help-13

[barren rose]

45 is by itself

And then we are subtracting the 3 events from x

45=x-(s1)-(s2)-(s3)

Mb typed it wrong the first time

So s1 we want 1/5 of x

[gentle shard]

no worries

oh ok

[barren rose]

So s1 can simply be (1/5)x

Then s2 we want 1/7 of what's remaining after subtracting (1/5)x

Which is (4/5)x

So we can do (1/7)(4/5)x

Then finally for s3 we want 3 times the original full price sales amount

so just 3(1/5)x or (3/5)x

So fill in the equation using those and simplify to find x

[gentle shard]

thank you let me try to find the x

[barren rose]

Lmk what you get then i'll send my work for it

[gentle shard]

ok

x=-255

[barren rose]

no

Not even possible

Send me your work

I'll see what you did wrong

[gentle shard]

ok

wait

75

x=75

[barren rose]

Still a no

Send your work

[gentle shard]

ok

[barren rose]

@gentle shard needs to be an x after 3/5

its 3/5 of the total

[gentle shard]

it just did not write it but solved i t

8/5

8/5x

[barren rose]

Ok let me send you my work and you can compare because I can barely read yours lol

[gentle shard]

my bad

[barren rose]

Np

@gentle shard

[gentle shard]

yes thank you for the help

i missed out on that 1/7

[barren rose]

ah

Alr well lmk if you need anything else

[gentle shard]

no worries thanks for the help

[barren rose]

@gentle shard Do ,close

[cedar kiln]

@gentle shard Has your question been resolved?

[cedar kiln]

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[flat grove]

Hello I need help with some algebra one Solving Systems Of The Equations By Substitutions the question is this y=-3x+5 5x-4y=-3

[flat grove]

I have not made any progress on the question and confused how to do it

I know in the other equations I need to get the y or x to which would equal a number for example let’s say this equation y=3x+10 and the x equals 0

[brittle anvil]

hello i can help

[brittle anvil]

yes thats correct

so you solved for y

[flat grove]

Yea

[brittle anvil]

are you trying to solve for x?

[lusty flax]

let's see what we can do here

[flat grove]

Ignore the pencil writing on the top but that’s the equation I need help with

[brittle anvil]

i think its asking for you to solve for y on one equation and x on the other

im not sure though

[flat grove]

Ok

[crimson sedge]

@flat grove what does y equal?

[flat grove]

-3x+5 ?

[crimson sedge]

Yep, do you know how to substitute?

[flat grove]

Would I put -3x+5 in front of 4 in parentheses that’s what I’m guessing

[crimson sedge]

you're subtracting 4y, what would -4y be in terms of x?

[flat grove]

Would I subtract 5-4

If not I have no idea then

[crimson sedge]

hmm, you know that y=-3x+5. What would you multiply y by to get -4y?

[flat grove]

-3 ?

[crimson sedge]

y*-3 would be -3y

[flat grove]

I’m I correct?

[crimson sedge]

yes this would work but make sure you subtract everything inside

since its -4y

so you would get 5x-4(-3x+5)=-3

[flat grove]

Ok let me write that down

Ok what’s next

[brittle anvil]

you solve it

first you would destribute

-4(-3x+5)

[crimson sedge]

Yep, solve for x

[brittle anvil]

after you solve for x plug that into y=-3x+5 and you would have the x and y coordinate that would be your answer

i think*

[brittle anvil]

nice now do the destributive property

[flat grove]

I’m confused how do I do that

[brittle anvil]

ok so

you would take the number outside the parenthesis

with is -4

which*

and you destribute it so you would be doing -4*-3x

and then -4*5

so you are basically doing two seperate equations

[brittle anvil]

oop you didnt multiply -4 by -3x

when a number is outside the bracket you multiply

[flat grove]

OH whoops I forgot that the star means multiplication sometimes let me redo that part

[brittle anvil]

lol

[flat grove]

Wouldn’t it be negative 7 tho ?

[brittle anvil]

no

lets get rid of the x

what is -4*-3?

you would carry over the x

[flat grove]

-12 ?

[brittle anvil]

it would be positive twelve

[flat grove]

Ok

[brittle anvil]

because when two negatives are multiplied they are positive

[flat grove]

https://tenor.com/view/edward-james-olmos-negative-times-gif-22341657

[brittle anvil]

so you would have 12x

yes

[flat grove]

I’ve got reminded of this one scene in Stand And Deliver but back to topic

What next

[brittle anvil]

ok so now you have 5x + 12x - 20 = -3

you see the numbers you solved for?

12x and -20

so now you can combine like terms

[flat grove]

Ok do I solve that ?

[brittle anvil]

yes

[flat grove]

0?

Wait

[brittle anvil]

lol

[flat grove]

Do I need to do 5 times 12

[brittle anvil]

no

[flat grove]

Since there’s a x ?

[brittle anvil]

they both have x and only x

so you can just add them together

that is what combining like terms is

[flat grove]

5 plus 12 is 17 then -20 equals 3

[brittle anvil]

remember there is still a x

17x - 20 = -3

you have that now

do you know what to do?

[flat grove]

No

[brittle anvil]

ok so we want to get the x term on one side

the x term is 17x

and we have a -20 on the same side of the equal sign

how do you make -20 become 0?

