help-0

[whole chasm]

.Open

5/2

[narrow hedge]

?

[neon kestrel]

omega= [-3 , 3] x [-3 , 3]
m(omega)= 36
Why?

[lone heart]

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[barren elbow]

Ik the definitions of injective and surjective with regards to functions but not affine maps. Pls could someone help me

[narrow hedge]

Maybe consider checking its kernel?

For injectivity

For surjectivity maybe check for the dimension of the image is dim(V)?

[barren elbow]

what is the image and what is V?

[narrow hedge]

Suppose the mapping is from V->W

Here V is R^4

To find dim(Im(V)) just do it’s RREF and find it’s pivots

[barren elbow]

what do you mean by the dim?

the rank?

[kind sparrow]

That particular mapping just happens to have a name and it's called an affine map. You can just replace it with the word "function" and do the problem as usual. That information is not important.

[narrow hedge]

Dimension

[neon kestrel]

@mighty cosmos @compact mirage

[barren elbow]

ok I found this on one of the slides

what is y in the question above

<@&286206848099549185>

[shut crest]

y is that notation trippin 😭

i guess the dimension of like

the vector space x is in

which doesnt make sense at all but idk maybe they r only over R^n

[barren elbow]

so dim y would be 3 here?

[austere locust]

Hey are we active right now, I figured out collatz conjecture and goldbachs and guess what it lead to? I solved L curve!

[barren elbow]

.close

[lone heart]

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[neon kestrel]

.reopen

[lone heart]

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[alpine sable]

for line l i made y = 2x + 6

[fleet crown]

When do 2 lines (curves) have common points?

[alpine sable]

uh

no clue

[fleet crown]

when their system has a solution

so for 2 curves to meet at 2 points

their system has to have 2 solutions

(real solution to simplify)

[alpine sable]

And how do I find that?

[fleet crown]

well, look at the system you have here
y = 2x + 6
y = 2x^2 + kx + 9

try to solve it

and see how k affects the number of solutions

[alpine sable]

make them equal to each other?

[alpine sable]

I'm genuinely confused

[fleet crown]

do this

and solve for x

[alpine sable]

how can we solve for x when k isn't found

i mean we can solve 1 side maybe

[fleet crown]

imagine k is a number

imagine you know it

[alpine sable]

2x^2 - 2x + kx + 3 = 0

and then

2x(x - 2) ...

[fleet crown]

no

don't do that

[alpine sable]

oops

[fleet crown]

2x^2 - 2x + kx + 3 = 0

you got this

good

[alpine sable]

yeah

[fleet crown]

now take x out of the middle two terms

2x^2 + (- 2 + k) x + 3 = 0

like this

this is a quadratic

solve it

[alpine sable]

2x^2 - 2x + kx + 3 = 0

2x^2 + (-2 + k)x + 3 = 0

multiplies to 6, adds to 1 (3 and 2)

2x^2 - 2x + 3x + 3

2x(x-1) - 3 (x - 1)

(2x-3)(x-1)
x = 3/2 and x = 1

[fleet crown]

ye

what

[alpine sable]

?

[fleet crown]

solve this for x now

use the quadratic formula

[alpine sable]

?

[fleet crown]

do you know how to solve quadratic equations?

[alpine sable]

yuh

[fleet crown]

then solve it

[alpine sable]

i did

x = 3/2 and x = 1

[fleet crown]

what?

how?

where did k go?

[alpine sable]

ah u meant

a = 2
b = -2 + k
c = 3

[fleet crown]

a = 2

not 2 x

[alpine sable]

ohhh

ye mb i clocked

[fleet crown]

so now plug hat into the quadratic formula

and simplify as much as you can

[alpine sable]

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

[ocean seal]

$a

[fleet crown]

ye that one

[alpine sable]

$\frac{{2+k \pm \sqrt{{k^2 - 4k - 8}}}}{4}$

[ocean seal]

$a

[alpine sable]

@fleet crown

[fleet crown]

looks about right

so, when is that gonna be 2 distinct numbers?

under what condition?

[alpine sable]

b^2-4ac > 0

[fleet crown]

good

so now solve k^2 - 4k - 8 > 0

[alpine sable]

u could've just told me to find the root thing

aight

2+2root(3) (+)
2-2root(3) (-)

[fleet crown]

if those are the solutions then ye

those are the k's you were looking for

[alpine sable]

that's so hectic

and so messy

lmao

@fleet crown did i do something wrong?

[fleet crown]

Sorry, don't have time to check numbers

recheck it

[alpine sable]

i see what i did wrong

i did - 8 and not - 20

4 * 2 * 3 = 24

lmao

Thank you, aki

.close

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[cloud drift]

hello, i'm working on finding the derivatives of composite functions for calculus, and i attached a picture of my professors work. i'm confused by the first step, in which the function is rewritten before the derivative is found. where does the 3 go? shouldnt it be 3(t^3-t^2)^-2

[karmic jetty]

Yes you’re correct, I think the prof made a mistake

[cloud drift]

mannnn ive been so worried i missed something

thank you very much for the assistance

much appreciated

.close

[lone heart]

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[unkempt willow]

We can represent the height of a current with h in meters at a given moment of the day.

