help-17

[hasty pulsar]

is it about kirchoff?

[near island]

What technique are you using?

Loop analysis or nodal analysis

[serene hazel]

Part A

[hasty pulsar]

Show their circuit

[serene hazel]

Where did the current come from

[hasty pulsar]

we need to see where they labelled the currents

[serene hazel]

Idk this just isn’t clicking

So how do you label it??

I think I need this explained to me like I’m 5

[hasty pulsar]

i2 and i1 are coming to the point

whilst i3 is going out

coming= positive, out = negative => i2+i1-i3 = i2+i1=i3

[serene hazel]

Ok sure

What’s the point?

[hasty pulsar]

above 5 ohm

[near island]

there's an error in that line right? i2+i1-i3=i2+i1 would imply that i3 =0

[serene hazel]

This is beyond confusing

[hasty pulsar]

change = to => then

i2+i1-i3=0
i2+i1=i3

[near island]

Look at it like this: when water flows through pipes and there is a junction, equal amounts of water flow into the junction as come out

Otherwise water would pile up

[serene hazel]

Ok yes that makes sense

[near island]

In electricity it's the exact same. The current into a junction (=node) and out of has to be equal

Otherwise charge would pile up which is impossible

[serene hazel]

Ok

[near island]

That's where the equation in A comes from

[serene hazel]

So how do you tell where it’s coming and leaving

[near island]

You don't know that yet.

You take a choice of arrow direction and continue to work with it

[serene hazel]

So is I1 the entire circuit??

[near island]

I1 here is the current flowing through the middle branch

[hasty pulsar]

from battery to the two resistors

[serene hazel]

Then why did this guy label it on the far right?

[hasty pulsar]

?

[near island]

There's no i1 label on the right?

[serene hazel]

Well why isn’t I1 labeled anywhere

[hasty pulsar]

i assumed the middle arrow was I1

[serene hazel]

Ok how did he know it was going that way??

[hasty pulsar]

we dont

[near island]

He didn't. Arrow directions are a choice

After the calculations we could find that i1 is negative

[serene hazel]

Ok

[near island]

which means it's actually flowing downward

[serene hazel]

Where did that number come from

To me it just spawned in

[near island]

Which number?

[serene hazel]

All the answers

[hasty pulsar]

have you watched any videos about kirchoff laws?

[serene hazel]

He didn’t match an equation to an answer

[hasty pulsar]

you got three equations and three unknowns

u can solve it either through algebra or matrices

[near island]

The method used assumes you have a solid grasp of the kirchoff laws

[hasty pulsar]

if you havent watched any video watch this

https://www.youtube.com/watch?v=6F_rmZ1nXFQ

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[calm hinge]

👋 Hi!
I'm writing some mathematical statements for my friends to disprove and, of course, I want them to be as clear as possible.

I have defined an operator m: A x B → y, where A and B are subsets of ℝ and y ∈ ℝ.
If A and B are subsets of ℚ, the operator will output a number in ℚ.

What's the best way to describe this?
I think something similar to the concept of set closure under an operator would be perfect, but A and B are sets, not numbers, so saying that ℚ is closed under the operator m is just wrong

[mild flower]

if you want mathematical language you can say that $m: \mathcal{P}(\R) \times \mathcal{P}(\R)$ and that $\mathcal{P}(\Q)$ is closed under $m$

[twin meteor]

Hayley

[mild flower]

where P is the powerset

[hard atlas]

well its not tho

[mild flower]

oh whoops

[hard atlas]

cause then you would expect that the result is a set again

[mild flower]

i can't read

just say it in words tbh

[hard atlas]

just write down this condition in words

its fairly short

doesnt need a name probably

[calm hinge]

Ok!

[worthy citrus]

you could say that the image codomain of m under the restriction to Q x Q is Q

and convert that to symbols if you really wanted, but i agree that writing it in words is probably best

[calm hinge]

Oh I see, that's fancy

Thank you so much people!

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[halcyon sluice]

I'm looking to determine the sup-norm of $\frac{nx}{1+n^2x^2}$ on $x\in[1,\infty)$ (here $n$ is any natural number). If the interval was compact, I would know by the extreme value theorem that the function obtains a maximum there. However, with $[1,\infty)$ I'm unsure. I could look for extreme values at the boundary points and critical points, but I don't know if the function actually obtains a maximum there.

[twin meteor]

sunside_

[worthy citrus]

you might want to look at the behaviour of the function as x tends to infinity

[halcyon sluice]

true

[worthy citrus]

and then boundary and critical points is a good idea

[halcyon sluice]

at x=1, we have n/(1+n^2) (this is a boundary point); at x=1/n, we have 1/2 (this is a critical point); in the limit as x tends to infinity, we have that the function tends to 0

[wraith venture]

x=1/n is out of the domain anyways

[halcyon sluice]

oh, yeah, I was wrong there 🙂

if I could establish that the function is decreasing for all x in [1,oo), then clearly x=1 would be the max

[wraith venture]

What's the derivative ?

,w d/dx (nx)/(1+n^2 x^2)

[twin meteor]

Results provided by WolframAlpha

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

Requested by mateo713#0

[wraith venture]

So it's just plain decreasing

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[halcyon sluice]

ok, thanks, I think I got it now

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[obtuse lintel]

im trying to solve this problem

and here is what i got, was I correct in treating Kmax as a variable in itself?

