help-33

[still crow]

integrating r^2 wrt x is not r^2

[agile kiln]

Oh

OH, is it r^2x?

[still crow]

yea

[agile kiln]

Okay lemme work that out

That worked thank you

.close

[marsh citrus]

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[dense igloo]

this is for the roots of a cubic equation, where did the negative numbers come from??

[dense igloo]

its been too long since ive done these i forgot everything 😭

[fathom ridge]

it's just to get the maximum possible number of solutions

since you can have two negative roots which multiply to a positive root

[dense igloo]

oh i see

is the list just a list of all the factors

cuz it doesnt look like theyre factors of a/factors of b??

confused\

[fathom ridge]

yeah possible integer factors

[thorn gale]

factors of the d value

the last value

6

[dense igloo]

oh

[thorn gale]

the one with no x

[dense igloo]

no need to do a?

factors of a

[thorn gale]

i don't believe so

pretty sure you don't but i may be wrong

[lone heart]

You need factors of a

[dense igloo]

uhhhh

this is the example im doing

$x^3 + x^2 - 4x + 2 = 0$

[elfin berry]

risa★

[lone heart]

In that case, a = 1

[dense igloo]

yes

and factors of 2 are just 1 and 2

so the list would be -1 1 -2 2?

[lone heart]

Yes

[dense igloo]

ok

and then i plug in each

to see which ones work

is that all?

[thorn gale]

it depends

[dense igloo]

aw man

[thorn gale]

if you're asked to factor fully then no

[dense igloo]

its compute all roots

[thorn gale]

after that you can do long or synthetic division with the factor that makes the equation equal to 0

then you get the solution from that, which would be second degree, and factor that

[dense igloo]

aa

[thorn gale]

unless the second degree has no solutions

[dense igloo]

i dont remember any of this 😭

i will look up the diviaion stuff

[thorn gale]

synthetic division is super easy

[dense igloo]

is it

good

[thorn gale]

it's pretty simple

and useful

[thorn gale]

especially if you're factoring simple equations like this

[dense igloo]

ok i plugged in and only x=1 works

[thorn gale]

so first you find the value which makes x = 0, which you already found

ok

so now

[thorn gale]

these are the coefficients

[dense igloo]

Mhm

Yes

[thorn gale]

from the equation from before

and on the left is ur factor

[lone heart]

Do you know polynomial long division?

[dense igloo]

no ;-;

i forgot

[thorn gale]

yeah you can do polynomial long division too

[dense igloo]

yes

[thorn gale]

cause sometimes synthetic divsion won't work

polynomial long divison is 100% more reliable

[lone heart]

Synthetic division is just a shortcut to long division

[thorn gale]

but this is how ou do synthetic division

you just add the number, multiply it by the factor

[dense igloo]

do i start on the right side

[thorn gale]

yes

[dense igloo]

okay

[thorn gale]

1 + 0 is 0

i mean 1

lol

[dense igloo]

lol

[thorn gale]

and then multiply that by 1

here i'll show you step by step instead

[dense igloo]

tyyy

[thorn gale]

there's nothing here so we just add

[lone heart]

https://www.youtube.com/watch?v=FxHWoUOq2iQ&ab_channel=TheOrganicChemistryTutor

[thorn gale]

yes thank you

[lone heart]

That should help

[dense igloo]

oop thanks

[thorn gale]

yeah watch that instead

of listening to me

[dense igloo]

Got it :0

[thorn gale]

nice

[dense igloo]

so that means x=1 is a solution indeed

is that it

oh no

theres 2 more

ahh

[dense igloo]

theres 2 more roots still

$1\pm\sqrt{3}$

[elfin berry]

risa★

[dense igloo]

do u know how this was found

[thorn gale]

probably quadratic formula

[dense igloo]

uh

[thorn gale]

this gives you two solutions, you solve one when -b + sqrt... and the other -b -sqrt...

[dense igloo]

oh i can do that

[dense igloo]

but where do i get the quadratic equation from?

oh is it the x^2+2x-2

got ittttt

tysm

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[spare peak]

hello

can someone help me with this question?

.close

[marsh citrus]

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[marsh citrus]

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[stiff whale]

can someone confirm if i got this right? "If x is in the 2nd quadrant with sinx = 4/5 and y is in the 3rd quadrant with secy = -13/5, determine sin(y-x)"

[stiff whale]

i got sin(12-3)

but i forgot to take a picture of the answer key so i couldnt confirm 😭

oh wait nvm

.close

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[still temple]

Hi.. I need to come up with a defination or equation for the following:

1. There are n objects on L levels and each objects needs to be assigned with a weight wi.

2. Each level can have different number of objects. Let's say the number of objects on level j is pj

3. The objects on the same level will have same weight value.

4. Each object in a lower level need to have large weight value than a higher level.
   Level 1 is the lowest level and level L is the highest level.
   So object in level 1 need to have a large weight value than level 2.

