help-0

[past wave]

0

Ahh

[weary wyvern]

I assume by loga you mean log_a

[past wave]

Yh

[weary wyvern]

Log_x(y) means x to the what power equals y?

a to the what power equals a^b?

[past wave]

uh

how do u know that

to get from a to a^b is just b ?

[weary wyvern]

Yes

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[vast mesa]

I have absolutely no idea how to turn this into an equation

[snow furnace]

So if he reduces speed by 3 every mile the acceleration would be -3

And the relationship between acceleration and velocity is?

[vast mesa]

I have no idea.

[snow furnace]

Ok, well what would the slope be

[vast mesa]

I got the a velocity of 2 6/25

The issue is the word problem.

I don’t think it’s correct though.

I did 28/v equals 3/(v-2)

All the equations I do give me different results.

Nvm. I found it out myself.

!close

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[daring tiger]

Hi I need to learn geometry

[fallow drift]

That’s vague

[daring tiger]

ik but i only need to learn it bc I'm already in 11th grade and only in algebra 1b

[fallow drift]

I mean you can do the khan academy module

[daring tiger]

i tried that and non of it made sense to me

.close

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[agile cypress]

Hi, I was working on a problem and I'm not sure to understand how to solve it. It's probability.

Let E be a set and A a fixed subset of E. Let F = {B subset of E such that A ⊆ B}. Determine the tribe G generated by F.

[lone heart]

@agile cypress Has your question been resolved?

[agile cypress]

<@&286206848099549185>

I already did this : let B in F, the A ⊆ B.
But the complementary of B ⊆ complementary of A.
Then A is not included in the complementary of B.

So B is not a tribe

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[quaint furnace]

yes

[lone heart]

@alpine sable Has your question been resolved?

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[finite flax]

the f***???

ok, this is a dogfood question

but what they are trying to say is

if you plug 20 into C (do F(C) where C=20)

and pretend you could divide what F(20) is by 0,

what is the limiting value of F

F = (9/5)(20) + 32

F = 180/5 + 32

= some constant

(I'm lazy and it doesn't matter)

[fallen verge]

Technically not of the answers are right

It should be undefined

[finite flax]

lim c -> 0 F/c = lim c -> 0 (some constant)/c

= infinity

+inf

[fallen verge]

Thats 0+

0- gives you -inf

Limit dne, f is undefined

[glossy junco]

there is no undefined option

[fallen verge]

I know

The question is written poorly

[finite flax]

being undefined doesn't mean a limit doesn't exist

understand that

[glossy junco]

also ik it’s undefined but I think they are asking to make it on one of these forms I guess

[fallen verge]

The limit doesnt exist either

[finite flax]

the limit wouldn't exist if the left hand and right hand limits don't exist

you are saying that? 🤔

you are right

[fallen verge]

No they dont match

[finite flax]

@glossy junco tell your teacher about this, they will be proud of you

[peak bough]

What is that round thing with the x in it

[fallen verge]

G) Undefined+bad question+L+ratio

[peak bough]

Is this some wheel theory shit

[finite flax]

so try this

[peak bough]

What is the symbol supposed to represent?

[finite flax]

and compare how it changes as ....

wait

I need a different function 😅

oh no, different divisor rather

[fallen verge]

Oh yeah i shouldve asked about that symbol

@glossy junco please clarify what that OX symbol is pls

[glossy junco]

inf/0

inf/0=OX

[fallen verge]

What class is this?

[glossy junco]

well idk but one of my teachers gived us this question and challenged us to solve it so.. idk where he got this question from

[fallen verge]

But what class are you in?

[glossy junco]

but he told us that OX = inf/0

[fallen verge]

Ok but what class are you in?

[glossy junco]

algebra

[fallen verge]

What kind of algebra

Like are you in secondary or uni

Im trying to see if your teacher gave you a fair question

[glossy junco]

uni

[fallen verge]

Ah

[glossy junco]

algebra & Trigonometry

I have been trying to solve this for three days now

[fallen verge]

🤨

Your teacher gbave you an unfair question

[glossy junco]

wt

why would he do that lol?

[fallen verge]

Idk

[rose sigil]

what the heck is that question 😕

[fallen verge]

I know right

[peak bough]

I don't really know how you would evaluate 68/0 with the definition that inf/0 = OX

[glossy junco]

I’m 100% sure there is an answer for this but I think it’s too advanced that no one can solve it( normal people like me and you), anyways I gave up I will ask him for the answer when I see him.

[finite flax]

not the same function but this is the idea of limits here

what is the value of f as we evaluate f(x) at c where c is getting closer and closer to 0?

we do this approaching 0 from the left side

and from the right side

notice that if you follow the branches

you arrive at different places as you approach 0 from the left and from the right

so the limit DNE here

[peak bough]

I don't really think limits apply the same way to this question

[finite flax]

they totally do

it's just a dogshit question

[peak bough]

We're talking about a system where division by 0 has been defined

[finite flax]

where is that definition

@glossy junco

[fallen verge]

Inf divided by 0 is defined

[finite flax]

...as what

[peak bough]

Inf/0 = OX

[glossy junco]

it’s not defined

[finite flax]

hugs and kisses?

