help-13

[deft gull]

ohh nice i haven’t taken that yet

i am in calc bc rn

no problem!

nice!

[mossy mango]

@crimson sedge gj

.close

[cedar kiln]

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[deft gull]

Let $ABC$ be a triangle such that $AB=13$, $BC=14$, and $CA=15$. Let $D$ be the point on $BC$ such that $AD\perp BC$. Let $\omega$ be the circle through points $A$ and $D$ such that $\omega$ is tangent to line $AB$, and let $O$ be the center of $\omega$. The line through $D$ tangent to $\omega$ intersects line $AC$ at point $P$, and $OP$ intersects $\omega$ at $X$, where $X$ lies between $O$ and $P$. Find $\frac{PX}{OX}$.

[wraith dagger]

RetroTurtle

[deft gull]

does anyone have a non coordbash non trig solution to this?

i am interested in finding one

[cedar kiln]

@deft gull Has your question been resolved?

[static edge]

seems like an olympiad geometry question

ill try later when i have time ._.

[deft gull]

i think it is around early AIME or late AMC level

[silent finch]

If the circle passes through A, it cannot be tangent to AB

[static edge]

it can tho

[silent finch]

Am I being dumb?

[deft gull]

it’s tangent to line AB at point A

[silent finch]

I was being dumb, lemme try it

Btw this best fits in #competition-math

[deft gull]

ohh sorry i didn’t know that exists

[static edge]

very inaccurate diagram but this should show it

[deft gull]

sorry i just joined today lol i’ll make sure to look there next time 😄

[static edge]

it's fine lol

[cedar kiln]

@deft gull Has your question been resolved?

[deft gull]

<@&286206848099549185> any ideas?

[dawn sand]

hmmmm

whats the ques

[deft gull]

oops wrong copy paste 💀

[dawn sand]

im in a class rn i will help u jus send me ques

[deft gull]

Let $ABC$ be a triangle such that $AB=13$, $BC=14$, and $CA=15$. Let $D$ be the point on $BC$ such that $AD\perp BC$. Let $\omega$ be the circle through points $A$ and $D$ such that $\omega$ is tangent to line $AB$, and let $O$ be the center of $\omega$. The line through $D$ tangent to $\omega$ intersects line $AC$ at point $P$, and $OP$ intersects $\omega$ at $X$, where $X$ lies between $O$ and $P$. Find $\frac{PX}{OX}$.

alright that’s fine

[wraith dagger]

RetroTurtle

[deft gull]

looking specifically for non coordbash non trig soltuion

[dawn sand]

non trig

idts non trig is possible-

well im a 10th grader idk if i can solve this

ima give it a shot tho

[static edge]

non trig is probably possible, there are many bizarre theorems out there that would be useful

but the prob is that i kinda suck at olympiad geo ;-;

[deft gull]

do you think menelaus would be useful???

bc there are a lot of lines you can probably use it for

but idk if it would work

or be useful

[solemn perch]

Hi

[deft gull]

hello

[solemn perch]

How are you

[deft gull]

I am doing well. Do we know each other?

[solemn perch]

No.. but we can introduce each other now

[deft gull]

sure, but i think we should focus on the math question here and casual chat in another place

[cedar kiln]

@deft gull Has your question been resolved?

[static edge]

idk, i tried angle chasing and found pretty much nothing

i mean i know AD=12 by heron's formula

[static edge]

wait i think i got it

[static edge]

took a while but it requires a couple of similar triangles and the use of power of point

i might have made some mistakes lol

[cedar kiln]

@deft gull Has your question been resolved?

[latent bloom]

Wow your handwriting is awesome!

[cedar kiln]

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[austere maple]

what do i do when my variable is negative

[austere maple]

for e.x,

-x = 3

[frosty thicket]

multiply both sides of the equation by negative one?

[latent bloom]

[austere maple]

u sure?

[frosty thicket]

yeah why not

[austere maple]

then its -3

[frosty thicket]

yeah

[latent bloom]

Yeah it's fine if x=-3

[austere maple]

alr tysm

.close

[cedar kiln]

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[bright agate]

great question

[cedar kiln]

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[tropic oxide]

what stage are you on?

1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[split crest]

Explain the outside angles as 180- the vslue

Then use the knowledge alfa beta gama is 180 inside

[marble mica]

i have not star

started

[split crest]

And u have linear equation with 1 unkown and calculate it

[tropic oxide]

@split crest your wording could be better, to be honest

[marble mica]

1

[tropic oxide]

@marble mica do you understand Fate's suggestion on how to begin, or should i say it in my words?

