help-49

[coral bobcat]

Wair but dont u turn the (1-sin^2) to cos^2

[dusty turret]

Was a mistake

Char. Thats why i was helping him with the first part

Him/her

But you said they had it, so i assumed it was discussed

[frozen plover]

substitute, factor out 9, divide the 3s

1 = sqrt(1-sin^2(x))

1 = 1-sin^2(x)

1 = cos(x)

[tribal temple]

Feel like the explanation might have been confusing tbh

[coral bobcat]

Wait why divide 3

[frozen plover]

0 = sin^2(x)
0 = sin(x)

[coral bobcat]

Wouldjt u take sqrt?

[tribal temple]

Try and do the rearranging again yourself

[frozen plover]

3 = sqrt(9 - (3sinx)^2)

correct?

[coral bobcat]

Okk 😭

[coral bobcat]

Yeahh

[frozen plover]

3 = sqrt(9 - 9sin^2(x))

factor 9

3 = sqrt(9(1-sin^2(x))

u can take out 9

[dusty turret]

You made a simple mistake you transposed sin and cos

[frozen plover]

3 = 3sqrt(1-sin^2(x))

divide 3s

1 = sqrt(1-sin^2(x))

understand now?

[coral bobcat]

OH OK I SEE

[dusty turret]

Over simplified

Computer left 3 =

Cant simplify more than that

[frozen plover]

simplest form is sin(x) = 0

or cos(x) = 1

[tribal temple]

!nosols for one

[midnight plank]

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.

[coral bobcat]

Would it be like this

[dusty turret]

No

[tribal temple]

What you had originally was fine

[dusty turret]

He confused you

Let Charbit help you and i will drop out

Nobody else HELP

[coral bobcat]

Okk 😭

[tribal temple]

This is technically right, but you can keep going

What's that 1 - sin^2(theta)?

[coral bobcat]

Cos?

^2

[tribal temple]

Yep, and you're taking the square root of that

[coral bobcat]

Ohh so cos?

So the answer would be 3cos?

[tribal temple]

"yes"*

(* there is a comment about square rooting things you've squared, but I think you don't have to worry about that here, and that they want you to have it as 3cos(theta) for their answer)

[coral bobcat]

Ohhh okk :DD

Its right! 🎉

:)))))

[tribal temple]

Well done good job

[coral bobcat]

THANK YOUUU 🥹🫶

[tribal temple]

(Also @dusty turret thank you so much )

[coral bobcat]

YESS THANKS EVERYONE SM :DDD

Okk cya laterr!

:))

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[coral bobcat]

Hii 😭

[coral bobcat]

Would it be Wsin(theta) / Wcos(theta)?

[tribal temple]

Oh wait

[tribal temple]

And yep, you can simplify that

[coral bobcat]

Wtan(theta)?

[tribal temple]

(I thought this was an actual mechanics question )

[coral bobcat]

LMAO ME TOO AT FIRST

IM SO GLAD ITS ACTUALLY EASY

[tribal temple]

Almost! double check this

[coral bobcat]

Hmmm

Would the W’s cancel?

:DDD

[dull yoke]

chartbit loves his mom

[coral bobcat]

THANKS CHARBIT UR THE BEST

[coral bobcat]

LMAO

[tribal temple]

You always gotta love your mom

[coral bobcat]

Yesss 😌

AKDJSKFJ TYSM :DDD

BYE BYEEE

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[last slate]

Using green theorem here, getting a pretty different answer than what it should be

am i applying this correctly?

oh wait no i am not

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[shy narwhal]

solve f(t)=2 for 3r+2t=18

[shy narwhal]

I don't know where to plug in f(t) or where to start.

[slender walrus]

is this the full original question

[shy narwhal]

The full questions asks :
Consider the relationship 3r + 2t = 18
c. solve f(t) = 2

[candid dock]

hello, can I ask something here? because I asked in another channel with my name but no one answered

[slender walrus]

what's f

[fast basin]

do they give you f(t) anywhere?

[slender walrus]

what's r

this also seems to be part c
what's all the stuff before this

[shy narwhal]

Consider the relationship 3r + 2t = 18
a. Write the relationship as a function r = f(t)
b. evaluate f(-3)
c. solve f(t) = 2

[slender walrus]

ok

that makes more sense

you have r=f(t)
so substitute r with 2

then solve for t

[shy narwhal]

That worked, thank you!

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[shy narwhal]

are all circles functions

[red crater]

no

no circles are functions

[feral sedge]

that depends on how you define the inputs and outputs of said “functions”

[gleaming latch]

wdym by that

[red crater]

they are relations

because they fail the vertical line test

a function has to only have one distinct y value for any x value

and circles have 2 y values for one x value

[shy narwhal]

I see, thank you

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[cold sleet]

can someone help me with this

[tribal temple]

What is "Theorem 5"?

[cold sleet]

not sure but i did a problem similar to this before but i forgot how to get the fraction

[tribal temple]

What's it with these things not specifying their theorems and stuff bruh

I'm making a guess that it would tell you how exactly to relate the coefficients to the inner product and the orthogonal basis vectors etc etc

Would be nice if they gave it to you, but it probably tells you the orthogonality of those basis vectors tells you that $c_i = \frac{\vec{v} \cdot \vec{u}_i}{\norm{\vec{u}_i}^2}$

[grand pond]

@tribal temple

[cold sleet]

i just cant remember how i got the fraction

i did this problem

im trying to figure out how got 5 and 45

@tribal temple are you still here?

