help-38

[lean quiver]

First explain

80percenf of 800

Omg

I need reasoning behind it

Not just what it is

Bro stop

U can’t help me

i need someone else

.close

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[snow saffron]

OMGG

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[fathom cave]

Hello

[marsh forum]

,rotate

[fathom cave]

Anyone tell me approach for average question

[arctic mountain]

do you know how to take a weighted average

e.x if 40 make $10k and 60 make $20k, what's the average

[fathom cave]

Just to know average as some of all upon total numbers of

[arctic mountain]

try this

[fathom cave]

Have heard it bro but now don't remember

[fathom cave]

How ?

From where to start

[arctic mountain]

do you see how it's
(40/100) x 10000 + (60/100) x 20000 = 16000

[fathom cave]

No

[arctic mountain]

ok let's use smaller numbers

what if 2 make 10000 and 3 make 20000

can you find that average

[fathom cave]

30 k by 5?

Or 30k by 6

[arctic mountain]

you can just add them up and divide
(10000 + 10000 + 20000 + 20000 + 20000) / 5, right?

[fathom cave]

Any simple approach ?

[arctic mountain]

yeah we're working up to that

that's the same as (2 x 10000 + 3 x 20000) / 5, right?

[fathom cave]

Nah

Can't understand

Comprehend

[arctic mountain]

10000 + 10000 = 2 x 10000
and 20000 + 20000 + 20000 = 3 x 20000?

all divided by 5

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[wraith hinge]

Hi

[wraith hinge]

I have some images uploading

Can u tell me if my answer is right

Bruh

Why is my answer taking so long to upload

<@&286206848099549185>

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[wraith hinge]

I got sin(2pi/15x) is that right

Question

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[wraith hinge]

so are we allowed to js plug in tangent and why

[rough pier]

this is a substitution

its taken into account by converting the start and end of the definite integral

it goes from $2\sqrt{3}$ to $\frac{\pi}{3}$

[solid kiln]

leo~

[wraith hinge]

how do u know what put in for x

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[valid bobcat]

The only thing I know is 0 and 2 is important

Idk what to do here

[wary basin]

Try substituting the given values of X and y into your equations

Or you could calculate whether x value outputs the correct y value based on each formula

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[valid bobcat]

I just got this

What if it's not consistently one for the x values

What would B be

[wary basin]

You seem to have the right answer

[valid bobcat]

How would they have gotten ten

Though

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[frozen ocean]

Can i have some help starting this problem? just the expressing it as a rational function part. I understand how to go from a function to a series but i dont get how to do the reverse

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[frozen ocean]

.close

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[versed obsidian]

.close

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[solid urchin]

Anyone able to explain why this equals e^3/3 i get the top half but not the bottom

Ik 12 goes into 36 3 times so is it just simplifying down

Yea ig so

Nvm then this

Not really sure where to start

[marble flare]

What is your goal here

[solid urchin]

Simplify

[marble flare]

What part are you confused on?

[solid urchin]

Basically where to start lol

[marble flare]

Have you seen that (ab)^n = a^n b^n

[solid urchin]

Im really just confused about what logs ln and e is and i cant find good resources

Ik e is a constant but

[solid urchin]

Not really sure what to do with that i dont really have a b value do i?

Is e the value of b in that

[marble flare]

Yeah and 3 would be the value of a

Like what would (3*x)^5 be equal to

[solid urchin]

15x^5

[marble flare]

No

[solid urchin]

No sorry

[marble flare]

Almost

[solid urchin]

3x^5 ig idk

243x^5

[marble flare]

Yeah whatever 3^5 is

So now what would (3e^(3x))^2 be

[solid urchin]

9e^6x^3 idk thats why im confused lol

[marble flare]

No

Let’s go over what (e^a)^b is

Do you know what that is?

[solid urchin]

e^ab

Or

e^b(a^b)

[marble flare]

Yes exactly

[solid urchin]

Oh

Ok

[marble flare]

Nah it’s the first one

There is a distinction: e^(a^b) is different!

This is really important

In this case we get e^(ab) though

Does that make sense

[solid urchin]

Yea

Look though

I already have an exponent of e so what am i supposed to do

bc what is (e^3x)^2?

it wouldnt just be e^6x

[marble flare]

Yes!

Exactly

So what do you get when you combine rhose

[marble flare]

It is just this!

[solid urchin]

9e^6x

[marble flare]

Yes!!!

Great job

[solid urchin]

Thanks

I mean im not closing this yet though lol i still have a bit of problems i dont understand

[marble flare]

👍

For now 🔠

[solid urchin]

Lol

Supposed to rewrite to y=a(1+r)^t or vise versa

[marble flare]

U sure?

