help-10

[next trellis]

like your method

[timid silo]

this is how I would come up with the function

[next trellis]

yes but can you write the function youre thinking of?

[timid silo]

sure

I think it's going to be $f:(0,1)\to (a,b)$ defined by $f(x)=(b-a)x+a$

[warm shale]

yume ♡

[next trellis]

ok so what made you put (b-a) for m and a for c ?

[timid silo]

it's like a high school algebra problem, finding a line that passes through two points

the two points are (0,a) and (1,b)

[next trellis]

I totaly forgot how to do that lol

[timid silo]

haha, happens

[next trellis]

whats the formula again?

is it like m = x1 -x2/y1 - y2

[timid silo]

uhh you can write f(x) = mx + c and replace f(x) with a and x with 0

then a = c

then do the same with (1,b)

[next trellis]

oh

I see thanks

[timid silo]

yea you could do that too but I remember my way easier 😛

[next trellis]

ok one last question, why exactly send a to 0, and b to 1? why didnt you say b to 0 and a to 1

[timid silo]

you could do that too

[next trellis]

okay

thanks

[timid silo]

@next trellis are you ok with all of this? do you have any more questions?

[obtuse pebble]

@next trellis Has your question been resolved?

[obtuse pebble]

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[low sand]

Hey

[hybrid gull]

[low sand]

This is Quadratic Equations

I just need to understand what it means by “factored over true integers”

[hybrid gull]

So you can factor the expression and not use any numbers that aren't integers

That is, use only integers in the factoring of the expression

[amber iris]

Eisenstein's criterion

[obtuse pebble]

@low sand Has your question been resolved?

[low sand]

Not to be like rude or anything but can you solve this im still so confused

[hybrid gull]

I can solve it, but I can't just give you the solution

[low sand]

Yes, I will try my best from there if you can do that please. 🙏🏽

[low sand]

is it bxy?

[hybrid gull]

Doesn’t seem like you’re trying so I won’t try either

[low sand]

.close

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[opal musk]

need halp

[opal musk]

ive got the perimeter=1/2 circumference

so perimeter = pi r

not sure what to do now

[obtuse pebble]

@opal musk Has your question been resolved?

[opal musk]

https://tenor.com/view/emoji-disintergrating-meme-emoji-disintegrating-meme-gif-21239978

someone please; help this fool with eyes and hair of coal.
for i have waited many a year and withered away my soul.

[obtuse pebble]

@opal musk Has your question been resolved?

[opal musk]

^^^^^^^^

[opal musk]

.close

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[crisp dune]

how do u find the area shaded in green?

[high lily]

integration

[crisp dune]

i know but i dont know how to do it for an example like this

like the shaded area isnt in the concave down region

[high lily]

consider area addition/subtraction

shaded = area under blue - area under red

(over that interval)

[crisp dune]

ok so

if it was this interval

and the graph was x^3 + x^2 for example (i guess) how would u write it out

could u please show me?

like the integration steps

[high lily]

that graph doesn't resemble x^3 + x^2

can you set up an integral that represents the area under the blue line from
-2 to 4?

[crisp dune]

no

[high lily]

do you have the equation of the blue line?

[crisp dune]

y=4 i guess

do u want me to have equations that perfectly resemble my graph or it doesnt matter?

[high lily]

ideally you'd have a specific example

[crisp dune]

ok i have one

so

ill paste the question

[high lily]

actually someone will have to take over, I gtg in a few minutes

[crisp dune]

damn

.close

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[onyx warren]

brain is melting

[onyx warren]

why are these expressions different???

isnt it just multiplying the expression by 1, both should be the same, no?

[obtuse pebble]

@onyx warren Has your question been resolved?

[onyx warren]

<@&286206848099549185>

[obtuse pebble]

@onyx warren Has your question been resolved?

[timid silo]

In the first expression you calculate the 80th power of .99 and then multiply by -1. But in the second (the correct one) you exponentiate -0.99.

But these two different orders don't give the same result.

[onyx warren]

yeah im just really confused as to why the hell that's a thing

whats the point of even having a negative sign

on a calculator

if it considers it subtraction unless there's a parenthesis

@timid silo

[timid silo]

I mean, (ab)^n = (a^n)*(b^n), thus (-1 * 0.99)^n = (-1)^n * (0.99)^n and (-1)^80 = -1.

[onyx warren]

yes, i get that, i was just really confused why negative .99 raised to the 80th

didnt by default evaluate as (-.99)^80

because otherwise why even have a separate negative function and subtraction function

[timid silo]

Honestly, I don't really "know" why the conventions are like that. Apparently some people just don't want to type an additional 0 on a calculator. (or write it on a piece of paper)

[timid silo]

Okay, sorry. I misunderstood your question.

