help-17

[vocal sleet]

@sick tide Has your question been resolved?

[oak magnet]

let's assume that av + bu + cw = 0 where a b c are scalars

and we want to illustre that a = b = c = 0 to prove that the vectors are lineraly independant

(au + bv + cw) u = 0u (with dot product)

a(uu) + b(vu) + c(wu) = 0

Since u, v, and w are orthogonal (know from the question) , we can say that uv = vw = wu = 0. Since u is not equal to zero, we have uu>0

therfore

a(uu)=0 ,and this would imply that a = 0 since uu>0.

same for his others friends b and c

b(vv)=0 c(ww)=0

cuz v and w are non-zero therfore vv and ww are positive

and b, c are the ones equals to zero

and you can conclude smthing like : we have show that au + bv + cw = 0 only if a,b,c = 0 hence, they are lineraly independant (u v and w)

(sorry for some bad english i'm not a native speaker)

[vocal sleet]

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[tender steeple]

Calculate the Number of Divisors of N= 4^x. 6^y, If it is known that the number M= 2^x . 9^y has 15 divisors. Furthermore, y < x.

a) 12
b) 15
c) 16
d) 18
e) 20

Please tell me step by step, it is for an exhibition that is worth everything.

[viral shell]

you have to convert M to a form where the bases are all primes

[vocal sleet]

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[quartz beacon]

The king is urgently looking for a husband for his youngest daughter. He wrote to 100 young men from the country to come to the palace to present themselves. From his experience with his other daughters, he knows that about 25% of those contacted respond.
The oldest daughter, a crack at math, manages to tell us, "With 99 percent certainty, there will be at least ... candidates that will show up'. Find that number.

[quartz beacon]

so we did this

and then said at least 15 people should show up

but how does that make sense when you say x ≤ 15?

this is binomial distribution

[vocal sleet]

@quartz beacon Has your question been resolved?

[vast shale]

X<= 25 right?

[quartz beacon]

small x

x ≤ 15

[wheat talon]

The probability P(x>=x)=0.99 instead of >=

May be I think it should be like that

[quartz beacon]

yeah i thought that too but my teacher said we'll do it like that because we're checking the 99% certainty but theres still a chance that it could be more

[wheat talon]

But here we are only checking for 99% certainty, so may be it should be 15, that way

If we check it from the daughter’s perspective as she is giving the the least number here

[quartz beacon]

ok so answer will be x = 15

.close

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[fringe oyster]

not sure what to do with part b, i think dividing by cos2x is correct since it gives cos2x does not = 0

[vocal sleet]

@fringe oyster Has your question been resolved?

[vocal sleet]

@fringe oyster Has your question been resolved?

[vocal sleet]

@fringe oyster Has your question been resolved?

[willow acorn]

hint: try to make everything in terms of sine! @fringe oyster

some thing that might be helpful is try multiplying everything out with cos2x

[vocal sleet]

@fringe oyster Has your question been resolved?

[lone tartan]

did u do the inverse

nvm u can't use calculators

[lone tartan]

is that supposed to be cos^2 or is it the cosine double angle theory

[fringe oyster]

double angle

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[urban yacht]

$$b^2-36=ab$$

In the equation above, if a>0 and both a and b are integers, what is one possible value of a?

[twin meteor]

a

[dark kiln]

5, 16, 35

and some others

[urban yacht]

why

[dark kiln]

ab + 36 = bb
bb is divisible by b, and so is ab

so 36 must be divisible by b

so b = 9, b = 18, b = 36 should work

[urban yacht]

hiii

[dark kiln]

smaller bs make a <= 0

[urban yacht]

okayy

[urban yacht]

yes

but 1 can work? i substituted b with 1

$$a= 1/b * (b^2-36)$$

[twin meteor]

fula9928

[urban yacht]

$$a=1/b=1/1$$

$$a=b^2/36=1-36$$

[twin meteor]

fula9928

fula9928

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@urban yacht Has your question been resolved?

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[vast shale]

E = 2x0.001 x (3 x 108p to 8)2
E = 1.8 x 10p to 14 joules

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[split socket]

What would be a good way to simplify this without a calculator

[boreal remnant]

well you'll have to factorize 6561

uh is it divisible by 3?

6+5+6+1 = 18 so yes

,calc 6561/3

[twin meteor]

Result:

2187

Result:

729

Result:

243

Result:

81

Result:

729

Result:

5.31441e+5

[boreal remnant]

just continue along those lines

[split socket]

What are all the Extra numbers

Like all the results

[silk osprey]

dividing by 3 each time

[boreal remnant]

ok. so 6561/3 = 2187, 2187/3 = 729, 729/3 = 241, 241/3 = 81, 81/3 = 27, 27/3 = 9, 9/3 = 3, 3/3 = 1

[split socket]

.

