help-28

[languid junco]

then you're done

[torn jolt]

Is that like the formula for the shape

[languid junco]

that's the area of a rhombus yes

[torn jolt]

I got 19,6? 🥲

The answer sheet says its 5,7

[torn jolt]

Is it 19,6

[languid junco]

well the answer sheet is definitely wrong...

the area of the square below it is 16, and the area of the rhombus is definitely more than that

[languid junco]

if you round it then yes

[torn jolt]

Okay thank you!!!!

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[torn jolt]

Trying to solve this integral, but this indeterminate block is preventing me from doing that. I try using L’Hôpital’s rule to fix this, but it only ends up with more indeterminate things. The answer is converging to -1, so this circled term is supposed to be equal to 0, but I don’t know why or how.

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@torn jolt Has your question been resolved?

[torn jolt]

Also, although a different question, my partial fractions is also in bad shape. Although I got the right numbers, I got their signs flipped. A is supposed to be positive and B negative, but I got them switched up. Why is this?

Hmmm, in my opinion partial fractions are correct

Okay, I’ll bring it up to my instructor. Thanks.

[torn jolt]

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[full forum]

Please don't occupy multiple help channels.

[timber scaffold]

ok sory

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[paper bane]

So Im trying to understand this example. I get up to the x-int point, but to get pi/8 its supposed to be 1/4th of pi/4, but wouldnt that be pi/16?

[paper bane]

to make pi/4 = pi/8 wouldnt that be half?

[glacial pasture]

im not sure what you mean

[paper bane]

I think I might have found what I was missing.

the x-int is the halfway point

[glacial pasture]

ig, thats not how id think about it

just always relate it back to tan(x)

look at the zeros of tan(x), then transform them

[paper bane]

Im trying to understand the transform part

isn't that the 1/4th and 3/4 part

[glacial pasture]

those are the coordinates with +-pi/8 in them

[paper bane]

How do I get pi/8 is what Im asking

[glacial pasture]

on a regular tan graph, youre looking at between -pi/2 and pi/2
a quarter of the way along is -pi/4, 3 quarters is pi/4

tan of pi/4 is 1, of -pi/4 is -1

2x=+-pi/4 so x=+-pi/8

theres a vertical stretch of 3, so +-1 is +-3

(-pi/8,-3), (pi/8,3)

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[exotic shoal]

I'm trying to do this problem at the moment. I went with an overestimate using a right hand sum. I got -92 but I'm not sure how to check if it's correct. I was hoping someone here could check that? My math was 2(7+0-7-15-18-13-0)=-92.

[sick berry]

as you see

the maximum value of the function is 20

and minimum is -20

[exotic shoal]

Could you explain how that's relevant please? I don't get it.

[exotic shoal]

ah. so then my answer is too small.

[sick berry]

do you have the answer?

[exotic shoal]

I was not given the answer to the problem sadly.

[sick berry]

integral should lie between

um

-160 and 80

[exotic shoal]

oh. then yay! -92 is between those :3

[sick berry]

i think by estimation they meant the range

[exotic shoal]

doesn't the integral just mean the Riemann Sum?

[sick berry]

Riemann?

whats that

[exotic shoal]

https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-2/a/left-and-right-riemann-sums

[sick berry]

oh this is called riemann sum

[exotic shoal]

yee ^W^

that's what I'm trying to use to find an estimate of the integral

[sick berry]

yup you can then

[exotic shoal]

we haven't started using limits to find the exact value yet in my class

^W^

thanks for ur help! ^^

[sick berry]

have a good day

[exotic shoal]

you too!

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[obtuse harness]

i need help on c and d

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@obtuse harness Has your question been resolved?

[obtuse harness]

<@&286206848099549185>

[midnight breach]

i think c should be approx 34 because you take normcdf and the lb is 22 and ub is 34 then plug mean and sd to get answer.

And I think d should be approx 143 people.

because the data should be the same cause its approx normal because n>30 so we can just use same statistical informaiton and then apply that to 150 and not 50

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[hearty onyx]

hey guys do i just plug in 3 for this?

[stuck storm]

you have to prove that the $\lim_{x\to 3}f(x)=f(3)$

[glossy valve]

The Great D

[olive dew]

there are 3 requirements for continuity

oh

[olive dew]

yeah this basically does all 3

[hearty onyx]

oh so should i factor

[stuck storm]

sure, to evaluate the limit

[olive dew]

for polynomials you can just plug in the value to obtain the limit, you can also prove it rigorously using an epsilon-delta proof

[viral jasper]

Technically, you can't just plug in the value to obtain the limit, as that is based on the fact that the polynomial is continuous, which we are trying to prove. So it sounds like OP is stuck with epsilon-delta proving this.

[olive dew]

yeah that makes sense

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[marsh field]

I dont understand why this problem does not have a practical interpretation of the vertical intercept

[stuck storm]

because t = 0 would imply we are at the year 0

[marsh field]

but it says that it starts at 1987

so wouldn’t year 0 be 1987, there was 923769 banks

[stuck storm]

I'm not sure

the question is not really stating that

[proven mango]

it says t is the year so i’m assuming at 1987, t=1987

that’s just my interpretation

the question doesn’t specify much

[marsh field]

but the function is based off years 1987 to 1997

[stuck storm]

yes thats ok

at year 1987 we would have 923769-1987*458 banks

which is 13,723

this is a much more reasonable number

[marsh field]

so the number of banks in 1987 is not the vertical intercept?

[stuck storm]

no

[marsh field]

so i just adjust my interpretation and make it so that t = year instead of t is year since 1987?

