help-10

[analog shadow]

.

[jade rune]

how do i figure out the "a maximum at x=3pi " part?
so far i have f(x)=8cos[ 2/9 (x+c)] + d

[obtuse pebble]

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[silent lichen]

leibnitz theorem differential calculus

[tardy epoch]

What's the index on y mean

nth derivative?

[obtuse pebble]

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[fervent dirge]

does any of u know the sol of these probs

[spice citrus]

For the first one try subtracting or adding the first two equations and see if you can do anything with the result

[obtuse pebble]

@fervent dirge Has your question been resolved?

[silent lichen]

yes

[silent lichen]

bro it didnt even get resolved

[obtuse pebble]

@fervent dirge Has your question been resolved?

[obtuse pebble]

@fervent dirge Has your question been resolved?

[fervent dirge]

I solved that one can u help with the second one

[spice citrus]

Sorry, don't know the answer to that one. I assume it involves bounding though, somehow you have to prove the divisibilities are actually inequalities

[obtuse pebble]

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[silk elk]

the graph of cos (x^2) starts off wider and decreases in frequency over time. Is there a way to create the opposite effect–a graph that starts off with a high frequency and decreases over distance?

[obtuse pebble]

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[hollow oak]

im not positive how to do it, i did get an answer earlier but they both flagged as incorrect

[obtuse pebble]

@hollow oak Has your question been resolved?

[sweet gulch]

ratios, I can do it for the 130degree one, as I don't remember how many radians is a full circle but, 360/130 = 18/x I think

no, 360/130 = x/18

this gives you total area of the circle

which area of a circle pi(r)^2

so x = pi(r)^2

than solve for r

@hollow oak

[hollow oak]

Oh thank you!

2pi radian

is a full circle

[sweet gulch]

ok

2pi/0.7 = x/18

[hollow oak]

oh i see tyy!

so we first find area of the circle

and from there find the radius

[sweet gulch]

yep

@hollow oak if you have no further questions, please close this channel

[hollow oak]

alrighty

ty!

.close

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[nimble pilot]

help me please

[nimble pilot]

im not sure how to do this

help please :c

save me

[obtuse pebble]

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[fallen monolith]

would my thing be right?

[nimble pilot]

woah

ur doing similar stuff as me

can you help me with this

im good with matrix let me see if i can figure that out

give me a sec

oh wow

no fcking clue what that is

gg goodluck

u got this

[fallen monolith]

[nimble pilot]

im good with basic matrix

[fallen monolith]

this one draw circles

[nimble pilot]

not whatever that T 00 is

[nimble pilot]

oh?? can you show me

[fallen monolith]

so u have two states

which is two circles right?

s0 & s1

[nimble pilot]

yes

[fallen monolith]

since f(s0, 0) = s_0, that just means when the input is 0, you go back to s0 state

and so u just draw an arrow going back to s0 from s0

and put a 0 beside the arrow indicating your input was 0

and u do the same things for the others

[nimble pilot]

i see

so its a diagram

[fallen monolith]

f(s0, 1) = s1 implies if you have input of 1, then you go to state s1

yeah

2 circles representing state

[nimble pilot]

similar to a matrix diagram

[slow salmon]

wait, what's going on in this chat?

[fallen monolith]

i needed help with my thing but i ended up helping someone doing their fsm diagram

[fallen monolith]

@slow salmon

[nimble pilot]

I GOT IT

W STUFF

thanks G

let me search ur question on chegg

see if i can find something for u

im sorry

i cant find any

they are all using diff numbers

and diff questions

[fallen monolith]

rip

[nimble pilot]

like wahtever this is

[fallen monolith]

spectral decomp fun

[nimble pilot]

is this how it would look

@fallen monolith

?

the guy on youtube has outputs that he wrote as well

but mine doesnt have those

i dont think

[obtuse pebble]

@fallen monolith Has your question been resolved?

[fallen monolith]

yeah

[nimble pilot]

yay

my bad for taking over ur thing by the way

:c

one sec

[upbeat plinth]

not quite

write $\alpha=\brc{a_1,a_2,a_3}$. then
$$[T]^\alpha_\alpha=\m{[T(a_1)]\alpha & [T(a_2)]\alpha & [T(a_3)]_\alpha}$$

[warm shale]

Roketsune Janiku

[fallen monolith]

oh i see u just solve for each vector/basis in alpha Ax=b, solve for b w/ rref in terms of the given basis,

turns out chatgpt got the correct answer im asuming

[upbeat plinth]

i would double check all gpt output

[fallen monolith]

ok

[obtuse pebble]

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[old bloom]

not sure what videos to watch to learn the content

cant find stuff on it, especially the first one

just a bunch of vague nonsense about Nonlinear dynamics

[obtuse pebble]

@old bloom Has your question been resolved?

[obtuse pebble]

@old bloom Has your question been resolved?

[obtuse pebble]

@old bloom Has your question been resolved?

[old bloom]

😦

[obtuse pebble]

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[obtuse pebble]

@old bloom Has your question been resolved?

