help-38

[wraith hinge]

ok one sec i see

wouldnt subtracting y get rid of it from both sides

[last rune]

$3x-6x=y+2$

[solid kiln]

.doc

[last rune]

This is what you had right?

[wraith hinge]

i had 3xy-6x=y+2

$3xy-6x=y+2$

[solid kiln]

Matt

[last rune]

alright

so subtract -y both sides

$3xy-6x-y=2$

[solid kiln]

.doc

[wraith hinge]

ok yes that makes sense so far

youre moving y to the lhs

[last rune]

$3xy-y=6x+2$

[solid kiln]

.doc

[last rune]

valid algebra till now?

[wraith hinge]

yes

wait but why is there 2 y's there now since ur subtracting it

y-y

@last rune

[last rune]

$3xy-y=6x+2$

[solid kiln]

.doc

[last rune]

This is what I have

[wraith hinge]

yes i see

but my question is now, how come you arent removing the y so its

$3x=6x+2$

[solid kiln]

Matt

[wraith hinge]

because youre subtacting y

[last rune]

I can’t remove y like that

[wraith hinge]

how come

[last rune]

let me try to explain

Is 3(5)-(5) = 3?

suppose consider y=5

[wraith hinge]

ohh

no its not

you cant subtract because its being multiplied

[last rune]

Yeah, so it’s not a legal algebra move

[wraith hinge]

ok yes im following so far

[last rune]

Always try to plug in numbers and get a sense of rules before doing

see if it actually makes sense with numbers

[wraith hinge]

i see

[last rune]

3(5)-(5)= 5(3-1) = 5(2) right?

[wraith hinge]

yes

[last rune]

if you see over here, notice i just remove one multiple of 5

[wraith hinge]

yes

[last rune]

so back to orginal problem

I can factor y over there

[wraith hinge]

so youre on $3xy-y=6x+2$ and youve moved y to the lhs

[solid kiln]

Matt

[wraith hinge]

and if you wanna factor it

itll be

[last rune]

yes, now y is common to both terms

So factor it

[wraith hinge]

$y(3x-1)$ ?

[solid kiln]

Matt

[last rune]

Legal

[wraith hinge]

ok so

$y(3x-1)=6x+2$

[solid kiln]

Matt

[wraith hinge]

and next wouldnt you

divide it by 3x-1 each side

[last rune]

yes

as long as 3x-1 is not equal to zero, you can do it

[wraith hinge]

$y=(6x+2)/(3x-1)$

[solid kiln]

Matt

[last rune]

yes, here you put a restricting x not equals to 1/3

It’s a good practice to start early

[wraith hinge]

i see

this makes much more sense now

tysm

[last rune]

do you see why?

[wraith hinge]

yes

because you cant divide by 0

[last rune]

yes, all of the algebra we did considering we didn’t divide by zero

$y=(6x+2)/(3x-1), x \neq 1/3$

[solid kiln]

.doc

[wraith hinge]

i see, it makes sense now

ty for the help

.close

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[toxic belfry]

i have no clue on how to start this

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@toxic belfry Has your question been resolved?

[toxic belfry]

<@&286206848099549185>

[hoary folio]

Well It looks like rectangle 3 has the same height of 6 as rectangle 1

[copper trellis]

6= 4x

Wait

6= k4x

[hoary folio]

and you could find x by doing 6/8=x+1/4x

[copper trellis]

Huh

[hoary folio]

sorry typo

[copper trellis]

Flip the rectangle 2

It will be similar to 1

[hoary folio]

OH SHOOT

[toxic belfry]

im still confused

what would be the firststep

[copper trellis]

lol

You want to find the ratio of the size of rectangle 1 and 2

[hoary folio]

Do what Chat Bot said

[toxic belfry]

and how do i do that

[copper trellis]

Let that be known

Oops

K*

Let the scale be 1:k

[hoary folio]

6/4x=x+1/8

[copper trellis]

Or you can do that

Which is definitely simpler

[hoary folio]

Once you find x it should be easy to find the perimeter

[copper trellis]

Yep

[toxic belfry]

what does this help me find

[copper trellis]

x

Multiply everything by 4x

And 8

Put all the terms on one side to get a quadratic equation

[hoary folio]

This helps you find x

[toxic belfry]

they arent equivalent so why would u do this

[crimson rover]

you have the find an x that makes it equivalent

@toxic belfry u agree with this?

[toxic belfry]

yes

[crimson rover]

if this was true it would mean x=x+1 which has no solution

i think it means that the perimeters arew the same

[hoary folio]

Oh

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[wraith hinge]

To find the line g(x) that is perpendicular to 2x+3=0 and passes through the point (3,-2) first I'd do

2x+3 = 0
2x = -3
x = -3/2

but where I get stumped is this doesnt look like y=mx+b form so idk how to solve for the perpendicular

[left oriole]

are you sure this is the right problem statement?

