help-13

[cedar kiln]

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[delicate condor]

I don’t understand this question at all

[brazen dagger]

Just assume first number as x then second will be x+1 third will be x+2

[delicate condor]

Huh I don’t understand

[brazen dagger]

Consecutive means one after another

[fair geyser]

divide by 3 adnd you get the middle number yeah

[brazen dagger]

3 4 5 are consecutive numbers for an example

[delicate condor]

Why

Oh

Cuz they’re after eachother?

[brazen dagger]

Yes

1 2 3 , 4 5 6 , 7 8 9

These are consecutive 3 numbers

[fair geyser]

the sum of any sequence is the average number times the amount

[delicate condor]

So do I need to divide 729 by 3

[brazen dagger]

@fair geyser explain maybe to her why we can do the divide way

[fair geyser]

in a consecutive numebr sequence, the average value is the one in the middle

8+9+10+11+12 = 10+10+10+10+10

[delicate condor]

Ohhhh

[fair geyser]

they told us there's 3 numbers

so we just have to divide by 3 and symmetrically extend that

[brazen dagger]

Yes do you know why we can do divide by 3 thing

And why does it work?

[delicate condor]

So it could be like 9+10+11=10+10+10

If I understand it right

[fair geyser]

no thunder, i can't explain it

[brazen dagger]

Want me to explain

First term =x-1
Second term = x
Third term = x+1
Now sum of them 
x-1+x+x+1 = sum
3x=sum
x=sum/3```

[delicate condor]

So do I divide 729 by 3?

[brazen dagger]

Yes but understand what I explained above

Do not blindly remember we can divide by 3

What if it was 4 consecutive numbers?

[delicate condor]

So wait is the answer 242+243+244=243+243+243

It was wrong

Wait I saw this thing

Ignore the text just look how they typed it

Am I supposed to do smt like that bcs im really confused of what that even is

[brazen dagger]

I already said at the beginning

[brazen dagger]

@delicate condor

[delicate condor]

Ohhh

So is the answer x and x+1 and x+2?

[brazen dagger]

You need to solve for x

It is given sum of them = 729

So

x+x+1+x+2=729

Do you know how to solve this ^^

[delicate condor]

I think I do do I need to add these () when i solve it

[brazen dagger]

Add them and find value of x

[delicate condor]

Okay so x+(x+1)+(x+2)=729

[brazen dagger]

Yes

[delicate condor]

Okay wait I’ll solve it on paper brb

[brazen dagger]

[winter brook]

Hi

[delicate condor]

X is 242

I think

If I did it right

[brazen dagger]

Yep

Correct

Other 2 numbers will be x+1 and x+2

[cedar kiln]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[delicate condor]

Ohhhh so

It’s

242 and 243 and 244

[brazen dagger]

Yep

[delicate condor]

I tried it but it says it’s wrong

[brazen dagger]

Show me

[delicate condor]

The part I’ve drawn a circle around is what I’ve tried

[brazen dagger]

They are saying answer with a space

Between numbers

Not comma

[brazen dagger]

Answer with a space between each number

[delicate condor]

Okay wait I’ll try

Omg

It was right

😭

[brazen dagger]

[delicate condor]

Thanks u🫶

.close

[cedar kiln]

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[cedar kiln]

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[cedar kiln]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[tight totem]

4

[odd verge]

What was your answer

[tight totem]

c

the one high lighted in red

[random swift]

working?

[tight totem]

cause we get the common ratio by dividing any term by the previous one

[random swift]

need to see your working if you want us to check it

[tight totem]

wdym

like why i chose the answer?

[odd verge]

Yes

[haughty osprey]

its correct

[odd verge]

It is

[tight totem]

i chose it cause we can get the common ratio by diving any term by the previous one and this case we're changing the formula up so we change it from divison to multiplication

[odd verge]

It's a geometric sequence

[tight totem]

okay thanks can i make sure these answers are correct please ?

im not sure about 12 i chose it cause it seems like its themost simple answer im not sure why it would be a or b or d cause they have negatives in them

[odd verge]

d seems correct

For 11

[tight totem]

and for 12 we get the common ratio of 2 from the geometric sequence formula and then we can play accordingly till we to ger 64

no reason to get any negatives right?

[odd verge]

What if the common ratio is -2

a=4
ar^4 = 64

Solve for r

@tight totem

[tight totem]

well thats why i used the geometric sequence formula

[odd verge]

Do this

[odd verge]

Yeah

[tight totem]

2, -2?

[odd verge]

64=4r^4

[odd verge]

Yes

The two acceptable answers

The remaining 2 are not necessary

[odd verge]

So you have 2 common ratios

Which means you have 2 geometric sequences

[tight totem]

this doesnt make sense to me

if i have 2 geometric sequencess the answers wont be the same

[cedar kiln]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[tight totem]

it will alternate the signs

[odd verge]

No, look at options with 2 sequences

[odd verge]

Yes 4, -8,..,64

[tight totem]

dude im stupid

how did i not see that

thanks for clariyfing

[odd verge]

No you're not 🙂

[tight totem]

gotchuuuuu

now it makes sense

cause theres 2 common ratios

[odd verge]

Yup

!done

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[tight totem]

so its a cause when we multiply -32 x 2 we get a positive 64

and the fsrt part makes sense

thx

.close

[cedar kiln]

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[cedar kiln]

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[indigo plover]

hi, i have a amc10 or algebra question

i did: x^2+y^2 = (x+y)^2 - 2xy
= (m+1/m)(-2m-1/m)
=(10)(10-3)
70

[hot crag]

it's m^3 + 1/m^3

cubed, not squared

[indigo plover]

oh, right

[hot crag]

do you know the factorization of a^3 + b^3?

it's (a+b)(a^2 + b^2 + ab)

[indigo plover]

ohh

okay

[vagrant elbow]

just cube m + 1/m

[dusk finch]

and expand that

[hot crag]

or this, yeah

[dusk finch]

and see what you get

it will turn out to be simpler than you'd expect

[indigo plover]

so (m + 1/m)^3?

