help-13

[crimson sedge]

231ABA=231(101A+10B)=
23.331A+2.310B= 4*(n^2)+B*(n^4)+A*(n^3+n+1)
I don't know if that would help

[lyric narwhal]

not sure what you wrote there

but that was similar to my approach

yep

[crimson sedge]

I did that multiplication

[lyric narwhal]

<@&286206848099549185>

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[untold coyote]

Yoooo i’m hella confused need help solving

[digital cliff]

have you tried drawing it?

[untold coyote]

No tehe question itself confuses me

Too many letters i cant draw it out

[digital cliff]

LN is a straight line with L at one end and N at the other. M lies in the center of that line equal distance from L and N

[untold coyote]

Okay thanks ill give it a try now

@digital cliff so to find a i would have to multiply to find what equals half of 64

By 8

[digital cliff]

could you explain what you mean more?

[untold coyote]

Mn is 8a

The whole line is 64 and it’s a biaector

Sector

[digital cliff]

MN is 4a not 8a, and LM=64, not LN

[untold coyote]

Oh i’m a idiot

I was doing the one under

Oops

Ill redo it

Okay so i multiply 4 by 16 and i get 64 which is the other half so a would be 16

And LN= 128

Is that right @digital cliff

[digital cliff]

seems good to me

[untold coyote]

Alr thanks for the help

[digital cliff]

np

[untold coyote]

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[atomic viper]

I'm not sure I'm doing this right

[regal pewter]

sorry, can you post the actual problem itself? like what youre supposed to be doing

[atomic viper]

Oh sorry, the problem is the first thing next to the number. I must simplify it

[regal pewter]

alright, you're on the right path

just consider, you have (x+2) in two places there. can you do anything with the (x+2)?

[atomic viper]

I can cancel them out?

[regal pewter]

yes

[atomic viper]

Now I have this

[regal pewter]

i think thats as simplified as youre gonna get unless im missing something

you could factor but i think that'd honestly be less simplified

[atomic viper]

I see

Could I factories them?

[regal pewter]

if you wanted to, im not sure if thats what your teacher wants you to do

[atomic viper]

Is there a way to write this in factorized form?

Is there a way to write this in factorized form?

[prisma fossil]

yes, you can write it just like you would any other factor

[atomic viper]

How tho

[prisma fossil]

so if x1 = -1 - sqrt(2) and x2 = -1 + sqrt(2) the factored form would be (x - x1)(x - x2)

when you plug in the values you get (x + 1 + sqrt(2))(x + 1 - sqrt(2))

[atomic viper]

Like this?

[regal pewter]

the top is fine not sure how you got the bottom

it might still be equivalent but that doesnt look like simplification to me

[cedar kiln]

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[green crow]

How do I solve this integral using complex analysis. I have never took a class in complex analysis, but I know Re(e^ix) =cosx and Im(e^ix)= sinx , this is not a homework problem

[mental trail]

I would suggest, instead of using Re(e^ix) = cos(x), use the euler formula cos(x) = (e^ix - e^-ix)/2

IF you are truly intending on solving it with complex analysis

[green crow]

If it was just cos(x) would using the other method work

It doesn't work for powers bigger than 1?

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[arctic magnet]

Simplify:

[arctic magnet]

I would appreciate a step by step explanation! have a math test coming up

[obsidian coral]

,tex .exp rules

[wraith dagger]

dldh06

[cedar kiln]

@arctic magnet Has your question been resolved?

[stiff totem]

what have you tried so far

[cedar kiln]

@arctic magnet Has your question been resolved?

[arctic magnet]

Distributing the ^3 but I’m lost at the division

[stiff totem]

it might be easier to distribute the ^3 later

first, you have 3(a^2)b divided by 2ab^(-3)

you have a in both, so you can cancel one of them out

and similarly, you can multiply through by b^3 to remove the b^(-3) in the denominator

[arctic magnet]

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[atomic viper]

I must simplify it

[cedar kiln]

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[atomic viper]

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[young spade]

I need help on how to solve these radius on number 7-8

[slate lintel]

where do you run into trouble? can i assume you did 1-6 already?

[young spade]

Yes the radius on 5 & 6 is still 20 and 5 right?

I cross their ² and √ so 20 & 5 remains

[slate lintel]

yes

[young spade]

In 7, i did (2√2) (2√) > (4√2²) >> cross out √ & ² so its 4x2 = 8

[slate lintel]

seems legit

[young spade]

Radius of no7 is 8 right

[idle sonnet]

No, that’s the square of the radius

[young spade]

all i need to do on those center is change their signs?

how?

[slate lintel]

yes sorry - when you said "radius" that's really the square of the radius, which is to say r^2

[slate lintel]

i don't understand this question

[young spade]

Like for no1 is (x-3)² + (y+2)² = 16

Since it says standard form

[slate lintel]

that's correct

[young spade]

Did i did this right? So its (x+5/7) + (y-2/2)² = 8?

[idle sonnet]

It should be + 2/7, but otherwise right

[young spade]

Oh yeaa havent noticed it

But the way i got 8, is it valid?

[idle sonnet]

(2 sqrt(2))^2 = 8

Your equation for the circle is right, once you change it to 2/7

[young spade]

(x+5/7)² + (y+2/7)² = 8

[idle sonnet]

This equation is now correct

[young spade]

Thank you, now im on 8

Searched it online and it says that the radius is 4/3, but i need to know how it got to 4/3

[crimson sedge]

2² (√3)²/3²

4 (3)/9

[young spade]

Ohh cross out it again?

[crimson sedge]

Yup

[young spade]

Its like distributive thing

Thank you!

