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[cedar kiln]

@brisk ferry Has your question been resolved?

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[misty talon]

is this correct?

[crimson sedge]

yup

,w 2/x-4=-3/2-1/(2x)

[wraith dagger]

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Requested by Toby#2275

[misty talon]

is there another way to do it?

can i start with multiplying everything with 2x as well?

[crimson sedge]

yeah, you can

[misty talon]

are they just as good

or

which ones best

[crimson sedge]

there is no best

do what you are comfortable with

[misty talon]

alright thanks

[crimson sedge]

yw

[buoyant latch]

I would say this is best in this case

Gets rid of all the fractions

Fractions are an arithmetic nightmare to deal with

[misty talon]

right okay

i realized its the same as what i did first tho tbh

i just split it into 2

.close

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[icy kelp]

Wrt to p-norm where p approaches infinity also known as $L_{\infty}$ it is known that it is equivalent to the max of the absolute value of its inputs. My question is if I want to differentiate the limit of the p-norm function can I swap the order of the derivative and the limit?

[wraith dagger]

bigbob

[crimson sedge]

you would need something like uniform convergence https://en.wikipedia.org/wiki/Interchange_of_limiting_operations

[icy kelp]

p-norm does converge uniformly on the max function

[crimson sedge]

then you are fine

https://math.stackexchange.com/questions/2696519/finding-the-derivative-of-the-infinity-norm

[icy kelp]

Jesus christ

[cedar kiln]

@icy kelp Has your question been resolved?

[icy kelp]

I'm not sure Mr Bot

Given that one can supposedly find the derivative of the infinity form can one use it in optimisation problems?

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[nova snow]

Does this notation as a domain make sense

[crystal raptor]

what does the x | mean out front

[nova snow]

x such that

[crystal raptor]

eh you really only use a pipe for "such that" if you're already in a set

you can just write "the domain is ${x \in \mathbb{R} \mid x \neq \pm 4}$"

[wraith dagger]

ΣAC

[nova snow]

Is | symbol called a pipe?

[crystal raptor]

one name for it yeah

[nova snow]

oh so if it’s in the middle of a set what does that indicate

you know how I wrote the x | on the outside

If it’s in the inside what does that mean

[crystal raptor]

such that

[nova snow]

Oh alright

so if it was on the outside would it change the meaning of the notation

[crystal raptor]

its just that its not used for "such that" outside of set building

[nova snow]

Ohh okay

[crystal raptor]

atleast not commonly enough that you should do it:)

[nova snow]

Alright thanks

[royal loom]

is it not used as within kind of

like I would read that first image as

for x within blah blah blah

[crystal raptor]

literally never

[cedar kiln]

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[olive heart]

how do you solve 20

[tame wraith]

,rotate

[tame wraith]

hmm, wouldnt you use polynomial long division as normal for this @olive heart

[static fern]

sythnetic division might also work?

[olive heart]

you would use synthetic

yea

but

idk how to start it since theres 3 terms

[static fern]

use synthetic to get $\frac{3x^3 - 7x^2 + 8x-2}{x-(1+i)}$ and $Q(x)$ and $R$ will be the quotient and remainder respectively

[wraith dagger]

FireBlazer

[olive heart]

wdym

[static fern]

do you know how to do synthetic division?

[olive heart]

yea

[static fern]

so you would write 1+i in the far-left place and proceed as normal with the synthetic divison algorithm

[olive heart]

oh alr

what happens to the x

[static fern]

synthetic division is basically analyzing coefficients so we wouldn't need x for the algorithm. we put the x's back into the quotient at the end

[olive heart]

ohh

alr

i see what u mean

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[latent flax]

can you help me understand interest and finance

[cedar kiln]

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[nocturne maple]

A pharmacist wants to mix 9% of zinc oxidide with 15% zinc oxide to make 30 grams of 10% zinc oxide. How many grams of 9% and 10% should he mix together?

[nocturne maple]

i got 25g for 9% but idk how to get 10%

[cedar kiln]

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[short sierra]

Hey can someone please explain to me how my answer is wrong? And how do you even get the right answer?

[short sierra]

Please

[royal loom]

Sure

the polynomial 2u^(1/3)

is only a scaled version of the polynomial u^(1/3)

and the polynomial u^(1/3) is an odd function

because an odd function tells you that f(-x)=-f(x)

this means that the left side of the y-axis will have the opposite total area as the right side of the y-axis

and in the case of your integral

the bounds are from -1 to 1

which has an even amount on the negative of and positive side of the y-axis

meaning the areas will cancel out

and give zero

[short sierra]

Could you please show me in calculations? Or is this just something I should notice and realize?

[soft flax]

The latter

[royal loom]

this is what the graph would look like

you should be able to notice and realize

no calculations necessary

your answer thus far was also correct, it would have evaluated to become 3/2-3/2=0

[short sierra]

So this is actually one part of a question, but when I tried to do the same thing to the other side I would have also gotten 0

let me try and show you

And I really appreciate the explanation

[royal loom]

so where did this new integral come from

the logic only applies to u^1/3

u^1/3 is an odd function

u^2/3 however is not

so the area under the curve will not be 0 for the latter part of the integrand

which you can see because -3/5 - 3/5 is not 0

[short sierra]

u^5/3 be an odd function?

[royal loom]

yes

[short sierra]

How does that become -3/5?

[royal loom]

if the power of the numerator is odd, it is an odd function

[short sierra]

Oh wait

Im so stupid

Ok one more question

is (-1)^(7/5)

Going to keep my -1?