[stuck storm]

thats my understanding of the question yes

[marsh field]

okay thank you

[stuck storm]

of course

[marsh field]

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[coarse swallow]

can someone explain how wolfram alpha got this i cant get past this step

[dusty scaffold]

,rccw

[dusty scaffold]

i think substituting u = x^2 does it

[coarse swallow]

what do i do with xdx

[coarse swallow]

is du = 2xdx

[dusty scaffold]

yes

that was the idea, the extra x vanishes into the du and you just end up integrating e^u/2

[coarse swallow]

thank you

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[obsidian barn]

I have a question about epsilon closures
so for the starting dfa state i came up with {q0,q2,q3,q4}
from there if i take 0 as an input would the next step include q0 again? making the next state look like {q0,q1,q2,q3,q4,q5}?

[obsidian barn]

the unlabeled orange arrows i drew are just the epsilon paths

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[obsidian barn]

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[shadow quest]

I’m a little confused on how to find the range

For f(x)

Before I start this problem

And to be honest the domain sell

As well

Because the denominator can’t be factored

So I guess the domain is all real numbers

<@&286206848099549185>

[silent quest]

U first need to do the substitution for the question needed

[shadow quest]

?

[silent quest]

Like 6a, u need to substitute x+4 into f(x) first

[shadow quest]

Oh

I think I see

[silent quest]

And simplify that thing into the format of a/b

Then find the values when b = 0

And that's the domain

Cuz in a fraction, the denominator cannot be 0

[shadow quest]

Ye but

That doesn’t help me I just tried

Even i plug it in

The denom can’t be factored

And the domain just stays all real numbers

Unless you could work out of the first one

And show me possibly

Because maybe I’m just doing it incorrectly

[silent quest]

Ys the domain will be real nunbers if the denominator cannot be factorized

[shadow quest]

But how would I determine the range

I know it would shifted 2 units down

But wouldn’t I have to determine the original range of f(x)

[silent quest]

Range is real numbers

[shadow quest]

Ye

[silent quest]

Done

[shadow quest]

???

But it’s not

[silent quest]

How

[shadow quest]

It’s not all real numbers

[silent quest]

How?

[shadow quest]

It has the same end behavior as

1/x

So it has to be as x approaches x infinity f(x) = 0

Right????

[silent quest]

Then, make the form of a+(b/c)

[shadow quest]

Here let me plug in demos

[silent quest]

a is the range of start or end

[shadow quest]

So it says it’s [-.256, .256]

But I don’t know how they got it

[silent quest]

Ys find the asymptotes

The horizontal asymptote

[shadow quest]

How would I find those

[silent quest]

Using derivatives

Have u learnt it?

[shadow quest]

I’m in pre calc

So no

[silent quest]

Umm

[shadow quest]

Maybe my teacher made this problem incorrectly

[silent quest]

Try this

[shadow quest]

And it should be x^x +x +6

[silent quest]

No, i remember wrong, this type of question no need dervivatives

[shadow quest]

Oh ok

[silent quest]

Make the form of a+(b/c)

[shadow quest]

Wdym

[silent quest]

Like g(x) = 1 + 2/3x

Make it to the form of this

[shadow quest]

I don’t think I follow

[silent quest]

Divide the numerator by the denominator

Have u learnt algebric division

[shadow quest]

Ye

[silent quest]

Then use it

Do it

[shadow quest]

I know this annoying but I have to go

My parents called for dinner

[silent quest]

Ok

Cya

[shadow quest]

But after I do long division

What should be my next step

[silent quest]

a is ur range

a is the asymptote

[shadow quest]

Ohhh ok

Thank you

[silent quest]

Gl

[shadow quest]

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[tired jay]

Statistics

[tired jay]

if you have ti 83 calculator

A recent survey reported that 49​% of adults in a certain country own tablets. Using the binomial​ distribution, complete parts​ (a) through​ (e) below.
c. What is the probability that out of the next six adults​ surveyed, at least four will own a​ tablet?

I tried binomcdf(6,0.49,3) and it gave me 0.6748

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@tired jay Has your question been resolved?

[tired jay]

<@&286206848099549185>

[gritty rose]

What is the issue

[tired jay]

how i do this with a ti 83

[gritty rose]

What's wrong with this

[tired jay]

ig its the wrong answer

atleast my hw telling me that

The probability is
  
0.3251.

[gritty rose]

What do you think binomcdf(6,0.49,3) tells you

In words

[tired jay]

fixed number of trials is 6, probability of success is .49, and 3 is the fixed number of success

binomial cumulative distribution

[gritty rose]

That doesn't explain what you think binomcdf returns

It's the probability of something

[tired jay]

cumulative probability

probability that x is less than or equal to x

[tired jay]

bro

-1

[gritty rose]

?

[tired jay]

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[plush egret]

what have you tried?

[austere slate]

my thing is I dont understand it

Ik how to find the transformation of two points

idk exactly what to do here'

[plush egret]

lets start with the projection matrix

do you understand what this matrix should do?

[austere slate]

I dont

[plush egret]

do you know what projection means?

theres a normal picture they like to put into peoples heads

like a flashlight, and a shadow

[austere slate]

yes I do

[plush egret]

so this just means

we shine a light on it from the right or left

the shadow gets cast onto the y axis

so like, say a point is at (4,5)

and we project it onto the y axis

wheres it go? what coordinate?

[austere slate]

I am not following anymore

[plush egret]

like in this picture

this would be projection onto the x axis

v is some vector, alternatively a point

lets say v is ... (4,1)

and then, we project it onto the x axis

and it becomes the red vector, alternatively coordinate

and now, its at (4,0)

because its just as far to the right as it was before

but we removed all the vertical information by projecting it

do you see what i mean?

