help-0

[pale kestrel]

u - 1

$$I=\int\frac{(u-1)^3}{u(u-1)}\dd{u}$$

[ocean seal]

Shuri2060

[pale kestrel]

right?

du/dx = e^x

du = e^x dx

du = (u-1) dx

then you'll be fine

[floral quail]

I believe I've got it, one sec

What would the integrated form be?

[floral quail]

I understand how you got to here now ty

But its set out different than the previous questions

Nevermind, go it

Thanks for the explanation @pale kestrel !

.close

[lone heart]

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[jaunty crane]

anyone knows

[jaunty crane]

if i got 2 equations

and they want me to find the range

0<m(x)<10

cuz i got 2 equation

first 1 is m(x)=3p-5/3

and n(x)=1/2x+5/6

any clues to get the range?

[echo socket]

m(x) = 3p - 5/3 ?

[jaunty crane]

should i draw a graph and find out how?

[jaunty crane]

ye

wait

nono

3x-5

/3

[echo socket]

Range of what tho? you gave 2 different functions

[jaunty crane]

wait

[echo socket]

For each of them?

[jaunty crane]

ill show u

[jaunty crane]

i cant continue write the questions

m(x) is ....

0<m(x)<10

[echo socket]

Okay so we have to solve 0 < m(x) < 10

By plugging in the expression of m(x)

0 < (3x - 5)/3 < 10

[jaunty crane]

owh

[echo socket]

And this will be domain of n as far as I understand

[jaunty crane]

the answer for the question is 0<x<10

how can i get that ?

any clues?

[echo socket]

"Define domain for n(x) when range for m(x) is ..." is this the question word by word?

[jaunty crane]

i actually translate

them

mine was in malay soooo

determine

[echo socket]

Is there any relation between m(x) and n(x)?

[jaunty crane]

i think

oh yea

its inverse relations

[echo socket]

Oh

[jaunty crane]

the question actually need the p and q from the equation

[echo socket]

So

[jaunty crane]

and i solved them

[echo socket]

Domain of n is the same as range of m

And range of m is (0, 10)

So domain of n is (0, 10) as well

Because they're inverses

[jaunty crane]

i dont rlly

[echo socket]

Thus n is defined for 0 < x < 10

[jaunty crane]

understand

[echo socket]

Which part

[jaunty crane]

this

and the rest

that u just said

[echo socket]

So n is inverse of m, right?

[jaunty crane]

yes

[echo socket]

So if m: A -> B, then n: B -> A

[jaunty crane]

oh

yes

[echo socket]

So

Look at range of m

It's the same as domain of n

[jaunty crane]

yes

[echo socket]

So if we know range of m, we also know domain of n

[jaunty crane]

yes

[echo socket]

And we're also given that 0 < m(x) < 10

[jaunty crane]

yes

[echo socket]

So the range (of m), B, is (0, 10)

[jaunty crane]

wait the 0<x<10

it means 0 and 10 are x?

or is it y

[echo socket]

0 < x < 10 means that x is in the interval (0, 10)

[jaunty crane]

what is interval

:d

[echo socket]

Part of the real number line

In this case, it's the segment of numbers between 0 and 10

[jaunty crane]

alright i think i understand now

thx

.close

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[jaunty crane]

the number of pupils in a school decreases by 4% each year. Find the period of time, in years, when the number of pupils in the school becomes less than 1/3 of its original number

[jaunty crane]

can someone help me with this question

its about arithmetic or geometric progression

[lone heart]

@jaunty crane Has your question been resolved?

[tawny orchid]

Geometric

[little drum]

Hint: Let the number of pupils in the school initially be x. Then the next year, the number of pupils remaining would be $x - 4%, \text{of}, x$. The year after that, this number would reduce to $(x - 4% , \text{of} , x) - 4% , \text{of} , ((x - 4% ,\text{of} ,x))$ and so on...

[ocean seal]

Ansh

[tawny orchid]

It's sort of like compound interest if you know that

[little drum]

Write number of pupils remaining in school after n year as $x_n$, then you have $x_n = x_{n-1} - \frac{4}{100}x_{n-1}$

[ocean seal]

Ansh

[tawny orchid]

You're subtracting 4% of it, then the next year, you're again subtracting 4% from the previous result

[little drum]

you can follow the same notion as from here

[tawny orchid]

For example, 100 -> 96 -> 92.16

[lone heart]

@jaunty crane Has your question been resolved?

[lone heart]

@jaunty crane Has your question been resolved?