[peak bough]

?

[fallen verge]

Its that one Ø symbol thing

[glossy junco]

that just a symbol for representing inf/0

[rose sigil]

i think mathematical rigor has left the chat

[fallen verge]

It wasnt there to begin with

[glossy junco]

Maybe idk

[peak bough]

Division by zero is not defined for real numbers, so unless he's working from another well defined system, Inf/0 = OX most certainly is a definition

[glossy junco]

if this symbol represents inf/0 then yes

[finite flax]

well he isn't doing that

[fallen verge]

We dont have inf/0 though

[finite flax]

he's doing a constant divided by 0

[peak bough]

^

[glossy junco]

in this question yes we do

so this is just a symbol

instead of saying inf/0 we sayOX that’s it

[fallen verge]

We dont have inf/0 in the problem tho
We have constant/0

[finite flax]

well it's whatever your professor thinks 78/0 is under this system

also, @glossy junco please include details like this in OP

[glossy junco]

that’s right so maybe the first three answers are wrong?

[finite flax]

it should be on the assignment

[fallen verge]

I dont know if thats true or nkt

[rose sigil]

.-.

this question and discussion is nonsense

[glossy junco]

if there is no inf/0 only constant/0 then it must be the last one

anyway I will send this to my teacher

maybe it’s the right answer who knows

[glossy junco]

lol

[peak bough]

How come, lol

🧍‍♂️🧍‍♂️

[rose sigil]

have you read it?? lol

🧍‍♀️🧍‍♀️

[peak bough]

Yeah

[lone heart]

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[soft tide]

1. Convert the following percentage into the simplest fraction : 36/5%
2. Daniel is a big fan of Spiderman while Jacob is a big fan of Batman. Jacob spends $800 per month on Batman products where Daniel only spends $300. how much more in percentage did Jacob spend more than that of Daniel?
3. Before Christmas, a box of candy cane was marked at $80. Due to low sales, the shop offers a buy 2 get 1 free promotion to customers. What is the discount percentage? How much is it for the discounted box of candy cane?

[soft tide]

help pls

[quaint furnace]

for the first one are you sure thats the question because that is 36/5

[soft tide]

yes

i could send a picture

[quaint furnace]

yes please

for the second one you do (800/300) then * by 100 to get it as a percentage

[soft tide]

thanks

[quaint furnace]

my only though is that it doesnt want it as a top heavy fraction which would mean its

but im not too sure if thats right

[soft tide]

do you think it could be a trick question and just write down 36/5?

[quaint furnace]

who knows does your school seem likely to set trick questions?

[soft tide]

idk, ill just write it like 36/5 = 7 1/5 like you did

what about the 3rd question

[quaint furnace]

i think the discount percentage would be 160/240 as usually it would cost 240 but because of the discount its 160 which simplifies to a discount of 2/3%

[soft tide]

alright

[quaint furnace]

and the discounted price of a box of candy would be $160/3 or $53.33

[soft tide]

okay thank you

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[pure spear]

I am trying to integrate the Maxwell–Boltzmann distribution with respect to x from 0 to infinity. The integral should be exactly 1, but I keep on getting sqrt(-1). The cause of this must be me not having a negative cancel another negative out, but I've tried solving it three times and each time I end up with the same wrong answer.

Here are my steps, i'm sorry if it's difficult to read.

[vague coral]

can you send the integral ?

[pure spear]

https://ibb.co/t4380Mp

[half mural]

jesus

[pure spear]

I show most of my work and try not to skip very many steps, but my writing isn't very good

[half mural]

you should tidy before sending, almost unreadable

[pure spear]

should I close this channel and then reopen a new one once i've attempted it again but with more care?

[vague coral]

$$4\pi \left(\frac{m}{2\pi kT}\right)^{\frac{3}{2}}\int_0^{\infty} v^2 \exp\left(-\frac{mv^2}{2kT}\right) dv$$

this ?

[pure spear]

that is correct

wait

sorry it's integrating with respect to v

[ocean seal]

Ｈｅｒｅｌｓ

[pure spear]

When I solved, it, I did it with polar co-ordinates, and so I end up replacing v with x in my steps

my process on paper probably isn't very helpful, so it might be a good idea to disregard it completely

[vague coral]

so huh, did you try integration by part ?

[pure spear]

I did integration by parts twice later in my process

I guess a spot I was sort of uncertain with was the changing of my double integral bounds from their cartesian bounds to polar

If my x and y integrals are from 0 to infinity, would my theta bound be 0 to pi/2, and my r bound be from 0 to infinity?

[vague coral]

$$u = v$$
$$u' = 1$$
$$w'= v\exp(\frac{-mv^2}{2kT})$$
$$w = -\frac{2kT}{2m}v \exp(\frac{-mv^2}{2kT})$$

wait...