[river kettle]

You’d probably have better luck asking this in a physics server

[obtuse lintel]

oh ok

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[solar lotus]

I have a few hard matrix questions

[solar lotus]

can i get some help?

[open spade]

I've seen you do this a few times, you should send your question first while asking for help. Don't wait for anyone to respond before sending you question.

[solar lotus]

i don't know where to start

<@&286206848099549185>

my bad

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[opal obsidian]

Is the set A={2,3,pi, 5/7} finite?

[thin vale]

What do you think?

[opal obsidian]

it has 4 elements, but the confuses arises when i think about decimal expansions

If I were to list it with decimals i cannot list it

[thin vale]

what's your definition of a set A being finite?

[opal obsidian]

I haven't gone through one yet, but bijection to N?

[thin vale]

that means it is countably infinite

N is not a finite set

but it is countable

[opal obsidian]

so the above set is countably infinite?

[thin vale]

Not what I said

the set N of naturals is countably infinte

{1,2,3,...}

[opal obsidian]

yes

[thin vale]

and a set that has a bijection onto N is countably infinite

can this set have a bijection onto N?

It can certainly be injective, but what about surjective?

[opal obsidian]

i don’t know if I can map ‘pi’ onto elements?

[thin vale]

Pi is just a constant

why would you not be able to

you could even define your function to take pi -> 1

the function could be something like f(a) with a in A, = 7a if a is a rational, and = 1 if a is not a rational

verify that this is injective onto the naturals

[opal obsidian]

I think my misconception is rooted because of misinterpretation of decimal expansion

[thin vale]

I agree

[opal obsidian]

7a if a is rational

and 1 if it’s not

?

[thin vale]

yes

mapping elements from your set A

to the naturals

[opal obsidian]

Map 2 to 1, 3 to 2, pi to 3 and so on

[thin vale]

well sure that works

the so on ends immediately though

after just 1 more term

map 5/7 to 4

and boom

now you have an injective function

is it possible for it to be surjective?

[opal obsidian]

for entire N?

Then no

[thin vale]

mhm

so no bijection

[opal obsidian]

only 4 elements in N is mapped

[thin vale]

so our set isn't countably infinite

[opal obsidian]

so rest are not mapped

[thin vale]

exactly

so we've deduced our set A is not countably infinite

the other options are it is uncountably infinite, or finite

which do you think it is?

[opal obsidian]

finite it is

[thin vale]

mhm

and there is a cool theorem here

basically saying that

if you can create an injection

but not a surjection

that set is 'less' in cardinality

than the other set

and the opposite holds

[opal obsidian]

less than alph nought

?

If N

[thin vale]

I don't know what you mean by that

[opal obsidian]

Idk, the cardinality of N

i just read about it

[thin vale]

we denote the cardinality of N as |N|

[opal obsidian]

so any number as it’s alone can be mapped to some element in N?

[thin vale]

sure why not

[opal obsidian]

but what does it mean to count pi?

[thin vale]

I don't know what you mean

[opal obsidian]

I think my understanding of numbers are fundamentally flawed

[thin vale]

I agree

pi is just a constant

[opal obsidian]

lol, mb

[thin vale]

because its decimal expansion does not terminate doesn't mean anything

for example

decimal expansions need not necessarily be unique

5.000.....=4.999999

but if we want to have every real number having one decimal expansion

then we always "pick" the repeating 9's

so 5=4.9999......

and we don't usual write 5=5.000....

and in both cases

the decimals don't terminate

just like in pi

[opal obsidian]

i thought when we say countable we can completely recite the number

[thin vale]

yet you wouldn't argue that 5 is infinite

countable is different terminology than infinite

and my point was the having an infinite decimal expansion does not change the fact that it is still just 1 number

5=4.999.... (infinite decimal expansion) pi=3.14159265358979323.... (infinite decimal expansion)

both are just singular real numbers though regardless

[opal obsidian]

yeah, sqrt(2) has a long repeating decimal but it’s has a clear position in number line

[thin vale]

that is because it is irrational, like pi

[opal obsidian]

when is a set uncountable?

as you pointed there is no bijection

[thin vale]

when it does not have an injective mapping onto the naturals

[opal obsidian]

So the set of Reals cannot be mapped to N?

[thin vale]

exactly

the proof of that is very cool

[opal obsidian]

how do I see that?

[thin vale]

(there are lots of proofs of that actually)

I can send you one I wrote up, but the typical proof is called Cantor's Diagonalization (that is what I used) and you could also just google it

[opal obsidian]

Any way related to cantor’s diagonal argument

?

[opal obsidian]

Please send

[opal obsidian]

I just know the terms in general out of being curious not the proofs

[opal obsidian]

Is sup the upper bound of set?

[thin vale]

the sup is the least upper bound of a set

[opal obsidian]

okie

[thin vale]

short for 'supremum'

[opal obsidian]

I have only read the term L.U.B

[thin vale]

same thing

[opal obsidian]

Thank you

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[silk bridge]

so like

is this true or false

cause if its liek circles on a graph

thats false

but like just circles in general

this question is confusing to me

[sly sierra]

if you take a circle and then move it, is it congruent to the original circle?

does it have the same center?