Main questions: what is t?

[solid urchin]

Yea thats the instructions

[marble flare]

There is only an x in ur oicture

[solid urchin]

Number 11

[marble flare]

Huh wow it really does

Anyways do you see that you need to write e^-0.5 in the form 1+r or 1-r

[solid urchin]

Well i'd assume + for the first 2

Or wait

No

[marble flare]

It depends on if e^-0.5 is > 1 or < 1

[solid urchin]

Ohh wait

Well ok

First one is a(1-.61)^t Second is a(1+.64)^t Third idk

That seem right so far?

[marble flare]

,w 1/e^(0.5)

[solid kiln]

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Requested by rbmuk

[marble flare]

Not quite

[solid urchin]

Well

I just rounded but

[marble flare]

Nah you don’t have the right exponent

[solid urchin]

x

^x

[marble flare]

Like the function is 0.61^t but you have (1-0.61)^t

[solid urchin]

Oh

Really?

Why would it want me to write it in that form?

[marble flare]

Not sure 🤣

[solid urchin]

Is the function for that actually .49^t

Bc thats when you actually subtract

[marble flare]

The r would be 0.49

[solid urchin]

Yea

[marble flare]

In the form (1-r)^t

[solid urchin]

Ohh yes

i see now

My bad

So then

#12 would be a(1+.36)^t?

And the rate would be .64

[marble flare]

,w e^(-0.2)

[solid kiln]

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Requested by rbmuk

[solid urchin]

-.2 not positive

[marble flare]

1/e^x is e^-x

[solid urchin]

Oh

[marble flare]

But it looks like it would be (1-.19)^x

[solid urchin]

Well

its 2e

[marble flare]

How did you get ur answer?

[solid urchin]

Not just e

[marble flare]

The 2 is not in the exponent

[solid urchin]

Oh

Thats my a

[marble flare]

This ends up being really important! (2a)^b is NOT 2a^b

[solid urchin]

Right

[marble flare]

Yep

Exactly

[solid urchin]

So r is .81

#13 5(1+.18)^t?

Orr

Growth rate would be 1.82 right

But then

I think thats just 82% for r

So then is it 5(1+.82)^t mmmm idkk

I have the answer key ig ill just look

Yea so its 82% r

Also

#12 we have flipped

Sorta

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[raven sand]

hello

[raven sand]

i need help

[prime crystal]

how do you find volume of cube?

[raven sand]

side x side x side

[prime crystal]

and here that is equal to?

[raven sand]

1000 / 3?

[prime crystal]

no, just 1000

[raven sand]

yes typo

[prime crystal]

still just 1000

not /3

[raven sand]

oh

[prime crystal]

side x side x side is equal to 1000

which is volume

now find the side

side x side x side = (side)^3

not 3(side)

[raven sand]

whats the side?

[prime crystal]

side of the cube?

[raven sand]

not sure

\😭

[prime crystal]

volume of a cube is side x side x side

and thats given 1000 cm^3 here

so side x side x side = 1000 cm^3

so (side)^3 = 1000 cm^3

[wary basin]

X^3=1000 since cubes have the same side lengths

[prime crystal]

now find the side

[raven sand]

1000 divide by 3?

how?

cube root

[solid kiln]

faiyrose

[raven sand]

wait

slow proccesing

@harsh surge 10?

i forgot

square

side x side x side

length x width

yes

6

yes

[solid kiln]

faiyrose

[raven sand]

how

[solid kiln]

faiyrose

faiyrose

[raven sand]

600?

is that the answer

why not cm^3

ohh yea

okayy thankyouu

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[raven sand]

.close

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[lavish tangle]

need help with my y10 specialist math on combinations and permutations, the idea is: you pick 2 card hands in 5 card poker. i have picked 3 of a kind and 4 of a kind, then you modify the game so the probability of getting those hands are higher. i have decided on adding 1 or 2 wildcards (jokers) to the deck. i feel like i am overthinking how complex this needs to be. please help 🙂

[lavish tangle]

is this the right idea?

im happy with the original hands but the wildcards are confusing me

<@&286206848099549185>

hepl 😦

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[glass olive]

hi

[lavish tangle]

g'day

im so confused on what to do for my modified game

[glass olive]

i do not know how poker works

[lavish tangle]

aw damn

algs thx anyways

<@&286206848099549185>

[glass olive]

wait actually i'm confused

what are we supposed to solve here

[lavish tangle]

in the total tab in the first picture am i supposed to be adding all of those

im like so lost

[late anchor]

idk if my math is wrong but im pretty sure it would be 316,251

[lavish tangle]

which part?

[late anchor]

for the top rectangle thingy

[lavish tangle]

that?

