help-28

[elfin lynx]

yeah done

[simple totem]

should look like this

[elfin lynx]

when I did that I got this

[simple totem]

perfect

did that answer your question of where the other 5(x^2+3)^3 went?

[elfin lynx]

yes

now I'm just trying to get

f'(x) simplified

5(x^2+3)^3 * 9x^2 + 3

[simple totem]

5(x^2+3)^3 * (9x^2 + 3)

can you factor anything out of (9x^2 + 3)?

yes

you had it

just move the 3 to the front of the expression and you've reached your answer

[elfin lynx]

3(3x+1)

[simple totem]

3(3x^2+1)

[elfin lynx]

what about the 15?

[simple totem]

5(x^2+3)^3 * (9x^2 + 3) = 5(x^2+3)^3 * 3(3x^2 + 1) = 5 * 3 * (x^2+3)^3 * (3x^2 + 1)

what is 5 * 3

[full forum]

@elfin lynx Has your question been resolved?

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[verbal valley]

Hii! I need with factoring, im the worst at it every since I learned it in summer school. Now I need it for geometry and im horrible at it. Please help me understand 😭

[hollow sable]

[verbal valley]

Im trying to figure out to type it give me oneee secondd

I cant figure out how to type exponents so ill do ^ for raised to its power

4a^2 b^2 - 16ab^3 + 8ab^2 c

[hot herald]

!status

[full forum]

What step 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

[verbal valley]

1

[hot herald]

are you familiar with the basic distributive property?

[verbal valley]

yeah

[hot herald]

pq + pr = ?

can you factorise that

[verbal valley]

isnt it just p(r+q)?

[hot herald]

yes

usually the first thing to look for when factorising is whether all terms have a common factor

similarly
pq + pr + ps = p(q+r+s)

can you try identify such common factor(s) for your expression and apply the same idea

[verbal valley]

yeah, theres 5 ab's

[hot herald]

wdym by 5 ab's

[verbal valley]

like there is five terms of ab

[hot herald]

wdym by 5 terms of ab

you're expression only has 3 terms

where's 5 coming from

[verbal valley]

oh you meant that mb

so, ab(16 +8)?

[hot herald]

no

apply the same idea as you would for

pq + pr + ps = p(q+r+s)

can you first identify the factor(s) common in all terms of
$$4a^2b^2 - 16ab^3 + 8ab^2c$$

[glossy valve]

ℝamonov

[verbal valley]

yeah i think, a, b, c, and ab

[hot herald]

no

c isn't a common factor here

just like when for asking the common factor of
pq + pr,
or
pq + pr + ps
its simply just p

[verbal valley]

okay

[hot herald]

a,b,ab are all common factors,
but there's no need to be repetitive,
ab already includes a and b

you should also consider the common factor of the constant coefficients when factorising

do 4,16,8 have a common factor?

[verbal valley]

yeah, 4 and 2

[hot herald]

what's the hcf/gcd

that's just be 4 right?

[verbal valley]

yeah

[hot herald]

so overall do you agree that 4ab is a common factor here

[verbal valley]

mhm

[hot herald]

i want a clear yes/no

[verbal valley]

yes

[hot herald]

can you try factoring that out
similar to how you went from
pq + pr to p(q+r)
and how
pq + pr + ps = p(q+r+s)

if needed you can consider
$$=4ab\cdot \br{\blue{\frac{4a^2b^2 - 16ab^3 + 8ab^2c}{4ab}}}$$
simplify the stuff in $\blue{\text{blue}}$; keeping the parentheses

[glossy valve]

ℝamonov

[verbal valley]

i dont even know where to start, ill just talk to my teacher on monday thanks for trying though

[hot herald]

i dont even know where to start,
i've just given you an idea of that

recall the basic property
pq + pr = p(q + r)

do you fully understand that

[verbal valley]

i guess yeah

[hot herald]

wdym by you guess,

do you have any doubts at all about it

[verbal valley]

idk, i just dont understand it that well with numbers when added to the terms

[hot herald]

p is just the variable representing a common factor

and q the variable representing the expression that when multiplied to p gives pq

[verbal valley]

yeah

[hot herald]

the same idea is applied with more elements present

also i forgot the ^2 on the b earlier,
the hcf/gcd is actually ab^2 here

lets consider another example first

would you be able to factorise
$$5ab + 25a$$

[glossy valve]

ℝamonov

[hot herald]

identifying any common factor is fine, doesn't necessarily need to be the hcf/gcd
you can factor that out first and identify/factor more if needed

[verbal valley]

okay

[hot herald]

so first step,
can you identify a common factor here

[verbal valley]

5

[hot herald]

ok

and applying

pq + pr = p(q + r)
can you factor 5 out of your expression

[verbal valley]

yeah

[hot herald]

what do you get after doing that

[verbal valley]

isnt it 5(ab+5a)

[hot herald]

yes, factoring out 5 gives that

now

you can factor that out first and identify/factor more if needed
5**(ab+5a)**
does the terms in the expression in bold: (ab+5a)
have any common factors

[verbal valley]

it wouldnt right?

[hot herald]

no

there is indeed a common factor there

[verbal valley]

just a then

[hot herald]

yes, a is a common factor

can you factor that out

[verbal valley]

yes

a(b+5) right?