[silk bridge]

yea ok thanks

so it is false

right

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[silk bridge]

ty ❤️

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[hidden ginkgo]

Is this C. y = cos (x - TT/2)

Just want to fact-check

[sly sierra]

no

[hidden ginkgo]

shit

[sly sierra]

if you plug in x=pi/2, that equation would give you y = what?

[hidden ginkgo]

uh

cos

[sly sierra]

no what number

[hidden ginkgo]

2

pi/2

[sly sierra]

cos is always between -1 and 1, it can't be 2

[hidden ginkgo]

o

so 0?

[sly sierra]

if x is pi/2 then what is cos(x - pi/2)

[hidden ginkgo]

ok fair

[sly sierra]

well one step at a time

if x is pi/2 then what is x - pi/2

[hidden ginkgo]

uhh

negative something

im not that good with pi 💀

[sly sierra]

it's simpler than that

[hidden ginkgo]

Just learned it an hour ago

idk

[sly sierra]

if x is 7 then what is x - 7

[hidden ginkgo]

7?

[sly sierra]

uhh

x is 7

[hidden ginkgo]

oh

0

[sly sierra]

subtract 7 from that

[hidden ginkgo]

my bad

[sly sierra]

yes

[hidden ginkgo]

that was slow

[sly sierra]

ok

so if x is pi/2, what is x - pi/2

[hidden ginkgo]

0

[sly sierra]

right

and what's the cosine of that

[hidden ginkgo]

so is y = cos x

[sly sierra]

no, what is cosine of 0

[hidden ginkgo]

1

[sly sierra]

right

so if you plug in x=pi/2

your y value for y = cos(x - pi/2) is 1

but what is the y value in your graph at x = pi/2?

[hidden ginkgo]

idk 😭

[sly sierra]

it's labeled right on the graph

[hidden ginkgo]

pi /2

i mean i see it

[sly sierra]

that's the x value

[hidden ginkgo]

idk what to call

[sly sierra]

what is the y value there

[hidden ginkgo]

1/1

1

?

[sly sierra]

the y value is below the x axis when x = pi/2, right?

so the y value is negative

(specifically it's -1)

[hidden ginkgo]

oh

[sly sierra]

whereas your equation says it should be 1

so your equation can't be the one that goes with the graph

[hidden ginkgo]

so D

?

its the same but not negative

[sly sierra]

nope, d is also wrong

[hidden ginkgo]

omg

[sly sierra]

again, try plugging in x=pi/2 and see what D tells you the y value should be

[hidden ginkgo]

OH DUH

it has 2

is B

ok

i swear im slow sometimes

[sly sierra]

yes, B is correct

[hidden ginkgo]

so

does that mean this one would be negative?

[sly sierra]

hmm, probably the easiest way to do this one is to shift the graph downward by 1

if you do that, what function do you get?

[hidden ginkgo]

y + 1 = cos x

?

[sly sierra]

nope

[hidden ginkgo]

OMG

this is embarassing

[sly sierra]

one thing to keep in mind

cos has a maximum at x=0

[hidden ginkgo]

but..

[sly sierra]

this graph doesn't

[hidden ginkgo]

oh

[sly sierra]

this graph has a maximum at pi/2

[hidden ginkgo]

ok

[sly sierra]

and so does sine

so you should expect this to be y = sin(x) + something

or equivalently,

[hidden ginkgo]

y + 1 = sin x

[sly sierra]

y - something = sin(x)

[hidden ginkgo]

oh

[sly sierra]

well y + 1 = sin x would be equivalent to y = sin(x) - 1

[hidden ginkgo]

y - 1 = sin x

[sly sierra]

yes

[hidden ginkgo]

oh duh

[sly sierra]

that's the one

[hidden ginkgo]

I siad it should be negative

i got confused

😭

but thanks

[sly sierra]

sure

[hidden ginkgo]

.close

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[waxen hawk]

assume that there's a polynomial cubic function f(x), and its leading term's coefficient is positive.
given that cubic function f(x) and the linear function g(x) been tangent to each other in x=1 and there's only one intersection point between the two function.

is x=1 where the Inflection point occur?

[waxen hawk]

how to prove it?

[split wind]

i think it's not necessarily the case?

[waxen hawk]

do you have any counter-example

[split wind]

lemme check with desmos real quick

[waxen hawk]

for sure

[waxen hawk]

do you notice there are 2 intersection points in your case

in which we can only have one as the question stated

[split wind]

ohhh

[waxen hawk]

I feel uncertain about whether it is true that the inflection point occurs in, as neither can i find any of counter example nor a way to prove it

can i use neither...nor in that way

is it grammtically correct?

[split wind]

i think its something related to this:
https://brilliant.org/wiki/cubic-discriminant/

[waxen hawk]

cubic discriminant

i know that stuff but i dont want to over-complicate the question

can't we consider it the geometry way, which is more intuitive in my perspective

[split wind]

i see

I don't really know then

can you do it with prove by contradiction?

just guessing

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[simple lark]

I don't know where to begin the equation

[paper depot]

@simple lark let x be the duration of Diophantus' life

write down each stage of his life in terms of x

[simple lark]

Yes I get that part

x = childhood, youth, bachelor

but I really don't where to begin in this kind of problem

[paper depot]

no bad

x is diophantus' entire life

childhood is one sixth of his life. how do we write that?