[tribal temple]

See here

[cold sleet]

i think i get it

[tribal temple]

Alternatively, think about the fact that {u1, u2} are orthogonal and nonzero so you can extract c1 and c2 out of those

[cold sleet]

let me try to do it

[shell vapor]

would it be possible for you to help me with my linear alg question in #help-39 ?

[midnight plank]

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[wanton shore]

Domain of (x-1)^x

[wanton shore]

Isn't this x>=1 -{2}

?

[feral sedge]

why is 2 being excluded?

also what about numbers like the non-positive integers 🤔

point being the domain should be a bit larger than that

[vale zephyr]

he's comparing it with exponential function i think?

[wanton shore]

Because x-1≠1

[feral sedge]

why is that a condition?

[wanton shore]

Yes a^x

[wanton shore]

a>0-{1}

[feral sedge]

but why can’t a be 1

[wanton shore]

Because someone defined it this way

[feral sedge]

maybe I’m just extremely tired 🤔

[wanton shore]

But wolfram is showing x>=1

I don't understand why

[feral sedge]

for (x - 1)^x… x = 2 results in an output of 1^2 which equals 1 hence the point is defined and 2 should be in the domain

i still don’t understand why it was excluded

[wanton shore]

Then why it is not defined when x-1<0?

It should be defined at discrete values

[midnight plank]

@wanton shore Has your question been resolved?

[midnight plank]

@wanton shore Has your question been resolved?

[fresh sparrow]

because you get a negative number to a fractional power, which only makes sense within complex numbers

[wanton shore]

X=-1
y=-1/2

[fresh sparrow]

yeah

[wanton shore]

So why it isn't defined at x=-1?

[fresh sparrow]

well try x=-0.9

[wanton shore]

Complex

[fresh sparrow]

yep

it's not continuous for x<1

[wanton shore]

But why it can't be defined for discontinuous set of values

[fresh sparrow]

try it, you technically can but i honestly don't know of any normal way to even write that

[midnight plank]

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[grand pond]

Sadie Carnot (η > 1)

[sage helm]

And now you take the first and add 0

[sage helm]

The value of the second integral here comes from here

And the third is easily done

[grand pond]

Sadie Carnot (η > 1)

Sadie Carnot (η > 1)

Sadie Carnot (η > 1)

[sage helm]

whole square inside the left integral

@rain wasp Hello

[sage helm]

*right side 🫠

[rain wasp]

hi

hm, is this your answer or?

[sage helm]

I mean the functions give -3.5 and 2.5 respectively

[rain wasp]

hm

[sage helm]

But neither of them satisfy the initial condiitons which I find strange

[rain wasp]

ill look into this later, in school rn

let me recheck the original problem

[sage helm]

Alright

Thanks for having a look

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[bleak pier]

A and d?

[last slate]

Is it 18?

[bleak pier]

No idea

[last slate]

What kind of technique have you tried?

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[keen basalt]

Directional derivatives

[keen basalt]

When looking at the directional derivatives with a point, and a vector u, does u always need to be a unit vector?

[odd solar]

Yes

[keen basalt]

Ahh fair enough, thanks!

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[stable halo]

hows he done this than

[tribal temple]

The way sinc(x) = sin(x)/x is defined

[stable halo]

yes

where'd the pi go toh

why not a sinc(piau)

[tribal temple]

Oh yea lol

Think it’s a mistake then

[stable halo]

i think he justy defined it differently

ahh

Normalised sinc function

[tribal temple]

Ahhh I see damn it, different definitions

[stable halo]

Yaaa

thx

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[stable halo]

why is it not du?

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[mossy rune]

Can i do that operation?
∫1/(sin(2x)) dx = -cot(2x)/2

[mossy rune]

oh, sin(2x) is not squared.

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[prisma frost]

i'm really struggling with exactly what to do in this question, i'm pretty sure i am on the right track with it but i'm not really sure why it's right

rn i have figured out that

cos(theta) = x/3

and then by differentiating both sides i get

-sin(theta) times d(theta) / dt = dx/dt

but i don't know if i am even doing this right

[olive matrix]

okay, and "how fast is the angle changing" is dθ/dt

and you know dx/dt

[prisma frost]

yeah dx/dt is 3m/s

but im stuck on how to make

−sin(θ)∗dθ/dt=3m/s

into something to solve for dθ/dt

[latent elk]

@prisma frost Do u have the solution

[prisma frost]

i do not thats why im asking for help 😭

[latent elk]

Is it something like -0.338529 deg/s

[prisma frost]

it has to be in radians, since its pi/6

[latent elk]

Too lazy to convert it to rad

Not sure if its correct though

Something seems off

[prisma frost]

yeah i still havent learned arccos yet, so i have no clue if this is right

<@&286206848099549185> ?

[latent elk]

Well, this is for arccos of x, use chain rule for arccos of u = x/3

[maiden zenith]

it seems right!

[prisma frost]

i'm just going to ask my teacher when i am back in class, she probably wont take arccos for an anwser since that isnt something we have learned

thanks anyways tho

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[latent elk]

@prisma frost Logically speaking, u can easily prove derivative of arccos

Well, this is the simple case of x

Use chain rule to consider the case of x/3

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[cyan haven]

If ABC company expects cash inflows from its investment proposal it has undertaken in time period 0, Tk. 2,00,000 and Tk. 1,50,000 for the first 2 years respectively and then expects annuity payment of Tk. 1,00,000 for the next 8 years, what would be the present value of cash inflows, assuming a 10% rate of interest?