[onyx warren]

@timid silo all g, thank you

.close

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[timid silo]

hi

[timid silo]

so basically

im doing my homework atm and the lesson is about polygons

the question is: Is there an octagon whose exterior angles are 115, 53, 26, 18, 48, 39, 32 and 29 degrees

<@&286206848099549185>

[copper marten]

it might be helpful to know what the total angle sum of an octagon is

[shell trout]

You can just count

iirc something like 180˚ * (n - 2)

[shell trout]

What have you tried so far?

[timid silo]

one min

i will try this

got it

.close

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[wintry stream]

how do i solve this?

[obtuse pebble]

@wintry stream Has your question been resolved?

[gloomy valve]

@wintry stream You could use an integral to estimate the upper bound of the error

[obtuse pebble]

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@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[fast fog]

This one is Hard(mayby)

[obtuse pebble]

@fast fog Has your question been resolved?

[fast fog]

I know, is hard. When I come up with an idea I't ends up wrong.

[timid silo]

so if you do x damage and the enemy has y hp
number of hits = y/(x*(1-percentage/100)

so if y=100 and x=1 and reduction = 0
100/(1 * (1-0/100)=100
but if reduction is 25
100/(1 * (1-25/100) = 133.333
if 100
100/(1 * (1-100/100)=100/0 undefined so cant kill

so multiplier is just new number of hits / original number of hits

[fast fog]

will the hit amount multiplayer will be diferent witha diferent x and y?

[timid silo]

the same as it simplifies

if you simplify you get multiplier=1/(1-percentage/100)

[fast fog]

So the multiplayer of hits needed is 2.(7)

.close

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[elfin kayak]

hey does anyone know what the q means by trace here?

[obtuse pebble]

@elfin kayak Has your question been resolved?

[obtuse pebble]

@elfin kayak Has your question been resolved?

[tardy epoch]

trace is drawing alpha(t) for t in [0, 2pi]

[obtuse pebble]

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[dry oak]

Anyone that could me identify the variable so I can solve the problem.

[devout solar]

Is that a test?

[dry oak]

Nop

Homework

Anyone else that could me is not a test.

[devout solar]

aand in english

[dry oak]

what you mean

.close

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[radiant lily]

need help trying to set this problem up.
A 10-foot ladder placed on level ground leans against the side of a house. The ladder reaches a point that is 9.2 feet up on the side of the house.
What is the measure of the angle formed by the ladder and the level ground? Round your answer to the nearest degree. Show your work.

[obtuse pebble]

@radiant lily Has your question been resolved?

[tardy epoch]

Draw a picture

[obtuse pebble]

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[versed zinc]

hi

[versed zinc]

this is not a question, idk if I can do this

but

can you give me a not that hard function to derívate

I want to practice

something easy 🥹

[blazing lodge]

differentiate?

[versed zinc]

wdym

[blazing lodge]

what does derivate mean

[versed zinc]

derivative

like

[blazing lodge]

Ok

[versed zinc]

ye

[blazing lodge]

e^(3ln(x^2))

[versed zinc]

ok

um

@blazing lodge can it be something easier?

I umm

just did this

[obtuse pebble]

@versed zinc Has your question been resolved?

[blazing lodge]

<@&286206848099549185>

[obtuse pebble]

@versed zinc Has your question been resolved?

[obtuse pebble]

@versed zinc Has your question been resolved?

[versed zinc]

can you give me a function to derívate?

I want to practice

[versed zinc]

um

f(x) = 1+x

?

no

I mean

I don’t know what logarithms are yet

so I am going to try to do f(x) = 1+x

ok ok

Im on my way home

I’ll do it in my house because notes on my phone is ugly

[obtuse pebble]

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[open lotus]

how is this true?

for 3 I get it but not sure for higher power

[obtuse pebble]

@open lotus Has your question been resolved?

[thick fog]

Using the truth of the case when k = 3, you can prove it using mathematical induction. Another way is to just argue that that x = -y is a root of f(x) = x^k+y^k for y constant. So x - (-y) is a factor of x^k + y^k.

[open lotus]

oh nice

I got that root argument

can you show the induction way?

couldn't go further from $x^{2*k-1}*x^2 + y^{2k-1}*y^2$

[thick fog]

I'm not going to go through a formal proof but I will sketch it for you

[warm shale]

sunho

[thick fog]

P(3) is true. Do what you just did

Since P(2k-1) is true, there exists a polynomial p so that x^(2k-1) + y^(2k-1) = (x+y)p or y^(2k-1)= (x+y)p - x^(2k-1)

substitute (x+y)p - x^(2k-1) in for y^(2k-1)

do some basic factoring and what not until you have (x+y) multiplied by some stuff

Obviously a bit sloppy but I think that should be enough for you based on what you said?

[open lotus]

yeah I'm getting c*(x+y)(x-y) + d(x+y)p if I do some factoring

seems to be working!

[thick fog]

yeah so you showed that x^k + y^k is (x+y)*stuff

[open lotus]

thank you very much

.close

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[versed zinc]

ok I know this is probably stupid

[versed zinc]

but

what do I do now

[hidden wigeon]

simplify the numerator

[versed zinc]

so

I remove the paréntesis?