[late anchor]

yes

[lavish tangle]

youve missed a zero or ive missed a zero

lemme check

nope, youve missed a zero

[late anchor]

I just did that

[lavish tangle]

its 54c5 isnt it?

ive finished the table, look right?

*4 of a kind at the end not 3 of

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[austere cedar]

Are you able to do part a?

[pearl plaza]

part a is provlem

i did part two correctly

[austere cedar]

For path 2
You want y = 2t

[pearl plaza]

why?

[austere cedar]

Should finish at (1,2)

[pearl plaza]

i did that but got-16/6 final answer

[austere cedar]

Another option is to integrate between 0 and 2 instead of 0 and 1

[pearl plaza]

isee

but now i have 5/6

[austere cedar]

Took a sec to look through that last integral, it seems well done

[pearl plaza]

seems, but it's not yes?

🤔

where is the mistake

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[pearl plaza]

.close

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[night arch]

Why is it -10x³

[prime crystal]

-5x^3 - 5x^3 = ?

[night arch]

Should it be positive ?

Cuz - - = +?

[prime crystal]

thats for multiplication

[night arch]

I see.

[prime crystal]

what would be -2 -3? apply the same above

[night arch]

-5

I understand now.

[prime crystal]

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[bitter sparrow]

Hey

[tough iron]

Hi!!!

[bitter sparrow]

Hello

How are you

[tough iron]

Im great just had lunch

[bitter sparrow]

Can you help me

[orchid wagon]

!da2a

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[bitter sparrow]

Two tanks were filled with water with 9 identical pipes. At first all the pipes filled the first tank, when it was 1/5 full the two pipes were switched to fill the second tank. At the moment when the first tank was completely full, the second tank was only 3/10 full. What is the ratio of the volume of the second tank to the volume of the first tank?

Lol

[tough iron]

So T2 was 3/10 full when filled with 2 pipes

And T1 got 4/5 filled with 7 pipes

Is that right? Did i read it correctly

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[bleak urchin]

Need help

[rough goblet]

what have you tried?

[bleak urchin]

Yea

I ain't getting tho

Pls help

[acoustic flint]

Do you know any logarithm rules?

In fact, in your working out line 3 just move x to rhs

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@bleak urchin Has your question been resolved?

[bleak urchin]

So what wrong i have done

[acoustic flint]

Just multiply by x to get rid of it

0*x = 0 on rhs

[bleak urchin]

Multiply what by x?

[acoustic flint]

Right now you have

[bleak urchin]

Can u pls show written soln of what ur trying to tell if possible

[acoustic flint]

$\frac{2-6x}{x} < 0$

[solid kiln]

shsgd

[acoustic flint]

So just multiply both sides by x

[bleak urchin]

Why but? Any reason behind it

[acoustic flint]

To get rid if the x on the denominator

[bleak urchin]

Cant we solve without get multiplying by x?

[acoustic flint]

Then you have 2-6x < 0

Why don’t you want to multiply by x?

[bleak urchin]

Final answer is (1/3,2/3)

But here i aint getting 1/3,2/3

<@&286206848099549185>

[acoustic flint]

Seems like it should be [-inf, 1/3)

You want to find the range of x such that 6x-2 <= 0

Since there is no lower bound, you just solve for when 6x-2 = 0 which is x = 1/3

So anything smaller than 1/3 will still be less than 0

[bleak urchin]

But the correct answer mentioned in book is (1/3,2/3)

Pls solve it urself and help me 😁

Or should i ping someone else

So that we can get help

@acoustic flint

[acoustic flint]

2/3 is not a valid solution

[ \frac{2-3x}{x} \geq 3 ]
[ 2-3x \geq 3x ]
[ 2 \geq 6x ]
[ x \leq \frac 13 ]

[solid kiln]

shsgd

[acoustic flint]

There is no lower bound on x

[bleak urchin]

Bruh

X belongs (1/3,2/3)

@acoustic flint

[acoustic flint]

Sub in 2/3 for x and tell me if you get a number greater than or equal to 3

[bleak urchin]

Why would i know in my exam that i have to substitute 2/3

[acoustic flint]

Can you send a photo of the solution then

[deep niche]

You shouldn't be multiplying throughout an inequality by x when x is of unknown sign.

Better is
(2-6x)/x >= 0
(3x-1)/x <= 0
draw number line or perform sign test.
Soln: 0 < x <= 1/3

Apologies I just saw the original problem.

[deep niche]

The first step is also incorrect since log(base 1/3) of x is a decreasing function, so it should become
0 < (2-3x)/x <= (1/3)^(-1)

The 0 in front is necessary because you cannot take the log of a non-positive number.

From here, break it up into two inequalities and solve separately, then take the intersection.

