help-39

[midnight haven]

this one

i have like two more questions

the rest i think i can do alone

[graceful tendon]

so

we're working backwards this time

so ur formula is Fg = mg

Fg is force of gravity

m is mass

[midnight haven]

what is g

[graceful tendon]

gravity is like

9.81

but obv ur gravity value isnt gonna be the same on earth and the moon

wait maybe it is

hang on

[midnight haven]

ok

[graceful tendon]

hang on one sec

so

on earth

your equation is

Fg = mg

981 = m * 9.81

100 = m

im kinda confused though because

idk why they gave you his weight on the moon

[midnight haven]

me too man

[graceful tendon]

does 100 work

i guess its a red herring

[midnight haven]

let me try

yez it worked

the question before this one i got wrong

i hope i dont have to do all over again

soo...

after this one the other i already did so 👍

hello?

@graceful tendon

[graceful tendon]

oo

ok

so

remember our formula

Fg = mg

the m is 150 kg

the g is 9.81

so multiply them

[midnight haven]

150 x 9.81?

[graceful tendon]

yes

Fg = m * g

so

Fg = 150 (kg) * 9.81 (m/s^2)

so

,calc 150*9.81

[jolly parrot]

Result:

1471.5

[graceful tendon]

theres your answer

[midnight haven]

ty

[graceful tendon]

i gotchu

do u get it now

did ur teacher never give u the kinematics equations?

[midnight haven]

i think so

[graceful tendon]

Fg = mg
Ff = μFn
ΣF = ma

i cant rember the fourth

[midnight haven]

can i do this one and you help me

[graceful tendon]

alright

[midnight haven]

fg = 3900 (kg) * 9.81 (m/s^2)??

,calc 3900*9.81

[jolly parrot]

Result:

38259

[midnight haven]

[graceful tendon]

yessir

[midnight haven]

really???

cause i got this one wrong once too lmao

i mean is the one i just got wrong

yess sir is right

ty

☺️

we just did this one but now is the moon not the earth

i will try this one alone just help if anything goes wrong pls

[graceful tendon]

alright

[midnight haven]

nah i need help

🥹

[graceful tendon]

its a trick question i thin

k

mass doesnt change if you change force of gravity

youre just changing the weight

mass and weight arent the same thing right

because weight just measures how hard the planet is pulling on the mass

so its mass on earth and moon are the same

therefore you should be able to enter in 100 like you did last time for earth?

i could be wrong

[midnight haven]

i just got wrong

before

but lessgo

[graceful tendon]

did it work

[midnight haven]

YES

wow

we done

ty

[graceful tendon]

WHOO

[midnight haven]

🤝

[graceful tendon]

🤝

[midnight haven]

appreciate your job here man

😂 🙏 you helped me so much tyy

.close

[pearl pond]

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[pearl pond]

@wet scroll Has your question been resolved?

[graceful tendon]

@wet scroll here i gotchu

I'll draw it out

So

Here's your lil trees

You've got 50 and an 80

You've got x and a 100

Let's set up the ratio

$50/80 = x/100$

[jolly parrot]

rollo

[graceful tendon]

I think you can solve from here right

[wet scroll]

62.5

[graceful tendon]

Alright

[wet scroll]

there’s more though

[graceful tendon]

So what times 50 is 62.5

[wet scroll]

1

[graceful tendon]

What

[fleet sky]

you heard him

it's 1

[graceful tendon]

oh

[fleet sky]

😛

[graceful tendon]

my bad

[wet scroll]

this is a no calc test

[graceful tendon]

you don't need

a calculator

to divide

62.5

by 50

[wet scroll]

it’s 1.25

[graceful tendon]

It's prolly 50/62.5 then oops

wait

No it's not I'm actually dumb don't listen to me

Ok so

wait am I stupid

[graceful tendon]

you take over now

good luck

[fleet sky]

so it's asking for the ratio between the difference between the two trees and the first tree (the first tree has height 50)

so I think it litereally means the difference

like subtracting

so $62.5 - 50 = 12.5$. So then we have $\frac{12.5}{50} = \frac{125}{500} = \frac{1}{4}$

[jolly parrot]

MellowDramaLlama

[fleet sky]

I think

so I think it's (A)

[pearl pond]

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[midnight haven]

.

Okay so Im watching a video and they are telling me to multiply the fraction by 1/x^n, where n is the highest exponent in the equation. (Like in the first picture). But in the second image, you can see that they only multiply by 1/x, even though there is 16x^2. I know that 16x^2 is in a radical, but when should I not use the highest power in the equation for n? https://www.youtube.com/watch?v=sjLFl7Z8W_I&ab_channel=TheOrganicChemistryTutor

[midnight haven]

do I use 1/x^1 here because the 16x^2 is in a radical? or do I use the highest exponent thats not in a radical

[honest bough]

You need to consider the radical as well

[midnight haven]

yes i am

I just dont know when you consider it

[honest bough]

(x²)^(1/2) = x¹

[ivory swallow]

Sqrt(16x^2 -8) is very loosely speaking, the same "order" or "degree" as x

[honest bough]

You consider all exponents, including fractional ones

[ivory swallow]

You can make that notion rigorous through asymptotics or some shit

[midnight haven]

so should I mentally make it so that sqrt(16x^2) is to the power of one?

then use that

[honest bough]

sqrt(x²) is x¹, yes

[midnight haven]

so in that case i would use 1/x

so in this, what would i use?

