help-13

[barren forum]

hmmm I must remind you linear independence is a property of a set😂, though yes, that's a quick way to spot it

say A = {(1,0), (0,1), (3,5)}, can you show me that A is not linearly independent? @errant wasp

[errant wasp]

hmm

it seems that hey are?

[barren forum]

a set is linearly independent when ....?

[errant wasp]

If the only solution is x = 0

[barren forum]

were you thinking "when no vector is a multiple of another"? That's not the definition 😝

[errant wasp]

yeah unfortunately

[barren forum]

you must have it in your notes?

[errant wasp]

If the only solution is x = 0 is linearly dependent

[barren forum]

taken out of context that's a bit confusing, but it's saying this:
let W = {v1, v2, v3, v4...} be a set of vectors, x1, x2, ... be numbers, W is linearly independent when x1 v1 + x2 v2 + x3 v3 = 0 has only the solution x1 = x2 = x3 = ... = 0

then we can for sure abbreviate (x1, x2, ...) as just x, (0,0,0,0...) as simply 0, and say that a set is linearly independent when it has only x = 0 as the solution

[barren forum]

It can be a bit abstract, because it's defined by the non-existence of something. To paraphrase, a set is linearly independent when there doesn't exist such not-all-zero x1, x2, ...,

Now can you tell me when is a set not linear independent?

[errant wasp]

at least one vector in the set can be represented as a linear combination of the other vectors in the set

[barren forum]

hmmm that's indeed an equivalent defintion...not the one I'm asking though

[barren forum]

but sure, do you understand what a linear combination is?

[errant wasp]

ax + by where a, b are scalars

and the solution is in the field

[barren forum]

yes, earlier I avoided the term. But x1 v1 + x2 v2 + ... is exactly a linear combination of the vectors v1, v2, ...

[barren forum]

Going back to this question, can you now try to represent a vector as a linear combination of the other two vectors?

[errant wasp]

(3,5)=3(1,0)+5(0,1)

[barren forum]

that's great, and thus you have shown A is not linearly independent.

It's really just by constructing linear combinations.

[barren forum]

It's especially easy in the cases when some are multiples of others, say {(3,4), (6,8)}. This set is not linearly independent just because (6,8) = 2 (3,4), the latter being a simple linear combination of one vector

[errant wasp]

its easier for me to see those examples tbh

[barren forum]

sure, here is another one

[barren forum]

a straight forward negation of the previous definition gives a equivalent form of "not linearly independent"

a set {v1, v2, ...} is not linearly independent when there exist not-alll-zero x1, x2, .... such that the linear combination x1 v1 + x2v2 + ... = 0

In this case, we can observe -3 (1,0) + -5 (0,1) + 1 (3,5) = 0

[errant wasp]

oh tats interesting

[barren forum]

You can see how the two definitons are equivalent, we just need to move a term and divide both sides by a coefficient to represent a vector by the other vectors.

-3 (1,0) + -5 (0,1) + 1 (3,5) = 0
can become
-5 (0,1) + 1 (3,5) = 3 (1,0)
and it becomes
-5/3 (0,1) + 1/3 (3,5) = (1,0)

and hence we represented (1,0) using a linear combination of the other two vectors

[errant wasp]

thank you this was really helpful

[barren forum]

no worries, linear algebra is fun 😃

[errant wasp]

can i ask another question?

[barren forum]

sure thing

[errant wasp]

can i show you what i did for a, and then ask about b

[barren forum]

I see no problem in a)

[errant wasp]

okay i dont think i fully understand b

[barren forum]

in b), you should understand what a projection is. The projection of a vector v, onto a space W, is roughly speaking, the part of v that's in W.

If v is already in W, then the projection is just v itself.

An example on a plane will be much helpful.

here, the green line is a subspace, say spanned by {(1,1)}

we have some purple vector

and the orange vector is the exactly the projection of the purple vector on to the green space. geometrically, it's obtained by drawing a perpendicular line to the green line starting from the arrow to the green line.

if a vector is already in the green subspace, projection onto the space just gives itself back.

[errant wasp]

because its already in itself?

[barren forum]

here is an animation I made, hope it help you better understand the concept of projection

don't mind the labels, I meant to show u but didn't

the orange vector u1 is the projection of u onto the green subspace

[errant wasp]

The idea is like you are already asking for something to be in something when it is already the something thus it is returned

thank you for that visual

[barren forum]

no worries, in general though, the subspace can be arbitrary, it can be a plane in R^3 for example, the same concept applies

take a random vector, shine a parallel light from afar perpendicular to that plane. The shadow on the plane is the projection of thatvector

if the vector is already on the plane, then the projection is itself

[errant wasp]

thank you so much

[barren forum]

no worries

[errant wasp]

whats an orthonomal basis

im sorta confused about that too

[barren forum]

it's a special basis sastifying the orthonomal condition.
Recall a basis is

1. linear independent
2. spanning
   but an orthonormal one has to be also
3. orthogonal
   and 4. normal

3 means, every pair of vectors in that basis is orthogonal to each other (their dot product is zero, or geometrically, they are perpendicular)

4 means, each vector in that basis has to be "normal", meaning the length of every vector has to be 1

[errant wasp]

this is so much to remeber

[barren forum]

the standard basis of R^2 {(1,0), (0,1)} for example 😉

[errant wasp]

P=A(A^T A) −1 A^T what if A is an orthonomal basis

then is it like euqal to its identity

[barren forum]

yeah the P = I actually. You have the trust the question that P is indeed a projection onto the column space of A. (This means for any vector v, P v is the resulting projection onto the space)

Since A is a basis now, the column space is just the whole space.