5. The sum of weights of all objects of level 2 need to less than the weight of an individual object on level 1.
   Similarly for level 3, 2 and so on

I need to find a equation to define this constraints. Please someone help me with this.

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[mild tendon]

How do you graph log2(8-2x) by chance?

[mild tendon]

I know you need to simplify although im not too sure where to start due to the 2x

[vernal forge]

try factoring out 2 inside the log

[sage arrow]

whats 2+2

[marsh citrus]

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).

[mild tendon]

Oh!!! Like log2(4-x)?

[sage arrow]

!occupied

[marsh citrus]

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).

[vernal forge]

kinda but not really

[mild tendon]

Log2(2(4-x)?

[vernal forge]

$\log_2(8-2x) = \log_2(2(4-2x))$

[vernal forge]

yes that

[mild tendon]

i see

[night mica]

wrong pic

[vernal forge]

bro's screenshotting my listening history

lol

[mild tendon]

[night mica]

yes you got me

[vernal forge]

now use the fact that log(ab) = log(a)+log(b)

[night mica]

i think you listen very good music

[vernal forge]

thank you

[night mica]

and you are only a cuttlefish

[mild tendon]

Would you do log2(4-2x) + log2(4-2x)?

[vernal forge]

no

[mild tendon]

Ill try again

[vernal forge]

pick a=2, b=4-x

[cloud kettle]

2*2=2^2

[mild tendon]

I see

[vernal forge]

oops i made a typo

$\log_2(8-2x) = \log_2(2(4-x))$

[elfin berry]

artemetra

[vernal forge]

like that

my bad

[cloud kettle]

I thought base was 10

Ok

[mild tendon]

Oh!!! Its my bad as well, i forgot about the multiplication factor

[mild tendon]

In that case log2(2) + log(4-x)?

[vernal forge]

yes

now log2(2) simplifies

[mild tendon]

1?

[vernal forge]

yep

[cloud kettle]

Yes

[mild tendon]

[vernal forge]

so now you have to plog log(4-x)+1

can't really simplify from there

[cloud kettle]

Or just plot log(4-x) and shift the graph by 1 unit upward on y axis

[vernal forge]

yep

[mild tendon]

okay, thank you guys so much!

I really appreciate the help

Would the graph start at the base #?

[cloud kettle]

What is base

[mild tendon]

Log2

[cloud kettle]

I didn't understand what you mean

[vernal forge]

same

[mild tendon]

Dont worry, i phrased in a bad way

[cloud kettle]

The graph will be a decreasing graph btw

[mild tendon]

I understand better now i think, ill make a chart before graph

[mild tendon]

keeping in mind

[cloud kettle]

Ok

[mild tendon]

.close

[marsh citrus]

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[lime cedar]

can i get some help?

[marsh citrus]

Please don't occupy multiple help channels.

[night mica]

.close

[marsh citrus]

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[marsh citrus]

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[still temple]

is the angle in quadrant 4?

for 19.b

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

<@&286206848099549185>

[dreamy shore]

I think the angle x is in the 3rd quadrant

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

can you explain why

cus my teachers answer said quadrant 2 but why is not 4?

[marsh citrus]

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[sand fable]

it is roughly half of that

[still temple]

.reopen

[still temple]

<@&286206848099549185>

can u help witht his question

[marsh citrus]

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).

[coral flame]

sorry but thats wrongg

@limpid ruin

[still temple]

man i was in this channel bruh

[coral flame]

area of Regular octagon ( A=2(1√+2)a2)

yes

[coral flame]

x is second quadrant

[marsh citrus]

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[limpid ruin]

can you reopen it now?
I'll also delete my messages

[coral flame]

@still temple

[still temple]

can you explain why

[marsh citrus]

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[coral flame]

Since the angle 2x is in the fourth quadrant and cos 2x is positive, we know that both the cosine and sine of 2x are positive. This means that angle 2x is located in the fourth quadrant.

[coral flame]

see where it lies

@still temple

[still temple]

wouldnt sin2x = 15/17 OH so its positive

cosine2x=8/17 also positive

so wouldnt it be quadrant 1

cus thats where they are both positive

[coral flame]

no wait let me show you smth

2x=cos raise to -1 into 8/17

when you solve this it approximates to

2x=61.93

and since 2x lies in the 4th quadrant

2x=360-61.93

x=149.04

did you undertsnad @still temple

[still temple]

yeah kind of what ur saying makes sense

why did u raise to the power of -1 tho?

