(fb))

it doesnt work

[marsh rapids]

$\frac{b-a}{2} (f(a) + f(b))$

[ocean seal]

mateo713

[marsh rapids]

Not a - b

[gloomy citrus]

ye

but that isnt the right formula

[gloomy citrus]

to figure out this qeustion

[marsh rapids]

Well that's for one trapezoid

[gloomy citrus]

yes

[marsh rapids]

Here they ask for 3 trapezoids

[gloomy citrus]

very true

does that change the formula?

[marsh rapids]

Of course

Now you got the sum for (a, b) in {(0, 1/3), (1/3, 2/3), (2/3, 1)}

[gloomy citrus]

ohhhhhhhhhhh

right

lemme try it

[gloomy citrus]

bruh

i got 1/2

nice

[marsh rapids]

Desmos gives 0.72

[gloomy citrus]

ye thats what the answer says too

[marsh rapids]

Which matches my expectations as it gives 0.724 > 0.721

[gloomy citrus]

did u just estimate?

[marsh rapids]

Desmos

[gloomy citrus]

wait what

how did u do that

[marsh rapids]

Google desmos trapezoidal rule

[gloomy citrus]

then

i got it bro

thanks

.close

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[storm dock]

what did I do wrong? isnt the formula f(b) - f(a)?

[dawn hamlet]

The first should be the -2 and the second one -4

[abstract fractal]

b is the top number

[storm dock]

always?

[abstract fractal]

Yep

[storm dock]

ty

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[alpine sable]

Haven't started this just yet. Wanted someone to solve it step by step

[wanton nova]

You could take 3 common factor

At 6a+6a^2

And simplify it with the 3a

[alpine sable]

So that would be after or before multiplying the 6a's with the 3a

[wanton nova]

Before

[hollow sparrow]

\begin{align*}
\frac{20a - 4a²}{3a}* \frac{6a + 6a²}{5a - 25}
&=\frac{4\cancel{a}(5 - a)}{3\cancel{a}}* \frac{6a(1 + a)}{5(a - 5)}\
&= \frac{-4(a-5)}{3} * \frac{6a(1+a)}{5(a-5)}\
&=\frac{-4\cancel{(a-5)}}{3} * \frac{6a(1+a)} {5\cancel{(a-5)} }
\end{align*}

[alpine sable]

Oh right you find a common factor

[molten pivot]

Yep then factor out a megative

[ocean seal]

Mehdi_Moulati

[alpine sable]

Whyd you put a negative by the 4

[molten pivot]

Tell me

What is 3-2

[alpine sable]

1

[molten pivot]

What is 2-3

[alpine sable]

-1

[molten pivot]

So would you agree?

(a-b)=-(b-a)

(3-2) = -(2-3)

[alpine sable]

Yeah

[tacit arch]

@alpine sable you should try doing the rest on your own

You'll learn better swimming than watching people swim

[molten pivot]

So (5-a) = -(a-5)

[alpine sable]

I got to 3(6a+6a^2)/-20

Probably the wrong step judging based off the answer

[tacit arch]

Show your whole work

[alpine sable]

What he did but I cross multiplied and expanded the 6a(1+a)

[tacit arch]

There is no cross multiplication happening here

That's for when you have two fractions equal to each other

[tacit arch]

I repeat

[alpine sable]

I might just not be awake enough

I'll try doing these questions later

.close

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[hybrid bear]

can i ask a question here

[tacit arch]

Yes

[raven rover]

[hybrid bear]

ok tg
At the Willowbrook Country Fair, Camille will pay an entrance fee in addition to any tickets she buys for rides. The total amount Camille will pay depends on how many ride tickets she buys.
This situation can be modeled as a linear relationship.
What does the y-intercept of the line tell you about the situation?

A. if camille buys 5 ride tickets she will pay $25 in all
B. the entrance fee is $15
C. each ride ticket costs $2
D. camille will buy 5 ride tickets
umm that's wat im doin

[waxen flame]

The x-axis indicates how many rides Camille has ridden. At the y-axis, x=0, which means Camille has not ridden any rides yet the cost is already $15.

[hybrid bear]

so b?

[weary quartz]

✅

[hybrid bear]

@weary quartz thank uuu

[weary quartz]

Your welcome

[hybrid bear]

can i ask another question?

