help-28

[sharp vine]

i.e. you cannot simplify such an expression further

In other words, like terms are algebraic expressions that have identical literal (variable) parts and differ only in their coefficients

And you're allowed to do operations only on them

[bitter python]

ah okay

so x^2 * x is NOT 2x^2, but x^2 * x itself is the expression

[sharp vine]

yes

[bitter python]

aka x * x * x

[sharp vine]

x^2 * x is x^3

[bitter python]

yeah

ah

so if it's 3x^2 + x, you can't simply it more?

[sharp vine]

because if you multiply powers with a common base, you add the exponents, not the coefficients

[sharp vine]

Exactly

[bitter python]

And if it's the same one, you can do it ( x^2 + x^2 = 2x^2 )

[sharp vine]

Yeah, you got it

[bitter python]

multiplying would be additioning the exponents?

aka x^4

[sharp vine]

Correct

[bitter python]

what about subtraction

x^2 - x^4

[sharp vine]

same as in addition

you can't simplify this any further

[bitter python]

ah okay

And division would just be x^2 / x^4 = 1/x^2

[sharp vine]

Yes

[bitter python]

Hmm think I got it

Gotta practice more though, thank you :)

[full forum]

@bitter python Has your question been resolved?

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[hollow meadow]

what do i do from here

[pearl sparrow]

provide the question

[hollow meadow]

?

[serene hedge]

Give the question you are solving

[pearl sparrow]

of nvm

[hollow meadow]

.

[pearl sparrow]

give the question which ur solving

[serene hedge]

o

eijeurdbusbsuxiocncfjsnejeie

[pearl sparrow]

i understand but like its a mess, so if u dont mind sending it from a clear page

[hollow meadow]

sure one moment

here it is

i started by diving by 2 and then transforming them into the degrees thingy

[sharp vine]

$\frac{\sqrt{3}}{2}\sin x +\frac{1}{2} \cos x = \frac{\sqrt{2}}{2}$

[glossy valve]

Modus

[sharp vine]

You have a nice formula for such an expression

[hollow meadow]

which?

[serene hedge]

Da one you did above

[hollow meadow]

ahhh

[sharp vine]

alpha = x, beta = 30 deg

[hollow meadow]

wait i got cos60*cosx on the second thing

did i do sum wrong?

I turned the 1/2 into cos60

[sharp vine]

sin(30) = 1/2

[hollow meadow]

YOU'RE RIGHT

my bad

[sharp vine]

I mean

cos60 * cosx is right, but not useful here

[pearl sparrow]

okk

[hollow meadow]

is there anything i should do with the sin(30+x)?

[sharp vine]

unless you use this one

[sharp vine]

No, now it simplifies to sin(30+x) = sqrt(2)/2 which is simple I guess

[hollow meadow]

but my teacher did some stuff with like pi and k

.close

[full forum]

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[sharp vine]

Just radians

Put 30 = pi/6 and that's it

[full forum]

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[still current]

how can i find the solution for 7^(x+2) = 1/2401

[still current]

7^(x+2) = 2401^-1

7^(x+2) = (7^4)^-1

[onyx glen]

(7^4)^-1

these brackets are obligatory

anyway you're most of the way there in fact

[still current]

x+2 = -4

x = -6

[neon basin]

opal you really gotta be more careful with these 'small' things

<@&268886789983436800>

[still current]

like what

[neon basin]

x + 2 = -4, x = -2

small errors that can cost you on the test

[still current]

yeah thats my fault lol

btw how can i find the solution or solutions for ln (2x-9) < 5

[neon basin]

How would you solve it if it was =

[still current]

i would do log e^(2x-9) = log e^5

but i think thats wrong

[neon basin]

What did we learn yesterday

[still current]

ln is natural log

[neon basin]

$\log_a(b) = c \iff a^c = b$

[glossy valve]

USS-Enterprise

[neon basin]

Here a, the base, is just e

[still current]

log e^5 = (2x-9)

[neon basin]

why log_e?

Just the base

e

[still current]

like that right

[neon basin]

no

$\log_2(8) = 3 \iff 2^3 = 8$

[still current]

log (e)^5 - (2x-9)

[glossy valve]

USS-Enterprise

[neon basin]

$\ln(2x-9) = 5 \iff e^5 = 2x-9$

[glossy valve]

USS-Enterprise

[still current]

e^5 = (2x-9)

[neon basin]

yes

[still current]

oh alr

[neon basin]

that's all there is to it

[still current]

yeah

but how do i find

x

the solutions

cause its an interval ]?,?[

[neon basin]

Yes

First solve e^5 = 2x-9 for x

[still current]

(e^5 + 9)/2 = x

[neon basin]

Great

But we need to solve $\ln(2x-9) < 5$

[glossy valve]

USS-Enterprise

[neon basin]

And the same intuition follows

$\ln_a(b) < c \iff b < a^c$

(if the base is >1)

here it is

[glossy valve]

USS-Enterprise

[neon basin]

sorry I wrote it wrong

it's flipped

it's not a^c < b

[still current]

wait but what does it do

[neon basin]

but b < a^c

[still current]

b is 5 right

[neon basin]

Well we do the same thing as here

no, b = 2x-9

[still current]

oh yeah

alriht

[neon basin]

but it's flipped

$\log_a(b) = c \iff a^c = b$

[glossy valve]

USS-Enterprise

[neon basin]

here it doesn't matter if we write a^c = b OR b = a^c right

$\ln_a(b) < c \iff b < a^c$

[glossy valve]

USS-Enterprise

[neon basin]

But here (when base a > 1), b < a^c

not a^c < b

So we get

$2x-9 < e^5$

[glossy valve]

USS-Enterprise

[neon basin]

And you just solve for x

[still current]

oh the equality swiped sides?

