help-38

[shrewd ridge]

i don't think that's right

[mossy breach]

yeah I am very confused

[shrewd ridge]

{{4,5}} has one element in common with B, {4,5} has zero elements in common

because we want {1,{1,2}} to exist we can't say every element must be disjoint, I'm just saying it's all confusing

[mossy breach]

so in {1,{1,2}} the element 1 is disjoint {1,2} is not?

[shrewd ridge]

yes

[mossy breach]

oh okay then that makes sense to me. and even though a = {1, a} looks like it has a disjoint element, it in fact has not because it's basically like saying a = {a}?

when a = {1, {1,2}} would not repeat itself that way?

[shrewd ridge]

i didn't understand that

[mossy breach]

this basically

[shrewd ridge]

if a = {1, a} , it means there's another set b = {a}
and no elment of b is disjoint from b

so i just use another set to show contradiction, instead of a itself

[mossy breach]

how is that the case?

[shrewd ridge]

because axiom of pairing

says {a} exists if a exists

i probably just don't get it tbh

[mossy breach]

I'm still thinking trying to understand

thanks for the help

[tribal jetty]

a={1,a} = {1}∪{a}

[shrewd ridge]

yes, that's also a way to show {a} exists

as a subset

[mossy breach]

and that subset is the problem?

[tribal jetty]

no

[mossy breach]

oh

[tribal jetty]

but look

1 ∈ a and a ∈ a

[mossy breach]

yes

[shrewd ridge]

like "all subsets of a set exist"

[tribal jetty]

a doesnt belong to {1}

[mossy breach]

yes

[tribal jetty]

i forgot what i was saying nevermind

[mossy breach]

okay?

[mossy breach]

I feel like this could be very helpful

but I fail to quite understand it

[tribal jetty]

Ok to proof the theorem you need three conditions

1. A ∈ A

2. {A} exists

3. A ∈ {A}

... Ok

1. from the picture A ∈ A

2. A = {1,A} = {1}∪{A} shows existence of {A}

3. A doesn't belong to {1}, so it must belong to {A}

[mossy breach]

ohh okay it think I am slowly getting it

how long did it take you to understand/learn all of that?

[tribal jetty]

i still dont get most of it

[mossy breach]

that is very reassuring 😅

[mossy breach]

thank you so much!

[shrewd ridge]

i never got a feeling I get something you don't @mossy breach

[tribal jetty]

what did i mess up

aight im out

[mossy breach]

thanks so much for the help again!

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[shy vault]

<@&286206848099549185>

[steady oracle]

do you know how to take a mean?

or do you know what does the term "mean" mean here?

[shy vault]

oh nvm its add; divide right?

[steady oracle]

yes

its the average of 15 numbers

[shy vault]

is the mode the number in the middle of a group of numbers or is it the median

[steady oracle]

median ig

check it on google or read their definitions

[shy vault]

ok

[scenic delta]

!15m

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[scenic delta]

^

[shy vault]

Oh im sorry

[scenic delta]

It's fine, just don't do it again

[shy vault]

No problem

[steady oracle]

do you not know the formulas?

[shy vault]

2 x pie x radius>

is that what i do here

[steady oracle]

circumference = 2 * pi * r
area = pi * r^2

[steady oracle]

yes

[shy vault]

adds to 3.14

[steady oracle]

also i would say that you should ask questions when you attempt the queston

[steady oracle]

its the circumference

calc area

[shy vault]

Okay hold on

2 x pie x radius x radius?

[steady oracle]

no

[steady oracle]

@shy vault

[shy vault]

oh ok

0.785?

[steady oracle]

wait lemme check

[shy vault]

okay

[steady oracle]

yes

[shy vault]

thank u for helping me your the first person to help me understand

[steady oracle]

np

[shy vault]

OMG

after days and days i finally know how to do the area of a circle

now can you help me learn this question?

[lone parcel]

Ez

[shy vault]

that doesnt help me rlly

[dapper swift]

It tells you that you can make a rectangle out of your parallelogram

Importantly, the base of the rectangle is the same

And the height is also the same

[shy vault]

so what do i do

how do i complete this question

cus ive been doing math for 8 hours

[lone parcel]

Well do the exact same thing you would do to calculate the area of a rectangle

There must be about 1000 videos explaining it look it up on YouTube

That shape is called a parallelogram

[shy vault]

sorry i was eating food

[dapper swift]

It literally says area = bh on the picture

base * height

[shy vault]

i got the answer

yall made it sound so difficult

i just had to muliply 25 by 12

[dapper swift]

Well it's worth understanding why at least

[dapper swift]

Yeah

[shy vault]

oh i see

this shit has to be 4

how tf isnt it 4 its the average number on the board

[dapper swift]

Did you add them all up and divide by 25?

