help-10

[languid valley]

with cosh or smth

[kind hawk]

not that weird

[languid valley]

lemme check I didnt even check

oh...

I see

it's a lot easier than I thought lol

well that makes me be able to solve the rest

thanks

.close

[obtuse pebble]

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[glad parcel]

help pls

[brittle blaze]

So if the number of 10p coins is n and the number of 50p is 5 more than the number of 10p. Its 5 more than n

[glad parcel]

also

[brittle blaze]

Set up an equation. 60 coins in total and we worked out an expression for the number of each

[glad parcel]

ok

n=19?

[brittle blaze]

Then you can solve for n and use that to find the value

[glad parcel]

ok

[brittle blaze]

Yeah n seems to be 19

So plug that into the expression to see how many 20p coins there are

And then multiply whatever you got by 20 bc it wants value not number of coins

[glad parcel]

17 x 20

is that right?

3.40p

[brittle blaze]

Is p pounds? But yeah that seems right

[glad parcel]

sorry, but also this

[obtuse pebble]

@glad parcel Has your question been resolved?

[glad parcel]

.close

[obtuse pebble]

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[pallid grail]

can someone please help me with my math

[hexed agate]

This is a math help channel what do you think?

[pallid grail]

(7^2)^5/{(7^x)(7)}^3

this is my question

[hexed agate]

$(7^2)^5/{(7^x)(7)}^3$

[warm shale]

Diegoro

[hexed agate]

This?

[pallid grail]

^ this symbol repersents exponent

[pallid grail]

yes

[hexed agate]

Do you know exponent rules?

[pallid grail]

no

[warm canopy]

$\frac{(7^2)^5}{((7^x)(7))^3}$

[warm shale]

ΣAC

[pallid grail]

I just started this unit and this is my homework which I got stuck on

[pallid grail]

yes it's exactly this one

[pallid grail]

ok

[hexed agate]

First what can you do with (7^2)^5)

?

[pallid grail]

mutiply 5 by 2

[hexed agate]

Yes

So it becomes ?

[pallid grail]

which gives us 7^10?

[hexed agate]

Yes

Now what's (7^x) * (7^3)?

[pallid grail]

is it 7^3x and 7^3?

for the bottom part

[hexed agate]

How?

[pallid grail]

an then you subtract 7^10 by 7^3?

[hexed agate]

(7^x) * (7^3) = 7^(x+3)

Get it?

It't the product rule in the chart I sent

[pallid grail]

ohhhh okkk

i see it now

yes

[hexed agate]

(7^10) ÷ (7^(x+3))

What can you do here?

[pallid grail]

subtract 10 by 3?

[hexed agate]

What about the x?

[pallid grail]

i think im stuck here

[hexed agate]

(x+3) is a power too

[pallid grail]

ok

[hexed agate]

7^(10-(x+3))

7^(10-x-3)

7^(7-x)

Get it?

[pallid grail]

yes

.close

[obtuse pebble]

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[primal jewel]

I have to simplify terms but dont understand how the result came to be. What rules were applied and how?
What happened to the "-2" from the bottom "s^b-2" and the positive "r^3*a+b" in the bottom in the last step

[tardy epoch]

1 + a + b - (3a + b)

b - (b - 2)

calculate those two and you'll see

[primal jewel]

okay i get now how the result came to be. By just subtracting them. But why? Is there a rule for when i subtract a bottom term with a top term ?

[tardy epoch]

https://www.cuemath.com/algebra/exponent-rules/

[tardy epoch]

you should have all these memorized

[keen cave]

by doing more problems that involves using it.

[drowsy girder]

Not memorised more like know what they mean and from where they come from

[tardy epoch]

i hear what you're saying. and i just disagree

[brazen saddle]

sigma male music starts playin

[drowsy girder]

Memorising math is a very bad idea

[brazen saddle]

Giving away ur race, RACE REVEAL

[keen cave]

not really, just do more problems.

[tardy epoch]

bad idea for what? grades? i don't think so

[drowsy girder]

Bad idea for actually understanding it

[tardy epoch]

i think doing more problems leads to memorizing

[fleet parcel]

Memorising and understanding aren’t mutually exclusive

😭

[tardy epoch]

Feel free to convince salber to be able to derive each of the exponent rules in the list and understand it

[primal jewel]

i know most of these bust still fail to see which of those rules state what i can do in my problem.

[drowsy girder]

If you dont care about understanding math you might as well just use wolfram alpha. I mean you will get good grades , right?

[alpine raven]

are you sure

[tardy epoch]

first and last one

[brazen saddle]

The logic is to be understood obv , but the the notation are simply just memorization work , and they are there purely for everyone to be able to understand

[tardy epoch]

there are some things you don't need to understand and better off memorizing. this is one of them.

[tardy epoch]

actually just the first one

[primal jewel]

ah great. i overlooked that. Thx for the help

[brazen saddle]

Ban the ÷ sign , it's useless

And it causes confusion sometime

[primal jewel]

.close

[obtuse pebble]

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[drowsy girder]

I agree with notation part you should just accept it

But this is purely logical

[obtuse pebble]

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[half perch]

what is the method to draw these constraints on a euclidian plane?

[half perch]

they are constraints on the feasbile region.

c2 i made a function x2=x1-6 but where do i cut if that function?

the c1 i turned into a function $x2=(2^(1/2)*(8 - x1^2)^(1/2))/2$

and then I should just plot them?