[flat grove]

Remove the 2 ?

[brittle anvil]

uh

what would you add to -20 to become 0?

[flat grove]

You said to make it a zero

[brittle anvil]

yeah

you cant just take away the 2

you would add 20 to -20

[flat grove]

If you remember the 2 from a 20 it’s just 0

[brittle anvil]

to make it 0

[flat grove]

Ahh

[brittle anvil]

lol

[flat grove]

Alright

[brittle anvil]

so you removed -20 from one side of the equal sign by adding 20

you have to add 20 to the other side

17x = -3

you would add 20 to -3

[flat grove]

Alright

Is it ok if I come back to this another time ? I’m getting confused and need to take a break I might come back tomorrow is that ok ?

[brittle anvil]

ok

ask your teacher for help

[flat grove]

Alright

I’m done

.close

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[radiant oxide]

sqrt(2) - sqrt(5)

[radiant oxide]

i don't know what to do

[dense wing]

it'd help to have instructions for one

[radiant oxide]

a) -3
b) -sqrt(3)
c) sqrt(10) * [(1/sqrt(5)) - (1/sqrt(2))]
d) sqrt(3) * [(sqrt(2)/sqrt(3)) + (sqrt(5)/sqrt(3))]
e) sqrt(5) * [(1/sqrt(5)) - 1]

i'll just send a picture lmao

[dense wing]

It'd still help to have instructions

Instructions aren't "here's a bunch of random numbers"

[radiant oxide]

instructions are "which of these is correct", basically

o

[dense wing]

🤨

[radiant oxide]

equals

sorry

[hazy hazel]

hes asking which one equals to the sqrt(2) - sqrt(5)

[radiant oxide]

which of the following equals

yes

thank you

[dense wing]

well obviously it isn't a

[radiant oxide]

yup lol

[dense wing]

I don't think I need to explain that yep

now is $\sqrt{a}+\sqrt{b}=\sqrt{a+b}$?

[wraith dagger]

Mosh

[radiant oxide]

uh

no?

[dense wing]

yes

so it's not b either

[radiant oxide]

it is?

[dense wing]

yes, it isn't

yes was in agreement to your no

[radiant oxide]

i see

[dense wing]

however $\sqrt{a}\sqrt{b}=\sqrt{ab}$

[wraith dagger]

Mosh

[dense wing]

likewise for division

so apply distributive property and that property of radicals to simplify c-e

[radiant oxide]

huh

[dense wing]

what?

[golden siren]

is the distributive property of radicals

[radiant oxide]

o

i don't see subtraction/addition there though?

how do i apply that to

my problem

[golden siren]

mk so in this situation it would probably be easiest to solve the choices to see which one matches sqrt(2) - sqrt(5)

we know its not A or B, so lets start with C

[radiant oxide]

hold on

is it C?

[golden siren]

it does seem to match. If you want to be really sure, you can even solve the other ones.

[radiant oxide]

i just realized i can check on the book

ITS C LETS GOOOOOO

[golden siren]

Lol good job!

[radiant oxide]

thanks guys i had no idea how to solve it :)

[golden siren]

Yw!

[radiant oxide]

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[sick sonnet]

Is the derivative correct?

[cedar kiln]

@sick sonnet Has your question been resolved?

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[crimson sedge]

Im having trouble understanding strong induction

[cedar kiln]

@crimson sedge Has your question been resolved?

[upper abyss]

Make sure you follow the steps exactly as they're written. Should start with the easy part - the base case.

That is, you want to show your statement is true for n = 1

[crimson sedge]

i have the base case down p(1) = 4 ,(1/2)(3+5)

i wish i knew how to use the bot

[upper abyss]

You're okay, I get you

Right, so both sides are 4, base case is proven

We assume the statement is true for n = k. That is, magically, this statement becomes true:
4 + 9 + 14 + ... + (5k - 1) = (k/2)(3 + 5k)

We want to prove the statement is true for n = k + 1.

[crimson sedge]

k +1 , is the next step?

[upper abyss]

You're very close in your picture. There's a minor mistake

On the left side, that (5n - 1) should read (5n + 4)

As that is the n+1th term

[crimson sedge]

could you explain why a bit more?

[crimson sedge]

ok i got it

ty for your help

.close

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[serene quiver]

Not sure if right and how to do the rest

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[opaque notch]

Please solve this question and explain the process

[opaque notch]

<@&286206848099549185>

[dire geode]

#rules

[opaque notch]

What rules?

[opaque notch]

What are the rules?

I just want the solving process. I am stuck with this math for like 3 days.

[uncut hedge]

@opaque notch develop f(f(x)). You will find something that simplifies really well, then solve the equation.

[opaque notch]

I get, f(x/x-1)=x/x-1

[uncut hedge]

You must have made a mistake in your calculations here

[opaque notch]

I know... but that's what I am finding. And, it's very frustrating.

[uncut hedge]

Show the steps that lead to this result

[cedar kiln]

@opaque notch Has your question been resolved?

[cedar kiln]

@opaque notch Has your question been resolved?

[opaque notch]

<@&286206848099549185> do you wanna help or what?

[dire geode]

Use the definition.
f(f(x)) = 1 + 1/(f(x)-1) = f(x) = 1+1/(x-1)

[cedar kiln]

@opaque notch Has your question been resolved?

