help-39

[severe flame]

y=mx+b?

ohh

yea

the y axis?

[bitter epoch]

Not quite, the y-axis is the entire axis we draw

[severe flame]

so how do we find that

[bitter epoch]

We were given y=-2x+9, what does the 9 mean? What happens when x=0?

[severe flame]

x=0 is the y axis

OHHH

so we have b

and the graident

[bitter epoch]

Yes, so how can we find "b" if we know the y-intercept?

[severe flame]

1/2x +9=y

[severe flame]

i dont understand is 9 not b

[bitter epoch]

Is that right? What is the correct y-intercept we want?

[bitter epoch]

What should our final "b" be equal to?

[severe flame]

to y?

[bitter epoch]

Reread the question

[severe flame]

the equatio

n

[bitter epoch]

Here

What does the question tell us

[severe flame]

wait we have the y intercept

so we have b?

[bitter epoch]

Indeed

[severe flame]

oh yay

[bitter epoch]

So what's our final equation?

[severe flame]

1/2x + 4 =y

[bitter epoch]

That's correct!

[severe flame]

YESSS

THANKYOU

[bitter epoch]

Remember the steps you went through here, and you can solve any problem that are similar to this

[severe flame]

can we do this if we have the line

[bitter epoch]

Which line? Do you have an example?

[severe flame]

yes one second

here

[bitter epoch]

Okay, first of all, what does the question want us to find?

[severe flame]

the equation of the line

[bitter epoch]

What form should it be in?

[severe flame]

y=-3 and x=-1

[severe flame]

itdoesnt say

[bitter epoch]

Then we can decide

What's the simplest equation for a line?

[severe flame]

y=mx+b?

[bitter epoch]

Sure, we can do that

What does the question tell us in terms of information to help us solve it?

[severe flame]

the y intercept and the x intercept

[bitter epoch]

Can we already find m or b just from that info?

[severe flame]

i mean if b is the y intercept wouldnt it be -3?

[bitter epoch]

Absolutely, that's right

[severe flame]

and can we also figure out m?

[bitter epoch]

We can

[severe flame]

how

[bitter epoch]

Well, what does the "gradient" actually mean?

[severe flame]

rise/run

like a thing sloping up or down

[bitter epoch]

Yep, and can we find a small interval where we know the rise and run?

[severe flame]

i dont know

do we make it a triangle

[bitter epoch]

Well, we know two points (the x intercept, y intercept)

Could we use the rise/run formula on these two points?

[severe flame]

ye

the run is 1 and the rise is 3

-1/3

[bitter epoch]

This step isn't correct, one of them should be negative

[severe flame]

run -1 and rise 3

[bitter epoch]

And how did you work out those numbers?

[severe flame]

by making a triangle

and counting

[bitter epoch]

When you do that, remember to "start" counting from the same point

[severe flame]

oh would it be -3

[bitter epoch]

Yes

[severe flame]

1/-3

[bitter epoch]

It's +1 to the right, and -3 since we go down

Remember it's rise/run, not run/rise

[severe flame]

yes

[bitter epoch]

What's our answer then?

[severe flame]

so would it be y=1/-3x+-3

[bitter epoch]

What's rise?

[severe flame]

ohhh

-3/1

[bitter epoch]

Yes

[severe flame]

so just -3

[bitter epoch]

Correct

[severe flame]

ohh

thankyou so much

[bitter epoch]

We don't need "+-3" at the end, just "-3" is good

[severe flame]

so itll just e y=-3x-3

[bitter epoch]

Yep, that's our final answer

[severe flame]

thankyou

,close

[bitter epoch]

No problem

[severe flame]

.close

[pearl pond]

Channel closed

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[pearl pond]

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[steady bluff]

Prove the statement is true using mathematical induction:

     2^(n-1)  ≤ n!

where n is an integer greater than or equal to 1

[chilly sigil]

what have you tried

[steady bluff]

ok so hypothesis is
2^(k-1) = k!
induction step is
2^k = (k+1)!
is this correct
2^k = 2^(k-1) + 2^k <= k! + 2^k
where 2^(k-1) = k!
so you just get
k! + 2^k <= k! + 2^k

[steady bluff]

I'm not sure if this is correct

[chilly sigil]

you need <= in hypothesis and you need base case

[steady bluff]

ok yea I meant

but besides that is it good

just replacing the = with <=

[chilly sigil]

what

I don’t think so

[steady bluff]

why

this shit confusing

the way I solved it just comes from this

https://i.imgur.com/larX0KE.png

I feel like its essentially the same way

[chilly sigil]

you have to get the end result though

[steady bluff]

I did

k! + 2^k <= k! + 2^k

[chilly sigil]

What that’s literally 0<=0

You would use the inductive hypothesis to go to the inductive step

si suppose 2^(n-1) < n! for some n>=2

2^(n-1) x 2 = 2^n < 2 x n! < (n+1) x n! = (n+1)!

[steady bluff]

ok so how about this

inductive step is

2^k <= (k+1)!

you could say

2^(k-1)+ 2^k <= (k+1)!