[hot herald]

yes, ab + 5a = a(b+5)

so overall you'll have
$$5ab + 25a = 5\underbrace{(ab + 5a)}_{a(b+5)} = 5a(b+5)$$

[glossy valve]

ℝamonov

[hot herald]

does that make sense

[verbal valley]

yeah

[hot herald]

if you can identify $\red{5a}$ as a common factor
$$\red{5a}\blue{b} + \underbrace{25a}_{\green{5}\cdot \red{5a}}$$
you can get
$$\red{5a}(\blue{b} + \green{5})$$
immediately

[glossy valve]

ℝamonov

[hot herald]

without factorising twice

[verbal valley]

alright that makes sense

[hot herald]

if you're struggling to identify the expressions in the parentheses, you can consider
$$pq + pr = p\cdot \frac{pq + pr}{p} = p\br{\frac{pq}{p} + \frac{pr}{p}}$$

[glossy valve]

ℝamonov

[hot herald]

applying that here
$$5ab + 25a = 5a\br{\underbrace{\frac{5ab}{5a}}{b} + \underbrace{\frac{25a}{5a}}{5}}= 5a(b+5)$$

[glossy valve]

ℝamonov

[verbal valley]

okay

[hot herald]

and the same idea can be applied to your question
$$\begin{aligned} &=4ab^2\cdot \br{\blue{\frac{4a^2b^2 - 16ab^3 + 8ab^2c}{4ab^2}}} \
&= 4ab^2\br{\blue{\frac{4a^2b^2}{4ab^2} - \frac{16ab^3}{4ab^2} +\frac{8ab^2c}{4ab^2}}} \end{aligned}
$$

[glossy valve]

ℝamonov

[verbal valley]

mhm

[hot herald]

would you be able to simplify
$$\frac{4a^2b^2}{4ab^2}$$

[glossy valve]

ℝamonov

[verbal valley]

could you do 8ab^4

[hot herald]

wdym by

do 8ab^4

[verbal valley]

like 8ab to the power of 4

[hot herald]

wdym by "do"

[verbal valley]

would it be an answer?

[hot herald]

it'd be wrong

consider
$$\frac{4a^2b^2}{4ab^2} = \frac44\cdot \frac{a^2}{a} \cdot \frac{b^2}{b^2}$$
would you be able to simplify now?

[glossy valve]

ℝamonov

[verbal valley]

idk none of this is making sense

[hot herald]

do you know the definition of exponents/raising something to integer powers
and/or exponent laws

would you be able to express a^2 in another way

idk none of this is making sense
which part exactly

[verbal valley]

all of it

[hot herald]

which part specifically

i've gone through multiple parts

[verbal valley]

i think im just gonna go to bed and give up for the night

[hot herald]

and you claimed that you understood them all

before i continued onto the next part

[hot herald]

ok, rest first

[verbal valley]

.close

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[celest fulcrum]

Idk if this question wroks here but its math soo

[celest fulcrum]

im trying to learn c++ and I cant get my clion to work

[rose rain]

wrong channel but

[celest fulcrum]

if anyone knows it would be extremely helpful

[rose rain]

let me answer this anyway

executable is not selected

[celest fulcrum]

thank you lol

Its not in the target

[rose rain]

wrong server too

[celest fulcrum]

and the file is not in the target

[rose rain]

this is a cmakelists issue

[celest fulcrum]

what server should I use lol

[rose rain]

figure out how to add it in

google "how to add main.cpp" into cmakelists

or just make a whole new project tbh

[celest fulcrum]

lets try that lol

ty ty

which c++ type is the best if im here rn

is 23 worth it

.close

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[fathom arrow]

#help-3

@jade bluff @atomic blade

[atomic blade]

Yeah it got closed

[jade bluff]

yes

[atomic blade]

Due to timeout

[atomic blade]

Did you read what I sent you earlier

[fathom arrow]

so it's (2x+6) - (x^2-3x)

that's the equation initially right?

[fathom arrow]

yes

[jade bluff]

@atomic blade can you help me in something

[atomic blade]

Well it should be for the integrand yes

[jade bluff]

when u finish this

[atomic blade]

Get your own channel

And no don't individually ask me to help you

[jade bluff]

why did u ping me

[fathom arrow]

i pinged you because i thought you were trying to help

but the channel was closed

[fathom arrow]

so then it is

2x+6-x^2+3x

[jade bluff]

yes yes

are u trying to find x

?

[fathom arrow]

and the defined integral is going to be from 6 to -1?

[atomic blade]

Why would it go backwards

[jade bluff]

😂

[atomic blade]

That makes no sense

[atomic blade]

@jade bluff just get your own channel

[fathom arrow]

6 would be on top

[atomic blade]

-1 to 6

[fathom arrow]

and -1 on the bottom

of the integral

[atomic blade]

Bounds always left to right for dx

It would NO sense to go backwards

Otherwise it would be negative

The area between two curves should never be negative. It measures absolute area

[fathom arrow]

$\int_{-1}^{6}2x+6-x^{2}+3x$

[glossy valve]

cynicalcola

[fathom arrow]

this is what i mean

[atomic blade]

Oh then you're fine

But that's not 6 to -1

That's -1 to 6, which is how you're supposed to do it

[fathom arrow]

hmmm

[atomic blade]

Lemme show you what happens

[fathom arrow]

i integrate

[atomic blade]

,w integrate x^2 0 to 2

[fathom arrow]

then i substitute by 6

[glossy valve]

Results provided by WolframAlpha

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Requested by Umbraleviathan#3607

[atomic blade]

,w integrate x^2 2 to 0

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Umbraleviathan#3607

[fathom arrow]

and then subtract "from" substitution of 6 the substitution of -1

f(6)-f(-1)

[atomic blade]

Yeah that's FTC

[fathom arrow]

so is that correct?