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[visual wing]

i know nothing abt root so help me in this question pls

[near vessel]

rationalize the denominator of the 2nd part

we want to rationalize the denominator to get rid of the sqrt in the denominator

such that it becomes 5, which is perfect because you can combine it with the left fraction easily

[oblique bay]

rationalisation

Basically we are not allowed to work with fractions which have roots in the denominator

[near vessel]

,,\frac{\sqrt{20}}{5}-\frac{2}{\sqrt{5}}

[solid kiln]

Renz

[oblique bay]

so if you multiply a root with itself, it gives you the number under that root

[near vessel]

we multiply the right one by sqrt5/sqrt5 like this

,,\frac{\sqrt{20}}{5}-\frac{2}{\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}}

[solid kiln]

Renz

[oblique bay]

The first term with root 20 is fine, as the denominator is an integer

But the second term is not

[visual wing]

ohh ok tysm

/close

.close

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[bleak urchin]

Tysm broo. I request u to pls provide a written soln if possible

.reopen

Help

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[severe fulcrum]

I'm a bit confused on how I'm supposed to solve this

[warped plaza]

Is that sht linear or quadratic

[true bane]

trial and error

you can kinda see that it isnt linear

[warped plaza]

For linear we have y=ax+b

Damn

[true bane]

there is also some potential symmetry at x = 0

so i would bet money on D

[severe fulcrum]

alright ima mess around on a graph

and check that out

thx for ur advice

[nimble stone]

if you dont want to guess or do it by testing each one, you can see by symmetry that the vertex is (0,1)
so y=ax^2+1, then use another point for a

[severe fulcrum]

makes sense, thanks guys

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[glad smelt]

I'm struggling to see how I might be able to solve this further from what I've done:

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[real gale]

Try calculating (x y z)*A*(x y z)

[glad smelt]

like expanding it>

?

[real gale]

multiply matrices

[real gale]

what do you get?

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[glad smelt]

Ooo sorry

Are you suggesting to just compare coefficients?

[real gale]

,rotate

[real gale]

Yes

[glad smelt]

Alright, thank you very much

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[glad smelt]

.reopen

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✅

[glad smelt]

I'm not sure how to solve this

The simultaneous equations aren't solvable

[real gale]

,rotate

[real gale]

What do you mean

There are many solutions

[glad smelt]

Oh..

So we can just put any values that satisfy this in?

And there are no more constraints

Like b and d could be 1&3 or 2&2

[real gale]

A doesn't have to be symmetric so yes

[glad smelt]

Cool

Thank you

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[short spear]

Which is the answer?

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[short spear]

@help

<@&286206848099549185>

[lavish dew]

A is conditionally convergent

remember the alternating series test, if the limit is as n goes to infinity is 0 and if the serie is alternating it means it converges!!

and if the absolute value of the serie diverges it means that is conditionally convergent

use direct comparision test for A, compare it to 1/sqrt(n)

p = 1/2 which is less than 1 so it is divergent

but bc the function has an alternating factor it will converge

it means that the convergence depends on the alternating factor, so it is conditionally convergent

[short spear]

I dont think 1/n^(1/2) goes to 0

Thats also how i knew it diverged

[short spear]

Oh nvm then youre right

[lavish dew]

the limit goes to 0 ah

[short spear]

I was thinking about the p series rule again

[lavish dew]

hm I see

but remember that if the serie is alternating, it just needs its limit to infinity to be 0, no other test required

if you want to know if it is conditionally convergent, then remove the alternating factor and do your regular convergence tests

if you get divergent, it means the serie is conditionally convergent, if it is convergent as well it means is absolute convergent!!

[short spear]

Yep thanks for the help @lavish dew

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[lavish dew]

< 3

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[gentle lance]

This is a problem of U substitution. I got -1/12, but the answer should be 4. I set u=3q^2+2, and rearranged the variables such that the equation 196q(u^-3)dq. I took the derivative of 3q^2+2 to recieve 6q A.K.A. du, then divided the 196q by 6q to receieve 98/3, and lastly set this coefficient outside the integral. I took the integral of u^-3 as u^-2/-2, and then multiplied this factor of -1/2 onto the 98/3 to receive -49/3. Plugging the original equation in the place of u, and then plugging in the value of 2 in that equation gave me this answer.

[wraith arch]

your anti derivative seems correct so you probably messed up plugging in the numbers

can you show your work for that?

[gentle lance]

One moment.

-49/3(3(2)^2+2)^-2
-49/3(14)^-2
-49/3(1/196)
-1/12

[wraith arch]

you forgot to plug in 0

[solid kiln]

y0shi

[wraith arch]

this is the fundamental theorem of calculus btw

[gentle lance]

You're right. I mistakedly assumed F(0)=0

Thank you.

or rather, F(a)=0

But I missed the +2

[wraith arch]

all good

yw!