[opaque notch]

Just solve it already😓

[wraith crypt]

since you seem to not want to read the rules and i'm feeling kind, i'll read it for you: We are not here to do your homework for you, but we are here to help you learn how to do your homework.

so, no, we won't "just solve it already"

[opaque notch]

Who said it's a homework...

[wraith crypt]

regardless, it's your question, and the effort to solve it needs to be yours, and we are only here to guide you in your attempts

so, i suggest you start by not demanding answers, and actually trying out the hints the others have given above

[opaque notch]

But I did use them. And, I am not able to go far.

Okay, as a member you can't help. But as a friend, you can. I am adding you. Just make that math a simple equation, then I will solve it.😑 I am very frustrated.

.close

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[grim thistle]

can someone help with question 5?

I’ve expanded the brackets

[fallen heath]

compare the rational, irrational parts of both sides!?

[grim thistle]

27-9surd2+3asurd2-2a=23-bsurd2

[cedar kiln]

@grim thistle Has your question been resolved?

[sand ether]

,rotate

[sand ether]

Equate the part without surds to 23

and the one with surds to $-b\sqrt{2}$

[wraith dagger]

azeem321

[grim thistle]

So I subtract 23 from the RHS

and subtract 23 from the LHS

to get -bsurd2 on its own

[sand ether]

So you have $27-2a -9\sqrt{2} +3a\sqrt{2} = 23 - b\sqrt{2}$

[wraith dagger]

azeem321

[grim thistle]

yes

[sand ether]

So $27-2a = 23$ you see that?

[wraith dagger]

azeem321

[grim thistle]

yes

but how did we get there

we got rid of the -9 surd 2 and the 3asurd2 and the -bsurd2

[cedar kiln]

@grim thistle Has your question been resolved?

[crimson sedge]

At what natural values ​​of a will the fraction a / 11 be correct and the fraction 6 / a incorrect at the same time? how do i solve it

im confused rn

[sand ether]

So we have $27-2a -9\sqrt{2} +3a\sqrt{2} = 23 - b\sqrt{2}$. In order for them to be equal the the surd parts must be equal and the other part must be equal

[wraith dagger]

azeem321

[sand ether]

So... $27-2a = 23$ an $\sqrt{2}(3a-9) = -b\sqrt{2}$

[wraith dagger]

azeem321

[cedar kiln]

@grim thistle Has your question been resolved?

[cedar kiln]

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[crimson sedge]

is this question more complicated than it seems?

Two players take turns placing quarters on a rectangular table with dimensions of 3 feet by 6 feet. If a player can’t find an empty area to place a quarter (so that it fully fits on the table) they lose.  Which player has a winning strategy and what is it?  (Assume the quarter is a circle with diameter of 1 inch.)

[crimson sedge]

i would say the first player

its just taking turns

[livid hound]

wdym by taking turns

taking turns isn't a strategy

[crimson sedge]

i dont think there is a strat

right?

[livid hound]

that doesn't answer my question

if you don't have a strat then you aren't guaranteed to win

[crimson sedge]

One player is garunteed to win

Since they are just taking turns in order

[livid hound]

you can place coins in a random position on the table

and if you don't have a clear start then both players are pretty much throwing their coin on

then the chances of each winning is pretty much 50:50

[crimson sedge]

Hmm yes

But will that affect anything about the spacing

[livid hound]

yes

[crimson sedge]

How

[livid hound]

consider that crappy 3x3 grid i drew

optimally you could place 9 coins there

however you don't have to place a coin exactly in the square

[crimson sedge]

Ah

This problem is much complicated than I thought

[livid hound]

the solution is quite simple

i can reveal a hint later if required

[jaunty mural]

Where are these problems from?

Theyre not rlly math math

Most people are unable to do these kinds of problems the first time they see them

[livid hound]

its apparently part of a logic / critical thinking class

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

That is very me

[jaunty mural]

It's most people

So what is this class?

[crimson sedge]

These problems are way too hard

[jaunty mural]

Is it part of a math course?

[crimson sedge]

Logic and proofs

[jaunty mural]

Ok let me put it this way. You see these problems --- you shouldn't be expected to solve them

Some of them are ok

[crimson sedge]

Yeah

[jaunty mural]

But the one you just posted, probably not without outside help

[crimson sedge]

They don’t expect you to solve every one

[jaunty mural]

Ok that's more reasonable

The point of these problems is for you to get used to 'being stuck' I guess

When you're stuck, you can be quite methodical

In approaching the question

[crimson sedge]

Ok

I will try something myself

But I will focus on the square one

[jaunty mural]

for this one

How do I put it

These kinds of problems involving 2 player games

[crimson sedge]

I’m going to try without help

[crimson sedge]

Yes

[jaunty mural]

yh but I will tell u something

These kinds of problems with 2 player games... often they involve parity in the solution

in some way

I don't know if you have seen many

But usually, to argue for a winning strategy, they use parity in the argument.

[crimson sedge]

Ok

[crimson sedge]

Do you know if parity is here?

The only problem with this one is that there is so much room

[jaunty mural]

I didn't mention it for no reason

[crimson sedge]

Ok

Well thanks for the hint

[jaunty mural]

Alright so.