[warm shale]

Hamza

[half perch]

<@&286206848099549185>

I get this

but it is get plot

[obtuse pebble]

@half perch Has your question been resolved?

[obtuse pebble]

@half perch Has your question been resolved?

[half perch]

I drew it on python but I would like to know the correct way:

<@&286206848099549185>

[kind hawk]

the first equation is $x_1^2 + (x_2+4)^2 \leq 16$, that's the equation of a circle with radius 4 cented at (0, -4)

[warm shale]

Denascite

[kind hawk]

the second is just some half plane. $x_2 \leq x_1-6$, so everything below the line $x_1-6$

[warm shale]

Denascite

[half perch]

.close

[obtuse pebble]

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[timid silo]

This is from Paul's Notes:

Could we say that n is a negative number (n element of set Z-)
and then write the formula without negative sign?
Is there any difference?

[hazy marlin]

Nope

you definitely can

It's how its defined.

[timid silo]

So, it's a definition, and I can't change it.
But I can say it this way and the meaning would be the same, right?

[obtuse pebble]

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[tawny ore]

when i tried to solve this i did the exact same thing but instead of solving it as a quadratic for y i solved it as a quadratic for x and it stumped me

how can we determine for which one to solve?

[timid silo]

\int (e(x-3))/((x-1)^(3))

[timid silo]

k

[tawny ore]

never mind it works both ways

.close

[obtuse pebble]

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[tawny ore]

.reopen

[obtuse pebble]

✅

[tawny ore]

it doesnt work both ways lol

heres the whole solution (ask me to translate if you need context)

when you solve it a quadratic for x you cant complete the square like this

because you get -8y+1=λ^2

[obtuse pebble]

@tawny ore Has your question been resolved?

[tardy epoch]

you might as well translate so a future helper can answer your question without asking you

[tawny ore]

it pretty much explains that since we are searching for integers (since its asking to solve the equation in Z*) the discriminant(not sure if thats the right translation) for the quadratic has to be a perfect square, thats why it sets x(x-8) equal to a perfect square (λ^2)

i understand the solution but my only question is how do we decide if we are going to solve a quadratic for x or for y

[tardy epoch]

it really doesn't matter, just do it both ways and make sure you get the same answer

[tawny ore]

But i can't really create a square out of -8y+1=κ²

[obtuse pebble]

@tawny ore Has your question been resolved?

[obtuse pebble]

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[livid oasis]

I really need help with these questions and if i dont get then right i fail my course <@&286206848099549185>

this is calc btw

[obtuse pebble]

@livid oasis Has your question been resolved?

[livid oasis]

anyone there?

<@&286206848099549185>

[dusty flume]

what kind of help do you need?

[livid oasis]

i need help with the question i provided

[dusty flume]

yeah, but what kind of help

[livid oasis]

the answers

desperately

[dusty flume]

well, just giving the answers won't do

have you even tried to solve them yourself?

[livid oasis]

yes

but i gave up

[dusty flume]

then, which parts give you trouble?

[tardy epoch]

No one's just gonna give you answers. Attempt your problem and ask when you get stuck

[livid oasis]

and this is a assignment and if i fail it a fail the year

[dusty flume]

well, then better get to work

[livid oasis]

u wouldnt understand

[tardy epoch]

Yea guilting people isn't gonna work here

[dusty flume]

we can help you understand the question and guide you towards a solution

but just giving away answers to pass a course is not allowed

[livid oasis]

i am saying factual information

u do not know the situation i am in and i do not expect that of you

[dusty flume]

that is correct

[livid oasis]

it is

[dusty flume]

but this is a math help group, not academic counselling

[livid oasis]

fucking hell

[dusty flume]

that's the attitude

now question 1, what does it ask for?

[livid oasis]

i will send what i have done up to 2,a,1

and just couldnt get my head around the next parts

[dusty flume]

so 2b?

[livid oasis]

well 2,a,11

onwards

wait i havent sent all the phots

photos

ah well nvm

wait no i have just in the wrong order

[dusty flume]

so what's the problem

do you know how to compute the numerical integration?

[livid oasis]

my problem is that i am struggling on how to write a comparason between awnsers 2a1 and 2a11

[dusty flume]

well, you may find that the difference is very big or very small, and maybe try to see why

[livid oasis]

i am stuck

[dusty flume]

okay, so step by step

are the answers (more or less) the same?

[livid oasis]

not really

[dusty flume]

for analytic vs numerical integration

well, then we have one considerably bigger than the other, right?

[livid oasis]

yeah

now the reasoning for that is hmmm

more presision?

[dusty flume]

well, that's more or less it

by the way, which one is bigger for you

analytic or numerical

[livid oasis]

analytic

[dusty flume]

that sounds correct

can you get an idea of how this function would look like?

[livid oasis]

not really tbh

[dusty flume]

well, we have an exponential which usually affects the most, right?

and then we are multiplying by some things

[livid oasis]

yes

[dusty flume]

you can see that the t in the exponent is multiplied by -3000, a pretty big (or small if you want) number

so almost everywhere, this will be close to 0, right?