[atomic blade]

If f(x) is the antiderivative of (2x+6) - (x^2 - 3x), then yes

[fathom arrow]

ok

before i find the antiderivative

is it gonna be

2x+6-x^2+3x

-3x becomes +

-x^2+5x+6

antiderivative is

-x^3/3 + 5/2x^2 + 6x

[atomic blade]

That's def not the anti

Oh wait

Yeah my bad I saw the "2x + 6..." and thought that was your anti

,w Integrate (2x+6) - (x^2 - 3x)

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Umbraleviathan#3607

[full forum]

@fathom arrow Has your question been resolved?

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[celest pollen]

How does it get to 6%?

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[eager canyon]

i cant find the third root

[limber flicker]

which third root

[full forum]

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

[shell pulsar]

instead of factoring try to recall a formula for the coefficients in terms of the roots

[sharp vine]

if 5 + 2i is a root then also 5 - 2i is a root

use this

[eager canyon]

so i know two roots:
5+2i and 2-2i

but i dont know how to find the third one

[limber flicker]

typo?

[eager canyon]

yes i mean 5-2i

[limber flicker]

nicu

[sharp vine]

do (x - 5 - 2i)(x - 5 + 2i)(x - a) = f(x)

[eager canyon]

OH thank you!!!

[shell pulsar]

I'd just use that the product of the roots is - the constant coeffcient

[wide sundial]

surely you can use the pentagon thing

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@eager canyon Has your question been resolved?

[sharp flame]

the what

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[placid fiber]

what is a pentagon thing😂

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[wide sundial]

the angle thing

[wide sundial]

i forgot what it's called but cant u get the other roots from the angles

roots of unity

[placid fiber]

oh sorry, I didn't something like that too exists

I thought how can you relate a pentagon with a cubic

[wide sundial]

oh i misread it as a quintic

lol

[sharp flame]

Oh right yeah

The roots form a shape in the complex plane

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[wheat kernel]

Permutation group - group theory

[full forum]

What step 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

[wheat kernel]

5

[full forum]

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

[wheat kernel]

I am stuck at the part where they did a^-1 xa

[wheat kernel]

Also is the solution correct?

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@wheat kernel Has your question been resolved?

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[wet vortex]

The apex of the defined angle ABC is 13 dm away from the circle center O, and BC = 12 dm. Calculate the length of the circle.

[onyx glen]

!status

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6. None of the above

[wet vortex]

1

[onyx glen]

consider triangle BOC

[wet vortex]

ok

then what

[onyx glen]

BOC is a right triangle.

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[silver valve]

What does this question mean?

[rocky vale]

They want you to find a formula for the number of squares in the nth figure

[silver valve]

Oh

Alright so would it y=x+3

Nvm

What would the formula be

[narrow granite]

take the center alone

that's 1

notice how each element of the sequence divides itself into 3 sides of n - 1 squares

n is the rank of the sequence's element

so the number of tiles would be

Un = 1 + 3(n-1)

[silver valve]

I figured it out

It’s just y=3x-2 right

[narrow granite]

yes

[silver valve]

Ty

.close

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[undone axle]

how would i do this

[frosty geyser]

does this formula sound familiar?:\
$\Delta v^2=2\cdot \sin({incline ; angle}) \cdot s$

[glossy valve]

Jigglyproff

[frosty geyser]

where s is the distance the block moves, incline angle is the angle the slope has to the horizontal plane of gravity

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[wintry zealot]

A system of linear equations with integer coefficients and right-hand sides of the equations is given. how can i prove that if this system has a real solution, then it also has a rational solution?

[brittle steeple]

Are you allowed linear algebra stuff?

[wintry zealot]

yes

[brittle steeple]

you could think about a vector space over Q

the conditions on when the system is consistent

and if it isn't consistent over Q, can we find there's no real solutions either

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[full forum]

Please don't occupy multiple help channels.

[primal ermine]

can explain how to do

[stiff musk]

what's the goal, just simplify?

[primal ermine]

yea

[stiff musk]

what are the factors of 54?

[primal ermine]

27

2

[stiff musk]

yea

and how does 27 factor

[primal ermine]

3 and 9

[stiff musk]

yep

keep going

factor 9

[primal ermine]

3 and 3

[stiff musk]

yep

so 54 = 2x3x3x3

aka 2 x 3^3

that cube is good because it will cancel with the cube root right?

[primal ermine]

yea

[stiff musk]

so what do you get for the cube root of 54 if you simplify it

[primal ermine]

you mean this 3.7797.....

[stiff musk]

i was thinking if you want to leave it in exact form

$$\sqrt[3]{54} = \sqrt[3]{2 \times 3^3} = 3\sqrt[3]{2}$$

[glossy valve]

Bungo

[stiff musk]

for the rest you have the right idea but it's not quite right

$$\sqrt[3]{p^3} = p$$ is correct

[glossy valve]

Bungo

[stiff musk]

what about $$\sqrt[3]{q^5}$$

[glossy valve]

Bungo

[primal ermine]

my guess would be q cuberoot q^2

[stiff musk]

yes

in your screenshot you forgot the cube root on q^2

and you forgot the little 3 on the root of 2, so what you have written is square root of 2 instead of cube root of 2

[primal ermine]

i see

thank you

.close

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[paper lion]

how to prove it

like its hard to me to understand this first set

functions from Z set to set of functions from Y to X?