Its equation is h = 5 sin (30(t-5)) + 7

[unkempt willow]

They want us to find at what moments you get your maximum and minimum height

Maximum is 12 minimum is 2

Whenever I isolate 2 I get 2 hours but it’s apparently 4

I am replacing h in my equation with 2

Is it sin -1 or just sin

[uncut furnace]

It's when sin is equal to -1

Ie, sin(30(t-5)) = -1

Which will happen at infinitely many points but given the context of time during the day, you should be able to narrow it down

[lone heart]

@unkempt willow Has your question been resolved?

[jaunty galleon]

Someone could help me?

[lone heart]

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[green robin]

Can somoeons show me a sequence an, such that lim n-> oo |an - a(n+1)| = 0, but the sequence doesn't converge?

[green robin]

I was thinking of something like sin(n) + (-1)^n / n, but not sure if that checks out

[trim yoke]

gm

[kind sparrow]

The easiest example is something like the nth harmonic number

[green robin]

does ln(x) have this property?

[median oar]

wdym there's no n in ln(x)

[green robin]

there is lol

ok fine ln(n)

you know what I'm saying don't be sassy

[median oar]

so you mean lim n-> inf |ln(n) - ln(n+1)|

[green robin]

yeah

exactly

I guess we'd have to know growth properties of ln or something, tho

[green robin]

How would we prove that lim n-> oo |an - a(n + 1)| = 0 in this example?

[median oar]

i think if you look at the taylor series of ln you will get that it goes to 0 but does not converge

[green robin]

Yeah that's what I was thinking maybe harmonic number are easier approch, I'll look

[median oar]

$\lim_{n\to\infty}\left|\frac{1}{n}-\frac{1}{n+1}\right|$

[ocean seal]

Frosst

[median oar]

simplify it

[green robin]

1 / (n * (n + 1))

[median oar]

that's 1/n^2 something

[green robin]

ok nice

[median oar]

as n gets big

this goes to 0

[green robin]

but the sequence 1/n converges right?

to 0

[kind sparrow]

That is not the correct definition but you should be able to do this without asking someone else

[green robin]

the harmonic number is defined has Hn = 1 + 1/2 + 1/3 + ... + 1/n right?

[median oar]

did i say something wrong?

[green robin]

I'm pretty sure that's not the def of harmonic

that's just the sequence 1/n

I guess in the harmonic sum you would get a tone of cancellation

pretty much just leaving 1/(n + 1)

and so it goes to 0

[kind sparrow]

H_n is more than just 1/n, that's all.

[green robin]

And how do we know it diverges?

[kind sparrow]

The limit is the harmonic series, you should have seen this before?

[green robin]

first time

[kind sparrow]

Well, there are a lot of proofs online, it is a useful object to know because it is a source of many counter examples

[green robin]

ok good to know I found a few proofs online but they all use term-re-arrangement which I was warned can lead to incorrect results

[green robin]

Would this be good? https://math.stackexchange.com/questions/172504/why-do-we-say-the-harmonic-series-is-divergent

[lone heart]

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[west timber]

"The function g(x) is continuous on the closed interval [e, pi] and differentiable on the open interval (e, pi), If x = 3 is where the slope of the tangent line is the same as the slope of the secant line from x = e to x = pi, and if g'(3) = 2, g(e) = -2pi, find g(pi)"

[west timber]

This problem is very dissimilar to the others on my calculus review packet and I don't really know where to start

[uncut torrent]

write an equation expressing the tangent line being equal to the secant line

solve from there?

[west timber]

could you help me get a start on that

I don't remember ever comparing tangent and secant lines

I know they are saying the slope of the secant line is 2 at x=3

[glad night]

do you know mean value theorem

[west timber]

yes

[uncut torrent]

do you need that

[glad night]

the concept

yes

[west timber]

wait

its literally just mvt

and it equals 2

[glad night]

yes

yes

[west timber]

becuase f prime of any number in between is what the 3 is serving

[glad night]

i think so

[west timber]

ok let me get the answer and see if that's somewhere on the packet

its a circuit

[glad night]

?