Like does it matter what the fraction is?

[royal loom]

it does matter

if the numerator of the fraction is even, the result will be positive

if the numerator of the fraction is odd, it depends

[short sierra]

I ask this because, when I put it in my calculator I get an error

[royal loom]

that is because your calculator doesn't know if it should interpret it as $(-1)^{\frac{7}{5}}={(-1)^{7}}^{\frac{1}{5}}$ or as $(-1)^{\frac{7}{5}}={(-1)^{\frac{1}{5}}}^{7}$

give me a sec to get this to print correctly

[short sierra]

Ok

[wraith dagger]

AustinU

[short sierra]

Ohh okay that makes sense

Thanks a lot for your help, I really appreciate it

[royal loom]

no problem

[cedar kiln]

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[young tide]

Hi This is a math question on my math exam "Daniel walks 8km on a bearing of 015T. He then walks 5km on a bearing of 120T. Calculate his distance from his starting point, correct to 2 decimal places" Any help would be great thank you!

[crystal raptor]

Youre doing an exam right now?

[lethal jackal]

you have the right to remain silent

[young tide]

No its on my practice exam sorry. I have the exam tomorrow morning. Im in queensland Australia. Teachers have said that this will be same type of stuff as whats on the exam tomorrow

[lethal jackal]

what does 015T mean again

[young tide]

015 degree true bearing

[lethal jackal]

what's 0 degrees true bearing

I forgot

[young tide]

so just 15 degrees clockwise from the origin.

[lethal jackal]

what origin

you can't be clockwise from a point

[young tide]

the center

of the graph thingo

[lethal jackal]

that doesn't make sense

[young tide]

cartesian plane

[lethal jackal]

you can't be clockwise from a point

[young tide]

hold on ill put a diagram in

[crystal raptor]

Bearings are clockwise from North

[lethal jackal]

let me just ask

oh okay

so maybe you should draw a diagram of Daniel's path

[young tide]

yeah its drawing a non right angled triangle and then solving for the unknown distance. I just don't know what the internal angles of the triangle are

[lethal jackal]

well you could do it that way and you could figure out the internal angles of the triangle

it's not particularly hard

you didn't draw his second walking vector

but surely you've done something involving vector addition, yeah?

[young tide]

not really. this is a year 10 Australian maths question so obviously for you it will be very easy. Can I draw out the diagram and then you could quickly help me with the angles?

[lethal jackal]

these types of problems aer usually seen int he context of vector addition

but go ahead and draw it out

[young tide]

what we have been learning this term is some trig rules like sine rule cosine rule and area of a triangle. So nothing too tricky

something like this

[lethal jackal]

looks about right

[young tide]

do you know how to find the internal angles?

[lethal jackal]

yeah

one sec lemme draw it on the diagram

[young tide]

ok thanks so much

[lethal jackal]

if I point out to you that the angle with the arrow pointing to it is 120 degrees

can you figure out the red angle?

[young tide]

is the 15 alternate? then plus 60?

[lethal jackal]

wdym by "is the 15 alternate?"

[young tide]

the 15 down the bottom left

[lethal jackal]

wdym by alternate

[young tide]

and is it an alternate angle so it is the same on the red angle side

it’s a bit hard to explain

but could you explain your way that you would go about it?

[magic talon]

There is two triangle

Try to calculate the angle first

[lethal jackal]

calculate the angles you can figure out

it's not going to be just directly reading off the red angle

[young tide]

so the left triangle that you made?

[lethal jackal]

just calculate whatever angles you can figure out

nobody is going to tell you exactly what steps to do on a test, because we won't be there and won't see the questions

so you need to be able to figure some of them out

these are like little puzzles

for example

what can you say about the angle right next to the 120 degree angle?

[young tide]

60 deg?

[lethal jackal]

why the question mark

it's 60 degrees because they make a line together

[young tide]

good

[lethal jackal]

what can you calculate now

[young tide]

the other one in the triangle

105

[lethal jackal]

okay

so what's the red angle

[young tide]

75

i thjnk

[lethal jackal]

wasn't so hard to figure out after all, right?

now there's a relationship between the congruence relations and the solvability of a triangle

basically if you know SSS or SAS or AAS or ASA or whatever in a triangle

you can solve for all of the remaining sides

and angles

[young tide]

are these similar?

[lethal jackal]

no they're not

also don't see how that's relevant

I'm saying if you have a single triangle

[young tide]

The SSS and stuff we always learnt in similar triangles

[lethal jackal]

okay well I'm saying something else

[young tide]

sorry

[lethal jackal]

I'm saying that if you know three values of a triangle in that sort of pattern

then you can solve for the rest

[young tide]

yeah ok i can solve for the unknown with cosine rule?

[lethal jackal]

yes you can

you can solve for the unknown side with the law of cosines

[young tide]

ok give me 1 min

so 60-80cos75

[lethal jackal]

I wouldn't know

I'm way too lazy to actually figure it out

[magic talon]

I think you make some mistake?

[lethal jackal]

you should get something like 8.26405

whatever that is

[young tide]

i got 6.27

[lethal jackal]

did I put it into the triangle calculator correctly

I think so

[young tide]

oh #

wiat

i’m a idiot

8^2

[magic talon]

64

[young tide]

whoops

yeah it’s right 8.26

sorry

thank you so much

@lethal jackal @magic talon you are legends thanks so much

[lethal jackal]

yw

[young tide]

i’ve got a 1 more but i’ll bother other people

thanks so much

differnt type of question tho

how do i end

.close

[cedar kiln]

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[honest bobcat]

epsilon delta shittery is making me aDHUISDHUAHD

[cedar kiln]

Please don't occupy multiple help channels.