[lone heart]

@hybrid bear Has your question been resolved?

[tacit arch]

Do it

[lone heart]

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[wheat axle]

9x - 7i > 3(3x - 7u)

[wheat axle]

anyone hows how to do this?

sorry im sucks at algebra

[tacit arch]

What are u,i

[wheat axle]

an angebraic

[last ether]

They're just random variables

[wheat axle]

oh

[last ether]

We can't do anything if we don't know their values

[wheat axle]

7i 7u

[last ether]

I mean i guess you can express x in terms of u and i

[wheat axle]

oh

[wary stream]

It's that stupid math thing where if you solve for i, you end up with i < 3u

[wheat axle]

oh can u show me the steps

[tacit arch]

Pain

[wheat axle]

sorry my friend told me to do this but i have no idea what the hell is algebra

[tacit arch]

Yea it's just a joke

[wheat axle]

oh lol

what are the steps then

[wary stream]

You added the role of undergrad but you don't know what algebra is?

[wheat axle]

yeah basically i just copied my friend question and send it here

if theres steps i can understand it

[wary stream]

I suggest one, removing that role because that's basically you lying on your education level

[tacit arch]

,whohas graduate+

[ocean seal]

Members in Graduate+ (Page 1/16)
================================
1.   Artificial Intelligence#0001
2.   Ragadast#4681
3.   Cyraven#4370
4.   DMAshura#3583
5.   Blaxapate#3807
6.   Quantum#1625
7.   Whoever#1913
8.   ipso_cheeto#5392
9.   sheeppunk#9037
10.  the_struggler#4518
11.  TTerra#5291
12.  g.#8603
13.  phil#8937
14.  Abriel#1341
15.  Wormyjim#1508
16.  SingularityResearch#5396
17.  Angetenar#4727
18.  JohnDS#1149
19.  Multramate#5341
20.  Sascha#5877```

[tacit arch]

,calc 15*20

[ocean seal]

Result:

300

[wheat axle]

into?

[wary stream]

Two, it's a joke inequality

You can look it put all over the internet

[wheat axle]

oh

[hybrid bear]

On Zoe's sixth birthday, her brother began measuring her height. Over the next year, Zoe got taller and taller.
This situation can be modeled as a linear relationship.
What does the y-intercept of the line tell you about the situation?
A. zoe grew 5 inches every month
B. zoe grew 1 inch every 5 months
C. zoe was 45 inches tall on her sixth birthday
D. zoe was 47 inches tall 10 months after her sixth birthday

[lone heart]

Please read #❓how-to-get-help

@wheat axle Has your question been resolved?

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[vapid steppe]

guys

[vapid steppe]

how do i write the equation for this exponential function

how do i find out the base

[placid zinc]

Have you got any other parts of it?

[vapid steppe]

sorry the pic is blurry here’s a better pic

[vapid steppe]

i’m pretty sure the end is a -3

idk how to find out the base

[placid zinc]

Good call. That adjusts the asymptote

The y-int is 1 above the asymptote, which means there is no vertical stretch

[vapid steppe]

oh i didn’t know that

[placid zinc]

We're left with
y = b^x - 3

[vapid steppe]

wait can you explain how this works

i found out the base is 4

but i don’t know how the vertical stretches work

[placid zinc]

Let's say I stretched it by a factor of 2. Then that point would be twice as high.

[vapid steppe]

how do you know it hasn’t been shifted left and right

[placid zinc]

The y-int would be 2 higher than the asymptote

A shift left/right is the same as a stretch vertically

[vapid steppe]

i thought stretches were like in front of the exponential

so like 3e^x is a stretch by factor of 3

[placid zinc]

Because
4^(x + 2) = 4²(4)^x = 16(4^x)

So a shift right by 2 is the same as a vertical stretch by 16, when the function is 4^x

[vapid steppe]

oh

[placid zinc]

That is, there will never be a horizontal shift, when you are willing to make everything a vertical stretch

(For exponential functions)

[vapid steppe]

ooh so there is no h like in a•f(x-h)+k

[placid zinc]

a can always do h's role

[vapid steppe]

ok ty

.close

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[onyx marlin]

How to calculate the integral of x^x^x？

[raven rover]

Ya can’t

It’s nonelementary

[onyx marlin]

oh...thank!