[neon basin]

(you get the same as when we were solving = )

inequality

and it's not really swapping

we can look at this again

[still current]

ooh nvm

[neon basin]

$\log_2(8) = 3$

[glossy valve]

USS-Enterprise

[neon basin]

Okay, so 2^3 must = 8

[still current]

but we found it already no?

[neon basin]

Yes

that's why we did it first

[still current]

oh i see

[neon basin]

you just get x < (e^5 + 9)/2

[neon basin]

But, if we were saying

$\log_2(x) = 3$

[glossy valve]

USS-Enterprise

[neon basin]

Again, $x = 2^3$

[glossy valve]

USS-Enterprise

[neon basin]

But if we wanted to find $\log_2(x) < 3$

[glossy valve]

USS-Enterprise

[neon basin]

Then, logically, $x < 2^3$

[glossy valve]

USS-Enterprise

[neon basin]

right?

x must be less than 8, so that log_2(x) is less than 3 because log_2(8) = 3

[still current]

yeah

because log_2 (x) < 3

because when u do log_2(x) the number is smaller

[neon basin]

yes

anyway

[neon basin]

we got this

but this is the right 'barrier'

[still current]

wait

[neon basin]

we also need to look at $\ln(2x-9)$

[still current]

nvm

[glossy valve]

USS-Enterprise

[neon basin]

is it always defined

[still current]

no because its <

so it can be a series of numbers

[neon basin]

I mean just the logarithm itself

$\log(x)$

[glossy valve]

USS-Enterprise

[neon basin]

Is this defined for all $x$ in $\mbb{R}$

[still current]

it isnt

[glossy valve]

USS-Enterprise

[neon basin]

when is it defined?

[still current]

when its smaller than (9+e^5)/2

[neon basin]

yes, in our specific case, then ln(2x-9) is smaller than 5

But I am asking about the log itself

what is $\log(-10)$?

[glossy valve]

USS-Enterprise

[still current]

log_10(-10)

[neon basin]

yes

[still current]

log_10(x)

its defined

because its 1 number

[neon basin]

well but then tell me the value of log(-10)

[still current]

theres no value

[neon basin]

what do you mean?

[still current]

because it says math error

in my calculator

[neon basin]

yes exactly

[still current]

log(-10)

[neon basin]

We would be asking 10 to the power of what equals -10

$10^x = -10$

[glossy valve]

USS-Enterprise

[neon basin]

can you find such a number

[still current]

we cant

[neon basin]

exactly

Because logarithms are defined only when the argument (whatever is inside the logarithm) is strictly greater than 0

$\log(x)$

[glossy valve]

USS-Enterprise

[neon basin]

this log is defined only for $x>0$

[glossy valve]

USS-Enterprise

[neon basin]

we can pick x = 10, x = 4345, x = e^5, but we can't pick x = -34

So

[still current]

wait how do u know

[still current]

oh alright

ohh

[neon basin]

from this

[still current]

because log cant have a negative number

so its xer

it cant be -1 right

[neon basin]

It can't have a negative number inside the log

[still current]

yeah

[neon basin]

its value can be negative. try log(0.5)

[still current]

-0,30

[neon basin]

yes

[still current]

oh alr

[neon basin]

but log(-2) doesn't exist

So

When is $\ln(2x-9)$ defined

[glossy valve]

USS-Enterprise

[still current]

exactly so when we have to find solutions it cant have negative in the brackets

because its the x

[neon basin]

exactly

[still current]

it cant have -2

[neon basin]

yes

[still current]

alrighht

[still current]

when its x > 0

[neon basin]

yes

if x is 2x-9 in your mind

[still current]

x has to be bigger than 5

[neon basin]

how did you get that

[still current]

but theres ln

[still current]

because 2(4) - 9 = -1

2(5) - 9 = 1

x > 0

[neon basin]

how about 4.6

we aren't working with just integers

[still current]

oh alright

[neon basin]

2*4.6 - 9 = 9.2-9 = 0.2 >0

So $2x -9 > 0$

[glossy valve]

USS-Enterprise

[neon basin]

so $x > \frac{9}{2}$

[glossy valve]

USS-Enterprise

[neon basin]

right?

