help-0

[rancid blade]

how about 2*7

[alpine sable]

14

[rancid blade]

how about 1 * sqrt(3)

[alpine sable]

sqrt3?

[rancid blade]

yep

and sqrt3 * sqrt3?

[alpine sable]

sqrt3 squared?

[rancid blade]

which is what

[alpine sable]

sqrt 9

which is 3

[rancid blade]

yes

which is 3 right

so

[ocean seal]

grist bundle

[rancid blade]

this is FOIL

are you familiar with that concept?

[alpine sable]

i think i have heard of it

[rancid blade]

ok, do you know how to do this multiplication?

[alpine sable]

no

[ocean seal]

grist bundle

[alpine sable]

ax+bx

[rancid blade]

x distributes to each term

yes

so we can write this

[ocean seal]

grist bundle

[rancid blade]

do you agree with that?

i just distributed

[alpine sable]

i dont understand this

[rancid blade]

i did precisely what you did above

[alpine sable]

how?

[rancid blade]

1+sqrt3 = x

1 = a

sqrt3 = b

[alpine sable]

what?

[rancid blade]

i did exactly what you did above with (a+b)x = ax+bx

and i did nothing more than that

you need to read carefully and slowly

the only difference in what i did is that there are more things going on, but the exact same process is applied

[alpine sable]

ok

i forgot the asterix meant x

like times

[rancid blade]

i never used an asterisk

oh

the dot yes

[alpine sable]

its in the image

[rancid blade]

sorry

[alpine sable]

thats fine

[rancid blade]

so you agree with this then?

[wary stream]

$$(a + b)(c + d) = ab + ad + bc + bd$$

[ocean seal]

dldh06

[rancid blade]

i just applied distribution

now for each of these, you distribute again

[ocean seal]

grist bundle

[rancid blade]

carry out each multiplication...

[ocean seal]

grist bundle

[rancid blade]

this becomes

[ocean seal]

grist bundle

[alpine sable]

what the

[rancid blade]

which part is confusing?

[alpine sable]

this

[rancid blade]

ok

look here

[ocean seal]

grist bundle

[rancid blade]

do you agree?

it's just distribution again

[alpine sable]

isnt 1x(1 + sqrt 3) just 1+sqrt 3

[rancid blade]

yeah, it is.

you can use that fact and just skip to the chase

but

i'm just trying to make sure you understand how distribution works

my process arrives at the same end result as yours

but yeah

1x1 + 1xsqrt3 = 1+sqrt3

obviously

[alpine sable]

ok

[rancid blade]

i mean

it's all the same stuff

just read it carefully until you understand because i only applied a single operation

if there's a more specific question you can ask

[lone heart]

@alpine sable Has your question been resolved?

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[lone heart]

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[slate kayak]

Hi

[slate kayak]

So in my Chemistry lesson we need to do a certain math at the end that i dont quite understand.

Mol^8/Liter^8
———————-
Mol^7/Liter^7

= Mol/Liter

My question is what if the raised power would be 8 and the underside also 8 ?

[vale sapphire]

then you'd get a dimensionless quantity

[slate kayak]

Would it still be Mol/Liter

But would the Mol/Liter have raised powers

[vale sapphire]

Well, usually dimensions aren't written with their units, but with specific symbols

a mol has dimension N, and a liter has dimension L^3 (it's a volume, L is length)

[slate kayak]

Yeah

[vale sapphire]

so the dimension of what you're writing is (N^8*L^(-3*8))/(N^7*L^(-3*7))

some algebra later, you reduce to N*L^-3

which is a mol/L

[placid zinc]

x⁸ / x⁷ = x¹

[slate kayak]

Would this be: N^8L^(5) / N^7L^(4)

[slate kayak]

Wat about mol^2/L^2 divided by mol^6/L^6

My teacher said the answer is then L^3 / Mol^1

I dont get that either

[vale sapphire]

it's a multiplication inside the exponent

[slate kayak]

Oh

[vale sapphire]

what? that's wrong, unless you copied something wrong

[slate kayak]

Then what’s it supposed to be?

I think my teacher misspoke i copied what he said

[vale sapphire]

mol^6/mol^6 = 1 (btw, the dimension of something that has no dimension is written 1)

so it does not change any units

[slate kayak]

Sorry wrong thing i wrote

Yeah now its correct it was supposed to be L^6

[vale sapphire]

just do algebra like you'd do if they had other names

[slate kayak]

Okay wait

Can i name the mol “x” and the Liter “y” then?

For the algebra part

[vale sapphire]

yeah they're names

[slate kayak]

Okay

[vale sapphire]

recall, units are tied to physical variables

[slate kayak]

Oh thnx okay I’m doing it now the algebra part

[vale sapphire]

they're a way of saying what each variable represents

if you divide a distance by a time you're going to get a speed

so if you divide a distance unit by a time unit, you will get a speed unit

everything works, and that's the point

dimensions works

mol/L is, in fact, a mol divided by a L

[slate kayak]

Is it y^4/x^4 so L^4/mol^4 ?

