help-26

[merry hill]

do you know this formula? $\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}} = \pm\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$

[thorny flame]

Astar777

[merry hill]

($a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ are the two lines)

[thorny flame]

Astar777

[junior linden]

No

Idk this formula

[merry hill]

idk how else to do it

you could find the formula i mentioned above by finding equation of the locus of a point which is equidistant from both lines

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[wary escarp]

I’m stuck with problem 60 which asks to factor to a different form. I thought I should just make the b-a into a-b to factor it but I’ve realized that it doesn’t really help me or maybe I’m just missing something

[cunning kayak]

hint: (b-a)^n where n is odd power = -(a-b)^n

[night imp]

Wdym factor to a different form? And your approach is correct

[cunning kayak]

oh yes approach is correct

[wary escarp]

The question asks to factor each polynomial completely

[night imp]

Follow your approach and what Vansh mentioned above

[wary escarp]

So then I get (a-b)^5 4(a-b)^3 right

[night imp]

Yeah

[wary escarp]

I forgot to put in there mb

[night imp]

Yep

Proceed to factor now

[wary escarp]

Alright I have it factored and checked my answer but I’m just curious on why you can’t factor (a-b)^3

[night imp]

You can

What did you factor out?

[wary escarp]

I took out the (a-b)^3

[night imp]

Yeah

[wary escarp]

As a gcf

[night imp]

Correct

[wary escarp]

And then factored the (a-b)^2 -4

[night imp]

What did you want to take out?

[night imp]

Right

[wary escarp]

And got (a-b-2)(a-b+2)

[wary escarp]

And then that’s multiplied with this

[night imp]

Correct

[wary escarp]

But I don’t see why the answer says not to factor it

[night imp]

What is the correct answer as per your textbook or from wherever you got the question

[wary escarp]

(a-b)^3(a-b-2)(a-b+2)

[night imp]

Yeah you want to factor the (x-3)³ also?

I mean (a-b)³

[wary escarp]

Yeah I was wondering why that’s not an option

[night imp]

Because (a-b)³ can be written as (a-b)(a-b)(a-b)

All 3 are exactly same

if = 0 was there , then for all 3 case it would have been a=b

[wary escarp]

But what about to a polynomial?

[night imp]

[wary escarp]

Nvm stupid question

Thought it was a^3 - b^3 like an idiot 😭

[night imp]

Ohhh

No worries

Yeah if it was a³-b³

Then further factorization can be done

[wary escarp]

Yeah

Thanks for helping

[night imp]

No problem

[wary escarp]

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[sharp spindle]

can I please get help with this

[sharp spindle]

lets say P = 1,2 then S = 3,3. Q = -1,2 then R is such that

u[x,y] = -1

[topaz sinew]

@sharp spindle Has your question been resolved?

[dusty epoch]

is this linear algebra

[sharp spindle]

ye

[dusty epoch]

u in college?

[sharp spindle]

ye

[dusty epoch]

taking a linear algebra too right now in uni

no clue what that questions saying though

didnt you just start your fall semester

[sharp spindle]

its a mathematical physics course

@golden blade so sorry for the ping but can you help me if free?

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@sharp spindle Has your question been resolved?

[sharp spindle]

.close

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[dense osprey]

How do I do this question using a ti84 calculator. I can do this easily with system of equations method but my teacher is asking to solve it doing row operations. All of us find it really hard and I'm trying to figure out a way doing it with a ti84 calculator

[smoky sparrow]

https://www.youtube.com/watch?v=9PNCjHemIhI

no calculator will do the row operations for you

I would recommend you watch this video

understand what the different types of row operations are

and most importantly, the strategy
so you want to divide the first row by 2 so that row 1, column 1 becomes a 1

and then do row 2 - row 1 => row 2 for example

so that row 2, column 1 becomes 0

[dense osprey]

Ok

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[twin pollen]

hi

[twin pollen]

rlly am stuck on this. I want to be able to anaylze all the options but the fractions through me off

[agile harness]

should be pretty clear just look at the zero

what do you notice about the zero

look to the left of the zero in both f and g

[twin pollen]

0 is in same spot

@agile harness

[agile harness]

yea but what happens to it in g

see how it’s much sharper

notice how g is never negative

[sweet shard]

process of elimination since 1/0 isn't defined

[agile harness]

yea you could always just do that

[twin pollen]

but what does it mean in general when f(x) is a part of a fraction

[agile harness]

they’re just taking the reciprocal

consider f(x) = x+1

[sweet shard]

,calc 1/3

[thorny flame]

Result:

0.33333333333333

[sweet shard]

f(0) is about 3, so 1/f(0) = 1/3

[agile harness]

$\frac{1}{f(x)} = \frac{1}{x+1}$

[thorny flame]

knief

[twin pollen]

so it would be a smaller value

if 1 ontop

and if 1 on bottom than f(x) would be same

?