[wraith hinge]

^

[left oriole]

ok, i guess they mean the vertical line then

[pallid cedar]

That’s weird

[left oriole]

x = -3/2

[wraith hinge]

so how would i find the perpendicular

[left oriole]

well you know it'll be horizontal

what horizontal line passes through the point (3,-2)?

[wraith hinge]

im not sure

maybe -2/3?

[left oriole]

well the y value has to be -2, right?

otherwise it isn't gonna touch (3,-2)

[wraith hinge]

i guess so

this is a weird problem haha

ive never had to find the parallel of a line like this

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[wraith hinge]

For #12, would the answer be {{3},{2},{a}}?

[wraith hinge]

my reasoning being that the cardinality of the set X is 1. |X|=1

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[wraith hinge]

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[dusty cargo]

wagwan

[dusty cargo]

need help with my work

[nimble stone]

you cant just send 24 questions asking for help no ones gonna do all that

pick and choose some you need help with

[dusty cargo]

um

question 4

i would like to know how to solve them

[nimble stone]

could you get a better image of 4 then please

[dusty cargo]

sure

[nimble stone]

so what do you think the steps are for 4

[dusty cargo]

use the factor of 42?

[nimble stone]

-42 but yeah

you need two factors of -42 that sum to -1

[dusty cargo]

right

what would be next

[nimble stone]

what factors did you get

[dusty cargo]

.close

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[nimble stone]

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[wide charm]

P^2+p-1

My answer is 89

Am I wrong?

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[rough dew]

Currently stuck on this math problem, tried solving but got stuck and confused can someone help me solve this

[rough dew]

<@&286206848099549185>

[cobalt drift]

insert an e^ln

$\exp \lim_{x \to 0^+} x\ln \left ( - \frac{1}{\ln x} \right )$

[solid kiln]

jewels!

[rough dew]

did that

[cobalt drift]

right uh

maybe write x as 1/(1/x) and use L'Hospital's rule?

Before that reciprocate the -1/(ln x) inside the ln

[rough dew]

yeah I changed -1/ln(x) to ln(x)^-1 inside the ln

which left me with ln(ln(x)^-1)

I then used L'Hopitals rule on that and got 1/ln(x)^-1 * ((-ln(x)^-2)/x)

[cobalt drift]

$\exp \lim_{x \to 0^+} -\frac{\ln (-\ln x)}{1/x}$

[solid kiln]

jewels!

[cobalt drift]

$\exp - \lim_{x \to 0^+} \frac{1/x\ln x}{-1/x^2}$

[solid kiln]

jewels!

[cobalt drift]

ooo

[stuck solstice]

isn't it simply

[cobalt drift]

$\exp \lim_{x \to 0^+} \frac{x}{\ln x}$

[solid kiln]

jewels!

[stuck solstice]

$\lim_{x \to 0^+} x\ln \left (x \right )$

[solid kiln]

Mr_KEBABOUR

[cobalt drift]

no ln x in the denominator?

I assume you ommited the e^

[stuck solstice]

why would you need it if you do #help-38 message

[cobalt drift]

not sure I follow you, I think we've gone down different paths

[stuck solstice]

simply what you said but without additional exp and ln

[cobalt drift]

I see

how do you deal with the x in the power then

[rough dew]

question

did I find the derivative of this correctly?

[stuck solstice]

if $ \frac{1}{ln x} = (ln x)^-1, (\frac{1}{ln x})^x = -x ln x$

$ \frac{1}{ln x} = (ln x)^-1 \Rightarrow (\frac{1}{ln x})^x = -x ln x$

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[coarse vortex]

can someone explain taylor series to me? im only taught linear approximation and my professor expected us to be able to understand from just that @.@

[coarse vortex]

please ping me if youre here

[outer widget]

A taylor series is used to approximate a function close to a point same concept as with a linear apporixmation. In fact a linear approximation is a taylor expansion and I encourage you to go away and confirm this. However a taylor series allows us to generalise this approximation to a higher order polynomial thus making our approximation more accurate. Think about it like this, the more terms we add to a taylor series, the 'better' the approximation becomes. Watch this for a clear visualisation https://www.youtube.com/watch?v=3d6DsjIBzJ4 .

Now let's have a look at the formula

[coarse vortex]

i get the more power = more accurate, but how does that relate to linear approximation? is linear approximation limited to the first derivative?

[outer widget]

Here if we expand the sum we get: $$ T(x) = \sum_{n=0}^{N} \frac{f^n(a)}{n!} (x-a)^n = \frac{f^0(a)}{0!}(x-a)^0) + \frac{f^1(a)}{1!} (x-a)^1 +......$$. Notice here how we can keep going by increasing the value of n.

[solid kiln]

NoName

[coarse vortex]

yea

[outer widget]

so that is $$ T(x) = f(a) + f'(a) (x-a) + ...$$

can you see your linear apporixmation...

[solid kiln]

NoName

[outer widget]

.