[west timber]

yews

I got -2e

and that's the connected answer on a different problem

got it

thank you

.close

[lone heart]

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[glad night]

👍

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[quick sphinx]

i need help pls

[lavish rapids]

Please state the problem first

[quick sphinx]

i did this far

but my z seems to be wrong

[lavish rapids]

Alrighty

I'll just read through it quickly

[quick sphinx]

thank uu

[lavish rapids]

For the first system, you forgot to add a negative sign to the 2

Should be -4x + 2y -6z = -2

[quick sphinx]

OHH

OOPS MB

THANK U THO

and after this do i jsut plug the z into another equation

[lavish rapids]

Lmk if that fixes everything

[lavish rapids]

Yeah, just do what you were doing before and it should work out

Do you mean like after you find the value of z?

Another thing, for the second system, you mistaked z for a 2 and completely got rid of z

Should be 2x + 2y -2z = 8

[quick sphinx]

OH

😭😭

hmm it still seems to be off

it says

-6z = 20

the answer is z = -4

[marsh reef]

Is there any vc

[quick sphinx]

what a vc

[tawny frost]

Voice call

[quick sphinx]

oh im in a library rn so

[tawny frost]

Btw hi I’m new

[quick sphinx]

idk if i can😭

[quick sphinx]

hii

[tawny frost]

I joined because I’m studying for the ASVAB and I would like some good recommendations especially for math

And also I want to get into it

[quick sphinx]

can somebody help me w this pls

im solving for z

[lavish rapids]

Lmao I just realized another problem

During the second system you mistaked a 7 for a 1 and did 2 instead of 14

Should've been 4x -2y +6z = 14

[bronze narwhal]

it should only go up to intro level trig i believe

[tawny frost]

Well I’ve been using the ASVAB for dummies

The app

[quick sphinx]

wit nvm

i realized

[lavish rapids]

Wait no nvm, was looking at the wrong thing 😭

[quick sphinx]

the mistake

[lavish rapids]

What is it?

[quick sphinx]

i mistaked the -1 for just a 1

[lavish rapids]

Where?

[quick sphinx]

in the second system

[lavish rapids]

I don't see it

[quick sphinx]

wait

hm

[quick sphinx]

wait what

oh

[lavish rapids]

Ignore that, I was looking at eqn 3 accidentally

[quick sphinx]

lemme resend what i have again

i made osme changes

[bronze narwhal]

why didnt you do subraction with 1 and 2

and then with 2 and 3

[quick sphinx]

equation 1 and 2?

[bronze narwhal]

yes

at the top

[quick sphinx]

i was truing to cancel out the y]

oh

[bronze narwhal]

you can cancel out x

[quick sphinx]

does it not work if i try the y

or do u mean

its easier w an e

[bronze narwhal]

i mean its simpler

[quick sphinx]

*x

true

[bronze narwhal]

because see

you dont need to multiply

wait no i mean 1 and 3

mb

[lavish rapids]

I think the answer given to you is just wrong

It SHOULD be -3.333 repeating

[bronze narwhal]

you could get the variable for y by doing elimination with 3 by 1

[quick sphinx]

for z?

i tried eliminating the x and i got z=2😭😭😭😭

mayb i should j drop out fam

[lavish rapids]

The second one you sent removed an equal sign in place of a negative dude 😭

[lavish rapids]

Yeah, should be -10/3

[lavish rapids]

In the second system

[lavish rapids]

The -20/6 we had earlier was right

I even plugged it into 2 different online calculators for intersecting planes and got the same answer so I’m pretty sure the answer given to you for z was wrong

[bronze narwhal]

yea z=-10/3

x=35/9

y=-29/9

i used substitution

[lone heart]

@quick sphinx Has your question been resolved?

[lone heart]

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[carmine rapids]

Is there anywhere to get math practice problems and answers?

I sorta forgot everything after a week

[lone heart]

@carmine rapids Has your question been resolved?

[tight pier]

Have you tried any of them?

[carmine rapids]

Not yet I got more to this but I also want practice question

[lone heart]

@carmine rapids Has your question been resolved?

[lone heart]

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[alpine sable]

with?

[jolly widget]

🤡

not the place to ask

[jolly widget]

🤡

quit trolling and close this channel

🤡

type .close

[pale zodiac]

just call the moderators lol

[alpine sable]

says the one joining a maths studying server with their friend tryna act funny n shit

💀

[pale zodiac]

cringe af

[jolly widget]

Ehh was thinking this was a trivial thing not worth pinging mods with

[alpine sable]

Drop the Insta then

[pale zodiac]

fair enough

[alpine sable]

how bout U pull up sum
Why hide behind ur SpongeBob lone wolf identity

tf does that question even mean

[jolly widget]

dun respond lol

[alpine sable]

how long did it take u to realize ur a disappointment

[jolly widget]

trolls be like that

[alpine sable]

Cuz u haven't yet apparently

I bet yo parents r proud

[alpine sable]

Nah i love this

Best form of entertainment

shi i realized that when I was a newborn

[jolly widget]

fair

[alpine sable]

but i ain no insecure bitch

Unlike yo lil baby

[pale zodiac]

yo can someone atleast close this channel? people wanna study lol

[lone heart]

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[narrow hedge]

Thx @torpid vault

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[opaque python]

Please help.