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[chilly spade]

yo

[chilly spade]

how do i graph this equation

T(x)=√((x_1-0)^2+(y_1-0)^2 )/9+√((x_2-x_1 )^2+(y_2-y_1 )^2 )/5

[livid hound]

do you have a pic?

are x_1, y_1, x_2, y_2
just constants?

[idle sonnet]

where is x in the right side of the expression

[idle sonnet]

do you have a picture of the original question

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[crimson sedge]

what is does the dot symbol notate?

if a>0, a-b dot 0 is in S

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[idle sonnet]

What is a?

[crimson sedge]

a is an integer

[idle sonnet]

Is it in S?

Could you provide a bigger pic

[idle sonnet]

the dot is times

[crimson sedge]

really?

[idle sonnet]

Yes

[crimson sedge]

I mean from the next few lines the notation for multiplication is a-b(2a)

by parenthesis I mean

[idle sonnet]

Yes

[crimson sedge]

aight

thanks for the help

.close

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[wooden geyser]

Yo I need help with matrices pls

[wooden geyser]

Why is it when matrix X_32 x Y_44 = undefined

I see that the textbook says in a matrix the M must = N

For it to be defined

Why is that?

[idle sonnet]

What are 32 and 44

[wet fossil]

I think the order of matrices

[wooden geyser]

The elements

Rows then columns

[wooden geyser]

Whats that?

[idle sonnet]

do you mean the number of elements in each row and column?

[wooden geyser]

Yea

Sorry

[idle sonnet]

do you mean the left matrix is 3 tall and 2 wide or 2 tall and 3 wide

[wooden geyser]

3 tall and two wide

Eg) (f)

[idle sonnet]

you can think of matrix X as a transformation from 2D space into 3D space, and the matrix Y as a transformation from 4D space to 4D space, hence the height and width of the matrices

[wooden geyser]

Why would it be undefined

[idle sonnet]

so it doesn’t make sense to do them one after another

[wooden geyser]

I see

[idle sonnet]

matrix Y outputs into 4D space, and matrix X takes inputs from 2D space

so they don’t match

and you can’t do it

[wet fossil]

Fkr matrix composition to be defined, number of columns of 1st matrix must be equalt to the number of rows of 2nd matrix

[wooden geyser]

Are there any textbooks to read about this for an understanding?

[idle sonnet]

I highly suggest you watch these videos https://youtube.com/playlist?list=PL0-GT3co4r2y2YErbmuJw2L5tW4Ew2O5B&si=hg-XOl1la6KiQKkP

[wooden geyser]

Or online?

Thanks so much

[idle sonnet]

they provide good intuition

[cedar kiln]

@wooden geyser Has your question been resolved?

[cedar kiln]

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[wooden geyser]

How do I go about solving this?

[thorn lotus]

,rotate

[wooden geyser]

I dont want an answer please, just a hint if you know how

[wooden geyser]

No memo

[thorn lotus]

is this logic table?

[wooden geyser]

I have 0 clue

[thorn lotus]

This sounds like a logic problem. But I've never seen them done with matrices lol

[wooden geyser]

Ive been stuck for awhile

[wooden geyser]

What is a logic table? Like truth tables?

[thorn lotus]

like that sort of thing

[wooden geyser]

I see

[thorn lotus]

But yeah it would be interesting to try doing these problems with matrices

[wooden geyser]

I take it back I just want to see the solution s

[hearty arch]

i mean

you could just have a 3x3 matrix and have the rows be the husbands and the columns be the wives

and if they're not husband and wife, put a zero

[thorn lotus]

that's what i was thinking

[hearty arch]

if they are, put a 1

use where they live and where they work to determine who's not married to whom

each row and column can have a max of 2 zeros and 1 one so once you get a row/column with 2 zeroes, the last place must be a 1, so the row/column (husband/wife) must be married

[idle sonnet]

What if you call a logic table a “matrix”

[wooden geyser]

Im not seeing it sorry

[idle sonnet]

Whose husband is Mike?

[wooden geyser]

I dont know

[wooden geyser]

How do I implement the names into the 3x3 matrix?

[idle sonnet]

Are Mike and Lisa married?

[wooden geyser]

Mike is shorter than Lisas husband so I assume no

[idle sonnet]

What about Mike and Kelly?

[wooden geyser]

I dont know?

[idle sonnet]

Where do they work?

[cedar kiln]

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[peak minnow]

need a bit of help

[peak minnow]

this is my working , i need to find value of x z and y

but these values are wrong cuz if i use them in the equation , they dont give the same answer

can someone check hwere i did a mistake?

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@peak minnow Has your question been resolved?

[cedar kiln]

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[atomic viper]

Need guidance with question 9

I must simplify it

[simple bane]

So what is your doubt

[atomic viper]

I've done it up until here, but am unsure which step I should take next?

[atomic viper]

.

[simple bane]

You can try to add the top

[atomic viper]

Must I multiply the terms in the brackets first with the power?

<@&286206848099549185>

[cedar kiln]

@atomic viper Has your question been resolved?

[atomic viper]

<@&286206848099549185>a

[thorn lotus]

Do you still need help?

[atomic viper]

yes very much so ;-;

I have four more questions and I wanna finish em tonight

[thorn lotus]

so essentially you have a situation where $x + \frac{1}{x}$

[wraith dagger]

TooManyCooks

[thorn lotus]

can you multiply the numerator and denominator by $(1-x^2)^{1/2}$

[wraith dagger]

TooManyCooks

[atomic viper]

That would eliminate the denominator thus, making solving it easier, right?