[tame narwhal]

<@&268886789983436800> spam

[solar crown]

Chaos: Making a New Science" by James Gleick seems interesting: 'For an ecologist, the most obvious sort of function for population growth is linear—the Malthusian scenario of steady, limitless growth by a fixed percentage each year (left). More realistic functions formed an arch, sending the population back downward when it became too high. Illustrated is the “logistic map,” a perfect parabola, defined by the function y=rx(1 - x), where the value of r, from 0 to 4, determines the parabola’s steepness. But Feigenbaum discovered that it did not matter precisely what sort of arch he used; the details of the equation were beside the point. What mattered was that the function should have a "hump.”'

There is "Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering" by Steven H. Strogatz

https://math.stackexchange.com/questions/3532590/logistic-map-confusion

[obtuse pebble]

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[woven creek]

helo

[woven creek]

cans= someone explain these please

what is it checking here

[obtuse pebble]

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[timid silo]

how do i do this again

[timid silo]

.

how do i

send the question

@timid silo

SORRY

mom came in

sending it rn

\sqrt{16x:-:16:}-\sqrt{x^5-x^4}

idk how to like'

[
\s{16x - 16} - \s{x^5 -x^4}
]

$\sqrt{16x:-:16:}-\sqrt{x^5-x^4}

is this is it

no not x5

[timid silo]

x5 - x4

yeah that one

okay what's the objective

operation of radicalsss...?????

uhhh

simplify

simplify

i think

?

yeah

okay

great

do you know factoring

yes

i hate it tho so i forget half of it

can you factor 16x -16

16(x - 1)

okay great

so

[
\s{a\cd b} = \s a \cd \s b
]

[timid silo]

use this

.

you have $\s{16(x-1)}$

[timid silo]

use the above

uh....

uuuuuuhhhhhhhh

:D

16 is the a
x-1 is the b

basically split them apart

sqrt16 then

mhm and?

oh wait

yeah

sqrt16

then

you are missing something

sqrt(16) and what?

you split them apart remember

x - 1

right

so fully?

sqrt(16) times (x - 1)??

almost

remember it's sqrt for both

sqrt(16) sqrt(x - 1)

yes [
\s{16\p{x-1}} = \s{16}\cd\s{x-1}
]

[timid silo]

so what is sqrt(16)?

4

right

so now

what is Ur expression

uh

😭

[timid silo]

it's just the same here

but we made the sqrt16 be 4

take your time, don't feel pressured

can we still sqrt the 4 since its a perfect square

[timid silo]

no we sqrt it once

[
\s{16} = 4 \ \s{16}\cd\s{x-1} = \mathord{???}
]

[timid silo]

4 sqrt(x - 1)

[timid silo]

yes!

so

[
4\s{x-1} - \s{x^5-x^4}
]

[timid silo]

is this fine so far?

uhuh

so then

we

okay so can you factor $x^5-x^4$ now

[timid silo]

x2 sqrt(x - 1)

yeah great

[
4\s{x-1} - x^2 \s{x-1}
]

[timid silo]

now finally

can you see the last factoring you can do

uhhmM

if u wanna see it better

like

let u = sqrt(x-1)

we are hiding the sqrt behind u

so we have [
4u-x^2u
]

[timid silo]

does this help

no-

can you not find the GCF of both 4u and x^2u?

2

nope

oh

u

yeahhh

so if u factor u out

what do u have

😢

[
4u-x^2 u = u\p{???}
]

[timid silo]

where did we get this equation

which

the one we're doing right now

you have [
4\s{x-1} - x^2\s{x-1}
]
I set $u =\s{x-1}$ so you can see the factoring easier. Meaning: [
4u - x^2u
]

[timid silo]

how is that still simplify-able

wouldnt that be the final answer already

the answer is fine. We can still simplify it further by factoring it

w h a t

so thats basically the final answer?

it's just simply [
4\s{x-1} - x^2\s{x-1} = \s{x-1}\p{4-x^2}
]

[timid silo]

that's it

i

why is there more

[candid yarrow]

You could still factor it by difference of squares

[timid silo]

yALL ITS OVA

my tutor said

thats the final answer

[candid yarrow]

$$ \sqrt{x - 1} (4 - x^2) = \sqrt{x - 1} (2 - x)(2 + x) $$

[warm shale]

rysrobrgldvoelr👻ep>vneae=u

[timid silo]

what the frick

[candid yarrow]

It can still be factored a bit more using the difference of squares formula, but since your tutor said that was the final answer, I guess it doesn’t matter

[timid silo]

thank you anyway

[candid yarrow]

have you completed the questions?

[timid silo]

yes

[candid yarrow]

nice

[timid silo]

the final answer is this

[candid yarrow]

not this?

[timid silo]

nope

thats according to my tutor

[candid yarrow]

that’s weird, because there’s a clear way to simplify it further

[timid silo]

ah well

maybe im just not there yet

[obtuse pebble]

@timid silo Has your question been resolved?

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[void helm]

help🙏

[void helm]

help

[candid yarrow]

have you done similar problems before?