?

k! + 2^k <= (k+1)!

then find a way algebrically to make the left equal the right

[chilly sigil]

no

How would that work

[pearl pond]

@steady bluff Has your question been resolved?

[pearl pond]

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[silver oracle]

so i got y

i forgot how do i get x

do i need to plug y=1 in

also is y correct

[hearty ferry]

This is how i usualy do this things

[silver oracle]

is that the correct way to do it?

insert one of them

even though its

elimination

after you elimination the y of course

[hearty ferry]

this how they showed me in school

maybe it can be done diferent

[silver oracle]

I think it's okay

[silver oracle]

btw how do i write out my ansawer on my computer to enter it in

[hearty ferry]

I think y=1 and x= -2

[silver oracle]

idk how to do this one

dont i need t o mulitply the 2nd one by 3?

but then both y and x will =0

so idk what i need to do

.close

[pearl pond]

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[pearl pond]

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[loud drum]

😔 what should I do next

[wanton burrow]

what is the sqrt of 3/4

Also it should be positive or negative sqrt3/4

[loud drum]

0.8660254038? ( I put this in a calc)

[wanton burrow]

no

Sqrt3/2

no sqrt on 2

[loud drum]

O

OH

[wanton burrow]

And that’s a unit circle value

[loud drum]

😋 yup

[wanton burrow]

and it should be easy to continue

[loud drum]

Tysm 🙏

.close

[pearl pond]

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[fickle patio]

May I know how to solve this and get the critical points

[humble lintel]

In the first equation, it is a quadratic as a function of y with x constant, so y has two solutions for every x.

Symmetrically the second equation is a quadratic as a function of x with y constant, so x has two solutions for every y

Those generate 4 solutions as expected

[fickle patio]

I don’t know how to solve that

[humble lintel]

You don't know how to solve a quadratic?

[fickle patio]

I never solve two equations that both contain quadratic

I only know to to solve quadratic with one variable and here, it’s has xy

[humble lintel]

Solve one at a time and treat one parameter as a constant

For example, 24y - 12xy - 4y^2 = 0 implies y^2 + 3xy - 6y = 0 then using the quadratic formula or factoring, you get y(y + 3x-6) = 0 so y = 0 or y + 3x-6 = 0.

Your solutions for the first equation are a function of x

[fickle patio]

I don’t really und

I can get y=0

But the one that contains x and y

Is it compare with the another equation or how?

[humble lintel]

where did the -4y go

[fickle patio]

I thought need to substitute y=0 in if not how can I get （0,0) and (0,6)

[humble lintel]

Can you solve (x-2)(x-3) = 0

[fickle patio]

Yes

[humble lintel]

So what is x?

[fickle patio]

2, 3

[humble lintel]

Ok, so if x = 2

What you are doing right now is going

"I'm now looking at (x-3)"

"and I know x = 2"

"so (2-3) = 0"

[fickle patio]

Huh?

[humble lintel]

Do you agree that if you have ab = 0, then a = 0 or b = 0

[fickle patio]

Yea for this

But u can’t und why you sub 2 to x-3

[humble lintel]

and when you are looking at the value of a, you don't care abou the value of b?

[fickle patio]

I don’t und what you actually mean

[humble lintel]

You have the equation

y (24 - 12 x - 4 y) = 0

That is the product of two numbers

y times 24 - 12x - 4y

Similarly, you have no problem with (x-2) times (x-3)

which is also the product of two numbers.

[fickle patio]

So the working I sub y=0 previously is correct?

[humble lintel]

When you have (x-2) times (x-3), you know that if x = 2, you have zero times (x-3) which is all zero, and when x = 3, you have (x-2) times 0 which is also 0

Yet what you are doing here is taking y times (24-12x - 4y) and saying ok, y = 0 makes the whole thing zero, now I am going to plug in zero into the right side.

And that should not make sense.

[fickle patio]

Then why you sub x=2 here?

Okay, nvm, I think you’re referring what I’m doing

[humble lintel]

My point is that is what you are doing, which doesn't make sense

Your factored form is y ( 24-12x - 4y) = 0, you know this is a quadratic, so if x doesn't change, there are two solutions

One is when y = 0, the other is when 24-12x-4y = 0.

[fickle patio]

Okay, well, if I use 24-12x-4y and 24-6x-8y, I can only get (4/3, 2)

[humble lintel]

You need to use them all

y = 0 is a candidate

[fickle patio]

If x and y both =0

How to use

That’s why I can only get the last critical point here

[humble lintel]

So if y = 0 is a solution, you need to go back to the second equation and check what solutions it has when you set y = 0 there

Then you need to check what solutions it has when -4y - 12x + 24 = 0

[fickle patio]

Nvm

I think I’ll ask again later

Thanks for your help

Sorry I really can’t und what you’re trying to say

.close

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[snow latch]

just gonna latex it just a sec

[snow latch]

the graph of

,tex $ y = \log_4(x)

[jolly parrot]

suds
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[snow latch]

passes through the point (16,2). Find the image of the point (16,2) under the following linear transformations

I have no idea what "find the image of the point" means, never had that terminology used

first question linear transformation is

,tex $ \log_4(x) + 3

[jolly parrot]

suds
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sharp vigil]

we want to find the "corresponding point" after the transformation

[snow latch]

so it would be 19>

?

because we keep the x value?