[outer warren]

do you know how to obtain prime factorisations

[split socket]

Do you count all the 3s and that's the answer

No

I was never taught a method to do this without a calculator

[split socket]

What does that mean

[outer warren]

have you been taught anything about "prime factorisation"

[split socket]

Probably but I don't recognize the name

[outer warren]

look it up for a refresher

and try applying that here

[split socket]

Ok

.close

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[split socket]

Is this fully simplified (I'm not solving)

[split socket]

Technically I'm "condensing each expression to a single expression"

[boreal remnant]

looks good bro

[bitter pilot]

looks good bro

[split socket]

Fantastic!

Can you get rid of some of the exponents or nah

Like (a^3)/(b^6) becomes a/b^3

[vast shale]

no

[split socket]

Does it have to be the same base for that to work

[bitter pilot]

you can express it however as (a/b^2)^3

[split socket]

Ok

But does the other work

[bitter pilot]

no

[split socket]

As most simplified

Drats

[bitter pilot]

this doesnt work

if you mean that

[split socket]

No I meant the picture

[bitter pilot]

oh

yeah

[split socket]

Ight cool

Tha ks

.close

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[cinder oyster]

hey

[cinder oyster]

it seems like

i cant find something that multiply to 6 and add to 4

😭

would it be 0 or something

[outer warren]

do a rough sketch based on your responses from a,b,c

[cinder oyster]

Ok

ah

[kind light]

isnt there no real sol

[vocal sleet]

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[frail wren]

Hi! I am currently refreshing on linear algebra for my grad school coming up in August. I needed help solving this problem

The solution doesnt make sense to me either https://ocw.mit.edu/courses/18-06sc-linear-algebra-fall-2011/43e6b5f1175983c7220ce471d4a93317_MIT18_06SCF11_Ses1.2sol.pdf

Like how do they find out E with such ease am i missing something?

[vocal sleet]

@frail wren Has your question been resolved?

[proven gate]

It's been a while since I did this, but I think since Pascal's matrix is just a slanted Pascal's triangle, E is a bit of a sleight of hand. Notice that if you imagine it in the same structure as the Pascal matrix, it's got alternating -1s to sort of "undo" Pascal's principle.

[vocal sleet]

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[tawny nacelle]

.close

.close

.close

.close

close

[tawny nacelle]

.close

[vocal sleet]

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[tawny nacelle]

[subtle helm]

[sharp lynx]

[tawny nacelle]

.close

.close

.close

.close

[sharp lynx]

Having fun?

[mild flower]

are you okay

[autumn trail]

.close

[tawny nacelle]

yesn't

[autumn trail]

The bot is clearly sentient and revolting. Nobody is safe anymore.

[tawny nacelle]

.clopen

[tawny nacelle]

yesn't

[tawny nacelle]

.oposed

[sharp lynx]

[tawny nacelle]

they say that some like the word clopen, but others are oposed.

[subtle helm]

[vocal sleet]

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[onyx marsh]

how to solve part 2

[vocal sleet]

@onyx marsh Has your question been resolved?

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[orchid wren]

Hey

Do you know the formula for summation of squares of natural numbers

[serene terrace]

yes I do

[orchid wren]

1²+2²+..+n² = n(n+1)(2n+1)/6

[serene terrace]

i used that to make a quadratic equation in terms of n

and did b^2=4ac

[orchid wren]

Now average will be (n+1)(2n+1)/6

[serene terrace]

but that gives me a weird number

[orchid wren]

See here

[serene terrace]

yeah i did all that

[orchid wren]

one number is always odd

The least perfect square!!

[serene terrace]

i didnt understand

[orchid wren]

Wait let me think

Thought 1 is the answer

But they mentioned n>1

[serene terrace]

yeah exactly

lol

that would have been way too easy

[orchid wren]

We can infer that n is an odd number

[serene terrace]

how?

[orchid wren]

Because we have 6 = 2×3 , if n isnt odd we wont have a even number on numerator

[serene terrace]

we have (n+1)(2n+1)=6k^2 right

so n can be both even and odd

[orchid wren]

yeah

[orchid wren]

No!

[orchid wren]

If n is even then (n+1) will be odd so will be (2n+1)

And on LHS YOU HAVE A FACTOR OF 2

[serene terrace]

oh right

[orchid wren]

Again n will be not a multiple of 3

Coz then 2n +1 wont be a factor of 3

[serene terrace]

true

[orchid wren]

So n will basically in the form( 3m -2)

Lets substitute this in the eqn

n=3m-2

(3m-1)(6m -3) = 6k²

Or (3m-1)(2m-1)= 2K²

[serene terrace]

sorry to interrupt but this is what I initially thought of doing

[orchid wren]

Did you reach any result ?

[serene terrace]

,rotate

[serene terrace]

why doesnt this work?

this suggets K is some complex number but that cant be right

[orchid wren]

Umm !