[hidden wigeon]

c

what is (x+h+1)-(x+1) equal to

[versed zinc]

I am dying idk if I am supposed to multiply or sum

[hidden wigeon]

you subtract it

you subtract x+1 from x+h+1

[versed zinc]

so I only have h

[hidden wigeon]

yeah

and what's h/h

[urban ridge]

.unpin

[obtuse pebble]

✅

[urban ridge]

thanks

[versed zinc]

1

[hidden wigeon]

yeah you got it

[versed zinc]

but

is there another way I can subtract that?

because that was weird

[hidden wigeon]

you're using the definition of the derivative to do it, you'll eventually learn methods to do it faster

[versed zinc]

ok but

the

[hidden wigeon]

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

this video does a good explanation

[versed zinc]

how can I do this like step by step

I am having a stroke

[versed zinc]

saved. Thank you

[hidden wigeon]

if ur learning calc that should just be super simple simplification

[versed zinc]

yes

I know

but

I am 15

I shouldn’t be learning this

[hidden wigeon]

i dont think you should be learning calculus without decent algebra skills

[versed zinc]

explain that then 😞

[hidden wigeon]

what is -(x+1)

[versed zinc]

x+1

[hidden wigeon]

what

[mellow cipher]

[versed zinc]

-x-1

[hidden wigeon]

ok yeah

and what is x+h+1-x-1

[versed zinc]

because I change all signs

[versed zinc]

I change it to

x-x
1-1

and I have h

[hidden wigeon]

yeah

[versed zinc]

I did that mentally but yea

[hidden wigeon]

and what is h/h

[versed zinc]

1

[hidden wigeon]

yeah that's basically how you get that

[versed zinc]

oooooh

[hidden wigeon]

i dont think you should be doing calculus yet tbh

[versed zinc]

what should I be doing now

I have brilliant

[hidden wigeon]

what are you up to in terms of math

algebra?

[versed zinc]

I am learning algebra in school

look

I made this the afternoon my teacher teach me this

wait

https://github.com/Squishey/EquationSolver

it’s just

ax^2 + bx + c

[hidden wigeon]

yeah you honestly should not be doing calculus

[versed zinc]

what should I be doing

😞

[hidden wigeon]

probably

algebra

[versed zinc]

I don’t know if you are into coding but, what do you think about it?

[hidden wigeon]

i dont do coding

[versed zinc]

oh

well

so

álgebra

1 or 2?

is this correct?

[obtuse pebble]

@versed zinc Has your question been resolved?

[versed zinc]

wtf

[versed zinc]

this is 2x

[keen cave]

correct as in?

[versed zinc]

f'(x)

[keen cave]

Yeah.

[versed zinc]

like f(x) = x^2

[keen cave]

You just need to expand everything out.

You we end up factoring out a h which cancels with the bottom.

[versed zinc]

ye

look

[keen cave]

Yep.

[keen cave]

Yep 👍.

[versed zinc]

yeee

ty

.close

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[vapid oak]

sorry i have another question

[vapid oak]

my textbook says that between 0 and 2 it is concave down, however it looks like a concave up, am i missing something?

[flat anvil]

that's the graph of f''(x)

not the graph of f

f'' is negative on (0, 2) so it would be concave down on that interval

[vapid oak]

oh

dangit how did i not realize that

tysm man mb

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[versed zinc]

am I doing good?

[high lily]

no

[versed zinc]

💀

[high lily]

in the first line of work the fraction you have is only the difference quotient
for the derivative you'd also need the limit as h→0
and as for the line after that there was a complete disregard for algebraic rules

[versed zinc]

I didnt write the lim because im on my laptop

this is awkward

but

what was the complete disregard for algebraic rules

[high lily]

i see like 3 main issues

[versed zinc]

say em

[versed zinc]

say say say

[high lily]

oh you made some edits

[versed zinc]

yup

[high lily]

now its currently only one issue

[versed zinc]

say it

[high lily]

$a(b^2)\neq (ab)^2$

[warm shale]

ℝamonov

[versed zinc]

so

I should have factored (x+h)^2 first

instead of multiplying it by 6

[high lily]

*expanded

[versed zinc]

sorry, not a native english

person

ok so let me re do the whole shit

can I do common factor?

what do I do now

[obtuse pebble]

@versed zinc Has your question been resolved?

[obtuse pebble]

@versed zinc Has your question been resolved?