[errant wasp]

my notes say p = i is generall false tho

[barren forum]

probably only works for orthonormal ones?

[errant wasp]

okay so you cant like assume p = i unless its orthonormal basically

[barren forum]

oh wait

[barren forum]

sorry I was assuming "A is a basis of the whole space"

[errant wasp]

P projects any vector in the space to the column space of A i guess in what i sent?

[barren forum]

I think as long as A is a basis of the whole space, then P = I. The spanning property of the basis is the same as saying that the columb space of the basis is the whole space

[errant wasp]

thank you so much for answering my questions i really appreciate it

.close

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[unique radish]

Hello i need help with the question F

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[feral nacelle]

Hi I tried my best to try achieve the answer although I don’t know how and would appreciate some help

[cedar kiln]

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[limber rain]

I got
x_2 =2.84
x_3=2.52
x_4=2.42

.close

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[orchid goblet]

Please help me with MATLAB code!!

[orchid goblet]

I need to print out the iterates and the function values f(x_k) for secant method on the function f(x) = cosh(x/4)-x over the interval [-15,15] with the initial guesses x_0 = 0.5 and x_1 = 2.0

This is the code I have so far

format long
f = @(x) cosh(x/4)-x;

x_old = 0.5;
x_curr = 2;
tol = 1e-10; %error tolerance
i = 0; %iteration
N = 100; %max iteration

f_old = f(x_old);

while i <= N
f_curr = f(x_curr);
x_new = x_curr - f_curr*(x_curr - x_old)/(f_curr - f_old);

if abs(x_curr - x_old) < tol
    fprintf('Solution is %f \n', double(x_curr))
    fprintf('Number of iteration is %d \n', i)
    return
end

i = i + 1;
x_old = x_curr;
f_old = f_curr;
x_curr = x_new;

end

Can you help me display the iterates and the function values please?

[cedar kiln]

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Show your work, and if possible, explain where you are stuck.

[obsidian coral]

How did you get 1/2?

Why a^4?

Can you show full work?

And the next step would be?

You have two equations, with two unknowns, solve for them

How did you solve for them

When someone asks you to show your work, you should show all your work, from start to finish

Then it's hard to help you without know what you did fully

My question is how did you get a = 1/2

You have two equations, how did you solve for a

Yes

This is the equation you should have gotten

You got a^4 = 1/16 the first time

$\frac{1}{2^3 \sqrt{2}}$?

[wraith dagger]

dldh06

[obsidian coral]

Then no

[obsidian coral]

Because this doesn't have a cuberoot

So then what would a be?

$\frac{1}{2 \sqrt[3]{2}}$?

[wraith dagger]

dldh06

[obsidian coral]

You found a wrong

That's what I'm trying to point out

[obsidian coral]

This is what a is suppose to be

Because this is the proper equation

Yes

Yeah sounds right, but since you have a and C just plug those back in and see if it passes through the given points

[cedar kiln]

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[boreal flame]

the equation is just z=0 right?

[weary vessel]

Yes

[boreal flame]

ty

[cedar kiln]

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[thorn sluice]

hi

[thorn sluice]

can someone help with this

im unsure on how to start

[cedar kiln]

@thorn sluice Has your question been resolved?

[vernal shell]

Well first you have to write each the line in vector form 🤔

In a horizontal line, y is constant, and x is variable. What vector would represent that line? 🤔

[thorn sluice]

<0, x, 0>?

[vernal shell]

Wait that has 3 coordinates

[thorn sluice]

yeah

so is it just <0, x>?

[cedar kiln]

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[native lion]

,w plot 8 + 6x -3x^2

[wraith dagger]

Results provided by WolframAlpha

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

Requested by cronus69

[cedar kiln]

Please don't occupy multiple help channels.

[native lion]

is this horizontal at 1?

[neat dune]

U can verify by finding the vertex

[native lion]

isnt that infinity?

like both -infinity

[neat dune]

The vertex is the peak/bottom of the parabola basically

[native lion]

yeah

so infinity

[neat dune]

here it'd be the peak

If the parabola points downward the vertex is the peak, if it goes upwards it's the bottom

[native lion]

ok?

[neat dune]

do u know how to find a vertex

[native lion]

yeh

i just told you

[neat dune]

no

U got it wrong

The highest point means the peak

[native lion]

1,11

thats the vertex

[neat dune]

ok then yes it's horizontal at 1

[native lion]

only at 1?