[normal tree]

yeah basically

[paper lion]

any hints to find a bijective function between them?

[full forum]

@paper lion Has your question been resolved?

[normal tree]

I'll use some weird notation, but it's so things are clear

yknow

like

anyway

Let's say we have a function in the first set f: Z -> X^Y

okay wait a second lemme figure out how to say things without being wrong

[full forum]

@paper lion Has your question been resolved?

[severe linden]

Map $f\colon Z \rightarrow (Y \rightarrow X)$ to $(y, z) \mapsto f(z)(y)$

[glossy valve]

M8732

[paper lion]

what does it mean

[full forum]

@paper lion Has your question been resolved?

[devout valley]

Basically, given a function $f$ from $Z$ to $X^{Y}$, take that and then create a new function from $Y \times Z$ to $X$ as per mentioned (so basically you have that for any $z\in Z$, that $f(z) : Y \to X$, so then create the function $\phi : (X^{Y})^{Z} \to X^{Y \times Z}, \phi(f) : (y,z) \mapsto (f(z))(y)$

[glossy valve]

chartbit

[paper lion]

my brain processing

[devout valley]

That $\phi$ is what you want to be the bijection between them, if that's any clearer?

[glossy valve]

chartbit

[paper lion]

i have a problem ith it

like

[devout valley]

[btw reply to the bot so it doesn't close the channel ]

[paper lion]

why (y,z) is there

Φ (f) is fuction that do operations on function f right?

[devout valley]

Basically you want to get a function from $Y \times Z$ to $X$, given a function from $Z$ to $X^{Y}$

[glossy valve]

chartbit

[devout valley]

Yea it takes a given function but does as per above, if you can kinda see it?

[paper lion]

o yeaa

finally i start getting it

thank u

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[kindred viper]

then A have to be equal to

{{{0}}}

right?

wait

no

A = {{0}, {{0}}}

it have to be the correct answer

[rocky vale]

yes

well except those 0's are empty sets

[kindred viper]

i mean, yes, it was just to write it easier

thanks

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[rocky vale]

just making sure lol

np 👍

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[severe basin]

A four digit number has 7 in the thousands place and 3 in the hundredth place. This number is divisible by 3,7 and 9. How many different values the unit's place can take?

[onyx glen]

is "face value" the exact term used by the problem statement, and was the problem statement written in english, or is this translation bullshit

and are you sure you meant "hundredth place" and not "hundreds place"...?

[severe basin]

I made up the "face value"

[onyx glen]

well, ok, "hundredths place" would make no sense

we're talking about divisibility so the number in question is an integer

anyway

so your number is 73__ and it's divisible by 7 and 9

[severe basin]

The number can be expressed as 7000+300+10a+b, where a and b are single digit whole numbers.

[onyx glen]

yeah, sure.

is there a reason why you insist on writing 7000+300 and not 7300?

[severe basin]

Umm ok, it can be written as 7300, much simplified

[onyx glen]

anyway, continue.

[severe basin]

10+a+b mod 3
10+a+b mod 9
should equal zero.

[onyx glen]

may i stop you here again

do you understand that divisibility by 9 implies divisibility by 3, and thus we need not consider divisibility by 3 separately?

[severe basin]

Oh, yes

I tried working different numbers and found that only a+b=8 suffices

For divisibility by 7, there are like 2 rules I know but I suppose I'm not applying any of them correctly

[onyx glen]

if you are not explicitly instructed to apply a particular divisibility test for 7, i would simply search through all possibilities for a+b=8 manually and check which ones are divisible by 7.

[severe basin]

There are nine possibilities for a and b to equal the respective values

[onyx glen]

yes there are.

is that an issue?

[severe basin]

No, it's not

[onyx glen]

to save you the apparent finger-withering trouble of doing long division nine times,

7300+10a+b ≡ 300+10a+b ≡ 20+10a+b ≡ -1+10a+b (mod 7), so the last two digits must form a number congruent to 1 mod 7.

and checking which of 08, 17, 26, 35, 44, 53, 62, 71, 80, 89, 98 does so is not hard at all.

[severe basin]

7300+10a+b ≡ 300+10a+b ≡ 20+10a+b ≡ -1+10a+b (mod 7), so the last two digits must form a number congruent to 1 mod 7.
I do not understand this ^

[onyx glen]

which part?

the chain of congruences, or the conclusion i draw from it?

[severe basin]

the former

the chain of congruences

[onyx glen]

is it that you don't understand what i am saying, or why i am saying it?

[severe basin]

what.

[onyx glen]

7300+10a+b ≡ 300+10a+b (mod 7)

do you understand this? Y/N

[severe basin]

N

[onyx glen]

the two sides of this congruence differ by 7000.

[severe basin]

Oh

7000 is divisible by 7, thus remainder equals 0 so it doesn't make a difference?

[onyx glen]

yes, though i would have preferred that you saw that i appeal directly to the definition of modular congruence.

[severe basin]

Understood.

[onyx glen]

likewise 300+10a+b ≡ 20+10a+b (mod 7) -- the two sides here differ by 280
and in 20+10a+b ≡ -1+10a+b (mod 7) the two sides differ by 21

[severe basin]

Right

08 and 71 checks out

I also wanna how would I incorporate divisibility by 7.