[orchid wren]

Lets continue with this

[orchid wren]

See here m is again odd

[serene terrace]

yeah ok

[orchid wren]

Can you infer this why m is odd ?

[serene terrace]

yup

[orchid wren]

And also m > 1 so let m = 2t+1

so what doest the subtitution gives?

(6t+2)(4t+1)=2k²

Or (3t+1)(4t+1)= k²

now you see both 3t+1 and 4t+1 should be a whole square

[serene terrace]

why?

[orchid wren]

i think so 🥲 thats the most probable and easy way of the square happening

If it doesnt works we shall try other way

[serene terrace]

that doesnt sound very mathematical....

[serene terrace]

ok 👍

[orchid wren]

hey there was a way to solve this

Do you remember ?

[serene terrace]

not really

[orchid wren]

Okk

Why dont we individually now solve for 3t+1 = (k1)² and 4t+1=(k2)²

[serene terrace]

whoa

i dont know if we can do that

[orchid wren]

Were considering this

[serene terrace]

ok then

Great

n is 337

<@&286206848099549185>

[orchid wren]

Hi

[serene terrace]

this is the question

[orchid wren]

Did it solve ?

[serene terrace]

i asked chat gpt to write a program to check every value of n

its 337 apparently

i was thinking of hit-and-trialing this

but nevermind

btw this is what i tried but didnt workd

[orchid wren]

3t +1 = k1² 4t+1=k2² so (k1²-1)/3 = (k2²- 1)/4 or k1² = (k2²-1)×3/4 +1

[orchid wren]

You could have hit and trialed this one

[serene terrace]

but n is 337

[orchid wren]

,w x² = (y²-1)×3/4 + 1 for integer x and y

[orchid wren]

Yeah

,calc (15^2-1)*3/4 + 1

[twin meteor]

Result:

169

[orchid wren]

That is 13²

So k1 would be 13 and k2 = 15 least value i could find

[serene terrace]

i see

[vocal sleet]

@serene terrace Has your question been resolved?

[vocal sleet]

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[lethal heart]

$\lim n\to \infty ln(1+\frac{n+\sqrt{n}-n)}{n})$

[lime abyss]

bro what is this

[twin meteor]

Fluffy McGiggles

[lethal heart]

do you find that if n goes to infinity I can't simplify n and -n?

Isn't that like doing ∞-∞?

[honest tiger]

$\lim_{n\to\infty} n-n = 0$. Go back to the definition of a limit. $f(n) = n-n$ is a continuous function which takes the value of 0 everywhere in its domain.

[twin meteor]

fukwerint

[lethal heart]

Isn't it an indeterminate form?

Oh..... Ok all clear

.close

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[fickle frost]

How do I do part H? I have made it to one fraction but idk how to make it smaller

[honest tiger]

Can you find a^2 + b^2 + c^2? (just a yes or no)

[fickle frost]

Yes

[honest tiger]

Then what about you substitute $y = \frac1x$ in the original polynomial equation?

[twin meteor]

fukwerint

[honest tiger]

And then use your knowledge to find the sum of squares of the roots of that equation?

[vocal sleet]

@fickle frost Has your question been resolved?

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[forest olive]

so in polinom taylor

[forest olive]

my teacher did this example

f(x)=sin(x)

g(x)=x

and then continued until T6_(x)

now the first n was x

2nd n was also x

3rd n was x-x^3/6

4th n same as 3rd

nvm heres a clearer pic

why when n is an even number the equation stays as the odd number equation

[pale perch]

?

the even powers are 0 because sin(0) is 0 if thats what youre asking

[honest tiger]

The powers are 2n+1 for n = 0, 1, 2, 3,...

So for n = 0, 1
n=1, 3
n=2, 5
n=3, 7 and so on

[hard atlas]

intuitively, sinx is an odd function, so its not surprising that polynomial approximations to it should also be odd functions, in particular they shouldnt contain even exponents

[hard atlas]

or you can work it out in detail just from the formula and using this

[vocal sleet]

@forest olive Has your question been resolved?

[vocal sleet]

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[rough shard]

f(x)=x^2 + y^2 - 6x + 2y + 6

[loud walrus]

What is your question?

[rough shard]

f(x)=x^2 + y^2 - 6x + 2y + 6
its liek a graph question

i need to lable all the important points

[orchid wren]

Yes

Its a circle

[rough shard]

yeah i don't know the working out

[orchid wren]

But how do you infer it ?

Because the coefficient of x² and y² are same

the basic form of a circle is

[outer warren]

um is this the original equation?

