help-38

[dapper herald]

?

[uncut aspen]

convert the percent increase to a flat amount

there's really no good reason to do just a flat amount over a percent increase since it just involves more steps

[fiery cedar]

i want to always add the same flat amount, for example 10

[uncut aspen]

that goes counter to the fundamentals of math

and the defintion of percentage

[dapper herald]

That was it

[fiery cedar]

i cannot use the total of value2 in some way to counteract the fact that im only adding 10?

as in, the more difference between 10 and value2, the 10 becomes higher?

[uncut aspen]

wtf does it mean for the "10 to become higher"?

[fiery cedar]

haha

i want to add to value2 increments of 10, then a formula to grab that 10, and check how big value2 is, and multiply that 10, for instance

[uncut aspen]

please write this out in equation form

[fiery cedar]

so even though im just adding 10, after going to the formula it became 3822 so the % increase is the same

if i knew the equation form for it i wouldnt be asking here

[uncut aspen]

no as in using math notation

not using an actual working formula

just assume there is a working function, and call it something

and then use that to describe your problem

[fiery cedar]

a x b+10 =/= a x (bx2)+10, i want a function that gets my +10 and checks the total value of b to multiply my +10 so that a x b+10 = a x (bx2)+10(which is not 10 now)

the formula that i want is not to multiply a * b but to convert the same value (10) in relation to b, so that it goes up with b

[trim jolt]

@fiery cedar Has your question been resolved?

[uncut aspen]

dawg you just defined a percent change

[fiery cedar]

so

a x (a*0.1) x b ?

[uncut aspen]

percent = a(b+10)/ab
n = 2ab*percent - 2ab

n being the variable 10 you referred to earlier

[fiery cedar]

would a x ( (a*0.1) x b ) not work? every point of b is 10% of a

[uncut aspen]

no cuz adding 10 is not the same as increasing by 10% for all a and b

[fiery cedar]

how could i do that then? make every point of b be 10% of a

so if i add 10 to b, i would be adding 100% of a, and so on

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[livid yacht]

why is C-part notP??

[wraith hinge]

what level of math is this

im just curious

[livid yacht]

its a second year math course

intro to logic for computer science

[wraith hinge]

oh, are you a math major?

[rotund grove]

which language?

[livid yacht]

nope im a comp sci major

[livid yacht]

?

[wraith hinge]

yea makes sense

ok thanks

[livid yacht]

yeah np

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[loud cosmos]

how to do this?

[loud cosmos]

without using the values for sin60 and cos60

[rugged latch]

Lol I mean u could do it kinda roundabout and use sin(90-30) and same for cos

Or get tan(60) - sec(60)

[rugged latch]

Using 90-30 for either or both of those or none of those

[rotund grove]

,w tan60

,w sec 30

,w sec 60

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[outer cradle]

iv gotten to

sin (x/2) + cos (x/2)

like it simplifies to that

from there how do i solve

<@&286206848099549185>

[plucky sable]

Hmmm

seems like you can get an explicit value for x from the very first line

[nova pewter]

As far as I remember the answer is B correct?

[earnest nymph]

tell them how to get there, not the answer

[frozen plover]

find cot (x/2)

if you know cot x you know tan x

and if you know tan x you can find cot (x/2)

then draw a triangle

alternatively sin x + cos x = sqrt(1 + sin 2x) on some interval that you should find

[nova pewter]

I mean it is in the form of sin(A+B) and cos(A+B), simplifying this will give you something in cosx or sinx

[frozen plover]

he's gotten to that part already

[frozen plover]

.

[nova pewter]

Square this expression

[earnest nymph]

uhhhh

?

[frozen plover]

he's referring to this

you end up with that

[nova pewter]

exactly

as we know cotx, we can figure out the problem

[plucky sable]

it might be helpful to remember that
1/ tanx = cot x

but we already know what cot x equals to :D

once you find your value for x, just plug it in and simplify

don't overthink it and good luck!

[outer cradle]

its non calculator

so like

if it were calcualtor u could have just graphed both

and gotten

[outer cradle]

okay alr lemme try

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[rugged geyser]

4 3 5
Q2) Simplify √81 − 8 √216 + 15 √32 + √225.

[rugged geyser]

4 raised to√81 − 8 3 raised to√216 + 15 5 raised to√32 + √225.

this is complete question

yes sir what to donow

??

sir opposite

4 is power of √81

i mean like √81 raised to 4

all of them like this opposite

yes sir

but 4 is before √81

like that

[worldly wing]

$\sqrt[4]{81}-8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}$

[solid kiln]

kheerii

[worldly wing]

this is what OP means

[rugged geyser]

yes this

[rugged geyser]

this i mean

[worldly wing]

... alright

!status

[trim jolt]

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2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[rugged geyser]

1

[worldly wing]

how comfortable are you with radicals?

[rugged geyser]

that flew above my head i am very bad at maths that's why i am learning from experts like you

[worldly wing]

hmm okay

do you know what a square root is?

[rugged geyser]

yes sir

i know that

[worldly wing]

can you tell me the value of the last term?