[still current]

yeah

[neon basin]

great

so now we have

[still current]

]9,2, + infinite[

[neon basin]

$x < \frac{9+e^5}{2}$

[glossy valve]

USS-Enterprise

[neon basin]

and $x > \frac{9}{2}$

[glossy valve]

USS-Enterprise

[neon basin]

no

[still current]

wait how did u replace the 5 by 0

oh nvm

nvmm

[neon basin]

you are mistaking two things

[still current]

cause

yeah

[neon basin]

one is solving the inequality

one is checking when the ln itself is defined

[still current]

yeah cause we said that 2x - 9 > 0

[neon basin]

Anyway we must find the intersection of this

[neon basin]

And this

Because x must suffice both conditions right

It must be smaller than (9+e^5)/2 so that ln(2x-9) is smaller than 5

And it must be greater than 9/2 so that ln(2x-9) itself is defined

that's why it must suffice both conditions at the same time

that's why we find the intersection

[still current]

oh

i got it

[neon basin]

$\left(x > \frac{9}{2}\right) \ \cap \ \left(x < \frac{9+e^5}{2}\right)$

[still current]

yu[

[glossy valve]

USS-Enterprise

[neon basin]

so $x \in \left(\frac{9}{2}, \frac{9+e^5}{2}\right)$

[glossy valve]

USS-Enterprise

[neon basin]

right?

I see you use ][

for open intervals

[still current]

[ ] is included and ] [ is excluded

[neon basin]

yeah

we use [ ] for included and ( ) for excluded

both are correct

[still current]

oh

[still current]

wait so

oh i see cause x has to be smaller than the fraction

so thats the max

and x has to be bigger than 9/2

so its the minimum

[neon basin]

yes, x must be larger than 9/2 and smaller than (9+e^5)/2

[still current]

that makes sense

[neon basin]

and we know (9+e^5)/2 is larger than 9/2

so it must be in between

[still current]

so if i recapitulate

it was ln(2x-9) < 5

[neon basin]

yes

[still current]

but when its something like this

im not able to think that this is a^b =c

because theres no exponent

[neon basin]

logarithms are inverses of exponents

you just have to remember

$\log_2(8) = 3$

[glossy valve]

USS-Enterprise

[still current]

oh this is a logarithmic right

[neon basin]

so, 8 = 2^3

[neon basin]

yes

and just apply it here

[still current]

so its the opposite of exponent

[neon basin]

8 is 2x-9; 2 is e; 3 is 5

and you simply get $2x-9 < e^5$

[glossy valve]

USS-Enterprise

[neon basin]

that's all

[still current]

e^5 > 2x-9

[neon basin]

yes, that's the same thing

[still current]

yeah

then e^5 + 9 > 2x

[neon basin]

if $m > p$, then $p < m$

[glossy valve]

USS-Enterprise

[still current]

(e^5 + 9)/2 > x

[neon basin]

( )

👍

[still current]

so this fraction is bigger than x

meaning its the max

or what

[neon basin]

it's not that "it is"

it must be

that's the condition we are setting

[still current]

yeah

[neon basin]

(e^5+9)/2 must be greater than x

or, flipped x must be smaller than (e^5+9)/2

that's why it's the max

[still current]

but now how did u make the 2nd equation

to find the smaller

[neon basin]

$\log(x)$

[glossy valve]

USS-Enterprise

[neon basin]

Is defined only for $x > 0$

[glossy valve]

USS-Enterprise

[still current]

oh yeah

[neon basin]

And we have $\ln(2x-9)$

[glossy valve]

USS-Enterprise

[neon basin]

which is defined when whatever is inside the ln is > 0

so 2x-9 > 0

x > 9/2

$\ln(x)$ is the inverse of $e^x$

[glossy valve]

USS-Enterprise

[neon basin]

And if you use a graphing tool to graph e^x

you will see that it is always positive - the graph is always above the x-axis

and, because logarithms are inverses of exponents

[still current]

but the thing is

[neon basin]

whatever are the outputs ("y values") of e^x will be the inputs of ln(x)

and since no y-value of the function e^x is negative

no input in ln(x) can be negative

[still current]

alright

[neon basin]

no

[still current]

log(x) > 0

[neon basin]

also no

[still current]

so 2x -9 > 0

[still current]

ah

[neon basin]

log(x) is negative many times

on the interval $x \in (0, 1)$

[still current]

no i meant like the (x)

[glossy valve]

USS-Enterprise

[still current]

mb

[neon basin]

Ah

yes

This is the graph of the function f(x) = e^x

We notice it is always positive (above the x-axis), right

And we know ln(x) is its inverse

So the outputs of e^x are the inputs of ln(x)

[still current]

yeah

[neon basin]

therefore, no inputs in ln(x) can be negative

that's why we must say x > 0

because we don't know what ln(-800) is

e^x never reaches the y-value -800

[still current]

yeah

[neon basin]

Well, that's it

So once you find these two inequalities

[still current]

tysm bro

[neon basin]

you just find the intersection of them

as we said x must suffice both at the same time

no problem!

[still current]

exactly

.close

[full forum]

Channel closed

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[still current]

Here's 2 points of a exponential function in which the asymptote is y = 4 : (-1,14) and (-3,254) Find the equation of the exponential function

[still current]

f(x) = a(c)^x + k

f(x) = a(c)^x + 4

how can i find a or c

[gritty rose]

plug in your points and get two equations with two unknowns a and c

[still current]

ok

is it f(x) = a*c^x + k or f(x) = ac^x + k

like ac are together

.close

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[knotty verge]

Hi! I've been struggling with my math homework because I'm finding these problems and fractions really hard to follow and I feel a little insane for thinking so. i admittedly ran the problem through chatgpt to see where im going wrong but i want to get a human perspective. any tips?