[marble mica]

you say it please

[tropic oxide]

ok

you know that the sum of all interior angles of any triangle is 180°, yes?

[marble mica]

ye

[tropic oxide]

do you understand how to express each interior angle in your triangle in terms of x?

[marble mica]

not sure

idk what you mean

[tropic oxide]

ok, let me recreate your diagram

do you see the angles i've marked with question marks?

[marble mica]

yea

[tropic oxide]

i want you to calculate these angles in terms of x. do you understand what to do, and do you understand how to do it?

[marble mica]

i can try

[tropic oxide]

try it, then ping me when you're done with it or are stuck

[marble mica]

wait can i see your way of doing it

[tropic oxide]

i'd like you to try it yourself.

i don't want to just give you any answers here.

[marble mica]

oh okay

180-(x+3) and 180-(3x+2)

[tropic oxide]

correct

these can be simplified, but it's fine.

so now you have three interior angles for your triangle

35, 180-(x+3), and 180-(3x+2)

can you write down an equation which says these three expressions add up to 180?

[marble mica]

35+(180-(x+3))+(180-(3x+2)) = 180

[tropic oxide]

good

now, are you able to solve this equation for x?

[marble mica]

i can try

[tropic oxide]

ok, try it and report back

[cedar kiln]

@marble mica Has your question been resolved?

[marble mica]

so i did

52.5

siorry late reply

[cedar kiln]

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[crimson sedge]

A marble is drawn from a bag of 15 marbles and replaced 60 times. How many times do you expect a blue marble to be drawn?

[crimson sedge]

There are 10 blue marbles btw

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[crimson sedge]

I know that the determinant of this is 5

How can I calculate the determinant of this?

[earnest socket]

use the property that the determinant of the transpose is unchanged

and then:

[crimson sedge]

Got it, thanks! didnt know you could transpose without any effect.

[earnest socket]

you can see it because you can use a cofactor expansion along either rows or columns

[crimson sedge]

oh i usually just use sarrus or gaussian elimination to calculate the determinant, maybe I should learn it

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• Show your work, and if possible, explain where you are stuck.
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[crimson sedge]

A marble is drawn from a bag of 15 marbles and replaced 60 times. How many times do you expect a blue marble to be drawn?

[crimson sedge]

There are 10 blue marbles btw

[feral juniper]

so what is your idea on how to start

Sometimes it helps to simplify these questions as much as possible first:

What is the expected probability to draw a blue marble if you only draw a marble once

[crimson sedge]

i dont know how to solve it

thats the thing

if were to only draw once though

it would have the probability of

2/3

since there are 15 marbles in the bag and 10 of em are blue

i meant blue

[feral juniper]

yeah thats correct

so you want to calculate the probability for drawing 60 times?

[crimson sedge]

mhm

[feral juniper]

Lets say you want to draw 2 times would you know how to calculate that

[crimson sedge]

umm

15+15?

which equals to 20/30

and uh

yes that

if

im right

[feral juniper]

Im just asking if this is clear to you

What does it mean for a probability to be expected?

[crimson sedge]

the likely hood of something?

[feral juniper]

It means the most likely (most probably) outcome

If you throw a dice 100 times
On Average you would expect to throw a 6 100/6 times

But ofc you could also throw a 6 100 times in a row or 0 times

Its just not very likely (not expected)

If that's understandable

[crimson sedge]

yes

[feral juniper]

so if you throw a dice 10 times
How many times would you expect to throw a 3

With that you can understand what to do on your original question

[crimson sedge]

um

let me thing

think

1/6 = 1 time

10/60?

[feral juniper]

10/60 is the same as 1/6

[crimson sedge]

then idk

how

this is the part where im making mistakes

cuz idk

[feral juniper]

I can simplify the question even further

If you throw a coin 2 times
How many times do you expect to throw tails

[crimson sedge]

2 times?

2/12?

[feral juniper]

I mean throwing a coin with 50% probability of throwing heads and 50% of throwing tails

[crimson sedge]

i need help not simplification and clairification

oh

umm

UMM

[feral juniper]

$\frac{1}{6} = \frac{1}{6} \cdot 1 = \frac{1}{6} \cdot \frac{a}{a} = \frac{1}{6} \cdot \frac{10}{10} = \frac{10}{60}$

[wraith dagger]

tobi

[crimson sedge]

why did it suddenly turn into 10/10

and also

is this for the dice?