[neat dune]

yeah

[native lion]

alright

thanks

[neat dune]

Mhm

[native lion]

can you help with this?

[neat dune]

This is for calculus right

[native lion]

yeah

[neat dune]

take derivative, that's ur line's slope

Then u can use point-slope form for convenience

To write the tangent line

[native lion]

alright i cant do that, nevermind.

for this would i just plug in 35?

[neat dune]

No u gotta take the derivative also

then plug 35 into that

[native lion]

how do i take the derivative

[neat dune]

Uhhh why do you have homework that's making you do this if you haven't gotten to it yet just curious

[native lion]

idk im just not good at math

[neat dune]

this isn't about being good at math or not tho it's the barebones they teach u in Calc class

Do u know power rule?

like if u see $x^n$ alone

[wraith dagger]

992qqoloy

[neat dune]

Have they gotten to how u derive that yet

[native lion]

probs but i forgot

[neat dune]

Well whenever u forget things u can always open ur textbook

idk sorry if I'm being like this. I could just tell u the rule but this seems like a fundamental thing we can work on

So I guess either open ur book or google power rule that's what I usually do :p

[native lion]

so whats the derivative

[neat dune]

I can't just tell u the derivative either solutions are against the rules

Helpers in this server r just supposed to guide

[native lion]

is it -2x?

[neat dune]

Close

There's the 50x term also

[native lion]

-2x+50

?

[neat dune]

Yup

[native lion]

so do i plug 35 now?

[neat dune]

Yup

[native lion]

so -20?

[neat dune]

Yeah

[native lion]

is this -350?

2nd requires derivative?

[neat dune]

yeah 2nd requires derivative

[native lion]

did i do the 1st one right?

[neat dune]

no idk what u did wrong tho

U plugged in 250 for x into R(x) right

[native lion]

yeah

[neat dune]

Hmm even in the scenario where it would say 0.2 instead of 0.02 ur off by a 0

I think u misread the 0.02 as a 0.2 no biggie

[native lion]

oh wow

7750

[neat dune]

Yee

[native lion]

for the 2nd one what do i do

do i just find the derivative of the R(x)

[neat dune]

p much the same as the first one

Yeah

I mean not the first part of this but the earlier problem using C(x)

[native lion]

36-0.04x

[neat dune]

Yup

[native lion]

?

and then i plug 250?

[neat dune]

Mhm

[native lion]

26

awesome

is this the same thing as the last question?

[neat dune]

Yup

Just dont plug in for t=1992

Gotta read the problem to figure out what t is supposed to be

[native lion]

25?

[neat dune]

Yeah

[native lion]

174.5

and then for the 2nd one i have to find the derivative

[neat dune]

Mhm

[native lion]

1.4x-13

[neat dune]

yup

[native lion]

do i plug in 25

so then its 22

[neat dune]

Yup

[native lion]

do i need to find the derivative?

[neat dune]

yeah

For this 1 u can either expand and then derive each term

Or use product rule, but ull probably have to expand either way to put it in simplest terms

So I think it'd be easier to just expand it first

[native lion]

oh ok

that makes sense

(12x^{8}-8x^{7}+10x^{5}-18x^{4}+12x^{3}-15x)

is this correct when expanding?

[neat dune]

Yeah

[native lion]

ok

its the last one

96x^(7)-56x^(6)+50x^(4)-72x^(3)+36x^(2)-15

this is what i got

as the derivative

[neat dune]

Oh

If I'd known it was multiple choice like this I woulda said product rule :p

[native lion]

oh lmao

[neat dune]

Yeah last one

[native lion]

quotient rule?

[neat dune]

For this one there's the mnemonic "LowDeHigh minus HighDeLow over LowLow"

So $\frac{f(x)g'(x) - f'(x)g(x)}{g(x)^2}$

[wraith dagger]

992qqoloy

[native lion]

(x^(4)+6x^(2)-2x)/((x^(2)+2)^(2))

so its either A or D

[neat dune]

The one that has the minus

[native lion]

ok

plug in 40?

and then derivative and plug in for 2nd?

[neat dune]

Mhm

[native lion]

4.984

[neat dune]

how many more of these are left cus i gtg

[native lion]

only a few more

itll be done quick

(-200(x^(2)-5))/((x^(2)+5)^(2))

for the first question is it 4.984?

or do i have to multiply it be a thousdands

since it says the population p in thousands

[neat dune]

Actually that's a really good question

Problem doesn't say so I've no idea for sure

but yeah maybe multiply by 1000

[native lion]

ok

my 2nd answer is a negative decimal
-0.124

are we sure its 4984? or should we leave it as a decimal to three places
im gonna leave it as decimal i think

[neat dune]

Aight

Yeah leave as decimal

Ic now that it says that at the bottom

https://www.wolframalpha.com/input?i=400t%2F(2t^2+%2B+10)+derive put your derivative into wolfram alpha just to double check

[native lion]

(-20(6x^(5)+2))/((x^(6)+2x)^(21))

thats what i got

[neat dune]

yeah

[native lion]

im confused

what do i do after i found the derivative

since it isnt 1 of the answers but its close to all 4

oh wait is it C?

