help-10

[ruby elm]

Question 4.

[obtuse pebble]

@ruby elm Has your question been resolved?

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[mellow viper]

I'm currently doing frequency counting, and I'm trying to find an equation , wherein if I input 1 or 2, the answer will be 5, but if i input 3 or 4, the answer will be 8, can anyone help me find such an equation?

[mellow viper]

I tried 3n-1, but it only works for 2 and 3, it doesn't work for 1

[crimson breach]

this isnt surely a linear equation

its a deegre 3 equation at least

[mellow viper]

it isn't, actually I don't know if this is even the right place to ask,since it's a data structures and algorithms question but maybe someone knows, basically I need to count the number of times an operation happen, I tried to trace my code and I have no idea how to make such an equation

but i do know those 2 parameters, if it's 1 or 2, the answer should be 5, and 3 or 4, it becomes 8

[crimson breach]

its clearly imposible to find an equation based on just 4 points

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[marble jay]

Is u antisymmetric??

[drifting wraith]

no

yes

[marble jay]

But like it only returns itself right

So?

??

[obtuse pebble]

@marble jay Has your question been resolved?

[marble jay]

<@&286206848099549185>

So for it to be antisymmtric is that when there is a relation from one point to another that other point doesn’t return a relation back

However in this case there is no relation to another

So I’m confused

[obtuse pebble]

@marble jay Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

@dim copper Has your question been resolved?

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[formal turtle]

can

[formal turtle]

Would I walk through this using the definition?

[timid silo]

Find the eigenvalues and compare multiplicities

[formal turtle]

.close

[obtuse pebble]

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[boreal light]

I am doing online summer classes and my professor doesn't have a lesson on this section. The textbook wasn't much help. How do I do this?

[native inlet]

do you know how to convert arcminutes to degrees?

[boreal light]

@native inlet dividing it by 60?

[worn yoke]

http://zonalandeducation.com/mmts/trigonometryRealms/degMinSec/degMinSec.htm

[native inlet]

yep, so if $1'=\frac{1}{60}^{\circ}$ what can you infer about $40'$?

[warm shale]

XxMrFancyu2xX

[boreal light]

that .6666 is the degree of 40'

it goes on but yeah

[native inlet]

yes! so $\cot(20^{\circ}40')=\cot\left(20+\frac{2}{3}^{\circ}\right)$

[warm shale]

XxMrFancyu2xX

[boreal light]

okay so I would get 20.66

[native inlet]

now use a calculator to find cot(20.66...) and Bob's your Uncle!

[boreal light]

then on my calculator I can do tan(20.66) (which is .377). Next would I do 1/.377?

Sorry my calculator doesn't have cot

[heavy thicket]

Are you in degree mode?

[native inlet]

yea most dont

[boreal light]

yes I turned that on

[native inlet]

so you know that $\cot\theta=\frac{1}{\tan\theta}$

[warm shale]

XxMrFancyu2xX

[boreal light]

yes I learned that the other day :)

[native inlet]

so to do cot(20.66....) do 1/tan(20.66....)

[boreal light]

got it!

yay thank you

[native inlet]

wait

yw!

also

[boreal light]

I was struggling

haha

[native inlet]

one more addendum

[boreal light]

yes?

[native inlet]

never simplify tan then put in the denom

just put 1/(tan(20.667)) into the calc

[boreal light]

oh i can do that???

[native inlet]

you could drop a very important decimal

[boreal light]

i didn't know

[heavy thicket]

1/tan(62/3) is more accurate

[boreal light]

ive been doing it the long way

thanks y'all

[native inlet]

also put it in fraction

there's another decimal 😂

[boreal light]

very helpful people indeed :)

[native inlet]

yw man! :)) have a very good day

[heavy thicket]

No need to worry about decimals

[boreal light]

you too!

[native inlet]

well its on a computer it can be kinda funky

[boreal light]

.close

[obtuse pebble]

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[river siren]

how to do this

[runic void]

@river siren

[river siren]

hi

[runic void]

are you able to identify the distance ‘d’ as a side of the triangle?

[river siren]

no right? cuz the distance is a variable

it changes based off where they are on the cliff

[runic void]

right, but if you were to identify as a side say adjacent, hypotenuse?

[river siren]

opposite?

[runic void]

with respect to thetha

[river siren]

yeah opposite right

cuz the 230 thing is the cliff right?