is that right

[agile harness]

f(x)/1 = f(x) sure

but what if f(x)<1

would 1/f(x) be smaller or larger

[twin pollen]

smaller i think

[agile harness]

really

[twin pollen]

but that might be wrong

well 1 over x is smaller than x

[agile harness]

$\frac{1}{\frac{1}{2}} = 2$

[thorny flame]

knief

[twin pollen]

oo thas what u meant

o shoot

ok ya u right

so if f(x) is x squared than how would parabola look if it is 1/f(x)

@agile harness

or anyone

[agile harness]

$\frac{1}{x^2}$

[thorny flame]

knief

[agile harness]

,w graph 1/x^2

[thorny flame]

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[twin pollen]

why does it look like that

[sweet shard]

plug in some numbers and plot it yourself

like x=0.1, 0.01, -0.1, -0.01

[agile harness]

1/x^2 is a completely different function than x^2

[twin pollen]

is learning how to graph fraction functions like a whole new unit to learn

[agile harness]

sure

[twin pollen]

it is?

[twin pollen]

?

@agile harness

[agile harness]

bro please stop pinging me

[agile harness]

yes

go to khan academy

precalculus

[twin pollen]

ok no i have never learned that

thas y

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[neon iron]

Is this not just 12?

4c2 x 2?

[topaz sinew]

@neon iron Has your question been resolved?

[mint crescent]

No - you need to consider cases based on how the corners are colored

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[warm bough]

I entirely forgot how to do this problem. Am i supposed to add like terms from both sides?

[acoustic pecan]

theres only one side, but yeah, if the goal is to simplify then collect like terms

[warm bough]

okay

[lunar plume]

blud this is algebra 1 🙏

in 2024 🙏

[warm bough]

i failed 9th grade math im sorry 😭

[lunar plume]

ur chilling im js messin

[warm bough]

wait

so

[lunar plume]

show the whole equation lil bro

u cut off something

[warm bough]

nah

[lunar plume]

nah?

[warm bough]

its just "Find the sum of the given expression"

[lunar plume]

olh

[lunar plume]

theres only 1 side theres no both sides

[warm bough]

wait so 7x - -3x would be 10x right?

[lunar plume]

its just addition and subtrraction

[lunar plume]

why did u get --3x

how*

[warm bough]

wait

no

7x + -3x

right?

[lunar plume]

and what is that

the answer to that

[warm bough]

would be 4x

but

[lunar plume]

ez

but what

[warm bough]

one sec

[lunar plume]

?

[warm bough]

would 1x squared + 4x be the answer to the problem?

since 2-2 cancells each other out

[lunar plume]

yes

-x^2 tho

[warm bough]

like this yes?

[lunar plume]

yes

[warm bough]

got it

thank you

[lunar plume]

i learned this when i was in the womb

(joke)

[warm bough]

and this would just be -3x^2 times whatever comes out of the brackets right?

[bold gust]

Yea didn’t see u already got help

Mb

[warm bough]

still havent figured this out

[bold gust]

Distribute the -3x^2

You can’t simplify what’s inside the bracket

[warm bough]

how do i distribute?

[bold gust]

You have to do -3x^2 times every term

You know how to do that?

I can explain deeper

[warm bough]

yes please

i dont know how to do that

[bold gust]

(-3x^2 multiply 2x^2)+(-3x^2 multiply -5x) + (-3x^2 multiply -3)

It looks complicated without on paper

I can write it down in a bit

[bold gust]

But your multiplying -3x^2 to every term

do you understand that so far

[bold gust]

Yes

[warm bough]

the middle x is the times symbol

okay so

[bold gust]

you remember how to do that

[warm bough]

a negative times a positive is a negative correct?

so this would be -6x^2?

[bold gust]

Yes

uhh not quite

You add the exponents

[warm bough]

the square?

[bold gust]

if it were add then yes but it’s multiply

When you multiply 2 exponents you add them

yes

the small number to the right

Is an exponent

[warm bough]

squared or cubed is an exponent?

[bold gust]

yes

[warm bough]

okay

[bold gust]

You add the exponents

when you multiply

[warm bough]

so this would be 4?

or?