$sqrt(9x^9 + 2)/2x+1$

[jolly parrot]

4 Caplan

[midnight haven]

would it be 1/x^3

[honest bough]

√(x⁹) ≠ x³

[midnight haven]

what

[honest bough]

√(a^b) ≠ a^√b

Otherwise, √(x²) would be x^√2

Anyways, I typically prefer dividing by the highest power of the denominator. I know others say highest power in general, but I haven't seen a case where dividing by the highest power in just the denominator leads to any error

[midnight haven]

okay

what if there is a sqrt in the denominator

what power do i use

[honest bough]

√(a^b) = (a^b)^(1/2)

Then use exponent rules

[midnight haven]

okay thank you

.close

[pearl pond]

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[midnight haven]

How would I write this statement using exculsively quantifiers and mathematical notation?

[midnight haven]

no words

.close

[pearl pond]

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[pearl pond]

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[warm nebula]

A relation can be modelled using a quadratic relation when the first differences are constant. Is this true or false

[pearl pond]

Please don't occupy multiple help channels.

@warm nebula Has your question been resolved?

[pearl pond]

@warm nebula Has your question been resolved?

[pearl pond]

@warm nebula Has your question been resolved?

[tall tangle]

wdym? When you first difference the dataset, it's stationary?

quadrative fitting has no requirement on the stationarity of the dataset

[warm nebula]

it is a question on my assignment

.close

[pearl pond]

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[finite crown]

https://gyazo.com/ae7af39d8f30b3f833ad63847885ae24

[finite crown]

anyone any ideas idk what to do

[pearl pond]

@finite crown Has your question been resolved?

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@finite crown Has your question been resolved?

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[pearl pond]

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[stoic ether]

Describe as simply as possible the set of all real numbers x, for
which applies (the solution must be specified):

[stoic ether]

(e) (x − 4)(x + 5)(x − 3) > 0

dont really know how to solve it

can i divide it into (x − 4) > 0, (x + 5) > 0 and (x − 3) > 0?

than it would be x > 4 , x > -5 and x > 3

<@&286206848099549185>

[pearl pond]

@stoic ether Has your question been resolved?

[stoic ether]

pls help

[earnest ivy]

When is the product of 3 numbers positive?

[stoic ether]

-5 < x < 3?

[earnest ivy]

Nope, it's when all are positive, or only two of them are negative

[stoic ether]

i dont understand

[fast zealot]

(-a)(-b)(c) = abc

(a)(b)(c) = abc

(-a)(b)(c) = -abc

[stoic ether]

ah ok

still dont know what to do

cant really calculate that (x − 4)(x + 5)(x − 3) > 0

x > 4?

or -5 < x < 4?

[pearl pond]

@stoic ether Has your question been resolved?

[pearl pond]

@stoic ether Has your question been resolved?

[earnest ivy]

The graph might help

,w graph (x-4)(x+5)(x-3)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by ralvrz#3477

[pearl pond]

@stoic ether Has your question been resolved?

[pearl pond]

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[wheat blaze]

Need help checking over if i have added enough math terminology or not

this one too

[pearl pond]

@wheat blaze Has your question been resolved?

[pearl pond]

@wheat blaze Has your question been resolved?

[pearl pond]

@wheat blaze Has your question been resolved?

[pearl pond]

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[grim patrol]

im backkkkkkkkkkkk

what fn has a left but not right inverse

or vice versal

[timid spindle]

f : Z --> Z, f(x)=2x

[grim patrol]

sry i just realised thrs this too

[timid spindle]

Yeah

[grim patrol]

ah

i think i get it w ur eg

so for ur eg there exist a left inverse

being

[timid spindle]

Or yeah just pick any function that's injective but not surj

[grim patrol]

g: Z->Z, 2x in Z, f(2x) = x

[grim patrol]

so then this shld be surjective

and only have a right inverse right

[grim patrol]

wait no this is actually bijective?

i think g is bijective?

[timid spindle]

You've only defined g on even Z

[grim patrol]

hm

how does that conflict with injective or surjective

surjective is all codomain is mapped to

injective is for each input we get some unique output

but if its undefined for odd we just ignore it right?

or do we not idk

um

thanks

ill just reask the right inverse part or injective of g

i guess

.close

[pearl pond]

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[oak agate]

please can anyone help me solve this limit without l'hospital, just using generic rules

[oak agate]

<@&286206848099549185>

[honest bough]

!15m

[pearl pond]

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

[oak agate]

yeah it's because i had asked the same question in another channel, waited for more than that, nobody answered so i closed but now im doing it again

happy now?