[runic void]

as he proceeds to move down, think the triangles has same geometry but lengths with be reduced

[river siren]

oh what

isnt d distance from top to friend

[runic void]

right, it’s some distance below 230ft

[river siren]

ye

so then what

[runic void]

Visualise an arbitrary triangle whose base is 375 ft and altitude is (230-x)

[river siren]

k

[runic void]

hypotenuse is such that it holds the Pythagoras theorem

[river siren]

proteas theorem? idt i learned that

[runic void]

we need to connect the side (230-x): opposite, 375ft: adjacent to the reference angle thetha

[runic void]

typos

[river siren]

ohok

[runic void]

^

[river siren]

so its tangent

[runic void]

right

so can you now come-up with the equation of that side in that arbitrary triangle?

[river siren]

tantheta is 230-x/375

and then solve for x?

so like x= -375tantheta+230

?

[runic void]

right

@river siren

[river siren]

hi

sry was afk

k cool

thats the final equation?

[obtuse pebble]

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[obtuse pebble]

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[modest vapor]

hi

[heavy thicket]

Post question

[modest vapor]

I need help with this:

I did the conversion in polar and all. But Im not sure how to integrate

[heavy thicket]

I can’t read French

Please translate

[modest vapor]

Eavluate the integral. D is the plane under the line y=x and above the line y=-x and in between the circles xˆ2+yˆ2=2 and xˆ2+yˆ2=9.

From what I got this is the region

[rigid pine]

You should have -π/4 <= θ <= π/4.

[modest vapor]

Thanks ill change that

Im not able to integrate <@&286206848099549185>

[native mirage]

convert to polar

[modest vapor]

I did already

[native mirage]

sqrt2 <= r <= 3

-π/4 <= θ <= π/4

did you add 'r dr dθ' at the end

[modest vapor]

yes you can see my work here

[warm shale]

Broken

[native mirage]

did you get this

[modest vapor]

oh

[warm shale]

Broken

[modest vapor]

because sqrt(xˆ2+yˆ2) = r?

[native mirage]

yes

and x^2+y^2=r^2

[modest vapor]

No i didnt get htthis thank you

[native mirage]

np

the inside should just become r^2cos^2

[modest vapor]

can I cancel the bottom r with one of the top r?

[native mirage]

i just canceled it from the one outside

the one on top becomes r^2

[modest vapor]

got it

thanks ill try to finish

[obtuse pebble]

@modest vapor Has your question been resolved?

[native mirage]

did you solve it

[modest vapor]

Im getting this as the final result

[native mirage]

is it wrong'

[modest vapor]

I think its wrong. I dont have the answer its a homework. But Im pretty sure its not the right answer

ill do it again

[native mirage]

let me try it real quick and ill lyk what i get

[modest vapor]

okok thanks

[native mirage]

i think ur right

[modest vapor]

ok im doing it again to make sure thank you

I have another question I was blocking on earlier ill make a new post and tag you if you wanna check it out

[native mirage]

ok

yeah i got the same answer

[obtuse pebble]

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[carmine verge]

any idea how to do this?

[long anvil]

Do you know how to sum a geometric series?

[glass dagger]

Recognize that you have a geometric series with the ratio b/7

[carmine verge]

yeah

is the ratio 1/7?

if I use that ration

[glass dagger]

It might be easier to see if you write it as b/7 + (b/7)^2 + (b/7)^3 + ...

[long anvil]

Nope.

[carmine verge]

ah

the ratio is b/7

so it becomes

(1/7)/(1-b/7)

[long anvil]

Close.

$1/(1-x) = 1 + x + x^2 + \cdots$

[warm shale]

Pseudonym

[long anvil]

What you have is $x + x^2 + x^3 + \cdots$.

[warm shale]

Pseudonym

[long anvil]

Or you could think of it as $x \times \left(x + x^2 + x^3 + \cdots\right)$ if that is easier.

[warm shale]

Pseudonym

[carmine verge]

yes

[long anvil]

Erm...

[carmine verge]

so what is wrong with this?

[glass dagger]

notice that the first term is b/7, not 1/7

[carmine verge]

what's wrong with this

[long anvil]

$x \times \left(1 + x + x^2 + \cdots\right)$

[warm shale]

Pseudonym

[long anvil]

That

[carmine verge]

so it becomes

b/7 / (1 - (b/7))

[glass dagger]

exactly

[long anvil]

Yup

[carmine verge]

so then

it becomes

b / 7-b

right

[long anvil]

You could also have said $\frac{1}{1-b/7} - 1$

[warm shale]

Pseudonym

[carmine verge]

oh ok

[long anvil]

Either way works.

[carmine verge]

what does that equal to tho

to solve for b

[long anvil]

Well, that's the left-hand side. The right-hand side in the question is 7.

[carmine verge]

b = 49/8

i got -1/5 for the sum

[long anvil]

I plugged that into the equation and it seems right.