$\sqrt{225}$

[solid kiln]

kheerii

[rugged geyser]

15 sir

[worldly wing]

you don't have to say sir lol

but yes, it's 15

[rugged geyser]

ok

[worldly wing]

when you're finding the value of a square root, you're effectively answering the question,
"what number, when multiplied to itself, gives me the value 225?"

the other roots are completely analogous to this

[rugged geyser]

5 square is 25 and nears square to 2 is 1

so 15

[worldly wing]

ah, well, that's a shortcut that you have been taught

[rugged geyser]

i used this trick

yes

[worldly wing]

to find out the square root of any number ending with 25

can you tell me $\sqrt{36}$?

[solid kiln]

kheerii

[rugged geyser]

6

[worldly wing]

correct.

now, say we have $\sqrt[3]{8}$

[solid kiln]

kheerii

[worldly wing]

the way to interpret this is to ask the question,
"what number, when cubed, gives me the value 8?"

[rugged geyser]

2

[worldly wing]

right

so in your question you have $\sqrt[3]{216}$

[solid kiln]

kheerii

[worldly wing]

can you do that now?

[rugged geyser]

oh sorry it is not under cube the 3 is power of √216

???

[worldly wing]

what?

[rugged geyser]

is power 3 next to underoot is same as undercube??

[worldly wing]

uh, I don't know what you mean

$\sqrt[3]{}$ does represent cube roots yes

[solid kiln]

kheerii

[rugged geyser]

i think this should be √216 √216√216

multiply

[worldly wing]

nope

that would be $(\sqrt{216})^3$, which is different

[solid kiln]

kheerii

[rugged geyser]

ok i understood

thanks

thanks a lot

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[rugged geyser]

that would be $(\sqrt{216})^3$, which is different

[solid kiln]

cyclonmaster36

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[gaunt temple]

can somebody help me understand the difference between disks, balls, circles and spheres please?

[gaunt temple]

if i understood correctly a disk in R^2 is the set of points that are at a fixed distance from a given point (radius)

a ball is like a disk but in R^3

but if that's the case what are circles and spheres?

(for context, i'm studying topology in analysis)

[twilit latch]

As my analysis prof put it, the ball is an idea

[gaunt temple]

ok thank you!

[eternal adder]

the disk is just a special case of the closed unit ball centred on the origin

[twilit latch]

The disk is the exact same in R^3 as in R^2

Except you restrict it to a plane

[eternal adder]

and the circle is just the boundary of the disk

for other dimensions, use balls/spheres

[twilit latch]

It's like you are take a 2 dimensional slice of your 3d space

Then apply the old definition

[twilit latch]

This also seems to describe a circle, not a disk

To make it a disk you would do within a certain distance

Not at a certain distance

The concept of a ball remains important in pretty much all metric spaces

[gaunt temple]

what is the circumference then 💀

[twilit latch]

It's a very general thing

Not just Euclidean metrics on R^n

[eternal adder]

circumference is the length of the circle (or more generally, boundary of a disk)

you shouldn't really need to care about it in topology

since it isn't a topological invariant

[gaunt temple]

so the problem is that i've always talked about circles when i should've talked about disks. i thought the circumference was the boundary of a circle

[eternal adder]

circles are the bit on the outside; they're not contractible, and the circumference is its length

disk is the circle plus the insides (these are contractible, and simply connected)

[twilit latch]

Circumference is a number (scalar)

[gaunt temple]

if it's a closed disk

[eternal adder]

disks are always closed by definition (as in, if the term "disk" is used unqualified, you should assume that it is closed, so the term "closed disk" is unnecessary)

[gaunt temple]

if it's an open disk the circle isn't in the disk

[eternal adder]

yes, if you specify open then you exclude the boundary

[gaunt temple]

nice i've never seen it that way

[twilit latch]

Have you talked about metrics

[gaunt temple]

not yet

[twilit latch]

You should have the definition of a ball in terms of your metric

[gaunt temple]

so i think i get what disks are now

but i still dont udnerstand what are balls

[twilit latch]

Wait for that then

[eternal adder]

balls and disks are the same thing

except you should probably qualify if the ball is open or closed

[gaunt temple]

so i can use them interchangeably?

open ball = open disk, closed ball = closed disk

[eternal adder]

well

almost

[lavish sequoia]

hi

[eternal adder]

disks are generally centred on the origin, if taking them as a euclidean subspace

[lavish sequoia]

can I DM you?

[eternal adder]

balls can be centred elsewhere

[eternal adder]

if you have a question, please open a thread #❓how-to-get-help

[lavish sequoia]

But I did

and every answer I was given didn't satisfy my only question

[twilit latch]

Interesting, in what sense is a disk a eucledian subspace?

[eternal adder]

as in the set B(1,0)

[gaunt temple]

so a =b = 0

[eternal adder]

inheriting the subspace topology from R^n

[gaunt temple]

if a and b are not equal to 0 then it is a ball

(in R^2)

[amber python]

i mean whether a disc should be open or closed is not something thats set in stone

and depends on your conventions

[twilit latch]

It seems like we might be using different ball notations. Is that a ball of radius 1 centered at 0?