[coarse vortex]

i dont see how you got that

[outer widget]

If we add the n = 2 term, we get a quadratic approximation

[outer widget]

Well f^0(a) is saying we have the 'zeroth' derivative of the function evaluated at a. Essentialy f(a). Then 0! = 1 (known fact) and (x-a)^0 is clearly 1. I am sure you can figure out how I got the second term...

[coarse vortex]

oh sorry i didnt know the f1 f2 is a derivative

right?

[outer widget]

Now in order for a taylor series to exist at a point, the function must be infinetly differentiable at that point (usually denoted $C^{\infty}$).

[solid kiln]

NoName

[outer widget]

yes they are, just the order of the derivative.

[coarse vortex]

what do you mean by infinitely differentiable? like forever continuous and smooth?

[outer widget]

Yes

[coarse vortex]

and the function is $C^{\infty}$?

[solid kiln]

cake

[coarse vortex]

or the point

[outer widget]

This just means the class of infinity differentiable functions. At that point, the !function! must be infinitly differentiable.

Sorry I am taking a while to type, I apparently can't spell 🙂 only time I like the auto correct feature

[coarse vortex]

nono its ok take your time to type

i need time to digest too hahahaha

[outer widget]

watch that video I sent you, and if you still have a problem feel free to dm me 🙂

[outer widget]

btw, I forgot to pull you up on this before, differentiability $\implies$ continuity. If you study analysis, you will prove this 🙂

[solid kiln]

NoName

[outer widget]

For reference, an example of a function that doesn't have a taylor series at a point is ln(x) at x=0. Maybe you can understand why to gain some practise...

GL 🙂

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[fair relic]

$\cot^2(\frac{\pi}{11})+\cot^2(\frac{2\pi}{11})+\cot^2(\frac{3\pi}{11})+\cot^2(\frac{4\pi}{11})+\cot^2(\frac{5\pi}{11})$

[solid kiln]

Wranav बेटा

[zinc ginkgo]

did you try using cot(pi/2 - x) = tan(x)?

[fair relic]

yea

it just made everything into terms of pi/22

@zinc ginkgo

[zinc ginkgo]

i'm not sure that's useful

[fair relic]

yea

and now im stuck

@zinc ginkgo

[zinc ginkgo]

Looks like you'll have to do something slick like
https://en.wikipedia.org/wiki/Basel_problem#The_proof

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[fair relic]

im confused, is this the only way you were supposed to solve this?

[zinc ginkgo]

well idk about you. for all i know you're supposed to use a calculator

[fair relic]

and also i get 50

when the answer is 15

[fair relic]

well yea with a calculator it would be too easy

[zinc ginkgo]

so try it this way

[fair relic]

yea i did

with that i got 50

when i was supposed to get 15

oh wait

nvm

it is 15

thats actually really nice

.close

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[ornate glade]

If I have X randomly determined nonrepeating whole numbers selected in a finite range from 0 to N, how do I determine the probability distribution of the average difference between each consecutive number?

[ornate glade]

For example if we let X = 7 and N = 100, there would be 7 randomly determined numbers with each being >=0 and <=100

I could just calculate every possible permutation and brute force it that way but I'm hoping there's an easier way.

[left oriole]

by difference do you mean A-B or |A-B|?

[ornate glade]

Absolute value of difference

So if we have say, A < B < C them I would want the average of C-B and B-A

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[ornate glade]

<@&286206848099549185>

[dense breach]

You're looking for the probability distribution of the average?

So like if your numbers are 2,8,4,10 the average would be (6+4+6)/3?

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[ornate glade]

Right

Well wait

No it'd be (2+4+2)/3

10-8 then 8-4 then 4-2

[dense breach]

So (10-2)/3 essentially

largest-smallest

then, calculate the distribution for the max, calculate it for the min, and then use those to do max-min

so like, chance minimum is a 1: X/N

chance minimum is a 2: (1-X/N)(X/(N-1))

chance minimum is a 3: (1-X/N)(1-X/(N-1))(X/(N-2))

etc

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[wraith hinge]

Hi, can someone please help me understand trig-sub please

[wraith hinge]

How do I know when to use it

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[real saddle]

can i get some help with this

[naive crest]

factor x²+8x-9 first

[real saddle]

pretty sure i can factor it too

A/(x + 9) + B/(x - 1)

[naive crest]

yes

now put A/(x + 9) + B/(x - 1) on same denominator

[real saddle]

denominator? u mean A + B/(x + 9)(x - 1)?

[naive crest]

A/(x + 9) + B/(x - 1) = (A(x-1)+B(x+9))/(x+9)(x-1)

[real saddle]

oh okay

so

A(x - 1) + B(x + 9)?

[naive crest]

what's your question

[real saddle]

does it turn into that

[naive crest]

does what turn into what ?

[real saddle]

this

[naive crest]

A/(x + 9) + B/(x - 1) = (A(x-1)+B(x+9))/(x+9)(x-1)

= x/(x+9)(x-1)

[real saddle]

how do i solve A and B when its llike that then?