I got, 300+10a+b-7 mod 7 = 0

I have a strong feeling that it's wrong

[onyx glen]

I also wanna how would I incorporate divisibility by 7.
was this not what i have been talking about at length just now...

35 and 98 checks out
no

[full forum]

@severe basin Has your question been resolved?

[severe basin]

This way is more easy, but I thought of using divisibility rule for 7 and solving two equations

Will that not work?

[onyx glen]

i don't know what you mean by "using divisibility by 7" exactly.

[severe basin]

Divisibility by 7 :
Divide the number into groups of 3 digits (starting from right) and find the difference between sum of the numbers in odd and even places. If the difference divisible by 7, then the number is divisible by 7.

[onyx glen]

eugh.

given that your entire number is only four digits long, that's going to be more trouble than it's worth.

[severe basin]

...

This is where I got the above thing from.

Is that wrong?

[onyx glen]

there is a difference between "wrong" and "masochistic". your method is the latter.

you'll end up at the same thing as me one way or another anyway.

[severe basin]

I couldn't get to the answer using my method

That's where I actually needed help

[onyx glen]

might be a consequence of your unintentional masochism.

[severe basin]

Also, why are there two divisibility rules for 7?

[onyx glen]

what's the second one you're talking about

is it the one that involves doubling the last digit and subtracting it from the rest or something to that effect

[severe basin]

Yeah, that one

[onyx glen]

i'm not sure what angle to view your question from

do you mean that you think there should be one and only one divisibility rule for every number, and the idea of there being two for the same number is preposterous to you,

or do you want to know why the rest-minus-double-last-digit thing works

[severe basin]

Usually for every number I learnt, there are only one

It's not a crime to have more than two also

No problem

[onyx glen]

k

the groups-of-three-digits thing is really a divisibility test for 1001 in disguise. as a consequence, it also works for checking divisibility by any factor of 1001.

[severe basin]

Oh

[severe basin]

Is this right or wrong?

[onyx glen]

it is correct

[severe basin]

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[modest sail]

Why is it 2pi/(the value) for the period of cos?

[modest sail]

Whats the reason it’s 2pi

[grave elm]

2pi is the period of normal cos(x)

[grave elm]

Do you know unit circle?

[modest sail]

Ooooh

[modest sail]

Yes

[grave elm]

How many radians is a full circle?

[modest sail]

2pi

[grave elm]

Okay, so the point moves around circle, and once it gets to the same place where it started, it has traveled 2pi radians.

[modest sail]

Oh ok this makes sense nowp

[grave elm]

great :)

[modest sail]

Much appreciated

🙂

Have a great day

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[grave elm]

You too :)

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[wooden hamlet]

Y’all know how to find the integral of y/(y+1)?

[robust shore]

y/(y + 1) = (y + 1) -1 /(y + 1) = 1 - 1/(y+1)

[wooden hamlet]

thank you

god I hate calculus

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[wooden nymph]

could you guys please verify this list of foundation for beginners in mathematics?

• Discrete Mathematics
• Mathematical Proofs
• Pre-Algebra
• College Algebra
• Algebra & Trigonometry
• Calculus

[torn jolt]

Linear algebra is a must

• probability and statistics

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[slim lark]

This questions

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@slim lark Has your question been resolved?

[slim lark]

<@&286206848099549185>

[sly frigate]

nobody's going to help you if you don't translate it

[slim lark]

Oke

1-Find the positive integer quads that can be (a,b,c,d)

2-Find the x,y integer pairs

3-Find all the (n,k1,k2,...,kn) values ​​that make it

4-Write the solution path for all (a,b) integers.

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[slim lark]

1-Find the positive integer quads that can be (a,b,c,d)

2-Find the x,y integer pairs

3-Find all the (n,k1,k2,...,kn) values ​​that make it

4-Write the solution path for all (a,b) integers.

[full forum]

@slim lark Has your question been resolved?

[slim lark]

.

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@slim lark Has your question been resolved?

[slim lark]

<@&286206848099549185>

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@slim lark Has your question been resolved?

[slim lark]

<@&286206848099549185>

[shell pulsar]

1. Try to factorise parts of the LHS

2. The LHS 'grows' faster than the RHS - assuming WLOG x>=y helps

3. The AM-HM inequality is probably enough

Idk immediately about 4 - probably mod 3 then smthn

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[placid umbra]

can someone remind me again how to calculate vectors

it asks what is a+b lenght

isnt k missing?

isnt it -3+9 and 10+2 so it goes to 6 and 12

and then its sqrt of (6^2 + 12^2)?

[brisk obsidian]

That should be -3 + 8, not -3 + 9.

[placid umbra]

woops

but i was right no?

[brisk obsidian]

But yes, that's correct.

[placid umbra]

ok then it means its A

13

Thank god maybe im not that stupid

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[placid umbra]

BD=15cm, need to find BK

[placid umbra]

Isnt BD=BC?

Point M is in the middle of BC

[full forum]

@placid umbra Has your question been resolved?

[placid umbra]

<@&286206848099549185> feel bad doing this :/

[torn jolt]

Do this one with vectors

[drifting hatch]

I don’t think it’s necessarily true that BD = BC

[placid umbra]

I dont think this is supposed to be done with vectors

i think its something with parallelogram properties

[placid umbra]

i mean yeah ur right it doesnt even help me that bd=bc

[torn jolt]

Its 5cm by vectors

Easy soln if one knows vectors

[placid umbra]

Vectors do not really like me can you explain how you did it with vectors?