[amber python]

and also whether its centred on the origin or not

[eternal adder]

yes

[amber python]

i feel like you shouldnt be preaching these things just because they happen to be your conventions

[gaunt temple]

am i overcomplicating things

[amber python]

maybe

just follow the definitions in your notes

[twilit latch]

Oh I was getting confused for no reason sorry. For some reason I was thinking of vector subspaces instead of metric space subspaces lol

Someone earlier was asking about vector spaces

[amber python]

if you define a disc to be the same as an open ball good for you

if you define a disc to only be the object in R^2, good for you too

[gaunt temple]

okay

thanks very much everybody for the help

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[radiant portal]

I dont understand something, why is it 3pi/4 divided by 2pi and not 2pi divided by 3pi/4

[twilit latch]

Because the whole disk corresponds to 2pi radians

[radiant portal]

Yes I agree but im still confused

I thought thats why we divide it by 3pi/4

[twilit latch]

You want to see what fraction of the circle 3pi/4 radians is

[twilit latch]

That would give you something bigger than 1

[radiant portal]

Ah ok

I get you now

[twilit latch]

That is like saying a slice of a disk is multiple disks big

[radiant portal]

Yeye makes sense

Thanks I think that answers my question

❤️

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[twilit latch]

Np

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[glossy berry]

Hello , i need help for an exercice , i don't know how to start

[glossy berry]

I have to demonstrate this

<@&286206848099549185> please can anyone help me for this exercice

[lavish karma]

Sure

[glossy berry]

thanks

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[glossy berry]

please <@&286206848099549185>

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[alpine viper]

How to shorten this to the shortest?

(1-(((-0.25)-3*(0.375/7))+(0.375/7)))+((-0.25)-3*(0.375/7))*x+(0.375/7)*x*x

[unkempt smelt]

!original

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[alpine viper]

I made it.

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[tardy hemlock]

(moved to #help-29)

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[fair forge]

How do I reach the last line from the first line?

[fair forge]

Ik as is it’s not a real proof

i is a natural number

[craggy sage]

Ayanokoji asking a ques is weird

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[worldly wing]

<@&268886789983436800>

[rapid mirage]

Thanks

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[rugged geyser]

Teach me how to facorise x³-6x²+11x-6

[rugged geyser]

By a polynomial methord

[wraith hinge]

use factor by grouping

for the first 2 terms, take the GCF of them and so on and so forth for the 2nd pair of terms

[rugged geyser]

Pls tell me of level of class 8

[wraith hinge]

oh wait

[simple haven]

Rational root theorem. The roots, if rational are one of ±1, ±2, ±3, or ±6

[wraith hinge]

i just realized you can’t do factor by grouping

[simple haven]

Just test all possibilities by synthetic/polynomial division

[wraith hinge]

yes rational root theorem

[simple haven]

Also we know by Vieta's theorem that the roots sum to -6

So there's only one possible sum of all real and rational roots.

[rugged geyser]

Pls teach me rational root theoram I do not anything about it

In small

[simple haven]

Ok so

Look at the constant of the polynomial

[rugged geyser]

-6

Then??

[simple haven]

The problem of factoring a polynomial is the problem of turning something like ax^3 + bx^2 + cx + d into a(x - r1)(x - r2)(x - r3)

[rugged geyser]

Ok

[simple haven]

So we can see that -a*r1*r2*r3 has to be equal to d

Do you see why?

[rugged geyser]

Nope pls tell me

[simple haven]

Multiply out a(x - r1)(x - r2)(x - r3)

[rugged geyser]

We will get back all of that equation when we myltiply all of them

ax³+bx²+cx+d

[wraith hinge]

the rational root theorem: you take the coefficients of the first and last term and find the factors of each one

[simple haven]

You should get ax^3 - a(r1+r2+r3)x^2 + a(r1r2 + r1r3 + r2r3)x - ar1r2r3

Now we can identify a = a, b = -a(r1+r2+r3), c = a(r1r2 + r1r3 + r2r3), and d = -ar1r2r3

[rugged geyser]

Ok

[simple haven]

So because we have d = -ar1r2r3, we know that, if d is rational, then if r1 is also rational, we can say that r1 is a factor of d.

[rugged geyser]

Ok

[simple haven]

Up to sign

[rugged geyser]

Now how we will use it to factorise

[simple haven]

So because d is 6, we have r1, if it is rational, must be one of the factors of 6, up to sign, which are {1, 2, 3, 6}

Just use polynomial long division and see if it divides evenly

[rugged geyser]

Ok

[simple haven]

Aka synthetic division

[rugged geyser]

You tried to explain me well sir but I was not able understand it quite leave it I am sorry

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[simple haven]

@rugged geyser do you not understand synthetic division?

Because I can explain that

[rugged geyser]

I know long division but not that

[simple haven]

.reopen

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✅

[simple haven]

It's just long division

But instead of powers of 10, we use powers of x.

So instead of something like 52|733 we do 5x + 2|7x^2 + 3x + 3

Mechanically it works exactly the same

[rugged geyser]

Yes

[simple haven]

If anything it's easier because you do not need to worry about carry

[rugged geyser]

I know that

I got it

I call that sec thing long division only

[simple haven]

Sure.

Synthetic division is a different name for the same thing

[rugged geyser]

Ok I understood that

[simple haven]

I use both terms because I don't know which one a student is familiar with

[rugged geyser]

Ok sir but it will help me in factorisation

?