[austere slate]

a little yes

[plush egret]

say v is <4,1>, then the red vector in the picture is <4,0>

okay

say we projected on the y axis instead

what would v become?

right now its <4,1>

well, projecting on the x axis removes vertical information ...

so projecting on the y axis should remove horizontal information ...

[austere slate]

wouldnt it be 0,1

[plush egret]

it would not

it would be <0,1>

[austere slate]

thats what I meant my bad

[plush egret]

youre good

so, you see what we have to do, right

if we want to project a vector onto the y axis

lets make it more general

[austere slate]

yes I do now

[plush egret]

say the vector is <a, b>

what should it become?

[austere slate]

it would be 0,b

right?

[plush egret]

yea

only vertical information

[austere slate]

yep yep I see it now

[plush egret]

so, we gotta design a matrix that does this

[austere slate]

yah but idek how to put that into a matrix

[plush egret]

well, you can guess

good place to start is like this

1's and 0's

do you know how to do vector matrix multiplication?

[austere slate]

yes I do

[plush egret]

okay, so start from the identity

$\mqty(\imat{2}) \mqty[x\y]$

[glossy valve]

jan Niku

[plush egret]

problem is, we do this

we get <x, y> back

i guess as youd maybe expect

but we want <0, y>

[austere slate]

yes

[plush egret]

err

yea

so whats the natural thing to try?

[austere slate]

but isnt x negative

[plush egret]

here? no, its just the identity matrix

[austere slate]

no I am talking about my question

[plush egret]

we can address that next

lets get this one first

[austere slate]

but if I wanted to get <0 y> wouldnt I multiply by <0 1>

[plush egret]

you want a matrix

so start here

https://media.discordapp.net/attachments/903486480985489478/1208266633819201566/331558346333224964.png?ex=65e2a8fe&is=65d033fe&hm=1c4e04c9a73c253b8c3ab6d88f56f587277b1fceb39462a140e789cc2e8e99ef&=&format=webp&quality=lossless

and this about what we want

we want <0, y>

what can we try to change in the matrix

what seems reasonable to try

remember we are only guessing 1's and 0's

make a 1 into a 0

or make a 0 into a 1

what do you try?

[austere slate]

make the first 1 into a 0

[plush egret]

sure

$\mqty( 0 & 0 \ 0 & 1 ) \mqty(x\y) = \mqty( 0x+0y \ 0x+1y )$

[glossy valve]

jan Niku

[plush egret]

did it work?

[austere slate]

I mean we got 0 and y

[plush egret]

[austere slate]

so I would say yah

[plush egret]

okay halfway done

[austere slate]

oh gawd there is more??

[plush egret]

you still need the reflection about the line y=-x

this one is easier to do graphically

[austere slate]

would I just change the 1 into a -

[plush egret]

well

do you know what you want yet?

i mean like, before we knew we wanted <0, y>

can you tell where P and Q go

https://www.desmos.com/calculator/cegk6ttrtv

when we reflect across this line

looks like <1, 0> goes to <0, -1>

how about <2, 1>?

[austere slate]

<-2,-1>

[plush egret]

not quite

<-1, -2>

[austere slate]

oh hold up

[plush egret]

see the pattern?

sure

[austere slate]

yep I see it now

[plush egret]

and you can maybe just tell like

we have a vector <x, y>

[austere slate]

I wasnt following but now I see

[plush egret]

and the quation says y=-x

so we could just write <-y, -x> directly

so, we wanna start with <x, y> and go to <-y, -x>

i can give you a starting point with this one

you fill in the gaps

heres the start:

$\mqty( 0 & 1 \ 1 & 0 ) \mqty(x\y) = \mqty(y\x)$

[glossy valve]

jan Niku

[plush egret]

so close

[austere slate]

hmm

just change the 1s into negatives

right?

[plush egret]

yea

okay, so we got two matrices

lets call this last one R, for reflection

the other one P, for projection

https://media.discordapp.net/attachments/903486480985489478/1208263662486421555/image.png?ex=65e2a639&is=65d03139&hm=8bc50ded1f8961d616f15a4f9820ca2f55a4d34f63f259dd7863cc5489dc3ef6&=&format=webp&quality=lossless

how do we assemble the answer?

say you have some vector <x, y> and we wanna apply the transformation the problem says in that order

in what order do you put R, P, and <x, y>

[austere slate]

reflection first

then projection

[plush egret]

thats the order of application, right

[plush egret]

so assemble the pieces

[austere slate]

so wouldn't it row 1 <0 0> row 2 <-1 0>

idk how to write a matrix here sorry

[plush egret]

oh, i think this is how it ends up

I was looking for PR<x, y>

but yea i think this is what comes out

,w {{0,0},{0,1}} * {{0,-1},{-1,0}}

[austere slate]

idk I am sorry I was just tryna see if Ik how to do it

[glossy valve]

Results provided by WolframAlpha

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

Requested by janniku

[plush egret]

looks like you got it, yea

[austere slate]

man I appreciate you sooo much

[plush egret]

[austere slate]

fuck I wish u can tutor me throughout the semester 😭

[plush egret]

idk the server is here

youre in linear algebra?

