help-13

[lyric narwhal]

what i mean to say is

if you input n and n+22 into the expression, you will yield the same result (since 22 * pi/11 = 2pi)

you cannot generate that kind of periodicity with sqrt(2)

since it's irrational

[lament yoke]

oh,okay

basically here i'll have 2pi

if i input what u said,and that's the periodicity

[lyric narwhal]

for the first case, yes

[lament yoke]

yeah

[lyric narwhal]

but you can't get a value of n which gives you 2 pi in the second case

[lament yoke]

yeah

[lament yoke]

can you explain why n and n+22 are the same?

or u meant 22n

i am sorry if that's a dumb question

[lyric narwhal]

try it out

what would you get for n

and what for n+22

[lament yoke]

oh yeah,it's sth +2pi

[lyric narwhal]

yep

[lament yoke]

and where does that thing with 2kpi come from?I have seen it in a bunch of places

[lyric narwhal]

are you asking why sin(x + 2kpi)= sin(x)?

[lament yoke]

yeah,I mean i get the 2pi period,but why 2kpi?

does that imply a multiple of 2?

[lyric narwhal]

yeah

[lament yoke]

okay.i was unsure ab that

[lyric narwhal]

it basically means sin(x)=sin(x+2pi)=sin(x+4pi).... = sin(x-2pi)=sin(x-4pi)=...

so k can be any integer

[lament yoke]

yeah,yeah

total sense

thanks for explaining

[lyric narwhal]

no worries

you're one of the nicest helpees ive talked to lmao

[lament yoke]

and taking the time to really add detail to ur explainations and the patience haha:)

[lament yoke]

really?haha,thanks:))

[lyric narwhal]

yeah lmao most people are pretty ungrateful

but i just come here to waste my time so

[lament yoke]

I just think it sounds ab right

it's fair

[lyric narwhal]

yep

[lament yoke]

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[fresh imp]

how am i supposed to work it out with 2 unknowns

[nova glacier]

What do you mean

y = c/x^3 ?

c/a^3 = 28

c/(8a^3) = ...

[fresh imp]

why did it become (8a^3)

[nova glacier]

I mean when x = 2a

x^3 = 8a^3

[fresh imp]

oh ok

[nova glacier]

So I guess you should just divide by 8

Does it make sense?

[fresh imp]

yeah

so the answer is 28/8

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[lost whale]

help

[kindred bridge]

!status

[cedar kiln]

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

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[silk wasp]

does angle AEB equal angle EDC here, or are the two not co-interior?

[thin topaz]

they are equal

but not co-interior

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[marble sluice]

Based on this information how were you supposed to know that the major axis is x?

[smoky idol]

I believe that the major axis is always the one with the foci on

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[fervent pike]

correct me if I'm wrong, but I can essentially take the vectors, turn them into row vectors and form a matrix A, then if rank(A)=# rows of A, then the vectors are linearly independent

[short blade]

you would want column vectors

then if rank(A) = # of column vectors then yes they are linearly independent

granted, this is the same as taking the transpose and doing row vectors

[fervent pike]

ok, and to find a trivial solution because they are linearly independent, I can take these column vectors and turn into matrix, and use row operations to solve

that is the right idea, correct?

[short blade]

yeah

[fervent pike]

ok see now this is the weird part

I first determined that they were indeed linearly independent

then I took the column vectors, turned into matrix, but got no solution. Meaning the system is not even consistent

I checked my 'no solution' answer with an online calculator too btw

[short blade]

can i see where you computed this online

[short blade]

that's a different question

[fervent pike]

what do you mean?

[short blade]

if it says "no solution" to that, then it means the system is consistent

[fervent pike]

what?

[short blade]

that's asking if you can write the last element as a linear combination of the first three

[fervent pike]

oh so [A|B], B should be a 0 vector?

[short blade]

you don't have an augmented matrix

yes

you have a system of L.I. vectors if [A|0] has only the trivial solution

by the way

this method you described above

is kind of the whole motivation for linear algebra

solving systems of linear equations

that's why you learn row reduction

im saying this so as to stress the importance of the method

[fervent pike]

of which method?

[fervent pike]

this one?

[fervent pike]

or this?

[short blade]

both, they're kind of saying the same thing

[fervent pike]

oh ok, in which instance would you have [A|0] vs [A| another non zero vector]

[short blade]

"solving" [A | b] asks whether the column vectors of A can produce (as a linear combination) the vector b

that's like asking

is b in the span of the columns of A
is b in the column space of A

[fervent pike]

oh okok

[short blade]

doing [A|0] is the same thing

"how many different ways can we write the 0 vector as using the columns of A"

either you can just write it as 0*v1 + 0*v2 + ...

or there is some nontrivial solution

in which case your vectors are linearly dependent

an easy example is doing [I|0]

where I is the identity matrix, which has as columns the standard basis for R^n

this is already row reduced, and gives you that to make the 0 vector, you need to have 0 of every basis vector

(as we might expect from a basis, that is, a set of linearly independent vectors)

[fervent pike]

oh ok

Thank you so much!