[simple lark]

oh I'm wrong

should I substitute it to variables?

like A=1/6

[paper depot]

you should not overthink it

you should not make any new variables as there is absolutely no need for that

childhood = 
youth = 
bachelorhood = 
marriage before son = 
son's life = 
grief of son's death =

fill in the blanks here

don't jump ahead

only fill in the blanks as i tell you to

in terms of x obviously

[simple lark]

childhood, =1/6
youth = 1/12
bachelor = 1/7
Marriage = 5 years
son's life = 4

[paper depot]

no

[simple lark]

Grief of son's death = 1/2

[paper depot]

childhood isn't 1/6

it's not one sixth of a year

it's one sixth of his life

which i have instructed you to denote with the letter x

so it is not 1/6 but x/6

also even with this mistake corrected you mixed up the son's life and the 4 years diophantus grieved him

[simple lark]

yes later I can multiply it to the actual age of Diophantus

[paper depot]

no, not later.

you cannot just defer such things.

[simple lark]

oh I thought it will be x = 1/6

[paper depot]

no

x is the number of years diophantus lived in all

if you said x = 1/6 it would mean diophantus lived only two months and died an infant

[simple lark]

okay I understand does x will all be applied to this?

[paper depot]

no, the letter x won't appear in all of these entries, but it'll appear in most of them.

but that info won't really help you all that much.

[simple lark]

childhood x/6
youth x/12
bachelor x/7

Marriage before son 5
Son's death 4
Grief of son's death x/2 father's final age

[paper depot]

almost there...

1/2 father's final age

why not write this as x/2 tho

[simple lark]

I see that also includes

[paper depot]

also you're once again mixing up the son's lifetime and the grief...

which won't matter THAT much but yknow. it's weird.

[simple lark]

is the father's half the final age means the father of diophantus?

[paper depot]

no it means diophantus himself.

he's the father of his son

[simple lark]

oki

[paper depot]

yeah ok so like

[simple lark]

5 years after his Marriage

[paper depot]

oh sorry are you gonna argue against me on this or what

[simple lark]

no I thought of you for me to answer the question

[paper depot]

communication is difficult.

anyway! we now have the following breakdown of diophantus' life:

childhood = x/6
youth = x/12
bachelorhood = x/7
marriage before son = 5
son's life = x/2
grief of son's death = 4
---------------------------
total life = x

can you write this as an equation? Y/N

[simple lark]

yes, but not sure if I will be correct

[paper depot]

try it and i will tell you where you went wrong

[simple lark]

since of all of the given denotes the total of Diophantu's life, I think I should only add all of them and solve for x

the LCM of childhood youth and bachelor is 84

[paper depot]

just

write down

the equation

don't attempt to simplify it beforehand

you haven't written it down so you're trying to keep it all in your head

that's a bad idea bc you are guaranteed to fuck it up one way or another and not even know it

[simple lark]

I don't think I can write in an equation but I already understood the problem

[paper depot]

those two statements contradict each other

[simple lark]

x/6 + x/12 + x/7 = 33x

[paper depot]

if you understand the problem, then you are also ab-

33x????

where did 33x come from?

[simple lark]

I multiplied it by 84 to remove the fraction

[paper depot]

and you chose not to write down this multiplication by 84 on the left-hand side.

and instead chose to claim that a number is equal to 84 times itself.

[ornate ember]

https://tenor.com/view/billy-madison-gif-18245726

[paper depot]

also i told you not to simplify anything

and you broke that instruction of minetoo

[simple lark]

that's their common lcm

[paper depot]

FORGET about this "common lcm" bullshit.

you have only half-written down the equation.

i TOLD you NOT to simplify anything beforehand.

NOT simplifying anything beforehand also means NOT thinking about any denominators or lcm's or any such things.

ping me when you respond.

[simple lark]

x/6 + x/12 + x/7

I really don't know how to put it into equation
@paper depot

[paper depot]

you've written it down halfway

you've accounted for three of the six parts of diophantus' life; are you going to also account for the other three, or let them fall into oblivion?

[simple lark]

5x+4x = x/2

[paper depot]

congratulations you fucked up once again

you attempted to simplify some things, didn't you?

and you also lost the first bit.

[simple lark]

I kinda solve it my head at the same time

[paper depot]

DO NOT DO THINGS IN YOUR HEAD.

don't attempt to simplify it beforehand
you haven't written it down so you're trying to keep it all in your head
that's a bad idea bc you are guaranteed to fuck it up one way or another and not even know it

the equation i wanted from you was: $$\frac{x}{6} + \frac{x}{12} + \frac{x}{7} + 5 + \frac{x}{2} + 4 = x$$

[twin meteor]

Ann

[paper depot]

very simple if you just don't overthink

[simple lark]

I see just exactly adding them

Thank you so much for help I'm practicing this board exam problems

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[severe violet]

need help with this

[vocal sleet]

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

[mild flower]

ok the first thing i'd probably do is try to find h

[severe violet]

ok

using pyth

right?

[mild flower]

yeah

[severe violet]

Which triangle should we use

Shouldn’t we find the height of the triangle

So then we will get the hyp

[open spade]

You already have the hypotenuse

[severe violet]

What is it

Is it 20

[mild flower]

look at right triangles that involve h

(one of them is drawn in light dotted lines)

[severe violet]

Oh

Yeah

[vocal sleet]

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[opal obsidian]

is A={1,2,3} and B={{1}} subsets?