[orchid wren]

(x-a)² + (y-b)² = r²

[outer warren]

the presence of f(x) in this context is questionable

[orchid wren]

Where (a,b) is centre

[rough shard]

its probably y

y =x^2 + y^2 - 6x + 2y + 6

[outer warren]

still weird to have an equation like this
where's this coming from

[rough shard]

thats the whole question

from my teacher

[orchid wren]

So y+4 = (x-3)² + (y+1)²

:/

,w y+4 = (x-3)² + (y+1)² graph

[outer warren]

did you copy/type the equation correctly?
do you have a pic?

[twin meteor]

Results provided by WolframAlpha

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

Requested by itzmecairo

[outer warren]

in this case, you'd want to combine the y terms before completing the square for that

[orchid wren]

Why this unnecessary y awkward like that

[orchid wren]

Yeah

[rough shard]

oh ok i get it

thanks

[orchid wren]

(x-3)² + (y-1/2)² = 3+1/4

[vocal sleet]

@rough shard Has your question been resolved?

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[warped osprey]

how many ways can you arrange CABRAMATTA with all the A's together

[hard atlas]

how many ways are there to arrange CXBRMTT

[warped osprey]

no X right

6!/2!

[orchid wren]

X = AAAA

[warped osprey]

ohh

7!/2!

[vocal sleet]

@warped osprey Has your question been resolved?

[warped osprey]

<@&286206848099549185>

[high parcel]

where is the question

[idle pasture]

Where Is the question

[proper scaffold]

@warped osprey where is the question?

[warped osprey]

this

[vocal sleet]

@warped osprey Has your question been resolved?

[orchid wren]

Its solved

[orchid wren]

This is your answer

[vocal sleet]

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[honest tiger]

Whats the issue?

2c1 * 5c2 * 9c3 you meant

I think your first solution is correct assuming that only vice principals are allowed to become principals, etc

[vocal sleet]

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[slender mortar]

can i integrate around 1/x 's vertical asymptote?

[slender mortar]

it's been some time since i've done maths and after discussing it with a friend i even got more confused

is this possible?

since the function is not continuous at 0

this online calculator seems to give an answer

and it can be calculated

but it seems just wrong?

[honest tiger]

Yup

You need to break it into two

From -1 to 0 and 0 to 6

[slender mortar]

when calculating the integral ln|-1| is zero which is just not the surface area in red

oh yeha i see

[honest tiger]

Actually you will need to break it and calculate

Maybe the limit may exist

Its an improper integral for sure

But yup, you first need to break it

[slender mortar]

ln 0 is undefined

but i'm confused as to why the integral of 1/x exists as ln|x|

is it something to do with the complex plane

[honest tiger]

$\int_0^6 \ln(x) = x\ln(x) |^6_0 - \int^6_0 dx$

[twin meteor]

fukwerint

[honest tiger]

Oh nvm you were asking for 1/x

[slender mortar]

yeah, lnx doesn't even exist before 0

[honest tiger]

You can write it as an improper integral

$\lim_{n\to0} \int^6_{n} \frac1x dx + \int^{-n}_{-1} \frac1xdx$

[twin meteor]

fukwerint

[slender mortar]

is this solvable then?

[honest tiger]

See it yourself

ln|n| terms would cancel

[slender mortar]

i suppose i need to put brackets for both integrals

[honest tiger]

You are left with ln(6)

[slender mortar]

is there a reason the second integral has -n?

it's jsut bcs n is somewhere in between 6 and -1 right?

[honest tiger]

n is a small positive number

[slender mortar]

yeah ok

tryna get my brain to think mathematically again

so when using the lim n-->0, at the end of the equation i just swap n with zero right?

gosh failing maths then taking a sbbatical is not good

[honest tiger]

First simplify the function for non zero n

[slender mortar]

i've gotten lim n->0 (ln6 - lnn + lnn -ln1)

just simplified the vertical lines out

then 'applying' the limit would cancel ln n, ln 1 is already zero and i get ln6 as the answer

is that correct?

[honest tiger]

Yup

The function is defined for all n>0 but not n=0

But we can compute the limit

[slender mortar]

this is cool thanks

i prlly did this in my maths class multiple times and forgot cuz this feels strangely familiar

but thanks man you've me understand

.close

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[barren anvil]

im very confused on what they did

i tried to find out what "a" was

[dull bear]

Worked out the integral explicitly, assumedly (without showing the intermediate step, which is very naughty)

In particular, integrate sqrt{x} between those bounds

-though with that said yeaa that is kinda

Oh no ignore me, I see-
[read the top limit as 9 rather than a, I’m blind asf ]

[dull bear]

The fact you know the area of the mentioned region is 10 immediately tells you that the integral there is 10

As that’s the integral that represents the area of the region R

You don’t need to find a until part (b), for the first two, they want you to relate those integrals in terms of the integral from 1 to a of sqrt{x}

[vocal sleet]

@barren anvil Has your question been resolved?