[slate kayak]

Yeah

[vale sapphire]

yeah, seems right

[slate kayak]

Thnx

[vale sapphire]

in fact, you can do really powerful things with dimensions

[slate kayak]

O

[small swift]

Could someone take me through the chain rule for this? https://gyazo.com/13943745340b38719a6e48485fc7ea9f

[vale sapphire]

when you have a desired variable of a given dimension and some physical variables which you think it might depend on, you can declare "well, what if the variable is written as a constant times these other variables to various powers?"

[slate kayak]

Thank you yes I’ll use that from now on

[vale sapphire]

well

it's quite advanced, probably not something you'll have to use

and since two things must have the same dimension to even have the possibility of being equal, you can end up solving equations

[slate kayak]

I meant writing the subjects as variables and then doing it* sorry

[vale sapphire]

a British scientist used this principle in order to calculate the energy of the first nuclear bombs, much to the dismay of the US army, which kept that data classified

[slate kayak]

ah

[slate kayak]

Wow

[vale sapphire]

the gist of it is that he had access (like anyone who read the newspaper) to picture of the nuclear tests

[slate kayak]

So thats how he was able to use that principle?

[vale sapphire]

and so he hypothesized that the radius of the blast depended on : 1) the energy 2) time 3) the density of surrounding air

and so he wrote something like r=CE^a*t^b*d^c where a,b,c are arbitrary powers and C is a dimensionless constant

if you consider the dimensions of both sides, you can end up with a system of equations that is quite easily solved

[slate kayak]

Wow thats smart

[slate kayak]

Yeah i use that in dimensional algebra (idk how u call it in english im dutch)

[vale sapphire]

some solving and rearranging later, he reached E=K*(r⁵*d)/t²

all that was left to figure out was the constant

and G.I Taylor (the name of the scientist) had some lower-scale experimental data that indicated the constant was around 1

[slate kayak]

Oh

[vale sapphire]

with the data he had and this, he estimated E ≈ 10 kilotons of TNT

the actual figure was 18-22 kilotons of TNT

[slate kayak]

So thats way off

[vale sapphire]

it's still the same order of magnitude

[slate kayak]

Oh

[vale sapphire]

a factor 2 is important sure, but at the time people didn't know if we were talking a few kilotons or megatons or something

and besides, you can see how crude the reasoning was here

"let's hope it depends on these variables, solve and plug in values!"

for all I know there's an exponential somewhere in the actual expression and there are more variables

and in spite of all that, he was able to get a value in very much the right ballpark

with elementary dimensional analysis

and newspaper photographs

[slate kayak]

True yeah

Okay I’ll close it now since i’ve been helped

.close

[lone heart]

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[icy trail]

hmm so i need to solve this : Created an inequality such that the solution is {1, 2, 3} ∪ (4, ∞)

[icy trail]

ive made this for now: $\left(x-1\right)^{2}\left(x-2\right)^{2}\left(x-3\right)^{2}\left(x-4\right)^{1}$

[ocean seal]

uselessleaf

[icy trail]

but im not sure how to restrict it so that x=4 doesnt work

i tried this intead: $\frac{\left(x-1\right)^{2}\left(x-2\right)^{2}\left(x-3\right)^{2}}{\left(x-4\right)^{1}}$

[ocean seal]

uselessleaf

[icy trail]

but it makes it wrong

[gray isle]

you haven written an inequality

[magic geyser]

t

[icy trail]

i mean >= 0

at the end

[gray isle]

why do you think that your last attempt doesn't work?

[icy trail]

hm.

no clue tbh

[gray isle]

,W solve ((x-1)(x-2)(x-3))^2/(x-4) >= 0

[ocean seal]

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Requested by ramonov#0921

[icy trail]

hmm

graphing it ddidnt look right

but yea i thought it would be right. . .

.close

[lone heart]

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[narrow plume]

Could you help me in solving Throwback Question please?

[glass lichen]

,rotate

[narrow plume]

How many minutes will take to solve it?

<@&286206848099549185>

[gray isle]

@icy trail you didn't zoom out to see what happens after x=4

[alpine sable]

What type of function will this be

[narrow plume]

I believe it will be a sin func

f(x)= A(B sinx)+C

[alpine sable]

Yes

Do you know how to get a b and c

[narrow plume]

No and that is the problem

I am stuck here

[alpine sable]

Wait sorry you forgot another letter

It should be

I mean the wrong place

asin(bx)+c

So first of alll

Do you know what a means

[narrow plume]

OH!! you are right! I am sorry

[icy trail]

ah okay

[narrow plume]

the Amplitude correct?

[alpine sable]

Yep

You know the diameter of the Ferris wheel is 40

[narrow plume]

Yep

[alpine sable]

How can you make that the amplitude

Think of the minimum and maximum height

[narrow plume]

should I divide the 40 by 2?

[alpine sable]

Yeah

[narrow plume]

oh ok

[alpine sable]

so your a should be 20

Now let's go to b

Do you know what is b and how to calculate

[narrow plume]

Ok 👍👍👍👍

[narrow plume]

Idk what B is but I only know C and A

[alpine sable]

ok so tell me c first?

[narrow plume]

C is the vertical shift

[alpine sable]

and what would the value be

[narrow plume]

Ummm... 2?