[naive crest]

A(x-1)+B(x+9) = x by comparing numerators

[real saddle]

so A should be -10 and B should be 10 right?

or

[naive crest]

,w A+B = 1, 9B-A = 0

[real saddle]

-10x and 10x

[solid kiln]

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[real saddle]

9/10 and 1/10?

that doesnt seem right

oh

i see

nop dont think i do

[naive crest]

,w x/(x²+8x-9) = (9/10)/(x+9) + (1/10)/(x-1)

[solid kiln]

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[naive crest]

they said to simplify

[real saddle]

unforunately it doesnt seem to be in the right form

so

9/10(x + 9) and 1/10(x-1)?

[naive crest]

probably yeah

[real saddle]

it was

thank u very much

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[wraith hinge]

https://media.discordapp.net/attachments/853585747587891201/1161322818135400500/image.png?ex=6537e128&is=65256c28&hm=2aeb330ca34224413ef5a99e5c1f965c4ccd43fd5d33b7607642e7ebe1de8e83&

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[wraith hinge]

.reopen

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✅

[wraith hinge]

my brain ain't working show me how to solve the equation

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[uneven eagle]

since its asking for the number of solutions, my first instinct is to graph 2+3sin(x/2) and 5-2x and see how many points of intersection there are

[naive crest]

you can consider y = 1/2 x
and it becomes 3sin(y) = 3-4y
for 0 <= y <= 2pi
so you have less to look at

and you could write it 4y+3sin(y)-3 = 0
take the derivative and study the function

but if you allow yourself to just look at the graph

,w plot 3sin(y), 3-4y from y = 0 to y = 7

[solid kiln]

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✅

[wraith hinge]

.reopen

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[daring brook]

so i did h(1) = -3arctan(ln(e^(pi/4)))

and derived it

and its not working

[cerulean hazel]

product rule + chain rule right?

[daring brook]

yes

[fading lotus]

What r u getting as h'(x)

[daring brook]

0

[fading lotus]

It's wrong

What's the product rule

[daring brook]

a'b + ab'

[fading lotus]

So where is the ab'

I am seeing only the a'b

[daring brook]

so is a f(x) and b is g(x)?

[fading lotus]

No

Compare the equations

a is g(x)
And b is arctan(ln(f(x)))

[daring brook]

oh

[fading lotus]

Do u get it

[cerulean hazel]

but remember to chain rule

[daring brook]

1*arctan(ln(f(x))) - 3(arctan(ln(fx)))'

[fading lotus]

Yep

As PeterK said
U need the chains rule toh differentiate arctan(ln(f (x))

[daring brook]

arctan(ln(e^(pi/4))) - 3

ok 1 sec

[fading lotus]

arctan (π/4) isn't 1

[daring brook]

arctan(ln(f(x)))' = [1/(x^2+1)] * ln5e^(pi/4)

i thought it was a'b + ab'

g'(1) = 1

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[ivory spade]

Hey everyone, the memo for this exam paper and Wolfram have different answers. My answer correlates with Wolfram's, but I think there's a different way of doing it that I'm missing.

[ivory spade]

I used U substitution, I'm not sure what the memo did

[heavy lion]

first take t as 2x

[ivory spade]

Here's the memo

[heavy lion]

this is correct

[ivory spade]

I got this

I used U substitution. I took sec2x as U

[fading lotus]

What's du

[ivory spade]

sec2xtan2x dx

[fading lotus]

Yah

What did u do after that

[ivory spade]

I made the tan^5(2x) into (tan^2(2x))^2.tanx

then i got dx as du/sec2xtan2x

[fading lotus]

Why don't u take tan(2x) as u

Stuff gonna get much much simpler

[ivory spade]

would have made things harder would it not?

[fading lotus]

No
Lol

What's du if u is tan (2x)

[ivory spade]

your so right

sec^2(2x) dx would be du

But i think you still get the Wolfram answer

[fading lotus]

It's 2 sec^2(2x)

So it's simply integral of u^5 /2 du

[ivory spade]

oh yeah chain rule

So are both correct? Or is one more correct than the other?

Should I just stick to the less complex one?

[fading lotus]

Yah

Why make thing complicated when it can be done easily

[ivory spade]

okok that's true

Thank you for your help

.close

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[ivory spade]

.reopen

nvm i got another problem🤭

.reopen

i hope its not gonna close now

I used U substitution again, I took ( 2-x) as U

Then got x = 2 - U

And put that in to the original equation, but got a totally different answer than Wolfram and the memo

Nvm my wolfram answer is the same, just the memo that's different🤭

Wolfram answer

Memo answer

<@&286206848099549185> just wanna know if one is wrong and if so, why?

[ivory spade]

.close

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[wind saffron]

I don't know how to proof the surjectivity of the underlined function: If A is finite and B ⊆ A, then B is finite.

Proof. If B is empty, the conclusion is immediate. Suppose that
B is not finite. By the hypothesis, there exist n ∈ N and f: A → Jn bijective, and consequently,
f(B) ⊆ f(A) = Jn. By the previous lemma, f(B) is finite. Since
f|B : B → f(B) is bijective, then f(B) ⊆ B, and therefore, B is finite.