[steady oracle]

not "average"

[dapper swift]

I have a feeling you looked at the middle number on the board

[steady oracle]

mean = add everything / divide by total numbers

total numbers = number of terms

[shy vault]

nuh uh

[dapper swift]

Okay you know what to do now at least

[shy vault]

erm

meadian means the middle but idk what the answer is

[loud gale]

Arrange in ascending/descending and then find middle term that is definition of median

[shy vault]

im lost bud

[loud gale]

You have 25 numbers

[shy vault]

correct

[loud gale]

Write them all in a single line in ascending or desecnding order

Then the middle term in that sequence is median

[shy vault]

so highest to lowest?

[loud gale]

Yes or low to high, same result

[steady oracle]

ascending order will work too

[shy vault]

is it ok if there are duplicate numbers?

like if 4 was next to another 4

[loud gale]

Yes

[shy vault]

ok

[dapper swift]

When you have an odd number of numbers, the middle number will always be (n + 1)/2

Say n = 7 in the top example, then your middle number will be (7 + 1)/2 = number 4

So unfortunately you do have to write them all down / keep track of how many times each number is repeated

[shy vault]

its taking to long and i dont know how to line them up in a correct way on a notepad

can someone just do it for me

is it 5?

im in a rush so pls

[dapper swift]

Okay so how I do it is you just need the 13th number
3 appears 3 times
4 appears 7 times
5 appears 4 times

Yes it's 5

[loud gale]

Op bro

[shy vault]

its the same thing but its range

[dapper swift]

Yeah like if you have a dot plot or a stem-and-leaf diagram, this is the fastest way

[dapper swift]

Stem-and-leaf is where you group the numbers by their tens place

So if you know 5 numbers start with 4, 3 start with 5

It's the same thing all over again

[shy vault]

i need answer im in a big hurry

and i have 1 more after this

[dapper swift]

Range = largest number - smallest number

[shy vault]

no no no not again

[dapper swift]

~~Also please manage your time better ~~

We aren't under an obligation to do your work in a hurry

Range is easier luckily

[shy vault]

well its not my fault im in a hurry

and i have 1 more after this

[loud gale]

Are you asking help for a test?

It looks like it

[shy vault]

stfu

[loud gale]

💀

[shy vault]

its online

so it doesnt matter

but i still get graded on it

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[lost jay]

The question?

[solid kiln]

Muchacho

Muchacho

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[orchid wagon]

quadratic

wait, for clarification, can you translate what the question is asking?

[sudden parrot]

@orchid wagon alculate the zeros of the following generational functions. write down the smallest x value first

[chilly bobcat]

take $\frac{1}{4}x$ in common
$= \frac{1}{4}x(x+1)$

[chilly bobcat]

so now, either $\frac{1}{4}x=0$ or $x+1=0$

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[sudden parrot]

what is x1 and what is x2

[fleet pebble]

0 and -1

[fleet pebble]

because you equate both of these to 0 to find the "roots"

1/4 . x = 0 hence x is 0

x+1 is 0

so x is -1

[chilly bobcat]

the lower one is x1

x1 and x2 are roots of the given expression

[sudden parrot]

thanks

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[spring hull]

How do I show the first inequality here? I tried to multiply by (a+b+c) on both sides, then got 3 ≥ 2(a/(b+c) + b/(c+a) + c/(a+b))

[onyx smelt]

1/a+1/b >= 4/a+b

1/a+1/b+1/c >= 9/a+b+c

does it work?

[spring hull]

yeah that's true,but how does that help in proving the first inequality here?

it just proves 1/a + 1/b + 1/c \geq 9/(a+b+c)

[onyx smelt]

so we have:
1/a+1/b => 4/a+b
1/b+1/c => 4/b+c
1/c+1/a => 4/c+a

now u got it?

[spring hull]

OHH

yeah

makes sense

thanks

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[turbid leaf]

So a person wants to travel from A to B without passing through P or Q, provided that the person only goes either right or downwards, how many different routes are there?

[lament dragon]

whats the problem for? pretty fun one

[lusty delta]

u want principle of inclusion-exclusion here

[turbid leaf]

I am guessing it’s all possible ways - (P or Q)

[kind forum]

yes exactly

Can you find all possible ways?

[turbid leaf]

11 choose 5?

[kind forum]

yes

can you find the number of paths that go through P or Q?