[honest bough]

For all intents and purposes, this is a new channel

[oak agate]

wtf does intents and purposes mean

anyways would you be okay with answering my question?

cause i've been kinda stuck with this limit

💀

[honest bough]

I'm not okay with answering it

Mostly because I don't know how

¯\_(ツ)_/¯

[oak agate]

oh yeah dw me neither 😭

honestly its just easy with lhopital

but they aint letting us use it

[pearl pond]

@oak agate Has your question been resolved?

[oak agate]

not yet

😦

<@&286206848099549185>

[narrow briar]

Hmm. If you're allowed to be very hand-wavy, you could use a small-angle approximation.

[oak agate]

hey! how would that work?

[narrow briar]

tan(x) ≈ x for small x.

[oak agate]

for arctazn

would that be arctan(x) = x

[narrow briar]

Yeah. For any sufficiently small range (-eps, eps), arctan(tan(x)) = x, therefore x ≈ arctan(x).

[oak agate]

okay, so should i subtitute arctan x by x?*

that still didnt work for me

[narrow briar]

arctan(x)/x^2 - 1/x ≈ x/x^2 - 1/x at small x.

But this isn't rigorous at all. Perhaps it could be turned into an epsilon-delta proof, if you have some bound on |arctan(x) - x|, but that seems... annoying.

[oak agate]

why isnt it rigorous though

like x goes real close to 0

forever

so its indeed very small to the point that tanx=x becomes a lot more accurate

idk

[narrow briar]

It's not rigorous because you're (well, I'm) asserting that the lim x→0 arctan(x)/x^2 = lim x→0 x/x^2, without proving it to be true, just saying "it's really close". :P

Oh! Wikipedia reminded me that the squeeze theorem exists, and that can be used quite nicely to show lim x→0 arctan(x)/x^2 = 0.

[pearl pond]

@oak agate Has your question been resolved?

[pearl pond]

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[simple wren]

About z = -4

How is alpha pi

[buoyant panther]

-4 means Re(z) = -4 and Im(z) = 0

it's like you would have point (-4,0) in Cartesian plane

think about an angle

[simple wren]

Could you explain it on highschool level

Or write the steps that they skipped in the book

They skip a lot of little steps that i need to understand it

[buoyant panther]

= pi

[simple wren]

Oh the 0 = 180°

Would you mind explaining why 2pi = 360° or give me a link i remember i learned it once but i forgot

[buoyant panther]

u know what is measure of an angle in radians?

[pearl pond]

@simple wren Has your question been resolved?

[simple wren]

Kinda

.close

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[midnight haven]

Say I have the world coordinates of a player's head: https://i.imgur.com/2ruNjjn.png

If the player were to do a 180 degree turn, those coordinates would obviously change.

How would I calculate what the coordinates of the player's head would if be they had done a 180 degree turn, as shown here: https://i.imgur.com/60Y27Sl.png?

[little quartz]

broski pulled up with csgo ct player model

crazy

[plush bramble]

https://stackoverflow.com/questions/54369205/rotating-both-in-local-and-world-coordinate-system

[pearl pond]

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[summer current]

The sum of two consecutive integers is less than 55. Find the pair of integers with the greatest sum.

[summer current]

ive done the questoin, just want to double check my work

x+x+1 < 55

2x+1<55

2x<54

x<27

26<27

26, 26+1

(26,27)

[shrewd basin]

Yes

[summer current]

.close

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[summer current]

The sum of three consecutive odd integers is no less than 51 . Find the middle integer.

[jagged stump]

Hmm

Tke them as x, x+2, x+4

[summer current]

i did the equation then realized half way through it said three conseqcutive odd numbers

[jagged stump]

Then solve

[summer current]

not just consecutive

so x itself is odd

[jagged stump]

Assume so

[summer current]

an odd plus an even is always odd

[waxen crag]

so

if an odd number is x, then the next consecutive one is x+2

then the next consecutive is x+4

[jagged stump]

No. All given integers are odd

[waxen crag]

so [x] + [x+2] + [x+4] = 51

[timid spindle]

"No less than 51" are you sure?

[waxen crag]

or

[summer current]

inequality

[waxen crag]

oh

[summer current]

no less than 51

[jagged stump]

X=13

[timid spindle]

Can give you infinitely many consecutive odd integers who's sum is no less than 51

[jagged stump]

That should be the answer

Most appropriate one

[waxen crag]

yeah i believe by saying no less than 51, it actually is asking you to find these 3 numbers that are closest to 51 while being above it

which there exists 3 consecutive odd integers that are equal to 51

above or equal to it*

[summer current]

im guessing its the greatest sum of 3

[waxen crag]

it would be 15,17,19

[timid spindle]

There is no greatest sum

[jagged stump]

Less than

[timid spindle]

Where is the question from, it needs clarifying

[summer current]

let me do my work and ill check

[waxen crag]

it's saying "no less than 51"

[jagged stump]

Less than 51

[summer current]

can be 51

[jagged stump]

So answer is 13 15 17

[waxen crag]

no less than 51 means it can be 51 or greater

[summer current]

i got x > 15

after simplifying

[jagged stump]

Yes

The closest odd number less than 15 is 13

[summer current]

wait

so is 15 one of the numbers

cuz it can be 15

[jagged stump]

Doesn't it say less than 51?