[honest bobcat]

i watchged a bunch of videos

and i stil havent figured it out

.close

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[trail shale]

Hi, sorry if this is a stupid question but somehow I'm not able to google this.

How do I reverse a power? For example my formula is t^5, if t = 10 then it's 100000 but how could I go from 100000 to 10?

[dusk finch]

5th root

[wraith dagger]

MathIsAlwaysRight

[magic talon]

Base on the question

[trail shale]

Alright thanks

[magic talon]

If the question is x^5=100000

Then

You have 5 possible value for x

[dusk finch]

Only 1 of them is real tho

If he doesnt know roots he probably doesnt know complex numbers

[magic talon]

But if the question don't mention it, ignore the complex number will make you lose some marks

Just a reminder

[bold lotus]

thats not true at all, why would the questions mention a concept that has not even been introduced, are you expected to solve questions that have complex numbers as solutions when you are first introduced to the notion of a square root?

[magic talon]

I know, but since I don't know what have bloody learnt, so I'll just remind this thing to him.

[cedar kiln]

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[young gazelle]

I've solved a and b and i'm working on c

[young gazelle]

I'm not sure how to show that a linear operation on a convex set maintains that it's convex

Can I just use my proof from A that for any constant being multiplied by a convex set that the set maintains that it is convex and then show that since this is a linear operation that the set of resulting vectors are also also convex?

[crimson delta]

just check the definition of convex

and use linearity

[young gazelle]

The linearity thing is what I’m confused on

I’m pretty sure I can just use the definition and then say because of linearity this is convex still

But I don’t really understand why linearity means that

Is there somewhere I could read to understand linearity?

Maybe my question should instead be; in general how would I show that a set of vectors is convex? I’m not sure notation wise how to show this

Why does it follow since L is linear? Is my more precise question I guess

[crimson delta]

well linearity is pretty much exactly that condition used there

that you can move scalars and addition inside

a function L is called linear if L(a+b)=L(a)+L(b) and L(ca)=c L(a) for all vectors a,b and scalars c

(and then by induction of course for repeated addition)

have you not taken linear algebra?

[young gazelle]

I have I just took it in second year

And I’m in my 5th year

I’m pretty rusty on the concepts and they are pretty critical to this course 😬

I actually never knew that was called linearity but it makes a lot of sense that it would be

I did work for 2 years in a co op program mid degree and I feel like it hurt my knowledge of stuff

[crimson delta]

probably a good idea to check all your notes again

its super easy to forget stuff if you dont use it

[young gazelle]

Unfortunately I never took any notes, but I do have the textbooks and refresher materials from the course to reference

Thanks for the help, I assume proving that the inverse works is trivial since the inverse of some matrix is another matrix so it’s basically the same case?

[crimson delta]

well here the matrix doesnt have to be invertible

it's not an inverse, it's the preimage

(the proof is still not hard tho)

[young gazelle]

If for a generic matrix the proof holds then for some invertible matrix wouldnt the same proof work? Or am I missing something

[crimson delta]

if it is invertible

but here it isnt in general

its not even square

its unfortunate notation but the ^-1 means preimage and not inverse

[young gazelle]

Hmm

I’ve never heard preimage before gotta Google lol

[crimson delta]

not?

if $f:A\to B$ and $C\subseteq B$, then $f^{-1}(C) = {a\in A: f(a)\in C}$

[wraith dagger]

Denascite

[crimson delta]

its those elements that get mapped into C

[young gazelle]

ohhh

hmm so I have to show that the subset of inputs that map to the convex set D from A must be convex

or I guess the set x of all inputs that Ax is in D

[cedar kiln]

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[limber stone]

how did they simplify this

[lilac totem]

I think there is an error in the simplification

[limber stone]

How did u go from the 2nd to 3rd line?

[lilac totem]

Ok wait

[limber stone]

Thank u @lilac totem

[limber stone]

are u allowed to just multiply 2(x^2+1) on both denom and numer

[lilac totem]

Yes

You can multiply anything on the numerator and denominator except for 0

[limber stone]

Okay thanks

.close

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[eager remnant]

How is it sqrt2 and not sqrt6, -sqrt6 and 0?

Q5

[south tundra]

You should have solved for the intervals where the derivative is positive, not the function itself

[eager remnant]

What do you mean?

I solved intervals for Increasing and decreasing at zero points

[south tundra]

A function f(x) is increasing when f'(x) > 0, right?

[red pumice]

that the function is positive does not mean the function is increasing

[eager remnant]

Yes

[south tundra]

But you solved f(x) > 0 instead of that

[eager remnant]

OOOOH YOURE SO RIGHT

I don’t know how I missed that

Thanks

.close

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[dire sigil]

hello?

[dire sigil]

is chemistry applicable here?

[crimson delta]

you might have better luck by checking the #old-network

that said, sometimes chemistry questions do get answered here

[dire sigil]

oh, thanks I might look into it.

my chemistry teacher teach us about converting measurements but I get trouble answering this kind

[fickle trellis]

Did you try anything as of yet

[dire sigil]

im trying, but I dont know what to put.

[cedar kiln]

@dire sigil Has your question been resolved?

[cedar kiln]

@dire sigil Has your question been resolved?