[dusk finch]

Try expanding that

using binomial theorem

[indigo plover]

im not that advanced, idk what that is

[dusk finch]

do you know formula for (a+b)^3?

(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

something like this

Do you know that?

or does it sound familiar?

[indigo plover]

no, my teacher didn't explain that

im in 8th and we're basically just learning basic algebra equations, and this is practice

[dusk finch]

I see

which of the basic algebra equations do you know?

it might actually be best if you write $\left(x+\frac{1}{x}\right)^{3}=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{x}\right)\left(1+\frac{1}{x}\right)$ and do the multiplication manually

[wraith dagger]

MæthIsAlwaysRight

[indigo plover]

in school, we're up to functions and linear equations

but my tutor is teaching me amc10

[dusk finch]

Well, you have 2 options here

[dusk finch]

either do the expansion manually

or remember the formula $\left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$

[wraith dagger]

MæthIsAlwaysRight

[indigo plover]

got it

.close

[cedar kiln]

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[undone star]

uh i might be a little stoopid but idk how to diff

[cedar kiln]

Please don't occupy multiple help channels.

[undone star]

(different question)

$f(x) = (1+\frac{1}{x})^{x}$

[wraith dagger]

Dootud

[undone star]

the original question is to show that for any x>0, prove that f(x) is strictly increasing

I am highschool student so please do not pull out bazooka tier technique to nuke this

[peak minnow]

id just ln both sides and then chain rule

[undone star]

o real i completely forgor i can do that

well now i have

$f'(x)=f(x)[ln(1+\frac{1}{x} )- \frac{1}{x+1}]$

[wraith dagger]

Dootud

[undone star]

im assuming i would take a new function, say g(x) = ln(1+1/x) - 1/x+1 and diff again

chat im ngl this is not looking too good

[dusk goblet]

$\ln(y) = x\ln(1+\frac{1}{x})$

[wraith dagger]

knief

[dusk goblet]

you had this yes

[undone star]

yep

my justification was just exponentiation doesnt change monotone or not

although idek if thats necessary

[dusk goblet]

$y’ = y[\frac{-1}{x + 1} + \ln(1+\frac{1}{x})$

[wraith dagger]

knief

[undone star]

matches up to what i had

[dusk goblet]

yup

[undone star]

so we know y = f(x) isstrictly positive since x>0

so i just have to treat the inside stuff as a new function and diff

[dusk goblet]

yea

[undone star]

which i get 1/(x+1)^2 - 1/(x^2+x)

[dusk goblet]

well no you need to show it’s positive right

[undone star]

yeah

idk how to show ln(1+1/x) > 1/(x+1) without diffing

[dusk goblet]

is this a proof btw?

[undone star]

yes

prove that f(x) is strictly increasing

on x>0

[dusk goblet]

so we need to show $\ln(1+\frac{1}{x}) > \frac{1}{x+1}$

[wraith dagger]

knief

[dusk goblet]

for all x>0

[undone star]

is this an integral proof

i might try integration proof at least for itnegrating 1/x

RHS looks like lower rectangle

[dusk goblet]

no need we need to do algebra

$\ln(1+\frac{1}{x})^{x+1} > 1$

[wraith dagger]

knief

[dusk goblet]

since x+1>0 no change of sign

hmm

i mean it’s screaming e

because of the limit

well

we need to show its greater than ln(e)

because the right is ln(e)

[undone star]

yep

[dusk goblet]

so $\left(1+\frac{1}{x}\right)^{x+1} > e$

[wraith dagger]

knief

[dusk goblet]

well i mean clearly the limit at infinity is greater right

[undone star]

sure

[dusk goblet]

do you have any ideas

[undone star]

not really sry

[dusk goblet]

i tried x = 2 it works

this surely is true

now we just have to prove it

[undone star]

my curriculu mdoesnt really teach that much about limits besides the really elementary stuff so idk many techqiques for proving things like this

[undone star]

for what its worth i did end up proving this to be true with integration

[dusk goblet]

can i see?