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[cedar kiln]

Please don't occupy multiple help channels.

[elfin otter]

How do I continue?

[cedar kiln]

@elfin otter Has your question been resolved?

[cedar kiln]

@elfin otter Has your question been resolved?

[cedar kiln]

@elfin otter Has your question been resolved?

[elfin otter]

<@&286206848099549185>

[weary vessel]

You could have just differentiated the function

Then studied its sign

To know whether it is a maximum or minimum

[elfin otter]

Is this part correct?

[weary vessel]

Yes

[elfin otter]

Hold on lemme diff

P ' ' (t) = -80 < 0

[weary vessel]

Yeah, that indicates a maximum

[elfin otter]

How do I count the "how many bacteria present" part?

[weary vessel]

You have already done it

P(6)

[elfin otter]

...I...

[weary vessel]

lol

[elfin otter]

Cuz I thought

[weary vessel]

you did good!

but with an unexpected way

[elfin otter]

P(6) = 6th day

[weary vessel]

Using the definition of the derivative xD

Instead of directly differentiating and studying the derivatives :P

[elfin otter]

1 more ques

[weary vessel]

Allright, do it and I will take a look

[elfin otter]

Do I do + / - 1 hour
?

[weary vessel]

What

[elfin otter]

Uhh

Average rate of change

[weary vessel]

You just differentiate P(t)

Then evaluate at t=1

[elfin otter]

Is there any way of solving without using differentiation?

[weary vessel]

The definition of the derivative, like you did in the first exercise

We want P'(1)

So, as you know

[ P(1 + h) - P(1) ] / h

[elfin otter]

Ohhh

[weary vessel]

As h approaches 0

[elfin otter]

Wait how do I write that?

0 = [ P(1 + h) - P(1) ] / h
?

[weary vessel]

Using limits

You just evaluate this

It is P'(1)

I will let you try, tell me what you find

[elfin otter]

I think I did sth wrong

[weary vessel]

Show me

[weary vessel]

It is not - P(h) !!!

[elfin otter]

Ah.

HAHAHAHA

[weary vessel]

It is - P(1)

[elfin otter]

Okokok

Now I see it

[weary vessel]

😝

[elfin otter]

• 1

Negative 1

[weary vessel]

Correct!

Well done

[elfin otter]

So... the ans is 'moving back' ?

Or do I just put it at -1 ?

[elfin otter]

I need sleep frfr

[weary vessel]

No, -1 is the answer

if you want to be more precise, -1 m/h

Since it's the derivative of the movement (m)

[elfin otter]

"Moving back 1 m/h" ?

[weary vessel]

It should have a unit of m/h

[elfin otter]

Cuz -1 m/h kinda funny haha

[weary vessel]

Nooo, it's like that -1 m/h

or moving down

downward

[elfin otter]

Unless displacement?

[weary vessel]

by 1 m/h

[elfin otter]

Ah

[weary vessel]

You are used to have positive velocities

[elfin otter]

Sounds better

[weary vessel]

But the fact is that it can be negative

To indicate moving backwards/downward

in our case, simply -1 m/s

or 1 m/s downward

but this is math, not physics anyway

[elfin otter]

Aight I think imma close this for now
Thanks for the help!

.close

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[crimson sedge]

I need help

did i mark the currents correctly?

if i didnt I2 isnt 1A

[hoary chasm]

I mean u choose the direction of traversal and current initially ryt

And if the result is negative it's opposite of your choice

The direction of current

[crimson sedge]

can i make mistake while marking contours?

if yes, did i?

and what do i do with Is?

i made a mistake there should be 2 contours

one is combination of 2 and the other is independed

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[crimson sedge]

.reopen

[cedar kiln]

✅

[crimson sedge]

please

someone

how do i make equations

[cedar kiln]

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[vital sphinx]

How do I put this in the form of sinx cosx?

[south tundra]

The number inside the brackets is the same as $\sqrt2\cdot(\frac12\cdot\frac1{\sqrt2} + \frac{\sqrt3}2\cdot{\frac1{\sqrt2}})$, does this remind you of anything?

[wraith dagger]

A Lonely Bean

[south tundra]

(Consider what I said if you want to have only one trig function, but if you want to have 2 of them then just rewrite 1/2 and sqrt(3)/2 as cos(60) and sin(60) or sin(30) and cos(30))

[vital sphinx]

is tthere not a standard formula?