[versed zinc]

Im not in my house lol, I don’t have white normal paper

but um

<@&286206848099549185>

what do I do now

ok I’m so dumb

I swear

nvm

wait

idk what to do now

do I just delete the h

?

ye

.close

[obtuse pebble]

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[timid silo]

https://media.discordapp.net/attachments/886034590219780096/963285851859718144/unknown.png

[timid silo]

evening gentlemen

coukd i get some help on this pls

q4

[flat anvil]

what is confusing? find the mcclaurin series and then find the coefficient of x^(2022) and multiply it by 2022!

because coeff of $x^n = \frac{f^{(n)}(0)}{n!}$

[warm shale]

! Brontochad (RYC4blushysully)

[flat anvil]

(in a mcclaurin series)

[timid silo]

i can’t take derivative 2022 times

[azure lake]

Doesn’t it just mean (f(0))^2022?

[timid silo]

wut

[flat anvil]

no

[flat anvil]

and you aren't supposed to, find the mcclaurin series (pretty simple, its just a geometric series, but you have to multiply it by x^2), and then solve for the n such that the power = 2022

and then find the coefficient after doing that

[timid silo]

could u write this out for me? im afraid i dno what you’re talking about

[flat anvil]

$\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$

[warm shale]

! Brontochad (RYC4blushysully)

[flat anvil]

it will look something like this

[timid silo]

right

[flat anvil]

and what you are trying todo

[timid silo]

i understand that part

[flat anvil]

is solve for the n such that the power of x = 2022

then you can just take the coefficeint and multiply it by 2022!

[timid silo]

but how do i actually find the value of the numerator

no n is 2022

[flat anvil]

you find the mcclaurin series

well first find the mcclaurin series

then come back

[timid silo]

this is the mac series no?

[flat anvil]

not for your function

that's just the general form

[timid silo]

the stuff in front of x^n

[flat anvil]

do you know how to find a mac series

[timid silo]

no because i dno how to differentiate f(x) 2000 times

if n was 2 or 3 yeah

[flat anvil]

okay well then you should read up / review notes on how to find a general mcclaurin series

(or how to use a geometric series in this case)

here ill write it like this

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

[warm shale]

! Brontochad (RYC4blushysully)

[flat anvil]

do you recognize that the second term has a very simple series expansion?

(hint: geometric series)

[timid silo]

oh yeah

[flat anvil]

so we get this

[timid silo]

i forgot the exact form tho

[flat anvil]

$x^2 \cdot \sum_{n=0}^{\infty} x^{2n}$

[warm shale]

! Brontochad (RYC4blushysully)

[flat anvil]

geometric series is something you should know cold

so keep that in mind

anyways we can simplify this b/c x^2 is a constant in the n world

so we get :

$\sum_{n=1}^{\infty} x^{2n+2}$

[warm shale]

! Brontochad (RYC4blushysully)

[flat anvil]

so now here is where i was saying solve for when the power of x = 2022

so we get: 2n+2 = 2022 => n+1 = 1011 => n = 1010

right does this make sense?

[timid silo]

yes

[flat anvil]

okay so now we need to compare this to our form from before

[flat anvil]

this

but in this case it turns out that the massive coefficient is just 1!

[timid silo]

well im still lost on why u didn’t just substitute 2022 into n

[flat anvil]

$\sum_{n=1}^{\infty} 1\cdot x^{2n+2} = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$

[warm shale]

! Brontochad (RYC4blushysully)

[flat anvil]

cuz then the power of x (the thing we care about) wouldn't equal 0

anyways in this case it doesn't matter what n is because the coefficients are always 1

so now you just have to note that the coefficient of x^(2022) = f^(2022)(0)/2022!

and then from there you can solve what f^(2022)(0) is

[timid silo]

ok I’m gonna go pray they donkt ask this tmrw

[flat anvil]

well where are you confused

because this is something that is ALWAYS on a midterm

[timid silo]

well i have my final tmrw morning

ok yeah no

@flat anvil alright well i donkt understand this but thank you

.close

[obtuse pebble]

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[midnight dew]

(a-2)^2 = 12

[midnight dew]

hi im having issues on solving perfect square equations

[flat anvil]

what is confusing ?

[midnight dew]

just solving the problem, im not sure where to start or what to do

[flat anvil]

take square root of both sides

[midnight dew]

square root of both sides?

like square root of (a-2)^2/12 ?

.close

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[teal breach]

how do you find llv+wll vs llvll+llwll?

[elfin burrow]

depends on how the norm is defined

but the Cauchy-Schwarz inequality states that ||v+w|| <= ||v|| ||w|| when the norm is induced by an inner product

and there is equality whenever v and w and linearly dependent

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[crisp dune]

can someone explain how to do this?

[obtuse pebble]

@crisp dune Has your question been resolved?

[nocturne sun]

for f(g(x)) to be defined, g(x) has to be in [1, infty) so g(x) >= 1
for g(f(x)) to be defined, f(x) has to be in (-infty, 2] so f(x) <= 2

[crisp dune]

yes?