[thorn lotus]

Uh, which denominator are we talking about

[atomic viper]

1-x^2

wait

[thorn lotus]

No it won't

It's supposed to eliminate $(1-x^2)^{-1/2}$

[wraith dagger]

TooManyCooks

[atomic viper]

I'll try it first

[atomic viper]

Wait, so do I restart from the start. I already made that positive

Is this correct?

@thorn lotus is it ;-;?

[thorn lotus]

Yes

Don't be scared to try things out

just play with it until you see something

[atomic viper]

what did you meant by this?

[thorn lotus]

You see that x^2 term in the fraction?

it has a 1/(1-x^2)^1/2

[atomic viper]

mhmm

[thorn lotus]

This trick of multiplying by a fancy 1 will eliminate that

[atomic viper]

So do I cross out the stuff i can?

[thorn lotus]

Yes

please do

[atomic viper]

I have this now

[thorn lotus]

You cancelled the wrong thing. I wanted you to focus the numerator with the numerator

Focus on that

[atomic viper]

I should deal with these two?

[thorn lotus]

With the arrows? Yes

[atomic viper]

so that times that would be 1 right since they're the same

[thorn lotus]

Yes! That's right

[atomic viper]

Did I do it?

[thorn lotus]

You need to multiply the whole numerator

You skipped one part of it

The exponent of the first term should be 2/2 now

[atomic viper]

So I should've canceled (1-x^2)^1/2 with the other one first?

[thorn lotus]

$\frac{(1-x^2)^{1/2} + x^2(1-x^2)^{-1/2}}{(1-x^2)^{2/2}}$

[wraith dagger]

TooManyCooks

[thorn lotus]

That's your problem, right?

[atomic viper]

The denominator doesn't have 2/2

[thorn lotus]

what's 2/2?

[atomic viper]

It's (1-x^2) by itself

[atomic viper]

1?

[thorn lotus]

and do you agree that $a^1 = a$?

[wraith dagger]

TooManyCooks

[atomic viper]

Yes

[thorn lotus]

Okay, then $(1-x^2)^{2/2} = (1-x^2)$

[wraith dagger]

TooManyCooks

[atomic viper]

Yes that's right

[thorn lotus]

Cool, glad we agree

So

[atomic viper]

But I wanna ask

[thorn lotus]

Sure. what is it

[atomic viper]

Why place 2/2 there. It's right but why?

[thorn lotus]

Remember I asked you to multiply by 1 in a fancy way?

[atomic viper]

Oh so that's what you meant

[thorn lotus]

The multiply and divide by (1-x^2){1/2}

Yes

So now the denominator will have (1-x^2)^{3/2}

[atomic viper]

Wait so, you meant to say all the terms in there are multiplied by one, so that the ratio is no longer there, but it changes nothing to the other terms?

[thorn lotus]

$\frac{(1-x^2)^{1/2} + x^2(1-x^2)^{-1/2}}{(1-x^2)^{2/2}} \times \frac{(1-x^2)^{1/2}}{(1-x^2)^{1/2}}$

[wraith dagger]

TooManyCooks

[thorn lotus]

The one on the right is the fancy way of writing 1

$\frac{(1-x^2)^{1/2} + x^2(1-x^2)^{-1/2}}{(1-x^2)^{3/2}} \times (1-x^2)^{1/2}$

[wraith dagger]

TooManyCooks

[thorn lotus]

I just regrouped it

[atomic viper]

I'm confused again 😕

Can we go step by step

From the start

[thorn lotus]

Sure

You have this $\frac{(1-x^2)^{1/2} + x^2(1-x^2)^{-1/2}}{(1-x^2)^{2/2}}$

[wraith dagger]

TooManyCooks

[atomic viper]

The question just posting it

[thorn lotus]

To make things easier to look at let's rewrite it into $\frac{c + \frac{x^2}{c}}{c^2}$

I just hid the stuff inside c

oh sorry

wait

[wraith dagger]

TooManyCooks

[thorn lotus]

There

[atomic viper]

Why is the denominator squared?

Oh wait the numerator has a power of halves so that's why the denominator is 2 right?

[thorn lotus]

Because $(1-x^2)^{2/2} = (1-x^2)$

[wraith dagger]

TooManyCooks

[thorn lotus]

and we said $(1-x^2)^{1/2} = c$ just for now

[wraith dagger]

TooManyCooks

[atomic viper]

OK I'll go with it

[thorn lotus]

I'm not convinced

What's not clear

[atomic viper]

Like, I don't know why must the (1-x^2) have a power of 2/2 shown

I know it's one but I don't know why it's shown there and why a rationof 2 specifically

[thorn lotus]

Because I wanted to emphasize that

$(1-x^2) = (1-x^2)^{1/2} \times (1-x^2)^{1/2}$

[wraith dagger]

TooManyCooks

[thorn lotus]

We can just leave it like that if it helps

[atomic viper]

Leave it without the 2/2?

[thorn lotus]

Yes. If the two makes it unclear then let's not do it that way

[atomic viper]

Ok

[thorn lotus]

Alright

$\frac{(1-x^2)^{1/2} + x^2(1-x^2)^{-1/2}}{ (1-x^2)^{1/2} \times (1-x^2)^{1/2}}$

[wraith dagger]

TooManyCooks

[thorn lotus]

You agree with that?

That's your original problem

[atomic viper]

This is the original problem, which I must simplify

[thorn lotus]

So I take it you don't agree

[atomic viper]

No

Sorry

[thorn lotus]

Are you comfortable with exponents?

[atomic viper]

Kinda in the process of understanding

Algebra isn't really my strong suit

[thorn lotus]

If I asked you $a^2 * a^4$ what's the answer?