I would say cos 1/2x + jsin(sqrt(3)/2)x

or is the stuf between the ( )x in both the cos and sin

[south tundra]

Where did j come from btw

[vital sphinx]

imaginary number

I think i recognize eulers formula

no?

[south tundra]

You must have mentioned it here then btw

Anyway, if it's $e^{(\frac12 + i\frac{\sqrt3}2)x}$ and you want to represent it in its cartesian form

[wraith dagger]

A Lonely Bean

[south tundra]

Then you can rewrite it as $e^{\frac{x}2}(\cos{\frac{\sqrt{3}}2x} + i\sin{\frac{\sqrt{3}}2x})$

[wraith dagger]

A Lonely Bean

[vital sphinx]

Thats what I meant, thanks

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[young solar]

Is anyone able to help me with this question? (It has to do with rates of change at a yr 11 level):
A circle is inscribed in a square so that the sides of the squares are tangents to the circle. The circumference of the circle is increasing at a constant rate of 2m/s^-1, causing the sides of the square to increase, so that the sides remain touching as tangents.
(Image provided)
Find the rate at which the area of the square is increasing when its perimeter is 16cm
(For reference we were learning the chain rule at the time)

[young solar]

This is what i've done so far and im unsure of where to go from here

The answer says that its 16/pi m^2/s^-1

[ocean badger]

I can help you @young solar

For starters, what’s the side length of the square at the start?

[young solar]

uhm

it doesnt say

oh wait

sorry i meant

The side length is 4

[ocean badger]

Can you use that to find the radius of the circle?

[earnest anchor]

i have gottten to that, ill send workings

[young solar]

the radius is 2 since the sides of the square remain touching as tangents of the circle, meaning that the diameter of the circle is the same as the side of the square

and radius is half of diameter

OH WAIT

AM I MEANT

TO

MAKE ANOTHER CHAIN RULE BASED OFF

[ocean badger]

Now, what’s the area of a circle?

[young solar]

DA/DD

[young solar]

pir^2

[ocean badger]

Collateral seems to be taking over. Good luck! 😁

[young solar]

alright thanks guys haha :)

[ocean badger]

Ping me if you end up not being able to get it.

[young solar]

yuhp will do!

i understood it, thanks to the both of you :D

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[outer canopy]

stuck here

[buoyant latch]

Just cancel the 2x?

[young solar]

have you considered using double angle formula?

[neon moon]

Hint: Apply $\lim_{x \to 0} \frac{\sin x}{x} = 1$

[outer canopy]

Yeah I been thinking about that but idk how to make it like that

[livid hound]

wrong limit

[neon moon]

You have a sin 2x in numerator, what do you think should show up in denominator?

[wraith dagger]

jimmy1234

[outer canopy]

also 2x

[neon moon]

Same for that sin 4x

[young solar]

wait is the answer 0?

sterling?

i mean

infinity

wait no

0

im stupid

or ami

[outer canopy]

Yes. But we have to find ways to make it not 0

or dne

whatever

[earnest anchor]

is it sin2x or sin ^2 x?

[outer canopy]

sin2x

[neon moon]

There you go.

[outer canopy]

,rccw

[neon moon]

That’s a limit problem. The story here is 0/0 is not necessarily a 0. You may want to learn more on that on your own.

[outer canopy]

is this correct

[neon moon]

Sort of. And now do the same for sin 4x

[outer canopy]

there is no sin4x

[neon moon]

Well, not a strict equal

[livid hound]

make the bottom fraction line a bit longer

[neon moon]

Ohh soery

[outer canopy]

final answer is 1/2

is that correct

[livid hound]

yes

[neon moon]

How about this, when yoou group sin 2x / 2x tgt, what is left for you?

[outer canopy]

1

2x(1)/4x

2x/4x

1/2

[neon moon]

There you go. 😌

[outer canopy]

thank you

,rccw

[wraith dagger]

Couldn't find an attached image in the last 10 messages.

Jun 7, 2023, 12:57 PM

[outer canopy]

,rccw

[outer canopy]

@neon moon is this correct?

please correct me

[livid hound]

don't be lazy with writing those limits

[outer canopy]

wdym

[livid hound]

you're missing the limits in the second line and for the
5x/(2x)

[outer canopy]

mybad for bad angle

[buoyant latch]

That second line needs a limit

[outer canopy]

oh okay

[buoyant latch]

Otherwise the equality does not hold

[outer canopy]

so everytime I change something there should still be the limit

does 1/x equal to 1?