[nocturne sun]

yeah then you just solve the inequalities?

keep in mind that for g(x) >= 1 you also have x <= 2 and for f(x) <= 2 you have x >= 1

[obtuse pebble]

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[hexed summit]

<@&286206848099549185>

[hexed summit]

i don’t understand this

[obtuse pebble]

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[bright geyser]

I cannot get B
I found the forces on A = T - 2w_B
forces on B = w_B - T
and the accelerations of B and A are the same
so
(T - 2w_B) / w_A = (w_B - T) / w_B
w_B(T - 2w_B) = w_A(w_B - T)
and now this is already clearly super wrong
so idk what to do

im going fucking insane

[brave bramble]

Mind zooming in on the question? Very smol.

Well, okay I might have it.
The new normal force on the block is 2w(a)
The block is still pulled with gw(b)

Oh wait, duh. If the same coefficient of friction exists, and the block now just got heavier, then clearly the block won't move.

[obtuse pebble]

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[obsidian condor]

Hello I was looking at a solution. Why did square root on the top become a 1 ? The bottom also became a square root too. I didn’t understand the steps here. Any help is appreciated!

[modern kayak]

okay so

factor out the denominator

so that its the product of linear factors

which is x-2 and x+2

(with 3 infront as a scale factor of sorts)

you understand that part?

[obsidian condor]

Yep

[modern kayak]

okay so you see sqrt(x-2)

and the denominator

has x-2?

[obsidian condor]

Yes

[modern kayak]

so consider sqrt(x-2) = u, and x-2 = u^2

then when you cancel them out you get 1/u right?

[obsidian condor]

Oh I see

[modern kayak]

yep

[obsidian condor]

Tysm

[modern kayak]

np

[obsidian condor]

.close

[obtuse pebble]

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[modern kayak]

but idk why they did that lol @obsidian condor it wouldnt give a invalid answer

it would make the denominator 0

and n/0 = undefined

[obsidian condor]

Yeah the answer is 1/0

They want me to prove that it is undefined

[modern kayak]

ah alright

makes sense

[obtuse pebble]

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[gloomy cove]

Help frequency distribution

[obtuse pebble]

@gloomy cove Has your question been resolved?

[gloomy cove]

.close

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[strong dirge]

Hey everyone. I don’t know where to post this, but I am getting stuck on this proof in tao’s analysis 1. I’ve highlighted the portion that is bothering me. I’ve tried to do all kinds of “legal” inequality manipulations, and I have made no progress haha. I hope I can receive help on this. Thank you!

[obtuse pebble]

@strong dirge Has your question been resolved?

[novel knoll]

they fixed the epsilon such that |b_n-0|>epsilon for n big enough

[strong dirge]

Yes, but I don’t understand how he concludes |b_n|>= epsilon/2 from the triangle inequality

I’m able to get till here

[novel knoll]

if |b_n_0|>epsilon the only way |b_n_0-b_n|<=epsilon/2 is if |b_n|>epsilon/2

|b_n_0-b_n|>=| |b_n_0|-|b_n| |

[strong dirge]

This is the reverse triangle inequality?

[novel knoll]

yes

[strong dirge]

I still don’t understand how we need to have |b_n|>epsilon/2 for |b_n0-b_n|<=epsilon/2

[strong dirge]

?

[obtuse pebble]

@strong dirge Has your question been resolved?

[obtuse pebble]

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[grim plinth]

i dont see the pattern

[brave bramble]

So you are using the hint?

Let's say you've already got n lines down. If you add another, it has to intersect with all of them, producing n more intersections

[grim plinth]

ok lets say i have 2 lines, so i have 1 intersection

if i add one more line i get 3 intersections

i dont understand the plane divided into regions

[brave bramble]

A "region" is an area that is not broken by a line

A single line breaks the plane into 2 regions

[grim plinth]

i see

but i dont see how to use induction

[obtuse pebble]

@grim plinth Has your question been resolved?

[amber torrent]

you prove something by induction by adding one to the variable n (in this case it means adding one more line)

then use the formula to determine the new number of regions

if you come up with the same answer as you would by just substituting n+1 into the formula instead of n then you have proven that the formula is valid for any number of lines n

[obtuse pebble]

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[stray summit]

Golden Ratio Games is trying out a microtransaction model in their latest game. They expect that 1/5 of players who try their game will make some microtransactions and spend an average of $3.50. If the total cost to produce and market the game was $80000, how many people will have to play the game for the company to break even?

[hidden wigeon]

what have you tried

[stray summit]

i tried 1/5x * 3.50 = 80000

and got 114285.7

actually i think that is the answer

i do need help with this tho

The sum of two numbers is 40. If you added sixteen to twice the smaller number, you would get ten more than the larger number. What are the two numbers?

[hidden wigeon]

what have you tried for that then

[stray summit]

there are two variables so like im not sure

[hidden wigeon]

alright so

let's claim that the smaller number is x, and the larger number is y

the sum of these 2 numbers are 40

so x+y=40

you would do something similar for the other one

[stray summit]

Ok so?

[hidden wigeon]

can you write an similar equation for the other part

"If you added sixteen to twice the smaller number, you would get ten more than the larger number."