[alpine sable]

Correct

[narrow plume]

Oh cool!!

[alpine sable]

Now for b

[narrow plume]

is B the horzantil shift?

[alpine sable]

Nope

[narrow plume]

Oh then what is it?

[alpine sable]

b is the period of the function

[narrow plume]

oh ok!

[alpine sable]

So originally sinx has period of 2pi

But we need to modify it

We can do this by doing b= 2pi/10

[narrow plume]

so cz it takes 10 mins to be 1 revolution

[alpine sable]

Because 10 is the time taken to complete a revolution

Yeah

[narrow plume]

would it be pi/5?

[alpine sable]

Yes

[narrow plume]

Omg thank youuu so much!!!!!!!!!

[alpine sable]

You are welcome

You can also graph it to double check

I did that and it looks good

[narrow plume]

what about the x?

[alpine sable]

x is just x

[narrow plume]

Oh ok thanks I really appreciate

[alpine sable]

20sin(pi/5(x))+2

[narrow plume]

Thanks again I am so happy for real!!!!!!!

[alpine sable]

You are very welcome

Happy to help

[narrow plume]

Bye have a great lovely night/day

[lone heart]

@narrow plume Has your question been resolved?

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[slate rock]

solve 4sin²θ - 9cosθ - 6 = 0, where 0 ≤ θ ≤ 360

[slate rock]

I have no idea how to do this

[still vigil]

Convert sin^2 to cos^2 using the identity sin^2 + cos^2 = 1

then solve as a quadratic

[slate rock]

is fitaa equal to 104

[still vigil]

fitaa?

[slate rock]

θ

[still vigil]

there should be more than 1 answer

,w 4sin^2(x)-9cos(x)-6=0

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Toru#9912

[wary stream]

That's spelt theta btw

[slate rock]

omg that's difficult

[wary stream]

Not fitaa

[slate rock]

oh my bad

[slate rock]

I never saw this on my notes I don't understand at all

[still vigil]

What is your working right now?

[wary stream]

As stated before, use sin^2 + cos^2 = 1

To change that sin^2 to cos^2

[lone heart]

@slate rock Has your question been resolved?

[wind mist]

zombieman are you online

[lone heart]

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[whole lake]

Hi

[scenic sable]

i do not understand this

[lone heart]

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[whole lake]

Oh is this one taken?

[lone heart]

Channel closed

Closed due to the original message being deleted

[scenic sable]

can anyone tell me the working?

please

[strange rampart]

(x-2)(x+4)

[scenic sable]

ooooooh okay thanks

[strange rampart]

Np

[scenic sable]

@strange rampart sorry for ping but how do i get the area?

[strange rampart]

Try to express the area of the grey part using x

[scenic sable]

i dont understand

[strange rampart]

How do you calculate area of a square?

[scenic sable]

the length^2?

so whats the next step?

[placid zinc]

It isn't a square. But it is a rectangle

Identify what the side lengths are. Note the side lengths will include x

[scenic sable]

thats the part i do not understand

sorry

can you perhaps teach me how?

oh its okay i got it

[lone heart]

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[scenic sable]

thanks anyway

[scenic sable]

.close

[lone heart]

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[pearl solar]

Conormal points lies on a fixed curve that passes through h,k

(a^2-b^2)xy + b^2 kx - a^2hy

It is for an ellipse x^2/a^2 + y^2/b^2 =1

I need help proving it.

[cerulean galleon]

can someone please explain step by step how to solve this

sry thx for letting me know

[lone heart]

@pearl solar Has your question been resolved?

[pearl solar]

<@&286206848099549185>

[lone heart]

@pearl solar Has your question been resolved?

[long plover]

guys

[lone heart]

@pearl solar Has your question been resolved?

[sly comet]

why do i integrate to find area under something of an equation representing the rate of change

whereas sometimes i dont

[lone heart]

@pearl solar Has your question been resolved?

[pearl solar]

No

[versed dirge]

Is this calculus?

What does estimate mean in this case

[icy scaffold]

is x<−7 or x>−2 the same as −7>x>−2

[ocean seal]

KingOfDireWolves

[noble jackal]

you can try this

this is easily proved to be true by taylor expansion
oh btw I forgot approximation sign and put equals to instead

[pearl solar]

This channel was started for this question, please remain on topic and help me

[half skiff]

help

[fathom rain]

Write an expression for how many sticks with figure number n

[next chasm]

This channel is busy

[pearl solar]

..

[vagrant bronze]

[woven dock]

hello！I can finally lay my hands on this question.
I did spend some time on it 2 hours ago, but I can't get something useful.

My approach is to find the equations of the normal lines on the ellipse with (x_1,y_1) and (x_2,y_2) with their intersection be (h,k) and solve for h,k in terms of those x's,y's.

[pearl solar]

Alright after that

The normals would be y-y1 =(y1/x1)(a^2/b^2)(x-x1)

[woven dock]

Yea

But what is this
(a²-b²)xy+b²kx-a²hy
?

[pearl solar]

It is supposed to be the answer

The curve on which the conormal points lie

[woven dock]

But this is just an expression

Does it equal to 0?