[carmine verge]

oh

yeah that's right

thanks

.close()

.close

[obtuse pebble]

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[valid wave]

-x^2+4x+12 when you factorise when you divide 6 and negative 2 by negative 1 what do you put as the coefficient of x?

[tardy epoch]

What

Write out the math what you mean

[obtuse pebble]

@valid wave Has your question been resolved?

[obtuse pebble]

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[zenith yarrow]

Would this be done by the definition of a linearly independent set?

[zenith yarrow]

Like c1v+c2u+c3w=0

If not could someone give me a hint

[forest sinew]

huh?

you should start from a more clear definition

[brisk matrix]

if you're asking how you prove they are linearly independent, then sort of
you assume an equation of dependency, and show the coefficients are 0

so if you want to show x and y are L.I., you'd assume
ax + by = 0
and show that a and b must be 0

[zenith yarrow]

It's given that in your example that a and b are 0

since x and y are linearly independent

it's about the summation of the two

[brisk matrix]

no

i didn't state x and y are L.I.

i'm saying if you want to prove that they are

you assume ax + by = 0

and show the only solution is a = b = 0

that is what you should be doing for your problem

x and y could also be (a + b + c) and (d + e + f)

or they could be (u + v), (v + w), and (w + u), like in your example

[zenith yarrow]

ik but the problem states that they already are linearly independent

[brisk matrix]

no

it states u, v and w are

not (u+v), (v+w) and (u+w)

[zenith yarrow]

yeah that's what I meant lmao

x and y there would be linearly independent then no

[brisk matrix]

no

again

[zenith yarrow]

x+y is what you would be trying to prove

[brisk matrix]

my x and y have nothing to do with your problem

x and y are arbitrary vectors

i'm telling you that in general

if you want to show some set of vectors is L.I.

you assume ax + by = 0

and show a = b = 0

i didn't assume x and y are L.I.

now

you were given 3 vectors

(u + v), (u + w), and (v + w)

you want to show they are linearly independent

so what do you start with

my point is that x and y don't have to be u and v
x could be (u + v)
y could be (u - v)
and to show they are L.I. you'd do
a(u+v) + b(u-v) = 0

and show a = b = 0

[zenith yarrow]

I don't think there's any possible way for me to surmise that u+v u+w and v+w are linearly independent without knowing whether u,v,w are linearly independent

[brisk matrix]

i am aware

do you understand that to show some vectors are linearly independent

you need to show the only solution to
ax + by = 0
is a = b = 0?

[zenith yarrow]

oh lmao is that what you meant

yeah of course I do

[brisk matrix]

yeah i've said it like 4 times

[zenith yarrow]

I just meant in that specific instance I didn't really get it

[brisk matrix]

now

i also said

those vectors could be something else

like a sum of vectors

e.g.

u + v
v + w
u + w

[zenith yarrow]

I already figured it out cuz of a stack exchange discussion I found

[brisk matrix]

and if we wanted to show theay are linearly independent

ok

[zenith yarrow]

thanks though

:D

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[obtuse pebble]

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[obtuse pebble]

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[untold cloak]

hello

[untold cloak]

I need a help in a equation

rd sharma is killing me really badly

-5√3 + √27/ 4

[untold cloak]

how can I solve it

(i)-5√3 + √27/ 4
(ii) -2√5 - 4√5/ 6

I am having problem in finding the answer of these 2 question

[mellow glacier]

root27 can be rewritten as something else

which can then be added

[untold cloak]

in which form can i write it

[languid pelican]

3√3

[untold cloak]

how

[mellow glacier]

root27

What is the prime factorization of 27

[languid pelican]

27 = 9 × 3

[untold cloak]

9 and 3

[languid pelican]

9 is square of 3

[untold cloak]

ohhhhhh

[languid pelican]

Comes out root as 3

Yep

[untold cloak]

and then

[mellow glacier]

$\sqrt{3^2 * 3}$

[languid pelican]

Add

[warm shale]

Shockshwat

[untold cloak]

hmm

[mellow glacier]

= $3\sqrt{3}$

f

[untold cloak]

root 33

[languid pelican]

[untold cloak]

how

lol

[languid pelican]

No

[warm shale]

Shockshwat

[languid pelican]

Yes

[untold cloak]

ohh

i see

and what about the second one

[mellow glacier]

are you in 10th?

[languid pelican]

Which class, 10th?