[knotty verge]

its just. i feel like the common denominator shouldnt be 35 in the third question. there are a lot of questions like this in my homework because the numbers are randomly generated and i usually have to generate a new one so i can get one that forms rational numbers/fractions that can simplify

i feel its uncommon to have such large denominators in remedial math but im not too sure

[full forum]

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).

[silk bridge]

its common to have big denominators sometime

[knotty verge]

okay thank you

ill avoid using it

[knotty verge]

but for every solution?

[silk bridge]

you can use ai for maths if you have some knowledge first else ai can prove highly unreliable

[silk bridge]

what are you supposed to do here isolate y ?

[knotty verge]

i have a kind of decent understanding of this unit if i get a brief reminder beforehand

[knotty verge]

yes

the goal is to simplify as much as possible

[silk bridge]

hm.. do they allow to keep fractions ? else this can be simplified more
but if aim is to isolate y ig this should be it

[knotty verge]

they do allow you to keep fractions

in fact its required with this software if thats how the question is formatted

[silk bridge]

ok

this works then

[knotty verge]

alright so my initial answer to that question was correct

thank you!

[silk bridge]

np

you can type .close to close this channel

[knotty verge]

.close

[full forum]

Channel closed

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[frigid carbon]

A circle x² + y² + 2gx + 2fy + c = 0. AB is a variable chord subtending a right-angle at the origin O. P is a point lying on AB. OP is perpendicular to AB. what curve is the locus of P

[full forum]

@frigid carbon Has your question been resolved?

[frigid carbon]

<@&286206848099549185>

[tall onyx]

The circle is fixed right?

[frigid carbon]

what do you mean?

[tall onyx]

I remember doin this problem once hang on

[frigid carbon]

oh, it is a pyq hence

[tall onyx]

Welp

[frigid carbon]

oh, thx

too long chats but i'll read

[sour shore]

Lol tht was my fault

[tall onyx]

You can scroll down, my method is pure synthetic geo

[frigid carbon]

lol

[tall onyx]

but you probably need what above mine

[frigid carbon]

ok, i'll look but it seems to have vectors which i'm not good at

ty

.close

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[thick hedge]

I'm just wondering if the subset sign is the wrong way round

[rapid rain]

it's not

otherwise the union is trivially I_1

[thick hedge]

Okay, lemme try it then

[rapid rain]

and there would be nothing to say

[thick hedge]

yea, which is why I was hoping it's a typo 🙏

Oh, that's easy.

thanks

.close

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[undone arch]

Can someone check my work?

[delicate torrent]

hold on

[tall onyx]

the first line, it should have been -2-1/x

[undone arch]

switch 1 over

[delicate torrent]

yeah

[tall onyx]

oh

,w expand (6x^2-3x-1)^2

[delicate torrent]

,w calc (1-sqrt(5))/2 + 1

[glossy valve]

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[tall onyx]

,w factorize (36x^4-39x^3-4x^2+6x+1)

[glossy valve]

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[tall onyx]

Okay looks good

[undone arch]

,w 6x - (2x+1)/x=1+sqrt(3x+1)

[glossy valve]

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Requested by thomas2406

[silk bridge]

,w 6x - (2x+1)/x=1+sqrt(3x+1)

[tall onyx]

Tho you have to restrict value for x

since you're dealing with sqrt

3x+1>=0

[delicate torrent]

and x =/= 0

[silk bridge]

eh which value violates

[tall onyx]

idk, it's just a reminder

let me see

[delicate torrent]

no values violate

i checked

but yes, this is just a reminder

[silk bridge]

[tall onyx]

-1/4 and (1+sqrt(5))/6

[silk bridge]

then wolfram trippin

[tall onyx]

6x-3-1/x >0

[delicate torrent]

wha?

[tall onyx]

,w 6x-3-1/x >0

[glossy valve]

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Requested by fionna_fx

[tall onyx]

keep in mind $6x-3-\frac1x=\sqrt{3x+1}$

[glossy valve]

Fionna

[tall onyx]

the RHS>0

hence so is the lhs

[delicate torrent]

ooh

$\geq$

[glossy valve]

1 divided by 0 equals Infinity

[silk bridge]

oh

[delicate torrent]

so we gotta test our values right?

[silk bridge]

,calc (3 - sqrt(33))/12

[tall onyx]

the best way is just plug the value into the original equation and see which one is valid

[tall onyx]

Yeah

[undone arch]

,w 6x - (2x+1)/x-1-sqrt(3x+1) with x = -0.25

[glossy valve]

Results provided by WolframAlpha

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Requested by thomas2406

Result:

-0.22871355387817

[delicate torrent]

so anyways

after you got your x

if you do smth like squaring both sides

that does not work 2 ways

so you have to test the values

[undone arch]

oh ok

[silk bridge]

always shit like this when squaring

[undone arch]

thanks yall

.close

[full forum]

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[undone arch]

.reopen

[full forum]

✅ Original question: #help-28 message

[undone arch]

nvm i got more problems

[silk bridge]

that was quick

[undone arch]

i just realized

[delicate torrent]

?

[undone arch]

there's a back side to my homework

With whole a, b, we have p = a^3 + b^3 - 3ab + 1 which is prime. Find the largest p possible.

[delicate torrent]

$a^3 + b^3 - 3ab + 1$?