Let's go back

[crimson sedge]

To square?

[jaunty mural]

drawing

[crimson sedge]

Ok

[livid hound]

hint 1: ||kids do this all the time||

[jaunty mural]

https://i.imgur.com/RDm3mIp.png

[crimson sedge]

Which problem are you referring to?

[jaunty mural]

Ok that took me a while

[crimson sedge]

Good effort

[livid hound]

the two player problem

[crimson sedge]

I honestly don’t know what kids do

[jaunty mural]

@livid hound I don't get the hint either 😂

[livid hound]

hint 2: if the playing area is symmetrical and is sufficiently large to place at least 1 coin, size doesn't really matter

[crimson sedge]

Hmm

That’s better

[crimson sedge]

Size never reallly matters tho

[jaunty mural]

in the versions I've seen you are allowed to place coins that balance on the edge

so even if the playing area is smaller than a coin, it is ok.

[crimson sedge]

I think the winning strategy would be for the second player to place their coin symmetric to the second players across the origin

[jaunty mural]

Let's talk about this problem

why?

[crimson sedge]

My instructor told me that when they give us these types of problems

It’s usually has to do with symmetry

[jaunty mural]

yes symmetry is related to parity

[livid hound]

I think the winning strategy would be for the second player to place their coin symmetric to the second players across the origin

[jaunty mural]

But you're going to have to say what 'across the origin' means

[livid hound]

close

close but no

[crimson sedge]

Ok

[livid hound]

also you wrote second twice

[crimson sedge]

So I’m on the right idea I think

[jaunty mural]

are you sure?

crabbo might have it but just phrasing

Tell us what you mean by 'symmetric across the origin'

[livid hound]

they wrote second twice so i'm not sure exactly what hey mean

[crimson sedge]

It’s called point symmetry I think

[jaunty mural]

Ok, you want to use point symmetry

So what are the instructions

[crimson sedge]

Yes

The second player will mirror the first across the origin of the table

[jaunty mural]

I agree with Ramanov

Close.

Hint: you can't always do what you said.

[livid hound]

a bit clearer but the answer is close, but not quite

[crimson sedge]

When can you not?

[jaunty mural]

Think about it 😅

[livid hound]

have a think about it

[crimson sedge]

Oh wait

I have it

[livid hound]

if only there was a way for player 1 to prevent player 2 from doing that

[crimson sedge]

The first player puts a coin in the center, and then mirrors the second players moves across the origin

Bingo

[jaunty mural]

Right

If I'm not mistaken, I believe there are 3 winning strategies

[crimson sedge]

How much help did I get on this one

Lol

[jaunty mural]

not much

[crimson sedge]

That’s good

[jaunty mural]

although ramanov did give the symmetry hint

[crimson sedge]

It’s becoming less and less

[crimson sedge]

Did they? They just said size didn’t matter

And that made me think of symmetry

[jaunty mural]

Can you think of the other winning strategies

For player 1

[livid hound]

as for hint 1:
kids like copying (each other)

[crimson sedge]

Kek

Lol

[jaunty mural]

You didn't have to choose point symmetry

or wait maybe i remember wrong

ahahahaa

[crimson sedge]

I don’t think other symmetry would have worked

[jaunty mural]

no, i think point symmetry is right

hmmmmmm

🙂

yes

https://i.imgur.com/HaHaC4p.png

Ok, so for this one we have to do this agreed?

[crimson sedge]

Yes

[jaunty mural]

You can reverse the arrows

And we have the same problem

[crimson sedge]

Yes

[jaunty mural]

Right so looking at the 8x8 grid

However I chop it into 2

I have to keep at least 2 edges intact

Because the cut has to start at 1 edge

and end at another edge

There can only be one cut to chop this thing into 2 pieces

In this example the right and bottom edges are intact.

ok?

[crimson sedge]

Yes

[jaunty mural]

https://i.imgur.com/EigLykb.png

I can even try something like this

and cut from the inside

but it's quite obvious this will be impossible

[crimson sedge]

Yes

[jaunty mural]

Ok, so if 2 8x8 edges are intact

this tells me something for the 7x10 board

It narrows down the possibilities for where I must start/end my cut

on the outside

Can you see where they could be?

https://i.imgur.com/MvoCQlQ.png

For example, this is not possible

I cannot end up with any jigsaw with a side = 8 if I do this.

[crimson sedge]

Right

But one side will not be 8 tho

[jaunty mural]

Exactly

So where must we make our cuts

We will have 2 jigsaw pieces. We know 2 of their edges in total will have side length 8

Maybe one piece has 2 edges that are 8? Or maybe each piece has one

But you can use this information

[crimson sedge]

Hmm

I’m going to be afk for 15 minutes

Ok I’m back

[jaunty mural]

So can you see where we must start/end our cuts?

There are only a few possible places

[crimson sedge]

Not really

[jaunty mural]

Ok, everything I have said so far

have you made sense of it?

However we split out 8x8 grid into 2 pieces

2 edges must be untouched

[crimson sedge]

Yes

I understood that

[jaunty mural]

https://images-ext-1.discordapp.net/external/BhmubFUm-OqMVeWC3ATzM9pMyVw9mTByzli1xp-GMSA/https/i.imgur.com/HaHaC4p.png

So those red pieces

in total must have the 2 untouched edges

of length 8

When we translate this back to the 7x10 grid

It tells us where we must cut

Do you follow?