[turbid leaf]

P: 4C27C3=210
Q: 5C33C1=30
P and Q: 4C24C13C1=72
So P or Q:168

Yeah that was easier than I think

thanks anyway

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[copper raft]

can anyone help me with this

[copper raft]

think I'll get 4 values of x, but still not sure because double angles really f me up

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<@&286206848099549185>

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@copper raftdo you still need help?

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[calm bone]

this is recurrence relations, im stuck on part b, what would the particular solution ? i keep getting 1/3 but its apparently n/3?

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.reopen

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[serene junco]

May someone check number 18 for me

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<@&286206848099549185>

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<@&286206848099549185>

<@&286206848099549185>

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<@&286206848099549185>

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.reopen

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[nimble stone]

you mean just finding them?

cross section area * length = vol

surface area is just the sum of the side areas, 2 triangles and 3 rectangles

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[iron pelican]

can someone explain what type of graph this is?

[high spear]

Looks like a hyperbola

x^2 - y^2 = r^2

[iron pelican]

ah alright thanks

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[opal owl]

How to find the zeros of this cubic, I tested a few interger roots but none of them ended up being a zero of this cubic.

[cosmic topaz]

Maybe the roots arent integers

[opal owl]

how to find non interger roots

[cosmic topaz]

Well for cubics

The usual way is to factorise

Try and factorise

But

Not always

In this case it isn't working

So other thing is cubic formula

Which is fairly complicated

But if u want u can try that

[wooden plover]

there's rational roots theorem also

[cosmic topaz]

U can try completing the cube mahbe

[wooden plover]

hopefully there is one rational root for this polynomial

[opal owl]

what is completing the cube?

[cosmic topaz]

This particular

Polynomial seems unsolvable

[opal owl]

oh ok

imma go check if my determinant is correct then

idk what I did wrong here

[cosmic topaz]

Where did the - 2*2 come from

[wooden plover]

the -2*2 should be factored inside the (2-lambda)

[opal owl]

ohhh ty

[wooden plover]

yes it should

[cosmic topaz]

Both the 2s r getting eaten by the zeroes

[fresh gazelle]

Seems right

[wooden plover]

:/

[fresh gazelle]

He’s getting the det for A11, so need 2*2

[opal owl]

like this?

[wooden plover]

yea

[opal owl]

oh ok ty

[cosmic topaz]

W8 sorry my bad

[opal owl]

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[opal owl]

.reopen

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[wooden plover]

yea? @opal owl

[opal owl]

um how to unoccupy the channel

[wooden plover]

yeah .close

[opal owl]

ok

.close

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[wooden plover]

it just takes a bit of time

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[wraith hinge]

very confused

[marsh forum]

can you find the function when x>0?

similarly find the function when x<0

[wraith hinge]

well when x>0 then x^2 when x<0 -x^2

idk if thats wym

[marsh forum]

yes

now find the right hand derivative

and the left hand derivative

[wraith hinge]

2x and -2x

[marsh forum]

cool

now what's the value of the LHD as x approachs zero

[wraith hinge]

the limit would be 0

[marsh forum]

cool

that's your answer , then

[wraith hinge]

mm

okay then thanks

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[dense axle]

if a =/ b, whats the value of f(-1)

-3 < p <5 and 2 < q < 4, so ...

[knotty locust]

replace the h in your name with b and replace the n with m

[dense axle]

,,

[wraith hinge]

https://tenor.com/view/gym-cap-chris-bumstead-motivation-gif-22976644

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[jovial pecan]

6sin(pie/12(t)+47
is that right
the answer key is
12sin(pie/12(t-12)+47 but idk if this right

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[jovial pecan]

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[jovial pecan]

ok yall its the same thing

like my teacher anskjey key say its 13 as amp
but i belive it should be 6.5

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[celest storm]

if its 13 then the high and low temperature difference (peak-to-peak) should be 26

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[indigo socket]

hi uh- i dont understand

[indigo socket]

on why we have to do this ?

like the cross

like as u can see we did the x (2x+1) down

but why do we do the cross at the top?

and we plus it

[acoustic cipher]

They are making the denominator the same value

[indigo socket]

o

but i dont really get it that much on why we should?

bro he added more of this

algebra sucks

oo

but why we put the butterfly on top?

not between x (2x + 1) too

p

o*

that method?

[acoustic cipher]

To make the denominators the same, the entire fraction needs to be multiplied

An example without variables:
(9/8) + (8/9)
(9x9 / 8x9) + (8x8 / 9x8)
Both the denominators get 72, but the numerators need to be multiplied too
(I'm bad at explaining)

[indigo socket]

72 / 72?

wait

im lagging

💀

its so hard

[acoustic cipher]

(81/72) + (64/72)

Then you get 145/72

[indigo socket]

ik classic fractions..