[stark bison]

Can you explain what f(B) means?

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[wise pecan]

how do i solve sequences that arent arithmetic nor geomtric?

[desert osprey]

Do you have a question, cause it could be a recursive sequence

[wise pecan]

4,9,19,39,79

[desert osprey]

Is there any context as it’s very difficult to mathematically deduce a formula for the nth term with just a string of numbers

[wise pecan]

whats the eight term

[desert osprey]

639?

[wise pecan]

yea, but how did u get it

[desert osprey]

You just spot the pattern

[wise pecan]

how do i make a sequence tho

[desert osprey]

+5, +10, +20, +40, +80…

[wraith hinge]

to solve these in general:

[desert osprey]

,,,

[wise pecan]

but i need a sequence

how can i write a formula for it

[knotty locust]

!occupied

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[desert osprey]

Just look at it, that in itself is geometric

[wraith hinge]

Im helping them

they asked how to solve recursive in general

[wise pecan]

no that isnt

[desert osprey]

Yes it is

It has a common ratio

[wise pecan]

what

[desert osprey]

1 sec

[desert osprey]

But the addition is geometric, it has a ratio of 2

5 * 2^n

[wise pecan]

ohh, but how can i write a sequence for it?

[desert osprey]

Yea so the sequence would be Un = Un-1 + (5 * 2^(n-2))

You gotta just derive it from recognition

Then you can solve it as a standard recurrence relation

Do you get it @wise pecan

[wise pecan]

no

[desert osprey]

What do you not get

[wise pecan]

how you found that sequence

[desert osprey]

You just gotta spot it

If you’ve never done recurrence relations then you wouldn’t be able to derive that

[wise pecan]

because the person wrote a simpler one like a1=4

a(n)=2_an-1_ +1

[desert osprey]

What

-1 + 1?

Escape the asterisk

“\”

[desert osprey]

Yea that’s another sequence that gives you the answer

You just gotta derive it

By spotting it

There’s no mathematical formula (that I know of) that magically can get you there

But then you can go on to solve the recurrence relation

[wise pecan]

dude i got a test coming up tho

like i need help

[desert osprey]

Have you ever done recurrence relations

[wise pecan]

no

[desert osprey]

Then don’t worry about solving that, you just gotta get to that point by spotting

Practise makes perfect

[wise pecan]

i got one more tho

[desert osprey]

Try and spot it yourself

Post it here and give it a go

[wise pecan]

for b(n)

its 2 6 12

how do i do it?

.close

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[full dock]

The region under the curve y = f(x), $1\le x\le4$ has area 3, and the volume of the solid generated when this region is rotated about the x-axis is 8. Find the volume of the solid generated when the region is rotated about the line y = -2.

[solid kiln]

water beam

[full dock]

Not sure where to start

Except maybe the fact that if we add 2 to y= -2 and 2 to f(x)??

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[strange delta]

sorry, forgot to mention it's a derivative

[wraith hinge]

Can someone help with rational numbers? I get it but not on a numberline.

[split chasm]

!occupied

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[split chasm]

factor out 1/x from the stuff inside ()

and simplify x^cos(x)/x with exponent laws

[strange delta]

ah, I'm seeing it now, thank you

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[split chasm]

layout of the work is horrid btw

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[lament plinth]

https://en.wikipedia.org/wiki/Cusp_(singularity)

[lament plinth]

Does the union of two discs (which intersect) have a cusp?

Imagine something like this (but the whole area not just the blue colored one here)

[cinder swallow]

What is a cusp? Is there a cusp on this image?

[dull temple]

i don't think so, i think it needs to be "completely sharp" in a sense
it applies to a path though, not an area

[lament plinth]

well the derivative doesn't go to infintiy like in the first case

[dull temple]

eg the absolute value graph doesn't have a cusp afaik

[lament plinth]

I think you are right

[frozen plover]

what do they mean by reverse direction

like 180°?

[dull temple]

i think so, the direction has to fully flip

[frozen plover]

the wiki lays it nicely with parametric function

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[cinder swallow]

Can areas have cusps?

The curves of the two circles seem to have a cusp regardless of the areas they enclose, either in an intersection or a union.

"a cusp is a point on a curve where a moving point must reverse direction" - there seems to be a cusp at x = 0

[dull temple]

it's not reversing direction though is it

it's just changing sharply

[cinder swallow]

Oh I see, it has to be a change more than 180º, so if we travel on the curve of the intersection we reverse directions at the y axis.

While with the union we would travel on the outer curve and would pass through the y axis.

So in their union, the inner shape would not be part of the curve?

That leaf pattern.

[dull temple]

i have not understood what you said

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[wide charm]

My answer is 1/3

[mortal breach]

Show your working

[mortal breach]

?

