help-26

[north timber]

Transpose the matrices to obtain:

$$\left(\begin{bmatrix}a&b\c&d\end{bmatrix}\begin{bmatrix}1&1\2&0\end{bmatrix}\right)^\top=\begin{bmatrix}3&5\-1&4\end{bmatrix}^\top$$

$$\begin{bmatrix}1&2\1&0\end{bmatrix}\begin{bmatrix}a&c\b&d\end{bmatrix}=\begin{bmatrix}3&-1\5&4\end{bmatrix}$$

[thorny flame]

luke1337

[north timber]

You turn it into an augmented matrix:
[\left[\begin{array}{cc|cc}1&2&3&-1\1&0&5&4\end{array}\right]]

[thorny flame]

luke1337

[north timber]

Then apply guassian elimination to obtain:

[\left[\begin{array}{cc|cc}1&0&5&4\0&1&-1&-\frac52\end{array}\right]]

[thorny flame]

luke1337

[north timber]

We now have $$\begin{bmatrix}a&c\b&d\end{bmatrix}=\begin{bmatrix}5&4\-1&-\frac52\end{bmatrix}$$

[thorny flame]

luke1337

[north timber]

Transpose it:
$$\begin{bmatrix}a&b\c&d\end{bmatrix}=\begin{bmatrix}5&-1\4&-\frac52\end{bmatrix}$$

[thorny flame]

luke1337

[north timber]

Apply it:
$$T\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}5&-1\4&-\frac52\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}5x-y\4x-\frac52y\end{bmatrix}$$

[thorny flame]

luke1337

[north timber]

@brisk geyser ^^

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[brisk geyser]

sorry was afk

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[neon iron]

Can you start a graph at a different number which differs from the unit of the graph after the origin..?

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[neon iron]

No

No

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[neon iron]

No

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[karmic haven]

any hints? it should solve with Beta

[karmic haven]

i split in 2 the second one i B(2,1) but i don't know to solve first one

7 * Integral (1-x)^1/2

i got it

.close

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[karmic haven]

.reopen

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✅

[karmic haven]

my answer was 9 but it was wrong

[neon iron]

its not 9

do usub

[karmic haven]

i split in 2i

7/sqrt(1-x)

which is 7 * integral (1-x)^-1/2

which is B(1,1)

[ruby mural]

Solve with gamma? Wdym?

[karmic haven]

it's a way to solve integrals faster

and it's beta sry i was thinking of another one

like B(x,1) = 1/x

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[compact wolf]

having trouble with this, i think it’s a little easy but i’m so confused

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[whole otter]

Interpret the diagram like this
, you can observe that
BCDF looks like a parallelogram

Sum of two adjacent angle of parallelogram is 180°
So c°+d°=180°

[compact wolf]

ohh okay

[whole otter]

Now

b°=c° due to being alternate angle

Hence b°+d°=180°

[compact wolf]

OHHH

OML TYSM 🫶🏾🫶🏾

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[leaden ibex]

Hi, I am new here and i need some help with some probabilities.

[leaden ibex]

Jean finds an old watch in a drawer. It is 4 cm in diameter. It has a seconds hand measuring 2 cm which rotates at constant speed and a minutes hand of the same size which no longer works and remains stationary opposite the number 3. When Jean looks at the watch, let X be the distance between the two ends of the needles. a) Find F(x)=P(X<=x) the distribution function of X.

I know that x= 4 SIn (a/2): a= to the angle between 0 and 2pi

but for what following steps, I have no clues

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[unkempt grove]

It's really dumb and unless really into nerd stuff it's going to sound pointless.

So I'm trying to figure out how to calculate the rate of... well drop off. Not sure what you call it.

Basically you get good returns at the start but more energy you pump in less you get back. no idea how you calculate that.

So there is a character names Asura - video game dude - who can come back from the dead. Stronger he is faster he can come back from the dead.

When he had a max output of about 3.18e24 Joules, he was able to come back to life after 10,000 years.

Then when had an output of 2.487e32 Joules took 500 years.

Lastly when he had an output of 4.518e35 Joules took 1 week.
So I'm trying to figure out the rate of decay or whatever between energy and time here. Let's say X being energy and Y being time. So i can plug in X energy to figure out the time.

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@unkempt grove Has your question been resolved?

[unkempt grove]

no

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[unkempt grove]

nope

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[lyric crest]

Idk what is said here but keep in mind that speed is the derivative of your function and accelartion is the double derivative

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[forest bloom]

maybe dumb question, but how do i solve this system for r?

[vernal matrix]

You could e.g. expand (h + r)^2

[forest bloom]

that makes it messy

h^2 + r^2 + 2rh + 1 = 9

h^2 = r^2 - 1

r^2 - 1 + r^2 + 1 + 2rh = 9

2r^2 + 2rh = 9

[vernal matrix]

Oh yeah that does look painful a bit

[forest bloom]

yeah not sure how they got something for r "cleanly" lol

[vernal matrix]

Wait do you have any restrictions on what h and r are? Are they e.g. lengths?

[forest bloom]

yeah

[vernal matrix]

In that case, they have to be positive, so that second one gives you that $h + r = \sqrt{8} = 2\sqrt{2}$

[thorny flame]

@vernal matrix

[vernal matrix]

Rearranging for $h = 2\sqrt{2} - r$ gets you that $\cancel{r^2} - 4r\sqrt{2} + 8 + 1 = \cancel{r^2}$

[thorny flame]

@vernal matrix

[forest bloom]

hmm i think i see

thanks~~

[vernal matrix]

No problem

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[green gulch]

I’m stuck

[strange whale]

yeah

so what does that mean that g1 is in g2 K ?

what's g2 K even ?