I was wondering if I can use that to manipulate 3xtan(x)/sin(x)

for the numerator

[buoyant latch]

The limit should stay until you evaluate it

[buoyant latch]

Only if x = 1 or if you’re taking the limit as x approaches 1

[outer canopy]

so what can I do here

[livid hound]

same idea can be applied

just remember

The limit should stay until you evaluate it

[outer canopy]

,rccw

[outer canopy]

does this work

[buoyant latch]

It does but now you’re missing equal signs lol

[cedar kiln]

@outer canopy Has your question been resolved?

[cedar kiln]

@outer canopy Has your question been resolved?

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[dry cargo]

Can someone explain me this part (a) and what's a stand for ?

[crimson sedge]

a is what you are trying to find

not very well written tho as the 2 a's are different

[dry cargo]

the ans given is x/5 + y/3 = 1

and idk why and how

[brittle stag]

schrödinger a

[crimson sedge]

https://www.mathsisfun.com/algebra/line-equation-2points.html

[cedar kiln]

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[north fable]

Open

[north fable]

Can someone guide me where I am going weong

To find:
Distance of each pole from the point on common line,
AB and DC

[granite eagle]

last page

second line

[north fable]

Oh

Thank you so much

[granite eagle]

also

next time dont send such things

its good practise repeating the whole thing or checking

[north fable]

Ok

Will remember it the next time

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[crimson obsidian]

If f(x) is even function then is f(y) also even function?

[dire geode]

Are the x domain restrictions and y domain restrictions the same

[crimson obsidian]

There is no mention of domain, a functional relation is given

f(x+y) + f(x-y) =2f(x)f(y)

So I guess I assume that domain restriction of x and y are same

Ty then, I gotcha the problem

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[finite badger]

hey guys trying to figure out what i need to learn in order to understand what happened in the green

[finite badger]

i know how factorial works more or less, not sure why the n=n+1

[dawn junco]

uh sounds like the ratio test for series

there's no series anywhere on this screenshot tho

[finite badger]

https://www.youtube.com/watch?v=kkfILYpkJS8&ab_channel=TheOrganicChemistryTutor

yeah its just limit

this video should help me or i should find something more specific?

[dawn junco]

this sounds alright

you can't go wrong with organic chem guy 🙂

[finite badger]

lovely thanks @dawn junco

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[nova crypt]

In a group of four students, what is the probability that at least two have the same astrological sign?

[nova crypt]

four students

and 12 astrological signs

the answer key says that the answer should be 0.427

and this is how it got that

but im a bit confused

i know it should be 1 - something cause its at least and you can do it by finding the probability of some of them all having different signs

yeah im stuck on finding what number of people i should find having different signs if that makes sense

like if i find the probability of none of them having the same signs and then subtract that from 1

but its if theres at least two

(1/12)^4 right?

yeah thats what we've been doing in class for those sorts of problems

where its at least of something

13

yes

ohhh so the first one the probability that it'll be different is 12/12

then the second its 11/12

then 10/12

then fourth 9/12?

multiplication

990/1728

im confused on what thats the probability of

oh the probability that each will pick a different sign?

ohhhh and then at least two means that at least one will pick the same sign

so thats why its the complement of that right?

thank you so much for ur help that made it click for me

.close

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[hollow elbow]

Find the equation of the line parallel to the x-axis and 5 units above it

[hollow elbow]

it's like a series of four questions like this that i dont get and I have an exam tomorrow

is the answer y = 0?

[ember elk]

nope its y=5

[hollow elbow]

but

it's a horizontal line

[ember elk]

well thats what u need right,.....u want the line to be parllel to x axis

[hollow elbow]

yes, but i thought a horizontal line = 0

[ember elk]

which means horizontal to x axis

[vague rapids]

The line is parallel to x axis, think which of x or y co-ordinate would be fixed

[hollow elbow]

ok

i get it now

the next one is: the line with undefined gradient and x-intercept 3

so it's just slope = undefined

right

[ember elk]

yes thats right

[hollow elbow]

and then

"The line with gradient -2 and y-int 3

y = -2x + 3

here's one i def dont get:

"The line with x-intercept 4 and y-intercept -2"

that i def dont get

actually i figured it out

y = 1/2x -2

[ember elk]

can be little more specific of what u dont understand

[hollow elbow]

no i get it now

nevermind

sorry

.close

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[past narwhal]

anyone know how to do these?

i got a physics exam tmr and i know nothing about waves

[iron saffron]

Then you should study before doing exercises.

[cedar kiln]

@past narwhal Has your question been resolved?

[cedar kiln]

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[queen otter]

ok so area of a cone so r is 2 and height 5 now i did pi x 2 x 2 plus pi x 2 x 5

[queen otter]

and i was wondering if anyone could check if 43.96

is right

[median goblet]

yeas

[queen otter]

ty

[median goblet]

solve the equation for r

[queen otter]

what

[median goblet]

so we can say πr^2 = 43.96

solve for r

wait its a cone?