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

oof

think

it's a congruency exercise for sure

[undone arch]

<@&286206848099549185>

,w factor a^3 + b^3 -3ab + 1

[glossy valve]

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Requested by thomas2406

[delicate torrent]

okay

that's good here

[undone arch]

we know that a+b+1 is always > 1

[delicate torrent]

yep

so i think the only way this can be prime

is when $a^3 + b^3 - 3ab + 1 = a + b + 1$

[glossy valve]

1 divided by 0 equals Infinity

[undone arch]

or $a^2 + b^2 + 1 - ab - a - b$ is prime

[glossy valve]

Thomas

[delicate torrent]

well i don't think so

[delicate torrent]

i think this is more accurate

[undone arch]

so did you just not think abt tht or do you disagree w me?

[delicate torrent]

im not sure about that

im not disagreeing

you said that $a + b + 1 > $

[visual nebula]

I think we can assume from that, that $$a^2 - ab - a + b^2 - b +1 = 1$ or easier: $$a^2 - ab - a + b^2 - b = 0$

[glossy valve]

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

[delicate torrent]

a + b + 1 > 1 and if the other factor is prime, then the whole product is composite

[undone arch]

ohh wait tht must be 1 not prime

i mixed up

[delicate torrent]

$a^3 + b^3 - 3ab = a + b$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

,w factorize a^3 + b^3 - 3ab - a - b

[glossy valve]

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@undone arch Has your question been resolved?

[undone arch]

can't think of anything

.close

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[nova bolt]

Does anyone have maps of all the theorems in calculus 1?

[full forum]

Please don't occupy multiple help channels.

[tall onyx]

don't occupy multiple help channels please

[charred carbon]

.close

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[warped frost]

I was so close to instantly modpinging cuz I thought it was a spam bot

[tall onyx]

[loud minnow]

Not a spam bot

[warped frost]

problem is, I saw OP claim two channels in short succession

[loud minnow]

She is new to disc

[warped frost]

sure, but I've seen the typical behaviour of spam bots, so.

[silk bridge]

crazy bringing out the green one to close

multitasking

[tall onyx]

well yeah I can't close with w/o green

[tall onyx]

only takes me a min to change to that account so why not

[silk bridge]

fair

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[sturdy oriole]

hi there.

[sturdy oriole]

interesting one. I have found 80° with using trigonometry, but i am not sure about my answer. Is there any way to solve that without using trigonometry?

[raw heart]

want help?

[sturdy oriole]

yep, please :)

i tried something, such as drawing a circle around the triangle, but couldn't continue :d

may be, we have to construct a equilateral triangle on the exterior.

[quick thicket]

Does the property that the sum of all angles within a triangle equals 180° counts as using trigonometry in your class?

[zenith pilot]

Trigonometry involves the trigonometric ratios. The triangle angle sum property doesn't, and so it is not trigonometric.

[sick vortex]

i think this works actually

where did you stop?

[quick thicket]

Ok, well then you could name the angles as follows:
α = ADC
β =BAD
υ = DAC

Then those formulas follow from the angle-sum-property:

1. 40° + x + β = 180°
2. 30° + γ + α = 180°
3. x + α = 180°
4. 40° + 30° + (α + β) = 180°
   That are four linear equations with unkown variables, wich means you can solve that as a system of linear equations.

[sturdy oriole]

i meant without using the sine rule.

[tough salmon]

why would you wanna NOT use trig tho?

[grave elm]

if only they were linearly independent, which they most likely aren't

[tough salmon]

trig is goated

[sturdy oriole]

i wanna dive into another world.

[tough salmon]

i mean u cud use some weird ass geometry for that idk

[quick thicket]

Are they not linearly independent?

[grave elm]

oh, seems like they are. Weird (this usually doesnt work)

[sturdy oriole]

well, actually didn't do lots of steps. I mean there were several things on my paper, but most of them were useless.

[tough salmon]

they are

yeah thats a good solution ig

[grave elm]

2. is practically useless, since the other 3 contain no gamma. Then we can sum 1 and 3 to get 2x + a + b = 360 - 70
   2x = 360 - 70 - (180 - 70) = 360 - 180
   x = 90
   but its pretty sus

[sick vortex]

isn't x 80

[sturdy oriole]

90?

[grave elm]

yeah, seems like one of these is wrong

[sturdy oriole]

[grave elm]

which makes sense. It shouldnt be solvable without using the equal lengths property

[tough salmon]

wait yeah ig this overlooks something

also if this worked they wudnt have given tose lengths equal

[sturdy oriole]

don't know the correct answer.

[tough salmon]

meow

[grave elm]

the last one is wrong (should be gamma, not alpha)
And fixing that makes it linearly dependent

[sick vortex]

are they linearly independent tho? Isn't the angle sum of the big triangle a libnear combination of the angle sum of the left and right triangles

[sturdy oriole]

i guess the correct answer is here, @tough salmon .

[tough salmon]

i would solve it but im too lazy to walk all the way to my whiteboard

wait i have paint nvm

[zenith pilot]

It should be 80 degs.

Draw the circumcircle of ABC. Let the centre of the circle be O. Let AB = DC = x.
R = AB/2sin(angle ACB) = x/1 = x
Therefore, the triangle OAB is equilateral. Hence angle ABO = 60 deg and thus angle FBO = 20 deg.
Angle ADB = angle DBO + angle DOB = 60 deg + 20 deg = 80 deg

wait

a minute

Just ignore my work.