@green gulch

[green gulch]

it is uh

idk 😭

[flat kindle]

Do you know what Ker (phi) is

[green gulch]

kind of

i wrote it there

[flat kindle]

and are you familiar with subgroups

[green gulch]

yes

[vale furnace]

(g_1 \in g_2K) means (g_1) is in the same coset as (g_2) with respect to kernel (K). In other words, there exists some elelemt (k\in K) such that (g_1 = g_2k).

[thorny flame]

Pure

[green gulch]

How do i find this element

[flat kindle]

you don't need to

[vale furnace]

Not sure what you mean by finding but start by proving the forward direction; as you noted, ( \phi (k) = e_H ) then consider the expression
[
\phi(g_1) = \phi(g_2k)
]

[thorny flame]

Pure

[green gulch]

Okay

Then i can use the property

Phig2k = phig2phik?

And then

Phik is identity g2

So

Phig1 = phig2

Thanks

What about other direction

[vale furnace]

well start from (\phi(g_1) = \phi(g_2))

[thorny flame]

Pure

[vale furnace]

right multiply by something

[flat kindle]

do the same thing in reverse order

[vale furnace]

yeah you can do that

or something like

[green gulch]

Wdym by something

[vale furnace]

(\phi(g_1) = \phi(g_2) \rightarrow \phi(g_1)\phi(g_2)^{-1} = e_H)

[thorny flame]

Pure

[green gulch]

oh okay

thank you

wait why is phig2 the identity?

i mean

that thing

why is it the identity

[strange whale]

he just multiplied by phi(g2)^-1 on both sides

the left side is phi(g2) * phi(g2)^-1

[green gulch]

oh right yeah

my bad

thanks

im unsure of how that results in g2 in g2K

[strange whale]

so yeah what pure is going for is giving explicitly a k for which g1 = g2 * k with k in ker phi

[strange whale]

you can rework that right hand side a little bit

phi(g1)phi(g2)^-1

[green gulch]

so what phi(g1g2^-1)

[strange whale]

yeah

so g1 * g2^-1 is in ker phi

[green gulch]

ohh

okay i see

so g1 is in g2kerphi

[strange whale]

well technically this doesn't help us here

[green gulch]

i mean

g2k

[strange whale]

g2 * (g1 * g2^-1) is not necessarily g1

[green gulch]

bruh

[strange whale]

but what we have is almost correct

[strange whale]

just multiply on the left by phi(g2)^-1 instead of on the right

this gets us phi(g2^-1 * g1) = e_K

so g2^-1 * g1 is in ker phi

and then indeed g2 * (g2^-1 * g1) = g1

@green gulch

[green gulch]

okay i understand

thank you

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[forest bloom]

,, \frac{\frac{S+1}{S^2} + S + S^2}{4} \text{ ? } 2S

[thorny flame]

kanna

[forest bloom]

,, \frac{S+1}{S^2} + S + S^2\text{ ? } 8s

[thorny flame]

kanna

[forest bloom]

Then after some algebra i arriave at:

,, \frac{s+1}{s^2} \text{ ? } s(7-s)

[thorny flame]

kanna

[forest bloom]

at s = 7 we have that Quantity A > Quantity B and otherwise (maybe when s = 4) then Quantity B > Quantity A so the relationship can't be determined

Is there a "faster" or optimal way to do this question, or is what i did good enough?

[ruby tree]

Looks alright

Maybe a quicker way is to try S = 3 and notice that when S is very large, the S^2 term dominates

So when S = 3 you get A < B but when S = 10000 you get A > B

[forest bloom]

ohhh

why did you take S = 3 btw?

[ruby tree]

Because it has to be greater than 2

[forest bloom]

hmm right but what if it never helped

for instance, if the conclusion was still A < B for S = 3

[neon iron]

always try

[ruby tree]

Well then you'd try something else, like what you did

[neon iron]

for atleast 3 terms

or if you can, make a general term

specifically in SnS

[forest bloom]

I guess that brings up the question on how you know quantity A (whatever that is) grows faster than linear express in the long term

Hmm okay fair enough

[ruby tree]

All the terms in A are positive

x^2 grows faster than 2x

[forest bloom]

oh wait yeah

1/x^2 just contributes to the x^2

for some reason i thought some counteracting stuff was happening

okay thanks makes sense

[ruby tree]

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[rotund tusk]

i get that dy/dt = (dx/dt)(dy/dx)

[hoary kayak]

what is t here?

there must be extra information or else while putting the values we cant remove the t from the differential equation

[modern imp]

this is not a differential equaiton

you apply implicit differentiation on x^2+y^2=45

[rotund tusk]

.close

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[violet sand]

Is this 1?

[ruby mural]

!status

[topaz sinew]

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

@violet sand Has your question been resolved?

[carmine pelican]

yes

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[fluid bronze]

im on #29

thats what im supposed to get but instead i got

[neon iron]

the x should be 1

the derivative of x+3 is 1

[fluid bronze]

THANK YOU 🥰

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[daring oak]

in the solution they got for the error |x - 1| ^5 / 5

can't you replace x with 2?

and make the final answer 1/5?