[void helm]

no:(

oh wait we did but it was more easy to understand

but i still dont get

it wasn’t properly discussed so its really confusing

[obtuse pebble]

@void helm Has your question been resolved?

[void helm]

nour

[obtuse pebble]

@void helm Has your question been resolved?

[obtuse pebble]

@void helm Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

My doubt was about the differentiability of |x^2|

[timid silo]

What i know is that if we have, |f(x)| then the function is ND at f(x) = 0

but the other condition is, f(x)' should not be equal to 0 at the value of f(x) = 0

for x^2 = 0, we have x =0

f(x)' = 2x

x = 0 in f(x)' is 0

which means that the function is differentiable at |x^2|

[tardy epoch]

yes ?

[timid silo]

Is this correct?

[tardy epoch]

the function is ND
what's ND

[timid silo]

Non DIfferentiable

[tardy epoch]

Why do you think all functions of the form |f(x)| is not differentiable at 0?

[timid silo]

no

f(x) = 0

[tardy epoch]

oh

[timid silo]

with the added

SInce |x| is ND at x = 0

anything in |~| has to be ND at ~ =0

except the slope becomes zero, like x^3 which just turns into the graph of x^2 (approx) in | |

[tardy epoch]

well your function $|x^2|$ doesn't satisfy $f'(x) \neq 0$ at $f(x) = 0$

[warm shale]

riemann

[timid silo]

exactly

it is equal to 0

which means it is differentiable

right

because gpt and bard said no the first time

Even desmos agrees

[tardy epoch]

yea AI can't do math

[timid silo]

unless i am missing something

[tardy epoch]

don't rely on AI to do math

[timid silo]

it SUCKS

[timid silo]

yeah man

[tardy epoch]

right

don't waste your time

[timid silo]

so uhm

is my concept

correct

Looks like he is off

anyone else?

[old bloom]

.reopen

[timid silo]

???

tf

[old bloom]

hey why u taken my help thing

rude

[timid silo]

wtf

its mine 🤦‍♂️

[old bloom]

child

[smoky vigil]

@timid silo it also helps if you think of $|x|=\sqrt{x^{2}}$

[warm shale]

Moosey

[timid silo]

yes

thank you but can u verify my method

/concept

[smoky vigil]

|x^2|=x^2

[timid silo]

yes i do get it

but for x^n

oh

okay i got it

[old bloom]

lmao

[timid silo]

thanks moosey

[smoky vigil]

if f'(0)=0, then |f(x)| will be differentiable at 0

[timid silo]

thanks

that helps

.close

[obtuse pebble]

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[old bloom]

how to learn matlab before 11am GMT 13/12/23

[robust sleet]

what

[old bloom]

i have a test on matlab knowledge tmrw

any pointers?

[tardy epoch]

Google "Matlab tutorials"

[old bloom]

funny.

[tardy epoch]

Anyway, this isn't a math question

Ask in #discussion for study tips

[old bloom]

.close

[obtuse pebble]

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[final thunder]

Have you tried induction?

[tawny kite]

yes I dont understand the highlighted blue part?

[scarlet spruce]

a (n+1) term was factored out of the previous expression

[tawny kite]

the previous expression as in the expression it is equivalent to? right before the hhighlighted part?

[scarlet spruce]

yep

[tawny kite]

I see thank you!

.close

[obtuse pebble]

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[boreal plaza]

HELP

Find the smallest value of N for which we can say: among any N number of numbers, there are 18 numbers such that their sum is divisible by 18

[obtuse pebble]

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[static pebble]

could someone pls explain

[scarlet spruce]

if you expand out the product, only two terms will contain an x to the first power

$(5+nx)^2\left(1+\frac{3}{5}x\right)^n=25+100x+\cdots$\
Don't forget the parentheses around the expanded $5+nx$ term:\
$(25+10nx+n^2x^2)\left(1+\frac{3}{5}x\right)^n=25+100x+\cdots$

[warm shale]

lifefuel

[scarlet spruce]

you can use distributivity to proceed

[static pebble]

i dont get how to proceed tho 😭 i dont have value of n so i cant expand

[scarlet spruce]

maybe you can't expand the whole thing, but i saw you tried using the binomial theorem earlier

it's enough for the first term or two

take a look at https://en.wikipedia.org/wiki/Binomial_theorem#Statement if you need a reference

[static pebble]

lemme try again

[scarlet spruce]

in particular, the y = 1 case, outlined in the linked section, may be useful

[obtuse pebble]

@static pebble Has your question been resolved?

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[lethal grail]

The number x lies somewhere between the numbers 17 and 23.
x is p % greater than 17 and p % less than 23.
Determine x.

[lethal grail]

so i began thinking about where to begin

but since the 1,00-1,9999... (i dont remeber what it is called) and other one being 1,00-0,0000001... Then the number could litterly be anything how can i solve that?

[daring sorrel]

Set up an equation

[lethal grail]

but how do i set one up

i thought of that aswell but it didnt feel right

ill try and come back

[daring sorrel]

Think about what p% greater than 17 means algebraically

Do the same for p% less than 23

These numbers are the same so you can equate them and solve for p

[lethal grail]

like that or am i totally wrong?