[undone star]

well im not sure if my logic is 100% correct

i considered the integral of 1/x integrating from 1 to 1+x

my lower rectangle is x/(1+x) in terms of area

ie: (x+1-1)*1/(x+1)

this is the part im not sure

$\int_{1}^{1+x}\frac{1}{x}dx=ln(1+x)>\frac{x}{1+x}$

[wraith dagger]

Dootud

[undone star]

Apply transformation x=> 1/x and it yields $ln(1+\frac{1}{x})>\frac{1}{1+x}$

[wraith dagger]

Dootud

[undone star]

(would need someone to double check if i can transform x => 1/x like this)

@dusk goblet is my logic correct or is it kinda bogan

.close

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

There is an array of 4 numbers and can be chosen from 2, 3, 4, 5, 7, 8
Numbers can be repeated and the final result must be between x and 3700. There are a total of 216 probabilities, what is x

[crimson sedge]

so what i did was:
N(begins with 3) = 1 * 4 * 6^2 = 144
N (begins with 28 or 27) = 1 * 2 * 6^2 = 72
144+72 = 216

x is 2722

is there a more efficient way to solve this

and why isnt x 2721

[buoyant sail]

Why is this multipled by this, why i need third root of four, why jot just 3✓2*2 + 1

[crimson sedge]

sorry spot taken you can go to the available ones

[sinful bay]

#help-3 Has their own channel

[crimson sedge]

lol

[pastel vault]

oh right

[crimson sedge]

There is an array of 4 numbers and can be chosen from 2, 3, 4, 5, 7, 8
Numbers can be repeated and the final result must be between x and 3700. There are a total of 216 probabilities, what is x

so what i did was:
N(begins with 3) = 1 * 4 * 6^2 = 144
N (begins with 28 or 27) = 1 * 2 * 6^2 = 72
144+72 = 216
x is 2722
is there a more efficient way to solve this
and why isnt x 2721

<@&286206848099549185>

the initial message was 16 mins ago

[past wave]

I think this is the easiest way

[crimson sedge]

so this is the most efficient way?

[past wave]

Yesh

Yeah

[crimson sedge]

and one more question, why is x not 2721

the answer is 2722

but i want to know why it is not 2721

[past wave]

I think between means closed brackets

Like [2722, 3700]

[crimson sedge]

i see

[past wave]

Not (2721, 3700)

[crimson sedge]

thank you

.closed

.close

[cedar kiln]

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[frigid sinew]

What does 2D motion consists of

[frigid sinew]

does it incorporate

circular motions

[dry temple]

i think circular motion is usually taught seperately

[void sand]

I'm not sure what you're asking

[frigid sinew]

oh okay yeah im just asking

if theyre usually together but ty

.close

[cedar kiln]

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[void sand]

https://tenor.com/view/peach-goma-gif-19918534

weeeeee

[cedar kiln]

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[amber crest]

confused on this question

[amber crest]

I tried substituting v = dy/dt and v = d2y/dt2 but it didnt work

and Im not sure where to go from here

[dire geode]

The left side is the derivative of a product of two functions

Find those two functions

[wraith dagger]

ren

[hot crag]

use the product rule

[amber crest]

huh

im kinda confused tbh

wait ok

so one function is y' and the other is y

[hot crag]

mhm

[amber crest]

so were do I go from there

[hot crag]

work backwards to get f(x) times g(x)

set equal to 0

and use another identity there

[amber crest]

so y * y' = 0

can I have another hint lol

[hot crag]

sure

chain rule

g(x) is y
that's all you get

[amber crest]

that makes sense yeah so f'(g(x)) * g'(x)

[hot crag]

mhm

[amber crest]

and y' is f'(g(x)) and g'(x) is y

[hot crag]

no

g(x) = y

g'(x) = y'

[amber crest]

then where does f'(g(x)) play into this

[hot crag]

f'(y) = y

@amber crest you there fam

[amber crest]

yeah sorry lol just trying to think

[hot crag]

okay look

[amber crest]

Thats what ive gotten to so far but im not sure how to finish the problem

[hot crag]

f'(x) = x

what is f(x)

[amber crest]

1/2x^2 ?

idk

[hot crag]

yes

[amber crest]

anything with derivative x

[hot crag]

also use decimals

[amber crest]

oh

ok

[hot crag]

so you think that f(x) = 0.5x^2 + c

thus

f(y) must be 0.5y^2 + c?

and f'(y) is y

since g(x) = y

now you explain what we do next.

[amber crest]

uhh

well we integrated already

so is it not solved?

we have no more derivatives in the problem

[hot crag]

...

f(g(x))' = f'(g(x)) * g'(x)

use this

and the definitions for f(x) and g(x) i gave you

[amber crest]

.close

[cedar kiln]

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[wheat sigil]

can someone help with questions

[cedar kiln]

@wheat sigil Has your question been resolved?

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[alpine lark]

Looking for some help with this ODE question, I'm not really sure what's being asked of me here. I know acceleration is the derivative of velocity but I'm not sure how to make it so that v(t) is my unknown function, is it as simple as integrating both sides?

[cerulean sail]

"I know acceleration is the derivative of velocity"
they're basically asking you to use "the derivative of velocity" in place of "acceleration"

[alpine lark]

ah okay, so just integrate then?

oh wait, I see what you mean

change a(t) for dv/dt

[cerulean sail]

Yep, that's all they want from you here

(and of course what the initial condition is, i.e. the initial velocity)

[alpine lark]

okay, that part is a bit confusing to me, if velocity is constant wouldn't dv/dt just be 0. I guess I'm sort of lost on trying to compare this to other differential equations since v doesn't like, appear anywhere else and t is independent

[cedar kiln]