[ocean seal]

Ｈｅｒｅｌｓ

[fair rain]

I suck at math

[vague coral]

$$= \left[-\frac{kT}{m} v^2 \exp\left(-\frac{mv^2}{2kT}\right) \right]_0^{\infty} + \int_0^{\infty} \frac{kT}{m} v \exp\left(-\frac{mv^2}{2kT}\right) dv$$

[ocean seal]

Ｈｅｒｅｌｓ

[vague coral]

and thats something solvable I think

[pure spear]

What method did you use?

And it still looks like there's a guassian integral on the righthand side

I think i'll close this channel and reopen a new one when I eventually attempt it again

Because I want my process analyzed, and i'm doing it a specific way using polar coordinates

Thanks for the help

.close

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[alpine sable]

hi, How can I know if 3 vectors of r3 are linearly dependent with the geometric interpretation

[alpine sable]

i have 3 vectors u= (1,2,-4), w= (3,2,-4) y v = (-1,2,0)

[lean plover]

hi

[alpine sable]

hi

[lean plover]

can someone help about conic section heheheehe

[alpine sable]

this channel is ocuppied, sorry jaksajsk

[lean plover]

ayyyy

[alpine sable]

<@&286206848099549185>

[rose sigil]

see #❓how-to-get-help 🙂

oh wait sorry i just saw you asked somewhere already

[lean plover]

no responds yet from others hahahahah

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

.close

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[limpid maple]

Id like to know the answer please

[alpine nacelle]

note that x²-2xy+y² = (x-y)²

so you sum only zero or positive things to get 0

all these things are equal to 0
so x-y = 0
2y+6 = 0
-2x+y+z = 0

[limpid maple]

thank you so much

.closee

.close

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[astral grove]

I got a different answer to the textbook answers

[astral grove]

can someone tell me what they got to make sure im not going insane

[serene junco]

What did you get?

[astral grove]

I got 20412

nvm i hit the wrong number

im bad sorry

[tough jasper]

19926

[astral grove]

i did * .84 instead of .82

[serene junco]

All good lol

Not crazy

That's right

[astral grove]

.close

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[alpine sable]

i keep getting wackyass numbers with square roots and shit and i feel like im doing somethign wrong

[astral grove]

hey mate

[sour dove]

hi can you show your work?

[astral grove]

i did the same thing ngl

[shut pewter]

Bro I was about I wasl also about to do that

This is correct

[alpine sable]

You can also technically use the quadratic formula.

[astral grove]

this is right

if u use 15

15x16 = 240

15+16 = 31

240-31 = 209

so this is right

[alpine sable]

Must be positive

as per the question

[astral grove]

so 15 is ur answer

[alpine sable]

That would be the answer

yes

[cunning meteor]

hmm yes

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[harsh belfry]

Hello! Just had a quick question on what the 1/3 means here

[harsh belfry]

I've always seen it described as # revolutions per minute

I'm not to sure what the 1/3 means though. Can anyone offer some wisdom here?

[desert tendon]

well 33 revolutions and 1/3rd of a revolution per minute ig or maybe it has some other full form

[harsh belfry]

ahh ok

so 33.3 rpm

a little bit confusing as to why they'd do that

[desert tendon]

but wait for other advice

[astral grove]

prob means 100 rounds per 3 minutes

[desert tendon]

yes maybe

[astral grove]

which is 33 1/3 rounds per minute

[harsh belfry]

will have to look into it a bit more

thanks for the help

[astral grove]

is there any context for this?

[harsh belfry]

Theres a full question for it

[astral grove]

can we see?

[harsh belfry]

The answer is 0.556 rev/s

[astral grove]

ill be honest idk I cant help u

[harsh belfry]

so because 33/60 is 0.55

Im asusming 33 + (1/3)

is 33.3333

that divded by 60

is 0.556

[harsh belfry]

so I reckon was right, I just wanted a bit more context as so why its written this way or if its a rule

[astral grove]

thats 60 not 50 but the rest is correct mate

[harsh belfry]

typo

[desert tendon]

prob

[astral grove]

all good mate

well done

[harsh belfry]

ty again

.close

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[trail slate]

Hello cant solve pythogorean question thingy idk
the opposite is 6
the adjecent is 2
How do i get the hypotenuse
the formula is like square root a squared and b sqaured
I cant get question wont load

[lone heart]

Please don't occupy multiple help channels.

[astral grove]

ok

the pythagoream theorem is

a^2 + b^2 = c^2

where c is the hypotenuse in a right angles triangle

[trail slate]

k isnt there

like

a square root on a2 and b2

[astral grove]

later maybe

[trail slate]

k

[astral grove]

so it would be 6^2 + 2^2

[trail slate]

well since its squared, it would be 36+4=

40

[astral grove]

yep

which means the hypotenuse is √40

[trail slate]

how do i do square root of 40

is it like 20

[astral grove]

nah

just leave it as √40 mate

[trail slate]

oh

isnt that like incomplete

[astral grove]

nah thats fine

[trail slate]

k, let me bring up another onee

so what happens like during a algerba qusiton you get -x in between. like 15 -x = 6

[astral grove]

so add x to both sides

and subtract 6 from both sides

[trail slate]

so 15 = 6+x

?