Lol

[untold cloak]

yes

[languid pelican]

Karlo bete

[untold cloak]

it is my first time doing rd sharma

[mellow glacier]

second one is pretty straight forward

[languid pelican]

LCM lo

[mellow glacier]

just get the denominator same

[untold cloak]

bhai pehli baar rd shrma kar raha

[untold cloak]

smjha nahi

√5

[mellow glacier]

honestly just do ncert its more than enuf for boards

[untold cloak]

ans hai

[untold cloak]

arree mera tution teacher de raha karne ko

[languid pelican]

Mene math exam ke raat ko rd sharma karne ki kosis ki.
4 ghante soke gaya.

Ek bhi RD sharma se nahi aaya

[mellow glacier]

bhai add kardo na dono ko kya dikkat hai add karne me

[untold cloak]

hmm

ok mai karke dekhta hu

[untold cloak]

bhai mera math mai bahut kaam aya hai

45/100

💀

9th mai

[languid pelican]

If you're having difficulty, imagine the √3 as t

[mellow glacier]

fir to bhai tum ncert puri karo

[untold cloak]

hmm

[mellow glacier]

rd sharma ki aukaat nahi hai abhi

[untold cloak]

oh

[languid pelican]

NCERT nahi hua toh base taiyar nahi hua

[untold cloak]

ok

[untold cloak]

par mera tution teaccher de raha karne ko

mereko koi maan nahi hai woh karke ka

dukh seh raha hu mai

[languid pelican]

Sikh lo, jo sikh sakte

[untold cloak]

💀

[untold cloak]

hm

[languid pelican]

Ye karke dikhao hame

[mellow glacier]

tuition teacher ko nahi pata fir ki tera kya haal hai

use bol de ki bhai ncert to ho nahi rahi RD kyu de rha hai

[untold cloak]

-2√5 - 4√5/ 6 thoda karke dikha do ye wala

[untold cloak]

han

[untold cloak]

bhai usne ncert karwa liya hai

abhi rd karwa raha

[mellow glacier]

bhai ye 7th class ke question hai lekin

[untold cloak]

💀

[languid pelican]

Wait

[untold cloak]

ok

[languid pelican]

Galti

[untold cloak]

par answer tou √5 hai

[mellow glacier]

bhai tujhse denominator ka LCM nahi liya ja rha kya karega

[mellow glacier]

bhai

!nosol

[languid pelican]

?

Oh ok

Got it

[untold cloak]

bhai usme LCM kaha pe lena hai

💀

[mellow glacier]

!nosol

[languid pelican]

LCM tumhe lena aata he?

[untold cloak]

han bhai aata hai

[mellow glacier]

@obtuse pebble bhai kya ho gaya

[languid pelican]

Yes, abiding by your principle

[untold cloak]

par yaha pe kiyu lena hai

[rocky goblet]

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[mellow glacier]

denominator me

[untold cloak]

💀

samaj nahi aa raha kiya keh rahe

[mellow glacier]

what do you need to have for this command to work?

[untold cloak]

karke batado mai khud samaj jaunga

[languid pelican]

Tum root ko 't' consider karo naa.

Phir photo maarke bhejo

[untold cloak]

:skull

[rocky goblet]

it's nosols, not nosol

[mellow glacier]

oh ic

[languid pelican]

No life

[untold cloak]

💀

kaun se root ko t karna hai

[languid pelican]

@mellow glacier physics aata he?

[untold cloak]

mai second wala puch raha

[languid pelican]

√3

[mellow glacier]

konsi class ka

mai to 11th me hu

[languid pelican]

12th

[untold cloak]

usko t kiyu lu

[mellow glacier]

try kar sakta hu

[untold cloak]

aur t lene ke baad kiya karna hai

[languid pelican]

Are PEE aata he?

[mellow glacier]

full form?

Current electricity kar sakta hu optics thoda weak hai

[languid pelican]

Einstein's Photo Electric Effect

[languid pelican]

Never studied this thing 😂

[untold cloak]

bhai koi karke batayega 2nd number

💀

[languid pelican]

Tumhe karna chahiye

[untold cloak]

bhai karke batado mai samaj jaunga

💀

[languid pelican]

@mellow glacier

[languid pelican]

Try man

[untold cloak]

aree bhai

[untold cloak]

3 bar kar liya hai nahi ho rah

[mellow glacier]

wahi hota hai na electrons ko metal pe marte hai

[untold cloak]

karke photo bhejne mai kiya hota hai

[mellow glacier]

papa se help mang

[languid pelican]

Yes, show us the calc

[mellow glacier]

bohot tagdi help milegi

[languid pelican]

Yep

[untold cloak]

I am no helpable

[mellow glacier]

E = hv

[untold cloak]

mai chalta hu

💀

.close

[obtuse pebble]

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[azure anchor]

i found y' and y'', and plugged it into the d.e., but i'm not sure how to proceed.

it forms a quadratic kinda, and i can factor out e^rx, but i'm not sure how e^rx can ever be zero, so that leaves x^2 + x - 1 and the roots of that are 1/2 and -1

but those are the roots concerning x, which aren't what i need

so i differentiated with respect to x instead of r and those become the roots of r

but now on part b im not sure how to proceed

i thought about e^rx = ae^{r_1*x} + be^{r_2*x} but idk how i'd show that it's a solution

[obtuse pebble]

@azure anchor Has your question been resolved?