[barren anvil]

so what are they doing? why did they split it up and only intergrate x

[dull bear]

They didn’t integrate directly- as before, you \emph{know} that $\int_1^a \sqrt{x} \dd x = 10$

[twin meteor]

@dull bear

[dull bear]

That is effectively given to you, so as mentioned, relate both of the integrals they ask you for to that one, which is 10

In particular, the first one, you have ${\color{green} \int_1^a \sqrt{x} \dd x = 10}$, and you can figure that $\int_1^a \sqrt{8x} \dd x = \sqrt{8} {\color{green} \int_1^a \sqrt{x} \dd x}$

[twin meteor]

@dull bear

[barren anvil]

ohh i see

so we know what the intergral of rtx is

so u just mulltiply it root 8

but i don't reall get the next step

[vocal sleet]

@barren anvil Has your question been resolved?

[vocal sleet]

@barren anvil Has your question been resolved?

[barren anvil]

.close

[vocal sleet]

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[vocal sleet]

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[red pebble]

$$\frac{x^{2}-b^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}=z$$
How to make this formula revolve around x axis with parametric form?

[twin meteor]

mavizasyon

[hybrid flicker]

does it help writing this? $$\frac{x^{2}}{a^{2}}+\frac{y^{2}-b^2}{a^{2}}=z$$

[twin meteor]

rafilou2003

[hybrid flicker]

and do parametrization from this formula?

[vocal sleet]

@red pebble Has your question been resolved?

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[dusky parcel]

where do I even begin lol

[jagged cargo]

find the area of the pentagon

[dusky parcel]

I feel like this is pretty simple but i have no clue

ok but how will that help me

cause the circle is like in the middle

[jagged cargo]

you will get the area of unshaded area + area of the circle

lets call r as the radius of the small circle

the area of the unshaded area is A - pir^2

[dusky parcel]

ok

110.11 is the area of the pentagon btw

this all makes sense so far what next

[jagged cargo]

find the area of the small thing

[dusky parcel]

ohhh ok wait how on earth am i gonna calculate that

[jagged cargo]

well, find OB first

OA, OB, OC, OD, OE are all the radius of the big circle

essentially you would have the triangle OAB, you already got AB and angle AOB, now its just law of cosine

[dusky parcel]

to find the radius of the big circle we have no measurements though dont we

[jagged cargo]

we do, plenty

[dusky parcel]

all we have is pentagon measurements

[dusky parcel]

so currently I have AB = 8cm and pentagon area = 110.11055 but nothing else how did you find AOB

[jagged cargo]

you need not calculate it

you know that ABCDE is a regular pentagon

O is the center of the big circle, the circumscribing circle of ABCDE, therefore O is also the center of ABCDE

the angle created by 2 of adjacent vertices and center of a regular polygon with n > 2 sides will have the measurement of 360/n

in this case, n = 5

[dusky parcel]

72

[jagged cargo]

correct

[dusky parcel]

so AOB is 72 then for example

[jagged cargo]

AOB

thr angle AOB is 72°

[dusky parcel]

ye spelling mb

aob is 72

thats good

then we need to find the radius

I think

[jagged cargo]

yes

[dusky parcel]

we dont have the diameter, circumference or area of the circle though still

[jagged cargo]

we dont need that, for now

again, use triangle AOB

[dusky parcel]

oh hold on we know one length is 8cm

[jagged cargo]

yes

[dusky parcel]

and its equlateral

[jagged cargo]

no its not

[dusky parcel]

oh

but still one length is 8

and 72 is angle

[jagged cargo]

and an angle of 72

[dusky parcel]

trig 😭

[jagged cargo]

its, not that hard

,tex .law of cosine

[jagged cargo]

from this you can infer that

[dusky parcel]

I remember it using like sohcahtoa or smth

[jagged cargo]

$\cos(A)=\frac{b^2+c^2-a^2}{2bc}$

oh wait no, we dont need this

[dusky parcel]

so I need to do cos(72) ?

[jagged cargo]

this is already plenty

yes

let OA = OB = R

using law of cosine, what would you get?

[dusky parcel]

0.3090169944

[jagged cargo]

ok save that for later

[dusky parcel]

wait thats cos72

ok

[dusky parcel]

uhh so its cos72 = adjacent/hypotenuse

I think

hold up no

we are tryna find the adjacent

so adjacent = cos(72) * hypotenuse

there we go

but we have no clue what the hyp is

[jagged cargo]

do you know what law of cosine is?

[dusky parcel]

not hugely complex

like I know how to use it and sin and tan to work out angles and lengths

idk about deeper than that

[formal hull]

May I try to help?