[feral juniper]

yeah it just took me a while to write it wanted to show that 1/6 = 10/60

[crimson sedge]

oh yea

[feral juniper]

if you understand that then you know the solution for your problem immediately

[crimson sedge]

uhm

so

just

2/3?

thats literally when u simplify

10/15 is twice the time

which equals to 2/3

that..uhm

and even if i did it for example picked the blue marbles twice

it would be the probability of 20/30

[feral juniper]

Can we for a second go back to the coin example Ill explain it shortly if you don't know how to solve it (the coin example)

[crimson sedge]

which is exactly the same as 2/3

[crimson sedge]

yea

idk how to solve it

[feral juniper]

Throwing a Coin twice with 50% heads and 50% tails

[crimson sedge]

uhm yes?

[feral juniper]

If you throw once How likely is it get Heads once

[feral juniper]

I hope these questions are not annoying you but they are important for understanding this topic

[crimson sedge]

50%

[feral juniper]

50% is the same as 0.5

[crimson sedge]

yes

[feral juniper]

if you throw twice you need to multiply

2 * 0.5

[crimson sedge]

oh uhm

1?

which is 100%

[feral juniper]

that wouldn't be a percentage anymore

[crimson sedge]

oh..

[feral juniper]

I will show you why

sec

[crimson sedge]

okay

[crimson sedge]

yea uh

[feral juniper]

those are all the possible outcomes for flipping a coin twice if you add up all the posible outcomes you will get to 1

[crimson sedge]

thats basically just 1/2

[feral juniper]

If you throw a coin twice you would expect to have Tails Once and Heads Once

you could however have that in 2 different scenarios

[crimson sedge]

yeah but what if u have heads heads

[feral juniper]

gettings tails first then heads or first heads then tails

[feral juniper]

then the probabilty is (1/2) * (1/2) = 1/4

As it is for:

Tails and Tails
Heads and Tails
Tails and Heads
Heads and Heads

[crimson sedge]

ohh

soo..

how does that apply to 2/3

do I have um

2/3^60

[feral juniper]

almost

[crimson sedge]

what am i missing

[feral juniper]

if you throw twice you need to multiply

2 * 0.5 = 1

2 = Number of throws
0.5 = Probability for one throw
1 = Expected Value of throwing Heads (or tails)

thats what the numbers stand for

did that help

[crimson sedge]

it became more confusing

wait

hold on

so is it

2/3 * 60

whats that reaction supposed to mean

did i get it?

[feral juniper]

LMAO

yes

[crimson sedge]

YAYYYYY

wait

let me do the maths

wtf

i got 60

[feral juniper]

how did you multiply

[crimson sedge]

2/3 x 60

cross multiply

simplify 3 to 1

60 to 30

then multiply

2 x 30 =...

60

bruh

[feral juniper]

$\frac{1}{3} \cdot 2 = \frac{2*1}{3} = \frac{2}{3}$

[wraith dagger]

tobi

[feral juniper]

Thats how you multiply a number to a fraction

[crimson sedge]

yea but

isnt that just what i did

2/3 * 60 = 180/3 = 6/1

oh wait

its 6!

its

yay

wait what

bruh

how does that make sense

60 times

15 marbles

10 of them are blue

what the

[feral juniper]

what is 60 * 2

[crimson sedge]

oh wait

120

[feral juniper]

you wrote 180

[crimson sedge]

oops

120/3

4/3

yay!

i mean

wait

4/1

40

[feral juniper]

you dont seem to understand how to simplify fractions

[crimson sedge]

yes

40

sorry

[feral juniper]

Im currently writing an example wait pls

[crimson sedge]

my mind is uhmm

wait no i got it!

its 40!!!

[feral juniper]

would you agree with me that

40*1 = 40

[crimson sedge]

yes

[feral juniper]

but 1 is nothing else than a/a right

if a is not 0

1 = a/a

is that also understandable

2/2 = 1

3/3 = 1

[crimson sedge]

yes

sorry

[feral juniper]

pi / pi = 1

[crimson sedge]

i know this

yes yes

[feral juniper]

$\frac{2}{3} \cdot 60 = \frac{2 \cdot 60}{3} = \frac{120}{3} = \frac{40}{1} \cdot \frac{3}{3} = \frac{40}{1} \cdot 1 = \frac{40}{1} = 40$

[crimson sedge]

i was rushing

cause i kinda got excited

[feral juniper]

then this explanation should make sense

[wraith dagger]

tobi

[crimson sedge]

yes understood!

wait

why am i multiplying by 3/3

i meant 1

[feral juniper]

thats just to show how to simplify fractions

[crimson sedge]

ohh

[feral juniper]

and what it means for something to cancel out

[crimson sedge]

SORRY

i really do understand this part

[feral juniper]

do or do not?