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[daring oak]

<@&286206848099549185>

[neon iron]

I think not

Because the remainder is a function of x

But I don't know about the remainder of taylor series

So it's better you see with other person about it

[daring oak]

okay thank you

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[random meadow]

salam alaykum

[random meadow]

A block of mass 15kg slides down a line of greatest slope of inclined plane. The top of the plane is at a vertical height of 1.6m above the level of the bottom of the plane. The speed of the block at the top of the plane is 2ms^-1 and the spped of the block at the bottom of the plane is 4ms^-1

find the work done against the resistance to motion of the block

[analog valve]

who can solve this to me please

[rapid marten]

Draw a diagram

if you have lets see

[random meadow]

nah bro idk any of this stuff man

[rapid marten]

you dont know how to draw?

[random meadow]

that and the stuff in the question

[rapid marten]

sounds like laziness to me

[random meadow]

im tryna make money bro

this is some other dudes homework who is a year ahead of me

he said he will give me money to do it. so i took it

[rapid marten]

aint

no way

LOL

Tell him to get his fatass to the library

[random meadow]

aight

,close

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[old stream]

Hello, i need real solutions for X in this situation

(X²-X)⁴ = 4x⁴ -8X³ -4X² +192

We haven't learned polynomials or the quadratic formula, so please try to do this with basic 9th/10th grade math

[clear lodge]

you would start by expanding the left hand side. Can you do that?

[reef fjord]

(you can use pascal's triangle to help with that)

[clear lodge]

also, how tf are you being asked for the real solutions there without having learned polynomials or quadratic formula?

also also, can you use any tools, like a graphic calculator?

[old stream]

No, no graphics calculator. A few of my classmates have done it so it's definitely possible

[clear lodge]

ofc it's possible, but i want to know what you have avaliable

[old stream]

X⁸ -4X⁷ + 6x⁶ - 4X⁵ +X⁴ = 4X⁴ -8X³ -4X² +192

What next?

[clear lodge]

you would simplify that to a polynomial P(x)=0

and honestly, with the tools you have, i would say trial and error. Because this is way past 10th grade to do rigorously

[old stream]

Okey dokey, thank you

[clear lodge]

but at that point i would expect them to let you use a graphic calculator, because trial and error by hand with an 8th degree polynomial sucks

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[cosmic furnace]

Hello

[cosmic furnace]

i assume the radius here is a^2 because it's a cylinder?

x^2 + z^2 = r^2 is usually what I would do

not sure whether to replace r^2 with a^2 in all cases or only where x^2 + z^2 is present

it would be all cases, but then I can't integrate over 0 to r... can I?

🤔

or 0 to a

it's a cylinder with a height from -2 to 2, theta is from 0 to 2pi of course, but r is .. a?

how the heck do do latex

._. ?

assuming here x = rcosθ, z = rsinθ, x^2 + z^2 = r^2

i may need to parametrize instead but I think my solution may work but I don't have an answer key

answer would be a^3*pi*2/3 integrated over y from -2 to 2 = a^3*pi*8/3

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[cosmic furnace]

<@&286206848099549185> y'all know surface integrals of cylinders <3

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@cosmic furnace Has your question been resolved?

[cosmic furnace]

<@&286206848099549185>

[cosmic furnace]

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[cursive urchin]

Hi! I have a question on function extensions. It is alg 1.
I am currently working on this set,
g(x) = x^2 - 3x
h(x) = 4x + 4
And I am finding (g (of) h) (x^2).

[cursive urchin]

I am slightly confused, as it looks like I have to "FOIL", but I am not sure how.

[reef fjord]

it's composition, not foiling.

$(f \circ g )(x)=f(g(x))$

[thorny flame]

Moosey

[cursive urchin]

How would I do composition?

[reef fjord]

wherever you see 'x' in g(x), you plug in h(x)

[cursive urchin]

Don't I plug in (x^2)

[reef fjord]

yes, you plug in x^2 last in for x

[cursive urchin]

I plug it into any "x" I see, correct?

(sorry for late replies)

Ohh I forgot the circle didn't mean multiply

If I have 4(x^2), is it 4x^2?

I don't distribute the exponent because it is isolated with the x only

Wait let me try again this doesn't look right lol

The 16x^4 +8 looks correct

[reef fjord]

yes

:)

[cursive urchin]

Yesss

For the second part, do I "FOIL" it?

It's -3(4x^2+4)

Is it (-3 * 4x^2) + (-3 * 4)

Or (4x^2+4) * (4x^2+4) * (4x^2+4)

[cursive urchin]

And then turn the end product into a negative

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[mint raft]

If I am calculating my total grade, I add up all my assignments in the following way (see example):

Assignment 1: 8 points earned out of 10 points possible (8/10)
Assignment 2: 15 points earned out of 20 points possible (15/20)
Assignment 3: 20 points earned out of 25 points possible (20/25)

Total: (8 + 15 + 20) / (10 + 20 + 25) = 43 / 55 = 0.782

But I thought when you add fractions, you can't just sum up the numerators and the denominators. So what am I missing here?