[bold gust]

Yes

Exponent to the 4th

-6x^4

[warm bough]

okay

[bold gust]

yes

[warm bough]

this is a lot easier now

thank you

[bold gust]

then do that for others

you multiply by every term

Not just one

For example you did 3(x+4)

You distribute 3 to both x and 4

If u learned that and remember

[warm bough]

a negative times a negative is a positive right?

or is it still a negative

i mix that up with addition a lot

i think it stays a negative

so -3x^2 x -5x would be -8x^2

[bold gust]

Yes

negative times negative is a positive

[warm bough]

oh ok

so 8x^2

[bold gust]

If it’s 2 different signs then it’s always a negative

[warm bough]

okay

[bold gust]

and 2 same songs are always positive

[warm bough]

and if its the same sign its a positive

okay

[bold gust]

Ye

signs not songs mb

[warm bough]

and -3x^2 x -3 would just be -9x2

[bold gust]

and no you multiply

-3*-5

So it’s 15 and not 8

[warm bough]

right

[bold gust]

Then add the exponents

[bold gust]

Correct

[bold gust]

this one’s right

[warm bough]

so all of this is correct

[bold gust]

Correct ur missing an x tho

For the second one

[warm bough]

fixed

[bold gust]

Uhh last ones a positive

since its 2 negatives

[warm bough]

right

[bold gust]

and uhm

you still add the exponents

For the second one

x^2*x is what

[warm bough]

but theres only 1 exponent

[bold gust]

not x^2 because it’s x^2+x^1

X is basically x^1

Same thing

U gotta add that too

[warm bough]

oh so x^2*x is x^3?

[bold gust]

yes

Correct

[warm bough]

okay

[bold gust]

Everything else should be good

[bold gust]

Show me when ur done

ok

Loading on my aide

Side

[warm bough]

so in the end, the answer to the original problem would be -6x^4+15x^3+9x^3

[bold gust]

uhh 3rd one is just -3x^2 multiplied by -3

You probably got the mulitlly mixed up with x

[warm bough]

okay

so a simple -3 doesnt have an exponent

[bold gust]

yes

[warm bough]

only Xs do

[bold gust]

yes

[warm bough]

k

[bold gust]

Well it does but its 3^1

And your only adding the exponent of x

Not numbers

[warm bough]

okay

sorry for my writing, im doing it with a mouse

but this should be correct, right?

[bold gust]

Your missing an x

For the first one

[warm bough]

by the 6

sorry

[bold gust]

Tes

Yes

[warm bough]

but that should be it

with the x

[bold gust]

Yes

[warm bough]

thanks

and i have one more problem if u dont mind helping

what i especially failed in (because i missed it) is slope intercept

i never got graphing, never will

[bold gust]

You remember what the formula of slope is

[warm bough]

no

[bold gust]

y2-y1/x2-x1

Tell me what both y’s are

[warm bough]

2 and 0

[bold gust]

Good

What are your x’s

[warm bough]

0 and 4

[bold gust]

Plug it in

What do you get

[warm bough]

2-0/0-4?

[bold gust]

2-0 and 0-4

Gotta do it in order

Can’t do one first and the other one second

[warm bough]

okay

[bold gust]

also meant to be a division sign

y2-y1/x2-x1

plug that in and solve

[warm bough]

so it would be -2/4

which is 2

so the y intercept is 2

[bold gust]

uhm yes

but your slope is -2/4

[warm bough]

right

so -2

[bold gust]

-2/4

Is -1/2

So your slope is -1/2x

[warm bough]

and what about the y=mx+b?

[bold gust]

you know your slope is -1/2x

So plug it in

Y=-1/2x+b

B is your y interspe

U know it?

[warm bough]

its 2

[bold gust]

ye

so ur done

Plug that in

[warm bough]

y=-1/2+2

right?

[bold gust]

You keep the c

X

Otherwise it’s right

So -1/2x

[warm bough]

like this?

[bold gust]

Yes

[warm bough]

okay

thank you very much

im done

thank you

.close

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[lone stone]

how to find N(e)?

[lunar plume]

its simple

[lone stone]

ik

can u give a little bit more information?

[lunar plume]

no

[lunar plume]

what i can tell you though

is that there is an answer to it

you just have to solve the problem

[lone stone]

no

[sudden temple]

what's N(e)?

[lone stone]

abs(log(a,n)/n - 0) < e
and u must find N(e)
for any n > N(e) u can say first statement

0 is the limit of log(a,n)/n

[lone stone]

In mathematics
𝑁 (𝑒)
N(e) is usually used to denote the smallest natural number
𝑛
n, under which a given condition or inequality is satisfied. In this context, if we have an inequality of the form:

abs(log(a,n)/n) < e

<@&286206848099549185>

[prisma coral]

wym

what does it have to do with the limit

[lone stone]

chat gpt's definiton

[lone stone]

limit is not so vital here

[prisma coral]

whats ur question

[lone stone]

it is like sequence

an = log(a,n)/n

[prisma coral]

ok

go on

[lone stone]

how to find FUCKING N(e)

[lone stone]

here is definition of n(e)

[prisma coral]

so u just want to solve the inequality
| log_{a} (n)/ n|<1 ?

[lone stone]

actually no

u must solve | log_{a} (n)/ n|<e

[prisma coral]

ok do that

[lone stone]

where e can be any value

[hearty turret]

,w ln(n) <= epsilon*n

[thorny flame]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by lyheart

[prisma coral]

oh my god

r u just trying to prove

[hearty turret]

,w log_a(n) <= epsilon*n

[thorny flame]

Results provided by WolframAlpha

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

Requested by lyheart

[prisma coral]

log(n)/ n= 1

[lone stone]

no

[prisma coral]

limit of that

yes u are

since e can be anything

u have to show the inequality | log(n)/ n |< e

correct ?