[wide charm]

@mortal breach

[mortal breach]

You're missing one more possibility

@wide charm

[wide charm]

it is not written about elder or small

so why are we making it order

[mortal breach]

We have to. Its not their age, just a nunbwto differentiate the kids.
We know that there are total od three kids and one is a girl.
The other two can either be a boy or a girl but in both orders.
If child A is a boy and Child B is a girl, that is different than child A being a girl and Child B being a boy.

[wide charm]

Definitely wrong

[mortal breach]

?

You mean to say that it is 1/3?

[wide charm]

Yes

[mortal breach]

Let's look at this from a different angle.
If there are two kids, what wre the chances that one is a boy and one is a girl?

[wide charm]

@mortal breach

BB BG GG
1/3

[mortal breach]

Dont you think it should be 50%?

[wide charm]

Damn

[mortal breach]

Considering each kid a separate being

[wide charm]

But how to differentiate them?

[mortal breach]

You can plot every scenario or make a table

[wide charm]

You are right but

No no i meant this is correct but a little bit unacceptable why?

I want to differentiate GB and BG

How does this work?

Okay sure

[mortal breach]

Let me think of a easier way to explain

[wide charm]

Yes please

I am close to it

[mortal breach]

The way you calculate probability is
No of favorable outcomes/ total no of outcomes

When you write "GB" and "BG," they both represent the situation where one child is a girl and the other is a boy. But they both are a possible outcome

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[wide charm]

Thanks

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[pallid depot]

Hi, I have a problem: I can't solve algebrically this inequality:

|2x - |x^2 - 3|| < 1

Can u help me please?

[pallid depot]

I think this is not the right way to solve it

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[pallid depot]

<@&286206848099549185>

[pallid depot]

<@&286206848099549185> <@&286206848099549185>

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[wraith hinge]

.close

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[daring brook]

idk how to begin

[low gazelle]

Try using the chain rule

[daring brook]

a'b * b?

[low gazelle]

yeah

in h(x) what do u think a and b should be

[daring brook]

4e*2 x 2
8e x 2
16e

[daring brook]

f(x) and g(x)

[low gazelle]

Don't try plugging in the values yet

[low gazelle]

yeah but which function should be a and which should be b

[daring brook]

f"(x)g(x) * g(x)

a = f(x)
b = g(x)

[low gazelle]

it would be f'(g(x)) * g(x)'

[daring brook]

ok

[low gazelle]

to be clearer, this would be a'(b) * b'

[daring brook]

kk

[low gazelle]

actually I lied

I am gonna edit my 2 messages to fix them

[daring brook]

so i do g(1) and then put it in f'(1)?

ok

and then multiply by g'(1)

yes no?

[low gazelle]

I am confusing myself 1 sec

[daring brook]

ok then

[heady sedge]

ln(h(x)) = g(x)ln(f(x))

Is one way

[daring brook]

ln(h(x)) = g(x)ln(f(x))
ln(h(1)) = 2 * lne

lne = 0?

[heady sedge]

No

[daring brook]

is lne 1?

[heady sedge]

ln(h(x)) is not the same as h’(x)

But if you differentiate ln(h(x))

[daring brook]

so we e both sides?

[heady sedge]

You get h’(x)/h(x)

[heady sedge]

No

Differentiate both sides

So you get h’(x)/h(x) = d/dx(g(x)*ln(f(x)))

Right side you have to use quotient and chain rule

[daring brook]

ok

wait im doing it

slowly

OMG

I DID IT

[daring brook]

ily ❤️

ok i close now thx

.close

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[pale badger]

I got an answer of 0.83 secs but answers say 0.6182 secs

idk where i went wrong tho

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[brazen mist]

Hi

[brazen mist]

Need help with a weird integral

((lnx-1)/(1+lnx^2))^2

Integral calculator gives a very horrid solution involving the exponential integral. But this came in a high school calc 2 paper. So I’m kinda confused

[tiny dagger]

$\int (\frac{\ln(x) -1}{1+\ln(x)^2})^2 dx$

this right?

[brazen mist]

Yes this. I didn’t know how to use the tex bot or otherwise I would give latex

[wispy sparrow]

are you sure it's not this instead

[tiny dagger]

highschool? really?

[brazen mist]

Yup

I’m not kidding

[tiny dagger]

is it ln(x²) or [ln(x)]²

[brazen mist]

Second one

Whole square

My bad

[brazen mist]

This

[tiny dagger]

the way it wrote log²
makes me think that it's log(log()) lmao

[solid kiln]

epi

[wispy sparrow]

just sub u=lnx should be fine

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[wide charm]

My answer is A

[boreal dome]

why ?

[wide charm]

One angle will be positive and the other will be negative no?

[boreal dome]

true, if we look at the complex plane, if we plot z and z*, their angles are opposite

[wide charm]

So 0 is correct?

But given answer is D

[cunning cliff]

This depends on the definition of arg given in your textbook

Pay attention to its range.

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[wide charm]

@cunning cliff if we look at max then it can be 2π

[cunning cliff]

I've written everything you need to figure out what the error is.