[queen otter]

dont quit understand

yes

my first explanation was going to be a smooth circle triangle that is 3d

[median goblet]

so v = hπr*2/3

so we will plugin the values

[queen otter]

thats volume

im not going for volume

[median goblet]

then what r u asking

thats what u asked for

[queen otter]

if what i did was right

area of a yeah

cone

[median goblet]

what was the radius?

[queen otter]

2

height 5

[south tundra]

Surface area you mean?

[queen otter]

YWA

YES

[median goblet]

ok

[queen otter]

finally i know the name

[median goblet]

surface area is wayyy different

[queen otter]

isnt it

[south tundra]

It's pir^2 + pir * sqrt(r^2 + h^2)

[median goblet]

there u go

[queen otter]

pi r^2 plus pi r x h

[south tundra]

No

Pi r * l

Where l = sqrt(r^2 + h^2)

[queen otter]

yeah in norwegian its S

[south tundra]

Anyway, it's (4 + 2sqrt29)pi

Actually 2pi(2 + sqrt29) looks nicer

[queen otter]

legit dont get anything

so the r is 2

so i do 3.14 x 2 x 2

then 3.14 x 2 x 5

and plus together

[south tundra]

No

Again, it's not + pir * h

You do + pir * sqrt(r^2 + h^2)

[queen otter]

wait so i have to do pytagoras

[south tundra]

Yes

[queen otter]

for the hypotenus

ok so h^2 = 2^2 plus 5^2

[south tundra]

Which is then sqrt(29)

[queen otter]

whats that

[south tundra]

$\sqrt{29}$

[wraith dagger]

A Lonely Bean

[queen otter]

oh square root

wait so what do i do after

do the ¨2

^2

[south tundra]

No need to square anymore

Just put it in

4pi + pi * 2 * sqrt(29)

[queen otter]

ima just find an eksampel online and try to replicate it

is this correct

[south tundra]

Yes

[queen otter]

ok so i do 2 x 2 and 5 x 5

[south tundra]

Right

On the same side though

[queen otter]

then minus

so it is 21

[south tundra]

No subtraction

[queen otter]

whats that

[south tundra]

It's 2^2 + 5^2 = x^2

[queen otter]

i did that

4 and 25

then the next step is

[south tundra]

You did x^2 = 5^2 - 2^2

[queen otter]

when

i did x^2 5 x 5 2 x 2

[south tundra]

In no way that yields 21

[queen otter]

wait lets try again

i got h^2 = 2^2 plus 5^2

then i do h^2 = 4 plus 25

[south tundra]

That is right

[queen otter]

the next step is then

[south tundra]

h^2 = 29
h = sqrt(29)

Now plug this in

pi * 2 * 2 + pi * 2 * sqrt(29)

[queen otter]

OH

your doing from the other

im trying to focus on the pytagoras first

[queen otter]

why does it say subtraction here tho

[south tundra]

Because they wanted to solve for a side different from the hypotenuse

[queen otter]

oh

katet

[south tundra]

But here we needed to solve for the hypotenuse

[queen otter]

yes

so do i do 29 square root

[south tundra]

No, keep it for now

[south tundra]

Simplify this

[queen otter]

why

12,56

[south tundra]

Even if you are asked to approximate

You can do it at the end

[queen otter]

im kinda a step by step guy

causcious

[south tundra]

Cautious?

[queen otter]

i did not write that step by step

[south tundra]

What you said does not contradict my suggestion

Anyway, if you want to approximate now, then sure

[queen otter]

i want to do this basic

i dont know what approximate means

[south tundra]

$\approx$

[wraith dagger]

A Lonely Bean

[queen otter]

so that is like a wild guess kinda

around

what the anwer is

well im not going to do that

[south tundra]

Yes

Should not be a wild one though

[queen otter]

close guess

so the hypotenus

[south tundra]

Anyway, can you simplify this?

[queen otter]

currently we got 29

i dont know how

[south tundra]

What's 2 * 2?

[queen otter]

2^2

[south tundra]

As a single digit

[queen otter]

4

[south tundra]

Right

And 2 * sqrt(29) stays as it is

So the answer is 4pi + 2sqrt(29)pi

[queen otter]

yes

why dont we just do the sqrt now

[south tundra]

You can factor it as 2pi(2 + sqrt(29))

Now you may "do" the sqrt

[queen otter]

My laptop died

5.38

[south tundra]

That is approximating btw

[queen otter]

So i have to write down the infinite decimals

[south tundra]

You will not be asked to do that

[queen otter]

Yeah ok so how many decimals

[south tundra]

If you are asked to approximate, then do the 5.38

If you are not asked to approximate, then you leave it as it is

[queen otter]

Ok but were using really much time on one thing

[cedar kiln]

@queen otter Has your question been resolved?