[hearty turret]

i don't usually see N(e) being used to denote the smallest natural number for which we are epsilon away from our limit

[lone stone]

i have to find N(e)

[hearty turret]

generally in a proof, N(e) is just some natural number so that we can guarantee that we are within epsilon away from our limit

[prisma coral]

no u dont

[hearty turret]

N(e) is just written like that as a reminder that N depends on epsilon

[lone stone]

the task is find N(e)

[hearty turret]

anyway i plugged it into wolfram alpha and yeah there is no 'nice' formula for the smallest N

the answer depends on the lambert w function, which can't be expressed in terms of ur 'nice' functions

[prisma coral]

why do u have to find N(e) though the convergence proof does not depend on it

[hearty turret]

!nogpt

[topaz sinew]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[hearty turret]

^

[lone stone]

Actually it is used to but not in USA and Europe

[lone stone]

but u can think like this

[hearty turret]

oh interesting

[hearty turret]

which countries may i ask?

[lone stone]

i dont think that it is mistake

[lone stone]

kz

[hearty turret]

interesting

anyway the answer to ur question is disappointingly u have to use the lambert W function

if u know what the lambert W function then u can just rearrange and eventually get there

but just as a reminder that generally you do not need to find N(e) if ur proving the value of a limit is L, u simply need to find something >= N(e)

[lone stone]

tbh idk lambert W function but i can read it on internet it is not big problem

[lone stone]

yep ik it

okay ty im gonna read about lambert W function

[hearty turret]

nw

[lone stone]

.close

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[orchid geode]

No clue on how i would go on by solving this. i only know one vector with its magnitude and direction.

[ocean terrace]

let the center be O and the end of the force be F

from F, draw a line parallel to a. let the point it intersects with b be Q

[orchid geode]

triangle method or like the parallelogram law?

[ocean terrace]

we would have a triangle OFQ. we know one side is 50 deg. we know 2 lengths. we have to find another angle

[ocean terrace]

no idea what these are

[orchid geode]

oh okay. but how tho. bc in this case i dont know the angle under F

oohh okay

[ocean terrace]

u construct it parallel to a

and u know the angle a and b intersect at

[orchid geode]

yeeah got thanks!

[ocean terrace]

welcome

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[hollow pumice]

In each of the following cases, compute the derivative using derivative rules (or limits
of difference quotients, if you show all work). Explain what derivative rules you use at each step

[prisma coral]

what is 1+x+x^2 +.. x^n

hint( sum of a geometric series)

[hollow pumice]

theres a formula for it right?

[prisma coral]

yes

|| x^n -1 / x-1 ||

[hollow pumice]

so the first term is 1

the common ratio is the x after one right?

[prisma coral]

yes

[hollow pumice]

idk what to do with this. I have the answer, I just don't know how to do it

[prisma coral]

u got 1+x+..x^n = x^n -1 / x-1

did u get this

can u find d/dx ( x^n-1 / x-1)

[hollow pumice]

the derivative of that?

[prisma coral]

ye

[hollow pumice]

ohhh thats the answer ok

ty

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[verbal shadow]

Suppose that A is a subset of real numbers which may or may not be countable
Would I be allowed to say that A consists of a_1, a_2, a_3 ... etc?

[verbal shadow]

if not, how else would I make an argument concerning multiple elements of A?

[craggy haven]

this is mostly fine, you can talk about arbitrary elements of A (assuming it's not empty)

[verbal shadow]

mostly?

[craggy haven]

if you said like ...
A consists of a1, a2, a3, ..., and so we can associate elts of A with natural numbers,
then that would be bad

but if you were just like let a1 and a2 be arbitrary unique elts of A
then that's fine

what's your intention here

[verbal shadow]

I'm trying to construct a set B
where B = {a1, a1 + a2, a1 + a2 + a3, ...}

i don't think this is valid though

[craggy haven]

hmmm yeah this doesn't seem right

because that is asserting A is countable, or taking a countable subset of it

[verbal shadow]

I guess i have to change my approach to not assuming anything about A and just trying to construct it?

i hate uncountable sets ;-;

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[shell apex]

Suppose $X, Y$ are two normed spaces. if the closed unit ball in $X$ ($Y$) is $U$ ($V$), and we have $S(U) = V$ for a linear bijection $S : X\to Y$, how do I show that $|Sx| = |x|$?