C includes all of it

[neat dune]

Yeah

[native lion]

i didnt get 1/4

(x(2x^(2)-3))/(2(x^(4)-3x^(2))^((3)/(4)))

thats what i got

its A or D since 3/4 should be positive

yeah im confused

.close

[cedar kiln]

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[native lion]

can some1 help

[native lion]

in a rush

[digital cliff]

v=ds/dt
a=dv/dt

[native lion]

what

whats ds

and dt

[digital cliff]

derivative of s with respect to t

[native lion]

is it 14?

[digital cliff]

that is s when t=2

so no

[native lion]

so then what is it

[digital cliff]

do you know what a derivative is?

[native lion]

yeah

neither of them are 14?

[digital cliff]

no

[native lion]

alright

.close

[cedar kiln]

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[digital cliff]

why didnt you know what ds/dt means then?

[cedar kiln]

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[alpine forge]

I’m not sure how to start

[foggy sandal]

log(ab) = log a + log b

[violet brook]

^

so you can split it into log_4(4096) + log_4(x) = 64

[alpine forge]

Uh

[alpine forge]

That’s not what we used in class

I can show you a example similar to how my teacher did it?

[violet brook]

sure

yeah so what they did there

was change it into exponential form

[alpine forge]

yea

[violet brook]

if you did that for this question

youd be saying

4^64 = 4096x

if you put 4^64 into your calculator

it will not be nice

[alpine forge]

Is this what you meant by this

[violet brook]

yes

then you can solve for log_4(4096) using change of base

do you know the change of base formula?

[alpine forge]

log_c a / log_c b

[violet brook]

uh

[alpine forge]

That’s the formula u mean right?

[violet brook]

yes

but notice

the base is the denominator

yes

ok you edited it

so

use change of base to evaluate log_4(4096)

[alpine forge]

okay but what do I put for the log_a (C) in the formula ? log_4 (X)

[violet brook]

in log_4(4096)

[violet brook]

in this form

what is your b and c

[alpine forge]

4096 and x

[violet brook]

no

for log_4(4096)

b is the base
c is the argument

[violet brook]

so what is your base (b) and argument (c) for this

[alpine forge]

isn’t the base 4..? And the argument is 4096

[violet brook]

yes

okay

now changing the base

its better to change the base to either e or 10 calculators usually compute logs with those bases

so putting it into the formula

youll get

oh wait sorry

$\frac{\log_{10}(4096)}{\log_{10}(4)}$

[wraith dagger]

chef wang gang

[violet brook]

now you can use a calculator to evaulate it for you

(as like i said calculators usually compute logs with base 10 or base e which is the purpose of change of base)

[violet brook]

mhm

so you have

6 + log_4(x) = 64

so what do you think is next

[alpine forge]

isolate the x? Right using the exponential form

[violet brook]

yep

but in order to isolate the x you need to remove the 6 first

[alpine forge]

-6 from both sides correct?

[violet brook]

yep

log_4(x) = 58

now rewrite in exponential form

so you can get the solution for x

[alpine forge]

Ohhh so the answer is b

[violet brook]

yep x = 4^58

[alpine forge]

Thanks so much

[violet brook]

no worries

[alpine forge]

Sorry if I took a while to understand 😅

[violet brook]

oh no no its perfectly fine

[alpine forge]

Bye:)

.close

[cedar kiln]

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[patent pond]

hi,I'm having difficulty in this part, please help me.

[patent pond]

help

[solid juniper]

chronic?

[cerulean star]

pronic?

[upper abyss]

🌿

[patent pond]

The question was translated by Google Translate because I'm not very good at English

[cerulean star]

post the original question.

[patent pond]

original question in vietnamese:Cho x,y,z là các số thực không âm thỏa mãn x+y+z=6. Tìm giá trị nhỏ nhất và giá trị lớn nhất của: P

[cerulean star]

just says "non-negative real numbers"

what section of your textbook are you on

and what class is this for, Multivariable Calculus?

(Calculus 3)

[solid juniper]

competition math i would guess

[onyx dove]

looks like plain inequalities man

[patent pond]

It is for grade 9

[patent pond]

yeah

[cerulean star]

ok so I have an equation of the derivative

let Z = 6 - X - Y

then

P = sqrt(X+Y) + sqrt(6-X) + sqrt(6-Y)

differentiate:

$\frac{dy}{dx} = \frac{(6-X)^{-1/2}}{(X+Y)^{-1/2} (6 - Y)^{-1/2}}$

[wraith dagger]

Disorganized

[cerulean star]

um...

but we don't have Y only in terms of X

so idunno

else we could find the critical points

(maybe)

[patent pond]

What is dx,dy?