[pearl solar]

Yeah sorry

It is equal to zero

[pearl solar]

Yes

Occupied channel

[woven dock]

Did you get h,k in terms of x1,y1,x2,y2 from this?

[pearl solar]

Yes but the whole expression was a big mess

[woven dock]

Can you show it please, because I left my paper in my office.

[pearl solar]

Sure just give me a minute

[woven dock]

I'm reading.

[pearl solar]

Ok

[woven dock]

OK, fourth equation missing over x_1 and x_2

[pearl solar]

Oh i made a mistake, i am so sorry

Just give me one more minute

[woven dock]

h = (x1 x2 (a^2 - b^2) (y1 - y2))/(a^2 (x2 y1 - x1 y2))
I got this from wolfram for your reference.

[pearl solar]

Alright

[woven dock]

This is also what I got back then irrc. I saw it was a mess so I gave up and continue my work lol

[pearl solar]

Lol, so you have any lead? On how to approch further

[woven dock]

I was gonna use that and the normal equation to find k.
And we should have very similar result.

And you can notice that if we sub (h,k) into this equation, we can factorize it nicely.
(a²-b²)xy+b²kx-a²hy=0

[pearl solar]

We are substituting h,k in the normal's eqyation or what?

[woven dock]

(h,k) into (a²-b²)xy+b²kx-a²hy=0

It's like since we already know the final answer, we can make use of it.

[pearl solar]

Oh alright

But i was thinking

[woven dock]

Because we just have to prove it

Yea?

[pearl solar]

Is there a nicer way to go to that

[woven dock]

I am thinking too

[pearl solar]

Like if we did not know the answer we can sort of eliminate x1 y1 x2 y2 in a nicer way

Maybe i can try to take acostheta and b sin theta instead of x and y to make it easier

[woven dock]

Let's see.

[pearl solar]

I think it might be simpler as there would be only 2 variables to eliminate

Let me try

[woven dock]

Btw, you can take a=rcosθ and b=rsinθ if you are changing the variables

[pearl solar]

It is an ellipse

I dont get what you are trying to say

[woven dock]

This is k btw
k = (y1 y2 (a^2 - b^2) (x1 - x2))/(b^2 (x2 y1 - x1 y2))

[pearl solar]

I got some values from here

Any idea to eliminate the theta and phi

[woven dock]

No clue yet...

[pearl solar]

Ok

[woven dock]

(x2 y1 - x1 y2)
= (x1 x2 (a^2 - b^2) (y1 - y2))/(a^2 h)
= (y1 y2 (a^2 - b^2) (x1 - x2))/(b^2 k)

[pearl solar]

How?

[woven dock]

Btw why does this work?

[pearl solar]

The equation of the ellipse was x^2/a^2 + y^2/b^2 =1 so i saw that acos theta b sin theta lies on it

[woven dock]

Ohhhh

Cool

[woven dock]

I just change the subject, not sure if those works though

[pearl solar]

Oh alright i might just ask the prof then

[woven dock]

Is this question from any specific topic in mathematics?

[pearl solar]

It was just from the conic sections exercises

[woven dock]

Okay...

[pearl solar]

.close

[lone heart]

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[half skiff]

I don't understand what's going on

[torpid delta]

|x| = x if x >= 0 and
|x| = -x if x < 0

so |3x+2| = x -2 if:
(3x+2 = x-2 AND 3x +2 >= 0) or (-3x -2 = x -2 AND 3x+2 < 0)

[half skiff]

?

[torpid delta]

that 3x + 2 = 0 <=> x = -2/3

? what

clarify

don't be vague

[half skiff]

does I3x+2I means

(3x+2)^2

[torpid delta]

absolute value of 3x + 2

no

[half skiff]

cuz IxI means x^2

[torpid delta]

it doesn't

$|x| = \sqrt{x^2}$

[ocean seal]

TC159

[torpid delta]

$|x| \neq x^2, \forall x \not \in {-1, 0,1}$

[ocean seal]

TC159

[half skiff]

ok

so we try to make

x =2

and we do this by

((3x+2)^2)^1/2 = 2

[torpid delta]

nonono

it's an equation

for example how would you solve 3x +2 = x+ 2?

[half skiff]

well you takes the x

therefore

2x=0

therefore x=0

or 3x=x

3=1

wait

your example doesn't make any sense

@torpid delta

[torpid delta]

?

[half skiff]

explain it to me as you expain to a 3 year old child

[torpid delta]

3x +2 = x +2
cut out the 2's get
3x = x
x = 0 or 3 = 1 (which is impossible) so x = 0

remember $a\cdot b = a \cdot c \iff b = c \lor a = 0$

[ocean seal]

TC159

[torpid delta]

because if a = 0 you have 0b = 0c

which is true for all b and c

[half skiff]

ok

got it

[torpid delta]

and |x| is defined as the absolute value of x

if x is positive then |x| is just x since x is already positive

if x is negative then -x is positive so |x| is -x

[half skiff]

so i got

x=0 and x=-1

and i did it by

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

but why teacher said x> 2/3

@torpid delta

[half skiff]

hi

yes?

it's freakjng me out

much faster then normal

[noble sinew]

|3x+2|=x-2, for x>=-2/3 we have |3x+2|=3x+2, agree?