[summer current]

no less*

[timid spindle]

Send a picture of the question

[summer current]

The sum of three consecutive odd integers is no less than 51 . Find the middle integer.

[timid spindle]

Is it multiple choice?

[summer current]

no

[timid spindle]

Then your question makes no sense

[summer current]

huh

[jagged stump]

Then answer is 13

15 is not it

[summer current]

11 13 and 15?

are the 3?

[jagged stump]

Ayoo

[timid spindle]

111 113 115 also work, the question has no one answer

[jagged stump]

It's 13 15 17

[summer current]

but i have to find the middle number

[jagged stump]

The middle number would be 15

Most appropriate answer

[summer current]

when i simplify it down its x is no less than 15

[jagged stump]

Less than 15 means 13

[summer current]

no less

so 15 is the max

[timid spindle]

x is no less than 15 means x is bigger than or equal to 15

[summer current]

so its 17

[jagged stump]

At this point you trolling

[summer current]

what

[timid spindle]

Send a picture of your question

[summer current]

The sum of three consecutive odd integers is no less than 51 . Find the middle integer.

[timid spindle]

Does not look like a picture to me

[summer current]

well if you want the picture itll look the same

ill just ask my teacher tmrw

.close

[pearl pond]

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[lapis girder]

Wait its unclear

My question is why did it use 4q+1 or 4q+3

Isnt 2q+1 and 2q+3 the same thing

[jagged stump]

Why don't you try that yourself bud

Go ahead and replace 2q+1

[lapis girder]

I did but answer is different

4q(q+1)/8

[jagged stump]

I'll try

[lapis girder]

Heck u dont need 2q+3

Its just 2q+1

[jagged stump]

Lol

2q+1 is also true

They have just taken 4q for easier for students to understand

[lapis girder]

Can you send how you did it?

[jagged stump]

I'll write it neat dear

[lapis girder]

Alright

[jagged stump]

Any doubts?

It can be any integer really

Not just positive

[lapis girder]

i see

i think i havent understood this lemma fully since i thought the result you get should be divisible by the number

[jagged stump]

In maths just trust the process

It all comes down to final result

This should be a great lesson for you

[lapis girder]

how did u close this again

/close

[vestal tapir]

dot

[lapis girder]

.close

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[zinc oracle]

how would you use the fact that dz=(∂z/∂x) dx+(∂z/∂y) dy to find z here using direct integration, is it just solving every term with respect to both x and y?

[pearl pond]

@zinc oracle Has your question been resolved?

[zinc oracle]

<@&286206848099549185>

[solemn fog]

$ \int \left{\left(3x^2siny+2xsin2x+y^3\right)dx+\left(x^3cosy+3xy^2\right)dy\right}

@zinc oracle https://www.symbolab.com/solver/step-by-step/\int\left\{\left(3x^{2}siny%2B2xsin2x%2By^{3}\right)dx%2B\left(x^{3}cosy%2B3xy^{2}\right)dy\right\}?or=input

[zinc oracle]

just doesnt answer my question though

[solemn fog]

what should we find

[zinc oracle]

Finding z using direct integration and hence solving, so is it just the link that you posted then using x = pi/2 and y = pi ?

[solemn fog]

btw in which standard you are

maybe i cant solve this ques manually and also did'nt understood it fully coz i'm in 12th only

[pearl pond]

@zinc oracle Has your question been resolved?

[pearl pond]

@zinc oracle Has your question been resolved?

[pearl pond]

@zinc oracle Has your question been resolved?

[zinc oracle]

<@&286206848099549185>

[potent hedge]

What is C?

Is it the unity circle?

[zinc oracle]

integral over a curve?

[potent hedge]

But is C precisely defined or is it a random curve?

[zinc oracle]

i think ive solved it now regardless, its the curve defined by the integral

of which you can then set limits over

.close

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[deft jungle]

solve pls

[chrome lynx]

@deft jungle what did you try?

[deft jungle]

not sleeping

[buoyant shore]

Perhaps you could find a partttern

Pattern

[chrome lynx]

same question like in #help-38

[hidden anvil]

yes

just want answer

[deft jungle]

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[tribal shale]

A (vertical) electricity pole 7 m high breaks at a certain height from the ground and is left bent, hanging, so that the end of the broken piece falls 3 m from the base of the pole. At what height did it break?

[hallow remnant]

okay

so let say it breaks at x meters from the ground

so what will be the length of the 2 pieces

[hollow pollen]

Are you familiar with conservation of energy?