[wraith dagger]

TooManyCooks

[atomic viper]

it would be $a^6$

[wraith dagger]

Potato_Fries

[thorn lotus]

Good. What if I asked $(a^2)^3$

[wraith dagger]

TooManyCooks

[atomic viper]

It would be $a^6$

[wraith dagger]

Potato_Fries

[thorn lotus]

Cool. How about $(a^4)^{1/2}$

[wraith dagger]

TooManyCooks

[atomic viper]

$a^2$

[wraith dagger]

Potato_Fries

[thorn lotus]

Okay so you're fine then if I say $(a^{1/2})^2 = a^1 = a$?

[wraith dagger]

TooManyCooks

[atomic viper]

Yes

[thorn lotus]

Then $(1-x^2) = (1-x^2)^{1/2} * (1-x^2)^{1/2}$

[wraith dagger]

TooManyCooks

[atomic viper]

Don't understand why the two are being multiplied together

[thorn lotus]

Do you agree that $a*a = a^2$?

[wraith dagger]

TooManyCooks

[atomic viper]

Yes

[thorn lotus]

So how is this different: $(1-x^2) = (1-x^2)^{1/2} * (1-x^2)^{1/2}$

[wraith dagger]

TooManyCooks

[atomic viper]

Would that make $(1-x^2)^2$?

[wraith dagger]

Potato_Fries

[atomic viper]

Wouldn't *

Since 1+1?

[thorn lotus]

I'm multiplying exponents of 1/2 though

so it's 1/2 + 1/2

[atomic viper]

Oh, my mistake, I was adding it with x's square

I'm a bit sleepy, sorry ;)

Yes, I do agree with your equation

[thorn lotus]

Okay so let's call (1-x^2)^{1/2} = c just to simplify the notation

[atomic viper]

Ok

[thorn lotus]

Your original problem has this form $\frac{c + \frac{x^2}{c}}{c^2}$

[wraith dagger]

TooManyCooks

[atomic viper]

Correct

[thorn lotus]

Good

Now, I asked you to multiply and divide by c

$\frac{c + \frac{x^2}{c}}{c^2} \frac{c}{c}$

[wraith dagger]

TooManyCooks

[thorn lotus]

Can you simplify that for me??

[atomic viper]

Do I cancel c with c?

[thorn lotus]

Just give me an expression

I don't know what you mean exactly

[atomic viper]

Is this right?

[thorn lotus]

You can cancel like that

$\frac{c(c + \frac{x^2}{c})}{c^3}$

[wraith dagger]

TooManyCooks

[thorn lotus]

That's what you have

right now

I just need you to simplify

I don't want a 1/c

on that x^2

[atomic viper]

So no canceling yet?

[thorn lotus]

The only thing that should cancel is that c right below x^2

Everything else is your usual algebra

[atomic viper]

So I have this now, what do I do with it?

[thorn lotus]

I want you to bring the c inside the parenthesis

[atomic viper]

So multiply c with the terms inside the brackets?

[thorn lotus]

Whatever you think is right

[thorn lotus]

Great

so what is c^2

Do you remember?

[atomic viper]

(1-x^2)?

[thorn lotus]

Perfect

Now plug that in

What do you get?

[atomic viper]

As in replace the c with the actual terms?

[thorn lotus]

yes

[thorn lotus]

Good job

So what's the final answer?

[atomic viper]

That?

Idk

I don't think I can cancel them

[thorn lotus]

You have 1 - x^2 + x^2

[atomic viper]

Yes?

Wait that's the answer?

How?

[thorn lotus]

I never said anything about a final answer

What did you get?

[atomic viper]

Wait, since it's division

That means 2/2 - 3/2 no?

[thorn lotus]

Ignore the division

I want you to focus on the numerator

[atomic viper]

I'm not sure what I can do with the numerator at this point?

[thorn lotus]

maybe this will help

[atomic viper]

I open the brackets right?

[thorn lotus]

I need to go to bed. I'm way too sleep deprived on a saturday

[atomic viper]

Is this right?

[thorn lotus]

Yes

[atomic viper]

That's the final answer right?

.close

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[versed geyser]

how do i do this

[livid hound]

first and last line are illegible
also what exactly are you being asked to do

[versed geyser]

x+y=4
x^2+2y+y^2=12

solve for x and y

[livid hound]

assuming that last line says
x = 4 - y?

substitute that into your second equation

[versed geyser]

so x^2+2y+y^2-x=8+y?

[gritty galleon]

what?

$(4-y)^2 + 2y + y^2 = 12$

[wraith dagger]

ItzKraken

[gritty galleon]

thats what u get when u sub x = 4-y

[versed geyser]

okay thanks

.close

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[crimson sedge]

How to prove it ? I know that S = RUi_AUR^-1 is a solution but I can't find the proof of the theorem

[crimson sedge]

I don't know how to prove it is the smallest set

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[jaunty thunder]

Smallest set means that if $R \subseteq T$, $T$ is reflexive and $T$ is symmetric, then $S\subseteq T$.
@crimson sedge
(Remember to tag me if you reply, so I wouldn't miss your message)

[wraith dagger]

kelvinchan9786#0690

[crimson sedge]

Yes I know

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[feral basalt]

Let 𝑩(𝒙) be the statement “𝒙 is a barista,” 𝑪(𝒙) be the statement “𝒙 is a
customer,” and 𝑺(𝒙, 𝒚) be the statement “𝒙 served a drink to 𝒚.” If the domain of 𝒙 and 𝒚
consists of all people, express each of the following sentences in terms of 𝑩(𝒙), 𝑪(𝒙), 𝑺(𝒙, 𝒚),
quantifiers, and logical operators.
a. Some barista served every other barista a drink.
b. There are two different customers who were served a drink by the same barista.
c. No customer served a drink to anyone.
d. There is a barista who served a drink to exactly one customer.
Help please?