[simple haven]

Rational root theorem only tells you what the rational roots might be.

It doesn't tell you what they are.

You have to check each one

[rugged geyser]

Oh ok

I understand

Now I have to find it and directly divide that to factorise it??

[simple haven]

Yup

If you get a remainder of 0 then it's a factor

Otherwise you have to try some other value

[rugged geyser]

Ok let me check

If 1 came remainder

I checked from x-1

And 1 came reminder

???

[simple haven]

x-1 | x^3 - 6x^2 + 11x - 6

-5x^2 + 11x - 6

6x - 6

0

I get 0

[rugged geyser]

Wait let me compare

Oh I found my mistake

Ok after 0 came then

???

[simple haven]

Well, now we know that (x-1) is a factor

The result of your synthetic division is a quadratic equation

[rugged geyser]

Ok

[simple haven]

And you already know how to handle those

[rugged geyser]

The result will be x²-5x+5

[simple haven]

6

x^2 - 5x + 6

[rugged geyser]

Yes

Yes

I didn't corrected the previous mistake

Ok we find that then how will we get three of them

[simple haven]

The 2 hardest problems in mathematics are communication, naming things, and off by one errors.

[rugged geyser]

???

[simple haven]

A joke

Well, we know that x^3 - 6x + 11x - 6 = (x-1)(x^2 - 5x + 6)

So our other two roots of the cubic are in the quadratic

[rugged geyser]

Yes

[simple haven]

How do you solve a quadratic?

By factoring it or using the quadratic equation, right?

[rugged geyser]

How we will get third one??

[simple haven]

There are 2 in the quadratic

The third is in (x-1)

[rugged geyser]

Oh ok

Thx

I got it

[simple haven]

Yw

[rugged geyser]

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[manic jackal]

Prove that $\frac{\phi(m)}{m} = \frac{\phi(n)}{n} \text{ if and only if } m \text{ and } n \text{ have exactly the same prime divisors (possibly different powers)}$

[worldly wing]

I assume phi is euler’s totient function?

[manic jackal]

$\phi$ is the Euler's totient function

[solid kiln]

Halex

[worldly wing]

Yeah okay

!status

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2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
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6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[manic jackal]

Indeed

2

[worldly wing]

What have you tried

[solid kiln]

Halex

[manic jackal]

if $\frac{\phi(m)}{m} = \frac{\phi(n)}{n}$ then we get this

[solid kiln]

Halex

[worldly wing]

Indeed

[manic jackal]

How can we end up saying that each p_i and q_i are the same?

[worldly wing]

Now you need to justify that r=t and p_i=q_i

[worldly wing]

Hmm

Take common denominators of both

[manic jackal]

Oh yeah, this

[worldly wing]

And compare the numerators and denominators

[worldly wing]

Also here state somewhere that p1<p2<…<p_r

And same for q

So that you don’t have to deal with rearrangements of the primes at the end

[manic jackal]

Okay

You mean I should operate and expand terms?

Like 1 - 1/p_i = (p_i-1)/p_i

[manic jackal]

With this

[worldly wing]

Yeah, but you did that wrong

[manic jackal]

Fixed, but still don't see why r should be equal to t

[worldly wing]

Since both sides of the equation has fractions which must necessarily be in lowest terms (justify why?), you can directly compare the denominators of both fractions

From there the fundamental theorem of arithmetic does the trick

[manic jackal]

You mean, we can cancel the terms of the numerators and directly compare the denominators?

[worldly wing]

No, we don’t cancel anything

$\frac{a}{b}=\frac{c}{d}\implies a=c, b=d$ provided we know that all of a, b, c and d are positive integers and, importantly, that both fractions are in their lowest terms

[solid kiln]

kheerii

[worldly wing]

That is to say, GCD(a, b)=GCD(c, d)=1

[manic jackal]

$\left(\frac{p_1-1}{p_1}\right)\left(\frac{p_2-1}{p_2}\right) \ldots \left(\frac{p_r-1}{p_r}\right) = \left(\frac{q_1-1}{q_1}\right)\left(\frac{q_2-1}{p_2}\right) \ldots \left(\frac{q_s-1}{q_s}\right)$

[solid kiln]

Halex

[manic jackal]

I know that gcd(p_i, p_i - 1) = 1 but not sure about the whole product

$\left(\frac{\left(p_1-1\right)\left(p_2-1\right):\ldots :\left(p_r-1\right):}{p_1p_2\ldots p_r}\right):=:\left(\frac{\left(q_1-1\right)\left(q_2-1\right):\ldots ::\left(q_s-1\right):}{q_1q_2\ldots :q_s}\right)$

[solid kiln]

Halex

[trim jolt]

@manic jackal Has your question been resolved?

[worldly wing]

Hmm

[worldly wing]

Each of the terms (p_i - 1) must have a factor of 2

Unless p_1 = 2

[manic jackal]

from what I can see, it in fact holds

[worldly wing]

I think it does yeah

[trim jolt]

@manic jackal Has your question been resolved?