[silk pewter]

lcm

if the denoms of all the fractions are equal you can add the numerators

if not you have to make te numerator equal

[mint raft]

yeah I know

[silk pewter]

lcm is 150

[mint raft]

but when I calculate the final grade, how can I just add the numerators and the denominators like that

if that's not how you do it

[quick gorge]

because this isn’t quite adding different fractions
in this case you are trying to find the total points you got over the total possible points, so adding those numbers in the numerator and denominator makes sense
if we were to actually add fractions (such as assignment 1 plus assignment 2), it’s be 8/10 + 15/20 = 31/20, which is over 1.0 which doesn’t make sense anyway lol

[mint raft]

yeah, after thinking about it I came to this realization also

so in this case, we are taking the average instead of adding fractions right?

and I guess we weight it based on the number of points something is worth

hmm I guess "find the total points you got over the total possible points" makes intuitive sense as well if you think about it

thanks

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[quick gorge]

true but keep in mind I don’t think most grades are calculated this way

[quick gorge]

the problem with adding different denominators is it gives more weight to the ones worth more points

vs an equal average would be getting each assignment’s percentage, like 80% for 1, 75% for 2, and finding their average ((0.80+0.75)/2) which would actually be slightly different than the first approach

just thought I’d put it out there

[mint raft]

@quick gorge thanks, that makes sense

so basically they give assignments more points to give them more "weight", right?

so that's why you see big tests out of 100 points, etc.

[topaz sinew]

@quick gorge Has your question been resolved?

[quick gorge]

yeah ig

I haven’t seen it done that way here because they separate assignments into weighted categories (like tests, homework, etc)

but I’m just another american 🤷‍♂️

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[tardy schooner]

I'm in grade 12 and this is a chapter review question! it has to do with logarithms. I've checked over the chapter, and the only advice related to this question at all says to take a common factor out. In this case I thought that would be 4^x, but the -x is really throwing me off, and I don't know what to do. This is much harder than anything that has come before.

I have tried online algebra solvers to guide me in the right direction, but most of them don't even know what to do, and the only one that worked required money.

[flat kindle]

you can either take log on both sides

or let y=4^x

[exotic carbon]

that would be tiresome

[tardy schooner]

THANK U

YEAH THAT WORKED

I think that might be what the textbook was talking about with the common factor...? but this made it way easier.

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[chilly mauve]

Can Someone help me integrate

[acoustic pecan]

any ideas

[chilly mauve]

uh find the area under the graph?

[silk pewter]

do you know how |x| graph looks like?

,w graph y = |x| from x = -5 to x = 5

[chilly mauve]

yeah

[thorny flame]

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[chilly mauve]

^

[acoustic pecan]

hint: split the integral

[silk pewter]

so basically mirror image about y axis right?

[chilly mauve]

so id split the grapgh *and find the area underneath of using the L(Ram) and R(Ram)

?

and add both of them to get the total area?

[silk pewter]

yes

[torpid trout]

It’s symmetrical

[silk pewter]

instead since it is symmetrical, you can find for area from 0 to 1 and multiply the area by 2 to get the total area

[chilly mauve]

alr give me one sec

how many boxes should i use?

4?

[silk pewter]

box?

wt?

[chilly mauve]

to find the area, I divide the x axis into sepreate quadiants

and get its length times width

i add up all my quadrants (squares) to get an aproximate area

[torpid trout]

Just integrate x.

[chilly mauve]

how do you guys solve for area under the graph?

[silk pewter]

breh what

then what is the point of having integration

[chilly mauve]

im lost i thought we were finding the area from the intergrol (0,1)

[torpid trout]

Yes. Integrate x first

[chilly mauve]

ahh

1/2 x^2

[silk pewter]

,,\int_{0}^{1} x\dd x = \left[\frac{x^2}{2}\right]_{0}^{1}

[thorny flame]

Bettim

[silk pewter]

never seen this?

[chilly mauve]

no, never

[silk pewter]

wf

wtf

[chilly mauve]

im cal 1

[silk pewter]

no way

how else do you approach the integral operation

[neat surge]

you want to evaluate it using reimann?

[chilly mauve]

I was thinking jus integrating X, and the applyng f(b)-f(a)?

and multiply my answer by 2 to get its total area under the grapgh

[torpid trout]

Yes

[silk pewter]

how would you ""integrate X""

[chilly mauve]

antiderivite**

[silk pewter]

yea how would you do that

[chilly mauve]

1/2 x^2

right?

[silk pewter]

that is basically integration

[chilly mauve]

okay now all i do is f(b) - f(a)

[prisma isle]

rapturous

or bettim

can any of you help me with this

[chilly mauve]

theres channles

[prisma isle]

oh sorry

[chilly mauve]

[prisma isle]

bye

sorry

[reef fjord]

!occupied

[prisma isle]

my bad

[topaz sinew]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[chilly mauve]

so the answer is 1?

thats the area under X

What I did was get the anti dervitve of X, then plug in my intervols (0,1) to f(b)-f(a)

I got 1/2

i multiplied by 2 because it was only the right hand side of the grapghs area

leading me to get 1

is this the correct way of approaching it?

@silk pewter

@acoustic pecan

[silk pewter]

yes

answer is 1

but idk if it is the correct way to approach integration

[chilly mauve]

thank you pa

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[near schooner]

hello, how would i solve for x?

[outer delta]

Go ahead and single out the lnx

[near schooner]

so like lnx = 5-7-3

?

or put those together

whoops one sec

[outer delta]

That is 3 * lnx so it would be 5-7/3

Or -2/3

[near schooner]

oh right mb

okok

[outer delta]

The opposite to ln is e

So set both to base e (e is going to be risen to the power of whatever there is)

ln and e cancel out

[near schooner]

okok

[outer delta]

Let me try latex again, give me one sec

[near schooner]

oki 👍

WAIT I GOT IT

thank you so much

[outer delta]

$x = e^{-2/3}$

[thorny flame]

dragonbreath

[outer delta]

Yes

[near schooner]

YEAH!

yayay ty!!