[little drum]

ughh

[lone heart]

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[alpine sable]

A plane started a journey of 2500 miles at 2:00 pm and arrived at its destination at 7:30 pm.
Show that there are at least two times during the flight when the plane's speed
went 400 miles per hour

[alpine sable]

how do i solve this?

i am trying using lagrange in this

[mellow juniper]

I can help you with this

[alpine sable]

:0

[mellow juniper]

Though i dont know about lagrange, I would have to take some time to look into that first 😦

essentially we know that the flight took 5 hrs 30 minutes, or 5.5 hrs

and 2500 miles / 5.5 is 454.54 mph is the average velocity of the plane if it stayed at its avg velocity the whole time. So the plane for sure has to go faster than 400 miles per hour (thus 400 is not the max velocity and occurs twice)

[alpine sable]

but isnt asking to show 2 instants where it reached 400 miles?

[mellow juniper]

The above statement just proves beyond a doubt it has to reach 400 twice

[gray isle]

consider what a plane does irl when it travels between locations

[alpine sable]

yeah it makes sense, i can only get a instant velocity when they give me a function right?

[mellow juniper]

you could create a velocity function from the distance function via derivation but you'd need to know more than just (X, X0, and T) or else you'd simply be making an avg velocity function

[alpine sable]

exaclly so it gets a little confusing

[mellow juniper]

Do they give you more information like the acceleration at take off and landing?

[alpine sable]

nope

just that

[mellow juniper]

then they simple want you to use the concept of average velocity (change v / change t) to prove the plane must have a peak velocity faster than 400 mph ( and we know then that that value would have to occur twice when they sped up and slowed down)

[alpine sable]

some of this exercices are time consuming to understand and i cant use calculator so its messed up

[mellow juniper]

let me know if your having trouble with any others, I'm doing a mechanical engineering degree so this is right up my alley lmao

[alpine sable]

im doing the same

[mellow juniper]

Heck ye

[alpine sable]

then help me study for my math test!

;_;

[mellow juniper]

What course are you taking?

[alpine sable]

aircraft engineering

[mellow juniper]

dang haven't taken that yet 😦

[alpine sable]

can you jump on voice?

[mellow juniper]

not right now cause I'm in the library, but ill add you as a friend

[alpine sable]

okok

[stoic ridge]

2500/5,5=454
And you cant get an average of above 454 if you never get 454
And it has to be at least 2 because it is accelerating to above 400 so it hits 400 and it has to stop so it will hit 400 again

Oh the question was already answered

[alpine sable]

yeah, but thanks !

[lone heart]

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[alpine sable]

i have a problem

[alpine sable]

find the numbers a, b, g so that each of the following equations is true for every real number x

1. x²=a(x-1)²+b(x-1)+g
2. 19x²+11x+5=(a+2ab)+(b+3g)x-(2b+5g)x²

i cant find the numbers

[echo socket]

You can make the equations true for every number x by making the coefficients on LHS and RHS equal

For example in the first one we have x^2 = a(x^2 - 2x + 1) + b(x - 1) + g

Expanding the brackets yields x^2 = ax^2 - 2ax + 1 + bx - b + g

x^2 = ax^2 + x(b - 2a) + 1 - b + g

Alright

So here

Coefficient of x^2 on LHS is 1 and on RHS is a, right?

[alpine sable]

yes

[echo socket]

So they should be equal

a = 1

Because we need coefficients to be equal

Now let's do coefficients of x

On LHS there no term x by itself, so it's 0

And on RHS coefficient of x is b - 2a, which also is b - 2 as we just found out

So what would be the value of b?

[alpine sable]

2

[echo socket]

Yeah

Now can do you the same with the constant coefficient?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[arctic harbor]

I'm not sure if I did something wrong, the numbers don't make much sense

For context, here is the question

<@&286206848099549185>

[lone heart]

@arctic harbor Has your question been resolved?

[arctic harbor]

I'm asking if anyone can double check my work, because I don't think it's correct

[alpine sable]

there would be -18 not +18

[arctic harbor]

Oh, I see. That makes much more sense now

Thank you

[alpine sable]

np

[arctic harbor]

.close

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[vagrant falcon]

Hello! So this is a chemistry problem however this requires heavy mathematics which is why I came here. It is a half life problem. I have no idea (I did try to go on YouTube and find things but I couldn't) on how 400x(1/2)^20/5 became 25.

[vale wigeon]

$400 \cdot (1/2)^{20/5} = 400 \cdot (1/2)^4 = 400 \cdot \frac{1}{16}$...

[ocean seal]

Ann

[glass lichen]

heavy mathematics

[vagrant falcon]

Hm ok

damn i made a silly mistake thats why

thanks

idk what i was thinking and how I missed that

for some reason I did 20/5 = 4 * 1/2 to get 2

[vagrant falcon]

thank you very much

[vagrant falcon]

yea I was just tryna say that this more of a math problem than chemistry.

and i didn't want to get warned for doing a different subject

[lone heart]

@vagrant falcon Has your question been resolved?

[lone heart]

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[lost steeple]

I know colinear means to have the same direction

But for like [-a,-b] it's in the opposite direction

[junior vector]

Try to draw them all out

[lost steeple]

but is it colinear?