[manic jackal]

do you know how can I justify it?

been having trouble to do so

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[dim kite]

Hello

so i think

what im supposed to do here

is like make a row/column 0 and then determinant is 0?

or something

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[jade stirrup]

Im trying to figure out the beginning capital but it doesnt show the beginning capital in the trial balance

<@&286206848099549185>

[subtle lava]

Perhaps explain some of the terms: CR, DR ext.

Your course may have course specific abbreviations that others in this field aren't familiar with

[jade stirrup]

Is their a way to find Capital for the beginning of a month when it isnt given? im in acct 231

<@&286206848099549185>

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[desert tapir]

For questions 3, 4 and 5, can someone reword them / explain to me how i would go about solving them?

like what does f(1) or f(x) = 1 even mean??

[split chasm]

f(1),
on the graph, what is the y-coord when x=1

for: f(x)=1
on the graph, what are the x-coords when y=1

e) what are the set of values for x where y>1

[desert tapir]

ok, when you explain it that way it makes sense

is there a way to rememebr it?

functions always confuse me

f(1) = 0
f(x) = -1 , 1/2, 2
f(x) > 1 = -1, 1/2, 2, 4

? @split chasm

[split chasm]

f(x) = -1 , 1/2, 2
did you mean
f(x) = 1 when x= -1 , 1/2, 2

last one is just wrong

[desert tapir]

yes for the 2nd one

why is the last one wrong?

ohh

is it because its > 1

not greater than / equal to 1

so itd just be 4

[split chasm]

no

can you highlight everywhere on the graph where y>1

[desert tapir]

sure

lemme do it on my paint

the solid line means values that are y = 1

but it cant be those values

so its everything ABOVE the pink line

so itd just be (4,2)

x = 4?

[split chasm]

no

are you saying that (4,2) is the only thing above that pint line?

[desert tapir]

ah i see what ur saying

the lines

so would it be

(-1,1/2) U (2,4]

to account for the parts above the pink line?

[split chasm]

yes

[desert tapir]

i see, ok thank you

i have one more problem i completed myself already but i think i missed a step when doiing calculations

would u be able to just see what i did wrong? its a domain problem

i already did the work but the final answer i got was wrong so im not sure

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[desert tapir]

Can someone explain to me what I’m missing? In previous problems I was told to just ignore the numerator when finding the domain because it wasn’t irrational (such as the numerator being 5x for example) but now that the numerator is square root am I supposed to find the domain of the numerator as well?

I calculated the domain of the denominator and got all real numbers so I’m confused

[south latch]

The domain is all the possible inputs for the function that don't cause problems

So if there are parts of the function that could cause a problem, you need to identify those values and exclude them from the domain

Your function will have problems if the denominator equals zero (because of what x is) and if the inside of the square root is less than 0 (because of what x is)

To find those values, just solve this equation and this inequality:

x^2 - 9x + 8 = 0
2x-5 < 0

The x values that make these statements true must not be in the domain.

[desert tapir]

ahh so i need to also calculate the domain of the numerator because it is square root?

[south latch]

It's not the domain

It's just the solution of the inequality

The domain is all the values you are allowed to plug into the function

And it's not the numerator, it's the argument of the square root

That cannot be negative

So we solve to find out what x values make it negative, and exclude them from the domain

[desert tapir]

sorry can you try to use simple terms

im in precalculus and im having trouble understanding a lot

like examples may be easier to explain.. if thats ok

i understand the denominator can not equal 0

or negative

so we put the bottom >= 0

but the numerator has to be > 0 correct, because we cant square root 0?

[south latch]

"Argument" is what you plug into operators like sqrt(), sin(), cos(), etc

Arguments are not always single values, they can be expressions

The argument inside the square root is an algebraic expression

2x-5

The argument is not allowed to be negative if we want the result to be a Real number

That is why we are checking where 2x-5 < 0

The denominator is allowed to be negative

That won't cause any problem

[desert tapir]

ok

so for my problem i need to put

2x - 5 > 0 ?

or 2x - 5 =/= 0

[south latch]

I don't think (by inspection) that it can be negative though nevermind, it can (when x=3, for example)

[south latch]

Less than

You need to exclude those values from the domain

You can, I suppose, see what x values make that positive, and then remove the zeros of the quadratic from that set

But that's more info to hold onto

[desert tapir]

ok..

[south latch]

Imo

2x-5 < 0
--> x < 5/2
So x must not be less than 5/2

What happened to your attachment?

Oh, nvrmnd

[desert tapir]

what do u mean

oh im trying to solvei t right now

[south latch]

You got the quadratic factors right

So x cannot be 1 or 8, either

[desert tapir]

ok so i kind of understand what ur saying

but im just trying to do it the way my teacher taught me

im sry i dont mena to be rude

so ill show u my work

[south latch]

But we already need x to be greater than 2.5, so...

[desert tapir]

correct

so far i have this

for the numator

the interval is

[5/2 , inf)

and for the denominator

x cannot equal 1 or 8

so would the interval be

[5/2, 8) U (8, inf)?

[south latch]

Yes!

[desert tapir]

sry for making u explain all that

i was trying to follow along but tbh it was hard

i learn more from examples tbh

so i just tried solving it

[south latch]

The meaning of domain is very easy

[desert tapir]

isnt it just values that make the denominator 0?