[midnight plank]

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[cyan haven]

no ..... please help me

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[cyan haven]

no

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[cerulean temple]

is singature of quadratic form $\sum_{i=1}^{N-1} x_ix_{i+1}$ equal to $(0,0,N)$?

[grand pond]

Slowaq

[cerulean temple]

for example this is its matrix for N=5

all the ones cancel out i am am left with zero matrix

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[sturdy crystal]

hi

[sturdy crystal]

how do i find the sum of infinite series

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[twilit field]

This is $c(50,0)400^{50}+c(50,1)400^{49}+ c(50,48)400^{48}....$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

came until here

now what

[runic hamlet]

well most of these terms have a lot of zeros

find the one with the least zeros

[twilit field]

hmm, that would be the last term, no?

so just 1

[runic hamlet]

well ok sure but that doesnt tell you about the zeros next to it

find the term with the least zeros but which actually has zeros

[twilit field]

uh, that would be c(50,49)400

[runic hamlet]

which is?

[twilit field]

20,000

so the answer is 3

[runic hamlet]

why

[twilit field]

the last digit is 1

[runic hamlet]

you need to argue that the other terms actually have enough zeros to not matter

[twilit field]

and the next 3 digits are zeros

[stray salmon]

Pretty basic calculation 14362122643507669858966216033201181457942667669169554197971208203529754322055813539830291627210582268587414952574969199334596020001

jk

[twilit field]

that makes intutive sense, but formally?

I guess I could use the shape of the BD

yeah, that should work

thanks!

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[dawn prism]

What does this bilinear form do? p and q are polynomials

[dawn prism]

Does it take the derivative of p and multiplies it with q?

Or what

<@&286206848099549185>

This describes b but I still don't know what it does

[cinder fox]

are the polynomial functions in this case?

[dawn prism]

They are elements of a polynomial ring

And (R[x], b) is an euclidean vector space

[midnight plank]

@dawn prism Has your question been resolved?

[midnight plank]

@dawn prism Has your question been resolved?

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[zealous inlet]

Can someone help me with this problem?

[zealous inlet]

I factored the left side first which gives me (y-1)(y+2).
Than i multiply it to all of them to cancel out the denomitors. Which gives me 3y= 1(y+2) -2(y+2)(y-1)

but after that i am stuck for some reason. I cannot seems to make it work and i feel like I am messing up with my order of operations

When i foil out (y+2)(y-1) i get y^2 -1y +2y -2

but after that i mess it up. am i supposed to put the -2 over that? and than add the y+2?

[old storm]

when you get the thing you get

do all the operations till you have no more brackets

move 3y to the toher side

and solve the quadratic equation

[old storm]

okay

[zealous inlet]

what is the correct way to add these together than?

[old storm]

yeah

[zealous inlet]

-2 distribute over all of them?

[old storm]

y+2 - 2y^2 - 2y +4

[zealous inlet]

in the paranthases

[old storm]

yes

[zealous inlet]

how do i add the y+2?

[old storm]

3y = y+2 - 2y^2 - 2y +4 is what you get

move 3y to the other side

[zealous inlet]

oh wait yeah the 4

[old storm]

also its plus 4

[old storm]

you will get something like -2y^2 - 4y + 6 = 0

Divide everything by -2

you get y^2 + 2y - 3 = 0

solve that

[zealous inlet]

wait i think i missed something. you said divide by to. what do i need to divide by two?

or do you mean factor out a -2

[old storm]

the entire thing

just divide both sides by -2

since 0 / number = 0, everything is okay

[zealous inlet]

like this

so y=-1 and y=3?

according my answer sheet I think it is wrong. the answer is supposed to be -3

oh i see what i did wrong

i put -2 instead of 2 while factoring

y=1 y=-3 is the right answer

[old storm]

yes

[zealous inlet]

and 1 is most likely extraneous

but anyways. this is the correct wait to go about it, right?

[old storm]

yes

[zealous inlet]

thank you

ill close this section after writing down my answer. thank you for your help

[old storm]

yw yw

[zealous inlet]

.close

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[old storm]

y = 1 is wrong cuz 1 / y-1 would give you 0 btw

[zealous inlet]

yeah. cannot be zero

thanks for telling me though 😄

have a goodday mate

.close

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[spark nebula]

how they simplified to get that

[formal palm]

$9x^4+3x^2+9x^2+3 - 6x^4 -18x^2$ \newline
$\Rightarrow 3x^4 - 6x^2 + 3$
$\Rightarrow 3 ( x^4 - 2x^2 + 1)$

[grand pond]

penguin

[spark nebula]

yeah it's more clear now thanks for help

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[old storm]

So, I won't lie, I have 0 idea

@stiff heart

<@&286206848099549185>

Cmon someone, anyone

Someone fom the heavens of statistics

<@&286206848099549185>

<@&286206848099549185>

someone, anyone

[midnight plank]

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[old storm]

<@&286206848099549185>

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[digital crag]

How do i do this ?

[midnight plank]

@digital crag Has your question been resolved?

[midnight plank]

@digital crag Has your question been resolved?