9=x. oh

oh

k

then that is resloved

had lots of trouble on that one

[sour dove]

You FYI can also simplify $\sqrt{40}$ as $\sqrt{4 \cdot 10} = \sqrt{4} \cdot \sqrt{10} = 2\sqrt{10}$ but I'm not sure if that is within the scope of your question

[ocean seal]

MellowDramaLlama

[trail slate]

what

[sour dove]

you asked if you could further simplify sqrt(40)

I was showing you how

if you wanted ot

*to

[trail slate]

sure..?

i mean i dont really know how to solve those 40 square root thing

[trail slate]

kk

[sour dove]

sorry what do you mean by "solve" 40 square root thing? Do you mean find out what the value of sqrt(40) is?

[trail slate]

yea

i mean i dont really know how but it would be intresting to know how you solve a square root

[karmic solstice]

wdym

[sour dove]

ah so yeah. Knowing your perfect squares helps a lot.
So for example $\sqrt{1} = 1, \sqrt{4} = 2, \sqrt{9} = 3, \sqrt{16} = 4, \sqrt{25} = 5, \sqrt{36} = 6, \sqrt{49} = 7, \sqrt{64} = 8, \sqrt{81} = 9, \sqrt{100} = 10$. Then you can get a rough estimate from there

[karmic solstice]

solve a square root?

[ocean seal]

MellowDramaLlama

[karmic solstice]

ah

[trail slate]

yeah yeah i know those but how do you solve a 40

sqaure rot

[sour dove]

so for example $\sqrt{40} = 2\sqrt{10}$. Notice that $\sqrt{10}$ is slightly bigger than $\sqrt{9} = 3$, so our value for $\sqrt{40} = 2\sqrt{10}$ and $2\sqrt{10} > 2\sqrt{9}$. We know that $\sqrt{9} = 3$, so $2\sqrt{10} > 2(3)$, so $2\sqrt{10} > 6$.

[ocean seal]

MellowDramaLlama

[sour dove]

so $\sqrt{40}$ is 6 point something

in fact, $\sqrt{40} = 6.32455532034$

[ocean seal]

MellowDramaLlama

[trail slate]

thanks.

hm.

.close

[lone heart]

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[urban fjord]

q18

[desert tusk]

Try to draw out what the parallelogram looks like

[urban fjord]

yeah I had a go at that

I just dont know which way I should work from cause there’s two possible options

and idk how to specifically use vectors

[vale wigeon]

$\overrightarrow{WX} = \overrightarrow{ZY}$

[ocean seal]

Ann

[vale wigeon]

or i guess $\overrightarrow{XW} = \overrightarrow{YZ}$ might be somewhat less hairy to deal with

[ocean seal]

Ann

[vale wigeon]

not a big difference tho

[urban fjord]

how do you use that to find exact vertices

[vale wigeon]

XW = OW - OX

and YZ = OZ - OY

you know the position vectors of W, X and Y, and that of Z is what you're asked to find

[lone heart]

@urban fjord Has your question been resolved?

[lone heart]

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[fast tiger]

Can I have some help with this one?

[vale wigeon]

@fast tiger have you made any progress thus far or are you stuck not knowing how to begin

[fast tiger]

I know what factor form looks like, but I dont know where to start

[vale wigeon]

right

do you know the relationship between factors and roots?

[fast tiger]

No

Well, I know that the X-int are -1 and 3, while the y-int is -6.

That's about it

[sour dove]

okay so there's a few things to work with here.

so when we have a value that crosses from the positive to the negative y values, the degree will be 1. If we have a parabola at a point, such as this:

then we know our degree will be 2 there

so can you give me the roots and what degree those roots will be?

[fast tiger]

(x+2), (x+1), and (x-3)^2?

[sour dove]

close!

(x + 1) is also a parabola

[fast tiger]

Oh!

Then its (x+2), (x+1)^2, and (x-3)^2?

[sour dove]

yep perfect!

now

next thing

look at what happens as x tends towards infinity

is it positive or negative?

[fast tiger]

negative

[sour dove]

bingo

so we have a - in the equation

so then, so far, we have $y = -b(x + 2)(x + 1)^2(x-3)^2$

[ocean seal]

MellowDramaLlama

[sour dove]

now

to solve for that b

we know that our y-intercept occurs at (0, -6)

so how can we solve for b?

[fast tiger]

@sour dove Input the coordinate from the y-int into the problem?

[sour dove]

bingo!

aka plug in x = 0 and y = -6

then solve for b 🙂

[fast tiger]

b=-1/3 ?

[sour dove]

close!

b = 1/3

well

yeah i guess all together it's -1/3rd lol

but technically

$-6 = -b(2)(1)^2(-3)^2\$
$-6 = -18b\$
$\frac{-6}{-18} = b\$
$\frac{1}{3} = b$

[ocean seal]

MellowDramaLlama

[sour dove]

but there you go!!

look familiar? 😄

[fast tiger]

Yep!