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[woeful fiber]

hola

[woeful fiber]

alguien puede ayudarme

con un tema que no entiendo

[queen valley]

hey bro

how ya doing

[queen valley]

dime que tema

trata de hablar ingle aqui btw

[woeful fiber]

gradiente conjugado

es el tema

debo exponer sobre eso pero no entiendo la teoria y menos la practica

[queen valley]

que parte no entiendes

quieres un resumen general?

te estan pidiendo eso antes de entrar a la universidad? 💀

[woeful fiber]

no entiendo la teoria, o sea ya lei y vi videos pero necesito la explicacion, tipo clase

noo, ya estoy en la universidad, necesito exponer sobre ese tema

pero no logro entenderlo

[queen valley]

no creo q encuentres alguien que te de una clase por aca

menos en espanol XD

[queen valley]

lo decia por tu tag lol

[queen valley]

la pagina de wikipedia esta bastante buena

https://es.wikipedia.org/wiki/Método_del_gradiente_conjugado

[cedar kiln]

@woeful fiber Has your question been resolved?

[cedar kiln]

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[old skiff]

Hi!

[old skiff]

I have a math problem. I finished it, but I just need someone to check it

it's integration by trigonometric substitution

I've been trying to check it myself, but I'm not sure anymore

[cerulean sail]

,w int 1/(x * sqrt(4 - x^2))

[wraith dagger]

Results provided by WolframAlpha

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

Requested by chartbit

[cerulean sail]

Damn, can't be lazy

[old skiff]

I am confused huhu

[cerulean sail]

I was trying to be naughty and check final answer

Your work seems fine though, I haven't spotted anything that rings any alarm bells (but it's nice to have the sanity check )

[old skiff]

sooo, my work is fine?

[old skiff]

thanks, i'm going in spirals here. i even see calc problems in dreams

.close

[cedar kiln]

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[waxen fiber]

Does anyone understand the transformation for pt 3

[waxen fiber]

i dont get where the shift up would be? as well as the reflection

pt c**

[crimson sedge]

Can you rewrite $\frac{x}{x+1}$ in a way which would be helpful

[wraith dagger]

gautamdb

[waxen fiber]

let me try

(x+1)-1 / (x+1)

[fiery plaza]

Try putting in some x values, does that give you the same as if you put it in the original?

[cedar kiln]

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[mellow cape]

How do you find the range of this? I found the domain already. And for the range I got 1<=y but there should be another one

[mellow cape]

I have to find vertex

Domain is -2<=x<=1

<@&286206848099549185>

[cedar kiln]

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@mellow cape Has your question been resolved?

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[broken carbon]

can someone help me go through shell method and see if im doing this correctly?

[broken carbon]

.close

[cedar kiln]

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[crimson sedge]

Does this need to be simplified further for simplifying radical expressions?

[crimson sedge]

also separate from the first question is the factoring for this right?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

https://tenor.com/view/awkward-umm-what-what-gif-14694719

[crimson sedge]

😔😔

.close

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[sharp cloak]

What to do next?

[cedar kiln]

@sharp cloak Has your question been resolved?

[cedar kiln]

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[soft lark]

I'm doing a math unit on solving quadratic equations by factoring

[soft lark]

I've been factoring for years so I'm good at that

But...

I don't really understand word problems

Like how do I get this into the trinomial or whatever so I can factor it

[gusty ridge]

1 Area = length * width
2 length = x, width = x+10
3 Area = 6600
4 x(x+10) = 6600, solve for x

[soft lark]

Where is the part that I factor tho I'm confused

[gusty ridge]

...

solve step 4

[cedar kiln]

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[open cliff]

Can someone pls help me with this q

[undone star]

!status

[cedar kiln]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[open cliff]

1

[undone star]

do you know how to do partial fractions

i think thats the easiest way to dothis integral

[open cliff]

Yeah

Do I have to use that

[crimson sedge]

you can't do partial fractions here

[open cliff]

😭

[undone star]

you can partial fraction decompose this

just not with the standard form

u just kinda have to 'eyeball' it

[crimson sedge]

a more straightforward method is to sub u = x^2 and do a trig sub

[open cliff]

Uhhh can u pls show me the first two steps

[undone star]

why i say this is because you have x^3 and sqrt(1+x^4) as denominators for your partial fractions

it becomes very easy to find what they are just by inspection

if we realise that 1 = 1+x^4 - x^4

then $1= \sqrt{1+x^4}^{2}-x^{4}$

[wraith dagger]

Dootud

[undone star]

do you realise what we can do now? @open cliff

we want this algebra to take form of your integrand

if you find this too weird to follow then trying Qylo's approach might be more appropirate

[open cliff]

Sorry I still don't understand what u r trying to say

[undone star]

we can integrate this by breaking it down into something simpler right

realising that $1= \sqrt{1+x^4}^{2}-x^{4}$

[wraith dagger]

Dootud

[undone star]

we have to divide this algebra by something to get the form of our integrand

if we divide both sides by $x^{3}\sqrt{1+x^{4}}$ we get?

[wraith dagger]

Dootud

[cedar kiln]

@open cliff Has your question been resolved?