[indigo socket]

81 + 64?

[acoustic cipher]

Yes

[indigo socket]

i wanna die with quardartics

idek how to spell it

😭

as i forgot abt it

5/6

[acoustic cipher]

You’ll need to master how to do unlike fractions before moving to algebraic fractions and quadratics

[indigo socket]

so uh

so we only cross butterfly to put on the top of fraction

i mean like the x/y, only put the x?

[indigo socket]

im scared

and tired

😭

[acoustic cipher]

It's alright, it's quite easy when you get it

[indigo socket]

i hope i do

but like its hard to me

the algebra ones

[acoustic cipher]

Do more exercises. What you'll need to master most is subject-changing in both integers and fractions

[chilly bobcat]

$\frac{9}{x}+\frac{5}{2x+1}$\\
$\frac{9(2x+1)}{x(2x+1)}+\frac{5(x)}{(2x+1)x}$\\
$\frac{9(2x+1)+5x}{x(2x+1)}$\\

[chilly bobcat]

the person solving the problem did it in a shorter method@indigo socket

people call it 'cross-multiplying' or 'the butterfly method'

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[astral ore]

if a = 2 and n = 99, does this work?

[astral ore]

doesn’t look like it works to me lol even though 2 is relatively prime to 99

,w 2^99 mod 99

[solid kiln]

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[quartz juniper]

If you look at the power of a, n is being put through a function, so instead of 2^99, you need 2^lambda(99).

That function is the euler’s totient function

The function counts how many coprime integers there are between 1 and n

[astral ore]

oh wait i mixed it up cuz

it worked on another question

3^123 mod 100

i used the fact that 3^(100* lambda)= 3^20 = 1 mod 100

and so i had 27

,w totient function(99)

[solid kiln]

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[astral ore]

okay so 2^30 = 1 mod 99 right? @quartz juniper

[quartz juniper]

Should be 2^60

[astral ore]

nah the definition has 1/2 phi(n)

if your prime base is 2

and the exponent is greater than 3

or smth like that

[quartz juniper]

Mb, I thought we were in base 10

[quartz juniper]

No clue rlly for base 2

[astral ore]

oh no i meant 2^99

the base is 2 lol

anyway so like back to the euler thing

do i actually always have to count?

like phi(99)?

would i count all the numbers coprime to it?

[quartz juniper]

Yh, I’ve only dealt with small inputs tho if I have to count manually

[astral ore]

that doesn’t sound convenient at all

[quartz juniper]

Better used for proving theorems than for actual numbers

Unless u have python

[astral ore]

i’ve seen people whip out phi(100) = 40

like out of the blue

did they really count or something?

[quartz juniper]

Maybe they use it often and memorise the values?

[astral ore]

ah maybe

but they seem like beginners just like me

on stack lol

maybe they bruteforced it

okay then thanks

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[hollow warren]

8/(x+3)-3/(2-x)

[hollow warren]

=2

solve the equation

idk where i am making a mistake

[verbal idol]

send solution

[hollow warren]

wait

=

(16-8x-3x-9)/(2x+6-x^2-3x)=2

=

(7-11x)/(x^2-x+6)=2

=

7-11x=2x^2-2x+12

=

7-11x-2x^2+2x-12=0

=

2x^2+9x+5=0

then how will i factorise

idk if any of the above steps are incorrect or not

please check

[verbal idol]

ok

[verbal idol]

wouldnt it be -5?

[hollow warren]

I took the -1 common

[verbal idol]

how did this become +x2 ?

it should be -x^2

[hollow warren]

💀

I am so dumb

thanks btw

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[verbal idol]

np

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[umbral lily]

Can someone teach me this question?

So far i just obvious that x² + (48×101)² = (50×101)² does it work for the rest steps?

And did difference of square and it does not come out anything...

[split chasm]

show what you have after applying difference of two suqres

and/or show all the work you've done on paper

recognising/applying what you've mentioned is pretty much already 80% of the way there

[umbral lily]

@split chasm

[split chasm]

notation issues, and suboptimal way to apply difference of two squares

[umbral lily]

yea i forget to add the +

[split chasm]

the end goal is the find x, doing it this way doesn't help much

instead consider subtracting (48 * 101)^2 from both sides

$$x^2 = (50 \cdot 101)^2 - (48 \cdot 101)^2$$

[solid kiln]

ℝαμΩℕωⅤ

[split chasm]

and apply difference of two squares on the right,
you can also first factor out 101^2

[cosmic meadow]

Also, don’t recommend you to ise “x” as the product symbol

[umbral lily]

x^2 = 101(196)

for solve x sqrt(101*196) huh?