[junior vector]

Yes

Google says yes

Lol

If they're parallel

[lost steeple]

it's just if it's parallel?

hm

[junior vector]

Thats what Google says

[lost steeple]

ok thanks

.close

[lone heart]

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[unkempt umbra]

How can this be proved by induction?
I know that there are m^n functions since every time you pick an element from A you have m possibilites to assign it to an element from B

[unkempt umbra]

But how do I use induction for proving this?

[wanton junco]

so how many you'd have for n+1?

[unkempt umbra]

there should be m^(n+1)

[wanton junco]

looks like you induced it from your assumption that it's m^n

by using your induction hypothesis that it's m^n

does that make it clearer?

[unkempt umbra]

well when doing induction you should start from a base case and generally that would be 0
in my case m^0 = 1
that would mean that there is a function with domain the empty set

i don't know how this would be possible
so i think my base case should be n=1
right?

[wanton junco]

well the mapping from an empty set to another doesn't defy the definition of a function

after all how can you have a mapping of one element to 2 elements simultaneously , if you have no elements

I mean.. if I would prove it, i'd show it for both n=0 and n=1
since n=0 is sort of what is called, a trivial case

and then proceed

[pale kestrel]

f(empty) should be empty

if we're mapping subsets of the domain here

or did i misread

[wanton junco]

but isn't that still a function?

[pale kestrel]

im talking for a general function

as for a function with empty domain, no clue --- defn will say

[wanton junco]

I think we're mapping values from a given set tho

[pale kestrel]

https://math.stackexchange.com/questions/45625/why-is-an-empty-function-considered-a-function

it is.

[unkempt umbra]

well the definiton of a function is something along the lines of for every element from A, there is a corresponding element from B
and since A is the empty set this def is ok

because for every element in the empty set it is true

since there are none

[pale kestrel]

it can also be a set of ordered pairs

[wanton junco]

wait actually no not necessairly
it doesn't really matter either way

[pale kestrel]

Making {} a function

[unkempt umbra]

But how does
| Hom( {a1, a2, ... an}, {b1. b2 ... bm}) | = m^n TRUE
imply
| Hom( {a1, a2, ... an, a(n+1)}, {b1. b2 ... bm}) | = m^(n +1)
?

[pale kestrel]

ur adding a new element to domain each time

or codomain

[wanton junco]

how many new mappings do you have after adding 1 element?

[unkempt umbra]

in the domain? m times more

but how do I write this? like, mathematically

[lone heart]

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[supple tundra]

I've already solved the question. Was just wondering what an appropriate explanation for the boxed part would be?

I mentioned that "Since $a^2$ is a root and always non-negative, then the cubic in $u$ will always have a non-negative root for whatever value of b,c,p. Provided that the values of b,c,p obeys the first equation"

[ocean seal]

azeem321

[supple tundra]

It seems this wasn't the explanation they was looking for

[lone heart]

@supple tundra Has your question been resolved?

[lone heart]

@supple tundra Has your question been resolved?

[lone heart]

@supple tundra Has your question been resolved?

[lone heart]

@supple tundra Has your question been resolved?

[hasty elk]

i dunno, this looks like a perfectly valid explanation (as long as all coefficients are real)

[supple tundra]

.close

[lone heart]

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[magic terrace]

How do I reduce this further

[magic terrace]

$\frac{sin^2(x)}{cos(x)}$

[ocean seal]

Eyesonjune

[magic terrace]

I tried doing this

[ocean seal]

Eyesonjune

[magic terrace]

But I'm not sure how that would help

[wary stream]

Tan = sin/cos

[magic terrace]

So like uhh

sin(x) * tan(x)?

I need to reduce this into f(x) - cos(x) and find f(x)

[wary stream]

What's the original problem?

[wary stream]

Then this here is good, you can separate fractions

[magic terrace]

im an idiot how do i do that

[wary stream]

If you had 1/x - 2/x^2, the common denominator is x^2 so you have x/x^2 - 2/x^2 which then becomes (x - 2)/x^2

Do you think you can determine the reverse using your problem?

[magic terrace]

uhm

i am confused

like 1/x - 2/x^2?

Or 1/x-2 / x^2

[wary stream]

It'd be x/x^2 - 2/x^2

If you separated

[magic terrace]

so how does this relate to 1-cos^2(x)

/cos(x)

[wary stream]

You have a denominator, you can split it up

Same process with my example

Denominator is the same for the fractions, numerator is each term, and sign is in the middle of the fractions

[magic terrace]

yeah so you get

[ocean seal]

Eyesonjune

[wary stream]

Yes

[magic terrace]

which becomes

[ocean seal]

Eyesonjune

[wary stream]

And 1/cos is?