[coral flame]

oh sorry

dont know why i wrote that

i think it was in my previous clipboard or smth

If 2x is a little less than 300, but it's certainly bigger than 270 (indeed cos270=0) so 270<2x<300, which implies 135

[coral flame]

this is also a way to llok at it

look8

look*

[still temple]

how do you know these values btw

[coral flame]

solved i have a log book

[still temple]

ok thanks

.close

[coral flame]

.close

[marsh citrus]

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[marsh citrus]

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[surreal totem]

How would I get the u out of the ln in this case?

Would it be something like this?

I've kinda gotten used to automatically solving these that I forgot the actual steps lol

then the RHS would be like this

and then finally like this?

[desert dirge]

you go wrong here

$$ln\left(e^{e^y}\right)+ln\left(e^{C_2}\right)=ln\left(e^{e^y+C_2}\right)$$

[elfin berry]

AℤØ

[surreal totem]

well yeah but C2 is a constant

it's like...

it's like the C you get from integrals

like dy=dx
y = x + C

so e^C = C2

[desert dirge]

ah i see, wasnt clear you were redefining the constant

[surreal totem]

Yeah yeah

My bad

[desert dirge]

in that case, yeah i think youre fine

[surreal totem]

So the steps I would take are to add the ln(e^x) function to everything

and since ln(e^x) = x, I didn't change anything

[desert dirge]

would probably be faster to directly do e^{lhs}=e^{rhs}

would simplify in one step

[surreal totem]

Alright, thanks.

.close

[marsh citrus]

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[marsh citrus]

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[pseudo flame]

How do you find all permutations x of S7 with the property that x^4 = empty set?

[devout mauve]

well x^4=empty set makes no sense

(permutation) = (set) is nonsense

I assume that you mean x^4=id ?

[pseudo flame]

Oh, I'm sorry. That's a sigma apparently. Then x^4 = sigma.
Where sigma is the following permutation:

(1 2 3 4 5 6 7)
(2 5 7 4 6 1 3)

Or generally, how to solve this kind of questions?

[devout mauve]

you can first think about what the cycle structure of x could be

[marsh citrus]

@pseudo flame Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[tame pawn]

Pls could I have some help answering these questions?

[tame pawn]

is the answer to i) 1?

just {1, 1}

[devout mauve]

why not zero?

[marsh citrus]

@tame pawn Has your question been resolved?

[tame pawn]

but I thought I can only choose a number from 1 to n

<@&286206848099549185>

[dull cloak]

Yo

[craggy trench]

These are the elements of B. R is just a subset of B x B. The empty set is a subset of size 0

[tame pawn]

ah

ok, and for ii), there isn't a maxmimum. Are they asking for a formula in terms of n?

[craggy trench]

I’d assume they’re asking for a formula

[tame pawn]

Would letting n = 1, 2 and seeing how many elements there are in the subset of R x R be a good strategy here

trying to find a pattern

[devout mauve]

yes

[tame pawn]

cool, lemme give that a try

so if n = 1

B = 1, B x B = {1, 1}

R would contain {{∅}, {1, ∅}, {∅, 1}, {1, 1}}?

bit confused here

[spark siren]

is ∅ an element of B?

[tame pawn]

no

[spark siren]

so why should {∅,1} be in R?

[tame pawn]

What exactly is R

[spark siren]

a subset of B x B.

[tame pawn]

ok, so I'm trying to find the maximum number of elements a subset of B x B could contain when n = 1

[spark siren]

each element of B x B is something (x,y) where x is in B and y is in B. How many such pairs can you build?

[tame pawn]

1?

[devout mauve]

BxB is {(1,1)}, not {1,1}

big difference

[tame pawn]

ah

[devout mauve]

and B should be {1} btw

[tame pawn]

please could you give me a similar explanation as to what R contains

[devout mauve]

R is a subset of that

[spark siren]

its the same explanation-

[tame pawn]

so R is {∅, {1, 1}}?

[devout mauve]

no

[spark siren]

no.

[devout mauve]

neither of those are elements of BxB

BxB has the single element (1,1)

R can either have that or not

so either R=empty set or R={(1,1)}

those are the only options in this case

[tame pawn]

so the max length of R when n = 1 is 1 where R = {(1, 1)}?

[devout mauve]

yes

[tame pawn]

does it matter if we use { or ( for the pairs?

[devout mauve]

yes

[tame pawn]

ok

so When n = 2

B = {1, 2}

B x B = {(1, 1), (1, 2), (2, 1), (2, 2)}?

so is the max length of R 4?

[devout mauve]

size and not length

but yes

[tame pawn]

cool

so would the formula be n^2?