[shell apex]

i started by taking ||x|| = 1

we know ||S(x)|| \leq 1 now

[thorny flame]

CoolShot

[shell apex]

i understand i have to use the linearity of S here somehow

to show that ||Sx|| = 1

but i feel like im stuck

i can see that S is a homeomorphism as its both open and continuous

but i have no ideas on how to proceed

[craggy haven]

think about multiple points I reckon

or hmm

S is an equal opportunity transformer

[shell apex]

huh

[shell apex]

yeah, i tried taking y \neq x such that |y| = 1 too

and considering S(x - y)

but i still didnt really get anywhere

[craggy haven]

in my head is something about how if it shrinks one it has to shrink them all

[shell apex]

this ex i think is a standard FA one but i couldnt find anything on it

[shell apex]

sequence of points?

[craggy haven]

entire set I think

i haven't written it down

but I think you can quickly get a contradiction and show that the image won't be V

[shell apex]

well if i can show it for |x| = 1 => |Sx| = 1 then it does work for all points

[shell apex]

hmm im a bit confused on what you mean

[craggy haven]

imagine f took x to something with norm |x|/2

what will it do to y?

See if you can prove it for one dimension to start.

[shell apex]

hmm why would it affect the norm of the point y is mapped to

[craggy haven]

Well, if it's cutting the norm of X in half, then it stands to reason that it might do something to why as well, right?

Well, what if they have the same Norm?

[shell apex]

i guess it should be mapped to |y|/2 as well but im not sure what relation is there between an operator and the norm of the point its mapping

its continuity is probably used somewhere here

[craggy haven]

linearity

try it in 1D

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[calm merlin]

My teacher gave us a hint that the y intercept was 102

any clue on how to start finding the answer to this type of equation?

i tried plugging in a 0,102 into the equation but i got stuck there

im trying to do the second difference technique but its not working

the differences aren't constant

<@&286206848099549185>

[ornate osprey]

U need help representing an equation on graph?

[calm merlin]

ye

i need to create the equation and estimate the day 5

[calm merlin]

how do i do that

[ornate osprey]

Okay so first two draw a quadratic equation u need it's roots and it's minima or maxima

U know how to find the roots of the equation?

[calm merlin]

what is that

also the max looks like 8,155

i dont know if thats right though

[ornate osprey]

Roots are the values of x for which the equation is satisfied

[calm merlin]

I don't know how to find them then

[ornate osprey]

1&-5 are the roots

These roots will be the x intercept of the equation if u look at ur graph you'll see. That it passes thru these number in the x axis ..right ?

So we got how wide the graph will be now ...our next task is how long

[calm merlin]

Thats the second problem

i was doing a different problem

but sure lets do this one

sorry for taking picture of both and confusing

[ornate osprey]

Ohhh mybad

If u want I can solve the above one then

[calm merlin]

yeah

[ornate osprey]

Okay

[neon iron]

hi im in chapter 3 polynomial functions. and im having a trouble time how to solve real life functions or questions that are sly

[ornate osprey]

. occupied

.occupied

[calm merlin]

u have to go to the empty helps

the ones with no name

[neon iron]

my fault

[calm merlin]

king tier pfp

[neon iron]

luv u gng

[calm merlin]

💋

[neon iron]

❤️

[ornate osprey]

What u need to do is plot the points on the graph and join all these points which looks like a curve after that find the x intercepts

After finding the x intercepts ..which is the roots

You'll be able to estimate the quadratic equation

[calm merlin]

alright

[ornate osprey]

Let's say the roots are a & b then the equation would be (x-a)(x-b)

[calm merlin]

alright

[harsh flint]

Ok so basically, you use the equation of a quadratic which is: y= ax^2 + bx + c. You use 2 different points and the y intercept (giving c). so you can use the first two points given in the table: (1,115) and (3,135) and (0, 102). Then, you substitute y and x into the equation for the two different points.
So equation (1) would be for (1,115)
115 = a(1)^2 + b(1) + 102 ==> a + b = 115-102 = 13
Equation (2) would be for (3,135)
135 = a(3)^2 + b(3) + 102 ===> 9a + 3b + 102 = 135 ===> 9a + 3b = 33 then you can divide both sides by 3 to get 3a + b = 11

idk if that makes sense but then you basically solve the 2 equations simultaneously

[calm merlin]

wait what did you did you to get the last answer

[harsh flint]

for 3a + b = 11?

[calm merlin]

yeah

[harsh flint]

do you get how i get 9a + 3b = 33?

[calm merlin]

how you got this

^

[harsh flint]

oh ok so from 9a + 3b = 33

because 3 is the common factor you can divide both sides by 3

so itll be 3(3a + b) = 3(11)

and because theres 3 on both sides you divide by 3

and it cancels out to become 3a + b = 11

does that make sense?