[livid hound]

um wtf is with those questionable conversions

unit conversion is pretty much applying multiplication by 1

[cedar kiln]

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[jade sonnet]

I'm trying to sketch these loci, I've worked out that the arg one is just a vertical line because, but I don't know how I'd know If its at plus or minus 2, for the other one I've got the line y=x which is wrong but I'm not sure what I did

[cedar kiln]

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[polar sleet]

Find the area of a rectangular plot one side of which measures 35m and the diagonal is 37 m

[crimson sedge]

find another side by using pythagoras theorem 37 square = 35 square + breath square

[polar sleet]

I'll try

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[dusky citrus]

a box contains 4 red balls, 5 green balls, and 7 blue balls. in how many ways can 7 balls be chosen if there should be 2 red balls, 2 green balls and 3 blue balls in the selection?

[fair geyser]

it's simply the product of selections for each color

which two red, and which two green and which 3 blue

but also balls are kinda indistinguishable irl

so like, 1 way

if it was a trick question

[dusky citrus]

See this question is a Combination

[fair geyser]

yeah

that's very clear

[dusky citrus]

Oh so im right about that, the directions were identify if the equations are permu or combi then solve it like that

a box contains 4 red balls, 5 green balls, and 7 blue balls. in how many ways can 7 balls be chosen if there should be 2 red balls, 2 green balls and 3 blue balls in the selection?

[cedar kiln]

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@dusky citrus Has your question been resolved?

[spare fiber]

It's 4C2 * 5C2 * 7C3

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[prisma wadi]

how do you solve recurrence relations?

[prisma wadi]

i don't conceptually understand how to solve linear recurrence relations

like how would you do f(x+z)=f(x)+z

[cedar kiln]

@prisma wadi Has your question been resolved?

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@prisma wadi Has your question been resolved?

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@prisma wadi Has your question been resolved?

[prisma wadi]

<@&286206848099549185>

i watched a youtube video and i still don't fully understand generating functions

[dire geode]

show the actual problem in full. screenshot/picture

[prisma wadi]

i was watching a youtube vid and i was thinking about how you would find a closed form of f(x) with the knowledge that f(x+z)=f(x)+z

z being a constant

[dire geode]

don't invent your own problems

just do real ones

[prisma wadi]

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[deft spear]

not sure what to do next

ive done 7^(2K+1) + 5 - 7^(2k-1) -5
=7^2x7^(2k-1)-7^(2k-1)
= 13x7^(2k-1)

[long arrow]

wait

you're doing it by induction, right?

[upper abyss]

Very odd start. Remember to follow the rules of induction, at least

What's your base case?
What's your inductive assumption?

[deft spear]

ive done that just skipped to the hard part

[deft spear]

ye

[deft spear]

i have done the basis and assumption just stuck on the induction

[cedar kiln]

@deft spear Has your question been resolved?

[deft spear]

<@&286206848099549185>

[obtuse grail]

How far did you go on induction?

Oh nvm

I saw the message above

You don't have to take the difference

Use exponent properties to reduce the 2k+1 exponent back to 2k-1 and by your induction assumption it will automatically be divisible by 12

Does that help?

[deft spear]

Isnt that what i did

[obtuse grail]

Yeah, but you shouldn't subtract the assumption from the k+1 case

Just show that the k+1 case is divisible by twelve

[deft spear]

ye but whenever i try that method i just get lost lol

[obtuse grail]

Ok, um, try again with the k+1, but just the k+1 case mod 12

Feel free to write out your work here

[deft spear]

so
7^(2K+1) +5

7X7^2K +5

then im lost after that

[obtuse grail]

take it one step further with exponents

from step 1:

7^2 * 7^(2k-1) +5

can you continue from here?

the goal is to use the original assumption; 7^(2k-1) + 5 mod 12 = 0

[deft spear]

ohh i see

7^2x7^2k-7^2k?

gives me 48k^2 which is divsible by 12

wait no

[obtuse grail]

no you're right

thats also a way to do it

so now for any int k, 48k^2 will be divisible by 12

[deft spear]

is there not a way to do it from (k+1)-f(k)

[obtuse grail]

you mean a different way?

[deft spear]

ye cause ik theres a rule where u can still prove it by (k+1)-f(k)

the way i originatly started doing it

[obtuse grail]

Im not entirely sure what you mean, the only other way I can think of doing it is using modular multiplication properties on the 7^2*7^(2k-1) to separate the terms

[deft spear]

hmmm well atleat i got there in the end thanks for the heko

[obtuse grail]

no problem, im always he to heko lol

have a nice day

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[crystal raptor]

can you stop spamming nonsense in help channels

[cedar kiln]

@proud sparrow Has your question been resolved?

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[bold mauve]

How do I use the integral test to see if $$\sum_{n = 1}^{\infty}{\frac{1}{n^\sqrt{2}}}$$ is divergent or convergent

[wraith dagger]

donut
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[bold mauve]

I don't know what my lower bound of integration should be. Should it be x = 1, x = 2, ...?

Sometimes it is 1, sometimes it is not and I do not know why

[modern compass]

starting at n=1

[bold mauve]

when should I not choose n = 1

[modern compass]

usually the starting value of n

unless there's some reason that it's easier to start elsewhere, you can start anywhere past the starting value. So you could integrate from 10 to infinity if you wanted.

[bold mauve]

oh, okay. thank u

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[limpid zinc]

Sorry if it might be basic math but i've completely forgotten 😄

If i have two dice, both are fair - one has a total of 4 eyes and the other a total of 8 eyes, and i have to make a probability table over the possible sums you can get.