[drifting hatch]

Ohhhh this might help instead: Can you find any similar triangles?

[placid umbra]

i mean BAD and BCD are the same

[drifting hatch]

Right but what about 2 different triangles with the same angles?

[placid umbra]

@torn jolt thank you 🙏

and yeah i think

in the triangle BMK

and KAD

angles B and M = A and D no?

[drifting hatch]

Well angle M = A and angle B = D but I think that’s what you meant

[placid umbra]

yeah

im bad at explaining these

[drifting hatch]

And do you know the scale factor?

[placid umbra]

sadly not

[drifting hatch]

What’s special about M?

[placid umbra]

but im starting to understand where you going at

nothing much apart that its in the middle of BC

[drifting hatch]

Ding ding ding

So it’s 1/2 the length

[placid umbra]

oh wait

[drifting hatch]

And since it’s a parallelogram, length BC = AD 🙂

[placid umbra]

Yeah but i only know BD=15

nothing else

[grizzled marsh]

in the question m is a midpoint of BC?

[placid umbra]

yes

[grizzled marsh]

alr

[placid umbra]

i kinda want to do this with vectors

but

we never do them with vectors

[drifting hatch]

15 = BD = BK + KD = BK + 2BK

[grizzled marsh]

u tried doing some cons?

[placid umbra]

no?

[placid umbra]

oh yeah

now i get it

and it totals to 5

like with vectors

Thank you very much for the help

i appreciate it

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[supple sluice]

solve folowing recurencion:

[supple sluice]

I have problem with understanding what to do with the a) and on b)... why on earth does it return 0=-2

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@supple sluice Has your question been resolved?

[gritty rose]

you were probably given an initial value. go find it

either $a_0$ or $a_1$

[glossy valve]

riemann

[supple sluice]

Up there is written “solve folowing recurencion:”

That’s all the data I have

[gritty rose]

good so you know a0 = 0. use this to find a1

then a2, a3, ... until you find a pattern

[supple sluice]

how do i do it tho?

like

[gritty rose]

plug n=1 in here

[supple sluice]

Like this?

[gritty rose]

looks good

[supple sluice]

so is that all or do i go further?

like

it just looks like an = an-1 + new result

[gritty rose]

you're looking for a function $f(n)$ where you plug $n$ and get $a_n$

[glossy valve]

riemann

[gritty rose]

find as many $a_3, a_4, ....$ as it takes for you to find $f(n)$

[glossy valve]

riemann

[supple sluice]

I calculated a3 = 3/2 and a4 = 0

so now how do i turn iit into $f(n)$

[glossy valve]

Meph

[supple sluice]

bc i guess it would be like

Honestly im more interested in the B) @gritty rose and the a1, a2....

bc its confusing

[gritty rose]

Don't do hard problems until you can do easy ones

[supple sluice]

yeah... what would be fn in my case, its not a homework or anything, its an exam from last week and trying to understand what would be there.

honestly i just wanna know how to do it, already know what to do in B but have problem bc of weird values, next in A) i think now i will get more points from it, mainly i wanna learn B becouse its pass or not pass.

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[glossy obsidian]

Hello all,
Was getting help in another channel and was told to post here when I finished my work for confirmation

[glossy obsidian]

Basically, I was asked to evaluate this

My work came out to this

Going to assume I got it right this time around, just wanted some confirmation to see if I was correct or not

[ivory cairn]

there shouldn't be a t in your final answer anymore

[glossy obsidian]

Should be x, right?

Caught my eye right after I posted it, my bad

Here we go

Please, tell me this is right now
Getting tired, I keep making simple mistakes, I do apologize

[ivory cairn]

not just x

[glossy obsidian]

Oh, i see what I did wrong

Let me get this uploaded, hopefully for the last time

Changed f(x) to their respective values of x^2 and sqrtx

Alright, so that's right where it needs to be now?

Going to assume i'm good to go based on the checkmarks

[worthy tree]

,align
\frac{d}{\dd x} \qty( \int^{x^2}_{\sqrt{x}} f(t) \dd t )'
&= \frac{d}{\dd x} \qty(F\qty(x²) - F\qty(\sqrt{x})) \
&= x^2' * F'(x²) - \sqrt{x}'* F'\qty(\sqrt{x})) \
&= 2x* F'(x²) - \frac{1}{2\sqrt{x}}* F'\qty(\sqrt{x})) \
&= 2xf\qty(x²) - \frac{f\qty(\sqrt{x})}{2\sqrt{x}}

[glossy valve]

Mehdi_Moulati

[glossy obsidian]

Awesome, i'll revise my work to show, well, more work

Thank you so much!

[worthy tree]

anytime

[glossy obsidian]

Hey @worthy tree, not sure if I'm allowed to ping you or not, so I do apologize.
I cleaned up my work to hopefully show everything a bit more clearly
Look like i'm missing anything, or do I look good to go?

Getting my work

[worthy tree]

np bro

[glossy obsidian]

Please tell me this looks good 😄

I feel horrible for putting at least 6 helpers through my incompetence

This one just got me since there were no constants

[worthy tree]

i guess the second line is unnecessary

a(x) = sqrt(x) ....