[magic terrace]

sin?

noo

sec

[wary stream]

Yes sec

[magic terrace]

is sec^2(x) the same as 1/cos^2(x)

[raw shard]

yes

[magic terrace]

now i have this

I have reduced it to

[ocean seal]

Eyesonjune

[raw shard]

,w tan^2(x)+1 = sec^2(x)

[magic terrace]

how the hell does one get cot(x) from that

[ocean seal]

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[magic terrace]

oh

another rule?

so I shouldn't have gone down the 1/cos^2(x) route

[raw shard]

no

you shouldn’t have

[magic terrace]

frik

so

cot(x)(1-tan^2(x))

[raw shard]

no

[magic terrace]

wait that would be if tan^2(x) + sec^2(x) = 1

not like the sin^2 cos^2 rule

so it's cot(x)(1+tan^2(x))

[lone heart]

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[calm willow]

Write an equation for the polynomial graphed below

[calm willow]

so far i tried
(x+1)(x-2)

i forgot how to do this can someone refresh me please

u can use an unrealted polynomial

[lone heart]

@calm willow Has your question been resolved?

[calm willow]

<@&286206848099549185>

i think its been 15 minutes

yeah its been 18 minutes

[charred flint]

the problem is that (x+1)(x-2) is a quadratic with x^2

but the graph isn't U-shaped, so it's probably cubic

if the graph was slightly lower it'd look like this right?

2 zeroes really close to each other on the right

when a graph bends backwards at the x-axis like that, it's actually multiple roots

[calm willow]

so what should i do

[charred flint]

it's just (x+1)(x-2)^2 instead

since it's more like 2 roots at x=2

[calm willow]

i dont understand sorry

like the 2 roots part

i was doing this in class i my head and it i got it every time

[charred flint]

repeated roots are hard yea

[calm willow]

then what about this graph

(x+3)(x+2)(x-3)

[charred flint]

it goes back and forth twice, so it's probably a quadratic

but you only see three zeroes, so one of them is actually multiple

[calm willow]

ok how can i tell which one

[charred flint]

whether it bounces back

[calm willow]

oH

[charred flint]

like how the graph y=x^2 looks

[calm willow]

I SEE

ok

how do i get the a of the equation

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

[charred flint]

right

[calm willow]

?

thats not it is it

[charred flint]

oops I messed up there

it's facing downwards in the graph

so minus sign in front

[calm willow]

oh tru

[charred flint]

,w graph -(x+3)(x+2)(x-3)^2

[ocean seal]

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Requested by Plurmorant#1811

[calm willow]

but u still need to get the top offset

see

[charred flint]

top offset?

[calm willow]

like how it goes through -2

rather -60

on the y axis

[charred flint]

oh jeez

right that's the coefficient in front

[calm willow]

yeah i mean coefficitent sorry

[charred flint]

-(x+3)(x-2)(x-3)^2 at x=0

that's -3*2*3*3=-54 ok

[calm willow]

ok

[charred flint]

so you want to go from -54 and shrink to -2

so 1/27 in front

[calm willow]

so -2/54

then u get that

alright thanks

heere lemme do one with u just to make sure i got it

2/3(x+2)(x+1)(x-1)(x-3)

[charred flint]

right

[calm willow]

thank u so much

.close

[lone heart]

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[calm willow]

.reopen

[lone heart]

✅

[calm willow]

ok last one this time sorry @charred flint

oh nvm im just dum

oh no wait its still wrong

if its squared doesnt it become a positive four

so its 4 * 3 * -1 * 4

[charred flint]

yeah just off by a minus sign

that's tricky

[calm willow]

oh wait

u still have to add a minus sign if its going down @charred flint

[charred flint]

oh nvm

[calm willow]

i have a minus sign i put an = sign

i fixed it and its still wrong

[charred flint]

it's 4^2 at the end

so coefficient needs to be different

[calm willow]

yeah

[charred flint]

-1/96...

[calm willow]

i want to cry

how is it 96

[charred flint]

4 times 3 times 4 times 4

this hw is a big oof

oh that's 192

-1/192 aaaa

wait

[calm willow]

its 96

[charred flint]

yeah

[calm willow]

[charred flint]

aaaaa

192 to 2 means it's 1/96

ouch

[calm willow]

i got a new one

god these are so cursed

@charred flint

its to the degree of 5

[charred flint]

oh this is even harder wow

[calm willow]

does that mean -1/27(x-2)^2(x-3)^3

[charred flint]

it just keeps going

yeah then you need the coefficient

it's ^3 because it curves to being flat

[calm willow]

yeah i just got the multiplication part confused i think i got it

-1/27(x-2)^2(x-3)^3

[charred flint]

that looks good

[calm willow]

it was wrong but i got it

i just realized i need help on a different type of question

and its this one

im sorry lol

this is how i feel

[charred flint]

🌈 use a calculator 🌈

[calm willow]

(x-1)^2(x0)(x+1)

this is the question i sent above

right

what do i do with the point

<@&286206848099549185>

[glass lichen]

$P(x)=ax(x-1)^2(x+1)$

then solve for a.