My question is, is there a fast way to calculate this? Or is it the old table where i check each possibility and then see how many times there is a chance to get that result, and then make the probablity table based on that?

[upper abyss]

eyes? 👀

[limpid zinc]

Sorry, im not native english - thats just what we call them here 😆

[upper abyss]

Like, faces?

Or, sides?

[limpid zinc]

Like this for example, this would be 6 eyes

So i guess yeah, one would have 4 sides and the other 8

[buoyant latch]

One has a total of 4 eyes

There’s 6 sides

[upper abyss]

I think you do need to count them, yeah. Like, the number of ways to make a 6 is (1,5) (2,4) (3,3) (2,4)

[buoyant latch]

So either you have fractional eyes or sides that are empty?

[limpid zinc]

Right, i think i've just typed it wrong. One has 4 black dots in total, and the other has 8 black dots in total

There is no empty sides 😆

[buoyant latch]

Those are 2 dice right?

[limpid zinc]

Yes

2 Different dice

[buoyant latch]

So 1 dice has a sum of 4

For all of its sides?

[astral bay]

so as in
one produces 1, 2, 3, or 4 with 1/4 probability each
one produces 1, 2, 3, 4, 5, 6, 7, or 8 with 1/8 probability each?

[limpid zinc]

Yes, what bee said

[buoyant latch]

Oh

That’s not a sum of 4

[astral bay]

tbh i don't really know what an "eye" would be in the context of a dice

this is an eye: 👁️
most living things have two of them and use them to see

[limpid zinc]

Yeah im sorry, that is what we call them here the black dots, lmao

[astral bay]

well these dice wouldn't have a total of 4 black dots

one side has 1, one side has 2, one side has 3, one side has 4, so that's a total of 10

[limpid zinc]

Yeah i explained it poorly

But yes what you said before, one produces 1, 2, 3, or 4 with 1/4 probability each
one produces 1, 2, 3, 4, 5, 6, 7, or 8 with 1/8 probability each
and i have to make a probability table over them for the different sums you can get

[buoyant latch]

,w expand (x+ x² + x³ + x⁴)(x+ x² + x³ + x⁴ + x^5 + x⁶ + x⁷ + x^8)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Frosst#1111

[buoyant latch]

That looks about right

[astral bay]

...yeah that looks right

[buoyant latch]

There’s 2 brackets for 2 dice

Each term represents the value on each side of said dice

The expansion represents the number of ways to get the sum of the 2 rolls

Because when you expand they multiply and the powers add

Which is the same as summing the 2 results from the dices

That’s the simplest way I can think of to represent the situation you described

[upper abyss]

Ah nice, I need to try to remember these

[limpid zinc]

Let me try that, cheers ❤️

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[crimson sedge]

i know this is simple but couldn't find much online

[fair geyser]

if log_b(m) = 3 and log_b(n) = 9

then m is bbb and n is bbbbbbbbb

so n is m cubed

and log_m(n) is 3

so you divide 8 by 3 in this case

it doesn't seem to be hidden online for me

[crimson sedge]

thank you for your time and help

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[tired jasper]

I was asked to find the maximum length of a metal rod that could fit into a rectangular prism with 1.6m length, 1.1m width and 1.75cm height. I got 194.16cm but it says its wrong. Any one can help? Its Pythagoras theorem

[tired jasper]

(Sorry if its messy)

[lethal jackal]

is the rod of negligible thickness?

[tired jasper]

They didnt draw it so im not sure

[lethal jackal]

also is the 1.75cm a typo?

that would be a really flat rectangular prism

[tired jasper]

Its 16b

I got 16a right but b and c both of them way off

[lethal jackal]

you know the answer for b?

[tired jasper]

Yes they said 2.61cm

[lethal jackal]

2.61 meters?

[tired jasper]

No cm

[lethal jackal]

that's a typo/mistake by whomever wrote it

I guarantee you I can fit more than that in that box LOL

also I suspect I know what they calculated

[tired jasper]

If it wasnt a typo, is there any way that answer would make sense?

[lethal jackal]

yes there is

if you assume that the sides are 1.1 cm and 1.6 cm and 1.75 cm

then the Pythagorean theorem gives you 2.61 cm

the correct answer for the problem given is 194.17 cm

which is basically what you got

[tired jasper]

If u were to round it up yeah?

[lethal jackal]

I get 1.94172 meters

I think you have rounding errors in your calculation that make it off by the 0.01 cm

[tired jasper]

I converted all of them to cm and got 194.164. Does that make a difference?

[lethal jackal]

not really

it's close enough

it's definitely not 2.61 cm

[tired jasper]

Is 16c correct too in that case? The answer to it was 47.27cm

[lethal jackal]

I can't see all of the numbers in 16c

[tired jasper]

But i got 43.98cm

[tired jasper]

My mistake

[lethal jackal]

I get 47.27

[tired jasper]

I went over it and realised i made a mistake, i ended up with 47.27 too

Thank u !

Ill just ignore 16b and continue 💀

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[idle storm]

Hi, I'm confused about the mark scheme for question 6b. It just says 0. If its a root, it still has a gradient right?