[split chasm]

missing ^2 on the 101

[umbral lily]

thanks

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[wraith hinge]

@helper

<@&286206848099549185>

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[livid dune]

Is this correct?

[dapper swift]

https://www.desmos.com/calculator/pqc8kmaetw

First one is correct and second isn't

If those are left and right rectangular sums

Btw this is how you can check your calculations

[livid dune]

ok thx

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[spiral badger]

Can anyone help me? I want to find the limit of this question.

[dapper swift]

Knowing multivariable calculus, the limit probably doesn't exist

All you need to do is to find two different paths which result in different limits

Yeah also I think it should be e^(xy - 1) no

[spiral badger]

Can I use Khan Academy from elementary school to college?

[dapper swift]

They do have multivariable calc resources

But I think Paul's Math Notes are better

https://tutorial.math.lamar.edu/Classes/CalcIII/CalcIII.aspx

Ah it's interesting here cause only xy appears in the limit

So you can let u = xy and ofc u goes to 0

And there's definitely a vertical asymptote

[spiral badger]

They do have multivariable calc resources

[dapper swift]

<@&268886789983436800>

[dapper swift]

I know

[spiral badger]

can i be friends with you

[dapper swift]

Nah no thanks

[spiral badger]

I am a child and I am new to mathematics. Can I come to this channel to ask questions at any time?

[full hull]

yes (u can use .close for that)

[dapper swift]

You're over 13 right, just checking

Cause you have to be over 13 to use Discord

[spiral badger]

Yes I am eighteen years old

[dapper swift]

Okay then you're an adult.... weird

[spiral badger]

I have a problem

[full hull]

6x2x?=36

[dapper swift]

Lol do you have a kid or something; you were just asking about multivar calc

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[spiral badger]

yes

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[spiral badger]

.close

Is this how it is used?

[split chasm]

it was already closed,
you new message reclaimed the same channel
do you have another question?

[spiral badger]

No problem, I'm just curious if I just need to enter .close

[split chasm]

you clicked the check when the bot asked, that's sufficient
you should see the bot message saying Channel Closed

if you look a few messages up

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[spiral badger]

Thank you very much

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[high plank]

Why didn’t they include minus at the end

[marble wharf]

where should they include a minus

[opaque aspen]

last step?

[calm quarry]

x<0 so it's already minus

If you divide by -2x than sign won't change because -2x is positive as x is negative

[high plank]

So if I divide it by minus again the sighn still won’t appear

[high plank]

Shouldn’t the sighn be there to display it loositve

Positive

[calm quarry]

x is variable so it can be positive or negative. But in your question it's explicitly given that it's negative so you have to consider accordingly

[calm quarry]

Since x stand alone is negative so whether you divide or multiply it by some negative number you'll get a positive value

-x is positive, x is negative

[high plank]

Yh but it still can be -(4x)

[high plank]

If it 2 different thing why didn’t they display the minus

[high plank]

X is still negative tho it would just be -(x)

[calm quarry]

They have already said x is always negative

[high plank]

So it positive

But if I divide it again by minus then it would be a minus number

[calm quarry]

In -x the value of only x is negative but -x as whole is positive

[high plank]

So they just did - divide by - to get positive

[dusty sleet]

No

[split chasm]

whut?

[high plank]

X is negative then it get divided by -1

[dusty sleet]

They divided by x. Since x is neg you have to reverse the inequality

[dusty sleet]

Where??

[high plank]

Ohhhh ok that make sense

I thought they divided by -1

Lol mbbb

[dusty sleet]

In that case you would see the other signs changed, which is not the case

[high plank]

K thank I undetstan bow

Now

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[tender thistle]

Can I receive some help with this problem? I have to determine if:

1. The gradient of f will equal 0
2. The gradient of f will equal the gradient of g(x,y)= y + x/2
3. The gradient of f will be parallel to the gradient of g(x, y) = y + x/2
4. The contraint line will cross the graph of f

[tender thistle]

I was thinking that it is the parallel one

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[steel summit]

goddamn hi guys its my first time joining such a server are we just allowed to ask about questions in here or ?