[feral basalt]

I need help especially with part b

[cedar kiln]

@feral basalt Has your question been resolved?

[cedar kiln]

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[finite raven]

How do you prove that there is no natural n such that 0 < n <1 without the well ordering principle?]

[bleak gazelle]

1 = 0'

[finite raven]

agree

[bleak gazelle]

Peano axioms

[finite raven]

.close

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[pure wave]

I think I need to rationalize the denominator here because otherwise the function = 0/0?

[pure wave]

I know the limit is 5/4 because I graphed it, I just don't know how to prove it algebraically

[violet night]

have you tried multiplying by the conjugate

to try and get rid of that 5-5 = 0 on the denom

[pure wave]

I have, but maybe I'm not doing it right? every time I do I get an expression that equals 0 when I substitute 4 for x

[violet night]

what are you getting on the denominator after the process?

[pure wave]

the conjugate would be 5+sqrt(x^2+9) right?

[violet night]

yes

[pure wave]

so when I multiply that out I get 34-x^2?

[violet night]

wrong sign on the addition

[pure wave]

25-(x^2)+9= 34-x^2

[violet night]

it distibutes through the root

-(x^2+9) = -x^2 - 9

then think about what you have and if you can do anything with it to get rid of your 4-x problem

[pure wave]

wait okay
so when you multiply the denominator it = 25 -(x^2) -9

[violet night]

which simplifies to?

[pure wave]

(-x+4)(x+4)

[violet night]

or?

remember you have (4-x) on the numerator

[pure wave]

oh yeah I guess it makes more sense to write it as (4-x)(x+4)

[violet night]

double check

actually

its gud

ok now I think you can do it :D

[pure wave]

okay I got it! Thank you so much!!

[violet night]

❤️

[pure wave]

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[agile plume]

ok

[graceful island]

.close

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[runic dune]

What is wrong with this method of solving the problem? Did I make a mistake?

I know I can use the tangent addition or subtraction identity

[silent bobcat]

At the start, what is inside the tan ?

[runic dune]

Sorry

Tan(7pi/12)

[silent bobcat]

Also why do you think it's false ?

[runic dune]

Khan academy has a different answer

[dire geode]

Did you use a calculator to check it's different

6 = 2 * 3

[runic dune]

Calc says -root(3)-2

Khan academy says that

[dire geode]

Use a calculator to get a decimal

[runic dune]

-3.732050808

[dire geode]

It's that the same as khan

[runic dune]

No

Look at the pic

Or is it...

Lol it is

Ok so Khan academy just looks different

To be fair I haven't slept in like 20 hours..m

Thanks

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[crimson sedge]

so the sequence An is given by an+1 = 3 - 1/an

where a1 = 1

now i have to show that this converges

how can i go forth in doing that

[dire geode]

Induction could work

And Bolzano weirstrauss

[crimson sedge]

what we have done in uni so far is to show that the sequence is monotonic and bounded

[dire geode]

https://mathworld.wolfram.com/Bolzano-WeierstrassTheorem.html

[crimson sedge]

i need to show that an+1 >= an basically

[dire geode]

Yea that's basically this theorem

[crimson sedge]

ah so thats what it is called

but yeah i know about it but i dont know how to actually apply iy

it

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[rich sandal]

Hi I’m confused by this

[rich sandal]

It says I can integrate using u substitution

[surreal cave]

,rotate

[rich sandal]

I set 10-t as u

And the derivative of that is just -1dt

[surreal cave]

so if 10-t is u what does that make 40t?

[rich sandal]

I have to make that du

Idk how

[surreal cave]

just algebra, mein freund :)

[rich sandal]

I can put -40 outside the integral sign to make it -1t dt

[rain drift]

correct! Now there's one more thing you can do here. You said u = 10-t. So if you were to solve for t, how would you do that ?

[rain drift]

also hello friend!

[rich sandal]

t=10-u

[rain drift]

bingo!

so you can sub that into the numerator

[rich sandal]

I haven’t done that in forever forget that exists

[rain drift]

nifty little trick when you need it 🙂

[rich sandal]

Thank you!

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[versed fulcrum]

Not sure what to do

j = √(-1)

[cedar kiln]

@versed fulcrum Has your question been resolved?

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[lost agate]

Confused on how to continue this

[cedar kiln]

@lost agate Has your question been resolved?

[gleaming temple]

f(x)=1x−4,g(x)=2x+4c

For the following exercises, use each pair of functions to find f(g(x))
and g(f(x)).
Simplify your answers.

i do not understand

[lost agate]

<@&286206848099549185>

[cedar kiln]

@lost agate Has your question been resolved?

[cedar kiln]

@lost agate Has your question been resolved?

[cedar kiln]

@lost agate Has your question been resolved?

[hearty arch]

hi i can help i think

so if the function is continuous at x=-2, then as x approaches -2 from both sides, it must approach the same y-value for it to be continuous, right?

[lost agate]

This is where I’m at now

i think it cant be continous because the limits do not equal each other

[hearty arch]

well i'm not sure you need to actually use limits to solve this

shouldn't the very first part of the piecewise function also have to equal -6 when x=-2

$\frac{x^2-kx-2}{3x^2+4x+3}$ must also be equal to -6 when x=-2

[wraith dagger]

Wumbo

[hearty arch]

because $x^3+2=-6$ at x=-2

[wraith dagger]

Wumbo

[lost agate]

the first part does equal -6

or you talking about the first equation in the function?