[crimson sedge]

do

I do!!

[feral juniper]

thats great

[crimson sedge]

tobi

[feral juniper]

why sorry then

[crimson sedge]

cuz i was rushing earlier

and gave u the misinterpretation that i dont know

[feral juniper]

its fine

[crimson sedge]

but

can i

ask u

more questions

later

cuz i feel like

u genuinely taught me something

that will be very useful for me in the future

[feral juniper]

Thanks that's great feedback

And yeah you can DM me im just not sure If I will be online

[crimson sedge]

hmm

wait

whats ur level in maths

[feral juniper]

Im a Physics undergrad

[crimson sedge]

woa

[feral juniper]

my knowledge stops somewhere in real analysis when it comes to pure maths

[crimson sedge]

wdym

so

u cant complete every single math question?

but its ok

i suck at maths

im in algebra 1

im sure every qs i have is simple and ez for u

tobi?

[feral juniper]

ye?

[crimson sedge]

right?

[feral juniper]

If you are interested in probability
I can suggest binomial distribution

this is a picture of Throwing a Coin 4 times and the height represents how probably it is to get Heads
1, 2, 3, 4 times

The Area of the rectangle = Probability

The Area of the Whole Distribution has to be 1
Because as you see in the picture the probability of adding all the possibilities is always 1

Its like the probability of Probabilities

[crimson sedge]

uhm

okay

i have

another

question

now

[feral juniper]

yeah probably but you should understand these topics

[crimson sedge]

this question

wait bruh

here

[feral juniper]

Always start with given and Looking for

Try putting these questions in a formula that you can use

[crimson sedge]

well

$3 = per pound almond

$3.50 = per pound cashews

me buy : 2 pound of almond

$6.25 cashews

how many pounds i buy for each

[feral juniper]

I would write it like this if I am starting

Given:
Let A = $ per pound almond => A = 3
Let B = $ per pound cashews => B = 3.5

looking for:
Let X be the Value in $ that we are looking for then

X = 2A + 6.25$ of cashews

[crimson sedge]

kind of

wait

what is lf meaning

[feral juniper]

looking for

[crimson sedge]

ohh

but how do i calc that

its not an equation where i can just

yea

hmm

whats Y

[feral juniper]

Wait sry

missread the question

[crimson sedge]

oh

[feral juniper]

nvm I didnt

wow

Let Y be in pounds the number that we buy cashews, then

B * Y = 6.25

[crimson sedge]

BUT

whats b

and hats y

whats*

[feral juniper]

Let B = $ per pound cashews => B = 3.5

[crimson sedge]

oh

3.5*y = 6.25?

sooo

wait

let me get my calc

[feral juniper]

25/14

[crimson sedge]

equals

1.78571428571

oh

[feral juniper]

25/14 = 1.78571428571

[crimson sedge]

ok ok

so basically

1.9

[feral juniper]

approximately

the correct answer would be 25/14

or precise but whatever

[crimson sedge]

that many pounds?

but wait

didnt it say 3.5 per pound

[feral juniper]

the question is poorly written

[crimson sedge]

so what is 1.9 or 25/14

[feral juniper]

let me ask you

2 pounds of almonds and (cashews for 6.25 dollar)

has to be this one

[crimson sedge]

yes

i think

them together

wait

waittttt

thats impossible

[feral juniper]

yeah the question is poorly written

dont know where to put the brackets

[crimson sedge]

well

uh

i think it means

2pounds almonds and cashews together costs 6.25

which means uh

wdym

it says "for" which means its together tho

which means u did right

2A + cashrews = 6.25

[feral juniper]

plug in A

[crimson sedge]

2(3) + cashrews = 6.25

-6 cashrews = 6.25-6

wait

doesnt make sense

[feral juniper]

it does

continue

[crimson sedge]

u cant have cashrews for 0.25!

[feral juniper]

per pound

[crimson sedge]

oh

[feral juniper]

you can buy less than 1 pound

[crimson sedge]

how much is that

[feral juniper]

You know that 1 pound of cashews = 3.5 Pound
so

B = 3.5

[crimson sedge]

yes

[feral juniper]

Let Z be the number of pounds that you buy cashews for then

B * Z = 0.25

does that make sense

[crimson sedge]

yea

but

how do

OH

0.07

i mean

[feral juniper]

given values on one side and the values that you are lf on the other

[feral juniper]

1/14 is the exact answer

always leave it in fractions if the fraction isn't to complicated

[crimson sedge]

wait

do i still use pounds?

if 1/14 is the case

[feral juniper]

its 1/14 of a pound

[crimson sedge]

okay okay!