[stray summit]

k i think i understand

[stray summit]

what would the answer be?

[hidden wigeon]

lets break it down further then

"sixteen to twice the smaller number" is the same thing as "ten more than the larger number"

and we claimed that the smaller number is x, larger number is y

[stray summit]

equation is
x*2+16=y-10?

[hidden wigeon]

that's 10 less

[stray summit]

so it would be +10 instead of -10?

[hidden wigeon]

yes

[stray summit]

or do i not need to add 10 or subtract 10

oh ok

[hidden wigeon]

"sixteen to twice the smaller number is 2x+16 in terms of x (smaller number)
"ten more than the larger number" is y+10, as 2x+16 equals to 10 more than y

so now you have these two equations
x+y=40
2x+16=y+10

do you know how to solve systems of linear equations?

[stray summit]

i havent started it yet

[hidden wigeon]

h

[stray summit]

how would i do algebra when there is 2 different variables?

[hidden wigeon]

you would normally solve the system of linear equation

1 method is substitution

x+y is 40, right

[stray summit]

yes

[hidden wigeon]

what if we

make y=40-x

and we plug 40-x as y in 2x+16=y+10

[stray summit]

so 2x +16 = 10 + 40-x

[hidden wigeon]

yes

solve for x

[stray summit]

k

i got 34/3

[hidden wigeon]

you have your x value

x=34/3

now plug 34/3 as x into x+y=40

[stray summit]

and then i add 10?

k

i got 86/3

[hidden wigeon]

yeah

you solved it

the smaller number is 34/3, larger number is 86/3

[stray summit]

Ohh

i c

tyvm

[obtuse pebble]

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[short spruce]

use exponent rules

[timid yacht]

not rly

[tardy epoch]

Yes

But there are other properties you want to use

a^(bc) = (a^b)^c

[short spruce]

mainly just $\frac{x^a}{x^b}=x^{a-b}$

[warm shale]

a disappointing son

[short spruce]

how'd you get the base to be 1

you don't have 2/2

you have 2^(3x)/2^x

[timid yacht]

x= 2 a=3x b=x in our case

[short spruce]

it is 2^(2x), but that does not simplify to 4x

if you type 2^2x into an online calculator, it will interpret that as (2^2)*x

which is 4x

if you have more than one character in an exponent, surround it with parenthesis

[obtuse pebble]

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[stiff cipher]

I Need some help understanding submanifolds

[obtuse pebble]

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.close

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[indigo quarry]

i dont know how to solve this one, infact i only guessed the upwards normal vector.\

[obtuse pebble]

@indigo quarry Has your question been resolved?

[indigo quarry]

<@&286206848099549185>

[obtuse pebble]

@indigo quarry Has your question been resolved?

[indigo quarry]

<@&286206848099549185>

[astral ivy]

@indigo quarry for the second one, just think of it like a triangle, where the angle is 29° and the adjacent side has length 1. What is the length of the opposite side?

[indigo quarry]

oh simple triange cos equation with the angles and side lenght got it what about the shear factor?

[obtuse pebble]

@indigo quarry Has your question been resolved?

[astral ivy]

Nice

What did you get for the magnitude?

[indigo quarry]

@astral ivy 0.554

[astral ivy]

Ok are you familiar with bases and linear transformations

[obtuse pebble]

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[tardy epoch]

.close

[obtuse pebble]

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[obtuse pebble]

@broken owl Has your question been resolved?

[broken owl]

<@&286206848099549185>

[obtuse pebble]

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[river zephyr]

Hi how can I solve that?

[obtuse pebble]

@river zephyr Has your question been resolved?

[elfin belfry]

Show what?

[river zephyr]

Just realised my file didn’t upload bc of wifi 😭

[timid yacht]

your B is (2,9)?

@river zephyr

[river zephyr]

Wait

I think I got it

Oops

Could someone help me with this ?

[timid yacht]

normal means 90 degrees

also did you solve the previous one?

[river zephyr]

Yup

Got 2,1 I subbed into the wrong eqn

[river zephyr]

I’m lost after this step

[timid yacht]

so the point is (2,3+4-4)

u want to draw a line at this point with a slope of 1/2

[river zephyr]

Could it be down this way ?

I think I’m supposed to make the gradient -2

[timid yacht]

last part

its 2y-x

[obtuse pebble]

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[obtuse pebble]

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[late night]

.close

[crystal stream]

Hi there

[crystal stream]

I have a question in Calculus

It is a prove or disprove

I have a hard time with these.

(a) looks like lagrange

but idk how to do that

[obtuse pebble]

@crystal stream Has your question been resolved?

[crystal stream]

<@&286206848099549185>

[eternal bloom]

This is pretty hard. Try #real-complex-analysis and try to write more about your own attempt

[obtuse pebble]

@crystal stream Has your question been resolved?

[steel marsh]

@crystal stream You need to do this problem for your calculus class?