[austere slate]

yes I am

my thing is I dont retain any of the information that is taught because I am just a slow understander

[plush egret]

same

[austere slate]

but you understand this

[plush egret]

i think its pretty universal to not retain very much during a first course in linear algebra

[austere slate]

you are insane bro I'll give it to you

[plush egret]

ive taken the same class multiple times

[austere slate]

man I need help in this class in calc 2 and in physics its all bad

[plush egret]

different names

well calc 2 and lin alg they can help u here

youre in university? do you have a math lab

[austere slate]

yes I am and no we do not have math lab

[plush egret]

no tutoring center or anything

[austere slate]

at my last college they did

[plush egret]

thats wild

ive never been to a school that didnt have a big math tutoring room where all the math nerds hung out

[austere slate]

no tutoring for this level of math

[plush egret]

hmmmmmm

[austere slate]

well this is just my luck

[plush egret]

well hey im around ping me iyw

other people too

linear algebra and calc ii are popular subjects because math nerds have a soft heart for them

youre lucky

[austere slate]

I think Imma transfer colleges because this is bad

[plush egret]

real analysis, probability and multivariable stuff is harder to get help with

linear algebra you will have no problem

[plush egret]

to not have tutoring available for calc 2 is pretty poor im not gonna lie

but dont get too bent out of shape about it

there are tons of resources

[austere slate]

it sure don't feel like it

[plush egret]

well you can do what you can do

im just sayin stay positive

[austere slate]

I studied so hard for my calc 2 exam and still failed

[plush egret]

everything is doable if you put in the work

[plush egret]

relatable

[austere slate]

Idk how to stay positive'

with that

I have to get a B on my next 2 exams in order to pass this class

[plush egret]

well im around, so are others

[austere slate]

with the way I am retaining information I don't think its doable

[plush egret]

so if you have questions, we can help

[plush egret]

calc ii can be pretty grab bag

[austere slate]

look with calc 2 I swear I thought Ik what I was doing but still managed to fail

[plush egret]

its totally possible to not do well in the beginning and get better

maybe the test was hard idk

and retaking calc 2 is a pretty common thing too you know not the end of the world

but you arent even there yet so you know

anyways i dont want to waste all your time

feel free to ping if you got questions

i believe in you fwiw

[austere slate]

you arent like I said thank you man

you were very helpful

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[winged loom]

hi i asked this yesterday, but it got timed out so im posting again

i still dont know how 2 do it

[polar valve]

assume you know a and b, how would you solve p(x) = 0?

[winged loom]

someone suggested yesterday about vieta formula, if the roots are, say p, q
then:
a = -(p + q)
b = pq
also since p and q are distinct integers roots,
p(x) = (x + p)(x + q)
this is what i currently know

[polar valve]

as p and q are (distinct) integers and if you would know b, you could determine all prime factors of b and separate them to get all possibilities for p and q.

[winged loom]

yep but how 2 satisfy the other condition that p(60) is perfect square

while also getting the least possible value for b

[polar valve]

if p(60) is a perfect square it is some c^2, now you can write (c-60)(c+60)=c^2-60^2=60a+b, again you can find some criteria for divisibilty.

[winged loom]

i dont get the "now you can write (c-60)(c+60)=c^2-60^2=60a+b" part

[polar valve]

ok, lets try it another way. p(60) is a perfect square which means there is some integer c with c^2=p(60). and p(60)=(60-p)(60-q). so every prime factor of 60-q and very prime factor of 60-q must appear twice (or in an even count) in the product (60-p)(60-q) which is c^2..

[winged loom]

why it's (60-p)(60-q) and not (60+p)(60+q)

[polar valve]

if p and q are roots of a quadratic function, then this function can be written as (x-p)(x-q), now use x = 60.

[winged loom]

oh yea, also p,q will be < 0

[polar valve]

no one said this.

[winged loom]

i just figured since a and b should be positive

and since a = -(p+q) and b = pq
so p,q should be negative?

[polar valve]

ok.

[winged loom]

so it means (60-p) > 60 and (60-q) both greater than 60 ?

but heres where i get stuck cuz it's kinda hard 2 find a p,q that minimizes b and also makes (60-p)(60-q) a perfect square

[polar valve]

as i said before look at the divisibility. is p = -1 an option? 60-p would then be 61 this would give you that 60-q must be 61 times some square which means 60-q is at least 4 x 61 -> q is at least ... -> b is at least ...
no try p = -2 , ....

[full forum]

@winged loom Has your question been resolved?

[winged loom]

so (60-p)(60-q) must have even prime factors exponents

and since p != q, (60-p) != (60-q)

[polar valve]

i said this here

[winged loom]

so it's kinda like just bruteforcing til i find a suitable p, q

[polar valve]

you do not try all possibilites for p and q. you wont calculate all q = -1, - 2, -3, ... for p = -1. youre trying (some) p and make conclusions what q can be.

therefore i wont call it bruteforcing, its a distinction of cases.

[winged loom]

makes sense, because i found (60+660)(60+120)
p=-660, q=-120

[polar valve]

there is a soluton with much smaller (in absolute values) p and q.

[winged loom]

(60+192)(60+3)

p = -192, q = -3

[polar valve]

you could make some educated guess. the criteria that (60-p)(60-q) is a perfect square is fulfilled if 60-p and 60-q itself are perfect squares. what are the smallest squares greater then 60?

[winged loom]

(64)(81)

p = -4, q = -21

[steel veldt]

Why can't p and q be less than -60?

[polar valve]

why not?

[steel veldt]

Have you try it yet?

[winged loom]

a=25, b=84 if p = -4, q = -21

but is this the smallest b

[polar valve]

why should i? you need the smallest b which is p times q. so i need small (in absulut values) p and q. a p like -37654 doesnt help anything for the question. yes, such p and q are possible, btu you need a small b.

[winged loom]

i think it's the smallest

but it's hard 2 prove

i'll try submitting this answer now looks correct

omgggg

it's correct

thanks man this has been going on 4 days lol

anyways thanks man i'll close this now

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[high granite]

Does anyone know which textbook this comes from?