[queen otter]

No

[south tundra]

Why not?

[queen otter]

Im failing this quiz 100

Idk

.close

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[blissful drift]

Hi there! I would like some help here 🙂🙏

[cedar kiln]

@blissful drift Has your question been resolved?

[zenith nymph]

sin 15 is (sqrt(6) - sqrt(2))/4

something that helps to have memorized occasionally

[blissful drift]

ty!

[zenith nymph]

tan (x) - tan (y) = tan(x-y)*(1+tan(x)+tan(y))

which should help you for c)

[blissful drift]

is't c) tan(57-12)=tan57-tan12/1+tan57*tan12

[zenith nymph]

yep

[blissful drift]

oh

i see

thanks again

tomorrow I have a term test and

[cedar kiln]

@blissful drift Has your question been resolved?

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[tepid bramble]

hello guys, do u mind helping me how do i solve this number?

[neon moon]

Which part do you need help, a or b?

[tepid bramble]

uh both-

but i figured out OP

but just not sure what to do next-

[cedar kiln]

@tepid bramble Has your question been resolved?

[neon moon]

For (a), ST is an arc. Can you recall the formula for calculating the arc length on a circle?

[tepid bramble]

huh-

[neon moon]

And if you have OP, do you know how to use that to further find the radius of the larger sector?

Hint: Draw a line from P through O and touch ST.

[tepid bramble]

angle/360 x 2 x pi x r

[cedar kiln]

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[grand iris]

silly question but whats something like this called

[grand iris]

7x= y
7y=m
7m=r
7r=k
7k=o

is it just to the power of 7 ?

and so on n so fouth picking random letters

[upper abyss]

I see 5 equations

[iron saffron]

m = 49x
r = 343x etc...

[upper abyss]

It does happen to be true that o = 7⁵x

[grand iris]

the actual number of equations doesn't matter

its just an example

like in terms of real numbers lets start off with 7

and then if you get a set of 7 7's that counts as 1 in the next set

and then if you get a set of 7 like 3 sets deep that counts as 1 in the next set

like is there a way to describe that

is it just whatever number *7

[upper abyss]

and then if you get a set of 7 7's that counts as 1 in the next set

[upper abyss]

This lost me

[grand iris]

exactly

is that just to the power of 7

or is it something diffrent

[upper abyss]

That was a copy/paste of your message. Just wanted to show which message lost me

[grand iris]

oh

ummmm

[upper abyss]

What does it mean for 7 7s to "count as" 1?

[grand iris]

so you start off with just one 7

[upper abyss]

I'll care for it with my life

[grand iris]

and then if you get seven of those together

that counts as one set

it could be x's or y's or whatever

lets say chickens

you have one chicken

then you get 7 chickens

that counts as one flock of chickens

and then you get seven flocks of chickens

that counts as one herd of chickens

then you get seven herds of chickens

that counts as a gang of chickens

and so on and so fourth

im afraid none of this is making sense

i just used to do this as a child in my head with sets of 7's and i wanna know if there is a name for it

or is it just *x

[grand iris]

[cedar kiln]

@grand iris Has your question been resolved?

[grand iris]

<@&286206848099549185>

[cold knoll]

What’s the issue?

[grand iris]

the sets of 7 idk if the chicken explanation was too great

[cold knoll]

Alright

So that’s just known as variable substitution.

[grand iris]

so would it be a variable substitution to the power of 7

[cold knoll]

To the power of 7 would be
(This example we will do 3 to the power of 7) so it would be 3333333

Replacing with a variable

Would just be a normal equation

Instead of 5*3

If you don’t know that there’s a 5

You would instead have

Y*3=15

[grand iris]

so in your example 3 is the variable

[cold knoll]

In this case the variable is Y for 5

Yup

And multiplying itself 7 times is to the power of 7

[grand iris]

so kinda like logs

kinda

[cold knoll]

Powers are when you multiply the base number by itself the exponent times

So like 3

Would be multiplying the base by Itself 3 times

The base is the smaller number/first one

5^3

5 is the base

Does that make sense?

[grand iris]

yeah that makes perfect sense

so its x*7 in this example

but if it was sets of something else like sets of 5 it would be x*5

i think

[cold knoll]

In that instance

5 would be

The second number

So yes

[grand iris]

ok perfect ty v much

i just used to do it for sets of 7 as a kid in my head and i was wondering for years what it was called

[grand iris]

ty v much kind sir its been bugging me for years hahahah

!solved

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[rich blaze]

Algebra skill 📉

[wraith dagger]

ivoturi

[rich blaze]

I don't know where to begin

[wraith dagger]

ivoturi

[rich blaze]

these are the answers

[serene fern]

The bottom is 3(x+2)^2/3 so maybe start wirh that

[rich blaze]

nvm I figured it out

thanks though

.close

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[nova snow]

When working with trig proofs, are we allowed to change the RHS?