[livid oasis]

right

[dusty flume]

also, for t=0 we multiply by 0 so it is exactly 0

but what happens when the exponent is close to 1?

in that case, the function is not zero, and the exponential part does not really affect. and since it is all multiplied by a large constant, it gets quite big

so in short,

it's zero at 0, almost 0 in nearly all of the segment, but very big in a small section

this is why the numerical solution is very small

[livid oasis]

would that be a good comparison

[dusty flume]

where are the "sample points" of your numerical integration?

[livid oasis]

idk

[dusty flume]

well, we have 5 points between 0 and 10

then no points would be near the "big spike"

[livid oasis]

ok

[dusty flume]

in short, the comparison is just the analytic integral is way bigger than the numerical integral or something like that

the rest was to get an idea of why this can happen

[livid oasis]

ok

ty

[dusty flume]

and you were mostly correct knowing that it is because of accuracy

we would need a lot more sample points to have a good approximation

with this kind of function

now are you able to use computer software for numerical integration?

or is it given to you or something

[livid oasis]

idk but ty i am going to sleep now

[dusty flume]

good night!

[obtuse pebble]

@livid oasis Has your question been resolved?

[obtuse pebble]

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[limber latch]

I tried doing T= i / p * r. But it dosent give me a proper number

[limber latch]

t = 116.67/(5000*0.035) = 0.666685714

<@&286206848099549185>

[tardy epoch]

!15min

[obtuse pebble]

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.

[limber latch]

bruh is it cause interest is generalized in years? like that 3.5% is for the entire year

i did 0.035/12 and it worked, well re-added that part into the equation

[tardy epoch]

right

interest rate is always annual

so when you calculate "in months", you divide the rate by 12 months

[limber latch]

thanks boss

so it would be the same for this but instead they ask for principal

damn it was suppose to be 11/12

so it isint the same

wait it dosent matter

0.095/12 or 11/2 works

im confused now so i can divide anyone by 12?

[obtuse pebble]

@limber latch Has your question been resolved?

[obtuse pebble]

@limber latch Has your question been resolved?

[lucid smelt]

I notice this one doesn't give the compounding frequency, nor does it provide a way to infer it

[obtuse pebble]

@limber latch Has your question been resolved?

[obtuse pebble]

@limber latch Has your question been resolved?

[obtuse pebble]

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[timid silo]

On the average, how far are the heights of the teachers from the average height?

[fading birch]

What's the teacher's height?

[timid silo]

there's the table

[royal basin]

you are asked to find the mean absolute deviation

[timid silo]

158.8?

[royal basin]

how did you get that?

||that sounds like the mean itself, not the MAD that i mentioned||

that sounds like the mean itself, not the MAD that i mentioned

[timid silo]

oh yeah it is the mean itself because it says 'average height'?

wait a sec, ill get the MAD

[royal basin]

you were asked
on average, how far are the teachers' heights from the mean

[timid silo]

oh yeahh, wait a sec

[timid silo]

well, I keep getting 7.5. im pretty sure it's not right.

[royal basin]

how are you sure it's not right?

it looks reasonable to me, but maybe you did mess up your calculations

[timid silo]

is the MAD supposed to be the actual answer?

[timid silo]

I know for a fact that I did get the MAD right but it being the answer is quite vague.

[royal basin]

the MAD is what you were asked for

[timid silo]

it have to be 5. something--the answer

[timid silo]

oh

[royal basin]

why does it have to be 5.something? how do you know it has to be that?

[timid silo]

look at the table

• the book gave a hint haha 😉

[royal basin]

i fail to see what part of the table implies the answer has to be between 5 and 6...

what hint was given in the book?

[timid silo]

ive been going back and forth between all the formulas under frequency distribution but I still can't see how too

[timid silo]

wait a min

it says it's between 5.58 cm and 5.36 cm

[royal basin]

what says what?

[timid silo]

it's an after reading question

the book says the answer is between 5.58 cm and 5.36 cm

after you read the lesson

[royal basin]

right...

well let's see

let's go through the computations

[timid silo]

alright, i appreciate it

[royal basin]

,calc (172 * 1 + 167 * 5 + 162 * 4 + 157 * 8 + 152 * 6 + 147 * 1)/25

[warm shale]

Result:

158.8

[royal basin]

so this is the mean

[timid silo]

the average, yeah

[royal basin]

the deviations of each class from the mean are...

,calc 172-158.8

[warm shale]

Result:

13.2

[royal basin]

13.2, 8.2, 3.2, 1.8, 6.8 and 11.8

,calc (13.2 * 1 + 8.2 * 5 + 3.2 * 4 + 1.8 * 8 + 6.8 * 6 + 11.8 * 1)/25

[warm shale]

Result:

5.36

[royal basin]

so that is 5.36

idk how you got 7.5

[timid silo]

There it is

instead of 25, i divided it by 6?

wow, thank you so much!

[royal basin]

that would not have made such a small discrepancy tbh

[timid silo]

i did this

,calc (13.2 + 8.2 + 3.2 + 1.8 + 6.8 + 11.8 )/6

[warm shale]

Result:

7.5

[timid silo]

lol im so dumb

sorry for the word

thank you so much!!

ive been here for like an hour

[royal basin]

ah.

so you didn't take the freqs into account

[timid silo]

yeah...

im really sorry

anyway, thank you again

I'll close this one now, alright?

youre a life saver

.close

[obtuse pebble]

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[obtuse pebble]

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[sterile trellis]

need to find 2nd derivative

[hexed agate]

find the first derivative first

[sterile trellis]

ye but too complicated

How to use smart method?