Do I have to enter the extra parenthesis in there too?

[sour dove]

nope

that was an accident

the equation is $f(x) = -\frac{1}{3}(x+2)(x+1)^2(x-3)^2$

[ocean seal]

MellowDramaLlama

[fast tiger]

I got it!

I just need help with 2 more problems!

Note: Ik that the end behavior and leading coefficient are correct, I just need help with the degree of f(x) and leading term of the polynomial

[sour dove]

yep!

okay so yeah

so the leading term is the same thing as the degree yeah?

so we have an $x^{15}$, an $(x-10)^2$, and an $(x - 8)$. So we know we have an $x^{15}$, and $(x - 8)$ multiplying give us a $x^{16}$. Finally, we know that $(x - 10)^2$ will produce an $x^2$, so then we have $x^{18}$, so the degree is $18$

[ocean seal]

MellowDramaLlama

[sour dove]

then the leading term is the degree and the leading coefficent (I think)

so it would be $-\frac{4}{9}x^{18}$

[ocean seal]

MellowDramaLlama

[fast tiger]

That was right!

One more to go!

[sour dove]

so I haven't heard the term power function in a bit

but I think it's just your leading power and coefficient

so what's your leading coefficient and power?

[fast tiger]

Leading power=9 and coefficient is x?

[sour dove]

so the leading coefficent is 1

and the highest power is 6

so our power function (I think)

is $1x^6$

[ocean seal]

MellowDramaLlama

[sour dove]

Then we know that powers of even functions are "parabolic". We don't have a negative in front of the leading power

so it looks something like this:

[fast tiger]

So the bottom two answers are just infinity and negative infinity?

[sour dove]

nah

as x heads towards -infinity (aka to the left)

the values grow

postiveily

same with the right side

infinity and negative infinity would be something like this

[fast tiger]

Oh okay

I'm trying to figure it out, but its confusing me

[sour dove]

sorry tbh I'm starting to fall asleep I think I need to head to bed. You can ping the @Helpers or close and re-open the thread to push back up the help stack.

best of luck

[fast tiger]

Thank you for helping me!

.close

[lone heart]

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Channel closed

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[peak raft]

Hello.
I Have A Homework That I Need To Do And I Cant Understand It The Questions Are Circled 35 Can Someone Help Me Answer It.
Thank You.

[cold jasper]

Ermn can you rotate the picture before sending? 🙂

[peak raft]

Just For U

:)

[uncut summit]

wait am I in the right channel

[cold jasper]

You are not

[uncut summit]

sorry((

[peak raft]

is This The Help Channel I Cant Understand Any Thing

[cold jasper]

This is

[peak raft]

Ok..............

[cold jasper]

I can give you the answer or I can help you understand what you need to do

[lone heart]

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[finite gate]

How do I start solving this problem?

[scenic sierra]

first multiply the integers so 5x10

then you multiply the a terms together

[finite gate]

oh

Yeah

just the 5 and 10 confused me

50a^8b^4

[scenic sierra]

and they have exponents so when u multiply two things with the same base you add the exponents

[finite gate]

ty

.close

[lone heart]

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[peak raft]

answer it...

[cold jasper]

Oh I thought you closed the channel

You don't want it explained?

[peak raft]

why not?

send it

[cold jasper]

Ermn okay then, the answer is x=4 V x=9

[peak raft]

how did you answer it?

i can find the answer in photo math

the numbers i want to know how did you do it

[lone heart]

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[lime vapor]

had this question on a test today not sure what do i do

[vale wigeon]

P(4) itself looks not terribly hard to find

[lime vapor]

.close

[lone heart]

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[peak raft]

<@&286206848099549185>

[peak raft]

Can You HElp ME

[lone heart]

@peak raft Has your question been resolved?

[lone heart]

@peak raft Has your question been resolved?

[lone heart]

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[pastel bane]

For this, could I just place the condition of B being a subset of T? Would that be an appropriate response?

[rose sigil]

yea, but you could say something stronger too

[pastel bane]

what could I say that is stronger other than B=T?

[rose sigil]

B is a subset of the image of f

[pastel bane]

ohhh duh

that makes so much sense

thank you!

.close

[lone heart]

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[rose sigil]

wait actually

.reopen

[lone heart]

✅

[rose sigil]

sorry i misread

or more like i just messed up

@pastel bane i wanted to say the image of f is a subset of B

[rose sigil]

and this wasn't right

[pastel bane]

oh thank you! i meant to say that in my first message as well, but I appreciate it!

.close

[lone heart]

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[rose sigil]

yea sorry 😭

[lone heart]

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@alpine sable Has your question been resolved?

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[sand flame]

evaluate log₄2

[sand flame]

im so confused im so sorry

[golden canyon]

this basically asks the question: 4 to the power of what number gives 2?

4^x = 2

[lone heart]

@sand flame Has your question been resolved?