[alpine sable]

! status

[lone heart]

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

[alpine sable]

?

[alpine sable]

ok so

[visual gyro]

how do I factor this: a^2-2ab+b^2

[alpine sable]

use an unoccupied help channel

[opaque python]

get outa here

[alpine sable]

please

#help-16 is open

[visual gyro]

it says to post here

tf

[buoyant saddle]

use a different channel

[alpine sable]

IN A NON OCCUPIED CHANNEL

#help-16

[buoyant saddle]

chill

[alpine sable]

sorry

i'm sick and tired

[buoyant saddle]

it’s their first day lol

[opaque python]

...

[visual gyro]

mb guys

[alpine sable]

so

think

[alpine sable]

ok

[opaque python]

I know that they are supplementary

[alpine sable]

yes

ok

that is good

[opaque python]

❤️ is equal to 60 degrees

bruh

< 3

meaning that < 5

is equal to 60 degrees

because they are alternate interior angles

[alpine sable]

that image is not good

[opaque python]

and given that < 5 and < 3 are congruent

[alpine sable]

one sec

[opaque python]

the lines are parallel because of converse of the alternate interior angles theorem?

[alpine sable]

green and green, red and red, orange and orange, and blue and blue will only be the same if lines are parallel

if you can tell by that image

use that

[opaque python]

alr

[alpine sable]

and supplementary

[opaque python]

is it a theorem or a postulate?

[alpine sable]

theorem

[opaque python]

alr

wait what is it called when

angle 3

is congruent to 60 degrees

because of what

its something angles

linear angles?

oh nvm

vertical angles

[alpine sable]

this is wrong

or

not stated well

red and orange and blue and green are the same

[opaque python]

I answered it

alr so

we can see that

angle 2 and the angle that is 60 degrees is on the same line

meaning they are supplementary

making angle 2 120 degrees

and through alt ex ang theorem

it angle 2 equals to 120 degrees

then lines m and n are parallel?

Hello?

[rose void]

yea

[opaque python]

alr ty

how do I close this channel

[rose void]

.close :)

[opaque python]

.close

[lone heart]

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[lone heart]

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[gray monolith]

is fractal clip maker really recommended?

[gray monolith]

i really want a set zommer

zoomer*

<@&286206848099549185>

[ocean hawk]

ask in #computing-software

[gray monolith]

oh ok

sry

.close

[lone heart]

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[marble ibex]

Just wondering if my work is correct

[marble ibex]

Using Arc length formula

?

[pale glen]

yea

[spiral yacht]

!done

[lone heart]

If you are done with this channel, please mark your problem as solved by typing .close

[marble ibex]

Correct?

[lone heart]

@marble ibex Has your question been resolved?

[marble ibex]

??

[spiral yacht]

Which angle is 176

[marble ibex]

BA

[spiral yacht]

BA ?! It’s a line

[marble ibex]

Thats what my teacher called it

Wait it might be 32

[spiral yacht]

How?!

[marble ibex]

32 was the final answer

I thought it was 60 at first

but then I checked my notes and this problem was in there

and it says its 32

Which is why I said 32

What do you think?

Welp nvm ig

.close

[lone heart]

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[spiral yacht]

32 is not the answer I think

[marble ibex]

Yeah idk

[spiral yacht]

It’s not how it’s done

[marble ibex]

Its probably 60

[spiral yacht]

Umm

We can’t use that arc for angle C

reopen this channel

[marble ibex]

.open

.reopen

[lone heart]

✅

[marble ibex]

There

[spiral yacht]

Now, we can find angle AXC, AXB and we have to find AXO to find ACB

O is center

[marble ibex]

Im assuming we have to find AC ?

[amber elm]

AC is a line

And determining it is not necessary to solve the problem

[marble ibex]

I thought it was 60 at first because I did

1/2 (176 - 56)

[amber elm]

56 is not the measure of the arc XA

[marble ibex]

How do I find XA

?

[amber elm]

Since XBA is an intertior angle, the measure of arc XA is two times the angle of XBA

[marble ibex]

So it would be 112

[amber elm]

Yes

[marble ibex]

Oh so it is 32

[amber elm]

Yes

[spiral yacht]

No

[marble ibex]

1/2 (176 - 112) = 32

[amber elm]

It is 32 degrees

[marble ibex]

Is this correct?

I kinda guessed on it

[spiral yacht]

Why would you subtract arc AX and AB

[spiral yacht]

Opposite angles sum = 180

[marble ibex]

So its B

[amber elm]

Correct

[marble ibex]

(x + 16) + (6x -4) = 180

?