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[short trellis]

They turnd ABC isosceles triangle around point A with a 30 degree angle. B moved to B1 and C to C1. B1C1 line goes thro point C need to find ABC triangles cercumscribed circles radius if AC=10

[short trellis]

<@&286206848099549185>

[kindred skiff]

Dunno if this is the fastest way but you can solve for c1, solve for b1, solve for a-b/2, then get the cosine of that

Since ac and ac1 are the same length and <c1ac is given

[short trellis]

How do i solve for C1 what do i use

[kindred skiff]

^

[short trellis]

Don’t get if j get the same lengths but

[kindred skiff]

<c1ac is given as 30 degrees

<ac1c is (180-30)/2 since it's an isoceles triangle

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[flat wing]

can someone explain how this work

[flat wing]

i answered this question using differentiation but that a long way

why is t 4

since i get that makes y 0

but should y be able to be zero since

that at the bottom of thr mountain

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[spice spade]

why does this integrate to this?

[spice spade]

should it not become the integral of 10x^-1/2 + 3 which then integrates to 20x^1/2 + 3x + c ?

[wraith arch]

where did you get the 10 from?

the constant in front is 5/2

[near vessel]

you should move the constant, 5/2 outside of the integral, and you are left with the integral of 1/√x. Now youre integrating x^(-1/2), where you can use the power rule for integrals

[spice spade]

i thought i would first make it 5(2sqrt(x))^-1 which became 10

[spice spade]

I will try this thanks

[wraith arch]

well if you want to move the 2 up to the numerator

it would have to be 2^-1

since we still have to negate the exponent when moving things up and down fractions

[spice spade]

i see

[wraith arch]

itll still remain 5/2 regardless

[spice spade]

thanks i managed to acomplish the integral 😎

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[cobalt cloak]

if abs(x^2/2) < 1, how can i simplify this? i'm doing this for binomial theorem

[dapper swift]

Then -1 < x^2/2 < 1

[cobalt cloak]

tysm

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[dapper swift]

Npnp

[cobalt cloak]

.reopen

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✅

[cobalt cloak]

are my steps correct though

cuz ill get sqrt -2?

[dapper swift]

Can you zoom in on your writing?

[cobalt cloak]

Okay

My answer for (a) is correct

but the model answer for (b) is |x|<sqrt(2)

[dapper swift]

Yes, so x^2/2 can't be negative so you can disregard -1 < x^2/2

It's just x^2 < 2 then

Or |x|<sqrt(2)

[cobalt cloak]

is that the same as

[dapper swift]

Yes

[cobalt cloak]

Sorry to bother but how do i know? i genuinely haven't touched on this topic at all about absoluteness sorry

[dapper swift]

|x| < a means -a < x < a by definition of the absolute value

You can think of it as |x - 0| < a

Or the distance between x and 0 is less than a

If you draw it out, you do get -a < x < a as -a and a are the endpoints

[cobalt cloak]

yup

and in my case, at first |-x^2/2| < 1

so naturally |x^2/2| < 1

-1 < x^2/2 < 1

-2 < x^2 < 2

square rooting we can disregard -2

x < sqrt(2)

?

[cobalt cloak]

different from this

[dapper swift]

No, x^2 < 2 does not imply x < sqrt(2)

Cause for example, x = -1 works, (-1)^2 < 2

If you draw a graph of y = x^2 and y = 2

There's one negative and one positive solution for when they intersect

[cobalt cloak]

yup

[dapper swift]

Yeah so do you see how it's -sqrt(2) < x < sqrt(2) then?

Just from x^2 < 2

[cobalt cloak]

yes

oh i got it now

tysm

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[dapper swift]

Npnp

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[keen void]

$36x^2+9y^2-324 > 0$

[solid kiln]

Merineth

[keen void]

How do i easily draw this equation graphically ?

[vagrant prism]

by hand?

[keen void]

Yes

I know already that it'll become an ellipse but

Finding the extreme points of it seems very hard?

[rugged latch]

x^2 /9+ 0.25y^2 > 1

[sullen swallow]

Set y and x equal to 0 and find the points of intersection of the axis

[vagrant prism]

by extreme points you mean dy/dx = 0 ?

[marsh forum]

convert it into the standard form of an ellipse

[vagrant prism]

substitute y=0

[split chasm]

its centred around the origin so you shouldn't have that much difficulty finding where it intersects the axes

[marsh forum]

which is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

[solid kiln]

ƒ(Why am. I here)=I don't know

[keen void]

So what would be your advice on how to more easily draw it?

[split chasm]

sub x=0, y=0
for the intercepts

sub more values for additional points to make it look nicer

[keen void]

I'm not sure i understand

If i set x and y to 0

i get:

-324 > 0

What does this tell me exactly?

[vagrant prism]

no, first sub x=0, then solve for y
then sub y=0, solve for x

[keen void]

oh

[vagrant prism]

that's what they meant

[keen void]

ooh so when x = 0 then y = pm 6

okay that was fairly easy haha

tysm!!