[opal obsidian]

My guess is no, as {1} is different from 1

[thin vale]

subsets of what?

[opal obsidian]

is B subset of A?

[peak matrix]

B is not subset of A if that was your question

{1} would be subset of A

[paper depot]

no, they aren't subsets of each other.

[thin vale]

because B contains {1}, and {1} is not an element of A

and A contains 1,2,3 which are not elements of B

[peak matrix]

B is subset of A iff each element of B is also element of A

[opal obsidian]

so {1} can be element

of a set?

[paper depot]

yes, sets can themselves be elements of other sets, it's perfectly fine

[peak matrix]

yes, {1} is element of e.g. {{1}} or {2, {1}, {3, 3}}

[opal obsidian]

makes sense, So by this definition {1} element of B but not A, thus B not subset of A

thank you.

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[hidden ginkgo]

wtf is this shape 😭 im so lost

[vast shale]

its a hyperbola https://en.wikipedia.org/wiki/Hyperbola

[hidden ginkgo]

how do i determine

wheather greater than 1 or not

[vast shale]

use the definition of eccentricity https://en.wikipedia.org/wiki/Eccentricity_(mathematics)

[hidden ginkgo]

ah okay

thank you

[vast shale]

yw

[hidden ginkgo]

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[mint crystal]

Help

[mint crystal]

I have a problem I solved it through but I was just wondering if I was doing it correctly

Problem: I = E/R+r

I’m I’m solving for r

[paper depot]

$I = \frac{E}{R} + r$?

[twin meteor]

Ann

[mint crystal]

it’ll be R+r on bottom

Like this

[paper depot]

ok, so you should write E/(R+r)

[mint crystal]

Oh okay sorry

[paper depot]

ok, so you have your solution and you're about to send it here to get it checked, yes?

[mint crystal]

Yes correct

This was the solution I got

I can show you my work to see if the answer is wrong and see where I went wrong to correct it

[paper depot]

you should always show your work regardless

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[reef wave]

can someone texit bot this question for me please

[proven garden]

$\frac{x^5+3x^4-2x^2-3}{x^2+1}$

[twin meteor]

_wherewolf_

[proven garden]

is this what you want

[reef wave]

do u know how to do it in like polynomial division

long division

[proven garden]

calculate yourself

and ask if you need help

we don't give out answer here

[reef wave]

ive tried it but im not sure where im going wrong

[proven garden]

show your work please

[reef wave]

so i want to see the working

[worthy citrus]

It's a custom command that most people don't have, just Google a long division calculator if that's all you want

[proven garden]

you wrote it weirdly and that's why you made a mistake

[reef wave]

oh what

[proven garden]

try writing it like - (x^5+0x^4+x^3) instead of -x^5+0x^4+x^3

you made a mistake at -(-3x^3+0x^2-3x)

it should be +3x on the next line

[reef wave]

so remainder is 3x?

[proven garden]

yeah

[reef wave]

apparently this is the anwer

answer*

[inland canopy]

hi

[reef wave]

hello

[proven garden]

you copied the question wrongly

the question is -2x^2 but you wrote -2x^3

[reef wave]

..

ffs

well then sorry for wasting ur time bro

but thank you

[proven garden]

It's fine everyone makes mistakes

[reef wave]

.close

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[plain bear]

ABC is a right triangle, right angled at B.
D can be any point on AC (other than A and C)
Find the length of BD (integer values)

[paper depot]

so you want to find all possible integer values of BD?

[vast shale]

it seems like you would have to create a function for it and then find where the values of function are integers ig

[plain bear]

hmm

the way my friend did it was

[vast shale]

function could be with respect to AD:DC ratio or theta, theta = angle DBC I think

[plain bear]

minimum value of BD

[paper depot]

is that a yes or a no

[vast shale]

Ohhh

[paper depot]

if it's a no, then no goal has been clearly stated.

[vast shale]

Find minimum value of BD ?

[plain bear]

yes

[vast shale]

Yep

[paper depot]

so then we are to find all possible integer values of BD, got it

we will need to find the minimum possible value for that

which of course happens when BD is perpendicular to AC

and so can be found using the half-base-times-height area formula, perhaps

[plain bear]

really?

if BD is perpendicular to AC

then answer comes out to be 14.5

but that isnt right

[paper depot]

of course it isn't. i didn't say it was the smallest integer value.

but it's the lower bound.

[plain bear]

so

whats the answer

[paper depot]

word of advice:

never

ask "what's the answer" on this server

it is like a red cloth to a bull

anyway in our case we then know that if BD is an integer then it must be at least 15

[plain bear]

@paper depot hey you done?

[paper depot]

well, i did say what i wanted to say.

the ball's in your court now.

[plain bear]

welp

i see how it is

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@plain bear Has your question been resolved?

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@plain bear Has your question been resolved?

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[latent girder]

@plain bear the question is to calculate lenght BD ?

[plain bear]

yes

[latent girder]

ok

[plain bear]

minimum length of BD

[latent girder]

if you take (BC) as x-axis and (BA) as y-axis
then you find the equation of line (AC)
then you have coordinates of D and you can calculate BD², then find when BD² is minimum

but in your SS, it is not said to find the minimum of BD, but to find BD

[vocal sleet]

@latent girder Has your question been resolved?