[cerulean star]

nah nevermind

but the minimum maximizing values I think are X=Y=Z = 2 (by inspection)

[patent pond]

I also think x=y=z but it's hard to convert it to the maximizing values

[cerulean star]

oh

there's a few ways to do this with P(X,Y) = sqrt(X+Y) + sqrt(6-X) + sqrt(6-Y)

[cedar kiln]

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[honest bobcat]

practice exam

why is this the case

I undersatnd finding x if its minimal

but not largst

[vague rapids]

Product of LCM and HCF of two numbers is equal to their product

[honest bobcat]

ye

[silver parcel]

you want the largest x such that x =/= lcm(x, y)
but by definition, x divides lcm(x, y)
so x is the largest divisor of lcm(x, y) different from lcm(x, y)
since lcm(x, y) is divisible by 2, x is necessarily lcm(x, y)/2

since you can't divide lcm(x, y) by less than 2 without x being equal to lcm

[honest bobcat]

why can't x just be the lcm(x,y)

[honest bobcat]

oh

im a fuckiw

t

lmao

read the question challenge

[honest bobcat]

sorry for ping hope its ok but

similar question

but why is it 3^2 not 3?

[silver parcel]

because 3 is not in the gcd, so it divides only x or y but not both

so among x and y, one is divisible by 3² = 9

[honest bobcat]

hm

[silver parcel]

if x = lcm/3, y would divide 3, so be 1 or 3, but both are impossible

[honest bobcat]

nah im fucking dumb

okay

nAH IOn get it

[cedar kiln]

@honest bobcat Has your question been resolved?

[honest bobcat]

sorry for another tag i’m really tryna get this shit 😭

would you be able to elaborate a bit more

on maybe a general example too

nw if not

[silver parcel]

you want the largest x different from lcm(x, y)
but gcd(x, y) divides x and x divides lcm(x, y)
so can we choose x = lcm/2 ? no, because then 2² doesn't divide x, so gcd(x, y) wouldn't divide x
the gcd and lcm put constraints on our choices
could we take x = lcm/3 ?
no, because then 3 would divide x and y, but then it would divide gcd(x, y)

but 3 doesn't divide gcd(x, y)

you could say well then, I will take lcm/5, it's larger than lcm/9

but it still doesn't work for the same reason

5 can't divide both x and y

since it doesn't divide gcd(x, y)

[honest bobcat]

ahh this makes sense here, that would give x a factor of 3, but because the lcm contains another factor of 3 it has to come from y?

[silver parcel]

yes

but because 3 doesn't appear in the gcd

it's not possible either

[honest bobcat]

that’s not possible

and then 3^2 works cus

[silver parcel]

because the whole 3 as a prime factor get put into the x

so we don't have to make it appear in y

nor in gcd(x, y)

[honest bobcat]

i see

i will reread this until it fully settles thank you

[cedar kiln]

@honest bobcat Has your question been resolved?

[cedar kiln]

@honest bobcat Has your question been resolved?

[cedar kiln]

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[lavish jolt]

is this how we differentiate
im still not familiar with this notation

[cedar kiln]

@lavish jolt Has your question been resolved?

[cedar kiln]

@lavish jolt Has your question been resolved?

[crystal raptor]

Can you describe in words what you're trying to write? It's a bit confusing as written

[cedar kiln]

@lavish jolt Has your question been resolved?

[lavish jolt]

Im trying to derive the lagrangian density

A_μ is a vectorial field

thats one term of the expression

[cedar kiln]

@lavish jolt Has your question been resolved?

[cedar kiln]

@lavish jolt Has your question been resolved?

[cedar kiln]

@lavish jolt Has your question been resolved?

[cedar kiln]

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[naive vessel]

hi guys i want to get “v” from this equation, please help

[tropic oxide]

$mg(h-h_0) = \frac12 m(v^2 - v_0^2)$

[wraith dagger]

Ann

[tropic oxide]

!status

[cedar kiln]

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

[naive vessel]

3

[tropic oxide]

show work

(though i fear i might have a hard time reading it with this heavy cursive you've got...)

[tropic oxide]

what does this say

[naive vessel]

its not important

[tropic oxide]

anyway, line 3 seems to be wrong

or like

... even number of mistakes, i guess?

you added v0^2 into the fraction in a bullshit manner, and then you cancelled out the m's in a bullshit manner.

but you ended up with $2g(h - h_0) + v_0^2 = v^2$?

[wraith dagger]

Ann

[tropic oxide]

which seems correct to me

but then you square rooted both sides and

somehow $\sqrt{2g(h-h_0) + v_0^2}$ became $\sqrt{2g + v_0^2 (h - h_0)}$?

[wraith dagger]

Ann

[naive vessel]

Is this correct?

I changed last 2lines

[tropic oxide]

why did the $v_0^2$ get a 2 coefficient?

[wraith dagger]

Ann

[naive vessel]

i dont know i thinks its correct that way

thanks for helping btw

give me a sec i try something

This is my process😅

math is hard

[tropic oxide]

first line is wrong here

all of this has a lot of bullshit in it

[naive vessel]

and how is it correct then? i am lost

@tropic oxide can you please tell whats wrong?

[tropic oxide]

the one i can see and point out immediately is that you rewrite a/b + c as (a+c)/b

it is hard for me to point it all out since there is so much of it and it piles on top of each other

[cedar kiln]

@naive vessel Has your question been resolved?