[ebon juniper]

no

[obsidian coral]

Depends, on whether the LHS or the RHS is easier to manipulate. But you're not suppose manipulate both sides at the same time, you can only manipulate one side

[cedar kiln]

@nova snow Has your question been resolved?

[cedar kiln]

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[near shoal]

hi, i have a question

[near shoal]

its about 9th grade geomtery

accelerated

I got 3 A's and 1 B

If i failed the algebra 1 exam

can i not join Geomtery in 9th grade

[magic solar]

Just ask!

[lethal jackal]

that would depend on your particular school system

[hollow trail]

i don't think that's something we can answer, you'd need to ask someone involved with the school

[lethal jackal]

you should probably send an email to the relevant administrator at your school

[near shoal]

do you guys think i can or not

like an estimate before i ask someone

[lethal jackal]

are you asking whether you can understand the material?

or are you asking whether your school will let you?

[near shoal]

let me

[lethal jackal]

you'd have to ask your school

[rain drift]

that's up to the school unfortunately. I don't think anyone here can answer that

[near shoal]

have u guys gone to high school? if so then like what got u into geomtery

after that me done

[magic solar]

High school, that’s for people in matrix soz

[near shoal]

ok

[rain drift]

don't troll

[livid hound]

be aware though if you are struggling with algebra, you're gonna have issues with later topics

as pretty much everything involves algebra and the ability to manipulate / solve equations

[near shoal]

ok

i just ran out of time i didnt mange it well 😦

[livid hound]

did you just lose 5 minutes contemplating here?

[near shoal]

yes

[rain drift]

if it helps, reflect on this as a lesson. Time management (especially in your education) is a skill worth its weight in gold.

Learn to learn things in small chunks at a time and it'll help solidify the concept. That doesn't just apply in math but in almost all aspects of life

[near shoal]

mellow i have a question what college do u go to

[rain drift]

I'm graduated at this point. But I went to a state school in Texas

[near shoal]

Nice

[cedar kiln]

@near shoal Has your question been resolved?

[cedar kiln]

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[uncut flume]

Identify the equation corresponding to the following graph

[crimson sedge]

it looks like $y = \sqrt{x}$ with a few transformations

[wraith dagger]

seymourdavison22

[uncut flume]

I see

[cedar kiln]

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[uncut flume]

.close

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[sharp viper]

Prove using the rules of inference that the premises p ⇒ q, r ⇒ ¬q imply r ⇒ ¬p.

[sharp viper]

plz help quick

[royal loom]

Can you send a picture of the original question?

[clear berry]

you can do contrapositive of the second statement idk what's it called

[sharp viper]

i finishd that qn

im on 9 now

plz help

[royal loom]

what are these questions from?

[sharp viper]

from my hw

i hav test tmrw

[cedar kiln]

@sharp viper Has your question been resolved?

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[lime perch]

how many ways to choose a committee with an equal number of men and women from 7 men and 4 women

I'm supposed to use an idea with bijection or vandermonde to solve this

I considered letting x equal number of men chosen. Then that means there are 11cX ways to choose men and 11-Xc4 ways to choose women. But I didn't know where to go from there.

[rose swan]

your initial setup doesn't look right to me

if i read your question correctly, it's just asking how many ways can you choose a committee with something like nobody, 1 woman and 1 man, 2 women and two men, etc up to 4

[lime perch]

yes

oh wait lemme check my setup

hmm is it

sum of k from 0 to 4, (7ck x 4ck)

[rose swan]

there ya go

not sure why you needed a bijection or vandermonde

[lime perch]

does vandermonde speed the process up or smth?

[rose swan]

i mean the only vandermonde i know of is the matrix kind

[lime perch]

like can it simplify sum of k from 0 to 4, (7ck x 4ck)

[rose swan]

and i don't see how this is relevant

[lime perch]

no vandermonde's identity

it deals with combinatorics

and it has to do with sum of product of binomial coefficients

so

[rose swan]

ah i got it, i'm not super familiar with it

but the wikipedia page seems helpful

let me see if i can figure it out

[lime perch]

I remember seeing a solution to a problem similar to this

they set the women's k to a different number

to like something - k = women's choose

so they could take advantage of vandermonde's identity

[rose swan]

this is what i'm looking at, but your problem is different due to the restriction of equal men and women

[lime perch]

yea

I see that my thing is so similar

to the RHS of vandermonde

and turning it into LHS of vandermonde would essentially solve the problem

k from 0 to 4, (7ck x 4ck)

[rose swan]

maybe someone else can chime in here, but i'm not really seeing how we could apply the identity here

personally, i'd be satisfied with the answer we had, but you should look into your notes/related examples

[lime perch]

wait

@rose swan

nck = nc(n-k)

so can't we just do, sum of k from 0 to 4, (7ck x 4ck) = sum of k from 0 to 4, (7ck x 4c(4-k))

and then, using vandermonde's identity our m is 7, k is k, and n is 4 and r is 4?