[full forum]

@high granite Has your question been resolved?

[thorn pasture]

You expect us to know every math textbook in the planet?

Lol

[high granite]

naw, I've just given up because google won't tell me anything 😭

[thorn pasture]

Have u tried lens

[high granite]

is it a website?

I'll have a look thanks

[thorn pasture]

Google lens

Search by picture

[high granite]

oh I've tried that

[thorn pasture]

Oh

Well, if anyone has any idea, ill leave it to them

[high granite]

thanks

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[torn jolt]

I understand how to go from cos θ/sin θ/tan θ to cos2θ/sin2θ/tan2θ but i when i do the reverse, i get wrong answer

[torn jolt]

and i also don't get why cos2θ is positive if its quad 3

[fallen loom]

if theta in quad 3

then cos theta is negative

cos2θ may still be positive

can you show your working?

[torn jolt]

yeah 1 sec

i was attempting to solve for cosθ by plugging in the known value of cos2θ into the equation cos2θ = 2(cosθ)^2 and i ended up with -(2√5)/5

but when plug -(2√5)/5 back into the equation as cosθ, i don't get back the given value of cos2θ

so i guess basically im looking to reverse the double angle formula but im doing something wrong

is there an inverse function im supposed to use or is my current approach at solving for cosθ directly correct and im just doing something wrong during that process?

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[wraith river]

how do i find the derivative of 7/x^2 :?

[wraith river]

using the power rule ?

the answer is -14/x^3 but howd they get that

[wraith river]

ok so 7x^-2

and then u bring the negative exponent to the front and subtract the power by 1 ?

but then it would be 14x^-3

[random coral]

look at power rule again

[wraith river]

-14x^-3

-14/x^3

[random coral]

yes

[wraith river]

oh ok ty

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[robust slate]

$\frac{4}{1-\frac{1}{2}}=8$ and $\frac{1}{2^n}=2^{-n}$

[glossy valve]

Civil Service Pigeon

[robust slate]

Seems like just distributing tbh

$8\left(1-2^{-n} \right)=8(1)-8(2^{-n})$

[glossy valve]

Civil Service Pigeon

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[hollow loom]

idk how to do this

[sudden stirrup]

Forget about the latter requirement for the moment. Can you count the number of ways to assign the five people to the four rooms, assuming five people to a room is allowed?

[hollow loom]

4^5 isnt it?

[sudden stirrup]

Yup

Now how can we get rid of all the bad assignments?

[hollow loom]

subtract 4 i think?

idk

[hollow loom]

wait i just realised why cant it be 5^4

[sudden stirrup]

Sorry my phone just died

[hollow loom]

alr np

[sudden stirrup]

Yes there’s exactly four bad assignments - you put everyone in the same room, and there’s four possible rooms

[hollow loom]

alr but i was wonder why cant 5^4 be an option for my above answer

[sudden stirrup]

So you’re thinking of assigning rooms to persons instead?

Well if you try to count it that way, you might assign multiple rooms to the same person

Also you’d always miss at least one person in the assignment

[hollow loom]

ohhhhh

ok i see

yea alr

so the answer is 4^5 -4?

[sudden stirrup]

Yup

The general technique here is called “counting the complement”

[hollow loom]

alr

[sudden stirrup]

If you want to count all the things in a set of S objects that DON’T satisfy some property P, then you could instead count how many objects do satisfy P and subtract that from S

Sometimes counting this way is easier than the reverse

[hollow loom]

ohk

tysm!

[sudden stirrup]

Of course

[hollow loom]

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[rapid rain]

Look at f-g

Oh sorry wrong question

Did you graph the function

What does it give you

[full forum]

@torn jolt Has your question been resolved?

[rapid rain]

I would expand sin(x+2) and sin(x+1)

to get rid of sin^2(x)

[rapid rain]

sin(a+b)

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[unique fossil]

Why did I get part b wrong? The answer they got is 16

I juts followed the equation

[twin wolf]

What the fuck

sir or maam I think there may be something on your book

ok never mind

that’s in the page

💀

[unique fossil]

LMAOO

Do u know what I done wrong 😂

[limpid junco]

Temp increases by 2

You put temp = 2

You need to find the change

f(T + 2) - f(T)

[unique fossil]

What’s f

[limpid junco]

The function

That you got

[unique fossil]

Which is -24?

Idk what to do

Pls help

Still got it wrong

<@&286206848099549185>

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[craggy tapir]

if 2 linear equations are parallel, then what do they share?

[misty lynx]

the same gradient

[craggy tapir]

correct

so, does the second equation satisfy that condition?

[misty lynx]

yes

[craggy tapir]

correct

and does it pass through (1, 7)?

[misty lynx]

im not sure

[craggy tapir]

well, then calculate it

does (1, 7) belong to $y=-\frac57x-\frac{54}7$

[glossy valve]

FungusDesu

[misty lynx]

so i usubstitute

substitute*

[craggy tapir]

correct

and then?

[misty lynx]

7 = -59/7

so false

right?

[craggy tapir]

correct

therefore the original statement is false

good job

[misty lynx]

tahnks

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[mellow anvil]

Can s1 explain why it equals?

[sour egret]

a short answer is that sin is an odd function

[wild sleet]

a − b = −(b − a)

if you go the opposite direction you end at the same distance, but reflected vertically

sin is y

[mellow anvil]

Okay, so if I have sin equasion, I just can take the "-" and put it inside the angle value (inside the brackets next to the sin)

?

[wild sleet]

yes

[mellow anvil]

thx

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[hoary monolith]

How would i use contradiction to prove this??