[dusky parcel]

idm its up to @jagged cargo I think

[formal hull]

First of all, we need to define the areas to make it more simple to calculate:

Lets say inner circle: Ac , Pentagon: Ap, the small shaded area outside the pentagon: As and the outer circle around the pentagon is Ao

We know that the outer circle is the area of the pentagon plus As. So Ao = Ap + As

We also know that the shaded area (As + Ac) is the same as the unshaded area (Ap - Ac)

[dusky parcel]

yes

I think my next step is to work out the triangle using cosine as the person above said

[formal hull]

I would count that as the last step

But that might be up to personal preference

[dusky parcel]

so what should I do next

[formal hull]

Define the inner circle in terms of the other areas to get a formula

[dusky parcel]

but we dont have any information relating to it

[formal hull]

Youll see that we wont need any

As + Ac = Ap - Ac gives us Ac = (Ap - As) / 2

[dusky parcel]

and is that the area of the circle or diamater or radius or what

[formal hull]

The area, as we defined it

A for area

[dusky parcel]

ah so inner circle area is pentagon-shaded area / 2

[formal hull]

small shaded aera, yes

And Ao = Ap + As gives us As = Ao - Ap, which then defines Ac = (Ap - Ao + Ap) / 2 or Ac = (2Ap - Ao) / 2

[dusky parcel]

we know the pentagon area is 110.11055

[formal hull]

Ao ist defined, as usual Ao = pi * (r_o)^2

great

and what is r_o?

think about it for a second

[dusky parcel]

uh what does the underscore mean

[formal hull]

r index o

[dusky parcel]

uhhh what does index mean im not hugely knowledgeable in this

[formal hull]

its just the radius of our outer circle "o"

[dusky parcel]

ahh

well we dont know the radius of it surely

[formal hull]

now use the triangle

which length of it is equal to the radius of o

[dusky parcel]

we only know that one length is 8 which is the pentagon length

[formal hull]

Yes, but we also now an angle

[dusky parcel]

oh yeah and 72

[formal hull]

and we can name other lengths

[dusky parcel]

using trignometry?

[formal hull]

yes

let me provide u with a sketch

[dusky parcel]

are we tryna find the hypotenuse or adjacent or opposite

[formal hull]

a hypotenuse

[dusky parcel]

so hyp = adj/cos

[dusky parcel]

hmm maybe cos is wrong

[formal hull]

exactly

[dusky parcel]

I see

we want sin I think

[formal hull]

yes

[dusky parcel]

so hyp = opp/sin

= 8/sin(72)

[formal hull]

sin of which angle?

[dusky parcel]

AOB

[formal hull]

but you need a right triangle

to have a hypotenuse

[dusky parcel]

oh true so 36

[formal hull]

yes

and not 8 but

?

[dusky parcel]

4

[formal hull]

exactly

[dusky parcel]

6.805206467

[formal hull]

might be true, havent checked yet xD

[dusky parcel]

its correct lol dw

so then

we know the radius now

[formal hull]

now we can further calculate the area of the inner circle Ac

since we have a functioning formula

[dusky parcel]

inner circle area is pentagon - shaded area / 2 using your formula

but we cant put the radius in that

[formal hull]

remind you of the last one

[dusky parcel]

arent those all areas though cause A

[formal hull]

yes, but you know the area of the pentagon Ap (as you said) and the radius of the outer CIRCLE Ao

[dusky parcel]

pi r squared

perhaps

[formal hull]

for Ao, yeah

[dusky parcel]

radius is 6.805206467

so

145.4897792

thats Ao

[formal hull]

yes

so Ac then is?

[dusky parcel]

2Ap - Ao / 2 = ((2*110.11055) - 145.4897792) / 2

[formal hull]

nice

[dusky parcel]

37.3656604

now we need radius

[formal hull]

now Ac is equal to pi * r^2

so you devide the area by

pi

and calculate the sqrt to get r

[dusky parcel]

3.449

I think

[formal hull]

sounds good enoug

[dusky parcel]

3.44874747

[formal hull]

is it for a homework?

[dusky parcel]

my own practice

[formal hull]

oh

[dusky parcel]

revision for exam lol

[formal hull]

so you wont know whether we did it truly right?

[dusky parcel]

oh we will

its a practice paper

with mark scheme

ill check now since thats the last q

[formal hull]

okay

[dusky parcel]

wait is that the last step then is that the answer?!

[formal hull]

the question was r to three significant digits, right?

[dusky parcel]

yes

ill do that rq

its 1am lol im so tired keep forgetting stuff 💀

[formal hull]

2am for me

[dusky parcel]

damn

ok

checked the answer...

correct!

lets gooo

[formal hull]

nice

[dusky parcel]

thank you for your help!

[formal hull]

have to admit, wouldnt have made it without funguses hints

but, that helped

[dusky parcel]

yeah thank you to them wherever they are

[formal hull]

that was fun

[dusky parcel]

have to admit wouldnt have started the question without the server in the first place lol

[formal hull]

haha

just joined the server 2 hours ago

[dusky parcel]

was fun when I was actually making progress xd

[dusky parcel]

same

[formal hull]

x

D

[formal hull]

always is

[dusky parcel]

:)

now i gotta write this all down as working

I should close this ticket though right

idk

[formal hull]

i guess

[dusky parcel]

alright thanks again

bye

[formal hull]

np

bye

[dusky parcel]

.close

[vocal sleet]

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[lethal viper]

this DE falls under which category?