[gentle lance]

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[gentle lance]

thanks!

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[trim trench]

we are starting on a new unit so im kinda new

we are suppose to learn this in 2 days but I have nothing better to do and cant ask anyone for help on this problem

plzhelp

[ionic pendant]

what is the question?

[trim trench]

I just have to find the error

and fix it

bruh

[ionic pendant]

what kind of triangles are trig functions designed to work for?

[trim trench]

?

[ionic pendant]

there is a particular category of triangle that trigonometric functions describe

[trim trench]

?

I just have to find the error

and fix it

for 11 the correct answer is 25 /12

35

[ionic pendant]

would you use trigonometry to describe this triangle? why or why not?

[trim trench]

idk we are suppose to start on this lesson on friday or on monday

[gusty bone]

sohcahtoa only works on right triangles

is the triangle in the problem right?

[ionic pendant]

i would recommend watching a video for the concepts before leaping headfirst into problems:
https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles

[trim trench]

aight

is the answer like

18/21.9

tan 55 = 18 / 21.9

[ionic pendant]

this part is important

[trim trench]

aight

[trim trench]

wait so am I right?

the answer is 18 / 21.9

[ionic pendant]

is the triangle right?

[trim trench]

?

no like thats the problem

and I have to fix that issue

so tan 55 = 18 / 11 is wrong and I have to try fixing the issue

[ionic pendant]

do you know what a "right triangle" is?

[trim trench]

yes

[ionic pendant]

so is this a "right triangle"?

[trim trench]

no

oh

💀

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[wraith hinge]

Can someone help me with the intuition behind using the degree 3 Taylor polynomial?

[wraith hinge]

This is to find convergence using big o notation btw

[scenic fractal]

do you have the question please?

[wraith hinge]

Yes!

One sec

[scenic fractal]

alr this makes sense, you know the taylor expansion of cos x, they took the $\frac{x^2}{2}$ to the other side of the equation.
cosx on LHS does have a term of $x^4$ in its taylor expansion (its $\frac{x^4}{4!}$ and it is the third term of the taylor expansion of cosx). since their is a term of $x^4$ on the RHS we can compare both now

[solid kiln]

penguin

[scenic fractal]

that is intuition behind taking the "third" term of the taylor expansion of cosx

[wraith hinge]

so we only took the third term because it already was given that it is O(x^4)

[scenic fractal]

yeah, you can take the next terms as well but since taylor series problems are based on approximation i dont think you should.

but yeah if it were o(x^6) on the RHS then you have to consider the 4th term of the taylor expansion as well

[wraith hinge]

i see what you mean

so how would i go about solving a problem that asks me to find big Oh notation of cosx without giving me O(x^4) for example

[scenic fractal]

O(cosx) you mean? cannot be that because they'll always give you something like O(k*x^n) in order to frame a problem on this. while the LHS can be anything

[wraith hinge]

ok that helps, i was asking though if they just give me like f(x) = cosx, find big O(f(x)), but that can't even be, right?

[scenic fractal]

yeah. the approximations would be such that when you consider them, you can solve both sides to be equal. i hope i put that well in words

$f(x) = k \cdot x^n$

[solid kiln]

penguin

[wraith hinge]

i seeee

thanks for your help!

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[sand helm]

I need help with this, I forgot some basic algebra, I just don't get how the one with the three made it like that

[round mango]

a÷b = a * 1/b

[sand helm]

Oh I got it

I feel dumb now

Ty tho

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[quaint thistle]

Need help with this question. These are always pretty easy and I often use an infinite series of sqrts and then substitute x. But most methods I’ve tried seem slow. What’s the fastest method one can think of for something like this ?

[livid thunder]

yo

what is the question?

to solve for x in terms of a?

[livid thunder]

yo?

[quaint thistle]

Yes

[livid thunder]

aight too ez

first i'd square both sides

which should give us a-x=a^2-2ax^2+x^4

$a-x=a^2-2ax^2+x^4$

[solid kiln]

Galaxy

[quaint thistle]

Yep

[livid thunder]

then take everything to one side

[quaint thistle]

Ye got it thanks

[livid thunder]

aight

[quaint thistle]

I was wondering if there’s another way

Since this seems to take a while

[livid thunder]

not that i can think of, never solved these type of questions

[quaint thistle]

Aight, thanks though

[livid thunder]

all good

[quaint thistle]

<@&286206848099549185>

[lavish dew]

what is the question/

?