[outer delta]

Which you can either leave it like that or solve it

[near schooner]

thats all i needed for this question, thank you!

[outer delta]

,calc e^{-2/3}

[thorny flame]

The following error occured while calculating:
Error: Symbol or string expected as object key (char 4)

[outer delta]

:/

I don't know that

But anyways use .close when done

[near schooner]

yepyep

.close

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[remote jetty]

I am trying to figure out how long it will take in hours for an item in a game to drop

So the the drop chance of me getting a special item is 11% and in those drops it is 3.72% to get the chest item I want from it and then another 1.11% for the item to drop from that chest item. It takes between 5.60secs-10.50secs to drop the special item

So what is the equation I have to learn to find how many hours it would take for me to get the 1.11% item?

[outer delta]

I wanna say ${11% * 3.72% * 1.11%}/5.60secs$

$frac{0.11 * 0.0372 * 0.0111}{5.6}$

[thorny flame]

dragonbreath

[outer delta]

I don't understand this stupid latex

[alpine mist]

wait. you get a special item 11% of the time, of the special items you get a chest 3.72% of the time, and of the chests you get the specific thing 1.11% of the time?
and
you get a special item ever (roughly) 5-10 seconds?

or do you just get an item drop every 5-10 seconds?

[remote jetty]

I get an item to drop every 5-10 seconds and the chance of it being special is 11%

[alpine mist]

you need a backslash in front of the frac

[outer delta]

So basically a chest is given every 5-10 secs. In that chest there is an 11% chance of a rare item, there is a 3.62% chance of that rare item being the wanted item, and then I am assuming you open that item which has a 1.11% of dropping the wanted item

$\frac{0.11 * 0.0372 * 0.0111}{5.6}$

[remote jetty]

Yes, pretty much

[thorny flame]

dragonbreath

[outer delta]

Ah

[outer delta]

Thank you

[remote jetty]

Thank you so much for the help, it will help me in finding all the special item drop chances per hour now, thank you 🙂

[alpine mist]

so the probability of getting your drop would, yes, be that product

,w calc .11*.0372*.0111

[thorny flame]

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[alpine mist]

expected number of attempts until success is 1/p, so you get

,w calc 1/.00004542

[thorny flame]

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[alpine mist]

every 5 to 10 seconds

,calc 22017(5/3600)

[thorny flame]

Result:

30.579166666667

[alpine mist]

,calc 22017(10/3600)

[thorny flame]

Result:

61.158333333333

[alpine mist]

expected 30 to 61 hours, possibly more or less.

[outer delta]

Ouch

[remote jetty]

Understood, that helps an extreme amount, thank you 🙂

[alpine mist]

welcome to the free-to-play grind until you buy the booster packs model!

enjoy your battle pass

[remote jetty]

It's an idle game so it's intended to take that long, so it's all good 🙂

[alpine mist]

ah, that makes sense

[remote jetty]

Thank yous both so much 🙂

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[neon iron]

I need help with this problem

Perimeter of the triangle is 16 cm
a = b
both a and b are 2.9cm more than c

[marble gate]

okay, so you can write b in terms of a, c in terms of a, and then 16 in terms of a - right?

[neon iron]

i dont think we can write c in terms of a
becuase a and b are 2.9 cm more than c

a + b + c = 16

[marble gate]

but you've just said that c is <some relation> to a

which is writing c in terms of a 😛

[marble gate]

and we know something about b in relation to a
and something about c in relation to a
so we can write this as a + f(a) + g(a) = 16

[neon iron]

thanks for your attention, i just solved it

[marble gate]

[neon iron]

.close

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[summer bear]

hmm

I don't even know how to start

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[regal peak]

how do i solve the question under the question 6 graph

[regal peak]

this is a practice test btw

but the teacher didn’t post the answers

HELP!

MY UNIT TEST IS IN 30 MINUTES

😢

[summer bear]

well

csc x = 1/(sin x)

maybe you could draw the graph of sin x

[raw mirage]

let cscx=3=y

[summer bear]

how would that help?

[raw mirage]

you can separate it as y=3 and y=cscx

then look for how many intersections there r between the two graphs

under the given interval

[summer bear]

yh

just use the graph

ou can see its 3

[topaz sinew]

@regal peak Has your question been resolved?

[pallid phoenix]

1. If u r allowed to use a graph, add the equation on graphs and u ill get the neccesarry inputs in the following interval
2. Otherwise, get the values for which the eqaution alligns with, u will get 3 solutions maybe

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[stark oyster]

.close

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[stark oyster]

.reopen

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✅

[stark oyster]

Any idea why the multionomial distribution doesn't work for C. normally if I say P(C) = 9!(4!3!2!) * (1/9)^9

Is not the correct ans

Tho if i multiply it all by 3! I do get the current response

But I don t see why would I need to do that, does the multinomial not count all the possible combinations?

Or I can't apply it? But the events are independent and each event has the same probability( I think)

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@stark oyster Has your question been resolved?

[stark oyster]

<@&286206848099549185>

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[obtuse flint]

for any natural numbers n(15 x 15^2n) + (2 x 2^3n) is divisible by

[vale furnace]

Can’t really read your question but definitely divisible by 1

[obtuse flint]

wdym?