[crimson sedge]

But we don’t have that

We can’t cut a shape like that

[jaunty mural]

https://images-ext-2.discordapp.net/external/1zNgzGh14a0cXtK-gMSBupr_Uvnw5fSbYstwwk9JrGE/https/i.imgur.com/MvoCQlQ.png

if i did this

the pieces have edges of lengths

4

6

5

2

10

7

can you see what im referring to?

[crimson sedge]

No

[jaunty mural]

The red lines

are meant to be me using scissors

https://i.imgur.com/YQQKDc0.png

this is an example ok?

what is the circled length

@crimson sedge???

[crimson sedge]

4

[jaunty mural]

https://i.imgur.com/MqHnkyK.png

?

[crimson sedge]

6

[jaunty mural]

4
6
5
2
10
7

Can you see where this comes from?

I am just counting the outside edges

[crimson sedge]

I see

Th

[jaunty mural]

https://images-ext-1.discordapp.net/external/BhmubFUm-OqMVeWC3ATzM9pMyVw9mTByzli1xp-GMSA/https/i.imgur.com/HaHaC4p.png

Two of the 8x8 edges are untouched

[crimson sedge]

Yes

[jaunty mural]

this almost guarantees we need two 8

[crimson sedge]

Ok

[jaunty mural]

on the 7x10

[crimson sedge]

We can do that

[jaunty mural]

so do you see where the cuts could be now?

[crimson sedge]

Yes

But I can’t really show how

[jaunty mural]

just in words

[crimson sedge]

Bottom and top row

[jaunty mural]

yes but where

I can't just do anywhere

between which squares?

[crimson sedge]

I don’t know

[jaunty mural]

???

It's getting quite hard to know what you're thinking

if you're not voicing your thoughts

[crimson sedge]

I don’t know how to voice them

[jaunty mural]

you just type what u r thinking.............

[crimson sedge]

The cut needs to start on the bottom and end on the top

And they both have to have a length of 8

That’s all I have

[jaunty mural]

https://images-ext-1.discordapp.net/external/sMPa8YT-t7CdGbocS2PgUdYwY72EUduv3UfrWoHXyIo/https/i.imgur.com/MqHnkyK.png

Remember this?

4 6 5 2 10 7

Two of those numbers needs to be 8

Is what I'm trying to tell you

[crimson sedge]

Right

[jaunty mural]

There is a slight possibility this doesn't happen

But I will tell you now, it is very likely

Because of how the problem works

If you think about it, it should make sense to you

[crimson sedge]

I wish I could draw and show you where I would cut

[jaunty mural]

You could describe?

You told me somewhere in the top row

where in the top row

You could tell me '5 squares from the left, make a cut'

[crimson sedge]

2 squares from the left make a cut on the bottom

[jaunty mural]

Yes.

https://i.imgur.com/qwyjbgR.png

So these are the possible places to cut

And you need to make 2 cuts

You should be able to see this only leaves 2 possibilities

https://i.imgur.com/bsrSrrH.png

https://i.imgur.com/1zpXVWa.png

These are the only 2 cases to consider

do you agree?

[crimson sedge]

Yes

[jaunty mural]

Right so this is a good start

Furthermore, can you see why one of these is probably impossible?

[crimson sedge]

No

[jaunty mural]

https://images-ext-1.discordapp.net/external/rz4D_XxBIh3lRj8N3hlBA-AOvRX0zmZ7cVAUbZvEkLU/https/i.imgur.com/bsrSrrH.png

If we try this one

one of the jigsaw pieces

will have a 8-7-8 sides

Doesn't that seem problematic?

When we are trying to make an 8x8 grid

I need to fit something like this on top

[crimson sedge]

It does

[jaunty mural]

to make the 7 side into 8

But you can see that is basically not possible with the left piece

[crimson sedge]

Right

[jaunty mural]

So basically

[crimson sedge]

You would have extra

[jaunty mural]

We will start by trying this

https://images-ext-2.discordapp.net/external/_bk9ZcXVSd7bYo2rj8kKap2IEpuH9oNbaDXpn1s04hg/https/i.imgur.com/1zpXVWa.png

Now think about the left piece for a second

The edges around the outside are 2-7-8

ok?

[jaunty mural]

Similar to this, we need something of width 1

to complete the 7

[crimson sedge]

I just see 2 and 7

[jaunty mural]

if you see what I mean?

The bottom is 8

[crimson sedge]

Hmm

Ok

[crimson sedge]

We don’t have 1

[jaunty mural]

You don't know

Let me show

https://i.imgur.com/XVpb6gY.png

I need to do one of these 2 things

if that makes sense?

The other piece needs to fit like that

[crimson sedge]

Not really

Eh

[jaunty mural]

Can you say more... I'm not sure how to explain it otherwise

[crimson sedge]

Wdym by this

[jaunty mural]

The other piece has to fit onto the left piece

to complete the jigsaw = 8x8 board

To complete the side of length 7

I must basically add an extra bit of side 1

[crimson sedge]

Ah right

How do you add that extra side of 1

I just don’t see how this will all fit together

[jaunty mural]

It will come from the other piece, of course.