[obtuse pebble]

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[pale wedge]

for the function g(x) = 10^(x+3) determine the effect the transformation has on the parent function f(x) = 10

[pale wedge]

i’m pretty sure it’s transformation out of
A. stretch
B. reflection
C. translation
D. compression

also this i’m 99% sure it’s A

but lmk

[obtuse pebble]

@pale wedge Has your question been resolved?

[shell elbow]

its a

[obtuse pebble]

@pale wedge Has your question been resolved?

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[scenic hedge]

I'm having trouble understanding and solving problems like this.

if someone could explain how to solve such problems

[rich fiber]

Try to simplify it first

[scenic hedge]

The problem is I have been sick the whole time they did this in class

so I don't know even the basics

[rich fiber]

This should cover all the basics you need
https://youtu.be/9tutJ5xrRwg

[scenic hedge]

Im confuzed how you get a -2x7^2x+2

[rich fiber]

49 = 7²

[scenic hedge]

I know

the problem is see the top and bottom lines

I am very confuzed to how in the top we went from all that is coloured to the bottom coloured section

[rich fiber]

Ahh I see
To make it simple x=7^(2x+2)
X-3X = -2X = -2.7^(2x+2)

[scenic hedge]

🤦‍♂️ I am so gonna fail this test

[fierce lagoon]

Well, let's break this down

Using exponent laws, rewrite $7^{2x + 2}$

[warm shale]

Umbraleviathan

[scenic hedge]

As I said

I am failing

[fierce lagoon]

I'm well aware

But do you at least know how to break that apart

[scenic hedge]

break it apart?

no not really

[fierce lagoon]

Alright so

There's this property

Or I should say identity idk what it is

It's a thing you can do

$A^{a+b} = A^a\cdot A^b$

[warm shale]

Umbraleviathan

[scenic hedge]

aha, yes Ive seen this before

[fierce lagoon]

Which applies to this

[scenic hedge]

and subtraction is devision

[fierce lagoon]

Yeah

[scenic hedge]

to then that would be $7^{2x}\cdot 7^{2}$

[warm shale]

PoTeTo

[fierce lagoon]

Yeah, nice

Now

With $49^x$

[warm shale]

Umbraleviathan

[scenic hedge]

49x?

[fierce lagoon]

What's another way to write 49

[scenic hedge]

7x7

[fierce lagoon]

In terms of 7

Which is the same as?

[scenic hedge]

7^2

[fierce lagoon]

Ye

So there's another property

Thingamajig

$(A^a)^b) = (A^b)^a = A^{ab}$

[warm shale]

Umbraleviathan

[fierce lagoon]

So $49^x = (7^2)^x$

[warm shale]

Umbraleviathan

[fierce lagoon]

Rewrite the right side

[scenic hedge]

ok so 49^x is the same as 7^2)^x

[fierce lagoon]

Yeah

[fierce lagoon]

Use this law

[scenic hedge]

$7^2x$

[warm shale]

PoTeTo

[fierce lagoon]

Yeah

[scenic hedge]

ish

[fierce lagoon]

Btw it's gotta be {2x}

For LaTeX

[scenic hedge]

$7^{2x$}

[warm shale]

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

[scenic hedge]

xD

[fierce lagoon]

You can just plug it into desmos, highlight it, copy it, paste it

Since desmos uses Latex

[fierce lagoon]

But yes

Now this is gonna take me a bit but

Oh and

The denominator, simplify the base 7 exponent

[scenic hedge]

you can simplyfy it even more?

[fierce lagoon]

The 7^(2x+1)

[scenic hedge]

yes

thats its simplest form

isnt it?

[fierce lagoon]

No

[fierce lagoon]

This

[scenic hedge]

$7^{2x}\cdot 7^{2}$

[warm shale]

PoTeTo

[fierce lagoon]

Not quite

[scenic hedge]

oh

wait

[fierce lagoon]

It's being raised to 2x + 1

[scenic hedge]

exponents make no sense

[fierce lagoon]

You'll get used to it

[scenic hedge]

quadratic function and intervals are easyer

[fierce lagoon]

Let 2x = a,
1 = b

And 7 = A

[fierce lagoon]

Heh, I remember those

Wait until you find the slope of a quadratic

Then it's just pain

[scenic hedge]

Thats easy for me

exponents

are

a

nightmare

They don't make sense

or logic

[fierce lagoon]

They'll eventually

You just gotta get used to them

Before you end your session I'll list some stuff out

But for now, let's focus on the question

[scenic hedge]

slight problem

[fierce lagoon]

So 7^(2x + 1) = 7^(2x) • 7

[scenic hedge]

exam is tommorow....