[cedar kiln]

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[worthy basalt]

hey

[worthy basalt]

f(x)=ax^3-8x^2+bx+6

f(x) is divided by x+1

there is no remainder

f(x) is divided by x-2 the remainder is x-12

what is the value of a and b???????

[onyx dove]

do you know your remainder theorem

[worthy basalt]

no

what is that ?

[worthy basalt]

so what do I do ?

[dim tiger]

So first if x+1 divides f(x) then what can you say about -1 with respect to f(x)

[worthy basalt]

idk

but I think I don't really need to know this exercise

but ty

[dim tiger]

x+1 divides f(x) then f(-1)=??

Do you know this ?

[worthy basalt]

f(-1)=a(-1)^3-8(-1)^2+b(-1)+6

[dim tiger]

Yea and this is =?

I want you to give me a number

Independent of a and b

[worthy basalt]

-a+8-b+6

[dim tiger]

do you know what is the meaning of root ?

root of a function ?

[worthy basalt]

yea

root of a function are when f(x)=0

find the roots

the x intercepts

[dim tiger]

ok nice

now i am telling you that x+1 divides f(x)

so f(x)=(x+1)(something)

isnt this right

[worthy basalt]

yes

we need to find that something

[dim tiger]

that something is also a polynomial

[dim tiger]

now what is f(-1) from here

[worthy basalt]

-a+8-b+6
?

[dim tiger]

just sub -1

i asked from here

not from the original expression of f(x)

[worthy basalt]

f(-1)=(-1+1)(something

[dim tiger]

which is

[worthy basalt]

0

no remainder

[dim tiger]

ok nice

so what can you say about -1

[worthy basalt]

f(-1)=0

cuz y=0

it's a root

[dim tiger]

exactly

now f(-1)=0

[dim tiger]

why +8

[worthy basalt]

f(-1)=a(-1)^3-8(-1)^2+b(-1)+6

nvm it's -8

[dim tiger]

ok great

[worthy basalt]

-a-8-b+6=0

and we do another simultaneous equations ?

[dim tiger]

so f(-1)=-a-8-b+6=0

exactly

now we use the other part of the given to get another equation in a and b

[worthy basalt]

ahhh

f(x)=(x-12)(?)

[dim tiger]

we can translate second part of the equation to f(x)=(x-2)(some other thing)+x-12 so f(2)=??

[dim tiger]

do you know how i got to this ?

remainder of division of f by x-2 is x-12

[worthy basalt]

f(2)=(2-2)(?)+x-12

[dim tiger]

=??

[worthy basalt]

0

2-12

-10

[dim tiger]

ok you have 2 simultaneuous eqs now solve the system to find a and b

[worthy basalt]

ok ty

/close

[dim tiger]

np

its .close

.close

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[spark moat]

Wouldnt A1xA2 be {(0, 0）(0,1 }？

[warm bone]

I think so

Wrong answers probably

[spark moat]

Would this be correct answer for part f?

[crimson sedge]

yes

[cedar kiln]

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[frigid sierra]

i need help for finding if something is a subspace of R^n or not

Anyone know a resource to help me learn this. im really stuck

[frigid sierra]

please use reply to notify me

[fallen moat]

reply text:
https://textbooks.math.gatech.edu/ila/subspaces.html
reply video:
https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/subspace-basis/v/linear-subspaces

[frigid sierra]

uh well that's not what i meant by reply hah

but thanks

are you aware if this covers enough examples?

AMAZING

thank you so much

.close

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[fallen moat]

hello there Peter!

your g(x) is just f(x-1), which is only shifting f(x) to the right side by 1

My method would be first move to graph so that we can move x=1 to x=0, and do it like normal reflecting along x=0, and finally move it back

[cedar kiln]

@crimson sedge Has your question been resolved?

[onyx dove]

christ you're good at dutch, peter

[cedar kiln]

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[placid bluff]

How to write a closed formula for the following series

[placid bluff]

it's an arithmetic series but with a geometric series in the common difference

[livid hound]

,rotate

[crimson sedge]

$2, 2+7(1), 2+7(2), 2+7(3), ...$

[wraith dagger]

deltaG

[placid bluff]

but it's not that

[crimson sedge]

wait nvm I'm stupid

[placid bluff]

lmao no worries

[onyx dove]

||2+7*2^(n-1)||

[placid bluff]

could you explain how you got to this formula?

[onyx dove]

common difference changing by factor of 2

so use exponential base 2

[placid bluff]

i should mention that the first item in the series is u_0

[crimson delta]

the formula is close but not quite

[placid bluff]

such that n = 0

[onyx dove]

then of course it's different

[placid bluff]

yeah sorry i forgot about that until now

[crimson delta]

instead of trying it via the common difference, subtract 2 from the original sequence and divide by 7. then you'll obtain another sequence for which finding a closed form should be much easier

[placid bluff]

ok i'll try that out and then given an update if i still have any more doubts

[crimson delta]

(the numbers 2 and 7 being motivated from the common difference attempt, they arent completely out of nowhere)

[placid bluff]

yeah i got that, thanks!

ok wait

my problem got "reassigned"

i have a very similar sequence

and i've noticed soething else about this one

and i can make the general sequence:

2 + n(11*3^(n-1))

well that was wrong

anyway, thanks for the help!