[hallow remnant]

um this question doesn't have anything to do with conservation of energy

[hollow pollen]

This can be solved with it

[hallow remnant]

stop trolling

[hollow pollen]

The conservation of energy in this problem can be represented by the following equation for each half of the pole: E = mgh + pgy. In the case that x is a positive value (above ground), one of the pieces will be completely above ground and the other below ground, and if x is not positive, both pieces will be at the ground. Thus, the total energy used for a is: ma_above = 5m + mh + pgy where y_above is positive.

[hallow remnant]

please stop

[hollow pollen]

I'm sorry if you are unsatisfied and I have messed up. You can wait for other people to help.

[hallow remnant]

this is just a very basic x+y=7,x-y=3 problem

@tribal shale are you there?

[tribal shale]

is this good?

[hollow pollen]

While it is true that one of the methods of solving the problem is very basic and uses basic algebra, there are many other ways to solve this problem as well. Some more sophisticated methods include calculus and the law of cosines, which one normally learns much later.

Again, I'm sorry.

[hallow remnant]

stop being a troll dont act like you have any actual stuff ur saying

[hallow remnant]

okay so i mean it will be more of a clean brake

[tribal shale]

thanks for the help, but i have no study about conservation of energy

[hollow pollen]

Calculus

We can use calculus by differentiating the function for the length of the pole with respect to its height to get a function that represents the rate of change the the length with respect to the height.

We can then take the derivative again to get another function which is proportional to the rate of change of the change of length with respect to its height, and use this to find the point where the rate of change is equal and opposite, thus giving us 3m as the height where the break occurs.

Law of Cosines

We can use the law of cosines in a similar manner to the calculus method.

By representing the height of the pole with the variable h, and representing the two pieces as the variables a and b, we can use the formula:

d =√ ( h^2 + b^2 - 2hb cos ⁡ φ ).
by taking the derivative of d with respect to h, the variable representing the height, in order to find the height of the point where the break occurs.

[tribal shale]

maybe you can draw it?

nvm

[hallow remnant]

stop otherwise ill get the mods

[hollow pollen]

Sure.

You say I have no idea what I'm talking about, not read a single word, and then bash me for it.

[hallow remnant]

@tribal shale this one

[hallow remnant]

<@&268886789983436800> can you please look at what @hollow pollen and give him a warning or whatever

[hollow pollen]

😅, No problem, It just came into my head.

[hallow remnant]

okay @tribal shale so what is the height from the point at the end of the seocnd segment

[tribal shale]

Is x?

sorry if i act silly

[hallow remnant]

we are defining x as the length of the segment which is below the point where it breaks off

in other words, we want to solve for x

i drew the point where it breaks of in red

[tribal shale]

maybe 7-x=x-3?

[hallow remnant]

yes thats correct

we get 7-x=x-3

then you solve that equation

do you know how to solve it

[tribal shale]

4=2x

[hallow remnant]

um its 10=2x

(the Left hand side of the equation becomes 7+3 not 7-3)

does that make sense?

[tribal shale]

so 5=x?

[hallow remnant]

yep

thats it lol

no calculus

or conservation of energy

[tribal shale]

Haha, thanks you very much. sorry if If I write something wrong in English, because it's not my national language.

[hallow remnant]

yeah you wrote everything correct i believe

should we use ".close" now?

[tribal shale]

yea

.close

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[nocturne moat]

Hi there, I need help with a **calculus **question. I have trouble solving this improper integral because I get an indeterminate form and I tried to manipulate the expression to use L'hopitals rule but I cant seem to do it. I'll attach my working.

Thank you!

[round temple]

Maybe try u-sub with $u = 1 + x^2$.

[jolly parrot]

jimmy1234

[midnight haven]

The problem is not the integration part

But the evaluation

[nocturne moat]

yea i checked my integral is correct

[midnight haven]

The integral is correct yes

[nocturne moat]

Oh am I subbing in something incorrectly?

[midnight haven]

Not that I can see

There must be some clever trick to evaluate this

Wait

[nocturne moat]

I think my issue is not being able to change the indeterminate form to one where l'hopitals rule could be used

[shrewd basin]

It is not convergent I think unless we look at CPV

[nocturne moat]

sorry im not sure what cpv is?

how did you determine it doesnt converge?

[shrewd basin]

Cauchy principle value

-infinity + infinity seems like it goes to 0 but we don’t know how ‘big’ these infinities are (intuitively)

[nocturne moat]

do I use that to evaluate the integral?

[shrewd basin]

Look it up

[nocturne moat]

just checked doesn't seem to be part of class scope, perhaps another method is relevant

for context we just started improper integrals

[shrewd basin]

Then this wouldn’t converge to anything

,w integral from -infinity to infinity of (x)/(1+x^2)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by Pure#9389

[nocturne moat]

Ok, I'll redo the working. Thnx!

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[prime vector]

.close

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[grizzled lodge]

.close

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[grizzled lodge]

.reopen

[pearl pond]

✅

[midnight haven]

what have you tried?