[fast basin]

You need to try and understand what h, h', h'' represent. h(6) is the area under the curve between 0 and 6, for example

Then you can find h' explicitly: $h'(x) = \frac{\operatorname{d}}{\operatorname{d}x} \int_{0}^x(f(t) \operatorname{d} t) = f(x)$

[grand pond]

fwa

[fast basin]

So h''(x) = f'(x), the slope of the graph f at point x

use those facts to find an answer :)

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[misty marlin]

Greetings
I am again getting the same thing. Now i did everything step by step and i get something like that.
The solution in book is x1=-7/4 and x2=1. I don't know what to do

[last slate]

no u r totally right actually

if the question is to find x of this, then yes what u did is rihgt

make sure u copied the question correctly

[misty marlin]

Lemme check

Oh...

Yeah.

Its -11

[last slate]

yeah bingo i guess

always good to double check if you wrote it down correctly

[misty marlin]

Okay. Thank you ahhhhhh

. close

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[sour tusk]

hello, i need help understanding using midpoint method to solve an ODE. im not sure how to translate the equation for calculation when given an example. i have the ODE
$\frac{dP}{dt}=f(P)=kP$
where k = 0.01 and P(0) = 100

i need to use a step size of $\Delta t = 1$ to solve for P(3) using the midpoint method which states:
$y(x_i) = y(x_{i-1}) + \Delta x f\left(y\left(x_{i-1} + \frac{\Delta x}{2}\right), x_{i-1} + \frac{\Delta x}{2}\right)$

really, i need help reading and understanding the equation and then a step-by-step understanding of using the midpoint method to do the first step, to find P(1) using the midpoint P(0.5)
my main issue is understanding the $f\left(y\left(x_{i-1} + \frac{\Delta x}{2}\right), x_{i-1} + \frac{\Delta x}{2}\right)$
thank you!

[grand pond]

kaikonic

[midnight plank]

@sour tusk Has your question been resolved?

[pine wave]

okay, your $f$ is a function of $P$ and $t$ but $t$ doesn't really do anything

[grand pond]

Element118

[pine wave]

so firstly you need to evaluate $y(x_{i-1}+\frac{\Delta x}{2})$

[grand pond]

Element118

[sour tusk]

okay i was actually working through it using a code example of midpoint method and understand it after walking through the debugger, thank you for your help!

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[dusky loom]

how to do this

[dusky loom]

the highlighted exercise

<@&286206848099549185>

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[dusky loom]

<@&286206848099549185>

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[shut canyon]

What is the connection of vectors in vector algebra, and arrows in category theory?

[hard umbra]

they look similar if you draw them

[shut canyon]

That's true

[shut canyon]

This is why I wondered

[hard umbra]

just because they look similar visually doesn't mean there's any deep connection between them

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[shut canyon]

.reopen

[midnight plank]

✅

[shut canyon]

I don't yet know if a function is an arrow, or whether a vector is a function, nor clearly understand what makes a category.

A vector has magnitude, a morphism does not.

I think arrows are not necessarily functions, are they?

Composing an arrow with it's inverse is similar (the same?) to adding the complement to a vector.

[shut canyon]

There must be a deep connection, do you think there isn't any?

look!

[hard umbra]

these are just properties of monoids

[shut canyon]

Maybe monoids and vectors connect. Is the set of vectors a monoid?

[hard umbra]

the underlying additive structure of a vector space is an abelian group

which is a kind of monoid

thats not really a deep statement though

[shut canyon]

Like that of the natural numbers

[hard umbra]

the natural numbers dont form an abelian group

they form a commutative monoid though

[shut canyon]

What are deep statements?

I'm trying to create a sense of what we are looking for when looking for deep statements

[hard umbra]

well i wouldnt say these things are things people really think about a lot

its fundamental but its just in the background when people study categories and vector spaces

[shut canyon]

I just started studying categories and vectors, so these connections are still interesting to me

Thanks for your help!

[hard umbra]

sure, if it helps you relate vectors with arrows in categories, the monoidal structure is definitely the relation

[shut canyon]

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[night egret]

on the last line, can someone please explain where the + in front of the ln is from?

[last slate]

they factored out -1/2 entirely

so the negative as well got factored

,, -\412\8{\4{y^2}{6^2}} - \412 \6\log{2\pi6^2} = -\412\8{\4{y^2}{6^2} + \6\log{2\pi6^2}}

[night egret]

oh yh, the 6 looking thing is sigma

[last slate]

wow thats the worst written sigma i have ever seen

[night egret]

IKR

[last slate]

but anyways just change the 6s to sigmas

same thing

[night egret]

uh i still don't get it

[last slate]

its just factoring really

take a simpler case

,, -ab -ac = -a(b+c)

[night egret]

-x(y) * -x(z) right

[night egret]

yh

[last slate]

so whats confusing u

$\6\exp x \2\6\exp y = \6\exp{x+y}$ as well btw

[night egret]

i can't see it. how. i'm having a blockage. it makes sense backwards

uh

[last slate]

ok but that's the nice thing about equalities you know

they work in both directions

so -a(b+c) = -ab - ac

but also

-ab-ac = -a(b+c)

idk how to explain it better but if you feel this is bothering you look up a bit on factoring numbers

[night egret]

i'm thinking -x(y) * -x(z) = -x(y*z) 🧍

[last slate]

right but u have to understand thats not the situation u have rn

[night egret]

yh

[last slate]

you have -x(y) - x(z) not -x(y) * -x(z)

no multiplication between them

[night egret]

wait why not multiplication??? the line before is e^x * e^y

[last slate]

also this is wrong EVEN if it was that

[last slate]

read this

[night egret]

when you take out e, * still applies

@-@ oki

[last slate]

e^x * e^y = e^(x+y)

[night egret]

oh right, this rule

[last slate]

yes

so no multiplication

[night egret]

ohh ok, got it :D

ty!