[south latch]

Rigorous definitions make it sound complicated, or like you will mess it up if you do t follow approved procedures

[desert tapir]

lol thats how i know it

[south latch]

Domain is just all the x values you are allowed to plug into the function because they get you a Real number back

[desert tapir]

and if your denominator is 0, its not a real number right

[south latch]

Right, because then the implication is that the output is + or - infinity, which is not reachable

[desert tapir]

okay thank you

[south latch]

Np

[desert tapir]

appreciate the help

🫡

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[desert tapir]

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[rocky karma]

hey, maybe this is a weird question, but trying to understand..
if a function is said to be derivable twice at a interval, what can I take from it?

[rocky karma]

like i really dont understand what I can take from this info

[nova spire]

well it's useful for many things

for example you know that critical points of a function f are given by f'(x) = 0

but you don't know if they correspond to local maxima, minima, etc...

f''(x) yields some answers to that

if f"(x) > 0, then it's a local minima

if f"(x) < 0, it's a local maxima

otherwise we can't conclude

[rocky karma]

ah okay that is interesting

[nova spire]

another use of 2nd order

is approximating your function around some point

you've seen that with derivable once

f(x) around a point a is approximately f(a) + f'(a) (x-a)

this is the equation of the tangent of f at point a

it's the best line that approximates f(x) around that point

if you have twice differentiable function

you can not only find the best line to approximate f(x) around some point x0

but you can find the best quadratic to approximate f(x)

like ax^2 + bx + c

which values of a,b,c should we pick? well we get them using this:

$f(x) \approx f(x_0) + f'(x_0)(x-x_0) + \frac{f''(x_0)}{2}(x-x_0)^2$

[solid kiln]

rafilou2003

[nova spire]

this is the best 2nd order polynomial for approximating f(x) around x_0

example :

cos is twice differentiable over the real numbers

let's look at x_0 = 0

we have cos'(0) = -sin(0) = 0

and cos"(0) = -cos(0) = -1

[nova spire]

with this formula, we get that $\cos(x) \approx 1 - \frac{x^2}{2}$

[solid kiln]

rafilou2003

[nova spire]

(for values of x around 0)

see accuracy here

[rocky karma]

okay that is some great info, i need to sit on it a bit so it will absorbs haha

thank you very very much

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[dim kite]

,rotate

[dim kite]

hello, i dont understand what i did wrong?

the difference is in the 3*B^-1

i took out the 3 so it became 3^4

oh i just realized

its (3B)^-1

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[marsh forum]

Prove that for a positive integer $n$ and any integer $a$ gcd(a,a+n) divides n

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

so I'm unable to understand the question due to my poor english

I feel this means prove $n|(a+n)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

is that right?

[amber python]

Prove the following statement:

\medskip
If $n \in \Z_{>0}$ and $a \in \Z$ then $\gcd(a, a + n) \mid n$.

[marsh forum]

ah, ok

thanks!

ok, so we know $d=nx+(a+n)y$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

so $1=\frac{n}{d}x+\frac{a+n}{d}y$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[amber python]

if you've seen bezouts, surely you've seen the euclidean algorithm

[marsh forum]

no,that's the next subsection

eudlid's lemma

yes

not the algorithm though

[amber python]

oof

[marsh forum]

I guess I could consider it case wise

where n is 1,2....a

[fair bison]

you don't need to use bezout's theorem

just use the definition of gcd

[amber python]

they didn't tell you anything about calculating gcds?

[fair bison]

if d = gcd(a, a+n), then what do you know about d?

[marsh forum]

$d|a and d|(a+n)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

no

[fair bison]

yeah, and what is the definition of |?

[amber python]

okay hopefully you'll be able to see this next step then

[marsh forum]

a=kd$

and a+n=k_2d

[fair bison]

okay and then how can you get n from that

[marsh forum]

so n= (k_2-k_1)d

oh

[fair bison]

yes!

[marsh forum]

that's it?

[amber python]

ye

[marsh forum]

why don't I ever see the obvious things 😭 . Thanks a lot both of you!

[amber python]

you can write this something like uh

,, d \mid a, ; d \mid a + n \Implies d \mid (a + n) - a

[marsh forum]

d|((a+n)-a)?

[vagrant marsh]

yes

divides is additive

[marsh forum]

ah

thans

*thanks

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[amber python]

but i used subtraction

[vagrant marsh]

what is subtraction but addition if you could time travel

[amber python]

i cannot time travel

[vagrant marsh]

unfortunate

guess you wont get it then

[amber python]

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[small wagon]

can you help me solve this parabola problem

[hallow spruce]

!status

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7. None of the above

@small wagon Has your question been resolved?

[small wagon]

i just ended up going to do my other questions lol

i dont know what to do since i know that the directrix to vertex and vertex to focus is the same distance

and that entire distance is use to get to the first point by the focus

but i dont know how to do that with just a point

[hallow spruce]

i know that the directrix to vertex and vertex to focus is the same distance
this is true of all points on the parabola, not just the vertex, is it not?