[quaint thistle]

Solve equation for x in terms of a

[lavish dew]

and what do infinite series have to do here :p

[lavish dew]

square both sides and solve with the quadratic formula

be caferull with the domain!!

you have a square root of a - x

[quaint thistle]

$x= sqrt(a-sqrt(a-x))

[lavish dew]

if a is greater than x,then it is ok because we get a positive value

if x is greater we will get a imaginary value

so we are squaring both sides assuming a will be greater than x

okie?

[quaint thistle]

Ye

Thanks

[lavish dew]

then solve with cuadratic formula

remember to be carefull with the domain and 2 variables expressions and roots

[quaint thistle]

Another way I do it is try and make something like this (with a instead of 6)

Just for fun

[lavish dew]

lol that is cool

the you get x = sqrt(6+x)

square and solve!

[quaint thistle]

Yep

[slow dawn]

that'd be way slower, no?

[quaint thistle]

Well it works with habit and squaring always just adds more terms and more work personally

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[fresh gazelle]

Can anyone help pls, i got stuck and don’t understand how they got to step 2

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[burnt fossil]

how to simplify (2n)!/n!

[lusty delta]

that's already pretty much simplified

[burnt fossil]

how

[lusty delta]

well there's not really a better way to write (2n)!

ig u can obtain a nicer expression if u use the pochhammer symbol or the falling factorial but that's not usually something that u think of when u consider "simplification"

[burnt fossil]

my bad then

thanks

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[turbid coral]

Have I set this up correctly?

[grim saffron]

👍🏼

[turbid coral]

Okay so if I solve

(I-c)x=d and use matlab, the values I get are wrong

When I row reduce, these values are wrong

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[wraith hinge]

prehaps teh 34, 61, and 55 hours are meant to indicate the "total" amount

rather than like that time + extra

nvm that doesnt work lmao

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[wraith hinge]

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[wraith hinge]

Anyone know why this is wrong?

[grim saffron]

Try plugging that x value to Ax=0

You should get 0=0, otherwise the answer is wrong.

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[tardy hemlock]

I get something different

[wraith hinge]

5 is supposed to be -5

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[warm sorrel]

Task:
For all $n \in \mathbb{N}$ the open Ball $B_{1}(0) = { x \in \mathbb{R}^{n} | x_{1}^{2}+...+x_{n}^{2} < 1 }$ is homeomorphic to $\mathbb{R}^{n}$.
A homeomorphism is for example $f:\mathbb{R}^{n} \to B_{1}(0), x\mapsto \frac{x}{1+\sqrt{x_{1}^{2}+...+x_{n}^{2}}}$
Prove that f is a homeomorphism to prove that $\mathbb{R}^{n}$ is homeomorphic to $B_{1}(0)$.

[solid kiln]

Endless_Meadow

[warm sorrel]

My ideas is to first prove that f is a bijection $\to$ Prove that f is surjective and injective
f surjective:

[solid kiln]

Endless_Meadow

[warm sorrel]

f injective: I have no clue how to actually prove the last step for injective

Secondly, prove that f is continuous:
Absolutely no Clue whatsoever

Thirdly prove that the inverse function is continuous:
Continuing from surjective i took that the inverse function is $x = y \cdot (1+ \frac{| y|}{1-| y|}$ but got no clue how to prove that is continous as well.

[solid kiln]

Endless_Meadow

[warm sorrel]

?

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[fathom cave]

How to solve

Like is it ok to take x as comman?

Igore uppar parts just look last two lines

Sry , this ticket is aquired

My bad

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[grim saffron]

Plug the last one into the second one

Now, you know how to invert y=f(x) given y != 0. Now you just have to handle the case when y = 0

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[steady turtle]

Can anyone teach me how to use n-th partial sum to find its convergence or divergence?

Nvm, I found that solution from internet

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[wintry bone]

I'm a helper of grade 0-7

[wintry bone]

Cheater

[lusty delta]

extremely funny comment

[wintry bone]

😅

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[barren hearth]

Is this correct?

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[wraith hinge]

looks like a big jumble of unrelated equations to me

[wraith hinge]

impossible to tell without problem context

[lusty delta]

damn that’s like 100% a troll but i forgive it because of the integral at the bottom

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[broken reef]

i don't know boolean algebra at all, so is this reordering of this equation something that is possible? and if so, how could i do it?
$$NOT(a)\ OR\ b = TRUE$$
$$b = \ldots$$

[solid kiln]

declspecl

[vapid grove]

i dont get what b = ... means

the first says "what values of a and b satisfy this equation"

but then what is... what does b = mean

b in terms of a?