Sorry for handwriting

<@&286206848099549185>

[obtuse pebble]

@lethal grail Has your question been resolved?

[lethal grail]

Yes i used geogebra and solved it

[obtuse pebble]

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[lean mirage]

why is it D

[lean mirage]

should it be + 4 theta +3 cus the intial value is 4 and 3

[obtuse pebble]

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

[lean mirage]

i took integral

and used intial value for c

[tame narwhal]

what did you get for its first derivative

the initial value doesn't equal c

[lean mirage]

ooh shit

how do i do this then

[tame narwhal]

write out all the steps, carefully

first find the first derivative, including the constant

then plug in 0 and solve for the constant.

then find the original function, including constant. solve for that new constant. then you get the original function particular solution

[lean mirage]

ohhhhhhhh

i get it now

i got it

thanks

.close

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[fiery wadi]

Is this right?

[fiery wadi]

wait how do I do this line integral? I used green's theorem instead

[obtuse pebble]

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[hard minnow]

Could someone help me with this problem: I'm being asked to "set up a triple integral to describe: inside y = x^(2)+4z^(2) and between the planes y = 1 and y = 9."

[obtuse pebble]

@hard minnow Has your question been resolved?

[hard minnow]

<@&286206848099549185>

[obtuse pebble]

@hard minnow Has your question been resolved?

[hard minnow]

@analog vault

<@&286206848099549185>

[obtuse pebble]

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[candid rock]

hint: $16 = 4^2$

[warm shale]

PrettyPrincessKitty FS

[halcyon flume]

so 16^a=4^2a

[candid rock]

yep

$4^{2a}=(4^a)^2$

[warm shale]

PrettyPrincessKitty FS

[halcyon flume]

so then we need root 225

which is 15

that about right?

[candid rock]

sounds good

[halcyon flume]

thanks for the help mate

.close

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[strong barn]

hi! i'm having trouble with this physics problem - i drew the free body diagram but i'm not sure how to approach b and c

[obtuse pebble]

@strong barn Has your question been resolved?

[strong barn]

<@&286206848099549185>

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@strong barn Has your question been resolved?

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[narrow mesa]

Why is the ln(x) in the answer in absolute value form here?

[narrow mesa]

But then for this problem its not in absolute value form?

[obtuse pebble]

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[stoic marlin]

do I need to take reciprocal/ why is that a step

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[proper snow]

how would i solve this?

[scarlet gale]

!status

[obtuse pebble]

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

[proper snow]

1

[scarlet gale]

OK, so they're subsets of R^3.

But the question is whether they're subspaces.

So, they need to be a vector space for that.

[proper snow]

right

I checked for the zero vector and they all qualify

but whats next

[scarlet gale]

OK, so addition and scalar multiplication work.

But are those closed operations?

In other words, do any two elements in the subset sum to another element in the subset?

I assume that the scalar multiplication is closed, since you can just multiply the variables by the scalar.

[proper snow]

example?

[scarlet gale]

Well, one example with part I is [2, -1, 6] + [1, 5, -5].

Is that sum in the subset there?

[bronze mica]

I would... recheck that

[proper snow]

i get [3,4,1]

[scarlet gale]

OK, is that in that subset?

[proper snow]

im not sure how to check that

[scarlet gale]

You have three simultaneous linear equations.

a + b = 3, etc.

And Edward II is correct on rechecking it.

[proper snow]

i dont think ive got the fundamentals right

[scarlet gale]

Which part?

Checking for the zero vector?

[proper snow]

like here, why am i doing this?

[scarlet gale]

Oh, the subset definition has the first element in the vector as a + b.

You said the first element in the vector from the sum was 3.

So, for 3 to be in the set, it has to be equal to a + b.

So, like [3, 4, 1] = [a + b, 2a - 3c, b + 5c].

So, 3 = a + b, 4 = 2a - 3c, and 1 = b + 5c.

If you can't find an a, b, and c that work, it can't be in that subset.

All elements in the subset have to work that way.

Does that make sense?

[proper snow]

ok, but how do i actually solve?

say for 3 = a + b

[scarlet gale]

Well, you can do basic algebra simultaneous equations or you can do Gaussian elimination.

[proper snow]

and find RREF?

[scarlet gale]

Right.

[proper snow]

to find the a,b,c values and then plug them in the subset

to check

[scarlet gale]

Right.

[proper snow]

oh, and is that all i have to do?

[scarlet gale]

For that specific sum, yes.

You need to prove it for all sums, though.

[proper snow]

would i still need to check the zero vector?

[scarlet gale]

Well, part I has the zero vector if a = b = c = 0.

So, that checks out.

You need to recheck it on other parts.

[proper snow]

ok, for the first one im getting 2, 1, 0

[scarlet gale]

For a, b, and c?

[proper snow]

yea

[scarlet gale]

OK, so [a + b, 2a - 3c, b + 5c] = [3, 4, 1], so that works.