[open cliff]

I got this as the ans

Is this correct

[cedar kiln]

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[open cliff]

. reopen

[open cliff]

But there aren't any options for the ans

[cedar kiln]

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[open cliff]

I got this as the ans to the q given below...but there is any options matching to the options...can someone please help me

[rapid pumice]

If we take 1+x^4 = t

[open cliff]

Ok lemme check

[open cliff]

I did not get

I can't substitute if I take that as t

[uncut ridge]

$\frac{\sqrt{x^4+1}}{x^3} - \frac{x}{\sqrt{x^4+1}} \neq \frac{1}{x^3} - \frac{x}{x^4+1}$

[wraith dagger]

penguin

[uncut ridge]

I don't get how you thought of that

[rapid pumice]

I got only √(1+x^4), not option a and c

[rapid pumice]

By substitution

Method

[open cliff]

Divided by sqrt x^4+1

[uncut ridge]

divided but didn't multiply the entire equation by the same? the equation won't be the same if you divide it by something and not multiply it by the same

[open cliff]

Ohh ok thank you lemme check no wonder I was not getting the ans

So after this what should I do

[rapid pumice]

Because we have x^3 in denomi. We can

Ig so

But at last it doesn't work

[uncut ridge]

please make some sense

[open cliff]

?

[rapid pumice]

My bad

hell

[uncut ridge]

the best method in my opinion is factoring out $x^4$ in the denominator term $\sqrt{x^4 +1}$ and then it should be doable from there, try. $1 + \frac{1}{x^4} = t$ substitution would help

[open cliff]

?

[wraith dagger]

penguin

[uncut ridge]

do you not know how to factor out a certain term or what?

[rapid pumice]

Then after that tan-1x int

[open cliff]

Ok

[rapid pumice]

Ohh Ik this method

[open cliff]

I m stuck

[uncut ridge]

show

[open cliff]

Ok I got this as the ans

Is this correct

[uncut ridge]

nope, this is a mistake

[open cliff]

ok lemme check

[open cliff]

where?

[uncut ridge]

1/x^3 does not cancel out the x^3 in the denominator, they both will get multiplied

[open cliff]

ohhh

my bad

[open cliff]

then how should i proceed after the substitution

[uncut ridge]

$$\int{\frac{1}{x^3 \sqrt{x^4 \left(1 + \frac{1}{x^4}\right)}}} = \int{\frac{1}{x^5 \sqrt{1 + \frac{1}{x^4}}}$$

[wraith dagger]

penguin
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[uncut ridge]

@open cliff do you get this step?

[open cliff]

I got this as the ans is this correct

But still there isn't and option that matches with the ans I got

[open cliff]

The ans is A but I'm still not getting it

[uncut ridge]

I am also getting an x^2 in the denominator

[open cliff]

How?

[uncut ridge]

$$ - \frac{1}{4} \int \frac{dt}{\sqrt{t}} = \frac{-1}{2} \sqrt{t} = \frac{-1}{2} \sqrt{\left(1 + \frac{1}{x^4} \right)} = \frac{-1}{2x^2} \sqrt{1+ x^4}$$

[wraith dagger]

penguin

[open cliff]

Ohh thank you so much

For all your help

[uncut ridge]

no worries!

[open cliff]

6

[cedar kiln]

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[idle gulch]

Riemann sum

Won't it give you integration 0 to 1 sin(pi•x)

[dreamy void]

https://youtu.be/2pe1d31b8EM?si=F7ulXbaXhAOh1apQ

[cedar kiln]

@eager hemlock Has your question been resolved?

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[worthy hawk]

How to do b(ii)? I tried a low but got nothinf

[clear umbra]

remember any theorems about it?

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[crisp silo]

how do I simplify this?

[solid juniper]

,rotate cw

[solid juniper]

depends what 'simplify' means then

[crisp silo]

Look im tryna find this

[solid juniper]

under some meaning that's already simplified

do you want to expand it?

[crisp silo]

Yea

Expand

How do I expand that

[solid juniper]

for any numbers a and b we have
(a+b)^2 = (a+b)(a+b) = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2

so (a+b)^2 = a^2 + 2ab + b^2

[clear umbra]

did you really just 2(x-3)² = (2x-6)²

[crisp silo]

Tyy

[crisp silo]

Huh?

.close

[cedar kiln]

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[clear umbra]

this

did you distribute that? @crisp silo

[crisp silo]

Omg yea I didn't know that

[clear umbra]

you cant btw

2 and (x-3)² has different exponents

[crisp silo]

Ye i came to know now

[clear umbra]

you cant distribute the 2 into it like that

[crisp silo]

Tyy

[cedar kiln]

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[worthy hawk]

How to do b(ii)

[worthy hawk]

I tried quite a lot but couldnt

i I got roght

[hybrid relic]

do you have str

[clear umbra]

theres a theorem regarding this

[hybrid relic]

nah

what's the relation between str and spr

[cedar kiln]

@worthy hawk Has your question been resolved?

[cedar kiln]

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[rare heron]

with the question f (x) = x^2 - 2x - 3
do i combine the x^2 with the 2x?

[hot crag]

no

[hybrid relic]

complete the square

[hot crag]

i'm assuming you need to solve for x such that x^2 - 2x - 3 = 0?