[round ridge]

Input x = 5 for the first part of the question

And p(x) = 0

For the second one, let me think

[crimson delta]

P(x)=(x-5)^2*Q(x) for some other polynomial Q

can you differentiate that

[round ridge]

Q is $2x^{2}-5x+(a-50)$

[wraith dagger]

RomeNunt

[crimson delta]

dont actually calculate Q

[round ridge]

In case for the third question

[crimson delta]

also dont just give out solutions

[idle storm]

i dont get b

[round ridge]

You're not supposed to find P(5) then differentiate the value

[idle storm]

i get a and c though

[round ridge]

You find P'(x) then input 5 into the function

Maybe that's the mistake

idk tho

[idle storm]

but i dont know a and b

like the values of a and b

[round ridge]

I don't think you need to for b

for 6b

[idle storm]

when you differentiate it its still there though

[round ridge]

I know

It's $8x^{3}-45x^{2}+2ax+b$

[wraith dagger]

RomeNunt

[round ridge]

Wait

After dividing x-5 twice from the equation

We get $10a+b-375=0$

[wraith dagger]

RomeNunt

[round ridge]

Why is that important? Well when inputting x as 5, we get 10a + b - 125

[idle storm]

thats a similar relationship to the one i found in part a

[round ridge]

From here $10a + b = 375$

[wraith dagger]

RomeNunt

[round ridge]

So 375 - 125 = 250

Answer to 6b is 250

I hope this helps

[cedar kiln]

@idle storm Has your question been resolved?

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[worthy ridge]

What is $o(a)$?

[wraith dagger]

normalAtmosphericPa=101,325

[worthy ridge]

https://math.stackexchange.com/questions/3006836/calculating-the-order-of-an-element-in-the-group-u-27

order?

[crimson delta]

order

the smallest positive integer k so that a^k=1

[gilded elm]

oh man

I really hope they don't denote it as that

$|a|$ is much better notation

[gilded elm]

also since when did we start allowing abstract algebra in help channels

[worthy ridge]

.close

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[meager glade]

what do i do once i get this log

[meager glade]

so far, i got the equation, y=1/2(x)+1

then

log4(y) = 1/2(x)+1 ?

[tame wraith]

yh, where log4(k) = 1, and log4(a) = 1/2

[meager glade]

but how do i make this into the form they want to find k=, n=

n= 1/2

but idk how to get k

[tame wraith]

wdym

your question is kinda cropped

[meager glade]

oh sorry

hold on

[meager glade]

my terminology in maths is terrible, i meant just to find k and n, not the form part

the "form" is really just a equation i think

[tame wraith]

that is clearly an a

[meager glade]

oops didnt realise, the last question was y=kx^n

heres this question tho

so i know the value for x, where do i go from here

[tame wraith]

you literally had it

[meager glade]

huh

(i'm terrible at logs)

[tame wraith]

$y = k a^x$ , so $log_4{(y)} = log_4{(k a^x)} = log_4{(a^x)} + log_4{(k)} = log_4{(a)} \cdot x + log_4{(k)}$

[wraith dagger]

doctor99268

[tame wraith]

you had $log_4{(y)} = 0.5x + 1$

[wraith dagger]

doctor99268

[tame wraith]

compare coefficients

[tame wraith]

^

[meager glade]

ohh i see

that was a great explanation btw

so now i got to burn the logs right

[tame wraith]

burn?

[meager glade]

yeah, thats the term i was teached anyway

perhaps could call it, getting rid of logs?

not sure of the proper term

basically getting it to, y=4^(0.5x+1)

[tame wraith]

sure

[meager glade]

well i got, k=4 a=2, thanks so much

.close

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[keen scroll]

Suppose I have an initial value problem. It satisfies the Picard's existence and uniqueness theorem on some closed rectangle $D$. Then, I have a unique solution in some neighborhood around my initial time point. Question: can I extend the solution to the whole time line in $D$ by taking another point from the neighborhood?

[wraith dagger]

271828

[keen scroll]

.close

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[tepid grail]

how do you use integration to find area of this part?

even tho it looks similar, but the line below y<7 is a different function to the line above y>7

the function for the line 0<=y<=7 is **4.7 tan⁡(0.293x+4.215)+15.18 **

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[latent fox]

how to expand

[stray otter]

Let the 3/x-2 be some alphabet like y first

[latent fox]

no

[stray otter]

wdym

[latent fox]

theres a better way than doing that

[stray otter]

then why are you asking if you know a better way

[latent fox]

i dont

but ik your way is very roundabout

and not entirely helpful

[stray otter]

well then just open the brackets

[quartz frost]

you can multiply the left fraction by (x-2)/(x-2)

also you can write 3 as 3(x-2)^2/(x-2)^2

that way you can already combine everything into one fraction

but i think that is what the diamond hoe wanted to tell you as well

[cedar kiln]

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[signal mauve]

can some one help me with this one? https://gyazo.com/1ae1ea17a98098888be0071ba9772435

[south tundra]

Would consider applying the double angle identity for cosine

[signal mauve]

this is what I have so far

[south tundra]

Right

[signal mauve]

the problem is how I find tanx = -1

Is there any smart way to do that?

[south tundra]

Do you want all of the infinitely many solutions or just some interval?

[signal mauve]

ohh yeah, the interval is 0 degrees and 360

betweeen^

[south tundra]

tan is equal to 1 at 45 degrees

[signal mauve]

yes

[south tundra]

So it would be -1 at -45 degrees

Here you can add 180

Because it is the period of tan

Therefore tan is -1 at 180-45degrees

Or 135 degrees

[signal mauve]

how do I know that -1 is eqqual to -45

degrees?

[south tundra]

.