[proper snow]

it doesnt work for 2a-3c?

oh wait it does

[scarlet gale]

No, it gives four for that.

OK, so that one sum is in the subset.

[proper snow]

now, can i move to the next subset or do i have to check for scalar?

[scarlet gale]

Well, you have to prove it works for any sum of elements in the subset.

[proper snow]

how would i do that

[scarlet gale]

OK, so you leave things as variables.

[a1 + b1, 2a1 - 3c1, b1 + 5c1] + [a2 + b2, 2a2 - 3c2, b2 + 5c2].

So, the a in the first vector is a1.

And it uses the formula in the subset.

Does that notation make sense?

[proper snow]

yea but im not sure why were not using scalar multiples

[scarlet gale]

Oh, we're proving additive closure.

Which means any two vectors in the set added together produce a vector in the subset.

Vector spaces have it where if you add two vectors in the space together, you get another vector in the space.

So, subspaces are spaces, so they also have that property.

So, you need to show that if you add two vectors in the subset together, you get a vector in the subset.

[proper snow]

oh like u + v

[scarlet gale]

Right.

[proper snow]

now can i choose any two vectors

[scarlet gale]

No, you want to choose arbitrary vectors.

Like you want to say that any vector in the subset added to any vector in the subset produces another vector in the subset.

So, you can't just go like [2, -1, 6] + [1, 5, -5] like we did before.

That only proves it for those exact two vectors.

We need to prove it for all pairs of vectors in the subset.

The way we do it is to leave it in variable form.

Like, if we write [a1 + b1, 2a1 - 3c1, b1 + 5c1], all vectors in the subset are in that form.

Does that make sense?

[proper snow]

yep

ok so that checks (I) as a subspace of R^3

[scarlet gale]

No, this is just about additive closure.

Then, we need scalar multiplication closure.

Then, we need the zero vector.

[proper snow]

well that works

if a = b = c = 0

[scarlet gale]

Yeah, we got the zero vector.

So, now we're doing additive closure.

[a1 + b1, 2a1 - 3c1, b1 + 5c1] + [a2 + b2, 2a2 - 3c2, b2 + 5c2]

That formula works for any two vectors in the subset.

You can fill the variables in and get whatever vectors in the subset you want.

If we prove that's in the subset without filling in the variables, it has to work for all the vectors in the subset.

Does that make sense?

[proper snow]

yea but now i dont have to do anything further do i

[scarlet gale]

Yes, you need to add them.

See what you get.

[proper snow]

you get this

[scarlet gale]

No.

This one:

[a1 + b1, 2a1 - 3c1, b1 + 5c1] + [a2 + b2, 2a2 - 3c2, b2 + 5c2]

[proper snow]

how do i add a1 + a2

[scarlet gale]

You write a1 + a2.

It's like regular algebra. Two different variables added are just written like that.

[proper snow]

a1 + a2 + b1 + b2, 2a1 + 2a2 -3c1 - 3c2, b1 + b2 + 5c1 + 5c2

[scarlet gale]

Right.

Now, we want it to be in the form [a + b, 2a - 3c, b + 5c].

So, what we can do is:

[a1 + a2 + b1 + b2, 2a1 + 2a2 -3c1 - 3c2, b1 + b2 + 5c1 + 5c2]
[(a1 + a2) + (b1 + b2), 2(a1 + a2) - 3(c1 + c2), (b1 + b2) + 5(c1 + c2)]

Does that make sense?

[proper snow]

i think theres a typo?

but yea

[scarlet gale]

Where?

[proper snow]

oh nvm

i get it

[scarlet gale]

OK, so now we can say that a3 = a1 + a2, b3 = b1 + b2, c3 = c1 + c2.

We get:

[a1 + a2 + b1 + b2, 2a1 + 2a2 -3c1 - 3c2, b1 + b2 + 5c1 + 5c2]
[(a1 + a2) + (b1 + b2), 2(a1 + a2) - 3(c1 + c2), (b1 + b2) + 5(c1 + c2)]
[a3 + b3, 2a3 - 3c3, b3 + 5c3]

And that's in the subset.

[proper snow]

is there a shortcut for this? bc i dont think its supposed to take this long to solve a part of a MCQ question

[scarlet gale]

Oh, it doesn't take very long once you get the hang of it.

It's like four lines.

[proper snow]

i think (II) is not a subsapce?

[scarlet gale]

Like this:

[a1 + b1, 2a1 - 3c1, b1 + 5c1] + [a2 + b2, 2a2 - 3c2, b2 + 5c2]
[a1 + a2 + b1 + b2, 2a1 + 2a2 -3c1 - 3c2, b1 + b2 + 5c1 + 5c2]
[(a1 + a2) + (b1 + b2), 2(a1 + a2) - 3(c1 + c2), (b1 + b2) + 5(c1 + c2)]
[a3 + b3, 2a3 - 3c3, b3 + 5c3]

Why?

[proper snow]

cuz of the 3

zero vector

[scarlet gale]

OK, so there's no zero vector.

Now, on I, we have to prove scalar multiplicative closure.