[sour egret]

big hint, 21a + 30b = 3(7a + 10b)

[hoary monolith]

oh wait it’s cos 1234 is not divivisble by 3 right

[sour egret]

yep

[hoary monolith]

thanks👍👍

[sour egret]

3(7a + 10b) is a multiple of 3, can't possibly by equal to a number that's not a multiple of 3

[hoary monolith]

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[rose wind]

i was absent for a whole week i don't get this😭

[quaint prawn]

It’s very simple, read the lecture notes

Or Watch a YouTube video

You’ll end up using a calculator

For the problems btw

[rose wind]

i can't find anything for double shaded parts🥲

[proven jay]

use symmetry

[torn jolt]

just find one and then multiply that value by 2

[rose wind]

one?

I'M SORRY I'M ASKING SO MUCH

[torn jolt]

what is the area for this part

look up 2.65 in ur z table

[rose wind]

this?

[torn jolt]

yeah

there should be two pages

[rose wind]

49.5975%

ohh yeah

OUHH OMG I GET IT NOW

tyy

🫶🫶

[torn jolt]

alright

i think you need to subtract that from 0.500 though

[rose wind]

wait huh😭

shxhxyxh why did i have to be absent for a week

.00 or .01 or

[torn jolt]

one second

alright uh

,, \m\P{X\ge 2.65} = 1 - \m\P{X< 2.65}

[torn jolt]

subtract 1 from it, my bad

[rose wind]

ouhh ok tysmm

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[next basalt]

Given the two vectors from R4, u1 = (1, 3, 7, 1)T and u2 = (2, 7, 5, 5)T Find, if possible, two
linearly independent vectors v1,2 ∈ R4 such that v1 is perpendicular to both u1 and u2 and likewise v2 ⊥ u1,2

[next basalt]

Could someone remind me what linear depended/independant meant?

I am linearly dependant on chartbit

[devout valley]

[full marsh]

When you only have two vectors, they are linearly dependent if one is a multiple of the others

[next basalt]

So being linearly dependant means that they are paralell and can be written as a linear combination of each other?

[full marsh]

That's for two vectors

For more, they are dependent if one is a linear combination of the others

There are equivalent conditions to this, but it's important to understand the meaning

[next basalt]

$u_1 = (1,3,7,1) u_2 = (2,7,5,5) v_1 = (x_1, x_2, x_3, x_4) v_2 = (x_5,x_6, x_7, x_8)$

[glossy valve]

Merineth

[next basalt]

But i'm trying to find v1 and v2 where they are perpendicular to u1 and u2

As in the scalar product is 0

[full marsh]

The vectors u_1 and u_2 define a plane since they are linearly independent, do you agree?

[next basalt]

hmm

I guess so?

[full marsh]

Then, a vector is perpendicular to them if (and only if) it's perpendicular to that plane.

[next basalt]

oooh right that makes sense

[full marsh]

Now, the set of vectors that are perpendicular to a plane (in R^4) is also a plane. R^4 can be split into two perpendicular planes.

[next basalt]

so basically i can think of it as u_1 and u_2 make x and y axis respectively and in that case z will be perpendicular to both of them? Where x and y make the plane

[full marsh]

And from that set of vectors, you want to take 2 that are linearly independent.

[full marsh]

That's 3 dimensions

[next basalt]

Oh

[full marsh]

u_1 and u_2 define a plane. All vectors perpendicular to it are also a part of a plane. The solution set to u_1 * x = 0 AND u_2 * x = 0 is also a plane in itself.

[next basalt]

Wouldn't the cross method between u_1 and u_2 create a vector that is perpendicular to both of them?

[full marsh]

Yeah in 3D

[next basalt]

ah right

[full marsh]

We're in 4D

[next basalt]

Forgot it's specific to 3D

So i form an equation?
u_1 * v1 = 0
u_2 * v1 = 0

[full marsh]

Setting x = (x_1, x_2, x_3, x_4) and solving u_1 * x = 0, u_2 * x = 0 (a system with 2 equations) will give you an infinite number of solutions.

Yes

v instead of x, whatever

Also, we can verify that those solutions form a plane in another way: we have 2 equations but 4 unknowns, meaning 2 degrees of freedom, which is precisely a plane.

So once you have your solution set, which you will express in terms of 2 parameters, you can just choose any 2 linearly independent vectors out of it.

Of course you don't NEED to know the solution set is a plane to solve this, but it's good to develop an intuition for these things.

[next basalt]

oooh

so this will be solved using matrix multiplication combined with Gaussian elimination?

[full marsh]

Yeah

Converting it into an equation of the form Ax = b

[next basalt]

$\left(\begin{array}{cccc}
1 & 3 & 7 & 1
\end{array}\right)$ *
$\left(\begin{array}{cccc}
x & y & z & t
\end{array}\right)$

[full marsh]

Then using gaussian elimination

[glossy valve]

Merineth

[full marsh]

No

[next basalt]

No?

[full marsh]

That isn't defined

Transpose one of them

Like the first one

[next basalt]

ooh

so that is what the T stands for

[full marsh]

I recommend writing like x_1, x_2, x_3, x_3 and then using t, s for your parameters in the end

[full marsh]

Yeah

^T is transpose

So it's just a column vector

[next basalt]

1
3 * (x_1,x_2,x_3,x_4)
7
1

[full marsh]

They have bad latex so they don't use columns

Yes

[next basalt]

time to remember matrix multiplication

$\left(\begin{array}{cccc}
x_1 + x_2 + x_3+ x_4 \
3x_1+3x_2+3x_3+3x_4 \
7x_1+7x_2+7x_3+7x_4 \
x_1+x_2+x_3+x_4 \
\end{array}\right)$

[glossy valve]

Merineth

[next basalt]

Is this how would be multiplied?