[broken reef]

yeah sorry i just meant how could i rearrange the equation to get $b$ in terms of $a$ and the original constant

[solid kiln]

declspecl

[vapid grove]

i think it's b = a

not a or a = true

the only boolean algebra i studied is for circuitry and i didnt really touch on rearranging equations, only expressions

but from what i know b = a makes the most sense to me

:/ but i dont know how one would write the steps

[broken reef]

this would be the truth table for this equation if it helps, i don't think it could be a = b though because of the F, T = T case

| a | b | res |
--------------
| F | F | T |
| F | T | T |
| T | F | F |
| T | T | T |

actually i see what you mean but i know for my use case it wouldn't work, not sure why though 😦

[shrewd ridge]

b = a or c

seems to say the same thing

[vagrant prism]

oh, this looks like implication

$a\rightarrow b$

[solid kiln]

artemetra

[vagrant prism]

https://en.wikipedia.org/wiki/Material_conditional

[shrewd ridge]

you said "in terms of a and orginal constant" and that's what i got

ok you didn't mean invent a new constant c i think

[shrewd ridge]

actually what you really want is an inequality

b >= a

it's both like normal algebra and true

@broken reef

[broken reef]

how does an inequality work in boolean algebra 🤔

[shrewd ridge]

well this isn't boolean algebra if you want "b ="

they don't do that

[broken reef]

oh i see, so you're saying true >= false, so b is true if a is false and b is true or false OR a is true and b is true?

[shrewd ridge]

if you want equality it should be b = a or c

we make up a new variable that wasn't there

[broken reef]

that does line up with how i ended up splitting it, (a = FALSE) OR (b = TRUE)

[broken reef]

interesting so you would have to define that variable's value each time inchecking equality?

[shrewd ridge]

your original statement doesn't let you find b in every case

we simply keep that property by using c

we don't know it

so we don't know b, so it's equivalent

and if a is true we know b

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[dreamy heron]

How do I find all possible values of the variables in the cryparithm
"FAILURE
+FAILURE
=SUCCESS"
?

[dreamy heron]

Sorry, there should be 4 "FAILURE" being added. mb.

To reiterate:
FAILURE
+FAILURE
+FAILURE
+FAILURE
=SUCCESS

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[dreamy heron]

.close

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[glass olive]

to start, you want to find the combined speed at which the 2 cars are closing the distance

and that would be 75 + 75 = 150

with this, we can find the time it takes for them to collide

100km divided by 150 km/h

2/3 hours

we know the bird flies continuously at a speed of 120km/h

you can figure out the rest

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[glass olive]

yes

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@wraith hinge Has your question been resolved?

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[wraith hinge]

find reminder when 13^42 + 14^42 is divided by 730

[wraith hinge]

13^42 equiv 1 mod 2
14^42 equiv 0 mod 2

13^42 equiv 4 mod 5
14^42 equiv 1 mod 5

13^42 equiv _ mod 73
14^42 equiv _ mod 73

idk how to find last two

aplease ping when youre replying

[ebon kite]

you probably need to use repeated squaring

[wraith hinge]

idk how to do that ?

can you please show me some steps ?

or just ell me ill do it on my own just guide me

[ebon kite]

so if you want to calculate $14^{42}$, you can break it down like this, $$14^{42} = 14^{32} \cdot 14^8 \cdot 14^2$$

[solid kiln]

AlphaNull

[ebon kite]

and since those are powers of 2, you can calculate them quickly by squaring 14 mod 73

[wraith hinge]

lemme try

[wraith hinge]

49 mod 73 = 14^42
56 mod 73 = 13^42
are thesse correct

?

[ebon kite]

dont agree with ur 2nd one

first one looks good

[wraith hinge]

ok lemme do it again

[wraith hinge]

is it 24 mod 73 = 13^42

[ebon kite]

yes

[wraith hinge]

thanks lordthat you responded

thanks to you too

[wraith hinge]

i got all things can you help me in one more thing like how can i cmobine all these ?

[ebon kite]

yes, so you need to use CRT

[wraith hinge]

chinese reminder theorem ?

[ebon kite]

yes

[wraith hinge]

my lord i forgot that lemme recall

[wraith hinge]

13^42 + 14^42 equiv 1 mod 2
13^42 + 14^42 equiv 5 mod 5
13^42 + 14^42 equiv 73 mod 73 equiv 0 mod 73

[ebon kite]

why are you adding them

oh

i see nvm

[wraith hinge]

now please tell me the last step

[wraith hinge]

..

sorry for ping

[ebon kite]

so do you know how CRT works

[wraith hinge]

actually no

i only heard name

actually i only used fermat and modular arithmetic till now

[ebon kite]

if you have $$\begin{cases} x \equiv 1 \pmod{2} \ x \equiv 0 \pmod{5} \ x \equiv 0 \pmod{73} \end{cases}$$, you can uniquely determine $x \pmod{\mathrm{lcm} (2,5,73)}$