[half skiff]

x>=-3/2?

[noble sinew]

Go read rules, this channel taken (and saying I don’t understand algebra is not a question)

[half skiff]

i ddon't get that paort

[noble sinew]

If x>=-2/3 then 3x+2>=0 agree?

[half skiff]

so you ties both size by 3 then -2

times

[noble sinew]

?

[half skiff]

not tie

times

[noble sinew]

3x+2>=0 <=> 3x>=-2 <=> x>=-2/3 agree?

[half skiff]

yes

[noble sinew]

So definition of abs value is then if x>=-2/3 then |3x+2|=3x+2

[half skiff]

but

what i don't get is

how you have an inequlities

[noble sinew]

,w mathematicsl definition of absolute value

[ocean seal]

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[noble sinew]

Do you agree this is the definition of abs value?

[half skiff]

yes

-x for x<0

and x for x>0

[noble sinew]

So we are simply just solving for when the part inside the abs value is >0

And <0

[half skiff]

ohhhhhh

[noble sinew]

Since then we can use the definition

[half skiff]

but why does the question have equation ?

x>=-3/2

[noble sinew]

x>=-2/3

[half skiff]

|3x+2|=x-2

what's the point of having it equal to somethign

[noble sinew]

The question is find the values of x such that those 2 are equal

[half skiff]

aren't the intercept 0 and -1

[noble sinew]

Its an equation just like solve 3x=3 is an equation

No

[half skiff]

i think im staring to get it

[alpine sable]

Can I have some help or smth guys

[wary stream]

Not your channel
Read #❓how-to-get-help

[alpine sable]

what?

I thought it was free @wary stream

There was no one chatting for 12 minutes

[wary stream]

So, read #❓how-to-get-help

[alpine sable]

@wary stream It was free no one was chatting for 12 minutes straight

[wary stream]

It's under the occupied section

[alpine sable]

huh?

[wary stream]

Meaning it's occupied

So use a channel under the available section

[alpine sable]

why is everything occupied now it wasn't a few days ago

[wary stream]

Doesn't matter, new system

[wary stream]

There are 2 channels open

[alpine sable]

Alright

Thanks for informing me

I'm sorry for any inconvenients

[lone heart]

@half skiff Has your question been resolved?

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[split slate]

How/Why is the limit of this -∞ as x approaches ∞?

$\lim_{x \to \infty}f(x)=6-3x$

[vale sapphire]

imagine x is really big and negative

like -100000

what is -3x

[ocean seal]

HappyAlt

[vale sapphire]

it's really big sure

but what sign?

[split slate]

negative?

but doesn't the infinity sign in the x approaches... line without a negative sign mean only positive infinity?

[vale sapphire]

oh wait

i misread that

if x is really big and positive

then -3x is really big, but what sign?

[split slate]

negative

[vale sapphire]

correct

[split slate]

I see what you're getting at

But my question was that doesn't infinity kinda count numbers like 1 and stuff in it as well?

[vale sapphire]

so as x gets arbitrarily large (positively), 6-3x gets arbitrarily large negatively

[split slate]

so it wouldn't always equal negative

[vale sapphire]

not sure what you mean by this

[split slate]

like

x could also be 1

[vale sapphire]

ok i see what you're getting at

the point isn't that 6-3x is always negative or anything

[split slate]

generally negative?

[vale sapphire]

it's that, when x gets large enough, 6-3x gets really large negatively

limits as something tends towards infinity don't make statements about what happens everywhere

[split slate]

I see

[vale sapphire]

they just say something about what happens when the variable gets large enough

[split slate]

So it's a majority kind of thing

[vale sapphire]

i wouldn't see it as a proportion

more like

if you have a slider that controls x

and you veer it all the way to the right at full speed

where is the output slider gonna go?

[split slate]

all the way to the left?

(negative side)

[vale sapphire]

yes

and that's precisely what a limit of -∞ as x → +∞ means

[split slate]

i see

Thanks for explaining this clearly

[split slate]

and making me chuckle a little

[vale sapphire]

heh

[split slate]

ight, cya

.close

[lone heart]

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[cursive drum]

Hello

[cursive drum]

does anyone know why when I try to plot this it cuts off at 0?

MINE:

[wary stream]

What exactly do you mean by cut off?

[alpine sable]

x intercept

i presume

[cursive drum]

the function doesnt continue like it does on wolfram

they just stop at x=0 and y=0

[wary stream]

The x and y axes are the asymptotes

[fiery bronze]

Did you put a limit on the x axis? a limit up to 15?

[cursive drum]

yeah no i dont mean on the asymptotes

i mean on the other side of the function

maybe here its more clear

[wary stream]

The lines?

[cursive drum]

see how the orange function cuts off

doesnt continue in the negative space

for x<0

[wary stream]

Maybe that software identities there is an asymptote so it automatically cuts it off?

[cursive drum]

if i just plot one though it still cuts off

[cursive drum]

this is what it should look like

[fiery bronze]

did you insert the same equation in both of them?