[night egret]

🥴 heh

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[last slate]

-x(y) * -x(z) = x^2*y*z

fyi

[night egret]

probably. lazy to check :p

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[raw hare]

This has to be right, right?

4/4√3
4 • (4√3)

4√3 • (4√3) = 4² (√3)²

8√3/
16(3)

8√3/
48???

[raw hare]

Im supposed to rationalize 4/4√3 but idk if im correct.

Ok.

I might be stupid.

.close

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[tight axle]

I've solved a and b, but I'm not sure how to approach c. Brute forcing it isnt exactly scalable as a solution and a and b did not help

[tight axle]

I got that a) is rejected and b) is accepted for what its worth

[ionic forum]

|x^2+x-6|>6 hi I have this inequality with the module, among the results I only get -4 and 3 for the book asks for a result which is
-4 ≤ x ≤-1 V 0 ≤ x ≤ 3 how can I do?

[wary epoch]

!help

[midnight plank]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[wary epoch]

Pro tip: the execution seems be encoding item deletion and going back to start.

(If you want to interact with me, please ping)

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[dim shard]

Janice rides a Ferris wheel in Japan called the Sky Dream Fukuoka, which has a radius of about 60m and is 5m off the ground. After she enters the bottom car, the wheel roates 210.5 degrees counterclockwise before stopping. How high above the ground is Janice when the car has stopped.

[dim shard]

I got two different answers

116.69 and 95.45m

Idk which one is correct

someone pleaaaase help

[lament fox]

why do you have two answers?

what did you do?

[dim shard]

okay

one

i calculated after 180 degrees

it'll be left with 30.5

to make 210.5

and then i get cos 30.5 = A/H

knowing H is 60

A = 116.69M

BUT

If you calculated the other angle AFTER 210.5

which is the angle between 270 and 210.5

you fet 58.5 or smth

and using Sin 58.5 = O/60

i get 95.45

[lament fox]

it would help a lot if you make a drawing

[dim shard]

ohhhhh

sorry

i can but

i figured it out

i messed sin and cos up

[lament fox]

which one is it?

[dim shard]

116.69

[lament fox]

i dont think that is right

[dim shard]

I used COH instead of sin

WHAT

[lament fox]

think about it the wheel has a radius of 60

and 210 degress is poitning downish

so it has to be less than 60m

[dim shard]

ahaha

here's the thing

it starts from lower part

since its a ferris wheel

so it'll actually be upwards

[lament fox]

ahh that is true

i see

ok good job

[dim shard]

tyty

heres a pic

anyways thanks alot good to have a new perspective

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[hollow moon]

Heres as far as I got, the markscheme says the answer is 1:40000

[hollow moon]

Can anyone help me out?

[latent elk]

,rotate

[midnight plank]

@hollow moon Has your question been resolved?

[latent elk]

You could assume the lake is a square, then use 1 side to convert the problem from area to length base

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[mint blade]

Hey, could someone please help me with the following?

[mint blade]

<@&286206848099549185>

[prime rapids]

It might help to draw a line going down from P.

[mint blade]

Okay, I’ve drawn a line down from P

[prime rapids]

What is the length of that line?

[mint blade]

I’m not sure, I don’t think it’s drawn to scale

[prime rapids]

Well, if the y-coordinate of P is $\sqrt{3}$, then the line should have a length of $\sqrt{3}$.

[grand pond]

Calculustache

[mint blade]

Ohhhhhhhh I see I see

[prime rapids]

Then you also have another line: the horizontal line from the origin to the intersection of the x axis and the line going down from P.

[mint blade]

Do I have to calculate that new line?

[prime rapids]

Yes, you can apply the same reasoning to get the length of that line.

[mint blade]

So that would be 1 right?

[prime rapids]

Yeah

[mint blade]

So to calculate theta I would have to use tan?

[prime rapids]

Yes. It would look like $\tan \theta = \frac{\sqrt{3}}{1}$

[grand pond]

Calculustache

[mint blade]

So theta is 60?

[prime rapids]

Yes

[mint blade]

Tysm!

[prime rapids]

You're welcome

[mint blade]

How would I calculate OQ? Do I just multiple the x value?

[prime rapids]

OPQ will make a big 30-60-90 triangle, so you can use the Pythagorean Theorem to find OP, then scale it by a factor of 2.

[mint blade]

I’m sorry, I don’t follow.

[prime rapids]

So the 30-60-90 triangle has side lengths $1x, 2x, \sqrt{3}x$. This means that if you know one side, you can find the length of any other side.

[grand pond]

Calculustache

[prime rapids]

So if we know that OP is length x, then OQ is 2x.

[mint blade]

OP as in the diagonal OP or the new line created when I drew the line coming down from P?

[prime rapids]

The diagonal OP

[mint blade]

So how do we know that OQ is twice the length of OP?

[prime rapids]

Because OPQ is a 30-60-90 triangle, and in a 30-60-90 triangle, the hypotenuse is twice the side next to the 60° angle.

[mint blade]

How would I write all of this in an equation?