[full marsh]

Why do you have 4 equations

[next basalt]

i thought that was how i would do it

[full marsh]

u_1 * x = 0
u_2 * x = 0

2 equations, write them in terms of the coordinates

Like simplify what u_1 * x is, etc

Then, write that as matrix multiplication

[next basalt]

Yes i'm trying

But i can't remember how matrix multiplication works

[full marsh]

(There is actually a way to get the final matrix immediately but I'll tell you how after this)

[next basalt]

https://youtu.be/PUR83JAKlUo

[full marsh]

Go check in your book then

[next basalt]

i copied this

The book fucking sucks

[full marsh]

You want to write it as Ax = b

and x is (x_1, x_2, x_3, x_4)

You want a matrix times your vector

[calm hull]

@next basalt I muscle memory matrix multiplication by holding my hands up in the air. Left hand moves right, and right hand moves down

[full marsh]

Here is an example:

[calm hull]

And that's literally how you do matrix mult

[full marsh]

Just multiply each row by each column when you have A * B

[next basalt]

But that is what i did on this one

and it's wrong apparently

[full marsh]

Example:
2 equations, 3 variables:

2x + 3y - z = 6
3x - 2y + z = 8
x + y - 5z = -4

[calm hull]

Ok

What you're wrong so far is just notation

So I'll fix it for you

[full marsh]

You can write this system as:

[calm hull]

You should be writing the matrix multiplication like this
[\begin{bmatrix}1&3&7&1\end{bmatrix}\begin{bmatrix}x_{1}\x_{2}\x_{3}\x_{4}\end{bmatrix}]

[glossy valve]

fishwhale

[full marsh]

$\begin{bmatrix}
2 & 3 & -1\
3 & -2 & 1\
1 & 1 & -5
\end{bmatrix} \cdot \begin{bmatrix}
x\
y\
z
\end{bmatrix} = \begin{bmatrix}
6\
8\
-1
\end{bmatrix}$

aa

my latex

Wait

No

lemme fix

[next basalt]

No

[glossy valve]

RedstonePlayz09

[full marsh]

Here we go

[next basalt]

(1,3,7,1) Is in transposed form @calm hull

[full marsh]

Take this as an example

[full marsh]

.

Try to multiply it out and see why it's equivalent

[calm hull]

The transposed form is what you need to get a matrix multiple to get a single dot product
[\begin{bmatrix}1&3&7&1\end{bmatrix}\begin{bmatrix}x_{1}\x_{2}\x_{3}\x_{4}\end{bmatrix} = \mathbf{u}{1}\cdot \mathbf{x} = x{1} + 3x_{2} + 7x_{3} + x_{4} \in \mathbb{R}]

[glossy valve]

fishwhale

[next basalt]

But why are you transposing v_1

in the assignment it never says that

u_1 is transposed

not v_1

[calm hull]

It's transposed

Because we assume all vectors are columns

Which are $n \times 1$ matrices

[glossy valve]

fishwhale

[next basalt]

what

[calm hull]

This is the norm in terms of writing $\mathbf{A}\mathbf{x} = \mathbf{b}$

[glossy valve]

fishwhale

[next basalt]

When they are transposed aren't they supposed to be columns and if not then they are rows?

[calm hull]

Yes

So they are columns

[next basalt]

Then you have done it wrong?

[calm hull]

You do not dot rows with columns

Lol

[next basalt]

You are literally saying two different things

[calm hull]

The matrix multiplication is only defined this way

[calm hull]

This doesn't work

[next basalt]

$u_1 = (1,3,7,1)^T$

[glossy valve]

Merineth

[next basalt]

Right?

[calm hull]

Yes

[next basalt]

And it is transposed

[calm hull]

Yes

[next basalt]

so it's a column now

Yes?

[calm hull]

Yes

[next basalt]

Ok so

[next basalt]

Why are you writing it as a row?

[calm hull]

Because the dot product is written like that

[x \cdot y = x^{T}y \in \mathbb{R} ]

[glossy valve]

fishwhale

[next basalt]

So because the dot product is written like that we should just disregard the entire rule of transposing?

[calm hull]

What rule

Dot products do not accept row vectors

[x^{T}\cdot y = \text{Illegal}]
[xy = \text{Illegal}]

[glossy valve]

fishwhale

[next basalt]

You can't multiply a 1x4 * 1x4 matrix

[calm hull]

Am I multiplying a 1x4 by a 1x4 matrix?

[next basalt]

Like i'm soooo confused now

[calm hull]

Is that how you read this notation?

[next basalt]

you guys keep saying different things

Am i not supposed to

take u_1 and v_1. Apply matrix multiplication and then find the variables with Gauss elimination?

[calm hull]

What is the dimension of (\begin{bmatrix}1&3&7&1\end{bmatrix})?

[glossy valve]

fishwhale

[next basalt]

R^4

[calm hull]

Ok

What kind of matrix is (\begin{bmatrix}1&3&7&1\end{bmatrix})?

[glossy valve]

fishwhale

[next basalt]

1x4

[calm hull]

What kind of matrix is (\begin{bmatrix}x_{1}\x_{2}\x_{3}\x_{4}\end{bmatrix})?

[glossy valve]

fishwhale

[next basalt]

4x1

[calm hull]

Then where do you get 1x4 * 1x4?

[next basalt]

You don't have [1 3 7 1}

[calm hull]

What

[next basalt]

You have $[1,3,7,1]^T$

[glossy valve]

Merineth

[calm hull]

So you can't transpose it?