[ocean seal]

Mosh

[calm willow]

4410 =a-6(-6-1)^2(-6+1)

like this @glass lichen

[glass lichen]

a(-6), but yes

[calm willow]

ok i got 2940

would should i do from there

or is that it

[lone heart]

@calm willow Has your question been resolved?

[lone heart]

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[buoyant kayak]

what have you attempted?

[lone heart]

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[loud oriole]

how would i evaluate tan(-2pi/5) without a calculator

[loud oriole]

.close

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[native elbow]

hi i need help with these problems-

[alpine sable]

The first one :
if all the coins of are same value(when true)
if the coins are of different values , like 10 cents , 50 cents.(when false)

The second one :
if A = B (true)
A != B (when false)

[lone heart]

@native elbow Has your question been resolved?

[alpine sable]

let the fraction be
n / m
if (n > m) and you add the same value to both numerator and denominator the value decreases

[native elbow]

may i know what the exclamation mark means?

[alpine sable]

i didnt know how to write not equal to

$\neq$

[ocean seal]

Linus

[native elbow]

ohh okay okay

[native elbow]

is this when the statement is not true?

[alpine sable]

hm

[native elbow]

alright thanks !

.close

[lone heart]

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[strong wyvern]

find the smallest value for the range of the function g(x) = 1 - 5x and the domain is {x | -3 <= x <= 2}.

[bright hedge]

what have you tried?

[strong wyvern]

I've tried using the value (-3) lowest x value but I'm super confused

[bright hedge]

do you agree that g(x) is decreasing?

that is, as x increases, g(x) decreases?

,w graph 1 - 5x

[ocean seal]

Results provided by WolframAlpha

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Requested by Yottachad#6969

[bright hedge]

this is the geometric intuition

as we can see, as we move right, the y value moves down

[strong wyvern]

so I would pick the highest x value ?

and plug it in

[bright hedge]

yes but do you understand why?

[strong wyvern]

yeah I get it

[bright hedge]

alright

use .close to close the channel

[strong wyvern]

alright thank you

yeah it was 9

.close

[lone heart]

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[loud oriole]

when i take out the 2 it becomes 2(sin(arccos...))(cos(arcos...))

why does taking out the 2 create the cos at the end

oh nvm its an identity

[alpine sable]

sin(2x) = 2 sin(x) cos(x)

[loud oriole]

ye thx xd

.close

[lone heart]

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[ornate gate]

idk how to figure this out

[noble silo]

You can use sine to find x

[ornate gate]

soh?

[noble silo]

Yes

Sin(62)=x/100

[ornate gate]

u mean 52?

[noble silo]

Yeah sorry

[ornate gate]

all good

73.9?

@noble silo

[noble silo]

I got 88.3

Wait

78.8

[ornate gate]

https://cdn.discordapp.com/attachments/934709637142442065/935067618778447902/IMG_6086.png

is that right?

[noble silo]

52 right?

[ornate gate]

oh oops

yeah

98.6

[ornate gate]

so 79 ft?

[noble silo]

I got that with my actual calculator twice. Don’t know why you aren’t getting that

[ornate gate]

https://cdn.discordapp.com/attachments/934709637142442065/935068668054888478/IMG_6087.png

is this right?

[noble silo]

What calculator are you using?

[ornate gate]

photomath

[junior vector]

Is it in degrees or radians

[ornate gate]

ohh

idk how to figure out what iti s

[junior vector]

Lol

[noble silo]

I think Photomath is assuming in radians

[junior vector]

Whats sin of 90

if its 1 then its degrees

[ornate gate]

0.8

[junior vector]

They're you go

What is my phones autocorrect

[ornate gate]

should i just use normal calc

[junior vector]

Sin of (90 degrees) is 1

Sin of pi/2 radians is 1

Yeah prolly if you don't wanna use radians

[noble silo]

Desmos scientific calculator is what I use. It’s great. There are other calculators out there too. You also might want to consider using a physical calculator to get used to tests and stuff.

[ornate gate]

in my physical calc

it says 78.80

@noble silo @junior vector

[noble silo]

Nice

[ornate gate]

is that right?

@noble silo

[noble silo]

Yes

[ornate gate]

Thanks

.close

[lone heart]

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@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[little drum]

WHAAAAAAAAAAAAAAAAAA

We did this in the morning

Just do the same thing

Evaluate x'(0) and y'(0) in both questions and just equate to the value given, and x(0), y(0)

[lone heart]

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[warm nexus]

What would be a sensible way to go about solving this?

[warm nexus]

Or rather maybe this style rather because obviously you can solve this by eye

like this

[little drum]

so basically you're asked to evaluate $10^{100} \pmod 3$

[ocean seal]

Ansh

[little drum]

can't you do that though?