(19,2)?

[sharp vigil]

we have a point (x,y) [one of a collection of points satisfying the equation] and we applied a transformation to that point and its collection

what kind of transformation was it?

[snow latch]

a horizontal translation by 3

is it like the same "part" of the graph but with the adjusted equation

like say its a polynomial and theres a peak at (16,3) where the peak would be if it went through a linear transformation?

[sharp vigil]

well we started with y = log_4(x) and now we have y = log_4(x) + 3. did we change the "x" or the "y"?

[snow latch]

the y

sorry i meant vertical

[sharp vigil]

yes

[snow latch]

so is it 19,2?

bc we've adjusted the y

and for something like log_4(x-3) it would be 16, -1?

or would it be log_4(-1)

[sharp vigil]

we have (x,y) = (16,2) to start with

[snow latch]

yes

my question is basically, what stays the same?

because you can sub the y or the x value

[sharp vigil]

so if we shift "y" by 3, then we would expect the second component to change

[snow latch]

oh ok

so does that make it 19 = log_4(x) + 3?

sorry, im just a little confused

[sharp vigil]

we have that originally (x,y) = (16,2) satisfied the equation y = log_4(x). then we would expect (x,y) = (16, ?) to satisfy the equation y = log_4(x) + 3

[snow latch]

oh okay

wait

so the x value is constant?

(i mixed up my x and y before sorry)

but it would be y = log_4(16) + 3

so 16,5?

[sharp vigil]

yes

[snow latch]

does the x value stay constant in any transformation?

[sharp vigil]

we would expect x to change if we changed the part where x is. so y = log_4(x - 3) would be a shift of 3 to the right

[snow latch]

okay

so then we use the y value as a constant? or sub x

sub x and have a new x

[sharp vigil]

in that example x would change and y would not

[snow latch]

ok thank you

thats perfect

its pretty easy tbh but i just haven't ever heard the term phrased like that

[sharp vigil]

but we could also have y = log_4(x - 2) + 4 in which case both would change

[snow latch]

if both change would you solve for one transformation (say log_4(x-2) and then sub the new x value to solve for the new y?

[sharp vigil]

yes, although you may also observe by inspection that the point moves 2 units right and 4 units up

[snow latch]

yep

and -log_4(x) the y value is -?

[sharp vigil]

yes

[snow latch]

oh im so stupid i realise what it is now

its basically what x and y values do you need for the linear transformation to be at the same point as the original

[sharp vigil]

we can think of these equations as defining a set of points (x,y) that satisfy those equations. Then applying these transformations changes all of those points at once (in the same way as we found for individual points)

[snow latch]

like -log_4(-x) the coords are (-16, -2) because when you substitute it, you will get the original point of (16,2)

[sharp vigil]

yes

[snow latch]

thank you!

lots of help thanks

.close

[pearl pond]

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[remote ledge]

hello, how do you do this question?

[remote ledge]

@sharp vigil 🥺

[sharp vigil]

two vectors are perpendicular if their dot product is 0

[remote ledge]

I realised that part, dont know what else to do now

[sharp vigil]

find the dot product and set it equal to 0

[remote ledge]

ohhhh

so for r, x just equals p+4

i got confused on that part

[pearl pond]

@remote ledge Has your question been resolved?

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[winged blade]

Hey

[winged blade]

Im back

I still need help

For a geometric series where the first term is -6 and the common ratio is 4, determine S6. (2 marks)

a is the first term -6
r is the common ratio 4
n is the term number 6

That's all ik for now

[dusky scaffold]

use the formula

[rustic nacelle]

@winged blade You haven't answered my last question, it seems like you don't quite understand geometric series yet.
If the first term of a geometric series is 1 and the common ratio is 2, can you tell me the first 5 numbers in the series? If not, what's confusing you?

[winged blade]

I'm confused man

[tidal kayak]

hes clearly currently taking an exam (as shown by the fact that the copy pasted version of the question has "2 marks" next to it) and is being rude as well as academically dishonest. not only are the rules against helping him, he doesnt deserve the help either

[winged blade]

Substituting the numbers in 😭

It's not an exam 🤣

It's hw

[rustic nacelle]

homework can have marks as well, i dunno

[tidal kayak]

sure maybe

i mean if you want to help him then i couldnt stop u, but im clearly not as nice as you

ill leave u to it then

[knotty urchin]

i mean, fwiw i solved the sum just for practice

but if this is academic dishonesty

then im not showing how to do it

[rustic nacelle]

What are you confused about? Do you understand the terms "first number" and "common ratio" in this context?

[winged blade]

Its hw

I can take it home

It's just that I'm confused

[winged blade]

Yes

[rustic nacelle]

So if the first number is 1 and the common ratio is 2, what's the second number in the series?