[feral juniper]

because the questions asks you to answer this in pounds

[crimson sedge]

now i have

but first

thank u thank u a lot

but now

i have a new question

umm

yes

uhoh

[feral juniper]

If you understood what you did you could come up with an Idea on how to Calculate how many pounds you could buy with 6.25 $ if thats what the question was asking

[crimson sedge]

yea

[feral juniper]

Im still not sure

[crimson sedge]

that wasnt the question tho

its asking

how many pounds did u buy

for each nut

if the total amount of moni u spent was 6.25

and bought 2 pounds of almond

but anyways

[feral juniper]

yeah?

[crimson sedge]

my kool new question

i have no idea what its asking

[feral juniper]

do you know how to solve linear euqaions

[crimson sedge]

i think yes

oh wait

its asking that??

so basically

[feral juniper]

yeah but a can be any real number

[crimson sedge]

its ok

ay = -5x

y = -5x/a

ok yay

ty ty ty

hold on

[feral juniper]

im still calculating

oh

actually

you dont need to solve anything for that

[crimson sedge]

is

y= -5x/a

the ultimate answer

and add a lil explanation

?

[crimson sedge]

oh right

but u cant eliminate

if its ay

OH

WAIT

no nvm

hmm

but if a can equal to anything

then i can argue that

[crimson sedge]

this is true

wait let me write it

yay

that question is done

now

new one

[feral juniper]

wait a sec pls

[crimson sedge]

ofc

[crimson sedge]

uh

what is this

[feral juniper]

your problem

[crimson sedge]

oh

so is this just an estimation?

[crimson sedge]

i dont think i actually needed to solve it though

it only asked for explanation

[feral juniper]

did you get 8 for a

[crimson sedge]

i put 8 for a

[feral juniper]

$\text{Yes but you are only allowed to do that if } y \neq 0$

[wraith dagger]

tobi

[feral juniper]

maybe they dont care about that tho

[crimson sedge]

oh.okay

its confusing my brain

tbh

[feral juniper]

to solve for a

you have to calc this

you are not allowed to divide by 0

thats what you did in your head

[crimson sedge]

um

this doesnt sound algebra 1

[feral juniper]

ok sry

[deft gull]

Sorry for the late response, but that is the right answer! Thanks for the help!

[crimson sedge]

um

[crimson sedge]

we move to different question

[crimson sedge]

no idk tbh

[deft gull]

yep i am aware, i was just responding because i was asleep

[crimson sedge]

oh

wait

do u think u can solve

this

idk what its asking me to do

[crimson sedge]

this

[feral juniper]

thats my answer

[deft gull]

yeah it should be a=8

[feral juniper]

or actually

if y= 0

another solution exists

im stupid mb

[crimson sedge]

wow

my head is spinning

[deft gull]

i don’t think so?

if y=0 then you have -2x = 7 and 5x=0 which is impossible

[feral juniper]

-7/3 = x

then

[deft gull]

no

it needs to satisfy 5x=0 too

[feral juniper]

-2x = 7 and 5x=0

=> -2x - 7 = 0
=> -2x -7 = 0
=> -2x = 7
=> x = -2/7

?

[feral juniper]

hm yeah

[feral juniper]

if we assume 5x = 0 has a solution other than x = 0 it would work tho

[deft gull]

but it doesn’t

[feral juniper]

yeah a=8 for y =! 0 solves your equation

[deft gull]

i still don’t think you need to include the y =! 0 condition

[crimson sedge]

hm

okay!

so

[feral juniper]

hm why y could be 0

not mathematically written enough to know tho

[crimson sedge]

okay

i need help

with this

new

question

[feral juniper]

$\text{Yeah } A \Rightarrow B \text{ means that B is only true if A is true your right}$

[wraith dagger]

tobi

[crimson sedge]

umm

yes

but

lets focus on this new question 😬

@feral juniper ?

[feral juniper]

yeah just wrote it down myself

so how would you calc this

[crimson sedge]

the thing is

there is no real given set of nums for each bag

or even a certain amount of bag and their cost

[feral juniper]

you have 2 equations and 2 unknowns

[crimson sedge]

yea

[feral juniper]

you can eliminate 1 value

and solve for the other unknown

[crimson sedge]

ok so

A = 4R + 2F = $78

B = 3R + 4F = $71.50

WAIT

that sounds like elimination or substitution to me!

[feral juniper]

great!