[sand flame]

...?

[sand flame]

its not the answer tho

[vale junco]

wdym

solve for x and you get the answer

[sand flame]

there is no x in the question

lmao

just says

[vale junco]

but we made an equation

[vale junco]

here

and we claim it's solution is what youre looking for

[sand flame]

this topic makes me wanna cry my eyes out

[vale junco]

why

look

log_4(2)

meaning "four to the power of what equal 2?"

[sand flame]

wait nvm i checked fot eh wrong answer

it is 2

what about

log_6(4) - 2log_6(12)

[vale junco]

2 is not the right answer.

[sand flame]

someone here said it is 2

[lone heart]

@sand flame Has your question been resolved?

[quaint furnace]

the term log_4(2) just means what number do you have to raise 4 to the power to to make 2

log_3(9) would mean instead what number you have to raise 3 to the power to to make 9

[sand flame]

oh ok

[lone heart]

@sand flame Has your question been resolved?

[lone heart]

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[idle path]

trigonometry

[idle path]

lemme post it

simplify

i wrote cotg as 1/tg if nessecery

necessery

oof

[brittle ember]

so t is tan?

[idle path]

tg

[brittle ember]

so $(tg\alpha+\frac{1}{tg\alpha})^2-(tg\alpha-\frac{1}{tg\alpha})^2$

[ocean seal]

Dri111

[brittle ember]

?

[idle path]

yes this is how i wrote it, bc the original is

[neon hedge]

idk the answer am just too dumb

[craggy token]

i'd suggest to write tan = sin/cos and cot = cos/sin to see where you can go from that and if you can simplify anything.

[idle path]

yes i agree

[brittle ember]

you can try using (A+B)(A-B)=A^2-B^2

[craggy token]

because an interesting fact is that tanx + cotx = 1/sinxcosx = 2/sin2x

[idle path]

what ?

fr

[craggy token]

hold on ykw

(a+b)^2 - (a-b)^2

[brittle ember]

is 4ab

[craggy token]

why don't you try to expand the question in this form.

[craggy token]

yeah

[idle path]

can u help me do it

[brittle ember]

can you see the similarity with $(tg\alpha+\frac{1}{tg\alpha})^2-(tg\alpha-\frac{1}{tg\alpha})^2$

and (a+b)^2-(a-b)^2

[ocean seal]

Dri111

[idle path]

simplify u mean

square them up ?

[brittle ember]

so you can think a as $tg\alpha$ and b as $\frac{1}{tg\alpha}$

[ocean seal]

Dri111

[idle path]

nvm ill do it later, help me with this other exercise.. 0≤ x≤ 360

2sinx = 1

[brittle ember]

then sinx=?

[idle path]

1/2 ?

[brittle ember]

yes and do you know any common angles that sine value is 1/2?

[idle path]

0.42 ?
i googled it

[brittle ember]

i dont know if word is "common angle"

[idle path]

im confused

[craggy token]

i use special angles

[idle path]

tell me

[craggy token]

just like what Dri111 said, sinx = 1/2, so find the values of x in between 0 and 360 such that sinx = 1/2

i think that is quite trivial if you know your special angles value

[idle path]

lemme tell u how sm did it

sinx = 1/2

x1= 300

x2= 180-30=150

i dont get this

[brittle ember]

actually

I think x1=300 is wrong

[craggy token]

this one is wrong...

[idle path]

how do u know if its wrong or right

[craggy token]

300 is in the fourth quadrant making the value of sine to be negative

[idle path]

how do u tell

[craggy token]

instead, we want sine to be positive

[craggy token]

what do you know about trigonometry?

[idle path]

quadrants and the basic formulas

[craggy token]

cool

you know quadrants

[idle path]

how do i know it must be positive

[craggy token]

so you must know why x= 300 is wrong

[craggy token]

sinx = 1/2,

do you see any negative there?

[idle path]

no

[craggy token]

well that clears it up then

sinx must be positive

in which quadrants will give us positive sine values

[idle path]

1st and 2nd

[craggy token]

cool

in the first quadrant, what value of x will give us sin x = 1/2?

[idle path]

45 ?

[idle path]

.

[brittle ember]

try again

[idle path]

how

[brittle ember]

you memorized special angle table?

[idle path]

i just assumed first quad is 90 nd 1/2 of 90 is 45

[idle path]

dont need it

[craggy token]

that's not how it works...

[craggy token]

cool

[idle path]

then explain it to me nd ill try

bc idk how to

[craggy token]

solving trig questions without memorising special angles

very cool

[craggy token]

okay first of all memorise the special angle values

[idle path]

tbh , the teacher didnt tell us to do it, idk how she maybe uses an equation

[craggy token]

just because your teacher doesn't tell you to do it doesn't mean you shouldn't do it.

[idle path]

can we get back to it

so,, how do i solve it

[craggy token]

as I said before, look at the sine function values for special angles

then see for which x will sinx = 1/2 in the first quadrant.

this ought to help you

[idle path]

30 degrees ?