[amber elm]

1/2 ("Exterior" arc - "Interior" arc) is a formula for determining a secant angl;e

[amber elm]

Correct

[marble ibex]

.close

[lone heart]

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[spiral yacht]

Okay thanks

[lone heart]

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[gloomy talon]

not ssure how to do this

[lone heart]

@gloomy talon Has your question been resolved?

[lone heart]

@gloomy talon Has your question been resolved?

[gloomy talon]

<@&286206848099549185>

[coarse laurel]

honestly this is of a nice enough form that i think you can just straight up integrate them one var at a time

[gloomy talon]

oh rlly how would I do that

[amber elm]

Your bounds look a bit like this

[coarse laurel]

try using the sub u = x+3y and v = -x +5y

then u is between 0 and 6 and y between 1 and 8

the rest should be quite dooable from there i think

[coarse laurel]

to see this play with the equations governing the shape of the parlellogram (i cannot spell)

[coarse laurel]

the sub it motivated by noticing the numerator and two of the boundry conditions have the same form

same with the demonminator

[gloomy talon]

oh okay let me try that

[coarse laurel]

did it work out?

[lone heart]

@gloomy talon Has your question been resolved?

[lone heart]

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[prime seal]

$M=\left{(x, y) \in \mathbb{R}^2 \mid x^2+4y^2=4\right}$\
How can i show that this is compact space?

[ocean seal]

Slowaq

[mortal trellis]

show that it is bounded and closed

[prime seal]

well is it closed because its complement is x^2+4y^2<4<x^2+4y^2 and for every x,y from this complement i can find delta such that delta neighbourhood of x,y will be contained in complement of M thus complement of M is open thus M is closed?

and i can find this delta neighbourhood that is entirely contained in complement of M because there is < not ≤?

[mortal trellis]

is that a question or an argument

[prime seal]

xd more like argument and i am asking wheter its correct

[lone heart]

@prime seal Has your question been resolved?

[mortal trellis]

if you arent sure then fill in more details

[lone heart]

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[alpine sable]

hey guys I wanna get good at calculus

[alpine sable]

any resources/courses you can recommend

I know some basics but I dont have a strong grasp on it

[terse meteor]

Khan Academy

[alpine sable]

lovely! thanks!

.close

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[lone heart]

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[ocean seal]

alee

alee

[lone heart]

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[fallen rain]

Why my calculation for 2b is incorrect?

[hushed ether]

whr did ur sqrt go

[lone heart]

@fallen rain Has your question been resolved?

[fallen rain]

I thought my mistake

thank you

.close

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[undone falcon]

An angle was chosen and then x and y cords calculated with sin and cos.

I have their rounded values so that:

x = floor(x_real * 32) / 32
y = floor(y_real * 32) / 32

That means it has the precision of 1/32.

If i go back to the angle using this:

angle = arcsin(x/sqrt(x^2 + y^2))

To what precision is the angle rounded then? I need to know how many bits of data I can safely take while being guaranteed that they are just like the original angle.
Any help would be appreciated, thanks.

[tall shuttle]

Depends on which language is that

sqrt has a precision specification

division also takes precision

then arcsin

[lone heart]

@undone falcon Has your question been resolved?

[undone falcon]

i mean the programming language im using has a much higher precision than the rounding used above so it should be irrelevant

x = floor(x_real * 32) / 32

here it's rounded to 5 bits, the programming language can calculate in over 32 bits precisely, so it has no effect.

[tall shuttle]

Yeah, but you lost a lot of information

[undone falcon]

I don't get what you mean sorry. The data I received is rounded like this:

x = floor(x_real * 32) / 32

This means the x and y values are precise to 1/32.
I want to know how precise the angle calculated using arcsin would be, knowing these are rounded.

Sure, the arcsin isn't precise, but it calculates with a precision much greater than 1/32, so it doesn't round any useful information.

Maybe I'm wrong I just don't understand your point

[tall shuttle]

I get it now

You have $|x_r - x| < e$ where $e$ is the error

[ocean seal]

tales

[tall shuttle]

You have $f(x) - f(x+e) \approx e\cdot f'(x)$

[ocean seal]

tales

[undone falcon]

im a bit confused

[tall shuttle]

You want aproximate the error you get on arcsin

based the error on the input

[tall shuttle]

use this

[undone falcon]

yes

[undone falcon]

I want to approximate the maximum error I can get yes

So i can know what part of the number is 100% correct

[tall shuttle]

Use this + Lagrange error bound, or read the language especification

[undone falcon]

Alright thank you. Sorry for the stupid question, but how would that work with 2 inputs? x and y? Since the e is the same for both, can I just replace y with x?

How would that work with the derivative though

[lone heart]

@undone falcon Has your question been resolved?

[lone heart]

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[scarlet fox]

is that a 5 on the root 8
or 4

[gray isle]

hint:
$$2 = \sqrt[4]{16}$$

[ocean seal]

ℝαμΩℕωⅤ

[prime garnet]

4 on everything

this is where i got stuck

[gray isle]

consider geometric series

[scarlet fox]

step 1 is use his hint

aka sub the 2 for ...