[small wagon]

i think so like the point that are from the same line the focus is on is the same distance as the directrix to the focus

[small wagon]

but the other points depends

but idk how to solve it

[hallow spruce]

i think you ought to go back and look at the properties of the focus and directrix again

[small wagon]

ya i looked at the notes and stuff the directrix to focus is the same distance as the focus to the first 2 points on the parabola

they are both double the p value

[hallow spruce]

"first two points"?

[small wagon]

like if the vertex is (-3,4) and the directris is x=-6 and focus is (0,4)

than to plot

you can use the distance bewteen the directris to focus which is 6

and two point for the parabola you can plot is (0,10) and (0,-2)

like its just makes it easier to graph instead of finding a random decimal point location on the parabola

[hallow spruce]

i see

well unfortunately that particular construction won't be of much help here

[hallow spruce]

this was a rhetorical question. Every point on the parabola has the same distance from the focus as it has from the directrix

[small wagon]

wait how does every point have the same distance?

what about the point that are farther away how would those be the same as points closer to the focus its not like its a circle right

so if it is all the same how would i solve this?

[hallow spruce]

it's not that they all have the same distance

it's that the distance from the focus and the distance from the directrix are the same

[small wagon]

ya ik that the vertex is the same distance from the directrix as well and the focuw

[hallow spruce]

please stop limiting yourself to the vertex

every point behaves this way

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[wraith hinge]

i dont understand the question

[cyan zinc]

let me think this through

we have that:

A^(-1) = 1/det(A) C^T

[trim jolt]

@wraith hinge Has your question been resolved?

[cyan zinc]

let's rewrite as:

[wraith hinge]

hold up

[wraith hinge]

how?

[cyan zinc]

are you familiar with the adjugate?

[wraith hinge]

adjugate?

adjoint?

[cyan zinc]

yes

[wraith hinge]

yea i am

[cyan zinc]

the transpose of the cofactor matrix

[wraith hinge]

yea

[cyan zinc]

that relation should be familiar to you if you know how to take the inverse of a matrix

[wraith hinge]

shouldnt it be C^T/det(a)?

[cyan zinc]

is that not the same?

[wraith hinge]

oh wait i thought u have it on the denominator lol

[cyan zinc]

yep

let me just walk you through this

do you agree that:
A^(-1) = 1/det(A) C^T

[wraith hinge]

yea

[cyan zinc]

where C^T is the transpose of the cofactor (known as adjugate or adjoint)

[wraith hinge]

yea

[cyan zinc]

let's multiply both sides by A

I_n = 1/det(A) A C^T

so far so good?

[wraith hinge]

yea

[cyan zinc]

where In is the nxn identity matrix

[wraith hinge]

yea

[cyan zinc]

det(A) I_n = A C^T

[wraith hinge]

wait

${|A|}^{n-1}=|C|^{T}$

[solid kiln]

pun pun

[wraith hinge]

taking transpose both the side should get us the answer right?

[wraith hinge]

i got this result from this

[cyan zinc]

you don't need the T anymore once you take the determinant

you have the answer already

det(C) = det(C^T)

[wraith hinge]

but in my notes, there's this property that says det(A^T)=(det(A))^T

this might be wrong cus transpose of a scalar makes no sense

[cyan zinc]

yes. but det(A) is just a number

i mean it's technically true, numbers are just 1x1 matrices

[wraith hinge]

oh yea

oh yea we have det(A) here, i thought it was a matrix or something

ok ok thanks i get it now

thanks for the help, i was just having some trouble understanding the question

imma close now

thanks

.close

[trim jolt]

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[tired marsh]

∫(6x^2-4x+7)/sqrt(2x-1)dx

[tired marsh]

evaluate this

[trim jolt]

@tired marsh Has your question been resolved?

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Use .reopen if this was a mistake.

[tired marsh]

.reopen

ok so (2x-1)(3x-1/2) = 6x2-4x+1/2

and this can help evaaluating the integral

but is there another way? This factorising method gives you a long a messy process.

[proper kernel]

you can save some time by doing u = 2x - 1

you can always do u = mx + b with no consequences, so u = 2x - 1 is a safe option to simplify the sqrt(2x - 1) in the denominator to sqrt(u)

[tired marsh]

what about the part above

[proper kernel]

🤔

u = 2x - 1

so x = (u + 1)/2

[tired marsh]

oops 😓

ok

.close

[trim jolt]

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Use .reopen if this was a mistake.

[trim jolt]

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[marsh forum]

prove $6|a(a^2+11)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

where $a\in \mathbb{Z}$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

so to start I have to prove it's always even

[vagrant marsh]

whenever you have a divides problem with a small integer, what should you do?

[vagrant marsh]

oh that works too

[marsh forum]

Try to simpliify it to prime number|a

[vagrant marsh]

sounds good

[marsh forum]

is that actually doable though?

[vagrant marsh]

yeah

if you want to go snow mode, you can show 6|a((a^2-1)+12)

i would just go by cases for such a small integer

[marsh forum]

case 1 would be when a is odd

[marsh forum]

so this is the same as $6|a(a^2+5+6)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

so I have to prove 6|a(a^2+5)

[vagrant marsh]

try taking out another 6

[marsh forum]

hmm

ok

$6|a(a^2-1+6)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

so 6|a^3-a

is what I have to prove

[vagrant marsh]

leave it factored

[marsh forum]

$6|a(a^2-1)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

now if a is even

oh

riht

[vagrant marsh]

you could factor it more

[marsh forum]

$6|a(a-1)(a+1)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

atleast one of which is guranteed to be a multiple of 3

[pulsar dust]

Yes

[marsh forum]

and atleast one of 2

thanks!