[dusky scaffold]

https://tenor.com/view/can-you-please-explain-kbrown-mark-angel-tv-what-do-you-mean-what-are-you-trying-to-say-gif-7818327139435432435

[knotty urchin]

2

the second number would be 2

[rustic nacelle]

I'm asking arz

[winged blade]

Then 4

[knotty urchin]

oh no, sorry, i didnt see the name

[winged blade]

Cuz multiply by 2?

[rustic nacelle]

Yes

[winged blade]

Ok

[rustic nacelle]

So what are the first numbers in the series?

[winged blade]

So 2 4 8 16

Etc

[rustic nacelle]

Yes, 1 2 4 8 16

What is their sum?

[winged blade]

Okok

Sum?

[rustic nacelle]

Yes

[winged blade]

Idk

Wait

Wait

23?

We just add them

I forgot

Mb

[knotty urchin]

sum is just when you add them all together

[winged blade]

Yea yea

I forgot mb

[knotty urchin]

it wont be 23 tho

[winged blade]

Ik

Now what?

[rustic nacelle]

So what is the sum of these numbers?

[winged blade]

31

[rustic nacelle]

Yes, exactly

[winged blade]

Okok

[rustic nacelle]

So in the series with the first number of 1 and a common ratio of 2, the sum of the first 5 numbers is 31.

Does that make sense?

[winged blade]

Yep

[rustic nacelle]

"the sum of the first 5 numbers" is written down as S5.

[winged blade]

Ohhhh

[rustic nacelle]

So you need to do the same steps for your question:

• List the first 6 numbers of your geometric series
• Add them up

[winged blade]

Alr

So -6

× 4

24

× 4

-96

×4 = -3,984

And so on?

[knotty urchin]

-6 * 4 is not 24

[winged blade]

-24

R those steps correct tho?

[knotty urchin]

yea essentially. youre just gonna need to multiply it by 4

first term * 4 = second term
second term * 4 = third term

and so on

you can do all of this, and you get your 6 terms

then you just add em all up

[winged blade]

Ohh ok

[knotty urchin]

this isnt the fastest way by any means

[winged blade]

Just do that 6 times?

Yea ik

[knotty urchin]

but its the basic principle

[knotty urchin]

yea essentially

[winged blade]

Okok

Ty guys

[knotty urchin]

yw

once you learn this it would be so much easier

[winged blade]

Alr

[knotty urchin]

like lets say i have to find the sum of 5 terms of a geometric sequence that is 1,2,4,8,6.
Thats gonna be 1*(32-1)/1

[winged blade]

Ohhh ok

[knotty urchin]

which simplifies to 31

[winged blade]

Yea

[knotty urchin]

and is way easier than doing all the efforyt

especially if i asked you to find the sum of the first 1000 terms

[winged blade]

Yea true

[pearl pond]

@winged blade Has your question been resolved?

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[tall trench]

can someone confirm i did this correctly

[tall trench]

Idk how to use bot to rotate so

[light helm]

,rotate

[tall trench]

Thankyou

[light helm]

,calc 6000*(1+0.07/365)^(3*365)

[jolly parrot]

Result:

7401.9193249465

[light helm]

looks good

[tall trench]

Wheres the 3365 comin from

Oh nevermind I see it

Lol thank you

[light helm]

discord eating the *

[tall trench]

Yepyep I just noticed the slightly tilted 3

[light helm]

btw no need to write the ^ on paper

[tall trench]

I write it because I get it confused with multiplication otherwise

[light helm]

that's just text/coding notation to represent exponention

[tall trench]

I'll type x # instead of xy if I do it without the ^

[light helm]

try get used to aligning what you write

[waxen condor]

, rotate

[jolly parrot]

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

Jun 7, 2024, 5:14 PM

[pearl pond]

@tall trench Has your question been resolved?

[pearl pond]

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[desert solar]

An elevator in the Mariott Marquis skyscraper in New York has a total travel length of 190m. Its maximum speed is 5.08 m/s, while its acceleration is 1.22 m/s^2. a) Starting from rest, what distance does it take to reach maximum speed? b) How long does the complete stroke take, taking into account the acceleration and deceleration phases?

[desert solar]

how do you do point b?

[finite crown]

the time and distance it takes to accelerate from 0 to 5.08

is the same time and distance it takes to de-accelerate from 5.08 to 0

[desert solar]

Why?

[pale frost]

Point $a$ is $v(t) \approx 10,6m$

[desert solar]

Yes

[jolly parrot]

jandro

[pale frost]

Why?

Explain pls

<@&268886789983436800>

@proud wharf sent a random video Just tò troll

[finite crown]

the acceleration is 1.22 m/s^2 no?

that applies for both when it is acceelrating and de-acclerating

accelerating at 1 m/s^2 for 5 seconds gets you to 5 m/s, de-accelerating at 1 m/s^2 for 5 seconds at 5 m/s gets you to 0

[desert solar]

What if the deceleration had been different?

[finite crown]

Than yea

that doesnt work than

but it doesnt look like it is

[pale frost]

But in this case is the deceleration when the elevator is about to stop?

[finite crown]

Yes i would think so

[desert solar]

How are these problems addressed?