[azure anchor]

.close

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[wanton dagger]

in this function, as the value of n increases, the graph becomes compressed and as n goes close to zero , the graph stretches out right?

[fathom flicker]

By compressed, do you mean vertically or horizontally?

[wanton dagger]

horizontally

[keen cedar]

Yes, increasing n increases the frequency of the sine wave, so it gets compressed along the x axis

[wanton dagger]

right

ok thanks

.close

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[timid silo]

How do I do part b?

[unique nebula]

How do you think a force body diagram on the m mass is like?

[knotty gulch]

F = ma

[timid silo]

29.4-14.7=m(9.8-4.9)

is it true?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[wanton dagger]

what transformation is present between a sin and cos graph?

[fathom flicker]

What transformations do you know of?

[wanton dagger]

there’s amplitude, period, phase shift

amplitude and period are the same though

so I think it’s just the phase shift we should focus on here

but how would I comment on the phase shift between the sin and cos graphs?

i’m a bit stuck there

[gilded needle]

what's cos(0)?

[wanton dagger]

that would be 1

[gilded needle]

ok, and where is sin equal to 1?

[wanton dagger]

90’

[gilded needle]

yea, aka pi/2

[wanton dagger]

yep

[gilded needle]

so that sounds like it could be the phase shift you're looking for

you could try checking another point to be sure

[wanton dagger]

how do I write that though?

[gilded needle]

well so far you have that cos(0) = sin(pi/2)

so a working hypothesis could be:

cos(x) = sin(x + pi/2)

you could try another value of x to check, maybe x = pi/2

[wanton dagger]

i see

[gilded needle]

OR

if you want to be clever

you could use the trig identity for sin(x+y)

and apply it with y = pi/2

and see what you get

[wanton dagger]

sure

ok I got it

thankyou

.close

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[obtuse pebble]

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[shell hill]

what do I do

[thick chasm]

Let 2^x = y and show that d<0

[shell hill]

so y²+3y+8=0 ?

[thick chasm]

y^2+ 6y+8=0

2^x+1 = 2× 2^x

[shell hill]

why 6y

you said 2^x = y

[thick chasm]

3 × 2^x+1 = 3×2 × 2^x = 6×2^x = 6y

[shell hill]

ohh nvm

Thank you

.close

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[proven musk]

How to do the example 5 problem

[proven musk]

I showed you my work, but i’m unsure of how to handle this problem

[wet drift]

Rank of A is not equal to A agumented B so no solution

[proven musk]

Look at the second image the say 0=0

[wet drift]

I am answer on the basis of your solution

In black page

[proven musk]

But I tried it in the third image and it doesn’t from my attempt

[wet drift]

One

[proven musk]

The textbook(paul’s notes on linear algebra says this….

https://media.discordapp.net/attachments/903430332324405288/1110150473496932373/image1.jpg

[wet drift]

Okk so first tell me do you know agumented matrix?

[proven musk]

Yes

[wet drift]

Do you know the rules

When rank of A and A agumneted B are not equal there is no solution

[proven musk]

Yes they showed me this

and my school said this

and another textbook

[wet drift]

When rank of A is equal to A agumented B but not equal to number of variables then infinite solutions

Here is the case of when Rank of A and A agumneted B is equal but varibale numbers are not equal

Because in your question rank of A and A Agumneted B are 2

But variables are 3

So in this care 3-2 = 1 varibale will be independent

I.e you can take that varibale at t ( generaly they take t)

Or you can also call it free variable

[proven musk]

Why did I get it wrong?

Why did I get 0 = 1 instead of 0 = 0?

[wet drift]

May be you made some mistake in transformation

[proven musk]

Can you check?

I tried to already

[wet drift]

Ok

Let me check

Where is your solution?

That you tried

Check 2nd matrix when you made a31 zero

You make a mistake in last part

-7+4 = -3

Not 3

[proven musk]

Oh I see I did the operation right but inputted the variables wrong

Might I might have had to restart in order to fix that without you

[obtuse pebble]

@proven musk Has your question been resolved?