[plain bear]

hmm

ty @latent girder

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[tidal iron]

Help me with this problem; I tried but it appears that something went wrong.

[tidal iron]

so in inductive step, I supposed the statement is true with n positive numbers x1, x2, ..., x_n-1, x_n + x_n+1 s.t sum <= 1/3 , simultaneously It True for n+1 positive numbers cause sum <= 1/3 but the product does contain equal sign.

<@&286206848099549185>

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@tidal iron Has your question been resolved?

[quiet maple]

I couldnot solve with induction but bernoullis inequality works for this inequality (1+x1)...(1+xn)> or egual to 1+x1+...+xn if xi>-1 and they are all positive or negative

[quiet maple]

You can let xi=-yi and -1/3< or equal to y1+y2+...+yn then you can prove the (1+y1)...(1+yn)>1+y1+...+yn> or equal to 2/3

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@tidal iron Has your question been resolved?

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[broken nimbus]

Hello there I have a question in regards to the Sandiwch Theorem. Lets say we want to find the limit as x approaches 0 of the function xsin(1/x). We know that x oscillates between [-1,1], therefore we can state that
-1 <= sin(1/x) <= 1
-x <= xsin(1/x) <= x
But now we have an issue because when we graph it we see that it does not abide the definition of sandwich theorem where f(x) <= g(x) <= h(x)

Now if we apply an absolute value, we get.

-|x| <= |xsin(1/x)| <= |x|

Now we cannot change the original function and it can be proved through cases that - |x| <= xsin(1/x) <= |x|

Now what my confusion is from that transition where we had that absolute value on xsin(1/x) to getting it removed after proving the cases. I don't get how to prove those cases. if someone can hint me along the way as I attempt to show my thinking and advise me so on and so forth it would be greatly appreciated.

[regal bane]

"it does not abide by the definition of sandwich theorem where..."
Why not?

Sandwich theorem does work on that line.

[broken nimbus]

The final definition does what I was saying

was that this

-x <= xsin(1/x) <= x

itself

does not abide the sandwich theorem. An exception has been made for that thing.

As x approaches 0 for the limit.

[regal bane]

Why not lol

[dull maple]

-x can be bigger than xsin(1/x).

[dull maple]

thus f(x) <= g(x) doesn't hold.

[broken nimbus]

Yeah.

[regal bane]

Oh I see. Could do a right limit and left limit instead

Might be the "through cases" you were referring to

[broken nimbus]

Yeah

-|x| <= |xsin(1/x)| <= |x|

Now this is correct, however we negate the absolute value case through proving cases and also because of the fact we cannot change the og function in the middle or were essentially changing it which would be wrong.

Please don't tell me the answer

but here is my thinking

We need to prove that x < 0 and x >= 0

therefore

[regal bane]

You're correct that |xsin(1/x)| is not the original function, but there's still useful information in using that function instead

[broken nimbus]

For x >= 0 - |x| < xsin(1/x) = |x|

and -x < 0

-|x| = -xsin(1/x) < |x|

Does this prove the fact that we can drop the absolute value on the |xsin(1/x)|

Or am I missing a detail, and if so hint me towards it don't tell me directly.

hello?

[broken nimbus]

Hey you there?

[broken nimbus]

ello?

[regal bane]

There's two different ways to prove this. Using the absolute value, or using the left side and right side

[broken nimbus]

Mk, so I used the left side and right side right to prove mine? Is my proof incomplete?

[broken nimbus]

So we can exhaust the cases on the left and right hand side to prove my point. Or I could place an absolute value in the middle and just observe what would happen.

[vocal sleet]

@broken nimbus Has your question been resolved?

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[vast shale]

Hello again

[vast shale]

momento

$\begin{array}{l}y\prime=x+\frac{1}{y}\cdot y\prime\
\end{array}$

[twin meteor]

yaomri

[vast shale]

how do put the y prime at left

i know i gotta divide but what happens to the rest

[west raptor]

So you're trying to solve this for y right?

[vast shale]

yes basically we want yprime(1)

[west raptor]

Oh ok

This is a separable differential equation so we can just factor y' to one side

[vast shale]

yes so y' = 1/??

[west raptor]

y'=x/(1-1/y)

[vast shale]

kk wait

here is the original

$g\left(x\right)\ =\ x+\ \frac{1}{x+\frac{1}{x+\frac{1}{x+...}}}$

[twin meteor]

yaomri

[vast shale]

I have to calculate g prime (1)

[west raptor]

Oh ok

[vast shale]

I'm a beginner

[west raptor]

Hmmm... I'll try

[vast shale]

Okay

one thing is sure

is that 1/.....

basically infinity

is 1/g(x)

[west raptor]

Yeah I know that

[vast shale]

alr

and you have to use some sort of quadratic formula

[proven garden]

$y'=\frac{x}{1-\frac{1}{y}}$

[twin meteor]

_wherewolf_

[west raptor]

I got $y^\prime=\frac{\sqrt{x^2+4}\pm x}{2\sqrt{x^2+4}}$

[twin meteor]

math_is_fun

[vast shale]

Well hailey said it wasnt quite right

[proven garden]

umm we just need g(1)

[vast shale]

the answer is (5+sqrt(5 )) /10 = g prime 1

[proven garden]

to find g'(1)