[finite crown]

you would just have to do two different calculations

instead of 1 and just doubling it

[desert solar]

?

[finite crown]

you would have to do 2 different calculations

if de-acceleration was different

[desert solar]

But if they are different, how do I know at what time it starts to decelerate?

[pale frost]

So the distance traveled during decceleration is the same as the distance traveled during acceleration?

[finite crown]

it travels 190 meters

[desert solar]

Yes

[finite crown]

assuming the acceleration starts from meter 0, to get it to full speed takes 4.16 seconds

in this time the elevator covers a dsitance of 10.6 meters

[pale frost]

Yes

Correct

Its point a

[finite crown]

Yes

[pale frost]

Yes now?

[finite crown]

you have about 180 meters remaining

[pale frost]

We know that t_acc = t_decc

[finite crown]

Yea

so take of 10.6 meters again

170 meters remaining

[pale frost]

Yes

[finite crown]

at which that time its travelling at 5.08 m/s

which takes 33 ish secodns

[pale frost]

Wait

V_max Is 5.08

[finite crown]

Ya

[pale frost]

However, it doesn't mean that it will go at this speed for the entire run

Right?

[finite crown]

It does not

[pale frost]

Ok

[finite crown]

at the start and the end it accelerates ot that speed

and thus is not going it

but it does for most the run

you have to make some assumptions here like it will always accelerate to maximum speed ect

so basically no one presse the lift button at any other floor

only top and botrtom floor

[pale frost]

Wait

We have 170 remaining

Now

[finite crown]

+- yea

[pale frost]

Do we need to do

170/ v_max?

[finite crown]

mhm

[pale frost]

I am blocked with this 170 m

[finite crown]

how so

[pale frost]

Idk

[pearl pond]

@desert solar Has your question been resolved?

[pale frost]

@desert solar

[desert solar]

What

[pale frost]

Click x

[desert solar]

Yes

[pale frost]

@finite crown

[finite crown]

it travels

170 meters

at a speed of 5.08 m/s

170 / 5.08 gets you time

its original 190 meters

but it takes 10 meters to acceelrate

and de-acceralte

so takeaway 20 from 190

[pale frost]

Yes

So t_tot = t_acc + t_decc + t(of now)

@finite crown

[desert solar]

I think yes

[pearl pond]

@desert solar Has your question been resolved?

[pearl pond]

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[brazen bluff]

i need to find the Transitive reflexive closure for this relation

{(0, 0), (0, 1), (0, 3), (0, 4), (1, 1), (2, 1), (2, 2), (2, 3), (3, 3), (4, 1), (4, 4)}

[brazen bluff]

I Wouldve said i add (1,0) (2,3) (4,1) (2,1)

[austere tinsel]

I'm a little confused, isn't that relation already transitive and reflexive?

At least for {0,1,2,3,4}

[warm current]

reflexive? yes

[warm current]

why (1, 0)?

[warm current]

(2, 3) is already in the relation

[pearl pond]

@brazen bluff Has your question been resolved?

[pearl pond]

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[pearl pond]

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[harsh fern]

!help

[pearl pond]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[harsh fern]

help

the khan solution

why do they multiply by 2

what am i doing wrong

i followed the guy steps

ok

what am i doing wrong i got the answer wrong

how do i simplify

idk how

[jolly parrot]

Kebesque Integral

[harsh fern]

no

i suck at m ath

so im doing it right but

its just not simplified so thats why its wrong

technically its right but its not simplified so they mark it as wrong?

not sure

i guess for here u multiply by 0.5

so it cancels out the 73/0.5

the 0.5 in the 73/0.5

well i already 100% algebra 1

and i am like 50% done with algebra 2

so u multiply by 0.5

wait nvm

cause if u do 73 * 0.5 its 30 something

man idk

dwasdfgawe

WAFASDAS

[jolly parrot]

faiyrose

[harsh fern]

OMFGASHDHASD

SCRAECH

[regal herald]

[harsh fern]

man

i dont understand

what am i doing wrong

[sterile plover]

why are you using ln for base 10

[sterile plover]

here

[harsh fern]

YES

I FIGURED IT OUT

I FIGURED OUT THE STRATEGY

WOOHOO

[pearl pond]

@harsh fern Has your question been resolved?

[harsh fern]

HELP

WHATS THE ANSWER

[heavy snow]

youre close

it seems you have two typos, one per answer

[harsh fern]

ok what is the typo

oh i see it

the x and the extra .

ugh am so stupid

[heavy snow]

yep exactly

[heavy snow]

noo

[harsh fern]

wtf

the answer is wrong

[heavy snow]

which one

[harsh fern]

i guess my answer is wrong somehow

[heavy snow]

is this the correct solution

[harsh fern]

5/5 The solution is:

is the correct solution

[heavy snow]

thats just pedantry

[harsh fern]

the screenshot above that is what i typed

and i got it wrong

[heavy snow]

anyway next time just use ln i guess

[harsh fern]

i hav eused

log many times before

and everytime it seemed like it was fine with it

theres something wrong with khan academy

you see minor inconveniences like this

always marks me wrong and i always have to restart

wasting my time

[heavy snow]

these things happen

nothing is perfect

[harsh fern]

absolutely unbelivable

ive been stuck on this same problem for like 2 hours now

why does khan academy not understand this..