[calm merlin]

and whats next after that

[harsh flint]

ok so

you have the 2 equations which is
(1) a + b = 13
and
(2) 3a + b = 11

you can make a the subject in (1)

so it becomes a = b - 13

and basically b - 13 is your value of a for now

so you substitute that into (2)

which becomes 3(b-13) + b = 11

you expand the brackets so it becomes

3b - 39 + b = 11

gather like terms on either side so

3b + b = 11+ 39

4b = 50

b then equals 25/2 or 12.5 depending if your teacher wants exact or approximate

[calm merlin]

wouldnt it be b subtracting 13

[harsh flint]

OMG youre right i was wondering why i was going wrong

so very sorry

[calm merlin]

i was reading it and i was confused xD

[harsh flint]

ok so it only changes substitution over here

[harsh flint]

lolll mbbb

so instead it will be 3(13-b) + b = 11

so it becomes 39 - 3b + b = 11

so gather like terms

-3b + b = 11 - 39

so -2b = -28

so divide both sides by -2

b = 14!

do you get it upto this point?

[calm merlin]

wait where did this come from

[harsh flint]

ok so i substituted value of a from equation 1 which is a = 13- b

into equation 2 which is 3a + b =11

[calm merlin]

okay that makes sense

[harsh flint]

ok

(sorry was speaking to someone lol)

uhh

where were we

ok

so we have the value of b

so basically so far our quadratic equation looks like so : y = ax^2 + 14b + 102

now we have to find a

but its really simple because we know that a = 13-b

[harsh flint]

from here

but now that we have b we sub its value into (1)

so a = 13- (+14)

therefore a = 13-14 = -1

then we sub back into our quadratic

we have c = 102, b = 14 and a = -1

y = -1x^2 + 14b + 102

or y = -x^2 + 14b + 102

does that make sense ?

[calm merlin]

yeah

and then for the second part i just plug in x=5?

[harsh flint]

pretty much

[calm merlin]

also the 14 was supposed to be 14x not 14b

[harsh flint]

oh whoops

[calm merlin]

but yeah thanks man this is a god send

theres a place in heaven for algebra tutors 🙏

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[harsh flint]

allg

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[raw canyon]

how would i go from general form to vertex form for the parabola 2x^2+8x+6?

[raw canyon]

well, i found the answer but idk how this works

oh i forgot the graph

[raven sparrow]

✨ complete the square ✨

[raw canyon]

do u know why the green graph is to the side instead of being the same as blue and red?

[raw canyon]

i dunno how to solve it using complete the square

i get to a certain point where i can't factor

[alpine mist]

because the green has vertex (-4, -2), by seen by the vertex form. Where the blue has vertex (-2, -2)

[raw canyon]

hmmm okay i see that now

so do you know how to get from red to blue?

im using cts and i got to

2(x^2+4x+8)-10

and i checked that

its the same as red

but at this point im pretty sure

i can't factor x^2+4x+8

[alpine mist]

when you complete the square, you find the factor as you go. It is, by construction, a perfect square.

[raw canyon]

hmmm, i must be doing something wrong then

[raw canyon]

well the issue is the 2x^2

[alpine mist]

that's also not what you get when you complete the square.

[raw canyon]

do i have to factor out the 2?

[alpine mist]

factor the two out of the first two terms with x's

[raw canyon]

i can do that and leave out the 6 at the end?

[alpine mist]

yes.

[raw canyon]

ohhh wait

ah i see

okay that makes sense

i just didn't realize you could selectively factor like that

[raw canyon]

ok thanks for your help

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[neon iron]

When solving equations like this I want the bottom to be the same but can I multiply just the 2 (top and bottom) by x+5?

[neon iron]

Like this and not touch the rest of the equation

Here’s the entire equation I just need someone to verify if it looks okay

[onyx summit]

Looks right to me, but you need to be care because if you plug in x=-5 into the original equation you will end up dividing by 0

[neon iron]

Will there be cases where the factoring is right but the answer requires u to divide by zero?

[onyx summit]

You can’t divide by zero, so x=-5 isnt a solution

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@neon iron Has your question been resolved?

[neon iron]

Thank you 🙇‍♂️

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[graceful leaf]

,rcw

[graceful leaf]

Is this not the right set up

Sadly i keep getting it wrong

but i feel like its a good set up i got g/mol from the periodic table as well for magnesium

for this

I keep getting a number no where near 9.08

maybe i should use the fraction bar

instead of division on my calc?

i cant even think of a set up

better than that

must be bad set up tho

am i not allowed to round it

ohhhh

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[opal plume]

help

[opal plume]

no

im slow

[merry hill]

it says speed equal to 1

what you have is velocity

so |v(t)| = |<....>|

so |<....>| = 1 or -1

now use calculator to find t

[opal plume]

what?

[opal plume]

sorry 😭

[smoky sparrow]

the absolute value of velocity is the speed

[opal plume]

yea but how do i put that on desmos

OWA

WOA

[scenic depot]

The answer is. 1.268
2.249
2.575.
El peso está tan devaluado que ya ni el número aparece

[merry hill]

and similarly for -1

it doesnt show the points tho, so just zoom till the number of decimals you want to round it to

[opal plume]

okay thanks

pro desmos

[smoky sparrow]

it's the average not instantaneous acceleration

[opal plume]

i looked at the examples and i never learned tis 😔

[opal plume]

heres a similar example to that

how do they evenget these numbers?