[obtuse pebble]

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[shell hill]

can someone check if this is correct

[rich plume]

That is all correct

[west sierra]

You assumed y 5^x

[shell hill]

Yeah

[rich plume]

oh wait

yes

5^x=5,1/5

now solve for x

[west sierra]

It can't be negative

[shell hill]

Which negative?

[west sierra]

5.2-

You used quadratic formula there, right?

[shell hill]

Yes

[west sierra]

Nvm

They both are positive

[rich plume]

solve after this

[shell hill]

Its correct?

[rich plume]

5^x=5,1/5

[rich plume]

you have solved for values of y which is 5^x

[west sierra]

Why did you substitute y with x on using Quadratic formula

That's the only incorrect step

Because you assumed 5^x as y and you were solving for y.

[shell hill]

I just used a random letter

[west sierra]

so this is rough work

But that random letter is not random because there is alr a x used before it

[rich plume]

You assumed 5^x=y
Using the quadratic you solved for y
but you need to solve for x

[shell hill]

Dont I just need to find x

[west sierra]

So it caused confusion for you to finalize answer for y as x

Am I right?

[shell hill]

oh wait you mean y is actually 5^x , not x

[west sierra]

Yes

So it's y= 0.2 and 5

Not x=0.2 and 5

[shell hill]

x=1 x=-1

[west sierra]

Yep

[shell hill]

Ok thank you

[west sierra]

I'd suggest you try put those values into original question once

So you can be confirmed they satisfy the answer

[shell hill]

Thanks 🙏❤

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[deft pier]

I'm going through the book called 'Algorithms Illuminated - Part 1' by Tim Roughgarden. He mentions that it takes 2n operations to calculate one partial product. In my observations, it actually takes 2n-1 operations. Can someone validate the statement in the book, perhaps by providing suitable example(s)?

[deft pier]

For example, in 1234x5678, calculating 1234x8 requires 8x4, 8x3, 24+3, 8x2, 16+2, 8x1, 8+1 ⇒ 7 operations = (n-1) operations

[obtuse pebble]

@deft pier Has your question been resolved?

[obtuse pebble]

@deft pier Has your question been resolved?

[obtuse pebble]

@deft pier Has your question been resolved?

[obtuse pebble]

@deft pier Has your question been resolved?

[obtuse pebble]

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[static shoal]

help

[static shoal]

how do I approach this?

[knotty crow]

notice that

[warm shale]

Modus

[knotty crow]

and use the property

[warm shale]

Modus

[knotty crow]

(it works for more logs as well)

in fact this can be done in memory knowing this

without calculations

[static shoal]

i need to do this using paper and pencil 😅

[knotty crow]

but you see how it works, right?

[static shoal]

i got 2

is it correct?

@knotty crow

.close

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[silver plover]

for part b i have shown it is a basis

how can i now show its orthonormal?

orthonormal means inner product = 0 and all of length 1

[abstract beacon]

Calculate their inner products

Ur shown how to do this

<X,Y> = ½ tr(Xdag Y)

[silver plover]

but whats X and Y in this case

sigma 1 and sigma 2?

[abstract beacon]

every pair from the four sigmas

[silver plover]

bruh do i need to show this for every pair?

[abstract beacon]

1,2; 1,3; 1,4; 2,3; 2,4; 3,4

these should all be zero

1,1; 2,2; 3,3; 4,4

these should all be 1

[silver plover]

yh

do i need to show this

for every pair

tho?

cuz thats so long

[abstract beacon]

i think so but maybe not but i cant see how you could get away with it

[silver plover]

the MS said that

i cant tell from it if they want u to show it or not

[abstract beacon]

... a direct calculation shows that <oi|oj> = dij
Yeah your gonna have to calculate

[silver plover]

for all the pairs

bruh thats so long

ight thanks

[abstract beacon]

ur lucky its conjugate symmetric else youd have to check 12 pairs

[silver plover]

so no need to show 2,1 3,1 4,1 3,2 etc..?

[abstract beacon]

Yeag

bc 0* = 0 and 1* = 1

[silver plover]

ight cool

ill try part c now

[silver plover]

stuck for part c i

i got this so far

[abstract beacon]

Fun fact matrices commute inside a trace

tr(AB) = tr(BA)

not sureif this helps

[silver plover]

im getting somewhere

ill get back to u soon

@abstract beacon I got this

But at the end it doesn’t agree with the determinant

I can’t see what I’ve done wrong

[obtuse pebble]

@silver plover Has your question been resolved?