[rare heron]

but i write it as (x - 2x - 3) and (x + 2x - 3)

[hot crag]

no

[rare heron]

i am finding the x-int y-int and vetex

[hot crag]

you can factorize directly

for y-intercept simply substitute x = 0

the vertex is -b/2a (where a is the coeff of the x^2 term and b is the coeff of the x term)

and for the x-intercept set equal to 0, factorize, and solve for x

[rare heron]

is the x-int -3/2

[hot crag]

oh mb i misread

lemme check, hang on

no

i suggest you find the y-intercept and vertex first, those are easier

we can move onto the x-intercept after that

[rare heron]

what did you mean by substitute x = 0

[hot crag]

replace x with 0

[hot crag]

here

[rare heron]

so the y-int is -3

[bleak viper]

yes

[hot crag]

yes

astar's just been waiting to say that hasn't he

[rare heron]

so the vertex is -1/2

[bleak viper]

no

[rare heron]

-2/-4

[hot crag]

that's 1/2

simplify

[rare heron]

-1/-2

[hot crag]

and also, that's incorrect

yet again

[bleak viper]

where did u get -4 from

[rare heron]

-2/2(-2)

[hot crag]

wtf

[bleak viper]

and why -2?

[hot crag]

no-

[bleak viper]

-b/2a

[rare heron]

-b/2a

[bleak viper]

a is coefficient of the x^2 term

f (x) = x^2 - 2x - 3, whats the coefficient of x^2 here?

[wraith dagger]

ren

[bleak viper]

(also this is x coordinate of the vertex)

[rare heron]

a is x^2 b is -2x and c is -3

[hot crag]

what??

no

a is the coefficient of x^2, the number beside x^2

b is the coefficient of x, the number beside x

[rare heron]

f (x) = x^2 - 2x - 3

[hot crag]

yes...

[rare heron]

i don't see a number it's just x

[polar path]

use 1

[rare heron]

okay thanks

-2/2(1)

[hot crag]

yes

[rare heron]

-1

[hot crag]

mhm

[rare heron]

so -1 is the vertex. .

how do i find x

[hot crag]

that's the x-coordinate of the vertex

for the y-coordinate you need to substitute x = - 1 into f(x)

[rare heron]

it = 0

[bleak viper]

what is it

[hot crag]

oh btw you might wanna check this again

it's -b/2a

not b/2a

that's mb

[rare heron]

f (-1) = -1^2 - 2-1 - 3

the answer of -b/2a is -2/2(1) = -1

right

f (-1) = -1^2 - 2-1 - 3 = 0

[rare heron]

i tried doing this

[bleak viper]

what -1

[rare heron]

its the vertex

[bleak viper]

it should be 1

-b/2a = -(-2)/2(1) = 1

[rare heron]

OH

i see

[bleak viper]

b is -2

not 2

x^2 -2x - 3 = x^2 + (-2)x + (-3)
^b

[rare heron]

so x-int is -4?'

[bleak viper]

what x-int

[rare heron]

1^2 - 2(1) - 3 = -4

[bleak viper]

x coordinate of vertex*

also its y

[rare heron]

ok

[bleak viper]

not x

x coordinate of the vertex is 1

which when u put in the function

you get the corresponding y

so coordinates of vertex is (1,-4)

[rare heron]

y-int is (0,-3)
vertex is (1,-4)

[bleak viper]

yes

[rare heron]

how do i find x-int?

[bleak viper]

x^2-2x-3

roots of this equation is/are the x-int

[rare heron]

like the sq root?

[bleak viper]

no

x^2-2x-3=0

[rare heron]

what do you mean by roots

[bleak viper]

where it intersects x axis

like roots of quadratic equation (this is a quadratic)

[rare heron]

i don't know what to do

is anyone there

[marsh flare]

@rare heron hey!

[wintry juniper]

Show thy question

[rare heron]

hello again

[marsh flare]

what's your question?

[rare heron]

f (x) = x^2 - 2x - 3

[marsh flare]

mhmh what are you supposed to do?

[wintry juniper]

That's not it

[rare heron]

I am fining the x-int because i already have the y-int and the vetex

[wintry juniper]

Can u just take a pic

[marsh flare]

yeah

[rare heron]

yes

[wintry juniper]

Bryan you are very cool

[wintry juniper]

I aspire 2 b like u

[marsh flare]

just equate it to 5

and solve it as a quadratic equation

[marsh flare]

...?

why

[wintry juniper]

Idk

[marsh flare]

like take x²-2x-3=5

[wintry juniper]

Yessir

[rare heron]

why is it that I put a 5 there?

[wintry juniper]

Cuz

"where does f(x) = 5"

[marsh flare]

"where does f(x) equal 5" means that you have to find the values for x

which when substituted in the equation

gives you 5

[wintry juniper]

Uessir

[marsh flare]

does that make sense?

[wintry juniper]

Bryan I live you

Love

[rare heron]

no I don't know where the 5 comes from

I'm sorry

[wintry juniper]

I'll explain it 2 u clearly dw

F(x) is x^2-2x-3

It says f(x) = 5

[rare heron]

i think you are looking at the second problem...

[marsh flare]

oh shit

[wintry juniper]

Ya

What

[rare heron]

its f (x) = x^2 - 2x - 3

[rare heron]

i said that here

my question was how i find the x-int

[deep kelp]

So y-intercepts would be x = 0 and x-inte... y = 0

Is so just find f(x)=0

And f(0)

[rare heron]

is the x-int (3,-1)

.close

[cedar kiln]

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[cedar kiln]

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[fast forge]

I have made this question before but haven't been able to confirm and I'm not sure yet about my answer 😔 can someone help please?