[wide charm]

Lmao

Thanks for your suggestions

[ruby peak]

i'll do this

show u

[trim lichen]

!nosols

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As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[ruby peak]

try to put z et z_ on an graphic

[cunning cliff]

This fails to answer their query which I believe I've addressed.

They're not asking how to do the question, they're asking why their answer was wrong

[wide charm]

Yes please

[wide charm]

Wtf no solution is applicable for schooling kids

If someone is helping me clearly then do not stop them. There is no assignment. It helps me for understanding

[wide charm]

Bro just need to relax good night

[cunning cliff]

[wide charm]

As a user, i have mentioned 100 times about my language skills

Do not entertain the chat

[wide charm]

I did this and found 0 as I posted

[cunning cliff]

Did you try what I said.

If yes, show your working on your findings.

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[wide charm]

I am not going to show you anything because i am not interested to talk with you anymore

I hope you will not help me or come in questions. I know this is rude thing. But i am what i am

[cunning cliff]

Well, I'd recommend switching off reply pings if that's the case.

[wide charm]

Yes please leave join the riemann gang

<@&286206848099549185>

<@&286206848099549185>

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[wide charm]

<@&286206848099549185>

<@&286206848099549185>

[wraith hinge]

yea?

[craggy bolt]

What?

[wide charm]

Please look at the question

[craggy bolt]

This one?

So there is a line above the horizontal at angle theta and one below the horizontal at angle theta

[wide charm]

Yes

[craggy bolt]

Do you understand

[wide charm]

Yes

My answer is 0

I want to understand 2π

[craggy bolt]

Since angels are positive

One is the angle theta

And the other is 2pi-theta

Adding them up gets 2pi

Because 2 pi is the full circle

Do you understand?

[craggy bolt]

.close

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[trim jolt]

Please don't occupy multiple help channels.

[white basalt]

.close

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[trim jolt]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[placid valve]

1

[knotty locust]

what have you tried?

[tulip violet]

That is an application of Cramer's Rule, if you didn't recognize that

[placid valve]

im not sure what to try

[tulip violet]

Do you know Cramer's rule? As I mentioned, it's an application of that

[placid valve]

not familiar with it no, guess i'll look that up

[trim jolt]

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[drifting rose]

can someone quickly do this for a second

[drifting rose]

because I did it one way but how the answers do it seem to me it kind of bs

would you move the cos to the other side or would you guys think of dividing both sides by cos and divide both sides by sqrt3

[split chasm]

divide both sides by cos(x) allows you to immediately express the equation with a single trig function

[drifting rose]

so does adding cosx to both side also work

albiet, longer

[split chasm]

yes

[drifting rose]

god.

I would never think of dividing both sides

is it because theres a negative in front of the cosx because thats technically a -1 which allows you to divide cosx out

because my brain is telling me you can't just divide it out

.close

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[ruby sparrow]

A car travelling in a straight path is recorded to have traveled 8m after 1s, 20m in 2s, 32m in 3s, and so on. How long will it take for the car to travel 1380m?

[ruby sparrow]

😭

[slate eagle]

Would you be able to post the original question?

[ruby sparrow]

thats the original

[golden bison]

you already asked this

[ruby sparrow]

yes

I didnt get

a proper answer

[golden bison]

hmmm was it multiple ch

[ruby sparrow]

multiple ch?

[golden bison]

how certain it was incorrect

[ruby sparrow]

ohhh

I need more opinion cuz my classmate said the answer was a whole number

but my answer had a decimal on it

and is it possible for an answer to be 115.33 seconds?

the .33 is in mili right?

I found the answer online but I dont know how they got the answer

Can someone explain it to me?

a^1 = 8
d = 12
S^n = 1380

these are the only given

yet they got the answer idk howw

[slate eagle]

Just to double check, the problem has the points

• 8m after 1s
• 20m after 2s
• 32m after 3s
• and so on
  Correct?

[ruby sparrow]

yes

[slate eagle]

Hm

I can't say I'm certain how far the car has travelled after 4 seconds if this is the only information

[ruby sparrow]

after 4 second the car has travelled for 44m

[slate eagle]

So we should assume that "the car accelerates until it reaches a certain speed, and then speed remains constant"?

[ruby sparrow]

yes

[slate eagle]

In that case, this is very close to the answer; I would double check the math, though

[ruby sparrow]

cuz I saw an answer

saying n = 15

then I used the formula of getting the sum

and I did get the answer 1380

but I dont know how they got the answer n = 15

[slate eagle]

I can't say I'm familiar with what formulae are being used for this where an exponent is necessary; would it be possible to provide a reference for me to look over?

[slate eagle]

Ah, I forgot to include the 1 back from the first 8 seconds; this answer would be correct using the methods I'm familiar with

[slate eagle]

Ah, interesting

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[trim jolt]

@fiery schooner Has your question been resolved?

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[drifting rose]

I'm marking a test paper I'm doing and can someone vertify if they did this right? because I feel like they didn't

[austere cedar]

The first line is weird, and it seems like they didn't stick with it

[drifting rose]

yea thats what I thought

because they immediately went with the derivative version

which is wrong right?