[cursive drum]

yep

[wary stream]

Probably something on the software side for that plot

[fiery bronze]

cause it looks like in the the second graph the orange function starts at x axis 0

[wary stream]

Try using other graphing sites to see if it does the same thing

[fiery bronze]

yeah

[cursive drum]

ill have a look then

[lone heart]

@cursive drum Has your question been resolved?

[golden falcon]

There are two free channels in MATH HELP(AVAILABLE). Try asking your question there

[pale sable]

my apologies. I had that minimized, didn't see

[real crescent]

hi how do i do 10b thank u

[wet cove]

its one of them y=mx+b

[wary stream]

Read #❓how-to-get-help

[wary stream]

Read #❓how-to-get-help

Not your channel to ask in

[lone heart]

@cursive drum Has your question been resolved?

[lone heart]

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[upper escarp]

May I ask how does it simplify to the answer written there?

for H

[lone heart]

@upper escarp Has your question been resolved?

[drifting phoenix]

final answer is ~10.83852

[upper escarp]

yes ik, but i wanna know how to simplify the exact value to get the final answer written there

[thorn vortex]

You can't.

[woven dock]

The second step looks wrong to me...

[lone heart]

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[alpine sable]

<@&286206848099549185>

[lone heart]

@alpine sable Has your question been resolved?

[full cliff]

Uh

[lament roost]

Put x = 1 @somber spoke

[somber spoke]

i didnt realise that the rule section changed

check help-1

[lone heart]

@alpine sable Has your question been resolved?

[light sleet]

[alpine sable]

@light sleet this channel is occupied

Go on other channels

Channel 9 16 or 18

[lone heart]

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[alpine sable]

help

[alpine sable]

Can I have some help here guys

A #1 my answer is

[light sleet]

ok my bad

[alpine sable]

1.152x10⁷

@light sleet btw

The answer to your question

Is 1/10

[light sleet]

oh

[alpine sable]

wait

is that minus

[light sleet]

ye

[alpine sable]

nvm

Yeah 1/10

@light sleet answer is 1/10

[light sleet]

so multiplication?

[alpine sable]

Guys can I have help here my answer in #1 is 1.152x10⁷ is my answer correct

[light sleet]

[alpine sable]

@light sleet 5-4/10

= 1/10

[light sleet]

ohh okk

ty

[alpine sable]

Anyways guys my answer at #1 is

1.152x10⁷

May I ask if my answer at #1 is correct

The question is about scientific Notations

<@&286206848099549185>

[lone heart]

@alpine sable Has your question been resolved?

[full reef]

ñ

[edgy cipher]

does someone think they can help me with my exam im in grade 12 and its 8 units im planning on doing 2 units per day the exam is next week

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

[fathom field]

can someone help me w the diff equation

(1-x^2)y'+(1+x^2)y=x

<@&286206848099549185>

[astral dagger]

suppose (y(x)=\sum_{n=0}^{\infty}~a_n\cdot x^n)

[lone heart]

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[ocean seal]

SubGui

[fathom field]

ohhh ok

power series it is

i was thinking i should change it to the form y' + (1+x^2)/(1-x^2)y=x

[lone heart]

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[sonic wyvern]

i know the ratio between each term is -2/3

so i know i need to multiply each preceding term by -2/3

idk how to write that mathematically

i thought (-2/3)^(n-1)

but then if n = 1, you just get 1

[raw shard]

that’s right

that’s how you would write it

[sonic wyvern]

but what about this

you would get -2/3 ^ 0

which is 1

[raw shard]

wait it’s actually -3*(-2/3)^(n-1)

[sonic wyvern]

oh yeah i had an idea about multiplying times 3

i didnt know if it'd work

thanks

.close

[lone heart]

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[alpine sable]

What am i doing wrong thanks in advance

[alpine sable]

My program says its inf not - inf ...

[charred flint]

,rotate ccw

[charred flint]

,w limit of sqrt(n^2+2n) - n^2 as n goes to infinity

[ocean seal]

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[charred flint]

[alpine sable]

Found the error thanks to it ...😂

Its -n^2

Thanks

.close

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[lone heart]

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[sage tartan]

need help with the second part

[lone heart]

@sage tartan Has your question been resolved?

[bleak ridge]

Well you can imagine the circle

The least point would be one of the points from the origin tangent to that circle

Idk if there's some other fancy way to do it but a nice steaming plate of trig could be in order

[lone heart]

@sage tartan Has your question been resolved?

[sage tartan]

yes i thought about that

but its a 2 mark question, there is obviously some easier way to do it that we're missing

[sage tartan]

<@&286206848099549185>

[lone heart]

@sage tartan Has your question been resolved?

[tough vessel]

Can y’all help me

[jagged trout]

What's the problem?

[raw shard]

probably get rid of all the decimals

so multiply both sides by 2

[tough vessel]

You have to solve the equation for t

[raw shard]

then multiply the 10 and (2t-3)

[jagged trout]

Multiply everything out and seperate the t

[tough vessel]

Thnx 🙂

[steep flax]

hello i am a bit confused as to how i simplify this

like how is 3x^2+6x=3x(x+2)

[placid zinc]

Common factor

If you're still confused, try distributing:
3x(x + 2)

@steep flax

Oh, this isn't your channel. See #❓how-to-get-help

[steep flax]

oh my bad

[lone heart]

@sage tartan Has your question been resolved?