,align
\frac{d}{\dd x} \qty( \int^{x^2}_{\sqrt{x}} f(t) \dd t )
&= \frac{d}{\dd x} \qty(F\qty(x²) - F\qty(\sqrt{x})) \
&= \qty(x^2)' * F'(x²) - \qty(\sqrt{x})'* F'\qty(\sqrt{x})) \
&= 2x* F'(x²) - \frac{1}{2\sqrt{x}}* F'\qty(\sqrt{x})) \
&= 2xf\qty(x²) - \frac{f\qty(\sqrt{x})}{2\sqrt{x}}

[glossy obsidian]

Aside from that, I should be good?

[glossy valve]

Mehdi_Moulati

[worthy tree]

yes

[glossy obsidian]

Awesome, thank you so much!

[worthy tree]

np

[glossy obsidian]

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[pale plank]

this is supposed to be partially review but i literally don’t remember any of this, can someone explain how to do #5 and #4

[round gust]

,rccw

[round gust]

At 4 you can simplify the radicals

[pale plank]

uhh

im sorry i swear i did this last yr i js cant remember anything hdhsj

[round gust]

$\frac{\sqrt{14}}{\sqrt{7}}=\sqrt{2}$

[glossy valve]

Oogy Boogy

[round gust]

Use this example

[pale plank]

ohh

so sqrt3

[round gust]

Over 4

[pale plank]

oh ok

i get that

[round gust]

Good

[pale plank]

how would i do that w/x tho in #5

[round gust]

$\sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{b}$

[glossy valve]

Oogy Boogy

[round gust]

Use this

[pale plank]

3sqrt2 for 18?

i thibk

[round gust]

Yes

Then split the sqrt(2x)

[pale plank]

wdym

[round gust]

With the thing i wrote

Use jt

[pale plank]

ohh

[round gust]

It*

[pale plank]

so sqrt9 * sqrt2x

3sqrt2x?

[round gust]

Yes

Now split it

[pale plank]

split it?

[round gust]

Sqrt2x=sqrt2 * sqrtx

[pale plank]

ohh

[round gust]

Do the same thing for sqrt(x^5)

[pale plank]

sqrtx^4 * sqrtx?

[round gust]

Yes

Good

Now

Can you cancel something?

Are there any common terms on the top and bottom?

[pale plank]

sqrtx?

oh wait 3

maybe

[round gust]

Yes

Both, right?

[pale plank]

yeah

[round gust]

So cancel them both

And tell me what you gef

Get*

[pale plank]

sqrt2/x^2

[round gust]

Yes

Well done

[pale plank]

tysm i have a whole page on these and u made it so simple!

[round gust]

You're welcome

[pale plank]

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[round gust]

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[pliant lagoon]

How do I find the 0’s

[pliant lagoon]

Nm I didn’t read it’s mean theorem

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[storm oracle]

how do i factor/simplify

[soft pine]

Can you write it again on more clear

[storm oracle]

nvm we're good

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[warm vault]

$4\sin\left(x+\frac{\pi}{4}\right)=2\sec\left(x-\frac{\pi}{4}\right)$

[glossy valve]

spetnaz

[warm vault]

How to solve for x?

[twilit leaf]

!status

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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

[warm vault]

1

[twilit leaf]

Cofunction

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[warm vault]

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[torpid zenith]

The second image is the parabolic coordinates we're suppose to use. Could someone explain how they were derived?

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[hasty igloo]

Would anyone be kind enough to explain me why these are the same. Im very bad at algebra.
dx = x - xy / (y + dy)
dx = x * dy / (y + dy)

Its origin was (x - dx) ( y + dy) = xy "Find dx"

[rose rain]

\begin{align*}
\dd{x} = \frac{x - xy}{y + \dd{y}} \
\dd{x} = \frac{x \dd{y}}{y + \dd{y}}
\end{align*}

?

[glossy valve]

active mental mutilation liker

[rose rain]

ok simply treat $\dd{x}$ and $\dd{y}$ as variables and manipulate it

[glossy valve]

active mental mutilation liker

[hasty igloo]

Hmm i dont know how to manipulate it.

So i got this far:

(x - dx) (y + dy) = xy
x - dx = xy / (y + dy)
dx = x - xy / (y+dy)

But then they just showed ...
dx = x * dy / (y + dy)

Why is x * dy = x - xy ?

[rose rain]

from x-dx = xy/(y+dy) to the next line

that's wrong

you made an algebraic mistake

[hasty igloo]

Hmm now im rly confused, could you help me out?

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@hasty igloo Has your question been resolved?

[wanton pelican]

i'm pretty sure it's dx=x/y

[hasty igloo]

yes, but id like to understand the conversion
Is the math faulty?

[wanton pelican]

dx=xdy/y+dy

that's what i got

[hasty igloo]

dx = (xy + xdy - yx) / (y + dy)
dx = x * dy / (y + dy)
does this make sense?

does it work?

[wanton pelican]

could u tell me the name of this type of problem?

[hasty igloo]

idk what it is ( math wise)

[wanton pelican]

i need to look something up

[hasty igloo]

i need to write this problem in a not very readable way for a program

[wanton pelican]

it's differential equations but what topic?

dx=xdy/(y+dy)

[hasty igloo]

just a curve i guess
x & y change based on how much dx or dy is added, then dy or dx has to be taken away for how much was added because x * y should always be equal to some number

[wanton pelican]

this is just a hunch but im pretty sure that either x or y is equal to 0

[hasty igloo]

u was right

would you show me your process?