[vagrant marsh]

snow mode

[pulsar dust]

Snow mode fr

[vagrant marsh]

thanks snow

[pulsar dust]

When in doubt use the snow mode

[marsh forum]

another similar problem,$24|a(a^2-1)$ where a is an odd integer

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

so a=2k+1

so I re-write it as $24|(2a+1)((2a+1)^2-1)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

then 1 as 25-24

[pulsar dust]

Btw 8|n²-1 when n is odd

[marsh forum]

hmm, yeah

[vagrant marsh]

why not leave it factored

or factor it

when youre doing division, factoring is everything

[marsh forum]

this is the same as $24|((a+24)-24)(a^2-1)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

this iis the same as this

[pulsar dust]

Not really required

[marsh forum]

I mean using this proving 3|a is enough

were a=2n+1

which is false for n=2

[pulsar dust]

It doesn't work like that

[vagrant marsh]

since 3 is coprime with 8, you 3 can be a factor of either without worrying about overlap

but have you factored it

and then plugged in a = 2n+1

[marsh forum]

hmm

and that causes problems I pressume?

[vagrant marsh]

no

[marsh forum]

no?

[pulsar dust]

Not really it doesn't mean that 3|n²-1 is not a case also
since this expression is divisible by both 3 and 8 this it will be divisible by 24

[marsh forum]

ah, right

[marsh forum]

which is why I wrote it like this

[pulsar dust]

Sure it works but you are overcomplicating it

[marsh forum]

so I now just have to show $24|a(a^2-1)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

for a=2k+1

[marsh forum]

how else would I do it?

$24|(2k+1)(4k^2+1+4k-1)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[vagrant marsh]

oh my god youre not listening to me

[pulsar dust]

i would do it like this since 8|n(n^2-1) and 3|n(n^2-1) when n is odd
this implies that 24|n(n^2-1)

[marsh forum]

what am I supposed to do?

I may have missed what you suggested

sorry

[marsh forum]

oen minute

[vagrant marsh]

its fine, conv got you

[pulsar dust]

yay

[marsh forum]

huh

thanks so much

that was suprisingly easy

[pulsar dust]

lol yeah

[marsh forum]

should have thought of that smh

[pulsar dust]

but i knew before that 8|n^2-1 was true

[marsh forum]

👍

[pulsar dust]

or else i would have been stuck like you

any other doubts @marsh forum i may be able to help

[marsh forum]

maybe, i'll ping you if I have any?

[pulsar dust]

if im online yes

or else itll waste your time

[marsh forum]

ok

thanks

.close

[trim jolt]

Channel closed

Closed by @marsh forum

Use .reopen if this was a mistake.

[marsh forum]

.reopen

[trim jolt]

✅

[marsh forum]

given an odd integer show $12|(a^2+(a+2)^2+(a+4)^2+1)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

so to start i expanded $a^2+(a+2)^2+)a+4)^1+1$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

@pulsar dust , you wanted me to ping you

$\frac{3a^2+12a+21}{12}$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

now to prove $3|(a^2+4a+7)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

$4|((a+2)^2+3)$

what am I doing wrong

[tacit echo]

4 should divide it

[marsh forum]

wait, what

oh

right

[solid kiln]

ƒ(Why am. I here)=I don't Know

[tacit echo]

4|a^2 + 7

or 4 | a² + 3

this simplifies it

[marsh forum]

how did you get a^2+3?

[tacit echo]

a² + 4a + 7 = a² + 7 = a² + 3 (mod 4)

[marsh forum]

I don't know modular arithmatic

[tacit echo]

4 | a² + 4a + 7
iff 4 | a² + 7
iff 4 | a² + 3

[marsh forum]

wait

oh

right

yeah

makes sense

thanks

so now to prove 4|a^2+3

[twilit kraken]

Remember that a is an odd number.

[tacit echo]

now you can rewrite a as the odd form

[marsh forum]

right

yeah, that works

thanks so much !

.close

[trim jolt]

Channel closed

Closed by @marsh forum

Use .reopen if this was a mistake.

[vagrant marsh]

alternately you can change it to a^2-1 and FACTOR

[marsh forum]

ah

thanks

[trim jolt]

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[dusk bear]

in order to solve this do i haave to change the minimum bound on the z bounds to include a y instead of just x ?

[dusk bear]

or do i keepthe bounds as they are and just solve accordingly

[worldly wing]

I mean, $\int_0^1\int_0^x\int_x^{2x}7yz\cos(16x^5)\dd{z}\dd{y}\dd{x}$ should be easy enough to solve

[solid kiln]

kheerii

[dusk bear]

i solved it twice that way and im unsure whether or not i did it right

[worldly wing]

!show

[trim jolt]

Show your work, and if possible, explain where you are stuck.

[dusk bear]

the only thing i can think of is the bound because i think the z bounds need both x and y to be included?