[smoky sparrow]

oh wait yeah it's the mean value theorem for integrals, my bad

[smoky sparrow]

you want a rectangle with the same area as the area under the curve

so $(b - a) \cdot \text{height} = \int_a^b f(x) \ dx$

[thorny flame]

higher's secret twin brother

[smoky sparrow]

no wait I was right after all, that's something unrelated

[smoky sparrow]

yes it's just (change in velocity)/(change in time) = average acceleration

[opal plume]

yea im cooked

[smoky sparrow]

you don't need the integral formula

[smoky sparrow]

just like how (change in distance)/(change in time) = average speed

so the change in velocity is v(3) - v(0)

change in time is just 3

[opal plume]

okay

[opal plume]

thing is how did they get the 1/3?

[smoky sparrow]

so $\frac{v(3) - v(0)}{3 - 0} = \frac{1}{3 - 0} (v(3) - v(0))$

that's literally it

[thorny flame]

higher's secret twin brother

[opal plume]

oh okay

thanks bro

[smoky sparrow]

npnp

[opal plume]

do u think u can help me with related rates

[smoky sparrow]

sure

[opal plume]

its more english than math 💀

[smoky sparrow]

oh that's not related rates but anyways

[opal plume]

dang my teacher lied to me

[smoky sparrow]

you just need to differentiate 2 * 1.34^t for this part

do you recall the derivative of a^x?

hint: $a^x = (e^{\ln a})^x = e^{x \ln a}$

[thorny flame]

higher's secret twin brother

[opal plume]

no 😭

wtaf 💀

[smoky sparrow]

can you use the chain rule to differentiate e^(x ln a)?

[opal plume]

actually understand

[smoky sparrow]

okok

[opal plume]

since they wanted the rate of the pumped i got the derivative of M(t) and pplugged in 3 and because it was >0 it was increasing

its not so bad

[opal plume]

thank u for ur help tho!!!!!

10/10 teacher

[smoky sparrow]

yep yeah also from functions knowledge, 1.34 > 1 so the graph of 1.34^t will be increasing

so you didn't need calculus but oh well, the question wants you to use it..

[opal plume]

yea

[smoky sparrow]

awwww that's so sweet!

[opal plume]

this is so horrible tho my teaher didnt teach this at all

esp particle motion

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[winter cypress]

M={(x,y,z) | x,y,z are natural numbers but not zero, x²+y²-z²=2024} Prove that M is infinite.

Someone please help me try to solve and understand this I need help 😭

[winter cypress]

Also if you are able to, simpler language is appreciated as I am an idiot

[topaz sinew]

@winter cypress Has your question been resolved?

[opal vault]

what you could do:
try to show any odd number is the difference of two squares

thus, if you pick y odd great enough, y^2 - 2024 is the difference of two squares...

[cursive patrol]

[smoky sparrow]

the wording "M is infinite" is really bad wording

[wary tulip]

how is that bad wording

[cursive patrol]

does that help

[winter cypress]

idk thats what i was given

[opal vault]

huh

[cursive patrol]

this is not homogeneous

all u need to do is rewrite equation as x^2 + y^2 = 2024 + z^2

[smoky sparrow]

yeah why did I think the RHS = 0

[wary tulip]

another idea is to use the sum of two squares theorem

[winter cypress]

uh

[winter cypress]

how does that prove it?

[opal vault]

this isn't linear

[shut night]

Ah true

My bad

[wary tulip]

i think rafilou's solution is simplest

[opal vault]

i would have said this is easy enough

[long stirrup]

what rafilou said, any odd x makes odd x²
then y²−z² is odd, and any odd number is just two consecutive squares subtracted

[winter cypress]

ok let me think about this

[long stirrup]

4 9 16 25 36... the differences are 5,7,9,11

idk how to prove that

[opal vault]

||(k+1)^2 - k^2||

[winter cypress]

where did k come from

[opal vault]

k represents any natural number

[winter cypress]

oh

[long stirrup]

the cool part is you don't need x to be large

[winter cypress]

huh

[long stirrup]

it's any x

[winter cypress]

uh ok

hold on what does "M= x, y, z" mean
is it that x, y and z are all the same number or m is x, y and z added together or something

[odd pagoda]

it says "M= x, y, z" nowhere

[odd pagoda]

this is set notation

the {} and () are important

"M is the set of triples (x,y,z) such that x,y,z are natural numbers but not zero and satisfy the equation x^2+y^2-z^2=2024"

[winter cypress]

ah

i give up 😭

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[hazy bluff]

How can I calculate xG?

For the first triangle?