.close

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[crimson sedge]

Can I get help solving these? I’m can’t do the steps. The answer is 31,125

[tardy citrus]

How do we combine fractions with two different denominators?

[crimson sedge]

Make them the common one

I think

[tardy citrus]

Yep so what’s the lcd of 8 and 12

[crimson sedge]

96

Orrr idk

WAIT

if u divide

[tardy citrus]

Ok so how do we change the fractions to have that denominator

[crimson sedge]

The first multiplied by 12 and the second multiplied by 8?

[tardy citrus]

Close but you have do 12/12 and 8/8

So the top and bottom are both multiplied by 12 for the first fraction and the top and bottom of the second are multiplied by 8

So do that work and lmk what you got

[crimson sedge]

Is that right

[tardy citrus]

Yes okay now we can combine the like fractions

[crimson sedge]

?

[tardy citrus]

So it will just be one big fraction over 96

[crimson sedge]

Ohhh ok I’ll do that rn

[tardy citrus]

Since the denominators are the same you can add the numerators together

Lmk what you get

[crimson sedge]

Omg thank you!!!!

I got the answer

Ty

[tardy citrus]

👍

[crimson sedge]

.close

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[crimson sedge]

how do i go about finding the range

just where x is bigger than zero or what

[feral coyote]

range of f = possible ys

[long arrow]

notice that

2|3-x| >= 0 for all real x

in other words if it was f(x) = 2|3-x| its range would be [0, inf)

since modulus is nonnegative

[crimson sedge]

so i would say

hmmm

but how would you prove that

with working

[long arrow]

it comes directly from the definition

of absolute value

[crimson sedge]

and whats that

[long arrow]

and hence |x| >= 0

always

it's a non - negativity property

[crimson sedge]

so i would write ranges for f are x is bigger than or equal to 0

[long arrow]

range is about values of the function, so it's about "y"

not x

I guess you don't understand what I'm trying to say

look at this simple example

let's say we want to state range of |x| + 1

knowing that |x| >= 0 we can easily deduce that |x| + 1 >= 1

[crimson sedge]

so then in this case of 2|3-x|+5 >= ?

[long arrow]

yes same way to solve this one

try it on your own

[crimson sedge]

uhh i forgot how to do these

6-2x+5>=0 would be the next step right?

[long arrow]

ee not really

you don't have to solve anything to get the range

it isn't an inequality you're supposed to solve

[crimson sedge]

so what do i do

f(x) ∈ [0; +infinity)

this?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[clever kite]

.reopen

my mom wants to put wallpapers on the walls, we need 30 cubic meters, one wallpaper is 30x35cm how much do we need?

[rustic siren]

how much cm^2 is a wallpaper?

[rustic siren]

you mean square meters, right?

[clever kite]

yes

[clever kite]

30x35cm

idk the square meter for that

im sorry

[rustic siren]

and then figure out the whole wall in cm^2, and divide

you have to change both in the same form, so we could see the ratio

1 m^2 is how many cm^2?

[clever kite]

idk 😭

100

ig

30m2 is 3000cm2

[rustic siren]

why

[clever kite]

idk bro

i dont understand

[rustic siren]

1m*1m

is 1m^2

so now change this to cm

1m=100cm

[cedar kiln]

@clever kite Has your question been resolved?

[clever kite]

im dumb im sorry

could you solve it for me and not ask questions?

[rustic siren]

no.

[clever kite]

why not tho

[rustic siren]

we do not do that here

#❓how-to-get-help

[clever kite]

ok then help me solve it

[rustic siren]

substitute in

[clever kite]

ohhh

3000x3500?

is 6500cm2

so i need 65m2?

wait waht

no again

lemme thnik

that was dumb as hell

wait

30x35 is 1050

so one wallpaper is 1050cm2?

doesnt look right tho

[rustic siren]

it is

one wallpaper is indeed this many cm^2

[clever kite]

ok so thats 10.5m2

the wall is 30m2

divide?

so i need 2.8 wallpapers?

so i should buy 3?

doesnt look right more

[rustic siren]

no

[clever kite]

why not tho

it isnt logical

xd

[rustic siren]

1m * 1m = 100cm * 100cm

see?

i asked you to do this before too

[clever kite]

oh right

+00

100

[rustic siren]

now you just have to put the 2 info together

[clever kite]

im lost

[rustic siren]

Its just algebra from this point

Write an equation

You know how much cm2 the 30 m2 is

And what is the wallpapers area

[clever kite]

3000x3000cm is 30m2?

idk man

im dumb

[rustic siren]

yes

[rustic siren]

dont say this

if you do not see the solution from the start its not bad, you just need small steps towards the answer using what we know at that point (it doesnt hurt to make a list os these infos separately)

[clever kite]

i have this noted

[rustic siren]

how much is 3000*3000?

and how many of the 1050 do you need to fill it?