lmao look if i type in log it understands

if i type in ln it doesnt

but ln was the correct answer and log wasnt

[heavy snow]

yes ln is defined to be log_e

[harsh fern]

khan academy has lost its mind

[heavy snow]

log and ln arent the same

$\ln(x) = \log_e(x)$

[jolly parrot]

esca (@ with reply)

[harsh fern]

ok whats the difference

[heavy snow]

see above

[harsh fern]

theres an = sign though

and = means same

[heavy snow]

right but look at the base

ln is log base e, not just log

sometimes people will write log to mean ln, but usually best practice is use ln for log base e

[harsh fern]

so

if it has eulers number

you have to use ln instead of log

[heavy snow]

yeah

[harsh fern]

and if its not eulers

you use log

[heavy snow]

yep

ln is just shorthand for log_e

[pearl pond]

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[untold frigate]

Hi! I've been stuck on this question for over an hour and honestly, I'm going insane right now. It would be very much appreciated if anyone could help me on this. Thanks!

[untold frigate]

ive tried using the similar triangles to find a formula for the variable radius of the cone but it says its the wrong answer

this is what i did

[warm current]

Okay, so it sounds like the first thing you need is to know the radius of the frustrum at a given height. Let h=0 mark the bottom base, and h=12 the top. Can you find r, given h?

[untold frigate]

i found the radius at a given height, yes

i took the height as x though

[warm current]

What did you get? I'm not seeing it on the paper

[untold frigate]

r = (24-x)/2

[warm current]

Oh so x is like depth below the tank, I see

[untold frigate]

i used the similar triangles to find the radius by d/(12-x) = 6/12 where d is the radius at any height x

after that i used the density of water and the area of a 2 dimensional slice of the frustum to integrate

[warm current]

Okay, so you define $x$ as the depth into the tank. And you define the radius as $h(x)=12-\frac{x}{2}$

[jolly parrot]

SWR

[warm current]

you should use r, not d, which can be confused with diameter

[untold frigate]

yep

[warm current]

So the area of at any height $x$ is $A(x)=\pi\left(12-\frac{x}{2}\right)^2$

[jolly parrot]

SWR

[untold frigate]

yep

the area of any circular slice

[warm current]

I do not see how you got here

[untold frigate]

so the question gave me 62.5lb/ft^3 as the density of water

[warm current]

What is this symbol?

[untold frigate]

r. sorry lol my handwriting is bad

the radius is 6+d

where 6 is the radius of the cylinder thing

[warm current]

Okay, so 62.5 is your volumetric density (let's call it $\rho$ for now).

[jolly parrot]

SWR

[warm current]

Now here's the the problem. What is $\pi [r(x)]^2dx$? It's your differential volume, $dV$. So what does $\int \rho dV$ represent? Use dimensional analysis if unsure

[jolly parrot]

SWR

[untold frigate]

i thought i could solve it by integrating the area function from 0 to 12 to find the volume, and then multiply it by the density

honestly im lost

[warm current]

What is volume x density?

[untold frigate]

mass

[warm current]

So $\int \rho dV$ is just giving you total mass. You want work

[jolly parrot]

SWR

[untold frigate]

yeah

sorry im really confused rn

[warm current]

So let's work backwards. You want work: $W$. So, let's think of $\int dW$. How can we define the differential work $dW$?

[jolly parrot]

SWR

[untold frigate]

F.S

where s is displacement

[warm current]

Okay, so $W=\int \vec{F}\cdot d\vec{s}$

[jolly parrot]

SWR

[untold frigate]

oh yeah

[warm current]

Now here's the fun bit, which force are we acting against?

[untold frigate]

gravity

[warm current]

yup

And which direction does force of gravity go?

[untold frigate]

downwards, or negative i think

[warm current]

oops it should be negative of the integral

[untold frigate]

oh

[warm current]

Let me actually skip a few things, are you familiar with W=mgh?

[untold frigate]

yes

potential energy i think

[warm current]

So, how can we define $dW$?

[jolly parrot]

SWR

[untold frigate]

potential energy per dH

[warm current]

close

per dm

[untold frigate]

oh

wouldnt mass be ocnstant?

[warm current]

no. Differential mass will be a function of height (x)

[untold frigate]

oh ok

so i would have to integrate the area of a circular cross section of the frustum right to find the volume

[warm current]

No. You want the differential volume to get the differential mass

[untold frigate]

right. but i have its density for that right

[warm current]

Yes

Still need help @untold frigate?

[pearl pond]

@untold frigate Has your question been resolved?

[pearl pond]

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[pearl pond]

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[stoic imp]

The digits of a four-digit positive integer add up to $14$. The sum of the two middle digits is nine, and the thousands digit minus the units digit is one. If the integer is divisible by $11$, what is the integer?