[little drum]

or here, for example, $10^{10} \pmod 6$

[ocean seal]

Ansh

[warm nexus]

i mean a more methodical way for more complicated ones because i saw someone do it algebraically but i didn't really understand it

I need to give a harder example, both of these are nice

[little drum]

I don't know how hard you're talking but here's one if you want to try

Let the sum of digits of a number n be denoted as $S(n) \$
Find $S(S(S(4444^{4444})))$

[ocean seal]

Ansh

[warm nexus]

I think this is the one that was more difficult

[little drum]

It's still as simple tbh

[warm nexus]

how tho

[little drum]

$10 \equiv 3 \pmod 7$

[ocean seal]

Ansh

[little drum]

$10^2 \equiv 2 \pmod 7$

[ocean seal]

Ansh

[little drum]

$10^6 \equiv 2^3 \equiv 1 \pmod 7$

[ocean seal]

Ansh

[little drum]

so $10^{10^{10}} \equiv 10^{10^{10} \pmod 6} \pmod 7$

[ocean seal]

Ansh

[warm nexus]

How are you going from the first line to the second, and from second to third like that

wait lmao

im dumpy rart

actually i would appreciate it if u could still explain pls

[little drum]

Hmm >_<

SO uh

You understand when I write $10 \equiv 3 \pmod 7$ I mean, 10, when divided by 7, leaves a remainder 3, right?

[ocean seal]

Ansh

[warm nexus]

yes

and i understand why its 10^2 = 2 mod 7

and also part of the last line

But from the last line, can you just cube both terms like that

like go from
10^2 = 2 (mod 7)
to
10^6 = 2^3 (mod 7)

[little drum]

Yeah ofc

okay see here

(10^6) = (10^2)^3 right?

[warm nexus]

Yes

[little drum]

= (98 + 2)^3 ?

= (98^3 + 3(98)(2)(98^2 + 2^2) + 2^3)

leaves remainder 2^3 when divided by 98

:o

[warm nexus]

yooo

ive never had this explained to me

thanks!

.close

[lone heart]

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[limber shard]

Can someone please explain to me how to even start this question?

[lone heart]

@limber shard Has your question been resolved?

[limber shard]

<@&286206848099549185>

[lone heart]

@limber shard Has your question been resolved?

[lone heart]

@limber shard Has your question been resolved?

[limber shard]

.close

[lone heart]

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[fleet hearth]

I don't know what this question wants me to do

[fleet hearth]

I understand that the first half is basically defining k

but how do I show the top and bottom are the same

[severe sluice]

hint: you subtract k in the integral (you know what that means on a graph, right?)

[fleet hearth]

Something like this?

[severe sluice]

,rotate

[fleet hearth]

Or Am i overcomplicating it

[lone heart]

@fleet hearth Has your question been resolved?

[fleet hearth]

<@&286206848099549185>

[lone heart]

@fleet hearth Has your question been resolved?

[alpine sable]

By definition

[fleet hearth]

How do I show that the top and bottom regions will have the same area

[alpine sable]

So basically the average value of f(x), k, is on the interval a<=x<=b

[fleet hearth]

Yes

[alpine sable]

So the average value lies between a and b

So make a line parallel to x axis, y=k

Now see

[fleet hearth]

I understand this far

[alpine sable]

Diagram should look something like this

[fleet hearth]

Yep, that's roughly what I envisoned

[alpine sable]

Hmm

[fleet hearth]

This much is intuitive but I don't know how to "show" it

[alpine sable]

That's what I'm also thinking about too of how to go over it

Please give me a minute to analyze this

[fleet hearth]

Is there anyway to simplify this maybe?

[alpine sable]

Well you can say the integral to be equal to F(x) and say
$$F(x).(1-\frac{1}{b-a})$$

[ocean seal]

Pencil

[fleet hearth]

Hmm, but that doesn't take me anywhere useful

[alpine sable]

I have an idea

See this

Let the intersection of the line y=x and the perpendicular drawn from (b, 0) be c

Now we will deal with two graphs here

Also let the distance from b to c be (c-b)

[fleet hearth]

Yeah I'm following

[alpine sable]

For the first graph h(x), we can vertically shift it down to x-axis by subtracting (c-b)

And then we can find the area from k to c of this

For the second graph

[fleet hearth]

Mhm

[alpine sable]

Note we're still maintaining the continuity of both the functions, so whatever we're doing till now is acceptable

Wait

There

With this info we can proceed

[fleet hearth]

Yes, it looks like changing the boundaries was the step

[alpine sable]

Yes

[fleet hearth]

And then you can add/subtract k(b-d)

[alpine sable]

Mhm

And this will be true

Oops forgot the dx there at the end

Okay so we need to prove both areas are equal

[alpine sable]

I think this statement would be necessary to say we're just manipulating the area under the curve and maintaining the continuity of the function f(x)