[craggy token]

correct

so for x = 30, sinx = 1/2

that's for the first quadrant

[idle path]

yea

[craggy token]

now for the second quadrant, which value(s) of x will sin x = 1/2

[idle path]

idk

idk how to check

[brittle ember]

did you learn that sin(180-x)=sinx?

[craggy token]

i thought you learned about

the basic formulas

[idle path]

nope, doesnt say anything on my book

@craggy token can u help me

lol

[craggy token]

sin(180-x)=sinx

what happens if i put x = 30, a solution we already know?

[idle path]

so 180 - 30

[craggy token]

yes

[idle path]

oh

sin^2x - 2sinx + 1 = 0

@craggy token what abt this

[craggy token]

notice something familiar with that equation?

[idle path]

yes

sin^2x = 1 - cos^2x

[craggy token]

that was not what i meant

[idle path]

lol

[craggy token]

hint: quadratic equation

[idle path]

then no

[craggy token]

what if i rewrite the equation with the variable x instead of sinx?

what would it be?

[idle path]

x^2 - 2x +1 ?

[craggy token]

yes

and you can factor that and find the values of x, then substitute sinx in again

wait change the variable (using x will be kind of confusing)

let's use a

so a^2 - 2a + 1 = 0

factor that and solve for a

keeping in mind that a= sinx

[idle path]

im stuck now, idk

[craggy token]

have you actually attempted from what i've just typed?

[idle path]

to factor it ?

[craggy token]

yes

[idle path]

ye but idk how to factor it.. a(a-2+1) its wrong

[craggy token]

...

you don't know how to factor quadratic expressions?

[craggy token]

this would result in a^2 - 2a + a instead, whereas we want a^2 - 2a + 1

[idle path]

i know

(a-1)^2=0

[craggy token]

yes

and set that to be equal to 0

then solve for a

[idle path]

?

[craggy token]

solve for a

find a

a = ...

[idle path]

1

[craggy token]

no

(a-1)^2 = 0

yues

[craggy token]

yes

but remember that a = sinx

so, we have sinx = 1

[idle path]

ye

[craggy token]

what are the values of x that satisfy the equation then?

[idle path]

bro what is satisfy in our case, english isnt my main language so i get confused

[craggy token]

what are the values of x that would make the equation sinx = 1 true?

[idle path]

can i get better at this on khan academy ?

[craggy token]

whatever helps you i guess

[idle path]

what do i search

can u give me the answer of it

so i know at least

@craggy token

hello

[craggy token]

what are the values of x that would make the equation sinx = 1 true?

[lone heart]

@idle path Has your question been resolved?

[idle path]

can u just drop the answer

[craggy token]

we don't simply 'drop' answers here.

understand the concept, then you'd be able to solve the question.

if you keep asking people to 'just drop the answer', you'd never understand the concept and would keep asking questions to ask the why's and the how's.

[idle path]

can we go back to my first question btw

[craggy token]

Dri111 earlier has tried to explain

expand, then simplify

[idle path]

ill watch other vids for that

expand in what form

like (sin/cos + 1/tg)² - (sin/cos -1/tg)²

[craggy token]

no

[craggy token]

just expand this

[idle path]

oh with the square

[craggy token]

yes

[idle path]

ok ill do in in a few . bc im doing smth else

do u mind helping me?

are u busy or smth i mean

[craggy token]

as long as you're willing to learn the concept and not simply asking me to 'drop' the answers, then yes.

[idle path]

yep thanks

[lone heart]

@idle path Has your question been resolved?

[idle path]

so

now i eliminate ?

[craggy token]

ye

and note that tanxcotx = 1

[idle path]

why

is it a formula

so i got 4tgcotg

[craggy token]

not a formula

logical thinking

cotx = 1/tanx

tanx*1/tanx = 1

[idle path]

huh

tg times 1 / tg = 1 u mean

[craggy token]

yes.

[idle path]

.

equal to 4

?

[craggy token]

yes

[idle path]

4tg × 1/tg

tg goes away

so its 4×1=4

[craggy token]

yes

[idle path]

btw im confused here

another exercise

cos 132

its cos(180-42)

132 is in 2nd quadrant so now cos(-42) = 132

?

but whether i put a minus in 42 or not its still positive

nd i need it negative

or not

[craggy token]

that's the exact question? to find the value of cos(132)?

[idle path]

no i mean 48 not 42 sry

ye but we know a method

so cos-48 =cos 132

[craggy token]

not quite.

cos132 = -cos48

watch your placement of the negative sign

[idle path]

yeaa ik lmao

sorry

but i meant that my confusion stands on the calc

whether i put a minus or not before cos 48 its still positive

[craggy token]

cos-48 = cos48
but -cos48 =/= cos48.

cos-48 meaning that you're measuring the cosine of angle 48 degrees measured clockwise

[idle path]

so it doesnt work on the calc ur saying

[craggy token]

-cos48 meaning you're measuring the cosine of angle 48 degrees measured counter-clockwise, then multiply it with -1

[craggy token]

wdym

well to solve this, i'd need to know what trig identities are you already aware of?