🫶

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[fathom helm]

hi could someone help me evaluate what the solutions to these limits are?

these are sequences

is there a way to estimate the value just by looking at the limit?

[magic bloom]

well L'hopital's rule for one

[fathom helm]

alright, thats what i did and the result is 0 right?

[magic bloom]

for which one?

the first one?

[fathom helm]

yes

[magic bloom]

lemme check

yes

[fathom helm]

alright

do u know if there is a way

to evaluate

like

estimate

the result

[magic bloom]

do u have a calculator?

for these questions

[fathom helm]

the simple one

u know +, -, /, * only

[magic bloom]

scientific calculator?

is it casio

[fathom helm]

no i mean

the basic one

the most basic one just four-function one

[magic bloom]

does it have log

[fathom helm]

no

they dont let us anything more

[magic bloom]

i see

then u'd have to use lhopital

[fathom helm]

i heard some method that if the

fraction has two functions

if one of them is fast growing

and the other one is slow growing

then u can say that the limit is 0 or like infinity

i saw it in some video in a similar example to the first one

[magic bloom]

yes but it is generally ambiguous as to which one is growing faster and which one is growing slower

like obviously u could say x^2 / (3x^3+2) will have a lim x->inf as 0

but when they combine other functions like trig and log, it becomes obscured

[fathom helm]

these equations are actually from an exercise about asymptotic notation

u know the O theta and omega

[magic bloom]

for instance, the second one here is very easily 0

u know the denominator is rising extremely quickly

[magic bloom]

via taylor expansion?

[fathom helm]

oh i had a problem with this

[fathom helm]

oh idk if its related to this, sorry english is not my first language and i have whole theory in my native language

[magic bloom]

like u express a function in terms of polynomials

sin(x) = x - x^3/3! + x^5/5!...

[fathom helm]

oh no, not that

its about upperbound lowebound and averagebound asymptotes

i have no idea if i said it correcctly

anyway for these questions we just calculate limits

thats why im asking about these ones

cuz i heard that the profeesor might give us some huge limits to do

and some people say that the only way to do it was estimating

like

[fathom helm]

like u said here

but im not sure i know the rules for that

[magic bloom]

u can intuit it through ur prior knowledge

like if u see n! or power of n, it is a very fast growing function

if u see √ or ln, the function slowly grows

https://www.youtube.com/watch?v=sm0BB_3qMlI

this vid might give insight

[fathom helm]

oo thank you so much

alright i understand that more

this is how someone from the past did the last one

@magic bloom do u know if its correct?

[magic bloom]

u cant pull the n down

in the first step

u can only do that if the n is the power inside the argument

the whole log is to the power of n here

[fathom helm]

oh wait ure right

so do u have any idea on how to calculate this limit?

[magic bloom]

u again wanna know which one grows faster

do u think the numerator grows faster or the denominator?

[fathom helm]

log2x right?

[magic bloom]

yups

[fathom helm]

it grows faster

so it would be 0 again?

what about the power?

does it do any affect

[magic bloom]

since both have the same power

it shouldnt effect

its 0 again

[fathom helm]

oh alright then

that makes sense

i have one question

if it comes to

the growth of the functions

is there any order that some functions generally grow faster than the other ones

i mean

do u know the general order?

im trying to find it on the internet

like for the most basic ones

nvm i just found it

i just found this

[magic bloom]

yups thats correct

[fathom helm]

i know that on my exam

i will have some exercise about

ascending functions in order depending on the growth

there will be like 4 functions

honestly idk how else i can put them in order rather than just follow up these rules

ill figure it out

anyway

thank you very much for help

ill close the topic

.close

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[magic bloom]

no problem! lemme know if u need anymore help

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[clever chasm]

my idea here was to multiply with sqrt(x^2+pi) + sqrt(x^2-e) on top and bottom

[marsh forum]

good, that's the key idea

[clever chasm]

ah nvm i see where it went wrong

i got pi-e/2 instead of pi+e/2

but i forgot to do the minusminus gives plus

xd

.close

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[true bane]

divergent?

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[sand obsidian]

$$z=\frac{(1-6i)(\sqrt{ 8 }+i)}{9i}$$ When you have something like this do you deal with the top first before you add conjugate?

[solid kiln]

Totalani

[unkempt gazelle]

get rid of the i in the bottom first by multiplying top and bottom by i

then u can deal with the i2 and the is u will get

[sand obsidian]

what?

[unkempt gazelle]

u dont really have to multiply by the 9 in this problem just the i

[sand obsidian]

so not the conjugatge?