[vast shale]

they just dont show how they did it

[vast shale]

from my souvenirs it was y prime = xy something

[west raptor]

So from that $g^\prime(1)=\frac{5\pm\sqrt{5}}{10}$

[twin meteor]

math_is_fun

[vast shale]

wait wait whaat

please explain

[west raptor]

From this

Ok I'll explain it from the beginning

[vast shale]

but how did you get this

kk

[proven garden]

you isolate y' to the left side

[west raptor]

First $y=x+\frac{1}{y}$
Then $y=\frac{x\pm\sqrt{x^2+4}}{2}$

[twin meteor]

math_is_fun

[west raptor]

Then $y^\prime=\frac{\sqrt{x^2+4}\pm x}{2\sqrt{x^2+4}}$

[twin meteor]

math_is_fun

[vast shale]

The question is

how did you get from y = x + 1/y to that

like what formula did you uuse

[west raptor]

Using the quadratic formula

Multiplying both sides by y gives a quadratic equation

[proven garden]

is the answer even right?

[west raptor]

Yeah

[proven garden]

it couldn't be -

[vast shale]

how did you simplify it

oh YEA

X = 1

[west raptor]

I used a computer

I'm kind of lazy

[vast shale]

since we ant g(1)

g prime

so 1 +5

sqrt(5)

1+4*

• 5

waitt

the /10 where is it?

[west raptor]

Multiply the numer and denomin by sqrt(5)

[vast shale]

2 x sqrt(5)

[west raptor]

The last question is that how do we know the answer is $\frac{5+\sqrt{5}}{10}$ and not $\frac{5-\sqrt{5}}{10}$

[vast shale]

oh the numer

[twin meteor]

math_is_fun

[vast shale]

it's both ig

+-

[west raptor]

A function can't have two values you know

It's the definition

[vast shale]

since it's on quadran 1

[west raptor]

g(x) reminds me of the golden ratio

[vast shale]

i'd say +

wait lemme do it

btw from x / (1- 1/y)

for the quadratic formula

how do you know where is b ,a c

[west raptor]

What is b ,a c?

[vast shale]

the quadratic formula

[west raptor]

Ok

[vast shale]

x = -b ect..

x = -b +- sqrt(b2 ..

[west raptor]

I think I just found a reason

[vast shale]

y' = x/(1-(1/y)) is not a degree 2

uc ant use quadratic

[west raptor]

Wait no

Never mind

[west raptor]

How did you get that?

[vast shale]

well what did you use

[west raptor]

Shouldn't it be y=x+1/y?

[vast shale]

yes but it's still not a degree 2 equation

[west raptor]

$y=x+\frac{1}{y}\implies y^2=xy+1\implies y^2+(-x)y+(-1)=0$

[twin meteor]

math_is_fun

[vast shale]

ok you squared i

t

[west raptor]

I found another reason

[vast shale]

yea?

[west raptor]

So the $y=\frac{x-\sqrt{x^2+4}}{2}$ is always negative

[twin meteor]

math_is_fun

[west raptor]

But g(x) can be positive

[vast shale]

hm

[west raptor]

Here's a summary of everything:
$$y=g\left(x\right)\ =\ x+\ \frac{1}{x+\frac{1}{x+\frac{1}{x+...}}}$$
gives $y=x+1/y$. Using the quadratic formula yields
$$y=\frac{x\pm\sqrt{x^2+4}}{2}.$$
But the $-$ solution is always negative, which is not the case for $g$. Differentiating gives
$$y^\prime=\frac{\sqrt{x^2+4}+ x}{2\sqrt{x^2+4}},$$
so
$$g^\prime(1)=\frac{\sqrt{5}+1}{2\sqrt{5}}=\frac{5+\sqrt{5}}{10}$$

[twin meteor]

math_is_fun

[vast shale]

so to get /10

you timed it by sqrt(5)

i maen 5*

[west raptor]

This one, not 5

[vast shale]

oh yes

but if you time back at the num

the squares cancel

don't they?

oh nvm got it

got it got it

thank you man for helping

appreciate it

[vocal sleet]

@vast shale Has your question been resolved?

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[plush shoal]

Is this what it wants me to do? To draw a graph right?

[proven garden]

yes

[plush shoal]

Okeyy

.close

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[low turret]

Hello, i was doing some practice problems that were given to us about Taylor Series Expansion. The attached image is the function about a = -3

[low turret]

I manage to get this answer

but the textbook's answer key says that the denominator should have a times 6 (so it says that the denominator should be 6•3^(n+4)

[median moon]

How can I proove : a ≡ d mod m
Given that,
a ≡ b mod m
and c ≡ d mod m

[proven garden]

stop spamming

!help

[vocal sleet]

Please read #❓how-to-get-help

[low turret]

based on the pattern that I got, i dont know how that 6 got there so I dont know if the textbook has a wrong solution or im just dumb asf

[vocal sleet]

@low turret Has your question been resolved?

[proven garden]

I am too lazy to calculate f^(n)(x) can you show your step?

[vocal sleet]

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[vocal sleet]

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[thorny mica]

this is personal HW but can someone help me
there is 1 coin and 1 dice for context
I dont understand
there should be one way
its like a type right?
*typo

[modest aurora]

whats the original question?