If tan(x) = y, then tan(-x) = -y

tan is an odd function

[signal mauve]

ok

[signal mauve]

I agree

what do I do next

[crimson sedge]

do you have a domain anything?

that restricts your answer to one value?

or two

[south tundra]

From 0 to 360

[south tundra]

Read what I said again, I provided you with an answer

[signal mauve]

tan x is either 135 or 215 degreees?

[crimson sedge]

so just π-π/4

and 3π/2+π/4

that should solve it

[crimson sedge]

do you know what the unit circle is?

[signal mauve]

yes

[south tundra]

215 isn't correct

[crimson sedge]

use that then

why aren't you using that mate?

[signal mauve]

tan(-45) = -1

[south tundra]

Right

[signal mauve]

so x = -45 + k*180

[crimson sedge]

yes

[signal mauve]

x = 135

[south tundra]

Now try k = 1 and k = 2

[signal mauve]

and 315

[crimson sedge]

good

[south tundra]

Correct

[signal mauve]

Ait I got it now

thx guys

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[stray otter]

But why not this?

[stray otter]

I thought this is the integration rule?

[cerulean sail]

Differentiate ln(cos(x)) (without forgetting ⛓️)

[stray otter]

Ok

Oooh

Ok

I see

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[heady salmon]

is there a short cut to doing this ?

[fair mortar]

$a^2 - b^2 = (a+b)(a-b)$

[wraith dagger]

bettim

[heady salmon]

i know how to factor, is there a way to simplify this equation faster?
cant i just square each term inside the parathesis?

[fair mortar]

no

$(a+b+c)^2 = a^2 +b^2 +c^2 + 2ab +2ac+ 2bc$

[wraith dagger]

bettim

[heady salmon]

k i'll try this out

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[gloomy verge]

if looking for a derivative of a function that has a given x value does it mean the answer is automatically 0?

[gloomy verge]

for example find the derivative of y=(X^3+9)^17/2 at x=-2

[crimson delta]

no

first take derivative, then plug x in

derivatives would be pretty pointless if that always gave 0

[gloomy verge]

ok thanks man

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[quartz salmon]

Hello everyone,

I am working on a wind power problem and need some help. Here is the problem:

During a windy day, the wind speed at a wind turbine can be described by the model:

v(x)=11sin(0,11x−0,89)+28,0≤x≤24
where $v$ is the wind speed in km/h and $x$ is the time in hours from midnight.

a) I need to find the maximum wind speed during the day.

b) I also need to determine how long the wind speed is over 36 km/h during the day.

c) Finally, I need to find the total amount of electrical energy that the wind turbine produces during the day.

I have attempted to solve this problem by finding the derivative of the function and setting it equal to zero to find the maximum wind speed. However, I am not sure if this is the correct approach.

For part b, I have tried integrating the function over the range where the wind speed is greater than 36 km/h, but I am unsure if this is correct.

For part c, I am planning to use the provided equation $P(v)=0,42 \cdot v^3$ to calculate the electrical energy produced and then integrate over the range of wind speeds.

I would appreciate any guidance on how to approach this problem and any corrections or improvements to my current attempts.

Thank you in advance for your help.

[wraith dagger]

dgh

[quartz frost]

for part a), you want a maximum

do you know how to get those?

[quartz salmon]

This can be done by taking the derivative of the function, setting it equal to zero, and solving for x. The resulting value of x can then be substituted into the original function to obtain the maximum wind speed.

[quartz frost]

👍

make sure to check the boundaries of the interval as well though

[quartz salmon]

$$v'(x) = 1.21\cos(0.11x-0.89)$$
We set v'(x) equal to zero and solve for x:
$$v'(x) = 0 \implies \cos(0.11x-0.89) = 0$$
$$\implies 0.11x-0.89 = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots$$
$$\implies x = \frac{0.89+\pi/2}{0.11}, \frac{0.89+3\pi/2}{0.11}, \frac{0.89+5\pi/2}{0.11}, \ldots$$
Note that we only need to consider the solutions that lie in the interval [0, 24].

[wraith dagger]

dgh

[quartz salmon]

Is this right

$$v(0) \approx 11.0$$
$$v\left(\frac{0.89+\pi/2}{0.11}\right) \approx 32.2$$
$$v\left(\frac{0.89+3\pi/2}{0.11}\right) \approx -32.2$$
$$v(24) \approx -11.0$$
We see that the maximum wind speed occurs at $x \approx 1.77$ hours after midnight, and is approximately 32.2 km/h.

[wraith dagger]

dgh

[quartz frost]

looks good

normally, we also verify if we got a maximum or minimum with the second derivative
but that is not needed here, since you compared it to other values and it is clear that 32.2 is indeed a maximum

hmm

i got something different

v(x)=11 sin(0.11x-0.89)+28
our x was x=(0.5pi+0.89)/0.11

the 0.11 and /0.11 cancel and then the 0.89-0.89 cancel

v=11 sin(pi/2)+28

sin(pi/2)=1

v=11+28=39

that is pretty important, since for part b) we want the interval for which v>36

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[candid granite]

Vu

[candid granite]

Im french my english is a little bad
I need help on arithmetic and geometric sequence, I don't know how to adopt which formula for his exercises and how to find if they are arithmetic or geometric

An exercice like that the 82

How i do if its arithmetical which formule i use

If one person can repond if i have a reccurence and i need to prove that is arithmatic i need to do vn+1-vn

If i need to prove that is geometrical i need to do vn+1/vn

[cedar kiln]

@candid granite Has your question been resolved?

[cedar kiln]

@candid granite Has your question been resolved?