[fierce lagoon]

Moment

You can use the stuff I list as like a quick study guide

[fierce lagoon]

From the property I showed before, you understand this right

[scenic hedge]

Yes

[fierce lagoon]

Aight so

And this is gonna take me a while

Because latex loves all this shit and what not and syntax skiddhdislakdkd

$\frac{7^{2}\cdot7^{2x}-3\cdot7^{2x}-2\cdot7^{2x}}{7^{1}\cdot2\cdot7^{2x}}$

That's not right

Uhhhh what did I miss

[warm shale]

Umbraleviathan

[fierce lagoon]

Alright there

@scenic hedge what's a thing you can factor out from the numerator

[scenic hedge]

7^2x from top and bottom

[fierce lagoon]

Ye

And then when you factor out 7^(2x), what also happens

[scenic hedge]

🤷‍♂️

[fierce lagoon]

Time to write another fraction

[scenic hedge]

$\frac{7^{2}-3\cdot7^{2x}-2\cdot7^{2x}}{7^{1}\cdot2}$

oof

[fierce lagoon]

Close

[scenic hedge]

the - got deleted

[fierce lagoon]

Not quite though

[warm shale]

PoTeTo

[fierce lagoon]

Not quite

[scenic hedge]

no?

[fierce lagoon]

The 7^2x needs to be factored from all terms

That's how factoring works

It has to take from every term

[scenic hedge]

🤦‍♂️

[fierce lagoon]

But you're very close

Oh

[scenic hedge]

im stupid

[fierce lagoon]

Yeah it doesn't let you use that word

[scenic hedge]

welp

[fierce lagoon]

But I can say fuck shit

[scenic hedge]

im fucking stupid

[fierce lagoon]

It's fine lol

[scenic hedge]

Now i remember

[fierce lagoon]

(That's how I fucked up the SAT today)

(Same exact mistake 😐 )

[scenic hedge]

if you got multiple separate "parts" i call it

[fierce lagoon]

Yeah, you take from all

[scenic hedge]

that are not compatible and separated by * then you take away from all of them

$\frac{7^{2}\cdot7^{2x}-3\cdot7^{2x}-2\cdot7^{2x}}{7^{1}\cdot2\cdot7^{2x}}$

[warm shale]

PoTeTo

[scenic hedge]

we start with this

[fierce lagoon]

Yeah

[scenic hedge]

$\frac{7^{2}-3-2}{7^{1}\cdot2}$

[warm shale]

PoTeTo

[scenic hedge]

uhmmm

I think its wrong

yeah its wrong

[fierce lagoon]

Nope

That is correct

Just simplify

[scenic hedge]

nope

its not

[fierce lagoon]

22/7

That's the same as 22/7

[scenic hedge]

anwser to this problem should be -50/7

[fierce lagoon]

Bruh what

[scenic hedge]

exactly

[fierce lagoon]

Oh I'm dumb, it's 49^(x+1)

I thought it said 49^x

Well then just multiply -3 by 49

[fierce lagoon]

Yeah just multiply the -3 by 49

I didn't see the + 1 in the exponent

[scenic hedge]

can we start from skratch?

[fierce lagoon]

Uh

I mean

It goes through the same process

[scenic hedge]

$\frac{7^{2}\cdot7^{2x}-3\cdot7^{2x}-2\cdot7^{2x}}{7^{1}\cdot2\cdot7^{2x}}$

[warm shale]

PoTeTo

[scenic hedge]

wait

[fierce lagoon]

Yeah so basically 49^(x+1) = 49 • 7^(2x)

[scenic hedge]

ill do it on paper and try again

[fierce lagoon]

I was blind and didn't se e the +1

But it goes through the same process

[scenic hedge]

7^2x+2 = 7^2x * 7^2

correct

?

[fierce lagoon]

Yeah

[fierce lagoon]

The work is right here

It's the same process

[scenic hedge]

I don't understand the 3(49)*7^2x

[fierce lagoon]

Yeah so

49^(x+1) = 49^x • 49

= (7^2)^x • 49

Which is 7^(2x) • 49

[scenic hedge]

oh

that makes sense now

[fierce lagoon]

Either way, do magic shit

-50:7

I'm assuming you're done, so I'll give you the list of stuff

The thing about latex is that you can save them as images

Separately, but they're images

But lemme give them very quickly:

$(A^a)^b= (A^b)^a = A^{ab}$

$A^{a+b} = A^a\cdot A^b$

[warm shale]

Umbraleviathan

Umbraleviathan

[fierce lagoon]

$a^A\cdot b^A = (ab)^A$

[warm shale]

Umbraleviathan

[scenic hedge]

i think i got it

[fierce lagoon]

Ye

Above your image are some properties

You can save them as images

There's one more

$(\frac{1}{a^b}) = (\frac{1}{a})^b$

[warm shale]

Umbraleviathan

[scenic hedge]

can we look at this one as well please?

[fierce lagoon]

I gotta go

But like

Maybe someone ale can

But it's gonna be the same thing

[scenic hedge]

@fierce lagoon just whats it diffent if its x+1 and x-1

[fierce lagoon]

Uh

x-1 will make it a^x • a^-1