[silver plover]

<@&286206848099549185>

[silver plover]

<@&286206848099549185>

[silver plover]

.close

[obtuse pebble]

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[sterile stratus]

I’ve already tried this but idk if it’s right

[bitter gate]

Can I have help pls with this I have been struggling for the past 30 minutes

[obtuse pebble]

Please read #❓how-to-get-help

[tawny fog]

It is right

[sterile stratus]

Thank you! What about this one

I don’t understand it at all 😭

[obtuse pebble]

@sterile stratus Has your question been resolved?

[tawny fog]

Yes I can but I don't have that much charging

Let's just say Volume = V

And rate of change of volume is ∆v

We need to calculate rate of change of volume first means ∆v

By using y-y1=m(x-x1)

v-113=∆v(x-3)
v-137=∆v(x-7)

My battery is at 8%

Btw x is the time here

At last you can write f(t)=∆v(x-3)+ initial volume of water

Ok I have to go now

[obtuse pebble]

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[supple creek]

how do you find the vector c1, c2 and c3, when you have the basis B = b1 b2 b3 and subspace b1 b2 b3 ?

[obtuse pebble]

@supple creek Has your question been resolved?

[obtuse pebble]

@supple creek Has your question been resolved?

[supple creek]

<@&286206848099549185>

[obtuse pebble]

@supple creek Has your question been resolved?

[obtuse pebble]

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[weary silo]

Hello, guys I need help with this de-composition of functions questiong

[weary silo]

Aren't there 3 possible solutions to this? Isn't
h(x) = x^3
g(x) = x + 5
f(x) = 3/x

Also a possible solution?

[keen cedar]

With this composition of functions we have f(g(h(x))). So take your h(x) and sub into g(x) for g(h(x)). Repeat by subbing this into f(x) for f(g(h(x))).

This should show you why these can't work as options for f, g, and h.

Specifically the problem is with g.

[obtuse pebble]

@weary silo Has your question been resolved?

[weary silo]

can f(x) be 3/squareroot(x)?

[keen cedar]

Yeah, that works

[weary silo]

thanks

.close

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[valid grail]

<@&286206848099549185>

[bold carbon]

hello

so there is actually a formula for finding the degrees per angle

[valid grail]

ok

[bold carbon]

so the formula is the number of sides, minus two, times 180

and then divide by number of sides

[valid grail]

ok

[bold carbon]

that should clear things up

[valid grail]

ok thanks i have another problem

[bold carbon]

oh

so what do all the angles of any quadrilateral add up to

[bold carbon]

divide 1800 by 12

the number of sides

then solve for x

[bold carbon]

solve for x

[valid grail]

?

[bold carbon]

so 150 would be the measure of any angle here

so then u set that equall to 9x+1.5

[valid grail]

ok

[bold carbon]

and then solve for x

[valid grail]

i don't get it

[bold carbon]

so since allt eh angles in the polygon have 150 degress

do u know how?

[valid grail]

what

[bold carbon]

so what is a regular polygon

[valid grail]

a shape that could have a infinite number of sides?

[bold carbon]

not quite

it is a polygon that has all of its sides equal to eacher other

and all the ngles are also congruent

make sense?

[valid grail]

not really

[bold carbon]

use this

https://www.mathsisfun.com/geometry/regular-polygons.html

[valid grail]

i see

[valid grail]

so would i do 9 x 150 + 1.5?

[bold carbon]

make sense?

[valid grail]

kinda

[bold carbon]

alright

now

[bold carbon]

not quite...

[valid grail]

what would i do

[bold carbon]

so now that you know what a regular polygon is, we need to find what the total degree measure of the polygon

so what do allt he angles add up to in a triangle?

[valid grail]

180?

[bold carbon]

yeo

p

and what about a quadrilateral

[valid grail]

240?

[bold carbon]

nope

not quitre

[valid grail]

idk

[bold carbon]

360 degrees

[valid grail]

oh

[bold carbon]

but the pattern is this:

(n-2) x 180

where n is the number of sides

this determines the total degrees in a shape

so

example

how many degrees would a pentagon

have

[valid grail]

540?

[bold carbon]

yep

now how many for the figure above?

[valid grail]

1800

[bold carbon]

yep

but now

we realize that this is a regular polygon

[valid grail]

ok

[bold carbon]

so if all the angles are equal

and they all add up to 1800

what would the measure of each angle be

[valid grail]

150

[bold carbon]

yep

but now we realize that one of the angle equalls 9x+1.5

but it also equals 150

so what do we do

[valid grail]

solve for x?

[bold carbon]

yes

9x+1.5 = 150

[valid grail]

is it 16.5

[bold carbon]

yep

good job

[valid grail]

YES THANK YOU

[bold carbon]

np

[valid grail]

so now what about this one

[bold carbon]

yep

so using the princibles of the first problem

what would the total degrees of the trapazoid equal

[valid grail]

360?