[torpid matrix]

for any natural number n that expression is definitely divisible by 1

[obtuse flint]

sooo 1's the answer for that?

i put 2 but i get prime

[torpid matrix]

???

[obtuse flint]

i put n = 1, and i get prime

[torpid matrix]

!original

[topaz sinew]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[obtuse flint]

i do not have it online 💀
all the question says is that
for any natural numbers n(15 x 15^2n) + (2 x 2^3n) is divisible by?

[torpid matrix]

take a picture of it?

are there answer choices?

why do you keep mentioning "prime"

[obtuse flint]

options are
7, 11, 13, 17

[torpid matrix]

so the question has more to it

can you please just post the original

also those are all prime, so you can just pick some n and then see if the expression is divisible by each answer

[obtuse flint]

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[golden edge]

@sweet shard if you're still around could you explain what you said in #help-41 I was eating my food sorry

[ruby tree]

then dividing by magnitude
It looks like you are thinking of orthonormal vectors

Orthogonal says nothing about the length

[sweet shard]

What about it didn't you understand

[golden edge]

Hmmm?

[golden edge]

How do I create a variable for it ?

[ruby tree]

Wait I didn't read, it does say unit vectors

[golden edge]

Yeah that's what I've got I think😅

[ruby tree]

What makes you think there are more than 2 then?

[golden edge]

I've just never done a question like this

And it says all

[ruby tree]

Well yeah, because there are multiple

[golden edge]

And yesterday when I spoke to people here i got like 5 different answers, someone said 5, 4 and 2

Then I asked a friend and he said he's made a general formula which I think is what @sweet shard is on about but I'm not sure the steps to do that

[sweet shard]

Good for them?

Nel is right there are only 2 unit ones

[golden edge]

Oh?

If I send my working would you be able to verify if you think it is correct for the question ?

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@golden edge Has your question been resolved?

[golden edge]

<@&286206848099549185>

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[cosmic plaza]

Hello everyone

[cosmic plaza]

This is the integral I am trying to solve

This is the solution photomath gives

And this is mine solution

Does anyone know what could I do wrong in my integration?

<@&286206848099549185>

[raven sparrow]

Pretty sure they are equivalent

[cosmic plaza]

Isnt mine somehow missing x^3-8

[raven sparrow]

You can just factor it as you did earlier and split the log in two it comes out the same

[cosmic plaza]

Thanks a lot

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[stable sinew]

how do i solve this equation for x?

[stable sinew]

.close

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[celest oracle]

anyone know how to do this

[celest oracle]

Have no idea

[celest oracle]

Please

[topaz sinew]

@celest oracle Has your question been resolved?

[celest oracle]

<@&286206848099549185>

[sinful pumice]

treat the units like variables

such as kg as a vriable

or m as a variable

[celest oracle]

My issue is the m-1 and m2 I forgot how you put them together

The answer should just be 1 kg m to the power of something which I don’t know

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[neon iron]

How do I make this plot the negative-x values too?

[neon iron]

#latex-help

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[rugged idol]

hey everyone!
Right now I am trying to show that R(C4, C4) = 6 , where R(G,H) is the minimal number n such that a graph on n vertices either contains G as a subgraph or the complement contains H as a subgraph. Showing that it is > 5 was manageable. Now I am trying the upperbound <= 6.
I am struggeling with finding a firm approach. I see why it has to be 6 but I cant really make a convincing argument of it.

I would really appreciate it if someone could give me a hint or a pointer without giving away too much.

[rugged idol]

I tried something like the standard argument for the friends and strangers problem but it didnt take me too far

.close

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[timid oar]

Can I cancel things grouped together as long as they "are the same"?

[wooden osprey]

only if they arent 0

for example $\frac{3x}x$

[thorny flame]

FungusDesu

[wooden osprey]

if you cancel x to get 3, you are assuming that x does not equal to 0

[timid oar]

alr

thanks

-close

.close

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[old plover]

(7x+42)/3

[old plover]

Can it get any more simplified?

[haughty mural]

you could write it as
7x/3+42/3=7x/3+14

depends on you, whether you prefer that

[old plover]

Yeah, but mine’s still technically as simplified as it gets, right?

With a single fraction and all?

[haughty mural]

i guess, yeah

I'd say it depends on what you prefer

[old plover]

Cuz on the test, I needed the simplest answer @haughty mural

@haughty mural

U there?

[haughty mural]

yes

[old plover]

So I’m correct, right?

[haughty mural]

i would say so

[old plover]

Yeah, I think it’s as simple as it gets

Thanks

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[shut spindle]

2cos(x) = 1 . When finding all solutions for X here should I start buy dividing both sides by 2 and get cos(x)=1 then proceed to find the angle and the subsequent angle?

[pastel salmon]

not cosx = 1

but

cosx = 1/2

[shut spindle]

yeah ofc haha

silly mistake

thank you

[pastel salmon]

🙂

[shut spindle]

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[haughty talon]

#❓how-to-get-help & thats a physics question

[warm drift]

a common related rates problem in a calculus course, so no

[haughty talon]

The former part of my comment still holds x

[warm drift]

yeah hold on

why is this wrong

ignore the units

[ivory wave]

No

[warm drift]

what

[ivory wave]

That's not a calc

[haughty talon]

It is tho

[warm drift]

its a related rates problem

[haughty talon]

Anyway, im super fatigued so i cant rly figure out where you went wrong but here's my approach:

You know dh/dt = 10ft/s with h being the left side of the triangle, you also know through basic trig that tan(x) = h/100 so that x = arctan(h/100). Then you calculate dx/dt which results in (100*h'(x))/(h(x)^2+10000) if i did the math correctly

you know h'(x) and you also know h(x) by integrating h'(x) (h(x) = 10ft/s * t) where t is the time

[warm drift]

right, you could use tangent, but I decided to use sine/cosine instead

why is it different?