[sturdy oriole]

[zenith pilot]

The same notation, different diagram. DC = AB = the radius of the circle.

As found earlier, angle CBO = 20 deg. Since OB = OC, angle BCO = 20 deg. Now DC = OC = radius. So angle CDO = angle DOC = 90 - angle C / 2 = 90 - 20/2 = 80 deg
Hence angle DOC = 80 deg. But angle AOC = 2 * angle ABC = 2 * 40 = 80 deg.
Hence angle DOC = angle AOC. So OD must lie on AO. But since D also lies on BC, the point of intersection of AO and BC is concluded to be D. Hence the earlier diagram was correct indeed even though assumptions were made. Now you can find the angle ADB easily.

[full forum]

@sturdy oriole Has your question been resolved?

[sturdy oriole]

thanks for your effort, imma gonna analyse it rn.

[sturdy oriole]

FBO?

probably you mean CBO here :)

@zenith pilot liked your solution.

but i guess, we ate the prohibited rule here, we used the sine rule also in the circle :D

[zenith pilot]

Well

You don't need the sine rule. angle AOB = 2 * angle ACB = 2 * 30 = 60 deg. OB = OA. So the triangle AOB is equilateral.

[sturdy oriole]

ah.

appreciate your work 🙏.

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[lofty totem]

hi guys

[lofty totem]

can somebody help me do my homework?

i dont understand it

[tall valve]

whats 2 squared

[lofty totem]

this is my homework

[void nova]

4! / 6

[tall valve]

ty bro

[void nova]

Don't troll though

[lofty totem]

guys the teachers coming

[tall valve]

i wont

[lofty totem]

to collect it

i need hel[

[void nova]

<@&268886789983436800>

[true hill]

welp bros gone

idk who did that, maybe himself

.close

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[lofty totem]

no im not

[true hill]

oh that was just an error? love discord

well dont troll

[devout valley]

We'll see about that

[void nova]

This means your age is (very) below 13, right?! @rquinn44

Oh already banned

Nice!

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[verbal sky]

hey

[verbal sky]

idk where i went wrong

i got x^2 + 4y^2 = 12, but i'm supposed to get (x+y)^2 + ay^2 = b

the problem is that (x+y)^2 produces a 2xy term, which i dont have in my x^2 + 4y^2 = 12

this is the correct answer

[verbal sky]

yeah i see it

[stiff musk]

where's the cross term that results from squaring the right hand side of the top one?

[verbal sky]

idk why i did that

[stiff musk]

it even has a name

https://en.wikipedia.org/wiki/Freshman's_dream

[gritty rose]

,tex .freshman

[glossy valve]

riemann

[gritty rose]

[verbal sky]

ty guys

.close

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[stiff musk]

ooh nice

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[cinder frost]

Prove or give a counterexample: If 𝑉 1, 𝑉 2, 𝑈 are subspaces of 𝑉 such that
𝑉 = 𝑉 _1 ⊕ 𝑈 and 𝑉 = 𝑉_2 ⊕ 𝑈, then 𝑉_1 = 𝑉_2

[cinder frost]

This is tough

I am not sure if V_1 and V_2 can be the same vector space

supposing V = F^2

[glossy valve]

andre

[lime trellis]

V is over F^2 or V=F^2?

F^2 is not a field

[cinder frost]

it is a vector space

over F

[lime trellis]

okay

[cinder frost]

which is assumed to be a field in my textbook

[lime trellis]

consider U = {1,1}

[cinder frost]

Yes

[lime trellis]

check if its in direct sums with span(0,1) and span(1,0) as you proposed

[cinder frost]

it isn't, @lime trellis

[lime trellis]

why?

sorry i meant U=span(1,1)

[cinder frost]

There are direct sums

[lime trellis]

yes

so we found V1, V2, and U that are subspaces of F^2 such that F^2 = V1 \oplus U and F^2 = V2 \oplus U but V1 ≠ V2

[cinder frost]

It's not clear to
me what I was supposed to look out for

What does all this mean

V = V1 oplus U and V = V2 oplus U

I don't think I know what a span is

[lime trellis]

span is just all linear combinations generated by the elements

[cinder frost]

okay

[lime trellis]

oplus is just the direct sum notation

V1 + U = V such that V1 intersect U = {0}

[cinder frost]

Yes

How is V1 not equal to V2

Can't U just be {(0,0)}?

[lime trellis]

V1 and V2 can be equal to each other

but they need not be

the question asks is it always true that V1 and V2 are equal in such conditions

[cinder frost]

OH

Okay, thank you

😓

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[cold sluice]

Can I get help finding the coordinates of point d?

[brittle sun]

Are those lines medians?

[cold sluice]

Yes

no

i mean

Yea it's given, my bad. I was wondering if it was given or not

[brittle sun]

There is a very quick formula to find the coordinates of a midpoint

[cold sluice]

I haven't learned it yet

please don't tell me it

[brittle sun]

Oof

[cold sluice]

Ty

[brittle sun]

Sorry

[cold sluice]

Np

[brittle sun]

Well do you know how to calculate distance

[cold sluice]

My strategy is that I know the distance between the x and y components of line segment ba

the x length is a-b, and the y length is b-a

[brittle sun]

Huh how did c appear in the x length

[cold sluice]

Oops!