In an athletic competition, the following conditions were given:
a. The competition was exciting if and only if one of the following two events occurred: either the "Blue" team won most of the medals, or the "Red" team did not participate.
b. The "Red" team participated if and only if the "Green" team did not win most of the medals.

Based on this information, answer the following questions:

1. Was the competition exciting?
2. Did the "Blue" team win most of the medals?

Here's what I did:

propositions
p: The competition was exciting
q: The "Blue" team won most of the medals
r: The "Red" team participated
s: The "Green" team won most of the medals

a. Can be written as (p↔(q∨~r))
b. can be written as (r↔~s)

I can join both this way (p↔(q∨~r))∧(r↔~s) , correct?

Then, can I make the questions in this way?

1.(p↔(q∨~r))∧(r↔~s)→p
2.(p↔(q∨~r))∧(r↔~s)→q

I just would like to confirm if the translation of 1. and 2. makes sense

[random swift]

the translation makes sense

can you not substitute b. into a. to get (p↔(q∨s))?

[cedar kiln]

@fast forge Has your question been resolved?

[fast forge]

thank you! hmmm not sure about that, I'm still in a basic course, I think I haven't seen how to do that specifically

do you think both (p↔(q∨s)) and (p↔(q∨~r))∧(r↔~s) would be correct then as the final expression?

[random swift]

yes because r↔~s if and only if ~r↔s

[fast forge]

I see, that makes sense

thank you! that's it 🙏 just wanted to confirm if I'm making sense

.close

[cedar kiln]

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[crimson summit]

LImits for 2 variable functions

[crimson summit]

Calculate the limit or determine that it doesn't exist

$\lim_{x^2+y^2 \rightarrow \infty}\frac{x^2y^3}{x^4+y^4}$

[wraith dagger]

binnet

[crimson summit]

Completely out of my depht here, need some start to finish guicance. Guessing I might have to use to polar co-ordinates here maybe?=

[dreamy void]

I would but before that I would use u = x² and v = y²

[crimson summit]

Okay, good suggestion, I'll start by re-writing it in that form

$\lim_{u+v \rightarrow \infty}\frac{uv\sqrt{v}}{u^2+v^2}$

[wraith dagger]

binnet

[dreamy void]

From here I used polar coords but you can also try the paths u = v and u = 0

[crimson summit]

Can you show a solution?

[dreamy void]

No

Show what you tried

[crimson summit]

As I said in the initial question: I honestly don't understand this and need some guidance. I have 2 more similar problems to solve afterwards, hoping that I can do those after I see a solution for the first one.

[dreamy void]

I gave you some start

guidance != handing out solutions

If you have no idea what I am talking about, you should review the topic with some YouTube video

https://youtu.be/MFF4mvyhAyA?si=H0jjUaJq7dD6FJx_

[cedar kiln]

@crimson summit Has your question been resolved?

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[vale ginkgo]

Can anyone help me on the approach to these 2 questions

[cedar kiln]

@vale ginkgo Has your question been resolved?

[pastel vault]

for the first question you can just note that (2, 4) lies on both y = x^2 and y = x + 2

so sub (2, 4) in and check if it satisfies

similarly for the second question

pick a point, say (2, 1/2)

it gets reflected to (2, 3/2)

and then to (-2, -3/2)

there is a proper way to approach it, so $y \mapsto 2 - y$ and so $2 - y = 1/x \implies y = 2 - 1/x$ at first

then find the inverse function and then do $x \mapsto -x$ to get the reflection across $y = -x$

I just think you can use the options in a clever way

[wraith dagger]

higher's secret twin brother

[vale ginkgo]

What do you think the answer is

And what is this method called?

[pastel vault]

21 is A

and I think 23 should be E

[pastel vault]

it's a common strategy for multiple choice questions but I don't think it has a name

oh wait

[pastel vault]

yeah I forgot to check C

so it's 23 C

it clearly can't be 23 E cause that's going to across y = x and then you shift 2 right

[cedar kiln]

@vale ginkgo Has your question been resolved?

[cedar kiln]

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[versed lynx]

z

[hot crag]

???

[worldly chasm]

gotcha

[versed lynx]

Ty

[worldly chasm]

context is an economics question

Can you repost the question

[hot crag]

ahk

[versed lynx]

Me?

Or

[worldly chasm]

yes

[hot crag]

yes you

[versed lynx]

:-:

[hot crag]

idek the context how am i supposed to repost the question

[versed lynx]

It’s on

[worldly chasm]

[
P = A \frac{{(1+i)}^n - 1}{i{(1+i)}^n}
]

[wraith dagger]

OmnipotentEntity

[versed lynx]

Well. I posted the question in the last thread they just told me to open this

[worldly chasm]

So we have P, the present value, and we want A, the annualized value.

[versed lynx]

The formula they’re telling me to use in class is like totally different

[worldly chasm]

what is the formula they told you to use in class?

[versed lynx]

Resembles what I have on my paper minus the bottom part

My problem is, I understand the formula but when putting it into calculator I get error or some far off number

[worldly chasm]

Can you show the formula, and also show what you input in the calculator?