[clever kite]

9million

so 9000000cm2 is 3000x3000

9000000 divide by 1050?

idk

[clever kite]

@rustic siren help pls

im confused

[rustic siren]

9000000 = 1050*x

find x

[clever kite]

but whats the x

[rustic siren]

unknown

the number of wallpapers in this case

you dont know how to solve equations?

[clever kite]

no

i said im stupid xd

[rustic siren]

and i say ur not

yes, 9000000/1050

what grade are you in?

[clever kite]

highschool

3rd xd

[rustic siren]

u should have met with equations then, no?

[clever kite]

yes

there is no way i need 8571 wallpapers

[rustic siren]

oh, sry

my mistake, i forgot the original Q

so 30 m^2

and we know 1 is 10000 cm^2

so 30 is 300000 , the square is on the cm not the number

see? thats why we need to always doublecheck : D

[clever kite]

im still confused

[rustic siren]

with what? please dont just say youre confused and leave it there like that

[clever kite]

everything is confusing me

i wish i could tell you with what

but my only response is everything

[rustic siren]

oh boy

[clever kite]

yea

[rustic siren]

im sorry then

i cant help with everything

[clever kite]

xd

[rustic siren]

but if you line them up i can try to go through the math related ones

anyway the answer is 30 * 10000 / 1050

[clever kite]

still stuck with that the 1 wallpaper is 1050 square cm and the wall is 30 square metres

[rustic siren]

30 mˇ2 is 300000 cm^2

[clever kite]

still the answer is 8571

[rustic siren]

how??

[clever kite]

oh

i calculated it wrong

[rustic siren]

look at it, it should not be higher than 300

[clever kite]

okay thank you sir

appreciate it

[rustic siren]

glad if i could help

next time try to sit on it a bit, and come here with more specific questions, it should make thing faster/easier

[cedar kiln]

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[golden thicket]

The height of the base of a regular triangular prism is 2√3cm. Find the area of ​​the lateral surface of this prism if the diagonal of the lateral face of the prism forms an angle of 30◦ with its lateral edge.​ need help with a solution

[fresh hound]

hello

that's easy

[golden thicket]

How?

[fresh hound]

alright i'll do it for you

96^2 cm

[golden thicket]

it can not be true

[fresh hound]

easy peasy : )

XD

i am the best at maths

[subtle yew]

Okay, man...

Please actually try to help the dude if you know how

Don't waste his time and disappoint him.

[subtle yew]

What does this mean by "height of the base"?

Oh okay, I see it. The yellow was too light

[golden thicket]

height of triangle

[subtle yew]

Which one is the lateral surface of the prism?

A1B1AB?

[golden thicket]

yes

[subtle yew]

Okay.... and that's the area we need to find, right?

Gotcha...

Do you know basic trigonometry?

[golden thicket]

i think yes

[subtle yew]

Okay, so if you look at the triangle that the 30 degree line forms...

What would the adjacent and opposite lengths be? (Don't need an exact value, just give me the expression itself)

like Length(A, B1) * sin (angle)

Something like that

[golden thicket]

sin(30)*ABB1?

[subtle yew]

nope.

Okay, so you claim to know basic trig, right?

[golden thicket]

sin(30)/ABB1

[golden thicket]

Yes

[subtle yew]

What's:
length(A, B)?
length(B, C)?
...in terms of length(A, C)

[golden thicket]

(A,B) = tg

or not? I don't quite understand your question

[subtle yew]

Okay, I'm writing it in a way more complicated way.

I'm asking:
What's length of AB?
What's length of BC?
Give me an equation for each of them, using length AC

[golden thicket]

AB=BC/AC=cos
BC = AB/AC=tg

like this?

[subtle yew]

firstly... what the frick is "tg"? 💀

[subtle yew]

No...

[golden thicket]

tan

[subtle yew]

Why would you call it that????

[golden thicket]

This is how we say "tg"

instead of tan

[subtle yew]

y'all are funny

Anyway

[subtle yew]

You know the soh cah toa rule?

[golden thicket]

what do soh cah and toa mean???

[subtle yew]

Sin(angle) = Opposite/Hypotenuse
Cos(angle) = Adjacent/Hypotenus
Tan(angle) = Opposite/Adjacent

remember this: opposite means the angle doesn't touch the line. Adacent means the angle does touch the line

[golden thicket]

ok

[golden thicket]

I understand it

[subtle yew]

Okay, wonderful.

So sin(angle) = opposite/hypotenuse can be rearranged as:
hypotenus * sin(angle) = opposite
if you know how to do a bit of algebra.

[golden thicket]

yes

[subtle yew]

Similarly,
cos(angle) = adjacent/hypotenuse
so
hypotenus * cos(angle) = adjacent

[golden thicket]

yes

[subtle yew]

So this becomes:
AB = AC * sin(12)
BC = AC * cos(12)

Understand?

[golden thicket]

yea

[subtle yew]

Okay, now apply this to your question.