Update, went even further, any help?

<@&286206848099549185>

[topaz sinew]

@hazy bluff Has your question been resolved?

[hazy bluff]

<@&286206848099549185>

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@hazy bluff Has your question been resolved?

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[simple perch]

i need some help with Logarithmic equations ( im new to it )

[iron blaze]

let ln x = u

[simple perch]

what do you mean?

[iron blaze]

let ln x equal u

then

the equation becomes 3u^2 + 1 = 2u

you can solve for u

and then solve for x

[neon iron]

take ln x = t

solve

in the end

x > 0

verify the answer

you will get t(x)

[simple perch]

is it possible to solve it with e instead?

[neon iron]

yeah

lnx^6 = lnx^2 - lne

[iron blaze]

ig

[neon iron]

x^6 = x^2/ e

easily

[iron blaze]

but e^ln^2(x) is x^lnx

[simple perch]

what is the diffrence between those? i have only used e

[neon iron]

approaches

[iron blaze]

wdym solve with e tho

[iron blaze]

can you solve this?

[simple perch]

no i have never used Substitution Variable

i have only learnt natural logarithm

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[lusty ocean]

I have a question about vectors, im asked to determine a vector v which stands perpendicular to both vector a = (2,3,-1) and b = (1,-2,3). also, vector v*(2ex-ey+ez) = -6.

e stands for the unit vector.

I know that I can take the dot product of a and v, which results in 0. Same thing with b.v, i also know that the cross product of a and b would be parallel to the vector v.
However, i have no clue how to solve this problem.

[strong viper]

does v*(2ex-ey+ez) = -6. stand for dot product or cross product

[fair thorn]

if two vectors are parallel (they lie along the same axis) but opposite directions, the dot product will be negative

dot

probably

cross product can't give out numbers

unless you're in 2D
which doesn't really work

[fair thorn]

also note that a dot product is

[lusty ocean]

dot product

[thorny flame]

ren

[fair thorn]

if theta = 180, cos theta = -1

so the dot product is simply -1 times the magnitude of v times the magnitude of w

[flint stump]

v is a vector along the direction of a cross b, but you dont know the magnitude for that vector. So you use the dot product relation to determine the magnitude for it

that uniquely determines v

[strong viper]

Find $$ \vec{a} \cross \vec{b} $$. WHat do you get?

[thorny flame]

RadMeerkat62445

[lusty ocean]

i get (7,-7,-7)

[lusty ocean]

how do i know its in opposite direction?

[fair thorn]

because the dot product is negative

otherwise it'd be positive

[lusty ocean]

ohh thanks

[fair thorn]

wait no actually hang on

[lusty ocean]

alr

[fair thorn]

first dot it with (2ex - ey + ez)

[lusty ocean]

dot what?

the cross product?

[fair thorn]

yeah

[lusty ocean]

and i can use cos180° for that?

im really confused rn

[fair thorn]

look i was wrong before

basically there can be a lot of x's and y's and z's

and the dot product isn't always going to be -6

bc yk they vary

so just dot it, set = -6, and that's your equation describing the relation b/w x y and z

[lusty ocean]

how do i dot it?

im not that smart😭

i know the formula

and yea i could get the magnitude of the cross product i just got earlier

i just dont really get the relation with those unit vectors

the 2ex-ey+ez

[fair thorn]

if a = (a1, a2, a3 ... an) and b = (b1, b2, b3 ... bn)

$\vec a\cdot \vec b = \sum_{i = 1}^{n} a_ib_i = a_1b_1+a_2b_2+\cdots+a_nb_n$

[thorny flame]

ren

[lusty ocean]

so id get something like 14ex+7ey-7ez = -6?
whats next

[topaz sinew]

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[prisma torrent]

idk if im bugging

but

sin + cos is tan

so (tan +cot)(tan)

so tan^2 +1

[charred moat]

?

[prisma torrent]

then that is cosec^2

[prisma torrent]

sin + cos = tan

no?

[charred moat]

sin theta/cos theta= tan theta

[fair thorn]

nononononon

[prisma torrent]

ohhhh

[fair thorn]

yes

[prisma torrent]

bruhhhh

no wonder

[fair thorn]

replace tan with sin t/cos t ig

[charred moat]

sorry

[fair thorn]

my guy.
you are not obligated to receive my help.
PLEASE stop advertising.

[prisma torrent]

then

sin/cos + cos/sin

ohhh

sin + cos + sin^2 +cos^2

hm

sin^2 + cos^2 is 1

so sin + cos + 1

hm

lemme do right hand side

1/cos + 1/sin

yh idk

[charred moat]

restart from this step

your addition is not correct

[prisma torrent]

i think i am dumb

(cos^2 + sin^2 )/cos sin

[charred moat]

correct

[prisma torrent]

so 1/cos sin

[charred moat]

correct

you're done after that just multipy the other part and simplify