[raven rover]

Looking it up will probably give you more information than I can provide

[onyx marlin]

okok

[last ether]

x^x^x goes crazy

,w integrate x^x^x

[ocean seal]

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[last ether]

Wait really, it doesn't even kick out any nonelementaries

@raven rover uhhhh

Are there functions that are unintegrable regardless if they're elementary or not?

[placid zinc]

We can fix it.
Let Kayne(x) = int x^x^x dx

Now there's a standard mathematical function

[last ether]

LOL

Isn't that how like erf(x) became a thing

[tacit arch]

The usual thing is to do a power series

Then integrate term by term

[lone heart]

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[cinder obsidian]

i forgot how to solve this and i need help

[cinder obsidian]

i completely forgot how to work this it was a year ago so i dont have any work to show

[pseudo ice]

Do you know how a function and its inverse are related?

If not, try reading up on that

In particular, ii will be instant, and i will not be that much work if you spot something…

[cinder obsidian]

yes i know that you put y (or any letter) instead of x and make y subject of the formula but i dont know how to make y subject while it is in power

when i know how then i put what ever i got equals 4

[pseudo ice]

Let me just say, you should not find the inverse function (or rearrange for y) in this question, as there’s no need to

But to do so, you can use logarithms if you wish

Can you tell me the definition of an inverse function please?

[cinder obsidian]

it is the exact opposite of function and it is the inverse operation in the reverse order?

[pseudo ice]

Yeah, that works, but do you have a more refined definition? If you have f and f^{-1}, can you tell me something about f(f^{-1}(x)) and f^{-1}(f(x))?

[cinder obsidian]

the do nothing they give the answer x?

[pseudo ice]

That’s pretty much it, so what’s the answer to part ii?

[cinder obsidian]

x

but still i dont know i

[pseudo ice]

Cool, perfect, now using that, for part i, we have j^{-1}(x)=4

Can you use what we’ve just said to find x, without working out what the inverse actually is?

[cinder obsidian]

81

it is correct but why and how

[pseudo ice]

Because j(j^{-1}(x))=x, just apply j to here

That leaves you with x=j(4)

[cinder obsidian]

what do you mean with apply j to here

[pseudo ice]

j^{-1}(x) = 4, so j(j^{-1}(x)) = j(4)

The first bit just becomes x

So x= j(4)

[cinder obsidian]

Thank you soo much i really appreciate your help 🙏🙏

[alpine sable]

Quick question what level of math is scatterplots and what usualy comes after scatterplots thanks.

[lone heart]

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[alpine sable]

.reopen

[cinder obsidian]

u should not do it here but ok

.reopen

[lone heart]

✅

[cinder obsidian]

@pseudo ice

[cinder obsidian]

idk which system you are on but i am in igsce( international british system) i got it in statistics after probability

about 10 or 9 grade

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[magic cove]

How to evaluate ln(e/2)?

[last ether]

ln(a/b) = ln(a) - ln(b)

[magic cove]

Oh

I forgor

Thanks

[last ether]

https://tenor.com/view/i-forgot-i-forgor-meme-memes-kinemaster-gif-22374063

[magic cove]

.close

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[wanton nova]

What do you need help with

I cant read

[tribal osprey]

zoom a bit bro

[wanton nova]

The quality is bad

[tribal osprey]

now which one?

[wanton nova]

Can you show your work so far?

Ok...

[tacit arch]

Man when I ask for help I sure wouldn't troll people volunteering their time

[tribal osprey]

B

(2b^2/a) * (9a^8 b^10/ 4a^4 b^2)
= (2b/a) * (9a^4 b^8/4)
= 18a^4 b^10 / 4a
= 9a^3 b^10/2

[wanton nova]

Something you will never understand

[lone heart]

Please read #❓how-to-get-help

[kind sinew]

only ping once

[shut crest]

dude what the hell is this lol

[wanton tusk]

[shut crest]

your like a 6 year old

y e s

you have a "tatto"?

[vague coral]

<@&268886789983436800>

[night geyser]

.close

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[alpine sable]

What are the intercepts for a point circle

[kindred anchor]

Intercepts on the axes?

[alpine sable]

Mhm

This to be more precise

[kindred anchor]

Right

Have you gone through the text?

Where do you not understand?