[hearty arch]

the first equation in the function

there's a k-value that exists such that the first two equations approach the same value as x approaches -2

[lost agate]

maybe there is a misscalculation in this equation that i am missing?

correction here after factoring

[hearty arch]

where did the k go

and you don't need to factor it

just plug in x=-2 into the equation and make the equation equal to -6

$\frac{x^2-kx-2}{3x^2+4x+3}=-6$ when x=-2

[wraith dagger]

Wumbo

[hearty arch]

find k

[lost agate]

so you get down to the equation
$\frac{-2^2-k(-2)-2}{3(-2)^2+4(-2)+3} = -6$

[wraith dagger]

dirt😶

[hearty arch]

yep!

[lost agate]

$\frac{4-2k-2}{48 - 8 +3} = -6$

[wraith dagger]

dirt😶

[lost agate]

after simplfying

[hearty arch]

the 48 should be a 12

[lost agate]

why is that?

oh

nevermind

$\frac{4-2k-2}{12 - 8 +3} = -6$

[wraith dagger]

dirt😶

[hearty arch]

yes

[lost agate]

$\frac{2-2k}{7} = -6$

[wraith dagger]

dirt😶

[lost agate]

sstill correct?

[hearty arch]

yep!

[lost agate]

this is a stupid question but i forget how to solve for k

[hearty arch]

lol. no worries

multiply both sides by 7 to get rid of the denominator

[lost agate]

$7(2-2k) = 7(-6)$

[wraith dagger]

dirt😶

[lost agate]

$14-14k = -42$

[hearty arch]

wait

[wraith dagger]

dirt😶

[hearty arch]

you multiplied both sides by 7, and there was a 7 in the denominator already

so it gets rid of the 7 on the left side, and you multiply by 7 on the right

$2-2k=7(-6)$

[wraith dagger]

Wumbo

[lost agate]

$2-2k = -42$

[wraith dagger]

dirt😶

[hearty arch]

right

[lost agate]

$-2k = -44$

[wraith dagger]

dirt😶

[hearty arch]

yes

[lost agate]

$k = 22$

[wraith dagger]

dirt😶

[hearty arch]

yes

[lost agate]

so k = 22 when x = -2?

[hearty arch]

yeah. so if k=22, then as x approaches -2, both equations will approach the same value, -6

which makes it continuous

both equations being the first two equations of the piecewise function

[lost agate]

gotcha, do you know about the 2nd part of the question? the discontinueites?

[hearty arch]

yeah i just googled it to refresh myself. a removable discontinuity occurs when there's a point in the graph with a hole in it. if you were to graph this piecewise function, then when x=0, the graph ends up jumping to a different point so that it looks like there's a hole in the graph.

and in order to prove that it's a discontinuity we must prove that it would be continuous otherwise

so as x approaches zero from the left and right, the function should still be continuous

even though there's a hole at x=0

does that make any sense

[lost agate]

one sec writing down the equations from earlier

semi makes sense, i feel like I would understand it if i knew how to graph these

does this part include limitsS?

[hearty arch]

no it doesn't

so we're most interested in what's happening around x=0

so we should look at x^3+2 and 2/(1-x) when x=0

[lost agate]

so why 2/(1-x) if x should be 0

[hearty arch]

because we're interested in what's happening around x=0, from the left and right

we know already from the piecewise function that when x=0, the value of the function is 0, but if we were to remove this piece of information and find that the function is continuous, then this would be a removable discontinuity

a removable discontinuity looks something like this

[lost agate]

the white dot?

means open right?

[hearty arch]

weird that it pasted like this

yes

so if the function approaches the same value from both sides it's continuous usually

but since AT the value x=0 it equals something else it isn't technically continuous

but it would be a removable discontinuity

in order to prove this, we must prove that the function approaches the same value from both the left and right

so x^3+2 should be equal to 2/(1-x) at x=0

if it isn't, then it's not removable

[lost agate]

$0^3+2 = 2/(1-0)$

[wraith dagger]

dirt😶

[hearty arch]

so the 2 equations approach the same value at x=0, do you agree?

[lost agate]

I assume both sides come out to 2

[hearty arch]

yes

0+2=2/1

[lost agate]

so they do meet at the same value from both the left and right

[hearty arch]

they do

[lost agate]

which makes it a discontinuity

[hearty arch]

our piecewise function tells us, however, that at x=0, y=0

so the function approaches the same value until it actually reaches that value, at which point it jumps

which makes it a removable discontinuity

because if we were to remove that piece of information the function would have been continous

[lost agate]

at what part do you get y = 0?

[hearty arch]

the piecewise function says that at x=0, f(x)=0 right?

[lost agate]

ah yes

[hearty arch]

.

yea

but the function itself approaches the same value from the left and right as x approaches 0, even if it jumps to a different value when it is exactly x=0

this means that there is a removable discontinuity at x=0

[lost agate]

makes sense, i appreciate the help and explaination man

[hearty arch]

yeah i hope i helped!

[lost agate]

it's hard finding videos on these problems especially cause I dont know what these types of problems are called and when I look up piece wise functions I can never find any examples close to these questions I have

[hearty arch]

yeah i feel like it's because piecewise functions can differ greatly due to their nature

[lost agate]

that's for sure lol, you up for another problem?

[hearty arch]

yeah go for it!

[lost agate]

I actually have these last 2 problems

how similar is the process for 3 to the piece wise function?

is this a limits question??