You do the same sort of thing:

k[a + b, 2a - 3c, b + 5c]

Then, you do the operation.

[proper snow]

ok

I do have other questions that i have the solution to but want to check if its correct?

[scarlet gale]

It's correct that II isn't a subspace.

[proper snow]

no i mean like different questions

different topic questions

[scarlet gale]

What did you mean by it in you want to check if it's correct?

[proper snow]

i have the solutions for them, i need to verify if i have the correct solution

for e.g. this is 4

[scarlet gale]

So, you have a different question you did the work for and you also have the correct solution from somewhere else and you want to check if you did the work correctly?

[proper snow]

not somewhere else, from me

[scarlet gale]

Oh, OK.

Yes, 4 would work.

After RREF, it would zero out only one row.

You can check by filling k in, doing RREF, and counting the nonzero rows.

[proper snow]

5 and 2 for this

[scarlet gale]

That's the rank.

[proper snow]

yea

thats what i did

2 for both

[scarlet gale]

I haven't done much with null spaces, so I can't help with those.

For 26, you can write the columns as rows in a 4x4 matrix, then do RREF.

Or you can notice some linear dependence.

[proper snow]

nullity is bascially the linearly dependent vectors in a matrix

[scarlet gale]

Like u3 = 2u1.

[proper snow]

is the answer 2 for 26 tho

[scarlet gale]

Yes.

[proper snow]

this one is "a"

this one "c"

[scarlet gale]

Yes, those are correct.

If you just want to check your answer, you can do it with Wolfram Alpha. You write a matrix like {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}} and multiplication is *.

,w {{1, 2, -3}, {-1, 3, 1}, {4, 2, -5}} * {{-1, 2, 5, -3}, {4, -2, 1, 6}, {-2, 1, 3, 2}}

[warm shale]

Results provided by WolframAlpha

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

Requested by chai.t.rex

[scarlet gale]

You can also do RREF[matrix here] and such.

But only use that to check your answers.

[proper snow]

right

[proper snow]

for this i had a question: I did matrix A times 3rd col of matrix B but if i had done Matrix B times 3rd col of Matrix A that would be wrong, how would I know which way i have to do it?

[scarlet gale]

Don't trust its determination of the rank if you have variables, though.

[proper snow]

i get that i can verify by actually multiplying them but what if i dont have the time

[scarlet gale]

Well, you have the dimensions of the matrixes as R1xC1 * R2xC2 = R1xC2.

Does that make sense so far?

Like 1x2 * 2x4 = 1x4.

[proper snow]

yea

[scarlet gale]

OK, so the rows in the result correspond to the rows in the first matrix.

The columns in the result correspond to the columns in the second matrix.

So, you need to use the third column of the second matrix to get the third column of the result.

[proper snow]

ohh that makes sense

they have to match

or else i cant multiply

[scarlet gale]

Right.

Like if you had 1x2 * 2x4 and you wanted the third column of the result, you couldn't get it if you used the third column of the first matrix.

There is no third column there.

Oh, wait.

There.

[proper snow]

gotcha

oh and for this one i got "d"

[scarlet gale]

OK, so you can do a times the first basis vector, b times the second, and c times the third.

Then, you have like a + b + 2c = -7 and so forth.

[proper snow]

i just put them in an augmented matrix and found the values after finding the RREF

does that also work?

[scarlet gale]

Yes, that will work if you transposed the vectors into rows, I think.

[proper snow]

i didnt transpose

[scarlet gale]

Oh, wait.

,w RREF[{{1, 1, 2, -7}, {2, -1, 1, 4}, {1, 1, 1, -3}}]

[warm shale]

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Requested by chai.t.rex

[scarlet gale]

Yeah, I was wrong.

[proper snow]

ok so my method is valid?

[scarlet gale]

We can check.

[proper snow]

we just did

oh you. mean linear combination

[scarlet gale]

3[1, 2, 1] - 2[1, -1, 1] - 4[2, 1, 1]
[3, 6, 3] + [-2, 2, 2] + [-8, -4, -4]
[-7, 4, -3]

We just use the coordinates and see if we get the original vector.

[proper snow]

yea

ok so back to this, is it cool if i tell u what i got as an answer later on?

[scarlet gale]

Sure.

[proper snow]

DM ok?

[scarlet gale]

You can ping me if you want.

[proper snow]

that works

[scarlet gale]

If I'm not around, you can always do @Helpers.

[proper snow]

sure, thank you for the help

[scarlet gale]

You're welcome.

[proper snow]

.close

[obtuse pebble]

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[lone dome]

Identify the coefficient of x^2 in the expansion of (1+4x)^7*(2-x)^6

[lone dome]

Id like to know what the answer and figure solution out myself

[scarlet gale]

Do you know about binomial coefficients?

[lone dome]

Let mecheck

Nope

[scarlet gale]

Have you done factorials before?

[lone dome]

I dont know but ive done the ncr thing

I think im supposed to use it here

[scarlet gale]

OK, so nCr is a binomial coefficient.