[crimson sedge]

Ok HOLD

wait

is this elimination

or substitution

[feral juniper]

that doesnt matter

whatever you prefer

[crimson sedge]

yay

also

another question

[feral juniper]

did you solve it

[crimson sedge]

YES

1 rice bag = 16.9

1 flour bag = 5.2

[crimson sedge]

also

tis new question

[feral juniper]

yeah correct

[crimson sedge]

:D

[feral juniper]

thats a formula on how to solve every linear with

2 unknowns and 2 equation

but whatever

[crimson sedge]

bru

my eyes

[feral juniper]

just to show you there is a formula for this where you just have to plug in all your known values

and they let you calculate this a million times

[crimson sedge]

woa

[feral juniper]

how many questions do you have left I am gonna go eat after this one

[crimson sedge]

i have

hold on

well

i have too many

so you can go eat

i'll contact u once u finish okk?

okkkkk?

[feral juniper]

I have to study myself after that sry

But there are many people here that will explain things to you

Have a good one

[crimson sedge]

ya

ok

but wait

didnt u say

one last questionnnnnn

tobiii

D:

[cedar kiln]

@crimson sedge Has your question been resolved?

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[latent rover]

hello can u give me a hint on how to solve this: find cardinality of a set of all chains in poset (N-{0}, |)
I thought of upper bound P(N) and I cant figure out what would be the lower bound. usually id come up with an injection f: X \to Y where X is a set with cardinality of continuum and Y is a set of all chains in N-{0}

[latent rover]

<@&286206848099549185>

[cedar kiln]

@latent rover Has your question been resolved?

[latent rover]

i constructed a function $f: (\mathbb{N}-{0})^\mathbb{N}\to X$ where X is a set of all chains in $\mathbb{N}-{0}$ defined like this: $\ f(a_n) = {2^0 \cdot a_1, 2^1 \cdot a_1 \cdot a_2, \dots, 2^{n-1} \cdot a_1 \cdot a_2 \cdot \dots \cdot a_n, \dots}$. is it an injection?

[wraith dagger]

ta

[latent rover]

i multiply by 2 because without it there is counter example $a_n = {1,2,1,1,\dots}, b_n = {1,1,2,1,1,\dots}$ so $a_n\neq b_n$ and $f(a_n)=f(b_n) = {1,2}$

[wraith dagger]

ta

[latent rover]

can someone smart approve or disapprove this

[cedar kiln]

@latent rover Has your question been resolved?

[latent rover]

ill wait

[latent rover]

.close

[cedar kiln]

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[west dragon]

(2x+6)/sqrt(x-1) > 0

[west dragon]

I solved it as I thought was usual

Tho checking photomath the result turned out to be wrong

[south tundra]

Show your work

[west dragon]

N: 2x +6 > 0 --> x > -3

D: sqrt(x -1) > 0 --> x -1 > 0 --> x > 1

I then proceeded to create the graph to calculate the solution

[south tundra]

So it's just x > 1

[west dragon]

No

Why only x > 1

• • • = +

@south tundra

[south tundra]

Didn't you get that x > -3 and x > 1?

[west dragon]

Yes

[south tundra]

If x > 1, then obviously x > -3

[west dragon]

But this is a disequation where there is the unknown term also at the bottom

[south tundra]

sqrt can't be negative

[west dragon]

So I have to write the conditions of existence for the sqrt

[south tundra]

Writing x - 1 > 0 is just making sure that sqrt exists

(And that it's not 0 in this case)

[west dragon]

If the equation didnt have the sqrt at the bottom

and was just

(2x+6)/x-1

[west dragon]

The solution shown here

[south tundra]

Then it's a different story yeah

[west dragon]

Would be right?

[south tundra]

Yes

[west dragon]

So if I have in any place of the disequation

a root

I shall write the conditions of existence

And then adjust the solution so that they are respected

In this case, since the root cant be negative

I have to remove the x < -3

Right?

(I'm trying to revise for a test I got tomorrow so I must have a prepared algorithm to solve everything)

[south tundra]

In this case you removed x > -3 because it's implied from x > 1 anyways

[west dragon]

No

Didnt u say I removed x > -3 because of the root

[south tundra]

No I meant this

[west dragon]

Ok

[south tundra]

If x > 1, then obviously x > -3

[west dragon]

k

[south tundra]

So saying x > -3 and x > 1 is the same as saying x > 1

[west dragon]

Ye

If I had:
(2x+6)/x-1 > 0

Then the solution would have been
S: x < -3 V x > 1

Right?