[haughty talon]

so plug in 50ft into h(x) to find the t for that instant and you've got your rate at that moment

uhhh

again im v v tired so im not confident in what i say rn but i think the issue is that you started out your differential equations with the value for the hypotenuse

but it doesnt stay constant for the whole trajectory of the balloon

only at the instant where its at 50ft is it that value

While the tangent relation is true for the whole trajectory of the balloon

so you can find a valid diff eq to solve for the trajectory and then plug in the specific value of 5ft

50ft*

hope that made sense, if not someone else will probably come around to help you

[warm drift]

but isnt it at a specific moment?

when y is 50

so why does it matter?

since the question asks for the rate, essentially at the moment when the balloon rises to 50 feet, which has a corresponding hypotenuse and degree rate, no?

[haughty talon]

Well the way of finding rates at a specific moment is first finding the rate generally at any moment and then plugging in the specific moment you are looking for into the function you found. So first to find the general rate function we look at the relations that hold generally throughout the whole motion of the balloon and the one you proposed with the sine doesnt fit that

Because the hypothenuse isnt always that length throughout the whole motion

[warm drift]

so we have to relate it to a constant? such as the horizontal distance

[haughty talon]

the tangent relation is general and holds throughout the whole motion tho, so we find a correct general rate function using that and then boom plug in the value we look for

Hmm, kindof but i dont think you're getting it. My explenation wasnt that great, try @ helpers (its allowed after 15min) maybe they'll do a better job

ive been up for like 24hours working on this fking deadline, but gl to you mate!!

[warm drift]

thanks, using sine would have two rates, while tangent has a rate and the horizontal constant, thats the only way I can see how it makes sense

in simple terms

so I am agreeing with you

[haughty talon]

The issue is that you're trying to relate the rate of change of the balloon with the rate of change with the angle generally. You propose to do that by linking the two through trigonometry which is a good step but then you go for sin(x) = h/hyp.

So now you have the angle in function of the height AND the hypotenuse, if you want to figure out the rate of change of the angle now you'd ALSO have to work out the rate of change of the hypotenuse, but we dont have that info.

If you go for tan(x) = h/100 you only need to consider the rate of change of the height and thats exactly the info given

Anyway, im really gonna go now haha. Goodluck!

[warm drift]

makes sense, thanks

.close

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[polar blaze]

Let $ABCD$ be an isosceles trapezoid with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$.

explain what this means, please

[thorny flame]

Aurora

[polar blaze]

what does it mean by "whose angle at longer base AD is pi/3"

does it mean angle A and D are pi/3 or 60 deg

or what

[sonic dawn]

ye that

[polar blaze]

why does it mean that?

whose angle at longer base AD is pi/3 doesnt make sense

[sonic dawn]

isosceles trapezoid implies the bottom angles are identical

so you only have to specify one

[polar blaze]

yeah but it said "at longer base AD" not "angle D"

[spiral slate]

the angle that is formed by the longer base on both sides is pi/3 due to it being isosceles

[sonic dawn]

you can try interpreting this as “angles involving side AD”

but i do agree that it’s not the most straightforward wording

[polar blaze]

oh ok ty

.close

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[smoky sparrow]

!status

[topaz sinew]

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

[glass crescent]

2

[smoky sparrow]

alr, what have you tried?

[glass crescent]

@smoky sparrow

[smoky sparrow]

one sec

right, so the left fraction is $\frac{2 \sin x}{\cos^2 x - \sin^2 x}$

and $2 \cos^2 x - 1 = \cos^2 x + \cos^2 x - 1 = \cos^2 x - \sin^2 x$

[thorny flame]

south

[smoky sparrow]

now rewrite the 1 in the numerator as sin^2 x + cos^2 x

you have a * when you should be adding sin/cos + sin/cos

[thorny flame]

south

[glass crescent]

Like that?

[smoky sparrow]

alright, from the very beginning, multiply (2 tan x)/(1 - tan^2 x) by (cos^2 x)/(cos^2 x)

this gives $\frac{2 \sin x \cos x}{\cos^2 x - \sin^2 x}$

[thorny flame]

south

[glass crescent]

Ok wait

Like this?

@smoky sparrow

[smoky sparrow]

no

you still have (2 sin x) for some reason

the numerator is $2 \frac{\sin x}{\cos x} \cdot \cos^2 x$

[thorny flame]

south

[glass crescent]

I dont get which numerator

[smoky sparrow]

$2 \tan x$

[thorny flame]

south

[smoky sparrow]

in $\frac{2 \tan x}{1 - \tan^2 x}$

[thorny flame]

south

[glass crescent]

Can you write the while equation part?

[smoky sparrow]

I'm doing it step by step

[smoky sparrow]

you keep multiplying sin x / cos x and sin x / cos x

that's wrong

you should instead have sin x / cos x + sin x / cos x

[glass crescent]

So is the 1st picture wrong?