[craggy token]

of course cos-48 = cos48

because cosine is an even function and therefore, cos(-x) = cosx

[craggy token]

what do you not understand here, exactly?

[idle path]

its a minus bc its 2nd quadrant right

[craggy token]

yes

[idle path]

ok

so ill check the quadrant first then add a minus if its necessary

[craggy token]

yes.

[idle path]

ok.

is it ok that im asking different exercises on same help channel ?

[craggy token]

i suppose yes

[idle path]

or should i open another one?

[craggy token]

not necessary i think

[idle path]

its ok he said

the mod

can u give me 2 diff cos and sin like previous

so i check quadrant and tell u

[craggy token]

can u give me 2 diff cos and sin like previous
might want to clarify what you meant on this one

[idle path]

like this

previous one

[craggy token]

still don't quite understand

what do you mean by 2 diff cos and sin

[idle path]

2 diff exercises

lol

[craggy token]

ah

so like cos330?

and sin240?

[idle path]

cos(360-30) = cos 330 cos330= cos 30

its +

[craggy token]

yes

[idle path]

sin(360-80)=sin240 sin240= -sin80

3rd quadrant

@craggy token

[craggy token]

sin(360-80) = sin 280

not sin240

[idle path]

oh f

120

sry

-sin120

[craggy token]

yes

or you can rewrite it as sin(270-30)=-cos(30)

[idle path]

i prefer using 180 and 360

[craggy token]

what's easier for you

as long as the final answer is correct

[idle path]

can u give me a problem with a triangle using sinus or cosinus theorem

[craggy token]

cosine law
(yes it's from the internet bcs i cbf to make a whole new problem)

[idle path]

yea i shudve go to google lmao

[craggy token]

simple sine law

[idle path]

btw@craggy token if theres an exercise and doesnt tell which one to use sin or cos

do i use 1 of them and the answer is still correct

[craggy token]

the laws?

[idle path]

yep

[craggy token]

no.

say you have a triangle ABC with AB = c, AC = b, and BC = a.

if you have either two of the lengths and the measure of one of the angles that was formed by the two given lines, then use the cosine law.

for example, if you're given the value of a and b as well as angle ACB

then you can use cosine law

but if you have the measure of two angles as well as one of the lengths, then use sine law.

[idle path]

oh ok

i see

so 2 angles 1 lenght sin

2lengths 1 angle cos

[craggy token]

yes..

[craggy token]

idk about the calculation but the numbers you input look right

[idle path]

yea thats what matters in our case

2nd problem

i think i did it wrong

@craggy token

[craggy token]

why is b = 20?

in fact from where did you get b = 20?

[idle path]

idk lmao

which should i find

[craggy token]

find angle B first

you know the sum of all angles in a triangle is 180

[idle path]

idk smh i found it

[craggy token]

then use sine law to find b

[craggy token]

that does not explain how you got b = 20.
it was not stated in the question and therefore you shouldn't add something that the question did not state explicitly or indirectly.

[idle path]

are u good at derivatives by any chance

[craggy token]

quite

but it's night-time in my country right now and i would like to sleep

good night.

[idle path]

gn and thanks for ur help. can we work tomorrow again?

[idle path]

?

[craggy token]

depends.

[lone heart]

@idle path Has your question been resolved?

[lone heart]

@idle path Has your question been resolved?

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[tacit island]

How do you solve this problem?

[tacit arch]

this means parallel

[tacit island]

but how do i write r''(t)

[tacit arch]

how do you describe lines that are parallel to a vector function?

[tacit island]

@tacit arch the cross product is 0?

[tacit arch]

think more easily

maybe think about it in 2d first

[tacit island]

they just dont intersect

[tacit arch]

what do their equations have in common?

[tacit island]

oh the slope is the same

[lone heart]

@tacit island Has your question been resolved?

[tacit island]

what do i do with that information @tacit arch

[tacit arch]

Figure this out

E.g. y=x and y=x+1 are parallel

[lone heart]

@tacit island Has your question been resolved?

[tacit island]

yeah the slope is the same

[tacit arch]

Pretend k is 1 now

[tacit island]

so I just have to set k to 1, make r'(t) and then i can do everything else?

[lone heart]

@tacit island Has your question been resolved?

[lone heart]

@tacit island Has your question been resolved?

[lone heart]

@tacit island Has your question been resolved?

[olive elk]

how do I divide?

[lone heart]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[loud tundra]

Guys i Need help with n24

[loud tundra]

I don't really have any idea

[old hazel]

@loud tundra i got u

[loud tundra]

ok thanks

[old hazel]

okay so

what u do is

treat 2^x

as x

@loud tundra u get a quadratic

[worldly basin]

@loud tundra What do u even want to find out

[old hazel]

4^x = 2^2x

the range where the function exists

[loud tundra]

Yes

[old hazel]

because if u do 2^x = x

u get

1/(2x^2 -5x +2)