[grizzled lodge]

I’m not sure how to approach

[midnight haven]

maybe try mod 3 and get a condition on x and n first

[grizzled lodge]

Well

I’m mod 3 you have

-1=x^n

What does this mean?

[midnight haven]

think about what n must be first

hint: fermat's little

[grizzled lodge]

n must be odd?

[midnight haven]

yeah

so what must x be mod 3?

[grizzled lodge]

-1

?

[midnight haven]

yeah

these things reduce your search space a bit ig

[grizzled lodge]

Hm

[shrewd basin]

consider x^n + 1 = (x+1)(1-x+x^2-x^3+...+x^(n-1))

[grizzled lodge]

Hm how do you proceed from there?

[shrewd basin]

both (x+1) and (1-x+x^2-x^3+...+x^(n-1)) must be some powers of 3

and (1-x+x^2-x^3+...+x^(n-1)) >= (x+1)

[grizzled lodge]

Right yeah

[shrewd basin]

In particular we must have (1-x+x^2-x^3+...+x^(n-1)) = 0 = n in mod (x+1) meaning x+1|n hence n|3

[grizzled lodge]

Got it

That restricts the values for n

I got (2,2,3) as the only solution

Is that right?

[shrewd basin]

Yes then obviously we have the trivial solutions when x=1

[grizzled lodge]

Okay cool

Ty

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[quartz ermine]

What does the number at the end mean

[quartz ermine]

Like =1/-1/0

[mossy canopy]

sin(theta)=1 asks for which angle theta is sin equal to 1

for example sin(0)=0

[pearl pond]

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[cold socket]

Anyone who can figure this out and explain how i start this?

[hallow remnant]

find the dimensions of the frame

you have a frame which adds x cms from the picture on each side of each dimension

so what are the new dimensions @cold socket if to each side of each dimension we are adding x

[cold socket]

Thats what I dont get, I cant see how to start this :a

[hallow remnant]

okay so we take the hieght of the picture

x+4

and then now by the frame, we are adding another x height above it, and another x height below it

so the new hieght is 3x+4

meanwhile, for the length, we add another x length to the left, and another x length to the right with the frame

so our new length is 2x+8

so therefore, the area of the picture and the frame together has dimensions (3x+4) and (2x+8)

[cold socket]

oh ok

[hallow remnant]

so therefore, (3x+4)(2x+8)=120

and you solve that quadratic

[cold socket]

I see, thanks alot 🙂

[hallow remnant]

.close

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[candid gorge]

which is the easiest way to solve limits x->0 or oo of x^k * e^1/x^k?

[candid gorge]

I feel like doing l'hop for (e^1/x^2)/1/x^k is a lot

and that always lead me to mistakes

[ivory swallow]

$\lim_{x\to0} x^k e^{\frac{1}{x^k}}$

[jolly parrot]

Mr. Gamer

[ivory swallow]

?

[sick adder]

As is this limit would evaluate to something to the likes of 0 * infinity which is indeterminate

I think we would want to use LH

[ivory swallow]

exp(k log(x)) * exp(1/x^k)
= exp (k log(x) + 1/x^k)

= exp[k (x^k log(x) + 1) / x^k ]

now let's just look at that inner part

$\lim_{x\to0} \frac{x^k log(x) + 1}{x^k}$

[jolly parrot]

Mr. Gamer

[pearl pond]

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[errant flax]

37 = 20+200e^-0,12t

[errant flax]

i got it to 17/200 = e^-0,12

but what do i do next step?

[minor dune]

are you trying to find t?

[pulsar lark]

Natural log

[errant flax]

yes trying to find t

[minor dune]

Yea ln both sides

[errant flax]

so lg on both?

[minor dune]

ln

The natural log is the inverse of the exponential function

[errant flax]

ln (17/200) = ln -0,12t?

and what is after that?

[minor dune]

Well if they're inverses they cancel each other out

[errant flax]

what u mean?

[minor dune]

Inverses cancel each other out

so if you ln an exponential function is just gets rid of it

[tropic saddle]

ln(e^x) = x

[errant flax]

my english is bad sry what u mean like can u type it plz

[errant flax]

but how?

[tropic saddle]

wdym how

that's just how logarithms work

[errant flax]

I know

But I mean

It is gonna be ln (e^, 0,12t) =?

[minor dune]

look at the example provided

ln(e^x)=x

[errant flax]

So u mean it = 17/200?

[minor dune]

Okay

look at the equation

17/200 = e^-0,12

[errant flax]

Yea

U forgot the t

[minor dune]

17/200 = e^-0,12t

okay

so ln both sides

ln(17/200)=ln(e^-0,12t)

using ln(e^x) = x

what should the right side equal

[errant flax]

Uhhhhh so it's ln(e^-0,12t)= - 0,12t?

[minor dune]

yes

so the right side is equal to -0,12t

then just solve for t!

[errant flax]

But bro

The next step is?