[alpine sable]

how does this look

[alpine sable]

part b of the same question

i got this

[tight locust]

you can simplify further.

(a^2-1) = (a+1)(a-1)

sqrt(a^2-1) = sqrt(a+1)*sqrt(a-1)

sqrt(a^2-1)/(a+1) = sqrt(a+1)*sqrt(a-1)/(a+1) = sqrt(a-1)/sqrt(a+1)

= sqrt((a-1)/(a+1))

[alpine sable]

so sqrt(x-y) = sqrt(x)sqrt(y)?

[tight locust]

no

sqrt(xy) = sqrt(x)sqrt(y)

[alpine sable]

ah

[tight locust]

but actually nevermind that. putting everything under one square root limits the domain of the function

sorry didn't realize that

[alpine sable]

so what i got is the most simplified?

[tight locust]

yeah it's good

[alpine sable]

thx :)

[late parcel]

Is this channel open?

Number 6, what do I do from here?

How do I solve for Y?

Never mind

I got it

[lone heart]

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[jovial cosmos]

how do ou do this?

nvm

.close

[lone heart]

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[jovial cosmos]

.reopen

[lone heart]

✅

[jovial cosmos]

why do we set gradient vector as 0,0 to find critical point

[lone heart]

@jovial cosmos Has your question been resolved?

[jovial cosmos]

.close

[lone heart]

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[ripe harbor]

I need the solution steps for this question 😅

I have no idea how to do this haha

[bright hedge]

well you konw that

f(x) = x/(4x+1)

and

$f(x) = \frac{ax}{bx+c}$

[ocean seal]

not discordmod

[bright hedge]

so plug in like 3 points into f

so like find f(0), f(1), and f(-1)

and use it to solve for a, b, and c

then if g(x) = sqrt(x+2)

you can say

$x = \sqrt{g^{-1}(x)+2}$

[ocean seal]

not discordmod

[bright hedge]

then just solve for $g^{-1}(x)$

[ocean seal]

not discordmod

[bright hedge]

but remember that g(x) is only defined for x >= -2

[ripe harbor]

Ahh I see, I'll try solving it now, Thanks a bunch!

.close

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[night geode]

I'm working through chapter 5 of Spivak's textbook. Could anyone why he takes epsilon and y0 over 2 in his proof for this lemma?

[lone heart]

@night geode Has your question been resolved?

[night geode]

<@&286206848099549185>

[ripe gust]

huh

[lone heart]

@night geode Has your question been resolved?

[night geode]

No

[thorn sundial]

need help

[umbral orchid]

cos is even and sin is odd

[lone heart]

@night geode Has your question been resolved?

[rose glade]

Number 17

Is cos p/2 convert to sin?

[runic idol]

(28)(27)(27)(-12)=

i.need,it,for.something

[lone heart]

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[vale wigeon]

@alpine sable

still here?

$\nabla \times$ is a linear operator. part (i) requires you to understand that and apply the equation $\nabla \times \mathbf{f} = k\mathbf{f}$ twice

[ocean seal]

Ann

[vale wigeon]

well $\nabla \times (\nabla \times \bd{f}) = \nabla \times k\bd{f} = k(\nabla \times \bd{f})$, don't you agree?

[ocean seal]

Ann

[vale wigeon]

i suppose $\nabla \bd{f}$ refers to $\nabla \cdot \bd{f}$, i.e. the divergence?

[ocean seal]

Ann

[vale wigeon]

that was me trying to clarify something for part (ii).

i want to ensure i didn't misunderstand the question.

no

k^2 f

@rain ocean do not post in occupied channels.

i am trying to think of a way to avoid just writing out all the partial derivatives in one big messy system of equations

i'm drawing a blank so i guess messy system of equations it is

yes k^2 f is the final answer for part (i)

no surprises there

let $f_{ij}$ be the derivative of the $i$'th component of $f$ with respect to $x_j$ ($i,j=1,2,3$)

[ocean seal]

Ann

[vale wigeon]

and let $f_i$ be the $i$'th component of $f$

[ocean seal]

Ann

[vale wigeon]

now let's hope i don't screw up the signs here

we have $\begin{cases} f_{32} - f_{23} = kf_1 \ f_{13} - f_{31} = kf_2 \ f_{21} - f_{12} = kf_3 \end{cases}$

[ocean seal]

Ann

[vale wigeon]

i may or may not be talking out of my ass right now

our goal apparently is to show $f_{11} + f_{22} + f_{33} = 0$

[ocean seal]

Ann

[vale wigeon]

okay, maybe this notation is just bad

hold on

let's use superscripts for components and subscripts for partial derivatives instead

so $f^3$ will be the third ($z$) component of $f$ and $f^3_1$ will be $\pdv{f^3}{x}$

[ocean seal]

Ann

[vale wigeon]

we have $\begin{cases} f^3_2 - f^2_3 = kf_1 \ f^1_3 - f^3_1 = kf_2 \ f^2_1 - f^1_2 = kf_3 \end{cases}$

[ocean seal]

Ann

[vale wigeon]

ah, then it appears all it takes is to differentiate the first equation wrt x_1, the second wrt x_2 and the third wrt x_3 and add them all together.