[brittle sun]

Gonna need you to label the points you're talking about

[cold sluice]

i meant to say b

[cold sluice]

It's already labeled

[brittle sun]

ehhh technically they're labelled by coords but fair

[brittle sun]

The y length is wrong still

[cold sluice]

How?

I suppose here the length of the change in y is just b because a sits on the x axis?

Oh sorry, it's supposed to be c-0

right?

Yeah it's c-0

[brittle sun]

Yes

[cold sluice]

Now the ratio of the change in x to the ratio of cd to ca is (a-c)*(cd:ca)?

If I can find the ratios, I think I can find the coordinates

The change in x is just a-c

or c-a, it doesn't matter

No it's b

i'm so confused

i should've labeled stuff either way 🙁

Thanks for trying to help! I gtg

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[vivid dagger]

How to do this without using the tangential argument which turns the dot product into multiplication of magnitude?

[umbral dome]

if you really want you can give B an explicit coordinate representation as B = B(unit tangent vector)

[vivid dagger]

-y i + x j?

wait it must be normalized no?

so just divide it by sqrt(x^2 + y^2)

this is the field that's tangent to any circle in ccw?

[umbral dome]

yeah

[vivid dagger]

wait, shouldn't the magnetic field is weaker when you're farther away?

this is magnitude 1 everywhere

farther away from the current i meant

which is a straight line in this case

[umbral dome]

the magnitude is B

magnetic field B = B(unit vector you just came up with)

[vivid dagger]

ohh okay, i'll try that

brb

[full forum]

@vivid dagger Has your question been resolved?

[vivid dagger]

Srry for the long wait

I can't just integrate the B

my opinion (might be wrong): The problem should've said B is tangential and equal in magnitude along any arbitrary circle

[vivid dagger]

in fact, in this approach i did assume B has a constant magnitude

[vivid dagger]

it should be - sin t instead of sin t i miswrite (doesn't change the answer)

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@vivid dagger Has your question been resolved?

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@vivid dagger Has your question been resolved?

[sleek mulch]

as the question already mentions that B is along the tangent of a circle of radius r, and since a circle is rotationally symmetrical about its center, thus it is appropriate to conclude that the magnitude of B is constant for a particular radius r

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@hallow tartan Has your question been resolved?

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@hallow tartan Has your question been resolved?

[hallow tartan]

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[tall walrus]

Here at question e, it says whats the limit as x approaches 0, now the graph says that theres a solid dot at (0,0), does that mean the limit DNE or is simply just 0? Can both a defined function point and limit exists at the exact same place?

[ancient timber]

The existence of the function at a point has no bearing on the existence of the limit at that point

The only thing that matters is what is happening around that point

[tall walrus]

So they can just coexist?

[ancient timber]

Yes

[tall walrus]

Oh i see thanks

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[hallow walrus]

For $a,b,c,\in\bR$ show that [ \iiint_R xe^{ax+by+cz} \dd{V} = \frac{4\pi a}{\abs{\mathbf a}^5} \big((3+\abs{\mathbf a}^2)\sinh\abs{\mathbf a} - 3\abs{\mathbf a}\cosh\abs{\mathbf a}\big) ] where $R$ is the unit sphere and $\mathbf a = (a, b, c)$.

[glossy valve]

kheer257

[hallow walrus]

Not sure how to go about this, I know we need to use the rotational symmetry somehow but the x term in the integrand is messing things up

If I rotate the axes so that one of them points in the direction of a then the exponential becomes e^{|a| z} but then idk what happens to the x term

[sour shore]

Ahh my bad

[elfin stream]

What. the. fuck.

split and integrate HEHE

[hallow walrus]

that would take far too long

I think I've figured it out, you just take the unit vector in the direction of a and extend it to an orthonormal basis and use that as your coordinate change

and only the coefficient of the first vector in x matters since the rest cancel out by symmetry

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[pseudo roost]

Can someone help me with the k method

[white badge]

what would that be?

[sturdy valve]

🤔

[pseudo roost]

Its wrong

Idk what to do

[static remnant]

um, the middle term is incorrect.

[sturdy valve]

how did u write x^(1/4) = sqrt(x) ^2

[onyx glen]

your factorization is incorrect

[onyx glen]

and also this

[pseudo roost]

Oh

[onyx glen]

it sounds like you wrote all this down without thinking

[pseudo roost]

I did think about

[white badge]

perhaps it's worth slowing down to check every line and ensure each line follows from the previous.

[torn jolt]

$x^{\frac{1}{2}} = (x^{\frac{1}{4}})^2, k=x^{\frac{1}{4}}$ should be what you are intended with k method

[glossy valve]

Hi

[torn jolt]

do i always have to do \frac{}{] for fraction

[onyx glen]

yes

[white badge]

unless you have a preamble that does away with that.

[torn jolt]

ah ic

[pseudo roost]

I don’t understand

[white badge]

you put the square on the wrong term.

you expressed 1/4 as 2(1/2), which is incorrect.

you should have expressed 1/2 as 2(1/4) instead.

[pseudo roost]

Wht though

I don’t understand the whole process

[white badge]

look at the second line of your working.