4sqrt16

then you get a geo serie of 2 --> 4 --> 8 --> 16 and i have no idea how that relates
unless youve been taught smth related to it recently
im actually curious how to do this

[prime garnet]

part of my college prep lol, i did this problem with my private math teacher before but there was a lot of scribbling and i cant figure out the handwriting on it

[lone heart]

@prime garnet Has your question been resolved?

[lone heart]

@prime garnet Has your question been resolved?

[fallen oar]

=> ⁴√2+⁴√4+⁴√8+1 = u

=> ⁴√2+⁴√4+⁴√8+2 = u+1

u+1 = ⁴√2+⁴√4+⁴√8+2 = ⁴√2(1+⁴√2+⁴√4+⁴√8) = (⁴√2)u

=> ⁴√2u = u+1
=> (⁴√2-1)u = 1

u = 1/(⁴√2-1)

u+1 = ⁴√2/(⁴√2-1)

1/(u+1) = (⁴√2-1)/⁴√2 = 1-1/⁴√2
@prime garnet

here is the solution step by step

if you still can't figure it out, you can look at the solution

[prime garnet]

thank you so much

[fallen oar]

you are welcome

[lone heart]

@prime garnet Has your question been resolved?

[lone heart]

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[rotund bough]

i need help w this

[rotund bough]

can someone help please

[merry iris]

do you know about y = mx + b

[rotund bough]

yea

i just need the answer ive been stuck on it for days

@merry iris

[buoyant saddle]

!noans

[lone heart]

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.

[merry iris]

what is the slope?

[buoyant saddle]

find the slope

then use point slope form

y-y1=m(x-x1)

[rotund bough]

im confused still

[buoyant saddle]

find two points on the line

to find the slope

[rotund bough]

yea

i trieed

that

and still

[buoyant saddle]

what did you get

[rotund bough]

i got it wrong

[undone falcon]

Ok. When x=0, at what y is the line?

[rotund bough]

-2

no

fuclk

thats x

man i dont understand this

[undone falcon]

even there, it would be -3 not -2

look well at the line. When x is 0 (the horizontal line, when you look in the middle, that's where it's 0). Where is the line on the y axis (how high is it)?

[rotund bough]

2

it lines up

perfectly

[undone falcon]

I don't understand where you see it pass by 2

when x = 0, you're in the middle, so you can look at the middle vertical black line. It crosses the blue line at -3, not at 2

[rotund bough]

OH

yea

thats the

y intercept?

[undone falcon]

ok, now we know when x=0, then y is -3.

Let's look at the equation of the function:
y = mx + c

If x is 0, then we have
y = m * 0 + c
which just is:
y = c

Since y is -3 when x is 0, then we know that c, which is equal to y when x is 0, is also -3.
Now we just need to figure out the m. To do that, we need to look at another point.

Let's try to see where the blue line crosses the x axis (the horizontal line), let's use that point.

at what x value does that happen?

[rotund bough]

isnt is y=mx+c

[undone falcon]

you can call the letter whatever you want, we can use c if you prefer c

up to you

[rotund bough]

yea

[undone falcon]

ok we'll use c

[rotund bough]

its more usual for me tanks

thanks

[rotund bough]

um idk

man

im acc dumb

im only good

in history

and english

[undone falcon]

everyone has their strengths and flaws don't worry

[undone falcon]

it just means, where does the blue line cross the horizontal black line.

[rotund bough]

-2 or -3?

so thats the y intercept

which means

[undone falcon]

-3

[rotund bough]

-3x -3?

idk

yea

i knew it!

[undone falcon]

to get m it's gonna be a bit harder

[rotund bough]

oh no

wait can i just ask

is it

-2x +3?

[undone falcon]

nope

[rotund bough]

-3?

-2x -3?

[undone falcon]

no

[rotund bough]

ugh

?

can you please give me the solution with the answer cause this is a WHOLE HOUR worth of homework

who thinks of doing this man

[undone falcon]

to get m, we can either do it mathematically like this:

between x=-3 and x=0 we did +3.
between y=0 and y=-3) we did -3
We do -3/3 and we get -1. That's m.
If you want an explanation for why that's the case, I can explain if needed.

Or we can do it the same way, but a bit quicker.
We need to understand in what direction the line goes when we go to the right. You can see that if you go 1 box to the right, you go 1 box down. Then you apply the same concept above and do 1/-1 and get -1.

that means the solution is -1x - 3

[rotund bough]

so y=-1x - 3

.close ..

[lone heart]

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[lone heart]

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[tulip crown]

if a 5 number digit is aa7aa and is dividable with 6 what would the letter a be? are there multiple solutions?