...or

on second thought

we could've avoided all this altogether

my apologies

cause we know that $\nabla \cdot (\nabla \times \bd{f}) = 0$ lmao

[ocean seal]

Ann

[vale wigeon]

so $\nabla \cdot k\bd{f} = 0$

[ocean seal]

Ann

[vale wigeon]

https://www.google.com/search?q=divergence+of+gradient+is+zero

anything else you'd like to ask or can we close this channel fttb

[lone heart]

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[alpine sable]

Hello, I'm unsure about this with Calculus

[alpine sable]

I have d/dx sin(x) = cos(x)
But, in an example problem I'm seeing that d/dy [-3cos(y)] = 3sin(y)

Shouldn't it be -3sin(y)?

[ancient saddle]

Hello. The derivative of cos is -sin

[alpine sable]

So, was there an error in my lecture slides?

[ancient saddle]

no, they are ok

d/dx cos(x) = -sin(x)

[modern lotus]

The derirative of sin and cos goes sin cos -sin -cos sin cos…

[alpine sable]

I don't quite understand that
-cos(y) = sin(y)?

[modern lotus]

d/dy -cos(y)=sin(y)

[alpine sable]

I see

Thanks

.close

[lone heart]

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[trail moss]

Can anyone help me

[trail moss]

Can someone define cuts the curve

in magthjs

*maths

[stuck aspen]

#❓how-to-get-help point 2 "Don’t ask permission to ask a question..."

[lone heart]

@trail moss Has your question been resolved?

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[harsh belfry]

Could use some help with this problem

[harsh belfry]

I know I need to combine my expressions and isolate x, but I'm not too sure how to get there.

Would appreciate any help.

[alpine sable]

15 = 1/x²
Now, flip flop
1/15 = x²
Same as saying
x² = 1/15

[harsh belfry]

Ah and from that step I should then square the x and square 1/15

by breaking up the 1/15 into factors

Thanks

[alpine sable]

IDK what you mean by that.

1/15 cannot be broken down in factors.

I mean, you will be end up with

[harsh belfry]

checking by what my assignment wants as an answer it expects this

[alpine sable]

$x^2=\frac{1}{15}$\
$\sqrt{x^2}=\pm\sqrt{\frac{1}{15}}$\
$x=\pm\frac{1}{\sqrt{15}}=\pm\frac{1*\sqrt{15}}{\sqrt{15}*\sqrt{15}}$

[ocean seal]

TeamSeas.org

[alpine sable]

Your assignment didn't include the negative answer but that's what they did basically.

[harsh belfry]

Ahhh I see

Thanks

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[uneven dock]

hey

[uneven dock]

is the 1st question true

how many percent is 35 out of 280

[alpine sable]

,calc 0.125*280

[ocean seal]

Result:

35

[alpine sable]

Yes.

[uneven dock]

is it 35?

35%

but that doesnt give any sense

[alpine sable]

No, that's not the answer.

That's me checking something.

[uneven dock]

so 12,5% is correct

[alpine sable]

Yep.

[uneven dock]

what about the 2nd question

how many percent is 450 of 125

[alpine sable]

You think.

[uneven dock]

i dont know how to do it

[alpine sable]

Well, how much do you know about percentage?

[uneven dock]

360%?

well i know a little

[alpine sable]

,calc 3.6*125

[ocean seal]

Result:

450

[alpine sable]

Your answer is correct.

[uneven dock]

360?

[alpine sable]

But how did you guess?

[uneven dock]

i said 450: 125

*100

[alpine sable]

Yep, that's the method to do it.

[uneven dock]

ahh

and the 3rd is how many percent is 28 higher than 7

[alpine sable]

Percentage is basically fraction * 100.
And fraction = how much : original value
like in this case we had "how much" as 450 and original value as 125.

[uneven dock]

uh okay

[alpine sable]

You try first.

[uneven dock]

70%?

just a guess lol

400%?

idk

[alpine sable]

Yep.

[uneven dock]

wait wha

400?

yeess

[alpine sable]

I mean.

400% of 7 = 28.

[uneven dock]

yeah true

[alpine sable]

I'm not sure what the question is asking for though.

[uneven dock]

but i dont get the 4th one

how many percent is 37 lower than 148

isnt it around 20?

[alpine sable]

Try yourself. You have around got the hang of it.

[uneven dock]

25%?

is that correct?

you there

anyonee

[hearty hazel]

No, 37 is 25% of 148: therefore, it's 75% lower

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@uneven dock Has your question been resolved?

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[uneven dock]

so its 75%

[alpine sable]

kiss me 😫

I wAs PiNgeD 5 TiMes

[sly mantle]

knock it off

[alpine sable]

I GotTa CalL a MoD cAUsE I caNt HanDle It My Self

[alpine sable]

yeah ok

doing a great job sir btw

[sly mantle]

ok you've been a nuisance for too long. goodbye

[stuck jolt]

Please shut up

[alpine sable]

bye

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[alpine sable]

Confused with A. Not too sure where to start and how to write out the formula. Need a bit of guidance please.