[tropic saddle]

ln(17/200)=-0.12 t

ln(17/200) is just a number

[errant flax]

So what I typed was wrong?

[tropic saddle]

no

I just put the LHS again

[errant flax]

Okay

Uhhh so that's it just make - 0,12 to left side so t can be alone

@tropic saddle right?

[tropic saddle]

yes

[errant flax]

Okay thx brothers

[pearl pond]

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[hollow scaffold]

Hello

[hollow scaffold]

Can someone help me solve this, it’s about derivatives

[plush bramble]

• Ask your math question in a clear, concise manner.
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[pearl pond]

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[ashen dagger]

solving logs

wouldnt there be a negative 1 on the other side

instead of a positive one?

3x+1 = ***

[timid spindle]

Yes it would be negative

In the box

[ashen dagger]

so this is incorrect right?

[acoustic jasper]

I think it is

[ashen dagger]

what would the correct answer be?

[acoustic jasper]

You’d have $3x= \frac{log6}{log2} -1$

[jolly parrot]

ikraamampampam

[acoustic jasper]

Then you can divide by 2

3*

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[wheat charm]

ok

[wheat charm]

so

it says find x

and thy give use these 2 functions

did i get it correct

?

see this is the page from my teachers notes

but idk where the questions answers and explanations are

so

read it and i fu understand wht im looking for pls answer

[pearl pond]

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[shut linden]

How is (a,inf) in the event space?

Does this just occur when a=b

[timid spindle]

They're just dummy variables

Anything that looks like (number,inf) is in F

[pearl pond]

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[spring minnow]

hi

[pulsar lark]

Bye 👋

[spring minnow]

.close

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[limber oasis]

On the wikipedia page for minimal polynomials (of matrices), it says "The minimal polynomial is often the same as the characteristic polynomial, but not always". In what way is it almost always ? Like a dense open subset of the square matrices of order n ? Any intuition behind the fact it's rarely of a lower degree ?

[limber oasis]

It probably comes from most matrices having n distinct eigenvalues right ? Because the set of simply split polynomials over C is open and probably dense ?

It probably comes from most matrices having n distinct eigenvalues right ? Because the set of simply split polynomials over C is open and probably dense ?

Also am I right in thinking dense + open can be interpret as almost always or is that not really strong enough ?

[pearl pond]

@limber oasis Has your question been resolved?

[limber oasis]

.close

[pearl pond]

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[midnight haven]

Hi

[midnight haven]

The juice is mixed in a ratio of 1 : 6.

b) How much concentrated juice is in 7L of pre-mixed juice and water?

<@&286206848099549185>

[formal summit]

Hello @midnight haven will be happy to help!

The answer is 1L as according to the condition above 6L will be water

[split oriole]

yeah

You take sum of ratio = 7

so

Juice = 1/7 *7L

= 1L

and same for water

[pearl pond]

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[midnight haven]

Perfect thanks

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[hollow cobalt]

The fraction looks similar to this

I'd nto recommend that

[hollow cobalt]

Just use this

[tall tangle]

how?

[hollow cobalt]

Let x = tan(alpha) and a = tan(beta)

[tall tangle]

oh

[hollow cobalt]

Then f(x) = arctan((x + a)/(1 - ax)) = arctan((tan(alpha) + tan(beta))/(1 - tan(alpha)tan(beta))) = arctan(tan(alpha + beta)) = alpha + beta = arctan(x) + arctan(a)

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[misty pivot]

The question is show that :

h ≤ g ≤ m

I figured out g ≤ m but i cant find why h ≤ g

Ive tried subtracting h from g to see if the result would be superior to 0 but it didnt work

[pearl pond]

@misty pivot Has your question been resolved?

[misty pivot]

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[wind mantle]

Ok tape 2

So like using a right triangle and stuff

[buoyant panther]

$\DoubleSin$

[jolly parrot]

Modus

[wind mantle]

So like to get to that point tho

[buoyant panther]

use this formula then right triangle

[wind mantle]

oh alr

[buoyant panther]

and notice that cos(arccos(theta)) = theta

[wind mantle]

rightt

ok so like smth just aint adding up

actually nah Im good, thanks

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[midnight haven]

Can someone help me

[pearl pond]

@midnight haven Has your question been resolved?

[cunning pier]

Try the sentence of l'hopital maybe

So derive both the top and bottom of the fraction and see whether that gives you a proper result

[pearl pond]

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[vague fiber]

I need help getting a tangent line for an inverse function

[vague fiber]

Ik i been asking a lot of questions here sorry I just dont understand these concepts

[timid spindle]

post the problem

[vague fiber]

slope tangent to f inverse

[timid spindle]

either explicitly find the inverse function, or use the inverse function theorem

[vague fiber]

i got the derivative

[pearl pond]

@vague fiber Has your question been resolved?