[prime rapids]

$2 \cdot \sqrt{1^2 + \sqrt{3}^2}$

[grand pond]

Calculustache

[mint blade]

Tysm!!

.close

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[fallen mountain]

I did A but idk how to do b and c now

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[fallen mountain]

<@&286206848099549185>

[fallen mountain]

.close

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[little sage]

Pls

[quiet hinge]

Reflect it about the x axis

What graph do you get after doing that?

[little sage]

Nuh uhh

Just do it 😞

[quiet hinge]

No one will solve your questions for you here

[last slate]

!noans

[midnight plank]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[little sage]

😒

what was the quesrion

[dusky loom]

what does a sine graph look like

[dusky loom]

if you dont know this already, plug in 0, pi/2, pi and see what it looks like for sine

then do the same for cosine

[little sage]

Im so cooked

Idk man

[dusky loom]

is this cambridge?

[little sage]

say what

[dusky loom]

looks like an IGCSE question

[little sage]

Nah

[dusky loom]

but might not be its pretty generic

okay

[little sage]

No clue

Im in hs

[little sage]

(Im so cooked)

[dusky loom]

this is the graph of sine

follow along, im tryna help

[little sage]

Yes

[dusky loom]

so compare this sine graph

with this

[little sage]

Yes

[dusky loom]

now do you know what i mean when i say coordinate (x,y)

[little sage]

Jow do u find the the sin

[little sage]

Nah

[dusky loom]

okay when you draw a graph

how do you know what point is at the center of the entire graph

like

do you know what this (0,-1) is

[dusky loom]

if not let me know

[midnight plank]

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[cerulean linden]

For q 2.

I’m using maple

But what commands do I use on that application to find the area of the region bound if any1 is familiar with this application

[last slate]

what's Maple?

[cerulean linden]

It’s a math software

I dont even know how to use it

All I know is plugging the functions

[last slate]

maybe you should look up some tutorials

[cerulean linden]

Ig

[midnight plank]

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[normal zenith]

@wide goblet did you resolve your question?

[wide goblet]

no. The channel got closed im in another

[normal zenith]

alr

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[floral minnow]

I did part a) already but im not sure how to do part b)

[torpid garnet]

For b) use induction again

And use the "definition" of fn=fn-1+fn-2

[floral minnow]

wdym

srry i went to the restroom 😅

.close

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[toxic flower]

does this proof look valid?

since $u_1 ... u_r$ and $w_1...w_s$ are basis, they are also linearly independent. so $\sum_{i=1} ^ r {a_i \cdot u_i} = 0$ and $\sum_{i=1} ^ s {b_i \cdot w_i} = 0$, each $a_i = b_i = 0$ assume, for a contradiction, that $\exists v \in U \cap W, v\neq 0$. suppose that $u_1...u_r$ $w_1 ... w_s$ are linearly independant. $a_1 \cdot u_1 + ... + u_n \cdot v +... + a_r \cdot u_r + b_1 \cdot w_1 + ... + b_n \cdot v + ... + b_s \cdot w_s = 0$. (since v is in both U and W) this implies all coefficients are zero, from the definition of linear independence. so $a_n = -b_n$, take $a_n = 1, b_n = -1$. but this is a contradiction, as $a_n = b_n = 0$

[subtle blaze]

Why do the b_s w_s go to s

Oh

No why does the sum on line 2 go to k and not s

[toxic flower]

typo

my mistake

[grand pond]

lewis_f04

[toxic flower]

sorry about that

[subtle blaze]

I’m struggling to see what a_n + b_n are

[robust isle]

"sum ai ui = 0" what

you remember what linear independence means or no ?

[subtle blaze]

Your phrasing is a little poor tbf

does this proof look valid?

since $u_1 ... u_r$ and $w_1...w_s$ are basis, they are also linearly independent. so $\sum_{i=1} ^ r {a_i \cdot u_i} = 0$ and $\sum_{i=1} ^ s {b_i \cdot w_i} = 0$, implies $a_i = b_i = 0$ for all i (in its relative domains)

[grand pond]

Frosst

[subtle blaze]

It’s a little awkward to use a_i = b_i because there may be differing amounts of the coefficients

Since we don’t know r = s

[toxic flower]

it means that the sum of all the vectors with coefficients = 0, the coefficients are 0

[toxic flower]

i see, how would you say is best to do it? im quite new to all this still

just $a_i = 0, \forall i \in {1,..,r}$ and $b_i = 0, \forall i \in {1,...,s}$

[grand pond]

lewis_f04

[robust isle]

ah yeah I didn't even process the ai = bi = 0 afterwards, my bad

[robust isle]

yeah

[toxic flower]

that makes sense

[robust isle]

and I have the same question as frosst really

[subtle blaze]

You haven’t defined them you just pulled them out of thin air

[toxic flower]

i realise my mistake, i get what you mean now

since $u_1 ... u_r$ and $w_1...w_s$ are basis, they are also linearly independent. so $\sum_{i=1} ^ r {a_i \cdot u_i} = 0$ and $\sum_{i=1} ^ s {b_i \cdot w_i} = 0$, each $a_i = b_i = 0$ assume, for a contradiction, that $\exists v \in U \cap W, v\neq 0$. suppose that $u_1...u_r$ $w_1 ... w_s$ are linearly independant. $a_1 \cdot u_1 + ... + a_n \cdot v +... + a_r \cdot u_r + b_1 \cdot w_1 + ... + b_n \cdot v + ... + b_s \cdot w_s = 0$. (since v is in both U and W) this implies all coefficients are zero, from the definition of linear independence. so $a_n = -b_n$, take $a_n = 1, b_n = -1$. but this is a contradiction, as $a_n = b_n = 0$

[grand pond]

lewis_f04

[toxic flower]

does this make more sense now?