[hearty arch]

i don't think they're similar besides the fact that they deal with the notion of continuity

for number 3, instead of focusing on where it is continuous, it may be beneficial to see where it is not continuous

[lost agate]

how would the overall layout of this problem go? like where to start with x values

[hearty arch]

well i see there's an x-1 in the denominator

[lost agate]

it could potentially be 0 which could make it undefined?

[hearty arch]

correct. if a function is undefined at a certain value of x, is it continuous at that value?

[lost agate]

i think no? because it would never reach that value of x?

[hearty arch]

yes. in order for the function to be continuous at f(x), it must be defined at f(x)

and since the function is undefined at x=1, then the function cannot be continuous at x=1

[lost agate]

could this be true for all real numbers?

[hearty arch]

any other number that you plug into this function will spit out a defined value

so the function would be continous everywhere except for x=1

[lost agate]

how would you write that as an answer? [1, inf)?

thats probably the incorrect notation

[hearty arch]

(-inf,1),(1,inf)

is the correct notation

having the bracket at 1 means it's defined at 1, which it isn't

number 4 i'll leave up to you. looks like you'll need to make a table of values around x=2

[lost agate]

gotcha thank you, will you still be around if I get stuck on it?

[hearty arch]

Yeah I'll be around. Feel free to dm me if I'm not. I'm chronically online so

[lost agate]

@hearty arch does this require x to have a list like 1.9, 1.99, 1.999 and so on?

[hearty arch]

Yeah basically

Values greater than 2 may also be beneficial to put into your table to prove your point

[lost agate]

yeah x > 2 for postiive side and x < 2 for negative

assuming this list can go on infinetely what is a comfortable set of numbers to stop at?

[hearty arch]

As many as you think you need in order to prove that there's a vertical asymptote

You probably don't need more than 5 or 6 total x-values in your table

[lost agate]

@hearty arch Think I did this wrong

[hearty arch]

The x-values greater than 2....

Wouldn't you get negative values under the radical?

[lost agate]

yes was using online calculator for some reason did not put negative sing

sign

[hearty arch]

Oof

[lost agate]

trying to budget run through my associate degree lol

based off the data it seems like the values dont go above 2, so that means it does have a vertical asmpytope of x = 2?

[hearty arch]

I feel that

Right so real values of the function don't exist past x=2, and it's undefined at x=2, and as x approaches 2 from the left the values are increasing

[lost agate]

so this problem sshould be finished?

[hearty arch]

As long as you have the table and have justified why it's a vertical asymptotic, yes

[lost agate]

awesome, thanks man. What is your major?

[hearty arch]

I have my degree in applied mathematics

[cedar kiln]

@lost agate Has your question been resolved?

[cedar kiln]

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[rose anchor]

lim𝑛→∞
|𝑎𝑛| = 0 ⇐⇒ lim𝑛→∞
𝑎𝑛 = 0.

[rose anchor]

How can I prove it?

[crimson sedge]

Is it a/n or an?

[south tundra]

I think they meant $\lim_{n\to\infty}\abs{a_n} = 0 \Leftrightarrow \lim_{n\to\infty}a_n = 0$

[wraith dagger]

A Lonely Bean

[crimson sedge]

I see

[idle sonnet]

are you allowed to use epsilon delta?

[south tundra]

Looks like the definition's application

[rose anchor]

Yes

[rose anchor]

yeh

[idle sonnet]

then you can expand out each of the limit statements into it and see how far you get

[rose anchor]

Ok will try that and ask again if I don't find it out

But just do them both on their own kind off?

[idle sonnet]

What do you mean

[rose anchor]

The implication

Do I just disregard that ? and prove that an and abs(an) both go to 0?

or actually I have no idea how to do it

.close

[cedar kiln]

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[willow bridge]

I know that for a function $f:\left(0,:+\infty \right)\rightarrow \mathbb{R}$

[wraith dagger]

TubyconB

I know that for a function $f:\left(0,\:+\infty \right)\rightarrow \mathbb{R}$

[willow bridge]

it is true that $f\left(a\cdot b\right)=f\left(a\right)+f\left(b\right)$ and

[wraith dagger]

TubyconB

[willow bridge]

$:\lim _{x\to 1}\left(\frac{f\left(x\right)}{x-1}\right)=1$

[wraith dagger]

TubyconB

[willow bridge]

so $f:\left(0,:+\infty \right)\rightarrow \mathbb{R}, f\left(a\cdot b\right)=f\left(a\right)+f\left(b\right) and :\lim _{x\to 1}\left(\frac{f\left(x\right)}{x-1}\right)=1$

[wraith dagger]

TubyconB

[willow bridge]

how do I find the limit $\lim _{x\to a}\left(\frac{xf\left(x\right)-af\left(a\right)}{x-a}\right)$

[wraith dagger]

TubyconB

[willow bridge]

from this I found that $f\left(1\right)=:0$

[wraith dagger]

TubyconB

[willow bridge]

I am not really sure wether that is entirely correct

since I had something like $\lim _{x\to 1}\left(f\left(x\right)\right)=0$

[wraith dagger]

TubyconB

[willow bridge]

then I assumed that a=1, which is the point where I am not sure if I made a mistake or not

[willow bridge]

but it seems to be supported by this

since I get $f\left(b\right)=f\left(b\right)$, which is true

[wraith dagger]

TubyconB

[willow bridge]

and then the limit is easy to compute, but only if a is truly 1

sorry for the long introduction

<@&286206848099549185> kind of a long one, sorry for that

read from the top if you could

nobody?