[alpine sable]

I have

[alpine sable]

I was just wondering what would be the intercepts of a point circle that is something like x²+y²=0

[desert tendon]

That's just a dot

[kindred anchor]

Yeah

[desert tendon]

So 0,0 ig

[kindred anchor]

There are no intercepts, the center is (0,0)

There's no radius

[alpine sable]

Interesting

[alpine sable]

But what if the equation is something like

(X-3)²+(y-4)²=0

[kindred anchor]

Expand to get something in the form x²+y²+2gx+2fy+c=0

[alpine sable]

This also represents a point but with center (3,4)

[vale wigeon]

But what if the equation is something like
(X-3)²+(y-4)²=0
this equation describes a single point, (3,4)

[kindred anchor]

Then, use the results given here ^

[vale wigeon]

there are no intercepts on either axis

as should be pretty obvious

[kindred anchor]

^

[alpine sable]

But then according to the text the intercept should be 2 root g²-c

[kindred anchor]

I didn't notice the radius

[vale wigeon]

well ok let's do it formally

expand everything out

[kindred anchor]

,w expand (x-3)²+(y-4)²=0

[vale wigeon]

(x-3)^2 + (y-4)^2 = 0

x^2 + y^2 - 6x - 8y + 25 = 0

g = -3, c = 25

[ocean seal]

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[alpine sable]

Right

[vale wigeon]

you will get sqrt(g^2 - c) ∉ R

as you should

[alpine sable]

Alright thanks!

.close

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[kindred anchor]

Bye fellow bee

[alpine sable]

Ciao quantum bee

[lone heart]

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[azure kindle]

is anyone familiar with reading eviews output?

[azure kindle]

im currently doing a impulse responses in eviews but didnt know how to read the graph ...

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@azure kindle Has your question been resolved?

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@azure kindle Has your question been resolved?

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@azure kindle Has your question been resolved?

[paper hinge]

cengage

[lone heart]

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[final viper]

i want to separate the equation of planes

is there any alternative to do it without factorization

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[past gull]

how do i open a question channel

?

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[sturdy grail]

hi

[sturdy grail]

why is 2sin2x = 22sinxcosx ??

[gray isle]

double angle identity

[sturdy grail]

ah I understand. the last question in which I even asked above is this , I don't understand why they never integrated 1/cos^2x

[gray isle]

change of variables

1/cos^2(x) dx was subbed with dt

[sturdy grail]

yes but then later on they substituted tanx with t so we should get dx back

[gray isle]

no?

[sturdy grail]

why shouldn't we substitute dt with dx again it is respect to x?

[gray isle]

you no longer have any differentials after integrating

the sub and related work was done on the top right
changing variables gives you the integral in t in the second line
integrating gives the first section of the third line
then subbing t back for tan(x) gives the end result

[sturdy grail]

don't we have (tan^6x/3)+(tan^12*x/4) dx ??? oh is it because we we have already integrated

ah

my bad we get that dx = cos^2x * dt which removes cos^2x from 1/cos^2x

?

lol it was a while I did this, I first was confused, I was thinking there were two substitutions method I haven't heard, I thought T substitution was something else because they use of variable t. But they use U substitution method.

[gray isle]

its just a dummy variable

u and t are the most common

the use of t is more common when trig is involved

you could use a 🍌 if you wanted

[lone heart]

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[sturdy grail]

since no one here I will be asking here again, :D. how come he didn't integrate 1/x?

[echo socket]

Consider doing a u-sub, let u = lnx

Then you get that the integrand is 1/u * du (since du = 1/x * dx, which you already have there)

So it's ln|u|, which is ln|lnx|

[sturdy grail]

oh yeah

that's true

another question regarding generelized integrals, is it only when the area is between b and infinity?

which makes me confused because someone said that integrals from 0 to 1 (1/sqrt(x))) is generlized integral

[lone heart]

@sturdy grail Has your question been resolved?

[lone heart]

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[lilac iron]

help <:(

[lilac iron]

god im so bad at geometry, no amount of practice seems to work

does anyone got any tips on improving?

[lone heart]

@lilac iron Has your question been resolved?

[vale remnant]

what have you done so far

and what do you know how to do here

[lilac iron]

i have no idea honestly

nothing I try to do seems to have any significant impact

[pliant cedar]

u can use similar triangles

[lilac iron]

how

[pliant cedar]

u know what similar triangles are?