[hexed agate]

do you know the quotient rule?

[sterile trellis]

yes

[grizzled shore]

what do you mean it's too complicated

there's no way to the best of my knowledge you can skip the first derivative

[sterile trellis]

O

let me try

[hexed agate]

whats the derivative of the numerator?

[tranquil arch]

you an do partial fraction decomposition first

,w Apart[(x^3 - x^2 - 2 x + 3)/(x - 1)^2]

[warm shale]

Results provided by WolframAlpha

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Requested by 秋水#5244

[eternal fiber]

u cud split it

[eternal fiber]

like this

[sterile trellis]

the first derivative is like (x^3-3x^2+4x-4)/(x-1)^3

o yes

[grizzled shore]

you still need to ifnd first derivative to get 2nd though

[sterile trellis]

This help alot,

let me try

[obtuse pebble]

@sterile trellis Has your question been resolved?

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[agile skiff]

intergral xe^(-x^2) dx can be solved by both intergral by parts and subsitution

[agile skiff]

Why you people chose subsitution method

why not intergration by parts

[kind hawk]

seems easier

you substitute and can immediately integrate something very simple

wait how do you do it by parts

you diff the e^-x^2 and integrate the x?

what good does that do

[agile skiff]

like by formula

intergal uv = uv -intergral v du

[agile skiff]

uv or udv?

[kind hawk]

udv

so you have to integrate one of the factors and differentiate the other

here e^-x^2 is hard to integrate so we have to differentiate it

but that means we have to integrate x

and then with each iteration we just get more and more powers of x

and an integral of the form x^n e^-x^2

just getting more complicated

[agile skiff]

but can be done

please ping me iam learning, sorry for late replies

[kind hawk]

I am not sure if it can be done and actually get a solution

if yes, it's definitely more complicated than just u-sub

[agile skiff]

so i need to think if the problem is solvable by subsitution if not to by parts

[kind hawk]

subbing u=x^2 gives a very easy solution

because du=2xdx and we already have the x factor in the integral

so we are just left with integral 1/2 e^-u du

which is very easy

[agile skiff]

ok but

[agile skiff]

this approch is ri8

[agile skiff]

I would appreciate if you can write formula for intergal uv in latex because many people you different notation

[kind hawk]

the formula many people use is $\int u dv = uv - \int v du$

[warm shale]

Denascite

[kind hawk]

if you want to integrate $\int uv dx$, then that looks slightly worse

[warm shale]

Denascite

[kind hawk]

$$\int uv dx = u \left(\int v dx\right) - \int \left(\int v dx\right) \frac{du}{dx} dx$$

[warm shale]

Denascite

[kind hawk]

you integrate v and you differentiate u

you get the other formula by setting $g=\int v dx$, then $v=\frac{dg}{dx}$ and so
$$\int uv dx = \int u \frac{dg}{dx} dx = \int u dg$$
on the left side (where we "cancel" the $dx$ suggestively so it looks better) and
$$u \left(\int v dx\right) - \int \left(\int v dx\right) \frac{du}{dx} dx = u g - \int g du$$
on the right side (where we again "cancel" the $dx$)

[warm shale]

Denascite

[kind hawk]

the important part is that we integrate one of the factors and differentiate the other

@agile skiff

[agile skiff]

Now it is extremely complicated

this form one example

[kind hawk]

we integrate the dv to get v

and we differentiate the u to get du

[agile skiff]

ya

[agile skiff]

like example sum for this would help

@kind hawk Thank you so much for your help,

One thing when to use Subsitution and by parts

[kind hawk]

that comes with experience

we generally use by parts if there is one factor which is easy to integrate and that integration plays "nicely" with the rest of the integral. and also the other factor has to be nice to differentiate and we want that process to end at some point in some form

for example for integrals of the form x^n sin(x)

here sin(x) is easy to integrate

and by differentiating x^n we will eventually get the derivative 0

and the process will stop

we use substitution if we have some "inner function" that looks complicated and maybe have the inner derivative of that already in our integral

so for integral x e^(-x^2) the inner part -x^2 of the e^() part is complicated. but the derivative -2x is already nearly there in our integral because we have the factor x

this would be the same for example for integral x^3 sin(x^4)

the x^4 in the sin is complicated. but the derivative is 4x^3 and we already have a factor of x^3

so we would u-sub u=x^4 and then go from there

sometimes it's not easy to see which one of the approaches will work

and you may have to try several different things (using a different substitution or integrating a different factor)

or often you will also have to combine both tactics

all of that comes with experience

@agile skiff

[agile skiff]

@kind hawk thank u so much

You are such a great guy

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[scarlet locust]

$(I+A)^3 = I^3 + A^3 + 3I^2A + 3A^2I$ (Binomial theorem holds as IA = AI)

[scarlet locust]

is this reason correct?

[warm shale]

Swas.py

[kind hawk]

yes

if you want you can slowly multiple out (I+A)^3=(I+A)(I+A)(I+A)

[scarlet locust]

yeah I know its true

[kind hawk]

it's true for all A,B such that AB=BA

[scarlet locust]

alright

it's still called the binomial theorem right

[kind hawk]

yeah

[scarlet locust]

alright

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[timid silo]

how do i make it so that the peaks and troughs are in intervals of 1

[timid silo]

like my wording is kinda bad but i want the value in the screenshot to be 1,1

[scarlet locust]

when x=1, sin(ax) = 1

[timid silo]

i thought thats what sin was to begin with btw

[scarlet locust]

solve that

for a

found a?