It's illegal?

[next basalt]

Is [1,3,7,1] transposed here?

[calm hull]

[{a^{T}}^{T} \overset{?}{=} \text{Illegal?}]

[glossy valve]

fishwhale

[calm hull]

(\begin{bmatrix}1&3&7&1\end{bmatrix}) is just (\begin{bmatrix}1&3&7&1\end{bmatrix})

[next basalt]

?

[glossy valve]

fishwhale

[calm hull]

You always read (\begin{bmatrix}1&3&7&1\end{bmatrix}) as a 1x4 matrix

[glossy valve]

fishwhale

[next basalt]

But i'm given $[1,3,7,1]^T$ not $[1,3,7,1]$

[glossy valve]

Merineth

[calm hull]

[{a^{T}}^{T} \overset{?}{=} \text{Illegal?}]

[glossy valve]

fishwhale

[next basalt]

@devout valley

Please for the love of god help

[calm hull]

If you cannot answer this you have bigger problems than matrix multiplication

[next basalt]

WELL

MAYBE

If you instead of repeating everything

you could perhaps

i don't know

explain a little better

You aren't even answering

my questions that i have

So i'm just getting more and more confused

Like how am i supposed to understand when literally none of my questions get answered

[next basalt]

You type this

but still haven't answered why you are changing it from transposed form

to row form

[calm hull]

.

[next basalt]

????????????????

Like if i don't understand it the first time you type it, how do you expect me to understand it by replying to the same message with a "." ???

[devout valley]

[next basalt]

Yeah honestly

sometimes this server can be insanely infuriating

[devout valley]

Tried to catch up but I'm confused

[next basalt]

Imagine me then

$u_1 \cdot v_1 = 0 \
u_2 \cdot v_2 = 0$

[glossy valve]

Merineth

[next basalt]

I assume this is what i want to accomplish

[devout valley]

Yea and the other products u1.v2 and u2.v1 being zero too

[next basalt]

$u_1 \cdot v_1 = 0 \
u_1 \cdot v_2 = 0 \
u_2 \cdot v_2 = 0 \
u_2 \cdot v_1 = 0 \
\
(1,3,7,1)^T \cdot (x_1,x_2,x_3,x_4) = 0 \
(1,3,7,1)^T \cdot (x_5,x_6,x_7,x_8) = 0 \
(2,7,5,5)^T \cdot (x_1,x_2,x_3,x_4) = 0 \
(2,7,5,5)^T \cdot (x_5,x_6,x_7,x_8) = 0$

[glossy valve]

Merineth

[next basalt]

So basically like this?

[devout valley]

Effectively, though as per before, notation is a bit off

[next basalt]

Where?

[devout valley]

either you can write them as say
[
\pmqty{1 \ 3 \ 7 \ 1} \cdot \pmqty{x_1 \ x_2 \ x_3 \ x_4} = 0
]
or
[
\pmqty{1 & 3 & 7 & 1} \pmqty{x_1 \ x_2 \ x_3 \ x_4} = 0
]
or
[
\pmqty{1 & 3 & 7 & 1} \pmqty{x_1 & x_2 & x_3 & x_4}^T = 0
]
(noting the dot product for vectors $x,y$ is defined by $x\cdot y = x^T y$)

[glossy valve]

@devout valley

[next basalt]

Even if i'm given a transposed vector, is it ok to transpose it myself at will?

[devout valley]

(of course, strictly speaking, the bottom two are matrix products and get you a 1x1 matrix as a result, but effectively no difference between a 1x1 matrix and a scalar)

[devout valley]

And you can transpose stuff as you wish if it makes things more convenient, provided what you end up doing with them is legal

[next basalt]

I see

So how come you write ^ T on

Isn't basic form row form to begin with?

[devout valley]

For that one, so that you get a 1 x 4 multiplied by a 4x1

(which would return you to a 1 x 1, which we wanted)

[next basalt]

Ooh

=

these are equal

[devout valley]

Generally in text stuff gets written in the transpose notation $(x_1, x_2, x_3, x_4)^T$ to make it look neater on paper, so then when you do something like $x^T y$ then you end up with transposes cancelling out on the left one $x^t$

[next basalt]

Aaaah

that makes sense

[glossy valve]

@devout valley

[next basalt]

is matrix multiplcation and dot product equivalent to each other ?

$(1,3,7,1) \cdot (a,b,c,d) = 0 \implies a+3b+7c+d = 0$

[glossy valve]

Merineth

[devout valley]

Not quite "equivalent", but

the dot product can be considered as a matrix multiplication (see if you have $x, y$, then their dot product you can think of as $x \cdot y = x^t y$)

for matrix multiplication, you can consider each entry as something you get from a dot product, in a way

[glossy valve]

@devout valley

[calm hull]

Let $a_{i}$ be the $i$-th row of $A$. Then $Ax=b$ can be written as
[Ax=\begin{bmatrix}a_{1}^{T}x \ a_{2}^{T}x \ \vdots \ a_{n}^{T}x\end{bmatrix}=b]

[glossy valve]

fishwhale

[calm hull]

So knowing how and where the dot products come in multiplication is very useful

[next basalt]

$a+3b+7c+d = 0 \
2a+7b+5c+5d = 0
\
\left(\begin{array}{cccc|c}
1 & 3 & 7 & 1 & 0 \
2 & 7 & 5 & 5 & 0
\end{array}\right)$

[glossy valve]

Merineth

[next basalt]

And now i solve using Gaussian elimination to find the values of a and b

okay and that gave me a = -34c +8d , b = 9c+3d