[solid kiln]

AlphaNull

[ebon kite]

you know if 5 divides x and 73 divides x

since both are coprime, then 5*73 divides x too

[wraith hinge]

yes

[ebon kite]

then you can reduce the system to

$$\begin{cases} x \equiv 1 \pmod{2} \ x \equiv 0 \pmod{5 \cdot 73} \end{cases}$$

[solid kiln]

AlphaNull

[wraith hinge]

got it till here

[ebon kite]

if $x \equiv 0 \pmod{365}$ then $x = 365k$ for some $k \in \mathbb{Z}$, so then substitute that into the first equation to get

[solid kiln]

AlphaNull

[wraith hinge]

365 x odd no is ans

[ebon kite]

$365k \equiv 1 \pmod{2} \iff k \equiv 1 \pmod{2}$

[solid kiln]

AlphaNull

[ebon kite]

so now you know k and can backsubstitute

[wraith hinge]

no no im overthinking

[wraith hinge]

wdym by this line ?

k is an odd no

[ebon kite]

k is any number

ah

yes you are right

this just says k is 1,3,5 or so on

you can see that no matter what odd we take, 365k will always be congruent to 365 mod 730

so we can just take 1

[wraith hinge]

means reminder is 1 ?

[ebon kite]

k is 1

remainder is 365 times 1

which is 365

[wraith hinge]

ohh han

got it

[wraith hinge]

i have onne more thing to ask

suppose if x mod 5 =2
is it equivalent to 2 mod 5 = x
if it is equivalent then whats the case ?

[ebon kite]

yes that is equivalent

well, assuming you are taking mod 5 both sides

x mod 5 = 2 by itself is not equivalent to x = 2 mod 5

but x mod 5 = 2 mod 5 is

[wraith hinge]

ohh both are same thing

okok

thank you so much

please leave the thread open i wanted to repeat it once more for myself
if i have any doubt i ll share it with you

its done now

thanks again

.close

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[rapid horizon]

150x=35 mod 31

[rapid horizon]

I tried to put random values but got nothing

[stark bison]

Start by reducing 150 and 35 mod 31, then find the inverse of the coefficient of x using Euclidean algorithm

[rapid horizon]

30x=7mod31

X=(30/7)^-1 mod 31

@stark bison

[stark bison]

Why do you think 150 and 35 are congruent to 30 and 7 modulo 31?

[rapid horizon]

26x =4 mod 31?

[stark bison]

Yeah that's correct now

Now you can find the inverse of 26 modulo 31 by using Euclidean algorithm on 26 and 31 to obtain integer a, b such that 26a + 31b = 1

[rapid horizon]

Ahh i am getting 31=26+5

[stark bison]

Right, that's the first step, now divide 26 by 5 in the same manner

[rapid horizon]

31=26+5
26=5×5+1

1=26-5×5

1=26-5×(31-26)?

1=26-5×31+26×5
1=26×6-5×31
1=156-5×31

[stark bison]

Yes

1 = 26 * 6 - 5 * 31 implies 26 * 6 = 1 mod 31

[stark bison]

So the next step would be to multiply both sides by 6

[rapid horizon]

24

[stark bison]

Yes, x = 24 mod 31

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[wraith hinge]

so i proved (i)

like so

but now i cant prove (ii)

[edgy willow]

you have to give descriptions for what each side of the equation represents

2^i will come from the (i) proof, subsets with a max of i+1

[wraith hinge]

so that means my proof from (i) is bad

[edgy willow]

I like your (i) proof

I'm just shifting by 1 betweek i and k to match what (ii) says

[wraith hinge]

yes well we know that $\sum_{i=0}^{n-1} = 2^0+2^1+\dots+2^{n-1}$

[solid kiln]

Derivative

[wraith hinge]

and our last term there can be considered k

(ie the total amount of subsets where n is the largest number)

i can also do this

$\sum_{i=0}^{n-1} = 2^{1-1}+2^{2-1}+\dots+2^{n-1} = \sum_{i=1}^n 2^{i-1}$

[solid kiln]

Derivative

[edgy willow]

I'm not really following

the left side is adding up all the subsets with 1 max, 2 max, up to n max

[wraith hinge]

yes

i was just trying to transform the left side to make it equal to the right side

im going to need a combinatorial argument for sure

the left side, like you said, is adding all the subsets of length i from i=0 to i = n-1

[edgy willow]

uh max not length but pretty much

[wraith hinge]

yes max of i

since the set of (1,2...i)

[edgy willow]

and the right side, remember that 2^n is just the total number of subsets

so just explain the -1 and you're done