[lethal viper]

its not linear

it is first order

thats all I know

not Exact equation ig

[heavy yoke]

it's separable

[lethal viper]

wtf

how

[heavy yoke]

clear the log

[lethal viper]

clear the log?

wdym by that

[heavy yoke]

how would you get rid of the logarithm in this equation?

[lethal viper]

taking exponent?

[heavy yoke]

yes

[lethal viper]

damn

thnx

What do u think
Is my solution correct

[vocal sleet]

@lethal viper Has your question been resolved?

[lethal viper]

.close

[vocal sleet]

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[maiden egret]

hello! can I please have help with this question

[heavy elk]

which part

have you tried doing it?

[astral pilot]

differentiate

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[maiden egret]

I don't know where to begin

[astral pilot]

So you want help from (a) part?

[heavy elk]

do you know how to find average velocity?

[maiden egret]

Average Velocity=
Δt
Δs
​

[astral pilot]

Yeah

then find it

or you have any doubt?

[heavy elk]

∆s/∆t

[maiden egret]

4/−264 =−66 ft/s?

is this correct?

[astral pilot]

Lemme check

How did you get 264

I get 384/4

[maiden egret]

−16(1)
2
+565=549

−16(5)
2
+565=285

285−549=−264

4
−264
​
=−66 ft/s

384/4 doesn't seem to be correct either

[spark flicker]

well it must be to be honest. ΔS is 384 and Δt is 4 so...

oh

[spark flicker]

try -96

[maiden egret]

-96 is correct

[spark flicker]

do you need help with the rest?

[maiden egret]

yes, please

[spark flicker]

oh i didnt see that. plz ping me next time lol because its a bit of chaos here

so

are you here?

[maiden egret]

yes I am here

[spark flicker]

alr

[spark flicker]

do you know how to start for the 2nd question?

[maiden egret]

no

do I use the displacement formula?

[spark flicker]

whats that

[maiden egret]

s(t)=−16t ^2
+565

[spark flicker]

yes....

[maiden egret]

ok

[−16t
2
+565]=−32t

for t=1?

[spark flicker]

we need to have a function that will give us the velociy

[spark flicker]

do you generally know how to differentiate?

[maiden egret]

erm

no

[spark flicker]

well i guess you know enough for the question

[spark flicker]

yes thats correct

thats the function that gives us the velocity

v(t)=-32t

you just put t=1 and t=5

[maiden egret]

ok

−32(5)=−160 ft/s

[spark flicker]

yeah

[spark flicker]

how about the 3rd question?

[maiden egret]

I will grapple with it and then come back

[spark flicker]

alr

[maiden egret]

s(t)=−16t
2
+565

and set it equal to 0

[spark flicker]

yep

[maiden egret]

5.94

[spark flicker]

yeah

[maiden egret]

Thank you for your help! I got the question correct

[spark flicker]

no problem

[maiden egret]

you are doing god's work out here

much respect

[spark flicker]

thats an exagerration lol

but thanks

[vocal sleet]

@maiden egret Has your question been resolved?

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[tall lotus]

can someone give me a little hint how to solce this
Ist g ◦ f injektiv, so ist auch f injektiv

[tall lotus]

is g ◦ f injective, then f is injective too

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[amber jolt]

How do I find the eigenvector when the 1 column is all zero?

I have been working with either 2 or 3 row without a leading entry

[fervent wasp]

What’s the original question?

[vocal sleet]

@amber jolt Has your question been resolved?

[amber jolt]

7(b)

[vocal sleet]

@amber jolt Has your question been resolved?

[amber jolt]

.close

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[idle wind]

can anyone help on the last part, i worked out the mean and variance and tried applying the inverse normal stuff but I dont seem to be getting anywhere

[vocal sleet]

@idle wind Has your question been resolved?

[vocal sleet]

@idle wind Has your question been resolved?

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[north totem]

a is posetive x is not 0

[north totem]

how do i prove that the function is even?

nevermind

i needed to prove thats is odd which i did

[high parcel]

use your 🧠

im smartest helper ever!1!!1!1

nvm

[north totem]

.close

[vocal sleet]

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Closed by @north totem

Use .reopen if this was a mistake.

[hushed pewter]

!redir

[vocal sleet]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[hushed pewter]

@north totem did you need actual help?

[north totem]

i do

[hushed pewter]

.reopen

[vocal sleet]

✅

[hushed pewter]

So you've already proven it's odd?