[timid silo]

i guess a is 1/(pi/2) or something

but like

[scarlet locust]

no

[timid silo]

why is it like that

[scarlet locust]

just pi/2

sin(a * 1) = 1
sina = 1
a = arcsin1
a = pi/2

so is your function

[timid silo]

what is "arcsin"

[scarlet locust]

inverse function of sin

[timid silo]

im kinda dumb but is that cos

[scarlet locust]

no

if sinx = y then x = arcsiny

[timid silo]

oh

[scarlet locust]

also written as $sin^{-1}$

[warm shale]

Swas.py

[timid silo]

so inverse sin is how i was taught to say it

[scarlet locust]

thats correct too yeah

just diff names

[timid silo]

alright thank you for help

i swear to memory

sin1 is 1

maybe im tripping

[scarlet locust]

sin(pi/2) is 1

sin1 is not 1 in radians nor in degrees

[timid silo]

alright

[scarlet locust]

sin1 is close to sin(pi/3)

3.1415../3 is about 1

[timid silo]

circles are crazy man

[brazen saddle]

||Sin(pie(x)/2)||

[timid silo]

alright

thank you

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[hexed trail]

Hey is there any way to tell b coordinates knowing all angles, lengths and sin of triangle xya?

[visual onyx]

yea

I think so

[hexed trail]

I think so too, I don’t know how tho

[brazen saddle]

No

[visual onyx]

but what is the relationship of b and a

[hexed trail]

I want it to be certain distance away from a

[brazen saddle]

Oh yeah then it's possible

[hexed trail]

Oh : P

So how could I do that

[brazen saddle]

How far would u like it?

[hexed trail]

Anything would do

[brazen saddle]

Well however far from a

Let's say at a distance k

Also the distance is to be measured on the circle or purely from a?

[hexed trail]

On the circle

[brazen saddle]

Okay so let's say it's k distance away measuring on the circle

And the line xb makes an angle £ with x axis

And all angles are in radians

So k = r(£-theta)

With me?

[hexed trail]

Give me a second to process that : P

[strange stump]

Most probably

[obtuse pebble]

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[thick oracle]

A student reads a book page on the first day of vacation. Then he reads double the pages of previous day every single day. After how many days did the student read 1,023 pages?

[real horizon]

you can calculate the number of pages the student has read after the first few days

[thick oracle]

ok

i got it

thx

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[timid silo]

How these two are same?

[compact shadow]

Just calculate those determinants

[timid silo]

Got it.

Thank you 🙂

@compact shadow

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[hollow oar]

Hi, may someone help with this - part b. I get that part a looks like this

[hollow oar]

but im really stuck for h in terms of theta

i was thiking it might have something to do with SOHCAHTOA

but looking at it again, it is more likely to be a graph transformation

but ima not too sure

[scarlet locust]

this is correct

[hollow oar]

yup, that is for part a which I managed to get ok

but its' part b that ima struggling with

[scarlet locust]

alright

[scarlet locust]

it would indeed be a transformation of sin

[hollow oar]

yh, that is what i was thinking but i haven't the best idea on how to do this, i thought it was a sin transformation as well, but looking at teh answer:

I'm not too sure how this works

[scarlet locust]

cos is a transformation of sin

so thats not an issue

[hollow oar]

oh, so 1-cos theta = sin theta?

[scarlet locust]

no

sin(90-theta) = cos(theta)

or cos(90-theta) = sin(theta)

[hollow oar]

aight

what does the number outside the bracket do to the graph? does it 'squish' it?

[scarlet locust]

instead of understanding the answer wouldn't it be better to derive it yourself?

its easy to derive

[hollow oar]

ok, i agree

what do you think the best way to start tackling the q is?

[scarlet locust]

construction

[hollow oar]

srry, what is the symbol on the hypotenuse?

[scarlet locust]

r

its the radius

[hollow oar]

kk

[scarlet locust]

now from CAH,
cost = (r-h)/r
rcost = r-h
h=r-rcost
h = r(1-cost)

see now?

[hollow oar]

ima read it through now

[hollow oar]

i get it all up to h = r(1-cost), how did you go from h = r-rcost to h = r(1-cost)?

nvm

[scarlet locust]

factoring r

[hollow oar]

i get it srry ye

dumb ques

[hollow oar]

ty thanks very very

much

[scarlet locust]

[obtuse pebble]

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[mental otter]

I'm having trouble with octave and the bisection method algorithm for tridiagonal symmetric matrix. I do know how to do bisection for nonlinear functions but that's about it as far my knowledge goes with bisection.

[obtuse pebble]

@mental otter Has your question been resolved?

[obtuse pebble]

@mental otter Has your question been resolved?

[mental otter]

<@&286206848099549185>

I am sorry for pinging but the bot pinged me so... 💀

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@mental otter Has your question been resolved?

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@mental otter Has your question been resolved?

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[sterile pewter]

How did they get this answer?

[tardy epoch]

do you know parametrization

[sterile pewter]

no i don't yet

[tardy epoch]

Do you know what dot product is?