[jolly parrot]

studying_calc_real_analysis

[lost flax]

!status

[pearl pond]

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

[stoic imp]

1

well

a + b + c + @lost flax = 14

[lost flax]

And a - d = 1

[stoic imp]

ye

[lost flax]

And b + c = 9

[stoic imp]

right.

now what?

[lost flax]

So you have three equations and four variables

[stoic imp]

is this true tho?

a + b + c + d = 14

[lost flax]

The fourth condition is not an equation but it almost is

[lost flax]

Yes

[stoic imp]

1000a + 100b + 10c + d = 14

[lost flax]

No

a+b+c+d

[stoic imp]

?

[lost flax]

a+b+c+d is 14

1000a + 100b + 10c + d is not 14

[stoic imp]

oh my bad

11 | 1000a + 100b + 10c + d

I meant

[lost flax]

11

11 divides the number

[stoic imp]

wdym

[lost flax]

Not 14

[stoic imp]

I mean that

my bad.

[lost flax]

Well

Do you know the divisibility rule of 11?

[stoic imp]

the given number can only be completely divided by 11 if the difference of the sum of digits at odd position and sum of digits at even position in a number is 0 or 11.

@lost flax

@scarlet raptor

[lost flax]

Or -11
Or 22, et cetera

But yes, that is the idea

[stoic imp]

?

[lost flax]

For example

93819

[stoic imp]

what about it

[lost flax]

This number is divisible by 11

[stoic imp]

how so

[lost flax]

However, if you do the difference

9+8+9 = 26

3+1 = 4

26-4 = 22

Which is not 0 nor 11

It is a multiple of 11

[stoic imp]

can you explain for dummies

[lost flax]

It is exactly 11*8529

[lost flax]

The true rule of divisibility of 11 is the following:

[stoic imp]

,calc 11 * 8529

[jolly parrot]

Result:

93819

[lost flax]

The given number can only be completely divided by 11 if the difference of the sum of digits at odd position and sum of digits at even position in a number is 0, 11, 22, 33, 44... or -11, -22, -33,...

[stoic imp]

a multiple of 11

[lost flax]

Yes

[stoic imp]

what about it

[lost flax]

^

[lost flax]

^

[lost flax]

Not this

[stoic imp]

11 | 1000a + 100b + 10c + d

ok how do I continue

[lost flax]

Then, because of the rule I told you, 11 | a - b + c - d too

[stoic imp]

oh yeah

so what?

[lost flax]

So now you have four equations

Solve

[stoic imp]

which equations

a + b + c + d = 14

a - d = 1

[lost flax]

a+b+c+d = 14
a-b+c-d = 0, or 11, or -11, ...
b+c = 9
a -d = 1

[stoic imp]

"a-b+c-d = 0, or 11, or -11, ..."

[lost flax]

The second one is actually not an equation

[stoic imp]

how is that an equation

??????????????????

[lost flax]

But an infinite number of equations

You have to solve every system of equations

[stoic imp]

how to solve a system of infinity-equations

I dont get that part

[lost flax]

You have to see that there is no solution for most values

For example

[stoic imp]

can you elaborate on this?

[lost flax]

Yes

a-d=1

You can substitute that

In a-b+c-d = 0, or 11, or -11, ...

And obtain -b+c+1 = 0, or 11, or -11, ...

[stoic imp]

1 - b + c = 0 or 1 or -11, . . .

so what?

[lost flax]

And then substitute b+c = 9

Add it

-b cancels with b

[stoic imp]

I cant

no

[lost flax]

Why?

[stoic imp]

$1 - b + c = 0$ "or" $=11$ "or" $=-11$...

[lost flax]

Yes but use the other one instead

b+c = 9 --> b = 9-c

[jolly parrot]

studying_calc_real_analysis

[stoic imp]

I cant

[lost flax]

So -b+c+1 = 0, or 11, or -11, ... --> -(9-c) + c + 1 = 0, or 11, or -11,...

[lost flax]

Do you agree?

@stoic imp ?

[stoic imp]

no

I dont get it

b = 9 - c?

[lost flax]

We knew that from before

Before the infinite equations

[lost flax]

The sum of the two middle digits is nine

b+c=9

b=9-c

[stoic imp]

okay

[lost flax]

So you substitute that b

[stoic imp]

ye

[lost flax]

Here -b+c+1 = 0, or 11, or -11, ...

[stoic imp]

what then?

[lost flax]

-(9-c) + c + 1 = 0, or 11, or -11,...

[stoic imp]

==> -(9-c) + c + 1 = 0, or 11, or -11,..

[lost flax]

2c - 8 = 0, or 11, or -11,...

Now, c is a digit

[stoic imp]

==> 2c - 8 = 0, or 11, or -11,...

[lost flax]

So it is between 0 and 9

[stoic imp]

oh yeah, c is digit.

[lost flax]

So for example if I choose 2c - 8 = 1100

That makes no sense

And the same applies for almost all possible equations

[stoic imp]

okay

[lost flax]

Except for one

[stoic imp]

ok

[lost flax]

Can you see which one?