[pseudo roost]

How do Ik which to use

[white badge]

1/2 is exactly 2(1/4), so.

2(1/2) becomes 1, not 1/4, so that's totally out of the question.

[pseudo roost]

I d sad till dont get it

[white badge]

how did red become green?

[pseudo roost]

But it gives 1/4

When I square

[onyx glen]

$(x^{1/2})^2 = x^{\frac{1}{2} \times 2} \neq x^{(1/2)^2}$

[glossy valve]

Ann

[white badge]

$(a^b)^c = a^bc \neq a^{(b^c)}$!

[torn jolt]

sniped

[glossy valve]

Nicole

[onyx glen]

$(a^b)^c = a^{bc} \neq a^{(b^c)}$ !!

[glossy valve]

Ann

[white badge]

oops, my bad.

[pseudo roost]

Oh ok

Ok

*ok

Wait but how do I continue

[white badge]

what do you have right now?

[pseudo roost]

Do I do this before saying let k bc the number

[white badge]

wait. from x^(1/2) to x^2?

[pseudo roost]

Oh

[white badge]

how did that happen?

[pseudo roost]

Mb

I didn’t write the full thing

[white badge]

also, why did you arbitrarily square the x^(1/4)?

you're supposed to turn the x^(1/2) into (x^(1/4))^2 so that you can use your k-substitution here.

[pseudo roost]

I do t understand

[onyx glen]

you can't magically transmogrify x^(1/4) into (x^(1/4))^2

that ^2 appeared out of nowhere. strike it out.

erase it.

[pseudo roost]

Ok

[onyx glen]

also x^(1/2) should not have become x^2.

the correct next step would be: $(x^{1/4})^2 - 3(x^{1/4}) + 2 = 0$ and after this you put $k := x^{1/4}$

[glossy valve]

Ann

[torn jolt]

-# lmao never heard that word

[pseudo roost]

Oh ok

Bus how dys u do Tht

[white badge]

by realizing that 1/2 = 2(1/4).

[pseudo roost]

Oh ok

But why did she change both x’s

[delicate torrent]

transmogrify 😭

[delicate torrent]

hold on

[white badge]

Ann only changed the first x.

the second x, she merely bracketed it.

[pseudo roost]

Huh

[delicate torrent]

wait wait

[white badge]

this helps with identifying what you need to substitute for k later.

[delicate torrent]

so the original problem was x^(1/2) - 3x^(1/4) + 2 = 0 right?

[onyx glen]

i didnt change shit

i REWROTE them using actual valid/legitimate index laws

and yeah the second term just got dressed in a pair of brackets that's all

[delicate torrent]

the k method that you say here is the substitution method

you define another variable to make calculations easier

[onyx glen]

i mean, any letter can be used for the new variable name, but since OP is clearly struggling a lot and seems to cling on to k as the One Correct Name, then maybe we can accommodate that.

[delicate torrent]

Ann realizes that your equation can be turned into a quadratic when $k = x^{\frac14}$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

okay

[pseudo roost]

Ok

[white badge]

I think most of OP's confusion revolves around not getting 1/2 = 2(1/4), if I'm assessing this correctly?

[pseudo roost]

Do we make one the same as the other

[white badge]

please clarify what you mean.

[delicate torrent]

idk if i can make some golden rules for this lol

[white badge]

do you mean make the two x have the same exponent?

[pseudo roost]

Do we make them have the same fraction

But one can be trurned it o it’s former

[white badge]

same exponent, but yes.

[pseudo roost]

So we’re making the first term 1/4

[delicate torrent]

first term into $\left(x^\frac14\right)^2$

[glossy valve]

1 divided by 0 equals Infinity

[white badge]

yes, but remember, we cannot do that by just changing the exponent directly.

[pseudo roost]

Are the rules for this please

[delicate torrent]

second term into $3\left(x^\frac14\right)$

[glossy valve]

1 divided by 0 equals Infinity

[pseudo roost]

Oh

[delicate torrent]

,exponent rules

oh come on

[white badge]

here you are using the power of a power rule.

[delicate torrent]

texit didn't summon

[white badge]

because it's ,tex .exp rules.

unless you have a different preamble.

[delicate torrent]

,tex .exp rules

[pseudo roost]

Ok

[white badge]

either way I sent it already.

[delicate torrent]

really?

[white badge]

unless you have a different preamble.

I am going based off the preamble I see people using that command use.

[delicate torrent]

i guess i can't summon rules through the bot now

[white badge]

if you want to talk about that, this is not the right place to, I don't think.

but if you are that keen on helping, I'll step back.

[delicate torrent]

in this case, you can use !redir, i'll stay back for you to help

[white badge]

no. you take over.

[sour shore]

-# nice punctuation work.

[delicate torrent]

anyways how's your work doing? show us your work please

[pseudo roost]

I kinda understand

[delicate torrent]

now replace each $x^\frac14$ into $k$

[glossy valve]

1 divided by 0 equals Infinity

[pseudo roost]

Ok

Wai to Ddont get the tik

Rule

What rule would be applied here again pls

[delicate torrent]

.

[white badge]

what do you have at the moment?

[delicate torrent]

to apply into what?

[pseudo roost]

16 and 1

As x

[white badge]

huh?

[pseudo roost]

Idk the rule that applies here

[white badge]

actually sorry, I'll step back.