[lilac iron]

of course, but I can't see anything related to them in this case

[pliant cedar]

what two triangles are similar

[alpine sable]

💗

[lone heart]

@lilac iron Has your question been resolved?

[lilac iron]

I only see afb ecb

[pliant cedar]

there are other triangles similar to them

try marking the angle BAF in the diagram as theta or something

and try to find other angles using that

and look for what triangles have the same angles

[lilac iron]

what about afb and efc

even so.

adc and afb are similar too I guess

thought again... that doesn't give me anything, does it

[lone heart]

@lilac iron Has your question been resolved?

[alpine sable]

wrong ig

[lucid marten]

I found its 9 cm

[alpine sable]

yeh urs is ryt ig

[lucid marten]

Hope you understand this: Don't know how to write using the bot

Let's say the following angles are declared since triangle ECB and triangle DCA are similar:
Angle EBC = Angle DCA
Therefore,
Tan EBC = Tan DCA => EC/CB = DA/DC
Let EC be x then the equation becomes:
x/6 = 6/(5+x)
Cross multiplying and simplifying you get 
36 = x^2 + 5x
Which becomes
x^2+5x-36 = 0
Solving for it's roots
x = 4 or -9
Removing the negative
x = 4 cm
Hence DC = AB = 5+4 = 9cm

[lone heart]

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[empty plover]

any clue what this equals, and why?

[novel gazelle]

You mean the first equality?

[empty plover]

no the second one

the first equality is given

[novel gazelle]

there is no equal sign... did you crop the pic correctly?

[empty plover]

it's in french, it says just calculate f^ng-gf^n

given that fg-gf=f

[tacit arch]

Differentiate this equation a few times using chain rule

[ruby current]

what does the notation f^n mean?

is it f composed with itself n times?

[novel gazelle]

Is it linear algebra?

[tacit arch]

@empty plover what does your book's notation mean for f^n? Power or derivative or composition?

[empty plover]

composition

linear algebra

[novel gazelle]

Do you need a neat repsonse, otherwise if you can stick to a "brute" answer you'd just have to compose by f^n-1 once from each side in the first equality then substract

[ruby current]

how does that work?

f^(n-1) o (f o g - g o f) o f^(n-1) = f^(2n-1)

[novel gazelle]

no that is not what i meant

[empty plover]

wdym

then

[novel gazelle]

But this is a stupid method

[empty plover]

well the answer is nf^n as it turns out

and theyve done it recursively. they didnt elaborate how they got to that form

[lone heart]

@empty plover Has your question been resolved?

[lone heart]

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[brisk leaf]

can someone help with this question

[brisk leaf]

A biathlon event involves running and cycling. Kim can cycle 30 km/h faster than she can run. If Kim spends 48 minutes running and a third as much time again cycling in an event that covers a total distance of 60 km, how fast can she run?

what i tried to do was assume x = kmph running

my equation was 16(x+30)/60 + 48x/60 = 60

[limpid turret]

Looks like the right setup

[brisk leaf]

i put that into symbolab and got 195/4

but the answers say its 15kmph

[limpid turret]

Running 15km/h at 48 minutes is 12km.

Biking at 45km/h for 16 minutes is 12km

[brisk leaf]

so the book is wrong?

[limpid turret]

That don't add up

Can you send a photo of the question?

[brisk leaf]

alright

[limpid turret]

Ah okay you wrote it exact

Sorry I just had to be sure

[brisk leaf]

alr

.close

[lone heart]

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[limpid turret]

Okay

I found the issue

@brisk leaf

[brisk leaf]

oh

yes?

.reopen

[lone heart]

✅

[limpid turret]

That phrasing "a third as much time again" sounded so weird

Specifically the word "again"

So I considered "again" meant she spent a third of the time "extra"

That is, she spends 4/3 of her running time biking

[brisk leaf]

oh

[limpid turret]

Try 4/3 instead of 1/3

Having said that, the problem is very badly worded

I would never ever have considered 4/3 over 1/3

[brisk leaf]

it got 15

[limpid turret]

Yeah

[brisk leaf]

thanks

[limpid turret]

So that's probably how the solution was found

Having said that, bring this problem up to your teacher, because the wording of awful

[brisk leaf]

alright

.close

[lone heart]

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[limpid turret]

Yeah I'm sorry you had to suffer through a problem like that. Good luck to you

[wary stream]

4/3 where?