[stoic imp]

no

0

[lost flax]

Well, if you choose -11, or -22, or something negative

You have 2c - 8 = that

2c = 8 + that

And (8 + that) will be a negative number

Do you agree?

[stoic imp]

so what?

[lost flax]

8-11 = -3
8-22 = -14

[stoic imp]

I know

[lost flax]

So these are impossible

c is between 0 and 9

[stoic imp]

okay

[lost flax]

It can't be negative

[stoic imp]

right.

[lost flax]

In a similar way

If you choose 22 or something larger

8 + the number will be at least 30

[stoic imp]

okay

[lost flax]

2c = something that is at least 30

c = something that is at least 15

Impossible again

Only two possibilities are left

2c-8=0
2c-8=11

2c=8

[stoic imp]

okay

[lost flax]

2c=19

[lost flax]

And this one gives c=9.5 which is absurd again

[stoic imp]

why?

[lost flax]

Because c is a digit

[stoic imp]

so only 0 is good

[lost flax]

Yes

[stoic imp]

like I was saying 15 min ago

but was due to pure luck

[lost flax]

^

[stoic imp]

so what?

[lost flax]

So 2c-8 = 0

c = 4

[stoic imp]

c = 4

right.

[lost flax]

And you can solve for the rest of digits easily

[stoic imp]

????

[lost flax]

^

Use the equations

[stoic imp]

a+b+c+d = 14

c = 4

b+c = 9

a-d = 1

[lost flax]

Yes, that is very easy to solve

[stoic imp]

b = 5

c = 4

a + d = 5

a - d = 1

a = 3

d = 2

[lost flax]

How can you make the text to look like this?

[stoic imp]

# and space

[lost flax]

text

Interesting

[stoic imp]

3542

[lost flax]

Yes

[stoic imp]

okay, thanks, I am closing

.close

[pearl pond]

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[unreal ridge]

Help with partial factoring

I don’t fully understand how to do it

Like explain how to do it

[pearl pond]

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[dusk nimbus]

How’d they find m in this equation?

[vital crescent]

geometrically speaking, and as in the solution, no solution implies distinct, parallel lines. and infinite many solutions would imply that they are the same lines

what can you say about the gradient if two lines are parallel?

[dusk nimbus]

The same

Wait are they just solving both equations for y and equating them to each other to find m?

[vital crescent]

so they made an equation that is "gradient of 1=gradient of 2"

and solved for m

[vital crescent]

i dont quite understand what you mean

[dusk nimbus]

Like for example 2m + 3x= y and 6m - 4x = y so 2m +3x=6m -4x to solve for m

[vital crescent]

in this case the lines will never be parallel since the gradient of the two lines are distinct

[dusk nimbus]

Ok like for example is that what they’re doing to solve for m

[vital crescent]

yes, because m is used to find the gradient

the logic is as follows:
If the lines are parallel, then the gradient are the same,
the gradient are 2-m and -m/2, and they must be the same when the lines are parallel

hence if the lines are parallel: 2-m=-m/2

is this part clear?

[pearl pond]

@dusk nimbus Has your question been resolved?

[pearl pond]

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[dapper kraken]

how can i split this up?

[toxic pebble]

Oh u use AoPS

[dapper kraken]

8x^3+2x+2x+2x+2x+2x+1/4x^2+1/4x^2+1/4x^2+1/4x^2

can this work

wth is your problem man

[dapper kraken]

i got it to the minimum value of $10\sqrt[10]{\frac{1}{8}}$

[jolly parrot]

Skill_Issue

[toxic pebble]

lol is grapher allowed

[dapper kraken]

dunno

[toxic pebble]

buy one then

[dapper kraken]

what

[toxic pebble]

buy a graphing calculator

[dapper kraken]

for??

[rough stream]

If you put them over 3, this is pretty easy to see as the arithmetic mean of the 3 terms. That get anywhere?

<@&268886789983436800>

[acoustic path]

is it supposed to

[vital crescent]

it works, minimal value can be achieved by solving 8x^3=2x=1/4x^2

[dapper kraken]

thanks yaal

.close

[pearl pond]

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[next barn]

With this diagonal matrix do I need to do something to show that I can calculate the norm as if it were a vector?

[next barn]

Since there's only values on the diagonal, I can treat it like a vector

[pearl pond]

@next barn Has your question been resolved?

[pearl pond]

@next barn Has your question been resolved?

[next barn]

.close

[pearl pond]

Channel closed

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Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[visual canyon]

what is going on with those symbols...

[bitter epoch]

Which symbols? If you mean the Greek letters, then $\mu$ is the mean, and $\sigma^{2}$ is the variance, $\sigma$ is the standard deviation

[jolly parrot]

X13warzone

[visual canyon]

ohhhhhh

yep thats what i mean

[bitter epoch]

No worries then, you know how to solve the problem now?

[visual canyon]

if the weird U is mean then why do they ask for expected value...

[bitter epoch]

It's been a while since I did stats, but usually expected value, mean, mu mean the same thing

[visual canyon]

the expected value formula is wild tho

[bitter epoch]

It's just a convention that mathematicians use only symbols in the more mathy sections