[thorny mica]

I know how do do it

its easy

I juast diont get what was meant

in the since statement

@modest aurora

[modest aurora]

oh its not a typo

[thorny mica]

oh ignore the last message wrong server 😦

[thorny mica]

rlly what happened

[modest aurora]

so its the common of all 3

cards, dices and coins

for cards there is 52/4

wait

omfg

wait

yea no i dont understand this shit

i was thinking 2x2=4

so 52/4 gets cancelle by 4 and only 52 remains

but probability for dice is 1/6

i think its a typo

👍

[thorny mica]

sorery i had internet issues

rweading it now

[modest aurora]

nah dont

[thorny mica]

ita finwe thx

[thorny mica]

i read it thanks for the effort 🙂

have a nice day/night

[modest aurora]

same toy you

[vocal sleet]

@thorny mica Has your question been resolved?

[vocal sleet]

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Remember:
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[white coral]

Could I get confirmation on my answee

[sharp lynx]

Looks good

[white coral]

Ty

Could you also confirm one more thing, someone one else did but I just wanna make sure

[sharp lynx]

Sure

[white coral]

I had two separate answer 190 and 10

Isnt 190 the closer point compared to 10

[sharp lynx]

190 is closer to 170 than 10 is, sure, but that's not necessarily relevant

cos 530° is also equivalent to cos 170° so distance isn't exactly a relevant metric

But 190 is correct

[eager dune]

this is correct

[white coral]

Alright thank you for the informative explanation appreciate it

Thx

[sharp lynx]

Np

[dreamy viper]

i literally computed the values for you

i would think the engine is enough confirmation

[vocal sleet]

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[neon mango]

Can someone help me with this?

[neon mango]

I'm pretty sure I know the answer but idk how to phrase it

Pls ping when replied

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[quartz jungle]

Give it your best shot

[vocal sleet]

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[vast shale]

Why did they minus theta?

[split wind]

to be honest, from reading these alone, I don't know.

[rugged orchid]

Can you show part I and the whole question

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

Oh sorry my bad

[split wind]

i think it's because of BA

draw a straight line on B and parallel to real axis

then consider the angle between line and BA

then we can find that the angle is pi/3-theta

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[sudden cape]

Hi

[sudden cape]

I have a quick question

Can a matrix be positive definite if it is not symmetric?

Or must it be symmetric?

I know there's a formula but its for like all x in R^n that isnt 0

so im not sure how to test that case for nonsymmetric matrices exactly

without brute forcing to find a failure

[neon mango]

x . y (dot product of x and y) = x1y1 + ... + xnyn >= 0 for all real values of x and y.

So if x . y = 0 then either x or y = 0 for the dot product to = 0.

If x = 0 then //x + y // = sqrt(y x y) = y

//x - y// = sqrt(-y x -y) = y

If Y = 0 then // x + y // and //x - y // becomes // x //

I can't put the double lines because discord hides it

and it gets confusing

[molten willow]

use \|

\|\|x + y\|\| gives you ||x + y||

or just use the latex bot

[neon mango]

Idk how to use latex

[molten willow]

😎

[vocal sleet]

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@sudden cape Has your question been resolved?

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[cosmic gust]

Find all values of the parameter (a \in R) for which the following inequality is valid for all (x \in R):

[1+log_5(x^2+1) \geq log_5(ax^2+4x+a)]

If the range of values of a can be expressed in the form of (A, B], then find the value of A + B.

[cosmic gust]

So 1 can be represented as (log_5(5)), afterwards we use the logarithm addition to product thingy and finally get (5(x^2+1) \geq ax^2+4x+a) which can be represented as ((a-5)x^2+4x+(a-5) \leq 0). We get one equation here and another one is just due to property of log that (ax^2+4x+a > 0). After that, to proceed I have to use the discriminant inequalities, as per the solution we take (D \leq 0) in the first equation, and also (D < 0) in the second equation. I don't understand why, I took (D \geq 0) in both equations assuming the roots to be real due to the inequality and got a different answer.

[twin meteor]

usopper.

[cosmic gust]

Why should the roots be complex, because that is the only situation when we take D less than 0 right

[twin meteor]

usopper.

[vocal sleet]

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[vocal sleet]

@cosmic gust Has your question been resolved?

[vocal sleet]

@cosmic gust Has your question been resolved?

[signal drum]

don't see a problem here

if you take D>=0 and D>0 you won't get any a

the answer should be (2; 3]

[cosmic gust]

Ah

Let me try

[cosmic gust]

You do. [-2,2] U [3,7]

Same answer as before, but it isn't the correct one. What is the reasoning behind taking D < 0

[signal drum]

When you take D<0 we find parabola that has no solution

if parabola is looking up then D<0 will get parabola that has values bigger than zero

[signal drum]

You don't sum up it

Look at this

[cosmic gust]

Yeah, I get it now. You want it to be above or below the x-axis depending upon the inequality, otherwise it wouldn't be satisfied (>0 or <0 inequality in this case). So D cannot be > 0, correct?

[signal drum]

you get a є null set

[signal drum]

yes

there is no such point a that will belong to each region

[cosmic gust]

Oh my god

How did I not see that...

True

That is some insane tunnel vision from my side lol, thanks. It is crystal clear now

.close

[vocal sleet]

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[vast shale]

Hello, i dont understand how to do this at all. sorry, but can someone help me guide through this so i can understand it please?