Not from nowhere

[crimson sedge]

Right

Well, I wouldn’t know how to make the piece tho

[jaunty mural]

You have to basically do some smart guessing

There isn't any routine method

https://images-ext-2.discordapp.net/external/_bk9ZcXVSd7bYo2rj8kKap2IEpuH9oNbaDXpn1s04hg/https/i.imgur.com/1zpXVWa.png

Start with these cuts

Try to continue the left cut

and see if it makes sense

For example, does going to the left make sense?

Doing something like this makes it see harder to me

Maybe I should give a bigger hint

https://i.imgur.com/psxfKlD.png

@crimson sedge I will tell you now that this looks helpful

Can you see maybe why?

[crimson sedge]

not really

[jaunty mural]

I am trying to do this

This is to complete the side of length 8 on the left

https://i.imgur.com/dZ7A8sf.png

I have a tip

[crimson sedge]

hmm

[jaunty mural]

Try to draw the cuts on BOTH boards at the same time

[crimson sedge]

why?

[jaunty mural]

Compare the 2 boards

To see what I am trying to do

[crimson sedge]

you are trying to make this

but how

[jaunty mural]

Yes exactly

Can you not see how the 2 correspond?

If I start with that cut on the top one

And fit them together like that

I will result in this

[crimson sedge]

ah

i see

still dont know hwo to continnue tho

[jaunty mural]

You need to draw these 2 grids on paper (using pen)

Then try different cuts in pencil

Draw the cuts at the same time on both grids

[crimson sedge]

how?

[jaunty mural]

exactly like i've done

I have drawn the cut at the same time on both grids

[crimson sedge]

i dont know what to try

[jaunty mural]

For what I am trying to do

Can you see the red line I've drawn on the grids correspond to each other?

I'm like trying to communicate between grids

[crimson sedge]

i dont know how to communicate bewteen them tho

[jaunty mural]

ok look

I am trying to link the grids like this

From the left grid

Add 1 to the number

And I am saying that will be exactly the same square after the puzzle is rearranged

1 -> 2

3 -> 4

5 -> 6

...
29 -> 30

[crimson sedge]

why do some get pushed down and other dont

also, im super sorry

but i gtg

i can continue this

later

im so sorry

but i honestly dont know where to make my cuts

[jaunty mural]

https://i.imgur.com/sJ6byzK.png

Ok new numbering

Basically another way of describing what we're trying to do

Is we need to create a bijection between the squares of each grid

========
This the bijection I am trying right now. I don't know if it works. But I am trying to do this with those cuts.

1 -> 1
2 -> 2
...
22 -> 22

@crimson sedge when you're back look at this ^

[crimson sedge]

ah

interesting

[jaunty mural]

https://i.imgur.com/qYwFIuN.png

Ok?

I think you should be able to continue from here.

[crimson sedge]

i took screenshots of it all

[cedar kiln]

@crimson sedge Has your question been resolved?

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[fossil trail]

could someone tell me if this is true or false

im thinking false but im not sure how to come to that conclusion

[cedar kiln]

@fossil trail Has your question been resolved?

[cedar kiln]

@fossil trail Has your question been resolved?

[cedar kiln]

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[tired mesa]

I would really appreciate it if someone is able to explain to me how this answer was found

[mighty drift]

limit of product = product of limits

[tired mesa]

can you explain how to get each individual limit

[mighty drift]

writing / 0 is not very rigorous

[tired mesa]

this is my professors work, I am just trying to understand it

[mighty drift]

13x+1 is pretty trivial. Now for sin x, you know sin(pi) = 0 so 1 / sin x is not defined, hence we're only approaching it from one side. This way we can see that we're approaching 0 from the negative numbers (looking at the graph of sin) so 1 / sin x is approaching 1 / 0 = inf from the negative numbers, so it goes to -inf. Then (13pi+1) * (-inf) = -inf

[tired mesa]

is there a way you could explain it slightly less easier, because i am really terrible at understanding math talk

basically, what I know is that sin(pi) = 0

but we are approaching pi from the right hand side

because there is a plus sign on the pi symbol

but how am i supposed to know that it is negative infinity at this point

[mighty drift]

near pi+ (so a little above pi), 1/sin x is defined and negative, and as x approaches pi, it goes to -inf because it stays negative but its amplitude goes to infinity

[tired mesa]

ok so i see that 1/sin x approaches negative infinity (at pi), but how would I know this without looking at the graph

[mighty drift]

for x slightly above pi, sin x < 0, so 1 / sin x < 0. Since sin x -> 0 as x -> pi, 1 / sin x is a 1/0 situation at pi. This means it goes to infinity. The sign of that infinity is decided by the sign of the zero. Since it was always negative, it's negative, so you get -inf

[tired mesa]

ok thanks alot

[cedar kiln]

@tired mesa Has your question been resolved?

[cedar kiln]

@tired mesa Has your question been resolved?

[fallen heath]

I'd like to ask the same questions as I did before...

what value is x approaching?

[tired mesa]

pi?

[fallen heath]

pi from the right

Do you understand how that makes a change!?

why I'm emphasizing on pi from the right?