[fleet hearth]

Hmm

[alpine sable]

So we need to prove this

[fleet hearth]

Yes

[alpine sable]

Alright so c and b are both y values

Since it's downward shift

So b will have y value 0

Since it lies on the x axis

[fleet hearth]

Yep

[fleet hearth]

Is this useful at all? Or nah

I think this is for a few steps before

[alpine sable]

Well

Wait let's try solving that

[fleet hearth]

This definition of the regions looks easier to work with

[alpine sable]

Yes

Mine looks a bit complicated cuz of the shift i did

[fleet hearth]

So what do we do next

[alpine sable]

Our k interval is a<=x<=b

So we can say

I just replace a with a weird symbol lol

Alright so by wording

k is the average of the continuous function x

[fleet hearth]

That makes sense but what happened to the (b-x)

[alpine sable]

Sorry

Alright so in the LHS, can we say the integral is equal to the sum of the integrals F2(x) + F1(x)?

We surely can right

[fleet hearth]

Hm

[alpine sable]

We can split b-a as (x-a) + (b-x)

[fleet hearth]

Yes

[alpine sable]

So

How can we simplify further now uuuh

[fleet hearth]

This is a really tough question

[alpine sable]

Yeah

[alpine sable]

Can we find something here to ease our working

Wait let's do this part again

[alpine sable]

This

[fleet hearth]

Alright

[alpine sable]

Say the integral in the LHS to be F(x)

[fleet hearth]

Mhm

[fleet hearth]

Alright, we've divided both sides

[alpine sable]

Yes

We substituted in place of 1/b-a the answer we got

Wait

[fleet hearth]

I don't think that achieved much

[alpine sable]

k value

Take square root on both sides

[fleet hearth]

Yep

[alpine sable]

We're interested only in positive values

[fleet hearth]

Yes

[alpine sable]

When F(x) = k(b-a)

Okay i think I'm overcomplicating it

[fleet hearth]

Yes

I think we're close tho

[alpine sable]

I just substituted our value of F(x) in this

[fleet hearth]

Does that accomplish something?

[alpine sable]

I'm looking onto it

[fleet hearth]

It's currently 2:30am for me so my mind is working very slowly now

[alpine sable]

Oh.... Sorry for taking ur time...

[fleet hearth]

It's fine, it's better than no one showing up

Then I'd have gone to sleep without getting any closer

[alpine sable]

Logically we can say F2(x) > F1(x)

Resulting a positive area

When that area is halved

And divided by the length (b-x)

We get the rhs

Hmmm

Oh we made a lot of progress

Let's connect the dots

I think we can accomplish this

So basically

So that means we need to find f2x and f1x

[fleet hearth]

Yep, we are trying to prove they are equal

[alpine sable]

So we need 2k(b-a)(b-x) to be 0

[fleet hearth]

They must be non-zero though

[alpine sable]

Like when they result 0, F2(x) = F1(x)

The area of the two for every known continuous function isn't always equal right. Sometimes they can be unequal

We can reject b=a case

since b≠a

b=x is acceptable

[alpine sable]

Wait let me check

[fleet hearth]

Wasn't the Defintion of x that it is a Point between a and b

[alpine sable]

Yeah

Wait

[fleet hearth]

I just realised this but if our example graph had just been a square then x would be unneeded

In that case, x = b

[alpine sable]

So that would be acceptable

[fleet hearth]

So I guess it works

[alpine sable]

Yeah it will

So we proved that by our example graph

If k=0 or b=x, or both, then F2(x) = F1(x)

Because obviously, it will not always be true for the two areas of every continuous function to be equal

Like if f(x) = sqrt(x)

[fleet hearth]

But this makes me curious of if the calculation becomes a lot more simple had we used a rectangle as our example regions

[alpine sable]

Hmm

So you're saying a graph like this?

[fleet hearth]

Yes

[alpine sable]

But area above the line doesn't exist

So it would be f2(x) = 0 + f1(x) since b=x

The does not makes sense

So b=x is also incorrect 🤔

Because it would mean 0 = f1(x)

[alpine sable]

So if k=0, then only f1(x) = f2(x)

[fleet hearth]

Maybe it doesn't apply to a rectangle because it isn't continous

Since it has vertical lines

[alpine sable]

Let me check pls give me a min... Sorry for consuming your time....

[fleet hearth]

It's a one to many relation at the ends

So not a function

[alpine sable]

But a linear function is continuous

Because it has no sort of discontinuity

Like if u take the limit from right or left it stays the same

[fleet hearth]

Yeah it's continous

I'm just going to read this again

[alpine sable]

If the f(x) is equal to the line y=k, then b=x, which implies that f2(x) doesn't exists

[fleet hearth]

Rectangles are not functions tho, their relations, I think that's why it doesn't work for them

[alpine sable]

I believe triangles would work

If k=0

Like k has to be 0 for the relation to make sense

[fleet hearth]

K has an asymptote at 0

[alpine sable]

For b=x, it only appears if f2(x) doesn't exist

[fleet hearth]

It can't be 0 unless f(x) = 0 I guess

So k can = 0

Nvm

[alpine sable]

But how is k asymptote?