[gloomy lynx]

Exact, si v(n+1) -v(n) ne dépend pas de n, c'est arithmétique, si v(n+1)/v(n) indépendant de n, c'est géométrique. La relation 2 du 82 c'est aucune des deux formes

[candid granite]

In my teacher correction the firsy its geometrical

Cause of vn+1 - vn the raison dont change

[candid granite]

Ty

@cedar kiln

<@&286206848099549185>

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[tawdry crater]

I'm confused on the conditions for this question, how is it possible to cut any shape in a manner such that both resultant shapes are similar to the original?

[tawdry crater]

I think I'm understanding the question wrong

[buoyant latch]

If you cut an a4 paper down the middle lengthwise

They become A5 would has the same ratio of sides as an A4

[tawdry crater]

I'm not sure what a4 and a5 means

lmao

[velvet mortar]

Kids these days ..

[buoyant latch]

You don’t know what A4 paper is?

[tawdry crater]

it's a paper size right?

[buoyant latch]

It is

What country are you in

[tawdry crater]

the U.S

[buoyant latch]

Printer paper

[tawdry crater]

do you have a drawing or something haha

I promise I'm not trolling I don't have a printer

[buoyant latch]

Ok what do you know about paper sizes

[tawdry crater]

okay so you're saying if we take a rectangle and cut it through the middle you get the same ratio of sides?

[buoyant latch]

Ok let’s make it more simple

[buoyant latch]

No

That property is special to the A4 paper ratio specifically

Another example is a right angled triangle with the same side lengths

Take a square, cut it along the diagonal

Now this triangle you have

If you cut from the right angle to the hypotenuse

You’d get another right angled triangle

[tawdry crater]

okay

so both of the smaller right-angled triangles are similar to the original right triangle right

[velvet mortar]

Which software are you using?

[tawdry crater]

is what we're getting at?

[buoyant latch]

Doodle

[buoyant latch]

Yes

I’m just giving you examples where you can cut the shape and get 2 same shape but smaller

You should notice for your question you must have 270deg corners

Since if you just put 4 90deg together you get a square

Not a hexagon

[tawdry crater]

yeah yeah I understand that

[tawdry crater]

just thinking about this for a sec

[buoyant latch]

A better picture if you will

[tawdry crater]

wow okay yeah that makes sense LOL

alright cool that works tyyyyyyy

for some reason that was super unintuitive to me lol

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[velvet ferry]

stuck with a

[velvet ferry]

this is what i have so far

[vast pike]

the last fraction its wrong ... should be the inverse

[velvet ferry]

ty

[vast pike]

idk if this could help..

[velvet ferry]

if any helpers see this is my general solution

[cedar kiln]

@velvet ferry Has your question been resolved?

[velvet ferry]

<@&286206848099549185>

[cedar kiln]

@velvet ferry Has your question been resolved?

[cedar kiln]

@velvet ferry Has your question been resolved?

[velvet ferry]

<@&286206848099549185>

[cedar kiln]

@velvet ferry Has your question been resolved?

[cedar kiln]

@velvet ferry Has your question been resolved?

[crimson sedge]

anyone there?

[cerulean sail]

Looks like you did a lot more than part a (though the characteristic equation looks to be miswritten, though the factored form seems to agree with what it should be)

,w expand (x-4)(x-3)^2

[wraith dagger]

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Requested by chartbit#3171

[cerulean sail]

That’s all what part a wanted, no?

[cedar kiln]

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[copper crest]

Is rate of change of displacement and rate of change of position the same thing

[copper crest]

If yes/no...how

Xf -Xi /t
And
(Xf'-Xi) -(Xf-Xi) /t = (Xf'-Xf) /t

[copper crest]

They're not the same ryt

[bold lotus]

no

rate of change of displacement is velocity and rate of change of distance is speed

im assuming by position you mean distance

[copper crest]

No by position I meant position vector

xi+yj+zk

[bold lotus]

displacement is a position vector

so then they are the same

[copper crest]

Displacement is change in posN vector

[bold lotus]

displacement is the vector showing how far you from the origin

then assuming the origin is 0i +0j + 0k then we get the same

[copper crest]

Change in displacement is change in change of position vector

[bold lotus]

what

i think we are working from different definitions

[tame wraith]

yh they should be the same thing

rate of change only cares about change

[copper crest]

Xplain

[copper crest]

Oh kk

[copper crest]

That's position vector ryt

[tame wraith]

displacement is a vector, so is position vector

change in displacement

should give the same vector

as change in position vector

[cedar kiln]

@copper crest Has your question been resolved?

[copper crest]

How

[copper crest]

Agreed

[tame wraith]

im pretty sure the only difference between those two is the reference point

which shouldnt matter when we are doing change

[soft flax]

finally found you

[cedar kiln]

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[fair mortar]

let $v=2^{\sin x}\
\log v = \sin x \log 2\
\frac{1}{v} \frac{dv}{dx} = \left(\frac{\sin x}{2} + \log 2 \cos x\right)\
\frac{dv}{dx}= 2^{\sin x} \left(\frac{sin x}{2} + \log 2 \cos x\right)$

[wraith dagger]

bettim

[fair mortar]

this is what i did

[tame wraith]

when you differenitated

sinx log2

you do know log2 is just a number

[fair mortar]

so constant?

[tame wraith]

yh

[fair mortar]

thanks

[tame wraith]

also you couldve just used chain rule for this anyway

[fair mortar]

so its just $\log 2 \cos x$