[vague fiber]

<@&286206848099549185>

[midnight haven]

feed the derivative into the inverse function theorem now https://en.wikipedia.org/wiki/Inverse_function_theorem#Statements

[vague fiber]

I got f'(4)

With the inverse

which is 10

but it says its wrong

1/f'(f^-1(25))

[vague fiber]

ok i found a way but it has nothing to do with that fancy formula

find the inverse then just take the derivative of it

omg im so stupid i got 10 but i forgot it was in the denominator

wow

[vague fiber]

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[pearl pond]

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[wraith saddle]

quick question, what is the 5th dimension axis usually called

[wraith saddle]

like the 1st 2nd 3rd and 4th dimension's axes are called the x y z and w axes

what about the 5th

or is it just called the 5th dimension

[hollow cobalt]

I really doubt that it has a universal name, but when the amount of dimensions gets large I've seen the them denoted as x1, x2, x3 and so on

[wraith saddle]

makes sense, i thought that might be the case

thanks for helping!

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[obtuse summit]

What’s the answer if I need to find an algebraic formula for this arthmetic progression: a+(a+1) + … + (n-1) + n

[obtuse summit]

@solemn fog

You gave me the answer for this one

What is the answer for this one:

If a = an odd number, and n = an even number. And also N>A

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[unreal sun]

Hi! I'm trying to solve the differential equaiton $(1+x^2)y'+xy=x+x^3$. I started off by multiplying both sides with $\frac{1}{2}$ which gives me $\frac{(1+x^2)y'}{2}+\frac{xy}{2}=\frac{x+x^3}{2}$, one can note that the left hand side is simply $D(\frac{(1+x^2)y}{2}$, integrating both sides and multiplying by 2 now gives $(1+x^2)y = \frac{x^4}{4}+\frac{x^2}{2}$, and solving for y gives $y = \frac{x^4}{4(1+x^2}+\frac{x^2}{2(1+x^2}$

[jolly parrot]

Chippotle Maths

[unreal sun]

However this is the wrong answer. It seems that I can get the correct one using integrating factor, however, I am expected ot solve it without using that method as we have yet to cover it. This is the correct answer

Where did my thinking go wrong?

[plush bramble]

your approach is identical to integrating factor

[unreal sun]

Is it? I have yet to check the method, my hw states that one can solve it using the method, however, it is expected to do without it, therefore I have yet to actually learn the method

That's amusing though

[merry carbon]

"one can note that the left hand side is simply $D(\frac{(1+x^2)y}{2})$"

[jolly parrot]

chartbit

[merry carbon]

Is it? Maybe I'm being slow but I don't seem to get that it is?

[plush bramble]

,w d/dx [(1+x^2) y(x) / 2]

[jolly parrot]

Results provided by WolframAlpha

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

Requested by riemann#2529

[plush bramble]

oh off by a factor of 1/2

[unreal sun]

You're not the one who's slow... that's me! I should've doubled checked that... Riemann and you are correct

Oh well, another attempt!

[pearl pond]

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Please don't occupy multiple help channels.

[coarse dawn]

Stop opening multiple channels

.close

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[coarse dawn]

You already have a channel open, just wait for people to help

[pearl pond]

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[mortal pawn]

use the convolution property to find x(t):

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@mortal pawn Has your question been resolved?

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[brittle wyvern]

hey

[plush bramble]

[brittle wyvern]

ignore the right answers that was just my lazy ass copying the answer key

but i decided to ask for help

[fleet sky]

for which problem? #8?

[brittle wyvern]

7 and 8

[fleet sky]

is there more to 7? Is 7 just asking you to write out the piecewise function?

[brittle wyvern]

piecewise function

[fleet sky]

ah okay

so since they're straight lines, we just need to find two separate y = mx + b and limit their domains

so for the first straight line (the one with positive slope)

we have two points

what are the two points?

[brittle wyvern]

yep

one sec

(4,-2) and (8,0)

[fleet sky]

wait that's for 6. You said 7 and 8 lol

but the right idea 🙂

[brittle wyvern]

oh mb i thought it was like 7 a, b

but yes 6,7,8

[fleet sky]

okay

so for #6

that's correct

so what is the slope of that line?

[brittle wyvern]

1/2?

im not good w/slopes

[fleet sky]

no worries

so the slope of a line between two points is

$m = \frac{y_2 - y_1}{x_2 - x_1}$

[jolly parrot]

MellowDramaLlama

[fleet sky]

so you just plug in your points

[brittle wyvern]

oh boy

soo

[fleet sky]

oh wait these are backwards. It's supposed to be (-2, 4) and (4, -2)

sorry I said that was correct but it wasn't

[brittle wyvern]

say what

[fleet sky]

so we have 3 total points on #6

[brittle wyvern]

so we're looking for the decreasing now

[brittle wyvern]

ya

[fleet sky]

the first two points (going left to right) is (-2, 4) and (4, -2)

yeah the first part is decreasing

[brittle wyvern]

alr

[fleet sky]

so then plug in those values here 🙂

$m = \frac{-2 - 4}{4 - (-2)} = \frac{-6}{6} = -1$

[jolly parrot]

MellowDramaLlama