[sterile pewter]

yes

[tardy epoch]

take F and dot it with $dr = (dx, dy)$

[warm shale]

riemann

[sterile pewter]

is $dx = \frac{3x^2-2}{2dy}$ and $dy = \frac{3}{2}x^2-1dx$

[warm shale]

Heavy

[tardy epoch]

naaaahhhhh

dx is just dx

[sterile pewter]

how?

[tardy epoch]

https://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsVectorFields.aspx

[obtuse pebble]

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[sterile pewter]

Thankyou

[obtuse pebble]

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[obtuse pebble]

@desert pecan Has your question been resolved?

[limber latch]

copy and paste the question for me i can chegg it for you 🤣

[obtuse pebble]

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[sterile pewter]

Are these true?

[timid silo]

any help on this plz

btw this is in french

[tardy epoch]

this channel's occupied. read #❓how-to-get-help

[timid silo]

<@&286206848099549185>

[tardy epoch]

stop spamming other people's help channels and stop pinging helpers

[sterile pewter]

.close

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[dense ingot]

Is there not a factor of (p-2) - ti missing when they expand the modulus squared

[obtuse pebble]

@dense ingot Has your question been resolved?

[tardy epoch]

There is. They multiplied it with the positive term

[dense ingot]

👌

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[chilly idol]

I'm pretty sure this is wrong but I can't think of a counterexample for my life. Could anyone help?

[brave bramble]

Think simple. Let's try letting C be the empty set.

Then the IF is always true, and we just want to know if
f(∅) ∩ f(D) = ∅

Is that true for any choice of f and D?

Actually, we can make it even simpler. Let both C and D be ∅. Then we are asking if it's always true that:
f(∅) = ∅

[chilly idol]

so D is an intersection witht the empty set assuming C is the empty set?

[brave bramble]

Hmm?

[chilly idol]

like for this

[brave bramble]

D is a set, not an intersection

[chilly idol]

is the set D intersecting with C?

Im not quite sure how to word it lol

[brave bramble]

If we let C = ∅, then the question transforms into:
"Is it true that, for any function f and for any set D, f(∅) ∩ f(D) = ∅?"

[chilly idol]

uhh sorry im trying to think

cause empty set is a subset of all sets

[brave bramble]

Well, that was my first idea. My second one is better. Let both C and D = ∅. Then the question transforms into:
"Is it true that, for any function f, f(∅) = ∅?"

[chilly idol]

Is it false?

[solar trellis]

You want to think of an example where f(C) and f(D) share atleast one element

[brave bramble]

Wait, at first I thought it was false but now I'm having doubts haha

Ignore me giannis sounds like they have a better plan of attack

[solar trellis]

Pretry sure it's always true

[ancient marlin]

C and D might not share any element but for some c in C and d in D you can have f(c) = f(d)

say C = {1,2} and D = {3,4}

C intersection D is empty set

f(1) can equal f(3)

heck, a constant function should be a trivial example

[solar trellis]

The original statement is true iff f is injective

[chilly idol]

Don't thihnk we've learned about injective and surjective so I dont think we would be using those

[ancient marlin]

don't need it

anyway, any constant function is your counterexample

f(x) = 0

[chilly idol]

ok hold on let me just confirm something im really dumb

if you have a set C = {1,2,3} then f(C) can be any of the values within the set

Is that right?

[ancient marlin]

you can think of f(C) as the set of values that the elements in C get mapped to

if f(1) = 2, f(2) = 5, f(3), = 9

f(C) = {2,5,9}

ofc if f(x) = 0 , f(C) = {0}

f of a set is a set

[solar trellis]

Not necessarily

[ancient marlin]

it's called the image of a set

[chilly idol]

and C would be, C = {2,3,9} ? The same as F(C)?

[ancient marlin]

that depends on f(2), f(3), and f(9)

f(C) = {f(2), f(3), f(9)}

[solar trellis]

In the example lems gave, C= {1,2,3}

[chilly idol]

ahh and F(C) is the image of C

[ancient marlin]

yes

C is the set

you think of applying f to every element in that set

[solar trellis]

$f(C) = {f(c) : c \in C }$

[warm shale]

giannis_money

[ancient marlin]

and you make that into a set

[chilly idol]

I see

so if C = {1,2} and D = {6,7} the intersection is an empty set because nothing intersects.

does that also mean that the image of F(C) and F(D) also have no values that intersect?

[ancient marlin]

no

as I said

think of a constant function

f(x) = 0

do you understand that f(C) = {0} and f(D) = {0}?

[chilly idol]

yeah

[ancient marlin]

so, do you understand that f(C) and f(D) does have an element in common, which is 0?

[chilly idol]

yeah I understand

[ancient marlin]

hence, you can have C and D be disjoint while f(C) and f(D) be the same set

absolute value is another example

f(x) = |x|

C = [-100, 0)

D = (0, 100]

f(C) = f(D), C and D are disjoint

[chilly idol]

god im so slow

are C and D not sharing the element 0?

[ancient marlin]

no

do you know open and closed intervals? @chilly idol

[chilly idol]

holy

[ancient marlin]

(0,100] means 0 is not in the set

[chilly idol]

im like actally braindead LOL

and blind

Ok that makes so much more sense

[ancient marlin]

👍

[chilly idol]

thanks

I understand now

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

open

[timid silo]

ok i believe i got this working

how to

<@&286206848099549185>

[quiet drum]

question four is 24.36

i rounded to the nearest two decimal places

is 15 divided by cos(52)

[timid silo]

thanks

[quiet drum]

you are welcome

[timid silo]

do you know these

[quiet drum]

question 1 is cos

2 is sine

[timid silo]

ty

[timid silo]

do you know this i feel so bad asking

[quiet drum]

168.45

would you like to add eachother on discord?