[tulip crown]

<@&286206848099549185>

[alpine sable]

wait 15m

god dammit

[tulip crown]

hahaha

didnot know sorry

[alpine sable]

" After 15 minutes, feel free to ping <@&286206848099549185>."

[tulip crown]

first time 😆

[alpine sable]

fuck

[alpine sable]

so

can

you tell me if my rewrite of the question if it is right?

[tulip crown]

okay sounds good

[alpine sable]

for a 5 digit number aa7aa divisible by 6, what could a be?

[versed ginkgo]

hm, I think from the problem info, we can know some things about this number

[tulip crown]

yes

[versed ginkgo]

so in order for a number to be divisible by 6, what conditions should it meet?

remember for a number to be divisible by 6 it needs to be divisible by 2 and 3

[tulip crown]

ooo thats right

[versed ginkgo]

so we know it needs to be:

n1 even

n2 divisible by 3

do you remember how can we check for divisibility by 3?

[tulip crown]

hmmm

[alpine sable]

so

[versed ginkgo]

if you add its digits 1 by 1

the answer needs to be divisible by 3

[tulip crown]

huh?

[alpine sable]

yes

so

[tulip crown]

what dighits? a?

[alpine sable]

if you add all the digits of a number

[versed ginkgo]

for a number to be divisible by 3, the sum of its individual digits need to be divisible by 3

[tulip crown]

aaa okay

[alpine sable]

yes

and

a will be even

[versed ginkgo]

we can start from the first clue

a is even

[alpine sable]

yes

[versed ginkgo]

it can be 0, 2, 4, 6, 8

and from the second condition, we need to add up the digits

aa7aa right?

[exotic belfry]

if 0 i wouldnt say its a 5-digit number

[versed ginkgo]

so a+a+7+a+a must be divisible by 3

[alpine sable]

yea

[versed ginkgo]

thus

[alpine sable]

2, 4, 6, 8

[tulip crown]

why is a even? we have a 7 right?

[alpine sable]

the last number

must be even

[versed ginkgo]

if 4a+7 = multiple of 3

[alpine sable]

for a number to be even

[tulip crown]

aaaa okay

[versed ginkgo]

we can make it equal to 3 times some intiger n

4a+7 = 3n right

and just try and check

[alpine sable]

4a+7 mod 3 = 0

2 is 15

[versed ginkgo]

wait what did I write

[tulip crown]

but 3n isnt dividable with 6...?

[versed ginkgo]

no

srry

[versed ginkgo]

this yeha

u can write it like this, or just try plugging in values

so let a be 2

2+2+7+2+2

[tulip crown]

aaa okay

[versed ginkgo]

15

so 15 is divisible by 3

[tulip crown]

but wait

[alpine sable]

4 is 23

[tulip crown]

15 / 6 isnt dividable?

[versed ginkgo]

so what u can do is as I did

[alpine sable]

6 is 31

[alpine sable]

it's 3

[tulip crown]

no 2.5...

[alpine sable]

the property is with 3, not 6

[versed ginkgo]

4a+7 = 3n, try plugging values for a, and solve for n, if n is an intiger, then the value of a is correct

that is how I would solve it

[tulip crown]

aaa okay

now i see

[versed ginkgo]

so if u plug a=2, u end up with n=5

[alpine sable]

8 is 39

fo

[tulip crown]

aaa yes

[versed ginkgo]

5 is an intiger so 2 is one solution

[alpine sable]

so

2 and 8

[versed ginkgo]

yes

now, what about 4 and 6

well we try

4*6 + 7 = 3n

31/3 = n

it is not an intiger so it doesnt work

[tulip crown]

aaa

now i see

[versed ginkgo]

with 4 it would be

4*4 + 7 = 3n

23/3 = n

n is not an intiger

so the only solutions would be 2 and 8

[tulip crown]

okay

[alpine sable]

and actually, 14 also works

[tulip crown]

now i get it

tysm

[versed ginkgo]

it has to be a 1 digit number

I think

[alpine sable]

and 20

[alpine sable]

kinda

[versed ginkgo]

because is aa7aa

[alpine sable]

yea but

[tulip crown]

yeah

[versed ginkgo]

a is a 1 digit number

[tulip crown]

it would work, kinda

but not really following the rule of the question

[alpine sable]

,calc 141471414/6

[ocean seal]

Result:

2.3578569e+7

[alpine sable]

that's not helpful

[tulip crown]

hahaha

[versed ginkgo]

hm

[alpine sable]

it's 23578569

[versed ginkgo]

it is not exact

[alpine sable]

,calc 23578569*6

[ocean seal]

Result:

1.41471414e+8

[alpine sable]

see

[versed ginkgo]

so idk, if the problem considers a to be a 1 digit number

it works

just for 2 and 8

[tulip crown]

yes

tysm

.close

[lone heart]

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[lean rune]

!help

[lone heart]

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

[lone heart]

Please don't occupy multiple help channels.