[lone dome]

Ohh

Alright

[scarlet gale]

OK, let's pretend it was only (1 + 4x)^7.

To get the (x^2) coefficient, you'd do (7C2 ((1)^{7 - 2} (4x)^2)).

[warm shale]

Chai T. Rex

[scarlet gale]

There we go.

You have 1 + 4x.

So, you put 1 in parentheses. You put 4x in parentheses.

You take the (4x) to the second power because you want x^2.

And then you take the (1) to the 7 - 2 power.

Does that make sense so far?

[lone dome]

Yup

I gotta write that down

What else is next?

[scarlet gale]

,w Binomial[7, 2]

[warm shale]

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[lone dome]

And then 4^2 is 21

[warm shale]

Chai T. Rex

[scarlet gale]

Does it make sense how I got that?

[lone dome]

Ohh yes that formula

Yup

[scarlet gale]

OK, so that's how to do it when you have one binomial to a power.

You have two binomials to a power multiplied together.

So, how can you get 2 by adding numbers from 0 to 2?

You can get 0 + 2, 1 + 1, or 2 + 0, right?

[lone dome]

Okay

Yup

[scarlet gale]

OK, so if we get the x^0 coefficient of the first binomial and the x^2 coefficient of the second binomial, x^0 times x^2 = x^2.

[lone dome]

Alright

[scarlet gale]

Also, if we get the x^1 coefficient of the first and the x^1 coefficient of the second, we have x^1 times x^1 = x^2.

Also, if we get the x^2 coefficient of the first and the x^0 coefficient of the second, we have x^2.

Does that make sense?

[lone dome]

Yup

[scarlet gale]

OK, so we need the x^0, x^1, and x^2 coefficients of both, and then we multiply them together to get those three ways of getting x^2.

Then, we add those final coefficients together.

[7Ca((1)^{7 - a}(4x)^a) \cdot 6Cb((2)^{6 - b}(-x)^b)]

[warm shale]

Chai T. Rex

[scarlet gale]

We can do a = 0, b = 2. Then, a = 1, b = 1. Then, a = 2, b = 0.

[lone dome]

Alright

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

hello I am 1 mark off from an A. I would appreciate if anyone could look over why I lost 1 mark on this working for why the triangle has two possible triangles. Sorry for the messy working. The second picture I provided is a clearer idea of what I was trying to prove.

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[hollow summit]

hi i need help with this qn

[hollow summit]

<@&286206848099549185>

[wispy bramble]

well

have you drawn a venn diagram?

as the question suggests

[obtuse pebble]

@hollow summit Has your question been resolved?

[hollow summit]

nope

i dont understand how it works

[obtuse pebble]

@hollow summit Has your question been resolved?

[hollow summit]

<@&286206848099549185>

[thorny axle]

uhh

i can try to help

ima just simplify the question for you rn hold on

total schools: 120
(i) fire exit sign faults
(ii) fire extinguisher faults
(iii) fire alarm system faults
schools that have at least (i): 18
schools that have at least (ii): 24
schools that have at least (iii): 15
schools that have (ii) and (iii): 1
schools that have only (iii): 8
schools with any two faults: 5
schools with all three faults: 4

does that help organize the data more? @hollow summit

[hollow summit]

mm i think so

but how do i draw the venn diagram out

[thorny axle]

well you have three circles, one for each type

and they all intersect each other

outside that big cluster you have the rectangle, the total

here gimme a sec

[hollow summit]

ohh

so i just add in the numbers and minus off the intersection?

[thorny axle]

venn diagrams work like this

for odd numbers of intersecting areas, you add

for even numbers of intersecting areas, you subtract

in this case, you add all the schools that have AT LEAST 1 type, subtract the amount of schools that have EXACTLY two types, and add back the numbers of schools that have EXACTLY three types

[hollow summit]

ohh

[obtuse pebble]

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[lament atlas]

can someone tell me how we go form the first line to the second one

[low pollen]

Multiply the eqn with root2/root2

Something like this

Then apply the cos(a+b) identity

Btw 1/root2 = sin(π/4) =cos(π/4)

[lament atlas]

thank you

.close

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[paper kraken]

can someone check this please

[paper kraken]

i found 36

is it correct

[warm canopy]

,w 19^53 mod 503

[warm shale]

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Requested by icaird

[paper kraken]

omg

ok ill try agai

if it was mod 53

,w 19^53 mod 53

[warm shale]

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[olive monolith]

Hello

[olive monolith]

((X(x^2+1))/(x-1)

How do you solve this using long polynomial division?

Because it has multiple

[robust raven]

$\frac{x\left( x+1 \right)}{x-1}=\frac{x^{2}+x}{x-1}=\frac{x\left( x-1 \right)+2\left( x-1 \right)+2}{x-1}=\\=x+2+\frac{2}{x-1}$

[warm shale]

Joanna Angel

[robust raven]

x + 2 is a quotient, and 2 is a remainder

x - 1 is a divisor

[olive monolith]

No x^2

X(x^2)

[robust raven]

ah ok, you wrote so small

for me

wait then :

[olive monolith]

I’m sorry

[olive monolith]

How can I write like this?