[south tundra]

Yes

[west dragon]

Right?

[south tundra]

Yes

[west dragon]

Perfect

Thanks a lot man

[west dragon]

I would have made the mistake of putting this solution in tomorrow's test if it wasn't for you

.close

[cedar kiln]

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[stuck cradle]

https://media.discordapp.net/attachments/918852002761551883/1060610751485706360/image.png

[stuck cradle]

I just need help understanding what happens in the black box

[supple badge]

OP is OA + AP

Which is the a + 1/4(-a+b)

Expand

Collect like terms

[stuck cradle]

hmm alright

thanks

[supple badge]

You get where it’s come from?

[stuck cradle]

a bit

would you have to add it with AB

[supple badge]

AB is the -a+b

The only use for that is to work out AP

Which is 1/4 of that because P is 1/4 of the way through

[stuck cradle]

yep

[supple badge]

They aren’t easy to wrap your head around those gcse vector questions

Just takes practice so you have that instinct of what to do

[stuck cradle]

alright thanks!

[supple badge]

.close

[stuck cradle]

yep

I just need help with how this part works

i dont where the 3/4 a comes from

yep..

yep

so far

for PB = 3/4 and AP = 1/4 but shouldn't it be 1/4 a + 3/4 b ?

this part

the circled part

yep

then im confused after that

oh

a

a

AP= 1/4

ohh

so its 1/4 with -a + b

alr

alright

yep

Good news , i nearly understand everything but how do you figure out k

like where do you get 1/4 from

I see

Oh that makes a lot of sense

yep

I fully understand it now

thanks a lot

k = 1/4

[cedar kiln]

@stuck cradle Has your question been resolved?

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[inner basalt]

I’m trying to find the limit as n approaches infinity. But I don’t know what to do with the square root before continuing

[cedar kiln]

@inner basalt Has your question been resolved?

[cedar kiln]

@inner basalt Has your question been resolved?

[inner basalt]

.close

[cedar kiln]

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[wet junco]

For the function f(x)=x^2 + 9, find the slope of the secant line between x = -4 and x = 4

[wet junco]

This is just 0 right

[muted bear]

Indeed

[latent bloom]

Yep

Mainly because this function is symmetrical around the y-axis

[muted bear]

Even function, as it were

[latent bloom]

Yep

[wet junco]

Alright coolio

[latent bloom]

[wet junco]

while I got u guys would u be able to double check one of my questions

[latent bloom]

Sure

[muted bear]

Sure

[wet junco]

just x=-1 ?

[muted bear]

Theres plenty more

[latent bloom]

Check at x=-4

x=-9

[muted bear]

Yeah it could be s shaped

[wet junco]

6 is a no tho right

cause its a sharp corner

[latent bloom]

x=3

[dull oxide]

This one is a maybe

Depends on how you define the domain

[muted bear]

6 does not count

[latent bloom]

It's included

[dull oxide]

Yeah but if domain includes x < -9, then the function is not continuous there. Thus, it cannot have a derivative/slope

Oh derp, we're both wrong Vulcan

Interval is given as -9<x<9

[latent bloom]

[wet junco]

Yep

[latent bloom]

Yeah double checked

Ok so x=-9 is a nogo

[wet junco]

gotcha

[latent bloom]

Was gonna say x=9 too but that's also not included

[wet junco]

so it's just -1?

[muted bear]

No

[wet junco]

and 3

[muted bear]

One more

[wet junco]

hmm

[latent bloom]

So you only have x={-4, -1, 3}

[muted bear]

Let them get it

[latent bloom]

Oops

[wet junco]

lol

[latent bloom]

[wet junco]

sorry but how would -4 be it?

[latent bloom]

Do you know why x=-4 has a horizontal slope?

[wet junco]

no

why does that count it's not S shaped just flat

[muted bear]

Try drawing the tangent line

[latent bloom]

Flat means...?

Flat

Doesn't that ring bells?

🔔

[frosty thicket]

🛎️

[wet junco]

huh lol

[latent bloom]

If it's flat

[wet junco]

nvm i see what ur saying

[latent bloom]

Flat means what in this case?

[wet junco]

the curve

at -4

[latent bloom]

Yep

[wet junco]

it's straight or flat

[latent bloom]

Looks horizontal

[wet junco]

yes

oh shit

that's in the directions lol

"horizontal tangent line"

[latent bloom]

[wet junco]

got it now

appr it

.close

[cedar kiln]

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[trim crescent]

Hello, does anyone know the comparison test for improper integrals ?

[trim crescent]

I don’t quite get why it still is valid if the functions are both negative ?