[cedar kiln]

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[cedar kiln]

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[soft star]

i keep trying to do this problem but my cooridnates are always wrong

help

[flint plinth]

.rotate

hmm

,rotate

[soft star]

sorry

[flint plinth]

No prob, couldn't remember the command at first

[soft star]

yeahh im stuck

[flint plinth]

What is the goal?

[soft star]

i gotta chose the correct graph

[flint plinth]

Ah I see

[soft star]

mhm!

i dont need the answer i just wanna know how to work out these problems

cause my notes arent really helping lol

[flint plinth]

Well $1/4$ is less than $1$, so $(1/4)^x$ gets smaller as $x$ gets larger. Therefore $(-1/4)^x$ gets larger (but is always negative) as $x$ gets larger

[wraith dagger]

OurBelovedBungo

[flint plinth]

So that rules out all except one of the graphs.

[soft star]

ahh i see

but do u see a flaw in my work?

there obv is but

i cant seem to find it

[flint plinth]

Not sure I understand your work, there is reference to $b$, but what is $b$?

[wraith dagger]

OurBelovedBungo

[soft star]

b= 1/4

1/ 1/4 can be written as 4/1 im pretty sure

4/1 = 4

thats where i got my 4 from

[flint plinth]

And those points are supposed to be points on the graph? I think you're forgetting to subtract 3

[soft star]

subtract 3 from x or y ?

i think y since k = translate down or up

[flint plinth]

From y. The formula is $y = -(1/4)^x - 3$, so if you plug in for example $x=-1$ you should get $y = -(1/4)^{-1} - 3 = -4 - 3 = -7$.

[wraith dagger]

OurBelovedBungo

[soft star]

omg that works

you're a legend thank you!!

.close

[cedar kiln]

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[grizzled violet]

can i get some help on 27 and 31?

[grizzled violet]

ik how to find limits at infinity, but im just confused by e

for 27, ik the answer is 1, but im not sure if my thought process is adequate:
it becomes (infinity-small number)/(infinity-small number)

and for 31, it's just 0 because e^-2(infinity)=very very small number

[flint plinth]

Essentially correct

[grizzled violet]

is there another way to do the problem?

other than just plugging it in

[flint plinth]

Yes and cos(x) remains bounded between -1 and 1.

[grizzled violet]

wait are u saying yes to another way?

oh wrong msg

[flint plinth]

For 31, note that $|e^{-2x}\cos(x)| = |e^{-2x}| |\cos(x)| = e^{-2x}|\cos(x)| \leq e^{-2x}$ since $|\cos(x)| \leq 1$. Since $e^{-2x} \to 0$, this means that $|e^{-2x}\cos(x)| \to 0$ and therefore $e^{-2x}\cos(x) \to 0$.

[wraith dagger]

OurBelovedBungo

[grizzled violet]

ok that makes sense

[flint plinth]

For 27, you can divide the num and denom by $e^{3x}$ to get $\frac{1 - e^{-6x}}{1 + e^{-6x}}$. The num and denom both have limit equal to $1$ and therefore so does the fraction.

[wraith dagger]

OurBelovedBungo

[grizzled violet]

so for further cases

if the numerator has a limit of x, and the denominator has a limit of x, the limit of the entire fraction will be x

[flint plinth]

If the num has limit $a$ and the denom has limit $b$, then the limit of the fraction is $a/b$. This assumes $a$ and $b$ are real (finite) and that $b$ is nonzero. In your case for 27, both $a$ and $b$ are $1$, so the fraction has limit $1/1 = 1$.

[wraith dagger]

OurBelovedBungo

[grizzled violet]

thank you for your concise answers and keen insight

.close

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

What does x^i mean over here? or in any general probability case?

[pallid arrow]

it just means $p_i = x^i$

[wraith dagger]

xdk1235

[pallid arrow]

so $p_1 = x^1 = x$

[wraith dagger]

xdk1235

[pallid arrow]

$p_2 = x^2$

[wraith dagger]

xdk1235

[crimson sedge]

and what does that mean.. i mean pi is just a probability right? here x is any constant so that x^i should be equal to pi?

[crimson sedge]

does it mean x is such a constant we need to find out here

<@&286206848099549185>

What does x^i mean over here? or in any general probability case?

<@&286206848099549185>

[pallid arrow]

yes

[crimson sedge]

how is that even possible?

[pallid arrow]

it's just guessing if it'd work

if it works great, if it doesnt you try something else

[crimson sedge]

suppose my P is 1/9. then how could you find x for it

[pallid arrow]

what is p

like $p_i = \frac{1}{9} p_{i+1} + q p_{i-1}$?

[wraith dagger]

xdk1235

[crimson sedge]

so u meant for every pi, there would exist x^i?

how is that even possible? hard to get this thing

[pallid arrow]

no i is a variable too

it's just a sequence

[crimson sedge]

x would be same for all right?

[pallid arrow]

yes

[crimson sedge]

so p(i) = x^i.. this means p(1) would be x^1.. p(2) would be equal to x^2 and so on?

[pallid arrow]

yes

[crimson sedge]

but for this to happen.. p(2) should be greater than p(1) , right?

[pallid arrow]

idk what this is for but if all p(i) are probabilities

p(2) will be less than or equal to p(1)

[crimson sedge]

yea, p(1) should be greater than or equal to p(2) if this is to hold true , right?