@lean rune Has your question been resolved?

[lone heart]

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[quartz lichen]

Can somebody review my answers and let me know if they are incorrect? Thank you!

[valid token]

it looks right

👍

[quartz lichen]

.close

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[eager iron]

Hi, a quick question on Matrix algebra

[eager iron]

If I have B^-1 - XA = B^(-1)XA

Sorry

Is all of this ”legal”?

So I get the equation in the top left to begin with

[ocean hawk]

is A known to be invertible?

[eager iron]

And the Matrix A and B

the det isn’t 0

Same for B

So it should be invertible then right?

[junior vigil]

yes

[ocean hawk]

you have particular matrices?

[eager iron]

Yep

Top right of the photo

Solve for X

[ocean hawk]

then yes

[eager iron]

So What im wondering is when I factor out x from B^-1X + X

Can I choose whether I want to put X on the right or left side?

[ocean hawk]

it'd have to be on the right

[eager iron]

Aka can I choose whether to have (B^-1 + I)X Or X(B^-1 + I)

Oh it has to be on the right??

Is that always the case or is there a particular reason ?

[ocean hawk]

expand X(B^{-1} + I), what is it

[eager iron]

Oh

I see

XB^-1 + X

[ocean hawk]

yeah

[eager iron]

Ok so then I get

[ocean hawk]

you have to make sure B^{-1} + I is invertible, too

[eager iron]

And i know What happens when you have like AB and invert that But What happens in this case? Does it become I + B?

Uhh that sounds rough though, is there a quick way to check?

[ocean hawk]

no. (B^{-1} + I) is a single matrix. I don't know if there's a nice formula for inverting it

[eager iron]

Hmm i see

So this is as far as I can get with the algebra and then it’s just straight to counting?

And probably start with B^-1 + I to make sure it’s invertible

[ocean hawk]

yeah I think so

[eager iron]

Algebra isn’t the best word.. simplifying?

Is that it? :p

[ocean hawk]

either, I guess

[eager iron]

alright, will count on then, thank you!

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[uncut portal]

i need help with this and i need to know if i got anything wrong please

😁

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[uncut portal]

<@&286206848099549185>

.close

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[fallow solstice]

I need checking for this one

[fallow solstice]

amplitude: 1
midline: -2
period: pi
frequency: 2

general form:
y=sin(2x)-2

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[molten pivot]

Let a1, a2, ... a_n be an infinite sequence of positive integers such that each appears exactly once. Define S(n) = a1 + a2 + ... + an. Prove that for every positive integer k, there exists some n such that k divides S(n)

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[junior vigil]

!status

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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

[junior vigil]

!showwork

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@remote mulch Has your question been resolved?

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[patent moat]

I am in 8th grade and I am in Algebra 1, learning factoring

I have this worksheet and I was wondering if someone could help me better understand what I’m supposed to do

[molten pivot]

slow down

choose one problem

you can ask more later

[patent moat]

Question 12

[molten pivot]

$$x^2a + x^2b - 16a - 16b$$

[ocean seal]

Mr. Gamer 🇵🇸

[molten pivot]

right?

[patent moat]

Yes

[molten pivot]

let's look at the first two terms

$$x^2a + x^2b = ?$$

[ocean seal]

Mr. Gamer 🇵🇸

[molten pivot]

it's ok to say you don't know btw

[patent moat]

X ( x • a) (x • b)

[molten pivot]

no.

let's go over something called the distributive property of multiplication.

what is 6(2+3)

[patent moat]

30?

[molten pivot]

yes.

what's 6x2 + 6x3

[patent moat]

6x5

Wait that’s wrong

36x5

[molten pivot]

no you were right...

6x2 + 6x3 = 12 + 18 = 30

[patent moat]

Does the x represent multiplication or a variable in this scenario

[molten pivot]

it's multiplication here

mb

6*2 + 6*3 = 12 + 18 = 30

so the concept here is this:

6(2+3) = 6*2 + 6*3

does that make sense?

[patent moat]

Yes

[molten pivot]

ok. now let's look at our example again

can you say this out loud?

how would you read this in your head

[patent moat]

X to the power of 2 times a plus x to the power of 2 times b

[molten pivot]

great.

what do both terms have in common

[patent moat]

X^2

[molten pivot]

so we can factor it out.

[patent moat]

X^2 ( a + b)

[molten pivot]

yes

now let's look at the other two terms

$$-16a - 16b$$

[ocean seal]

Mr. Gamer 🇵🇸

[molten pivot]

read it out loud again

[patent moat]

Negative sixteen a minus sixteen b

Both terms share ‘-16’
So
-16 (a + b)

[molten pivot]

great

now we have

$$x^2 (a+b) - 16 (a+b)$$