[rose swan]

ah, nice catch

we should test this

since the sum should then be 11C4

[lime perch]

so that is 330

[rose swan]

did you check it with the sum we came up with earlier?

[lime perch]

checking it rn

sum of k from 0 to 4, (7ck x 4ck) is 1 + 7x4 + 21x6 + 35x4 + 35

and that is 330 as well

[rose swan]

awesome

[lime perch]

nice, so it works

[rose swan]

looks like we figured it out

[lime perch]

thanks

[rose swan]

np, i learned something new as well

[lime perch]

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[cedar kiln]

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[amber surge]

We need to find the numerical value of A

[vagrant elbow]

Do you know how to write sin 2A in terms of tan A

$\sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta}$

[wraith dagger]

NEONPerseus

[amber surge]

I didn't know that formula

Still lost as hell sadly

[vagrant elbow]

I see

Do you know the double angle formula for sine?

[amber surge]

2sincos

[vagrant elbow]

Yes

That one

Draw a triangle and figure out the values of sine and cosine

[amber surge]

We have tan so we can find like
4sin = 3cos

@vagrant elbow seems legit?

[cedar kiln]

@amber surge Has your question been resolved?

[vagrant elbow]

Well you should be getting two values

One in the first quadrant and one in the third quadrant

Oh nvm 180 to 270

So only the third quadrant

Yeah it seems to add up then

[cedar kiln]

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[half elk]

I am stuck here, can you help me w this

[half elk]

$\int_{1/8}^{1}\sqrt{1+x^{-2/3}}dx$

[wraith dagger]

Ñøïr

[vagrant elbow]

Pull out x^(-2/3) from the square root and substitute whatever is inside the root for u

[half elk]

Thanks

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[delicate patio]

I'm stuck on this problem

I am not able to approach it

[clear berry]

what have you tried so far?

[delicate patio]

i thought of proceeding like this

1 = 3 - 2

to get the next natural number

added 1 to both sides

so 2 = (3 - 2 )+ 1

we will always add the 1 to the first number (which is 3 in this case)

so 2 = 4 - 2

but since 4 and 2 dont have same number of prime factors

so we can write it as
2 = (4 + k) - (2 + k)

where k belongs to Natural numbers

now if i can prove that (4+k) and (2 + k) have same number of prime factors for some value of k

basically i need to prove (n1 + k) and (n2 + k) will have same number of prime factors for some k belongs to Natural Numbers

[sage forge]

It sounds very hard to generalize though. I think this line is already a good start. You basically want to get from 1 to n. How else can you do this besides adding 1's

[delicate patio]

I can multiply?

[clear berry]

It seems a bit unlikely that an induction argument would work as that will require some general facts to be true about number of prime factors of two different numbers

[sage forge]

Yes

[delicate patio]

i can multiply by lcm of 3 and 2

that gets me 6

[sage forge]

Yes, adding 1 in general cases makes prime factors very unpredictable

[sage forge]

Unfortunately, this doesn't help

[delicate patio]

2 = 6 - 4

i can multiply by any n on both sides

that will get me all evens

as 6 and 4 have same number of prime factors so it is valid

[sage forge]

Why don't you multiply by n right away?

[sacred grail]

why not odds?

[delicate patio]

at what step?

[sage forge]

At 1 = 3-2

[delicate patio]

because we will get all 2n ig ?

[sacred grail]

2, 6, 4 all share a factor of 2

just delete it?

[delicate patio]

ohhh wait

im so dumb

1 = 3 - 2

n = 3n - 2n

Thanks for your help guys ^^

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[vagrant elbow]

$f(x) = a + 2b\arccos x$. If $f$ has the same domain and range, and $b > 0$, find the value of $b - a$.

[wraith dagger]

NEONPerseus

[vagrant elbow]

Not a surjection

I thought of assuming the set on which f is defined to be [p, q], but I'm not sure if it's right to say that since the function is monotonically decreasing, f(p) = p and f(q) = q

[tropic oxide]

f(p) = q and f(q) = p

[vagrant elbow]

Oh other way round

Shit that makes sense

Lemme give it a shot