[fleet hearth]

It isn't

It's late

[alpine sable]

I think this condition should work, if k =0, f2(x) = f1(x)

[alpine sable]

Sorry for taking up ur time... If u want we can end it here... I'll go through whatever we did once again

[fleet hearth]

If your available tommorow then sure

[alpine sable]

👍

Gn o/

[fleet hearth]

Night

[lone heart]

@fleet hearth Has your question been resolved?

[alpine sable]

I just realized that i stated F2(x) = F1(x), but did not prove A1 = A2

[lone heart]

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[crude birch]

Here, why the graph of function is not different when we divide the function by 3?

[summer sail]

they are..

[crude birch]

Yes, but here the value of function is zero.

[summer sail]

yes, as you can see these two functions share the same zero of the function

[crude birch]

Yes

Thank you 👍

.close

[lone heart]

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[proud basin]

for a, b > 0 and a + 2b < 3 find the minimum of P

[little drum]

Classic QM-HM

[proud basin]

eu

[little drum]

eu?

[proud basin]

wait

im not really familiar with the protocols in QM HM

thats a^2 + b^2/2 > 2ab/a+b

[little drum]

More like

[proud basin]

oke

idk

[alpine sable]

idk

[proud basin]

linustechtips..

omgoodness

t^t

[proud basin]

oke what do we do from there

[little drum]

hmm?

[ocean seal]

User cancelled member selection.

Jan 24, 2022, 6:38 PM

[little drum]

sry lmao

[little drum]

You want the solution? Don't want to think at all?

[pale kestrel]

btw is partial derivative no good?

as well as check boundaries

[little drum]

XD

[proud basin]

idk

ive never really explored the rest

[little drum]

Just put x1 = √(a+3), x2 = √(b+3), x3 = √(b+3)

in RMS ≥ HM

[proud basin]

woah

[little drum]

And you'll have constant ≥ 3/P

[proud basin]

wait

lemme take notes

hEy

[little drum]

Hi ?

[proud basin]

🥲

im dumb

not blind

thx for your help anyways

.close

[lone heart]

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[ripe plume]

Can you guys help with this

[tacit arch]

what have you tried

[ripe plume]

I don’t understand this at all

And i need to get it done asap

<@&286206848099549185>

hello?

[noble sinew]

Check if its idempotent?

Look at (I-A)^2

[ripe plume]

I dont really understand it

[noble sinew]

Understand what?

[ripe plume]

Idempotent

Idk what that means in my language

[noble sinew]

It means a matrix, A, that fulfills A^2=A

[ripe plume]

Can you please answer this questions fully on paper thanks a lot in advance

[noble sinew]

No, you can try for yourself

[noble sinew]

Do this

[ionic trail]

Omg what

[ionic trail]

It's literally the first line

[lone heart]

@ripe plume Has your question been resolved?

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[frigid frigate]

i need help

[frigid frigate]

<@&286206848099549185>

[alpine sable]

length of the longer side of the parallelogram is 12-2 = 10

and height is 4

so 40?

[frigid frigate]

alr

thanks

.close

[lone heart]

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[hard mountain]

Hey, have been stuck on this problem for like 4 hours. Any help would be greatly appreciated 🙏

[noble sinew]

Stuck where?

Induction works for both things for example

[hard mountain]

Noticed that x_(j+1) < x_j when x_j > sqrt(2). But dont know how to continue from there

[noble sinew]

So you have showed decreasing or what?

[hard mountain]

tried showing that x_(j+1) > sqrt(2) if x_j > sqrt(2) by induction but got stuck in the induction step

ill post pic of work

[noble sinew]

I mean everything cancels in induction step?

You want to show x_{j+1}>sqrt(2)

Write def of x_{j+1} given

Use x_j>sqrt(2)

And you get x_{j+1}>sqrt(2)

[hard mountain]

its ok to just swap out x_j with sqrt(2) in the def?

[noble sinew]

We are doing induction

We prove base case

Suppose true for j=k

So we suppose x_j>sqrt(2)

So you get an inequality using that

[hard mountain]

ok ill try

[lone heart]

@hard mountain Has your question been resolved?

[hard mountain]

did you do it some other way?

[hard mountain]

sorry have to run. Thank you ScapeProf! Ill check this thread when i get home

.close

[lone heart]

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[novel urchin]

Can anyone explain to me why the before x after is either 1 below or above the Fibonacci square

[novel urchin]

If possible can u do it in

Term to term relationships

[solemn grove]

context

[novel urchin]

Maths project

Investigate relationship between

Lat two columns