[robust isle]

tbh the whole equality seems to come out of thin air

[subtle blaze]

How do we know it = 0

[robust isle]

and what are the a1, ... an, b1, ... bn you're talking about

you're saying, ok u1, ... ur, w1, ..., ws are linearly independent

[subtle blaze]

(Btw we could give you what you’re looking for but it’s better for your learning to struggle through and try word it yourself)

[robust isle]

but linear independence doesn't tell you the existence of anything

it just says "if you have a linear combination of u1, ..., ws which sums to 0, then all the coefficients are 0"

[toxic flower]

yeh, and since u_1 ... u_r and w_1...w_s are linearly independent, we know all a_i, b_i (the coefficients in the sum $\sum_{i=1}^r a_i \cdot u_i$ and $\sum_{i=1}^s b_i \cdot w_i$) are zero?

[grand pond]

lewis_f04
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[subtle blaze]

No no no

If … then …

[robust isle]

the problem is you have no linear combination to talk of at this point of the proof

[subtle blaze]

If they sum to 0, then the coefficients are 0

[robust isle]

you just know v is a vector in U and W

[subtle blaze]

You’re saying since they are linearly independent, the coefficients are 0

You’re forgetting the part where they need to sum to 0 first

[toxic flower]

isnt the only way for them to sum to zero for all the coefficients to be zero because were given that theyre basis vectors, and a basis must both be linearly independant and span the space? or am i just completely lost 😂

[subtle blaze]

But you didn’t say anything about summing to 0

[toxic flower]

i really appreciate your patience, this must be very annoying for you

[subtle blaze]

You just asserted the coefficients to be 0

[toxic flower]

isnt that this part is is that something else (the tex got screwed hold up)

$\sum_{i=1} ^ r {ai \cdot u_i} = 0$ and $\sum_{i=1} ^ s {b_i \cdot w_i} = 0$

[grand pond]

lewis_f04

[subtle blaze]

Saying $\sum_{i=1}^r a_iu_i + \sum_{j=1}^s b_jw_j = 0 \implies a_i = b_j = 0$ for all $i, j$ that exist

[grand pond]

Frosst

[subtle blaze]

Now this is the definition of the u’s and w’s being linearly independent

[toxic flower]

i get you. the definition wasnt clearly stated?

[subtle blaze]

You did not state this

You did not use this either

What you wrote was $\sum_{i=1}^r a_iu_i + \sum_{j=1}^s b_jw_j + a_nv + b_nv = 0 \implies a_i = b_j = 0$ for all $i, j$ that exist

[grand pond]

Frosst

[subtle blaze]

Which is not necessarily true

[toxic flower]

so to start the proof, since $u_1...u_r$ and $w_1 ... w_s$ are linearly independent, $\sum_{i=1}^r a_iu_i + \sum_{j=1}^s b_jw_j = 0 \implies a_i = b_j = 0$ for all $i, j$ that exist

[grand pond]

lewis_f04

[toxic flower]

and since $u_1 .. u_r$ are basis vectors, $\sum_{i=1}^r a_i u_i = 0 \implies a_i = 0$, likewise for $w_1 ... w_s$, $\sum_{i=1}^s b_i w_i = 0 \implies b_i = 0$ for all (i,j) that exist

[grand pond]

lewis_f04

[robust isle]

yeah

[toxic flower]

how now do i use these implications, because there is nothing to say that any of this is true?

[robust isle]

you have to get that these statements don't tell you any existence of the ai's or bi's

what you can do is, if you have a linear combination of the ui summing to 0 at hand

you can give it to that statement

and it will tell you that the coefficients of the lin combo are all 0

[toxic flower]

how do i get that the uis sum to 0?

i think thats what is throwing me

[robust isle]

well that's our job still

there's one thing you haven't used yet really

which is v

[toxic flower]

as in v that is in both u and w

[robust isle]

yeah

and you know bases for U and W

[toxic flower]

so, assume, for a contradiction that $v \in U \cap W, v \neq 0$

[grand pond]

lewis_f04

[toxic flower]

oh and $v = \sum_{i=1}^r p_i u_i$ since ui is basis

[grand pond]

lewis_f04

[subtle blaze]

You could remove the ≠ 0 restriction

And show it must be the 0 vector

It could be easier than doing contradiction

[toxic flower]

i see, so instead of doing it by contradiction, so that the only v in the union is zero

[subtle blaze]

Intersection

[toxic flower]

likewise, $v = \sum_{i=1}^s q_i w_i$ since wi is basis

[grand pond]

lewis_f04

[toxic flower]

yeh thats the one

[subtle blaze]

You also have that U ∩ W is a subset of U (and also then a subset of W)

[subtle blaze]

That’s how you can say this

[subtle blaze]

And this

For any v in U ∩ W

[toxic flower]

how come its because its in the intersection? doesnt it come from beinging in W and U?