[bold carbon]

yep

and this isn't a regular polygon necessarily

but

if all the angles add up to 360

but they also add up to

(8x+1)+(17x+4)+(9x-6)+(18x-3)

what would you do?

[valid grail]

so does each angle equal 90?

[bold carbon]

no

because they never specified that the quadrilateral was regular

if they did, then yes

[valid grail]

so is it irregular

[bold carbon]

yea

or no

we cant assume

but we can just add up all the angles and set that equal to 360 degrees

(8x+1)+(17x+4)+(9x-6)+(18x-3)=360

now, solve for x

[valid grail]

ok

[bold carbon]

yea

gl

[valid grail]

is that right

[bold carbon]

no

[valid grail]

do i divide that by 4 or 360

[bold carbon]

its not a regualr polygon

so you cant do either

you have to add everything up and set it equal to 360

[valid grail]

so what do i do

[bold carbon]

here

since everything adds to 360

we can set an equation to solve for x

add allt he angles = 360

(8x+1)+(17x+4)+(9x-6)+(18x-3)=360

make sense

?

[valid grail]

i think

[bold carbon]

good

so now we have to solve this equation

(8x+1)+(17x+4)+(9x-6)+(18x-3)=360

so you should combine like terms

[valid grail]

i can't do it

[bold carbon]

alright

its fine

so

[valid grail]

i give up

[bold carbon]

do

no

dont worry

its not that hard

so

do you know what i mean by combining like terms

[valid grail]

no

[bold carbon]

alright

if i asked for 2 fries

a big mac

3 more fires

2 more big macs

a soda

another soda

whats wrong with that

[valid grail]

why not add the soda together?

[bold carbon]

exaclty

thats the problem

now apply it here

(8x+1)+(17x+4)+(9x-6)+(18x-3)=360

wouldn't you just add the xs

and the constants

?

[valid grail]

52x^4?

[bold carbon]

no

not exactly

here

this is a good website

https://www.mathsisfun.com/algebra/like-terms.html

[valid grail]

so just 52x

[bold carbon]

yes

plus what

[valid grail]

-4

[bold carbon]

yes

so 52x-4=360

solve for x

[valid grail]

7

[bold carbon]

yes

now

to find each angle

plug in 7 for x

for example

angle m

is 9x-6

9x7 - 6

[valid grail]

ok

[bold carbon]

63 - 6

etc.

so just do that for all angles

also

really quick

what is the website called

i might want it for practice and stuff

[valid grail]

wdym

[bold carbon]

what is the website ur practicing on called

[valid grail]

its not practice its homework

[bold carbon]

ok

nvm

i think you can close the channel now

i think you've got i

t

[valid grail]

ok

[bold carbon]

good job

[valid grail]

how do i do it

!close

.close

[obtuse pebble]

Channel closed

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[timid silo]

can someone explain how they were able to factor out the x^a-1 and the (1-x)^b-1

[native inlet]

$x^a\cdot b(1-x)^{b-1}(-1)+(1-x)^{b}\cdot ax^{a-1}\ \implies x^{a-1}[x\cdot b(-1)(1-x)^{b-1}+(1-x)\cdot(1-x)^{b-1}\cdot a]$

do you see how they got those terms factored?

[warm shale]

XxMrFancyu2xX

[timid silo]

ehhhhh

[native inlet]

yea there's a lot happening in this problem ain't there

[timid silo]

ya

im just confused the first half doesnt have x^a-1

so how can it be factored

[native inlet]

$x^a=x^{a-1}\cdot x$

[warm shale]

XxMrFancyu2xX

[timid silo]

how is that possible?

[native inlet]

$x^{m}\cdot x^{n}=x^{m+n}$

[warm shale]

XxMrFancyu2xX

[native inlet]

let n=1 and m=a-1

hence $x^{a-1}\cdot x^{1}=x^{a}$ or $x^{a-1}\cdot x=x^{a}$

[warm shale]

XxMrFancyu2xX

[timid silo]

oh ok i see it now

one sec

so i did that

but for my crit number

i get a/a-b

the answer key says a/a+b

oh nvm

ok

i think ig ot ti

it

thx

.close

[obtuse pebble]

Channel closed

Closed by @sturdy dragon

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[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[stoic merlin]

I'm working on this problem, Im having trouble understanding why we used an alpha of 0.95 instead of 0.5 on the table

[stoic merlin]

When the graph is telling me that the left-tail is the one associated with 0.05

[obtuse pebble]

@stoic merlin Has your question been resolved?