[cedar kiln]

@trim crescent Has your question been resolved?

[steep meadow]

If both functions are negative

I assume you can just make them both positive

And then apply the comparison test

Pretty much mirroring the functions through the x-axis?

[trim crescent]

That’s what they say yeah

But when you multiply by -1 surely you change the sign ??

So now the comparison test doesn’t work ?

@steep meadow

[steep meadow]

Say you have a function f(x) and g(x) and f(x) <= g(x) <= 0 for x >= a. If we now flip the function around the x-axis we get -f(x) >= -g(x) >= 0. These are the only sign changes as far as I am aware

Meaning that you're still bounding the function g(x) between f(x) and 0

So now if f(x) converges, in either case g(x) should be smaller than f(x) and also converge

[trim crescent]

But we were initially studying f(x) ?

So we know g(x) converges

But now when we flip it, we know -g(x) converges

But can’t say -f(x) converges because it’s greater than -g(x) ?

[steep meadow]

That is correct

But flipping it does not change any implications

If f(x) converges to a real number say x_0

then -f(x) converges to a real number -x_0

Since g(x) <= f(x) for x >= a

Then whatever g(x) converges to

say y_0

It should follow y_0 <= x_0

[trim crescent]

-y_0 >= -x_0 though

[steep meadow]

And for the flip we get 0 >= -g(x) >= -f(x) which is still bounded between 0 and -f(x). We know then that if the lower bound is -x_0 that -y_0 should be greater

[trim crescent]

Lets say

[steep meadow]

So we get it bounded between 0 >= -y_0 => -x_0

That means that -y_0 converges if -x_0 converges

And hence -g(x) should also converge

[trim crescent]

You mean -g > -f > 0 ?

[steep meadow]

Nope

[trim crescent]

I wrote that wrong

[steep meadow]

If we flip around the x-axis we still bound -g(x) between 0 and -f(x)

I will make a simple graph for you

[trim crescent]

But wait i think you got yourself confused in you initial considerations

[steep meadow]

Say green is f(x), blue is g(x) and red line being 0

Clearly 0 <= g(x) <= f(x)

Right?

[trim crescent]

f(x) and g(x) are both < 0 so the opposite is >0

[trim crescent]

The problem is when both of them are negative though ?

[steep meadow]

Yes both of them being negative

Now you see the graph right?

I will now flip it around the x-axis

[trim crescent]

They’re positive though ??

[steep meadow]

What do we see?

0 => -g(x) => -f(x)

[steep meadow]

Yes I am showing that you can flip between positive and negative

[trim crescent]

Oh shit

Ohhh

It just clicked

[steep meadow]

Awesome!

[trim crescent]

Damn that was mind f-ing me for ages

Really needed a graph !!

Thanks man love you ❤️

[steep meadow]

Graphs always helps IMO haha

No worries 😎

[trim crescent]

Definitely !!

Have a great day/night @steep meadow

[steep meadow]

You too!

[trim crescent]

❤️

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[wooden pasture]

.]

does root only count for numerator or numerator and denominator

[long arrow]

do you mean when we have fraction (whole) inside the root?

[wooden pasture]

yea

[wraith dagger]

Modus

[wooden pasture]

ah ok thx

.close

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[warm jacinth]

How do I solve this? I am completly lost

[cedar kiln]

@warm jacinth Has your question been resolved?

[warm jacinth]

<@&286206848099549185>

[cedar kiln]

@warm jacinth Has your question been resolved?

[final crag]

break the sum up! What is $x_0 + x_1$ less than or equal to? similarly with $x_2 and x_3$ and go on, and then you'll have to figure something out for the end

[wraith dagger]

OssihLikesBlue

[cedar kiln]

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[heady spindle]

why d/dx of y^2 is equal to 2y dy/dx?

[livid hound]

chain rule

[heady spindle]

why it isn't chain rule on x^2?

[livid hound]

well you're differentiating wrt x

you can still explicitly unnecessarily attempt to apply chain rule if you want

[unique dew]

you're taking the derivative with respect to x (hence the dx on the d/dx)

[livid hound]

where you multiply by dx/dx which is just 1

[heady spindle]

Ooh okay okay

[heady spindle]

So it's the derivative of y^2 multiply by the derivative of y?

[livid hound]

$\dv{x} y^2 = \br{ \dv{y}y^2}\br{ \dv{y}{x}}$

[wraith dagger]

ℝamonov

[heady spindle]

thank you!

.close

[cedar kiln]

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[twilit jasper]

Could I get help with this problem