[unkempt gazelle]

theres not a conjugate in that problem

u do the conjugate when theres 2 terms

like

1-9i

[sand obsidian]

well it would be -9i no?

[unkempt gazelle]

u dont have to

9 is a whole number

not negative

i is the only term not allowed to be in the denominator

[sand obsidian]

yea

[unkempt gazelle]

youre just giving urself more work with simplifying if u multiply by 9i

[sand obsidian]

right so multiply with i on both sides then right?

[unkempt gazelle]

multiply i with the denominator and numerator yes

[sand obsidian]

Didnt think you could do that, let me give it a go

$$\frac{(\sqrt{ 8 }+i-6\sqrt{ 8 }i-6i^{2})i}{-9}$$

[solid kiln]

Totalani

[sand obsidian]

do I then distribute in i?

[rapid dock]

yes

[sand obsidian]

interesting, hold up

I feel a bit lost not sure I did this correct $$\frac{-1+6\sqrt{ 8 }+6\sqrt{ 8 }i}{-9}$$

[solid kiln]

Totalani

[sand obsidian]

btw the question was to find the absolut value of z

[unkempt gazelle]

whats z

in this case

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[sand obsidian]

#help-38 message

.close

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[jovial juniper]

Two objects together cost 520 lei. If the price of one object increases by 20% and the price of the other object decreases by 25%, then their values ​​become equal. a) Can it be stated that the price of the cheaper object is equal to 240 lei? legitimate the answer. b) If p% of the price of the more expensive object represents the price of the cheaper object, determine the value of p.

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@jovial juniper Has your question been resolved?

[outer rivet]

What have you tried

[stark pilot]

did you write the problem in terms of a system of equations?

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[vapid token]

hi, with horizontal point of inflection, is it when f’(x) = 0 and f’’(x) = 0 both have the same result when solved?

[vapid token]

or how does it work? because i have a question; f(x) = 2(5x-3)^3

asking me to show it has a HPOI

just struggling with making f’(x) = 0

30(5x-3)^2 = 0

[tardy hemlock]

Also f'''(x) is not 0 or f'' changes signs at x

[vapid token]

wym??

how do i show it has a horizontal?

[tardy hemlock]

For a point of inflection at $a$, we need $f''(a) = 0$ and \Big($f'''(a) \neq 0$ or $f''$ changes signs at $a$\Big). \ For showing it's horizontal, we need $f'(a) = 0$

[vapid token]

triple derivative?

[tardy hemlock]

Yeah

Or just second derivative changes signs

[vapid token]

we haven’t used triple derivative

[tardy hemlock]

Then you've probably used the second derivative changes signs condition, right?

[vapid token]

this is the notes we had on it

[vapid token]

yeah, table of values stuff right?

[tardy hemlock]

"and concavity changes" is the same as "f'' changes signs"

[vapid token]

yep

so what do i do to show its a horizontal?

[tardy hemlock]

Show f'(a) = 0

[vapid token]

f’(a) or f’(x) ?

[tardy hemlock]

f'(a) where a is the point of inflection

[vapid token]

so i’d do x=3/5 and plug that into the f’(x)

[tardy hemlock]

Well first of all take the first and second derivative, you'll need both.

[vapid token]

yes i did that already

[tardy hemlock]

What did you get?

[vapid token]

and got 300(5x-3) = 0

and then got that to x = 3/5

and now that i’ve plugged that into f’(x) it’s shown as 0

[tardy hemlock]

300(5x - 3) as the second derivative?

[vapid token]

yes

[tardy hemlock]

Yep

[vapid token]

first derivative was 30(5x-3)^2

[tardy hemlock]

Yep

[vapid token]

so now all i do is test the concavity change to prove it’s a POI

[tardy hemlock]

Correct. So now you need to show f'(3/5) = 0 and f'' changes signs at 3/5

[vapid token]

and then sub in the x coordinate to the original function to get y coordinate

[tardy hemlock]

Yep, and afterwards show it's a horizontal POI by showing f'(3/5) = 0

[vapid token]

yeah i just did a table of values

it’s the way we were told to do it in class

[tardy hemlock]

You don't really need a table, it's just 2 x-values you need to plug in to check if it changes concavity

[vapid token]

pick something above, and below relatively close and see if it’s positive or negative

[tardy hemlock]

Yeah

I'd pick x = 0 and x = 1

[vapid token]

it’s more just doing it so it’s similar to other things like stationary points and more complicated ones

[vapid token]

i just picked 0.5 and 0.7, but yeah i’d normally do 0 and 1

okay

i’ve gotten HPOI @ (3/5, 0)

which checks out on desmos