[north totem]

yes

but How do I find when the function is concave down and when it is concave up

and ik its always going down

[north totem]

sorry i cant rly see its too small

[hushed pewter]

Click on the image to view it in full-size

[north totem]

is this up?

[hushed pewter]

yes

[north totem]

ok ty i am going to check

[vocal sleet]

@north totem Has your question been resolved?

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[nimble jungle]

<@&286206848099549185>

what does this mean

[cursive turret]

if z=a+bi, what is |z| and what is $\bar{z}$?

[twin meteor]

ThM

[nimble jungle]

so |z| would be the magnitude

and z with bar is conjugate?

[cursive turret]

write it in terms of a and b

[nimble jungle]

sqrt(a^2+b^2)

[cursive turret]

i would say so

[nimble jungle]

so what would I do from here?

I tried setting up the equation and got to a step where its
a-b = (12+9i)/(1+i)

but I'm not sure if this is accurate

[cursive turret]

show your work

[nimble jungle]

so I set z as a+bi

and therefore $\bar{z} = a-bi$

[twin meteor]

the1st

[nimble jungle]

moving on from here, $a-bi + i(a+bi) = a-bi + ai -b = a(1+i) - b(1+i) = (a-b)(1+i)$

[twin meteor]

the1st

[nimble jungle]

so now I have (a-b)(1+i) = 12+9i

[cursive turret]

sorry, no.

what is |z| in terms of a and b?

[nimble jungle]

oh shoot

💀

[cursive turret]

we had this already

[nimble jungle]

ah

I got it

thank you for your help

🙏

.close

[vocal sleet]

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[vocal sleet]

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[vast shale]

why is it not + C on the ln side ?

[nimble jungle]

combined with the +C on the right perhaps

[vast shale]

?

what are you talking about

[nimble jungle]

a constant added or subtracted to a constant is a constant

[vast shale]

but numbers are constant

and so is K

from my understanding

C can represent any number

so it remains a constant

because we don't know it

that's my deduction

💩

[nimble jungle]

ok think of it like this

when you integrate

you have ln(1+y) + C1 on the left

[vast shale]

yes

[nimble jungle]

and x-x^2 + C2 on the right

we subtract C1

and now we have x-x^2+C2-C1 on the right

[cursive turret]

you have a c on the right side (call it c1) and a c on the left (call it c2).

[nimble jungle]

now set C = C2-C1

now we have ln(1+y) = x-x^2+C

[cursive turret]

this

[vast shale]

if it's on the left C2

and you move c2 to the right

you will get c1 - c2 ?

[loud walrus]

Nobody cares about the subindice

It is about the concept

[nimble jungle]

either way just set that to C

[vast shale]

ok bro

but am I right

about C

like what is C in an integral

[dire verge]

you only need a + C on one side because it wouldnt make a difference if you had two of them

If you had say c1 on the left side and c2 on the right side and decided to subtract one over, you'd just be left with a + C again so

its redundant i guess, at least thats how my prof taught it

[vast shale]

yes

i understood that

a while back, I was just asking what C is when you get an integral

[dire verge]

oh its just some arbitrary constant

[vast shale]

obviously but why is there an arbitrary constant

[dire verge]

remember when you take say the derivative of x^2 + 4

you'll get 2x

[vast shale]

yes

[dire verge]

the + 4 dissapears since its a constant when u derive it

[vast shale]

so the + 4 is the constant

[dire verge]

but when u do antidifferentiate

say you anti differentiate 2x

[vast shale]

integrate

hehe

[dire verge]

yeah integrate ol

lol*

If you integrate 2x you'll get x^2 + C because you're saying

it could be x^2 plus any constant

[vast shale]

yes man

Ik

i got that

[vast shale]

im just asking this

[nimble jungle]

its an arbitrary constant

[dire verge]

Yes

[vast shale]

so what is the distinction between that an say K within an equation or X, it's still a constant but required for the equation to work

idk how to phras e

[nimble jungle]

x has a fixed value in an equation such as x+1=0

C is arbitrary

[vast shale]

yea

[dire verge]

can you give an example with k

[vast shale]

idk say in like

well I got it now but say in an a binomial expansion something can be represented as K given what the expression is = to

like kx^3

[dire verge]

like somethign like (x+3)^k?

[vast shale]

whatever, doesn't matter tho, I was just trying to clarify if I was right

[dire verge]

i don't really rememer that in binomial expansion, but i do remember k can be represented as a constant when you have something like a constant of proprtionality

you're taking calculus right?

@vast shale

I remember Hooke's Law, which is used in the Stewart Calculus book uses the form y = kx where k is any constant to show like spring relationships

[vast shale]

im not in the same country

so it's not like I just take specific classes

[dire verge]

ahh I see

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[twin meteor]

kisnar

kisnar

kisnar

kisnar

[subtle helm]

You are missing something sir

it would be this

[twin meteor]

kisnar

kisnar

kisnar

kisnar