[willow bridge]

guys I really just want someone to tell me if I am allowed to do that, you don't have to solve anything else

this is supposed to be an easy high school question and I already feel bad having to ask for help lol

[idle sonnet]

this is not necessarily true for any function

[idle sonnet]

It’s a bit hard to understand your work, do you have a picture of when you wrote it on paper?

[willow bridge]

yeah one second

some things are in greek

I will explain

didn't mean to send it as pdf

@idle sonnet here

[idle sonnet]

What are you asked to find?

[willow bridge]

this

that's what Ν. β. means

also somewhere I say {something in greek} a=1, that is where I assume a=1

also β = b

[idle sonnet]

for any a?

[willow bridge]

I didn't use the fact that f is defined in (0, inf+) though

[willow bridge]

for any a and b that belong in (0, +inf)

yeah it's very iffy

still there?

[idle sonnet]

you could also have derived that f(1) = 0 by substituting a = 1 into the functional equation you were given

[idle sonnet]

Yes

But ping me if I don’t respond

[willow bridge]

my question is, am I even allowed to do that

[idle sonnet]

Yes, because it says for all a and b

so you can do it

[willow bridge]

it's very convinient to do so, but where is the proof

ok

so I solved it?

@idle sonnet

[idle sonnet]

Solved what?

[willow bridge]

the limit

[idle sonnet]

Which limit?

[willow bridge]

this

[idle sonnet]

Why do you think you solved it?

[willow bridge]

since a = 1

and f(1)=0

[willow bridge]

you can see it in the bottom part of this

I just substitude a = 1 immediatelly, sorry if it's a bit confusing

[idle sonnet]

I think you have

for the specific case a = 1

it remains to prove that this is the case for all a

[willow bridge]

ok how do I go about doing that

ping me if you want to discuss that

[cedar kiln]

@willow bridge Has your question been resolved?

[cedar kiln]

@willow bridge Has your question been resolved?

[willow bridge]

.close

[cedar kiln]

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[little junco]

Help, what is the autonomy of this ODE?

[cedar kiln]

@little junco Has your question been resolved?

[cedar kiln]

@little junco Has your question been resolved?

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[near wadi]

Someone help me proof m1m2=-1 in a geomtry way

[near wadi]

Someone help me proof m1m2=-1 in a geometric way

using a^2 + b^2 = c^2

That was a clue

of how

[crimson sedge]

m1m2 ??

Whats that?

[junior dome]

i think they mean the slopes of two mutually perpendicular lines.

[junior dome]

be more specific next time.

[near wadi]

The slopes of 2 lines that have a 90 degres angle times each other = -1

I need help proving that

[cedar kiln]

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[deep otter]

I'm confused on why we add the negative in front of the square root when we put /x^6 in the square root

[slate lintel]

it's explained on the right

we're working with negative numbers here so x³ is also negative

but x⁶ is positive

[deep otter]

oh. I was confused on its explanation. But what you said makes sense. Thanks!

Actually, I'm still a bit confused

[slate lintel]

when we divide by x³, it's going to flip the sign

so we need to make sure that's still true when we bring it inside the square root sign

√(x⁶) is positive though

so what we need is -√x⁶

[deep otter]

so, if x^3 at the denominator at the start of the problem is x^4, would we have to add a negative in front of it? if that makes sense

Like I know this problem doesn't ask that, but just giving an example

[slate lintel]

no, if we were dividing top and bottom by x⁴ instead we could leave it as-is since that's always positive anyway

notice how the denominator at the start is -inf, but after we divide by x³ it's +1

[deep otter]

-inf?

[slate lintel]

negative infinity

[deep otter]

ohhh.

I understand now

[slate lintel]

if i have the infinity symbol on my phone i can't find it lol

[deep otter]

Thank you for the help again. It's much appreciated

.close

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

Could someone look through my steps

Your solo on this one 💀

Wym

<@&286206848099549185>

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

It's written in numbers and the letter x, but somehow it still skips my mind when i try to read

Please explain what you tried to do and why

I figured it out

thanks

.close

[cedar kiln]

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[grave relic]

i don’t ynderstNd this

[grave relic]

<@&286206848099549185>

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[grave relic]

ah

[night atlas]

Point A or point B?

[grave relic]

there’s points ??🥲

[south tundra]

There are two questions, yeah

[gleaming cloud]

bruh which question

[south tundra]

You don't understand which bit specifically?

[grave relic]

ohhh

both

[night atlas]

a) and b)

[grave relic]

i’ve got like 10 more pages

i don’t understand

[grave relic]

both

[night atlas]

The second one you just have to solve the equation, roots = solutions

But you can easily find them in the graph

[grave relic]

ummm

[south tundra]

Let's start with a perhaps

[grave relic]

i failed my exams

[night atlas]

And the turning point is the minimum of the function

[south tundra]

Can you see the lowest point of the graph?

[grave relic]

yes

[south tundra]

Can you tell its coordinates?

[grave relic]

-4,1?

[south tundra]

The other way, (1, -4)

[grave relic]

oops

[south tundra]

x coord is usually written first

[grave relic]

so that’s the answer for a

[south tundra]

Yup

[grave relic]

and thennnn

[south tundra]

Now, when looking at a graph, you can tell where the roots of the function are by looking for the points where the graph intersects the x-axis

In this case the given graph has two such intersections, can you tell their coordinates as well?

[grave relic]

what’s an intersection

where the line goes thru?

[south tundra]

Yeah, the points where the graph and the x-axis meet

[grave relic]

i don’t understand

the line that has the dip?

also for other questions do i keep putting them here or do i open a new thingy

[south tundra]

Like the places where the graph crosses the x-axis

[south tundra]

Keep this one open unless you don't get a response for many hours or smth