[timid silo]

sure

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[sweet flint]

Is this the right way to use the algebraic method.

[obtuse pebble]

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[sweet flint]

.close

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[lime cypress]

Not sure what this is asking

B is (1,3)

[frosty river]

Is <u, v> a basis of R²?

[lime cypress]

Wait u = (2,-4) v= (1,-1)

[frosty river]

Try writing b as a linear combination of u and v

b = au+bv → (1,3) = (2a, -4a) + (b, -b) = (2a+b, -4a-b)

[lime cypress]

This is my linear transformation

[frosty river]

$\left{\begin{array}{rl}2a+b = & 1\ -4a-b= & 3\end{array}\right.$

[warm shale]

jnkena

[frosty river]

Then you can use

$T(b)=T(au+bv)=aT(u)+bT(v)$

[warm shale]

jnkena

[frosty river]

Sorry I used b for two different things

[lime cypress]

So then a = -3/2 and b = 5?

From my linear combination thing

[frosty river]

-3+5=2 so you're not right

-3/2 · 2 + 5 · 1 = -3 + 5 = 2 ≠ 1

[lime cypress]

Ok then I did the first part of this problem wrong

[frosty river]

Yes

[lime cypress]

Disregard my writing

[frosty river]

$\left{\begin{array}{rl}2x_1+x_2= & 1\ -4x_1-x_2= & 3\end{array}\right.$

[warm shale]

jnkena

[frosty river]

You wrote the coefficient matrix wrongly transposed

[lime cypress]

So then it becomes

|. 2. 1. 1 |
|. 0. 1. 5|

[frosty river]

Yes

[lime cypress]

X2 = 5

X1 = -2

[frosty river]

No

Yes

[lime cypress]

Ok then I put it into b = x1u+x2v

[frosty river]

Yes

And use the definition of linear transformation

[lime cypress]

So t(x) = Ax

But then what is a

Cuz x = b = (1,3)

[frosty river]

No simply use the definition of T

You don't need its matrix

(You could calculate it, but it is not needed)

[lime cypress]

Then I am not sure how

[frosty river]

$T(b)=T(x_1u+x_2v)=x_1T(u)+x_2T(v)$

[warm shale]

jnkena

[lime cypress]

Ohhhhhhhh

That makes a lot of sense

[frosty river]

You know x_1, x_2, T(u) and T(v)

[lime cypress]

So it becomes (-17, 10, 0)

I just want to make sure this is correct

[frosty river]

No

2-2=0, 4-4=0 okay but 14-7=7≠7 and -18+11=-7≠7

There are at least two mistakes.

[lime cypress]

I did this row reduction

Then from here i separate x2, x3, x4, x5 into their own vectors right?

<@&286206848099549185>

[obtuse pebble]

@lime cypress Has your question been resolved?

[obtuse pebble]

@lime cypress Has your question been resolved?

[lime cypress]

.close

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[slender kindle]

where am i going wrong? Kindly ping me while replying

[obtuse pebble]

@slender kindle Has your question been resolved?

[slender kindle]

<@&286206848099549185>

[royal basin]

what does "|b| = 3 along c = i+j+k" mean?

does it mean b points in the same direction as c and its length is 3?

@slender kindle

[slender kindle]

ig it means that the vector b with a magnitude of 3 points in the same direction as c?

[royal basin]

what a longwinded way to say "yes"

in this case, b is not 3(i+j+k). the magnitude of 3(i+j+k) is not 3.

[slender kindle]

is it 3 root 3?

so how am i supposed to proceed?

i got the magnitude of the other vector

and i have the vector a

[royal basin]

well youre supposed to find b correctly

|c| is sqrt(3), so by what factor do you need to scale it to get a length of 3?

[slender kindle]

root 3

[royal basin]

ok so do that

and then do the dot and cross product things

[slender kindle]

ohhhhhhhhhhhhhhhhhhhhhhh

got it got it

thanks a lot

:)

.close

[obtuse pebble]

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[timid silo]

ello got a question about cakes, I know the circle cake would be (3.14)(4)^2 and the rectangle one is 9 * 13

[timid silo]

What does it mean by "if the round cake is twice the height of the rectangular cake"? which variable is it changing for the circle? the radius?

[drifting wraith]

cakes have height

[timid silo]

right wait how do I calculate that, use the cylinder formula?

does it .. mention the height? I only know the w and l, not the h unless I'm dumb af (which is true)

[balmy raptor]

You know it's twice the one of the other cake

Think about it this way, if the taller cake gets taller, then also the shorter one does, as its height is half of the taller

[timid silo]

You mean I don't need to know the height? it's just x2?