[austere cedar]

They're implicitly rewriting the integral as:
∫ (1/3)^x dx

But then yeah, that's the derivative not the integral

Wait no but then they switch it back later

[drifting rose]

[austere cedar]

Wait no they don't lol they applied the log property weird

[drifting rose]

this answer paper sucks

[austere cedar]

That's supposed to be the correct answer? Oof

[drifting rose]

yea now I'm afraid the others are wrong now

[austere cedar]

,w integral of 3^(-x) between 0 and 2

[solid kiln]

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Requested by kaynex

[drifting rose]

they got it right?

[austere cedar]

LOL. Their method is wacky but they end up with the right answer

[drifting rose]

wait did they do it the wrong way and got the right answer?

wait wait wait

lets say we have that intergral

and decide to not put the - into it so it fits the idea of the intergrations version

can I just go and use the derivative version?

[austere cedar]

Line 2 is just incorrect. That does not evaluate to the answer they want.

[drifting rose]

some spirit possessed them lol

[austere cedar]

But it seems like something got them back on track for line 3

,calc log(1/3)(1/3)^2 - log(1/3)

[solid kiln]

Result:

0.97654425659388

[austere cedar]

,w log(1/3)(1/3)^2 - log(1/3)

[solid kiln]

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Requested by kaynex

[austere cedar]

Right. That's just not the answer at the end

[drifting rose]

I think its one of those times where they wrote it wrong but was thinking the right thing

[austere cedar]

Maybe a typo. Perhaps they always wanted to divide by log(1/3), not multiply by it

That would have been better

[drifting rose]

¯_(ツ)_/¯

guess I got to write down the proper version

thanks for the help

.close

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[next rampart]

can you solve 10 x (-2 x (-11)) (ten times minus two times minus 11) with by calculating associative properties?

[next rampart]

I don't know whether brackets in English mathematics its brackets or something else like ( )

<@&286206848099549185>

[indigo totem]

-220

[tame solstice]

I think its 220?

[indigo totem]

oh i skipped a -\

its 220

[next rampart]

can you show with how to solve it? with math language

[tame solstice]

[tame solstice]

Minus and minus becomes +

While u r doin multiplication

[next rampart]

hmm

okay

and

-21 x (4+2)

by calculating the distributive property

with how to solve it

need image

<@&286206848099549185>

[indigo totem]

always solve the bracket first

4+2=6

then multiply by -21

-126

[next rampart]

oh

ok

thankyou

thats it

[indigo totem]

you are welcome. You should close the question with the .close whenever you are ready

[next rampart]

.close

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[keen haven]

Hi, f(x)=f(z) where x and y both are real numbers can we say that f is constant?

[keen haven]

and if yes with what argumentation?

[trim lichen]

this needs SO much more context than what you're providing.

[keen haven]

in the end of solving I got exactly this equation

f R:R no more limitations I guess

what information do you need?

[trim lichen]

i need the entire problem in full

!original

[trim jolt]

Please show the original problem, exactly as it was stated to you. A picture or screenshot is best.

If the original problem is not in English, then post it anyway! The additional context might still help helpers help you. Do your best to translate.

[keen haven]

it is much more complicated than what I am asking but ok

in the end f(yf(x)+x)=f(tf(x)+x) where y is real and t is positive

if f constant then f(x)=0

<@&286206848099549185>

anyone?

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[keen haven]

<@&286206848099549185>

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[drifting rose]

I know they are increasing when f'(x) > 0
but what does that exactly mean?
does that mean when its above the x axis?
or the parts of the graph that would be going up

[serene briar]

f'(x) > 0 means the value of f(x) is going up

[drifting rose]

so when its above the x axis????

[serene briar]

what is "it" referring to

[drifting rose]

or when the line is going up

its referring to the graph

[serene briar]

f(x) itself or f'(x)

[drifting rose]

oh f'(x)

[serene briar]

if f'(x) is above the x axis, then f(x) is going up aka increasing

[drifting rose]

oh

so its not whether what parts of the graph is going up and down

<@&286206848099549185>

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[drifting rose]

.close

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[wide charm]

Is this solution correct?

[nova spire]

the function is differentiable at 0 if $\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}$ exists and is finite

[solid kiln]

rafilou2003

[nova spire]

f(0) = 0 easily

and $\frac{f(x)}{x} = \sqrt{x}-\sqrt{x+1}$

[solid kiln]

rafilou2003

[nova spire]

so the limit at 0...

[trim jolt]

@wide charm Has your question been resolved?

[wide charm]

Is 1 which is finite

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@wide charm Has your question been resolved?

[frozen plover]

you can also check the continuity of the derivative

[wide charm]

Thanks

.close

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[wide charm]

how to find value of (√2+1)^6

[wide charm]

Besides binomial expansion

[heady wolf]

Without Binomial its gonna be lengthy ig

[wraith hinge]

you can separate it into two cubes and then FOIL it?

idk how to not not use binomial

[trim lichen]

!nosols