[versed lynx]

this should be simple but i genuinely just don’t get it, i was up from 9-12 Yesterday trying to figure it out

[worldly chasm]

It's difficult to help if I don't know what you're trying to do, and also don't know what you actually did

[versed lynx]

My teacher gave resources but I may be doing something wrong

[worldly chasm]

🙂

[versed lynx]

Y I showed my paper😭

Can I just dm when I have it I’m not home in order to show you everything

I did

to get what I got

[worldly chasm]

I told you what formula I would have used, and you said that the formula you were told to use is different. This is strange to me, and I do not know why. And I also do not know what formula you were told to use.

[warped coyote]

You might have issues since there’s a lot of divisions here. Try using a lot of parenthesis

[worldly chasm]

I know what formula you did use.

But I have no way of telling if those two are the same.

so I have no way of checking your work

[versed lynx]

👍🏾

[worldly chasm]

So....

are you going to share your formula?

are you going to take a picture of your calculator?

[cedar kiln]

@versed lynx Has your question been resolved?

[versed lynx]

^

[worldly chasm]

You may, but I cannot promise I'll be available at that time

[cedar kiln]

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[cedar kiln]

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[slender sundial]

At the beginning of a board there are the numbers 1,2,3,4,5 and 6. I can now carry out the following operation as often as I want:
Erase two numbers a and b and write the two numbers a+b and 2 (a-b) there.
Determine the largest natural number k for which I can achieve, after a finite number of steps, that there are k zeros on the board.

[cedar kiln]

@slender sundial Has your question been resolved?

[rare vault]

at first glance it seems like choosing a and b as the largest and smallest numbers goes forever and is fastest growing

oh this isn't right because a+b could be the larger term for finite ks

[cedar kiln]

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[harsh vault]

I am trying to solve this with partial fraction decomp but the system of equations never works

[cerulean sail]

What are you choosing as your partial fractions?

[harsh vault]

I've tried (x+1) and (x^2 + x) as well as doing (x+1), (x+1), and x

[cerulean sail]

(x + 1) twice in the latter one?

[harsh vault]

yeah if you take the x out of (x^2 + x)

[cerulean sail]

So once you’ve taken the x out of that, your denominator becomes x(x + 1)^2, right?

[harsh vault]

yeah

[cerulean sail]

You know how to deal with repeated factors in the denominator?

[harsh vault]

well I just tried to separate it into two separate factors cause I thought that should still work

but I've also tried combining the factors too

into x^2 + 2x + 1

[cerulean sail]

If you want to write the fractions separately, for this you’d want your equations in the form A/x + B/(x + 1) + C/(x + 1)^2

[cerulean sail]

If you choose this as one of the factors in the denominator, then you would want your partial fractions in the form A/x + (Bx + C)/(x^2 + 2x + 1)

[harsh vault]

I think I did that but then my system of equations didn't work

[cerulean sail]

Ah, what did you get when you tried it?

[harsh vault]

I got a+b+c = 0, a + b = 0, c = 10

or wait I think on another try I also got A+B+C= 0, 2A+B+C = 0, and A = 10

[loud wigeon]

oh

[harsh vault]

oh wait I think I might'be figured out what I'm doing wrong

[quasi jungle]

can i pay one of yall to do my math hw

micro and macro economics 300 dollars a month '

my dms are open for people who are interested

[past wave]

300 that's a lot

I was not expecting hope to hit the griddy on me

[cedar kiln]

@harsh vault Has your question been resolved?

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[leaden elbow]

Hey I need help for number 4 and 5

[cedar kiln]

Please don't occupy multiple help channels.

[cerulean garnet]

i guess the issue comes from not knowing if the negative sign is like (-x)^3 or -(x^3)?

for the 4th one i mean

[leaden elbow]

Oh but how do I describe it cause it say transformation

[dire geode]

close one of your channels

[leaden elbow]

How

[dire geode]

type .close

[cerulean garnet]

if it's like (-x)^3 then reflection on the y axis

if it's like -(x^3) then reflection on the x axis

and then obviously you have to do something about the +2

[cedar kiln]

@leaden elbow Has your question been resolved?

[cedar kiln]

@leaden elbow Has your question been resolved?

[cedar kiln]

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[finite hemlock]

why is the limit of sin(6x)/sin(3x) = 2

[grand shale]

magic

[solar tiger]

magic

[drowsy tinsel]

this guy is unhelpful

i can attest to that

[clear igloo]

when x approaches 0?

lhopital

take derivative of both and evaluate

[drowsy tinsel]

wait broitha u know that help me out in help-16

(im poaching u)

[grand shale]

sin(x)/x =1

[clear igloo]

6cos6x/3cos3x

[grand shale]

lim

use that

[clear igloo]

huh

[solar tiger]

lim x ->0 sin(x) = x just use this rule

[drowsy tinsel]

squeeze therom

[solar tiger]

can someone help in help-26?

[finite hemlock]

thank you

.close

[cedar kiln]

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[late onyx]

Hey, im not really sure how to approach this problem (linear algebra)

[digital cliff]

can you split T([203,202]) into anything, knowing the properties of linear transformations

[late onyx]

do you mean like T(0, 203) + T(202,0)?

oh I see what to do for part a now

you can write is as a combo of v and w thanks lol, but still not sure about b

im inclined to say no because u an v cannot span r2

[cedar kiln]

@late onyx Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[uncut ginkgo]

What good algorithemsare there to create random symetric graphs?

[crimson sedge]

transformations?

[uncut ginkgo]

does not have to be 1:1 that just someting similar in symetric and density