[wanton pelican]

yes i will send a pic, one sec

hopefully, that helps

[hasty igloo]

thx man!

[wanton pelican]

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[buoyant wadi]

You can use the vertices the get the side lengths via the Pythagorean theorem

Points:
A = (a, c)
B = (b, d)

Line AB goes from point A to point B
len(AB) = sqrt((a-b)^2 + (c-d)^2)

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[strong dirge]

Just curious what everyone else thinks. Does this sentence make clear the given trajectories are ordered and should be used in that order?

[strong dirge]

Lost 75% of points on basically a correct answer just "incorrect" interpretation

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[slender pond]

Is this question asking to prove that 3^n(n+1) is divisible by 3

[thick minnow]

No, it asking to prove that the general nth term of the recursion would be 3^n(n+1)

[slender pond]

How would I do that?

[thick minnow]

Use mathematical induction to prove it

Assume that P(n) means that the given relation holds for n ie that un = 3^n (n+1)

Prove the base case, ie the simplest case here it would be 1 cause Z+ starts with 1

Then for induction hypothesis assume that P(k) is true and try to prove that P(k+1) is true

[slender pond]

Oh ok

Thank you for the help

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[green lily]

hi can somebody explain this step to me?

[onyx glen]

the 1/2's from each parenthesis were brought out front, the 2/4 was simplified into 1/2, and the product of sums of e^(...) was expanded.

[green lily]

"sums of e^(...) was expanded." how does this work?

lol

nevermind

slow day

thanks

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[swift fulcrum]

Can anyone tell me where at 0, it isn't negative?

[atomic blade]

(-2)^(4/5) is not the same as -2^(4/5)

[deft tulip]

why* it isn't negative? not sure i understand the Q

[swift fulcrum]

Yeah, why it isn't negative

[atomic blade]

Because this

[hazy wedge]

(-2)^(4/5) is not the same as -2^(4/5)

[atomic blade]

You plugged it in incorrectly

[swift fulcrum]

Ohhh

[hazy wedge]

ah whoops

[swift fulcrum]

But if enough like that

[atomic blade]

Thanks for the ping

[swift fulcrum]

Why though

[atomic blade]

[hazy wedge]

wrong reply

[atomic blade]

,w (-2)^(4/5)

[glossy valve]

Results provided by WolframAlpha

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Requested by Umbraleviathan#3607

[swift fulcrum]

WHERE DID YOU GET THAT EMOJI

OMG

[atomic blade]

(-2)^4 is positive

[swift fulcrum]

Yee

[atomic blade]

And then the 5th root of (-2)^4 must also be positive

[deft tulip]

how do you do 5th root in latex again?

i forget the syntax

[swift fulcrum]

\sqrt[5]{x}

[deft tulip]

thanks

[swift fulcrum]

Ahhh I see, thanks

[deft tulip]

$(-2)^\frac{4}{5} = \sqrt[5]{(-2)^4}$

[glossy valve]

Sooshon

[deft tulip]

so it's a positive number because the numerator of the exponent is even

[swift fulcrum]

How would I be able to put this as a piecewise function?

[deft tulip]

sorry, i don't understand the question

?

[swift fulcrum]

$\left(x^{2}-2\right)^{\frac{4}{5}}$

[glossy valve]

AirToastie

[swift fulcrum]

How can I change this into a piecewise function?

It's not differentiable at -√2 and √2

So it must have piecewise functions

[deft tulip]

?

it's a function and it's not differentiable, why would that be a problem?

you would have to post what the actual question you're trying to answer is, because i am not clear

it's a function and it's continuous at all x but yeah it's not differentiable at those x values of +-sqrt(2), but you never mentioned what exactly you are trying to answer \ find

[swift fulcrum]

I'm just trying to prove that it's not differentiable at those two points

[atomic blade]

You can just take the derivative

And then show where the derivative is undefined

[deft tulip]

right, if you check the derivatives from the left and from the right at those points they won't match, which means it isn't differentiable at those x

[swift fulcrum]

Ohh I should be using limit properties?

[rose rain]

(the idea here is that the derivative is a limit)

[swift fulcrum]

Not derivative rules

[rose rain]

$\derivative$

[glossy valve]

DerpZ

[rose rain]

well okay not quite this

differentiability at a point is different but same concept

[deft tulip]

so, do you know how to calculate the derivative of this function?

[swift fulcrum]

$\lim _{x\to -\sqrt{2}^+}\left(\frac{\left(\left(x^2-2\right)^{\frac{4}{5}}+h\right)-\left(x^2-2\right)^{\frac{4}{5}}}{h}\right)$

Like this?

[glossy valve]

AirToastie

[deft tulip]

ok umm why we using the definition of the derivative here

[swift fulcrum]

No?

[deft tulip]

im sure you've learned power rule and what not for take derivatives already?

you've learned?:

[rose rain]

we don't need derivative definition its overkill

just differentiate

[deft tulip]

f(x) = x^2 and f'(x) = 2x ?

[swift fulcrum]

$f\left(x\right)=\left(x^{2}-2\right)^{\frac{4}{5}}$

[glossy valve]

AirToastie

[swift fulcrum]

$f'\left(x\right)=\frac{4}{5}\left(x^{2}-2\right)^{\frac{1}{5}}\cdot\left(2x\right)$

Oh no

[deft tulip]

-1/5 for the exponent but yeah essentially that