[pallid arrow]

yes

[crimson sedge]

but how?

[crimson sedge]

but how?

[pallid arrow]

?? wdym

[crimson sedge]

why should p(1) should be greater than p(2)?

how could we conclude that?

[pallid arrow]

because $x = p(1)$ and $0 \le p(1) \le 1$

[wraith dagger]

xdk1235

[pallid arrow]

so $p(2) = x^2 \le x = p(1)$

[wraith dagger]

xdk1235

[crimson sedge]

ok.. now i've one doubt.. so this would mean.. p(i+1) = x * p(i)?

[pallid arrow]

yes

[crimson sedge]

is x also some kind of probability or what?

[pallid arrow]

it's just some value

[crimson sedge]

the thing that i'm still not getting is.. see, from i we can go to either i -1 or i + 1.. and if we went to i -1 and again we came back to i.. in this case, what happens to our p(i) is it still stays greater than p (i+1)?

[pallid arrow]

if the guess is correct and all p(i)'s are probabilities then yes

that depends

[crimson sedge]

.close

[cedar kiln]

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[wraith dagger]

3317

[cedar kiln]

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[umbral ravine]

Where did the cos(2x)^2 go?

[crimson sedge]

@umbral ravine

this is because (a+b)/b is same as a/b + b/b

[umbral ravine]

very helpful Ty

but tan?

[crimson sedge]

cus sin x divided by cos x is same as tan x

i recommend taking brilliant.org’s course on trigonometry they explain it really well

[honest frigate]

they make u pay :(

[crimson sedge]

i took the free 7 day trial then passed the course in the 7 days and cancelled the subscription lol

it explains everything visually and lets you play around with parameters its super nice for learning

second way to simplify it

sry for bad handwriting

[umbral ravine]

smart

I’ll prob do that

[crimson sedge]

but sin^2(x) + cos^2(x) = 1 because of the pythogorase theorem

that’s why it turned into a 1

[crimson sedge]

gl but don’t forget to cancel it lol

[umbral ravine]

yeah lol

Does sin2x+cos2x=1?

Or does it need to be in that exact format?

[crimson sedge]

sin(2x) + cos(2x) isnt 1
sin(2x) is same as sin(x+x) and that turns into 2sin(x)cos(x)
cos(2x) turns into cos^2(x) - sin^2(x)

the = 1 thing only happens when both sinx and cosx are squared so:
(sinx)^2 + (cosx)^2 = 1
x can be anything

cos^2(x) same as (cosx)^2 btw

it might seem confusing and a bunch of formulas to remember but brilliant’s course explains this super well so u don’t even need to memorize the formulas, u just naturally know what equals to what

[umbral ravine]

Ty

I will give it a go

[crimson sedge]

aight gl

[cedar kiln]

@umbral ravine Has your question been resolved?

[cedar kiln]

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[languid leaf]

4

Wee need y=some x and then solve

[sand ether]

wut

[languid leaf]

Like biquadratic

We need something like that y=some x

And then solve

[languid leaf]

Do you get it?

<@&286206848099549185>

[cedar kiln]

@languid leaf Has your question been resolved?

[languid leaf]

<@&286206848099549185> guys

.close

[cedar kiln]

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[analog wind]

No idea how to do this

wat i have so far

[cedar kiln]

@analog wind Has your question been resolved?

[analog wind]

<@&286206848099549185>

[cedar kiln]

@analog wind Has your question been resolved?

[slim peak]

Hello

This looks like a science question

But I don’t know

🤷‍♂️

Maybe that’s why people might not be able to answer the question

[analog wind]

Its math

The subject is called engineering mathematics

[slim peak]

Okay

But umm im probably not that advance to answer this question

<@&286206848099549185>

[analog wind]

idk what to do

[slim peak]

<@&286206848099549185>

[analog wind]

Answer from symbolab matches mine, i guess i'm just getting the initial conditions wrong.
i dont have x'(0).
i think x(0)=0.2

[analog wind]

<@&286206848099549185>

[analog wind]

bruh

@slim peak

i got the answer

i didnt change my calculator from degree

-.-

.close

[cedar kiln]

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[crimson sedge]

hey can someone help me with this question. the diagonals of a rhombus meet at the point(-1,5) and one of them is parallel to the line 2x-5y=3 a) find the equations of the diagonals

[crimson sedge]

how can you find the equation using the formula (y-y1)=m(x-x1)

[urban niche]

Is it asking for the other diagonal it meets?

[crimson sedge]

yes

[urban niche]

Okay

Have you tried to draw it?

[livid hound]

are you able to determine the slope of 2x-5y=3

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[late spoke]

help

[late spoke]

pls

im acc gettung weak with these questions

[zinc mason]

do u know sum of consecutive integers formula?

[late spoke]

sum of arithmetic series?

[south tundra]

One of more general version of the other

The sum of consecutive integers formula is 1 + ... + k = k(k + 1)/2

[late spoke]

ok...

[late spoke]

would this work?

[south tundra]

Yes but in this case you need the sum until n-1, not n+1

[late spoke]

ok, i got this as my first eq

but for the other quation

i cant really use it, can i?

because the starting position is now at n+1