[subtle blaze]

$U\cap W \subseteq U$ is true for all sets

[grand pond]

Frosst

[toxic flower]

right makes sense

[subtle blaze]

But you need this otherwise you can’t say the v can be written as a linear combination of basis vectors of U

[toxic flower]

right i see

[subtle blaze]

Since you wouldn’t know that v is inside U, (and hence can be written as … since the basis spans U)

[toxic flower]

sure that makes sense

im a bit stuck with how to move forward

i guess theres something obvious that im missing to be able to start eliminating the implications

[robust isle]

well you have two different ways of writing v

[toxic flower]

but i cant say anything about them right? because there sums

[robust isle]

they correspond to the same vector

[toxic flower]

$\sum_{i=0} ^s q_i w_i$ = $\sum_{i=0} ^r p_i u_i$

[grand pond]

lewis_f04

[subtle blaze]

That is indeed true

[toxic flower]

but i was told that i cant match them up inside the sum

so im not sure what i learn from this

[subtle blaze]

You certainly cannot do that

Well

[subtle blaze]

We want to use this

At some point

[toxic flower]

can we say that $\sum_{i=0} ^s q_i w_i$ + $\sum_{i=0} ^r p_i u_i$ = $2 \sum_{i=0}^s q_i w_i$

[grand pond]

lewis_f04

[toxic flower]

im just guessing here tbh ive never seen this before

[subtle blaze]

Absolutely not

But we do know that a = b implies a - b = 0

[toxic flower]

can we say that $\sum_{i=0} ^s q_i w_i$ - $\sum_{i=0} ^r p_i u_i$ = 0

[grand pond]

lewis_f04

[toxic flower]

so then if we bring the minus sign in the sum

$\sum_{i=0} ^r p_i u_i +\sum_{j=0} ^s -q_j w_i$ = 0

which is awkward

[subtle blaze]

Don’t put the + signs outside the $

That is indeed true

[toxic flower]

i guess we can now let a_i = p_i and b_j = -q_j for all i, j that exist

so now we can use that to eliminate the implications

[subtle blaze]

You wrote q_i

There’s no “elimination of implication”

[grand pond]

lewis_f04

[robust isle]

https://en.wikipedia.org/wiki/Modus_ponens

[toxic flower]

sorry we call it implication elimination in my classes, is that the same thing

[robust isle]

it's fine

even wikipedia says it

[toxic flower]

so now we can say that a_i = b_j =0 for all i,j that exist

and then i think the proof works from there?

or is it still iffy?

[robust isle]

you're almost there

[robust isle]

well you mean p_i and q_j prolly

[robust isle]

that was our lin combo after all

with p_i's and q_j's

but anyway

[toxic flower]

yeh mb

[robust isle]

what does that say about v then ?

[toxic flower]

hmm well i guess it means that v = 0 but i have no idea why

because really all we've said is that v+-v = 0

[robust isle]

you've shown all these damn coefficients are 0

thanks to linear independence

[toxic flower]

oh it must be 0

since

its just a sum of a shit load of zeros

[robust isle]

yeah

[toxic flower]

thanks so much for your help

this was the solution they gave

which i think matches up

[robust isle]

yeah

[toxic flower]

thanks for your help, my exam is tomorrow so i think i might be a little screwed still

[robust isle]

now you can do the reverse direction (if your book doesn't ask you that already)

[toxic flower]

but hopefully this doesnt come up

[robust isle]

my little remix

[toxic flower]

oh i see, i might try that now ik what it is to look like

[robust isle]

I'm pretty damn sure it's true, but I haven't checked in detail really

[toxic flower]

I'll have a go tomorrow, as its close to 1am now

thank you for your help

and thank you as well @subtle blaze

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[normal plover]

somene please help im the only one in the class who didnt take physics i dont understand what a vector is

[midnight plank]

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[normal plover]

why are they lining up on the x?

[shell vapor]

so first a vector is just a point in spce from the origin right

[normal plover]

oh ok

[shell vapor]

the way you add vectors is by adding them from tip to tail

and same thing with subtracting

[normal plover]

so the tail of the first to the head of the other?

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[runic thunder]

can someone help me go through this
Reflection over x axis,
Reflection over y axis
vert str by 2
horizontal str by 3/ -4

right by 1 unit, up by 2 units
right?

[cinder fox]

would say shrunk here

but your right

[runic thunder]

but im stuck

what would -24 be

turned to

can u help

[cinder fox]

use the steps u had correcrly given

[runic thunder]

so inside bracket its opposite so we turn right by 1, so wer have -23 now

-23*(3/-4) = 17.25 or -69/-4

so x is that right?

if thats correct do i reflect it now

[cinder fox]

it has been fliped

when u times by -3/4

[runic thunder]

yea

[cinder fox]

so no need to reflect again

[runic thunder]

so am i right that -24 is transformed to -69/-4

?

[cinder fox]

ya but write 69/4

instead

negs cancel

[runic thunder]

ohh

ok ok

ok so now the y

2*1 is 2 then -2+2 = 0

right?

[cinder fox]

yuh

[runic thunder]

"In the answer field, add the x
and y
-coordinates of the image point together and enter this value.
For example, if the coordinates of your image point are (-12,7), enter the value -5, since -12+7 = -5."

how do I add these points together?

its cause its an online thing and I have to write it there and my work sheet

[cinder fox]

by adding the togehter

[runic thunder]

but it a fraction

[cinder fox]

so?

[runic thunder]

uhhhhhh

[cinder fox]

and its 0 anyway

[runic thunder]

.