[smoky sparrow]

yes

[glass crescent]

I keep erasing my solution idk what to put first

[smoky sparrow]

do you understand this bit?

[glass crescent]

No that part is what i dont understand

[smoky sparrow]

which part don't you understand?

[glass crescent]

This solution

[smoky sparrow]

that solution i swrong

[glass crescent]

So what should I do FIRST?

[smoky sparrow]

this

we're multiplying $\frac{2 \tan x}{1 - \tan^2 x} \frac{\cos^2 x}{\cos^2 x}$

[thorny flame]

south

[smoky sparrow]

yes

[glass crescent]

For the right fraction...?

[smoky sparrow]

2 cos^2 x - 1 = cos^2 x - sin^2 x as you've noticed already

and the numerator of that fraction is just 1

[glass crescent]

So the left fraction needs to be simplified?

[smoky sparrow]

yes

[glass crescent]

How?

[smoky sparrow]

this is $\frac{2 \sin x \cos x}{\cos^2 x - \sin^2 x}$ right

[thorny flame]

south

[glass crescent]

Ah yes

[smoky sparrow]

and we're adding $\frac{1}{\cos^2 x - \sin^2 x}$

[thorny flame]

south

[smoky sparrow]

so the next step is to rewrite the 1: $\frac{\cos^2 x + \sin^2 x}{\cos^2 x - \sin^2 x}$

[thorny flame]

south

[smoky sparrow]

no

$\frac{2 \sin x \cos x}{\cos^2 x - \sin^2 x} + \frac{\cos^2 x + \sin^2 x}{\cos^2 x - \sin^2 x}$

[thorny flame]

south

[smoky sparrow]

now, let cos x = a, and sin x = b

we have 2ab + a^2 + b^2

what do you notice?

[glass crescent]

They can be factor out

[smoky sparrow]

yep

what's a^2 + 2ab + b^2?

[glass crescent]

(a+b)^2

[smoky sparrow]

yeah

so we have (cos x + sin x)^2 as the numerator

after combining the fractions as they have the same denominator

[smoky sparrow]

can you factor the denominator also?

[glass crescent]

Yes

(a+b)(a-b) right?

[smoky sparrow]

yeah

so we have $\frac{(\cos x + \sin x)(\cos x + \sin x)}{(\cos x + \sin x)(\cos x - \sin x)}$

[thorny flame]

south

[glass crescent]

So from here i will make it that?

[smoky sparrow]

yes

[glass crescent]

And then cancel like factors

[smoky sparrow]

yes

that's it

[glass crescent]

@smoky sparrow Can i also ask for help if this is right?
Someone told me to do this

[smoky sparrow]

yes

that's correct

[glass crescent]

Okay last, how do i prove this?

[smoky sparrow]

multiply the right-hand side by cos^2 x / cos^2 x

you can work from the right hand side to the left since the = sign works both ways

[glass crescent]

@smoky sparrow

[smoky sparrow]

you don't multiply again by cos^2 x

sin^2 x + cos^2 x = 1

then you can times $\sin^2 x - \cos^2 x$ by -1 * -1

[thorny flame]

south

[smoky sparrow]

to get $-(\cos^2 x - \sin^2 x)$

[thorny flame]

south

[smoky sparrow]

should look familiar now

[glass crescent]

Oh ok

And then what?

[smoky sparrow]

do you know what cos 2x is?

like how else can it be written

[glass crescent]

No i wasnt yet introduced with double sign

[smoky sparrow]

RIP

oh wait you can still do it

[smoky sparrow]

from here just sub in $\sin^2 x = 1 - \cos^2 x$

[thorny flame]

south

[smoky sparrow]

no need to multiply by -1 * -1

[glass crescent]

Ah ok i get it now

Ty

[smoky sparrow]

no worries

[glass crescent]

@smoky sparrow Sorry but is it ok to make u help me some more?

[smoky sparrow]

pls ask someone else

[glass crescent]

Ok

.close

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[misty niche]

guys

[misty niche]

if i have

sqr 100(x-1)^2

is the transformation

hd bafo 1/10

or 1/100

[rigid ivy]

[misty niche]

thanks !

i appreciate it

fianly

[rigid ivy]

!xy

[topaz sinew]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[misty niche]

guys whats the transofmration

for this

i dont know how to idetnfiy the transfomration of a rational

[misty niche]

no no u helped me

i get it

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@misty niche Has your question been resolved?

[royal nexus]

you see, the common factor of tye denominator is 2

then your function will be more like $f(x) = frac/{1}{2(x^2+1)}$

[thorny flame]

this_is_a_trap

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[topaz sinew]

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[hard folio]

Hi guys anyone could help :

[hard folio]

tried to calculate it but i do not get the answers from the option given

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[agile burrow]

Help pls

[agile burrow]

nvm i got it

but let's see what people think

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[dusk sigil]

If 60^a = 3 and 60^b = 5, find 12^[(1-a-b)/(2-2b)]

[dusk sigil]

No log used

[neon iron]

.reopen

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@dusk sigil Has your question been resolved?

[topaz sinew]

@dusk sigil Has your question been resolved?

[neon iron]

$$60^a = 3$$
$$60^b = 5$$
$$12^\frac{1-a-b}{2-2b}$$

[thorny flame]

Several people

[topaz sinew]

@dusk sigil Has your question been resolved?

[topaz sinew]

@dusk sigil Has your question been resolved?

[remote pulsar]

So wait why can’t you use a log to solve it?

@dusk sigil