[limpid turret]

Hard to reiterate succinctly, just read the conversation

DM if still need to know

[wary stream]

I know, I don't see 1/3 mentioned

Except in the problem

.reopen

[lone heart]

✅

[limpid turret]

You lost me. If 1/3 is mentioned in the problem, where are you not seeing it mentioned?

[wary stream]

You said try 4/3 instead of 1/3, my question is where in that equation was 1/3 used?

[limpid turret]

To get the cycling time

One third of the running time

"my equation was 16(x+30)/60 + 48x/60 = 60"

16 is 48/3

[wary stream]

That was just a weird way to phrase it because I was gonna say the problem said, she travels a third as much cycling so you would do 48 + (1/3)48 to get the total time she biked

.close

[lone heart]

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[lone heart]

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[wanton lily]

i need help with some sequences, some trig, and some logarithmic functions, i can translate the questions but im not sure exactly as to what to do

[wanton lily]

i haven't the chacne to actually go to class since i had to study for another uni exam than the one here

but i have school exams so

[limpid turret]

Looks like an arithmetic series

[fallen verge]

not quite

its def a sequence

[limpid turret]

Oh derp

You are right

[fallen verge]

quadratic?

[wanton lily]

well i got to the point of

a18 = a17 + 19 = 200

now i am stuck

i might have mistranslated, i believe it is a series yes

[fallen verge]

keep working backwards

[limpid turret]

nah that's tedious

[wanton lily]

are you sure there isnt a formula for this

[fallen verge]

i was hoping youd notice a pattern

[wanton lily]

17 19

two difference?

[fallen verge]

lets say you continued to work backwads to like
a18=a15+17+18+19=200

you can continue that all the way to a1

[wanton lily]

would you mind if we go into call?

[fallen verge]

i dont do vc sorry

[wanton lily]

alright

well i could but im sure theres a formula for this i just do not know it

since otherwise you cant do 40 ofthese in 40 minutes

[fallen verge]

ok so lets say you did go all the way to a1

it would look like
a18=a1+(3+4+...+18+19)=200

do you understand?

[wanton lily]

ohh

everrythign

after a1 is 170 is it not>\

so a1 would be 30

[fallen verge]

umm you might want to recheck that

[wanton lily]

alright will do

187*

[fallen verge]

yup

[wanton lily]

is it 187

so a1 is 13

[fallen verge]

indeed

[wanton lily]

here it asks

how many terms of an is an integer

and what is the addition result of said integers

sorry for asking basically most ofthe questions but genuinely i do not have an idea towards this class, ihad to study for another uni exam than the one that the schools teaching for

[fallen verge]

all good

[limpid turret]

Are you sure this is right?

[fallen verge]

yes

why?

[wanton lily]

wait

a1 + 2?

shouldnt there be a 2 there

[limpid turret]

Why is +19 there?

[fallen verge]

oh

you were right the first time

[wanton lily]

awesome

the +2 should be htere too rigt

[fallen verge]

yes

[fallen verge]

anyway for this one, the first thing you should do is long division

[wanton lily]

do i just divide until icant

[fallen verge]

yes

[wanton lily]

15 left over

8 is the result

[fallen verge]

yes you get 8+15/(n+2)

[wanton lily]

ahhh ok

then i just do it over and over again?

well no sicne i dont know n

[fallen verge]

no you stop there

[wanton lily]

23/n+2

[fallen verge]

no

its $8+\frac{15}{n+2}$

[ocean seal]

GarlicBredFries

[wanton lily]

ah

[fallen verge]

how do you make sure that is an integer?

[wanton lily]

wel

n has to be either

3

5

13

[fallen verge]

youre close

[wanton lily]

i mean

n 1

n 3

n 13

[fallen verge]

ok, so what numbers do you get when you plug it in?

[wanton lily]

5, 3, 1

then

• 8

right>

?

[fallen verge]

add each to 8, then sum it all up yeah

[wanton lily]

so 13 11 and 9

33 integers?

alright, how do i find the addition of said 33 integers

[fallen verge]

nono

you got the answer to a above

33 is the answer to b

[wanton lily]

ahhh ok

i see

[fallen verge]

there are 3 integers that work