[robust raven]

it is LaTex editor

[olive monolith]

Oh Okok

[robust raven]

$\frac{x\left( x^{2}+1 \right)}{x-1}=\frac{x^{3}+x}{x-1}=\frac{x^{2}\left( x-1 \right)+x\left( x-1 \right)+2\left( x-1 \right)+2}{x-1}=\\=x^{2}+x+2+\frac{2}{x-1}$

[warm shale]

Joanna Angel

[olive monolith]

Still don’t understand how did you get the number 2

Let me check again

Can you tell me how did you do it step by step?

[obtuse pebble]

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[warm trench]

O3) a part

[warm trench]

I did two methods and im getting different answers

[plain owl]

In the second method, you accidentally included 3 in the HCF calculation as 3 is not a common factor and in the LCM you forgot to include 3

Method one was performed correctly

[warm trench]

Ohh alright thanks

.close

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[high lily]

!15m

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[high lily]

everyone ping isn't quite helpers, but doing it is just as bad

nor should you even ping before actually asking a question

<@&268886789983436800> first post here advocating gpt

the question is a bit unclear, is that the original wording?

[hot hazel]

don't barge into help channels if you're not going to help.

[timid silo]

is this meant to be in relation to relative velocity

also you should clarify on what "80/100 hour and 40/100 hour" mean

[high lily]

it was corrected to

80/60
And 40/60
so i'm going to assume 80minutes, 40minutes respectively or equivalent

[timid silo]

the objective is still unclear to be fair. Post the original question, OP

[high lily]

but the part with

how much km/h would be
is very unclear

[obtuse pebble]

@timid silo Has your question been resolved?

[prisma smelt]

Can you take photo of the printed form of the question?

[obtuse pebble]

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[olive monolith]

How did the it turn into this?

[true summit]

comon power

so the r^2- x^2 went in the root

[olive monolith]

Can you illustrate?

[hexed gull]

X^a * Y^a = (X*Y)^a

where a=0.5 in this case

[olive monolith]

Still don't get it

[hexed gull]

you have (r²-x²)^0.5 * (1+x²/(r²-x²))^0.5

as in the integral

[olive monolith]

Yeah

[hexed gull]

k, then you can apply the power rule: X^a * Y^a = (X * Y)^a

where we have a = 0.5

X = r²-x²

and Y = 1+x²/(r²-x²)

[drifting roost]

,, \sqrt{1 + \frac{x^2}{r^2 - x^2}} = \sqrt{\frac{r^2 - x^2 + x^2}{r^2 - x^2}}

[warm shale]

kanna

[drifting roost]

,, \sqrt{r^2 - x^2} \cdot \sqrt{\frac{r^2 - x^2 + x^2}{r^2 - x^2}} = \sqrt{r^2 - x^2 + x^2}

[warm shale]

kanna

[olive monolith]

Oh okok

I'm feeling dumb

Thank you

[drifting roost]

[olive monolith]

.close

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[spiral bronze]

help pls

this is the calculation my teacher gave but i dont understand

anything

<@&286206848099549185>

[obtuse pebble]

@spiral bronze Has your question been resolved?

[obtuse pebble]

@spiral bronze Has your question been resolved?

[short wing]

This property is used here

This the the process

[obtuse pebble]

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[spring frost]

how

[cloud berry]

find normal that passes thru (5,0)

[spring frost]

What

I just graphed the graph and am not sure what to do after

[cloud berry]

it wont help

lets say a general point (h, f(h)) is closest to (5,0)

[spring frost]

ok

[cloud berry]

find dy/dx and put x=h in the result

dy/dx will give slope of tangent at that point

[spring frost]

so I find the derivative of the function and then put in x = h

but where does the x = h come from?

oh any random point will work?

[cloud berry]

we are assuming that point to be (h,f(h))

[spring frost]

ok, but what good does the slope do?

[cloud berry]

its slope of tangent so we now need to find slope of normal

its -1/m where m is the slope of tangent we just found

[spring frost]

ok

[cloud berry]

now for the equation of normal we get (we are using (y-y1)=m(x-x1) form of line)

y - f(h) = (-1/m)*(x-h)

satisfy equation with (5, 0)

f(h) = (5-h)/m

[spring frost]

ok

[cloud berry]

now its just bunch of creepy calculation

[spring frost]

ok, thanks for the help

greatly appreciated

.close

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[cloud berry]

here's the graph i made for your ease https://www.desmos.com/calculator/jfwsrkhpeu @spring frost

[obtuse pebble]

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[swift comet]

is the cartesian product of an empty set with itself i.E: A={}, A X A
(∅,∅)? Or is it just ∅?

[spice citrus]

It's just ∅

[final thunder]

If it was (\wrb{(\nul,\nul})) then you'd be asserting that (\wrb{\nul\in\nul}), which clearly isn't true.

[warm shale]

Pure

[swift comet]

Thanks! That helped me solve my Problem

.close

[obtuse pebble]

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[unique shuttle]

i dont understand the second line

of the explanation