[balmy raptor]

What this means, is that you have something common between the cakes. The thing which makes you able to solve the problem is that the height "grows" the volume linearly. Meaning that 2 times the height means 2 times the volume.
So what you have to do, is write down the formula for the volume by keeping h in the smaller cake as height, and 2h in the bigger one.
When you confront them, or their "price ratios", you will see they depend on height. But both depend on a single h

So if you compare them through a division, the h will "vanish" and you will be left with their ratio

[timid silo]

Wouldn't that mean it's x2 on the round cake? since the h is gone

[balmy raptor]

It is

if they had the same size of diameter

But that varies

[timid silo]

I mean I believe I can make the assumption that it's going to be 8 as the diameter

[balmy raptor]

What is going to be 8?

[timid silo]

the diameter of the round cake?

[balmy raptor]

Yup, the text says so

I don't know where this emote came from lol

Sry

[timid silo]

@balmy raptor Let me just conclude this, since height is doubled in round cake. Thus that would mean it's safe to assume the 8 inch diameter cake would get doubled and the 9x13 cake would still stay the same.

[timid silo]

what the

[balmy raptor]

Hold up

This is the logic

[timid silo]

of the among us chart or the cake question?

🤣

[balmy raptor]

Lol

[timid silo]

bro is typing the amendments

[balmy raptor]

You have a box (squareish cake) with a 913 base. This one is h tall.
So its volume is 913*h

Then you have a 8inch round cake
This one has a (8/2)^2 * π basis, but a height of 2h

They both cost the same, so you just care about which one is bigger.
You can do this 2 ways:

See if the volume of the first OVER the volume of the second gives you a number greater or smaller than 1,
Or by subtracting them, collecting h, and seeing whether you get a negative result or a positive one

[timid silo]

wait the volume would be 117 for the square no?

[balmy raptor]

I can't count sorry

[timid silo]

you think I can?

[balmy raptor]

And 9*13*h is the volume. 9*13 is the base

[timid silo]

Ahhhh ok ok

um

What if it wants us to use all 3 dimensions and just plug in some numbers of h and see which one results in more volume

Just asking

[balmy raptor]

But would it account for ALL cakes?

[timid silo]

shiiii idk

[balmy raptor]

See closely, the question is absolute: which gives you more cake/$

[timid silo]

right

[balmy raptor]

Which means that there is (most likely since you are in high school/middle school) an absolute answer

[timid silo]

I mean I figured, but I'm still lost 😭

[balmy raptor]

Do 10 of these excercises i'd say

You'll get the logic

[timid silo]

Cuz I was just doing x2, considering it being x2 height

[timid silo]

I did the ones similar to it, but none of them had the concept of "If the round cake is twice the height of the rectangular cake, which option gives"

Like there is this one

but it's the same thickness, so not that bad

[balmy raptor]

Here you simply have each plate with just 1h and none with 2h

[timid silo]

THIS on the other hand. ._.

[timid silo]

yup!

[balmy raptor]

What year of school is this?

[timid silo]

precalc

10th

[balmy raptor]

Not too late to get into math olympics

G2g now, mark the issue as solved if you see it like so

Have a good day!

[timid silo]

thank you

still confused 😓

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[gloomy crescent]

The question is: "Think of a string of at least 15 data. Write down what that data set might represent. Write one question that you would most benefit from a bar chart, one question that you would most benefit from a line chart, and one question that you would most benefit from a pie chart. Answer these three questions and justify your choice."
How is it possible that some data can be a bar chart and a line chart a pie chart at the same time.

[worthy comet]

If you have some data like the numbers 10, 4, 2, you can present them in a bar chart (using bars of length 10,4,2 ) or in a pie chart (using respective parts of the circle) or in a line chart.

[obtuse pebble]

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[void steppe]

I'm not sure how to start on d. This is what I have so far, but it doesn't seem that this works as I don't know what it converges to.
Recall,
$$f'_k(x)=2k^2x\exp(-k^2x) - k^4x^2\exp(-k^2x)$$.
Let $a \in (0, \infty)$ and $\varepsilon \geq 0$. Choose $N = $. Let $n \geq N$ and $x \in (a,
\infty)$. Then
\begin{align*}
S_n(x) -
\end{align*}

[warm shale]

FrankF

[crimson jackal]

Help me😢

[obtuse pebble]

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[tawny rune]

Need help finding the constraints

[tawny rune]

Got this as working out

not sure if this is correct

[obtuse pebble]

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[polar socket]

for this answer, why did they add 500e rather than subtract?

[brave bramble]

Integral of e^(-t) is -e^(-t)

So note there's a sign swap

[polar socket]

when they've done the formula for 0 it was subtract, when it was for 6 it was add. I don't see why it changed between those 2 values

[kind hawk]

we have $\big[F(t)\big]_a^b = F(b)-F(a)$

[warm shale]

Denascite

[polar socket]

yeah but if I it write it out in full
1/6( (25(6) + 500e^(-0.01* 6)) - (25(0) + 500e^(-0.01*0))) it should still be a plus?

[kind hawk]

-(a+b)=-a-b

[polar socket]

ah i see my error, thanks

.close

[obtuse pebble]

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[thick harness]

what did I do wrong?

I don't know what to do next?

@grizzled shore, pinging once cuz you msged like 5 minutes ago 🙂

[grizzled shore]

-x - (-x)

Is 0

[thick harness]

I'm braindead

ty I fixed it up

and now the remainder is 0

frosst are you good in english

I'm actually soo badd

english is just annoying

[obtuse pebble]

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[lime mural]

Is 2(sinx/cosx)(1/cosx) equal to 2sin(x)/cos^2(x)?

[warped agate]

yes