B x B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

when n = 3

and the size of this is 9

[devout mauve]

yes

[tame pawn]

perfect

thanks a lot

.close

[marsh citrus]

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[shut latch]

How can I simplify this expression

[spark siren]

write $\sqrt{3}+1$ as $\sqrt{3}-1+2$

[elfin berry]

ThM

[spark siren]

then use a^(b+c) = a^b x a^c

then use a^c x b^c = (ab)^c

x means multiplication.

[marsh citrus]

@shut latch Has your question been resolved?

[shut latch]

.close

[marsh citrus]

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[craggy trench]

Do you know what 8x+x equals in radians?

[sick sedge]

https://www.cuemath.com/geometry/180-degrees-to-radians/

[idle hinge]

no

how did you know to add?

i know how to convert to radians

but how did you know to add?

[sick sedge]

do you know how to find the answer in degrees

[idle hinge]

i know the degree formula which is 180/pi

[sick sedge]

yes or no question

can you find the answer in degrees

[idle hinge]

i don’t know what the question is telling me

[lone heart]

You're solving for x first

[idle hinge]

how am i suppose to get those numbers

[sick sedge]

wants you to find the angle of theta

[idle hinge]

like x = 8x?

[lone heart]

You should know that they are supplementary angles

[lone heart]

You can use that to find the value of x

[idle hinge]

i don’t know what are supplementary angles either

i think i forgot them

this is what i hate i have such a bad memory

[lone heart]

Google it if you forget things

[idle hinge]

and i don’t have much time

[sick sedge]

ill leave this in the hands of captainnova

[idle hinge]

thank you anim

[lone heart]

Google is a useful resource

[idle hinge]

yes it truly is

alright i just searched it

supplementary is when a line equals 180

two angles

[lone heart]

So it's two angles that sum to 180, as hint stated, that problem, you can find x using the concept of supplementary angles

Can you write what the equation would be?

[idle hinge]

i see

180 = 8x + x?

[lone heart]

Exactly yes

Can you solve for x from there?

[idle hinge]

s 20

[lone heart]

Yes

[idle hinge]

which is 20 degrees

and 1/9pi

[lone heart]

Therefore theta is how many degrees?

[idle hinge]

20 degrees?

[lone heart]

Theta as pointed out, is this part

[lone heart]

No, that's the value of x

[idle hinge]

160

[lone heart]

Good that the value of theta in degrees

Now convert that to radians

[idle hinge]

8/9pi

[lone heart]

(8/9)pi or 8/(9pi)?

[idle hinge]

(8/9)pi

[lone heart]

Good

[idle hinge]

captain nova

i need help with this one aswell

do i do 90/3?

since each box is 90

so 210?

[lone heart]

As theta?

[idle hinge]

yes

[lone heart]

Yes that's correct

[idle hinge]

i see what does congruent mean?

i get mixed up with the other one

congruence and the other one i forgot the name

[lone heart]

Remember what I said? Google is useful

[lone heart]

I'm assuming "similar"

[idle hinge]

i think so

how did you get so smart captain nova

and how do you retain

my time is always limited

[lone heart]

Learning and researching

[idle hinge]

i see

how much math do you know?

is it like a muscle memory to you know?

what resources do you use?

[lone heart]

Google is a big one

[lone heart]

If you do it enough times, you retain that info

[idle hinge]

i see

do i search google for worksheets and do that to retain?

[lone heart]

This is good info too

[idle hinge]

i see

[lone heart]

If you can learn it and then teach to others, then it helps you retain the info better

[idle hinge]

hm i see

i will need to get super smart so my teacher can get off my butt

she barely helps us

[lone heart]

You don't need to get super smart, it's just how you learn and retain that info

People "look" smart because they can retain and research the info well

[idle hinge]

yes i need to start getting better at retaining

thanks @lone heart

[marsh citrus]

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[cerulean oxide]

if i have a linear transformation from T : R^4 -> R^4

[cerulean oxide]

then i find the relative standard base for R^4

i find it to this

now im asked to find the dimension of the im(T)

for a finitely (R^4, R^3 etc) im(T)=rank(T) = dim(V) - nullity(T)

if you solve for the nullity, you'll end up with this

now i have two questions

why is the nulity of T = 1 when there is a 4 dimensional vector

and two: why is the dimension of V equal to 4 instead of 3, when you can RREF it and find the last row is useless

[marsh citrus]

@cerulean oxide Has your question been resolved?

[static quarry]

V is R^4, it is the "input" vector space (domain), its dimension is 4 if the matrix is 4x4, regardless of the matrix entries

[marsh citrus]

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[still temple]

ya

[still temple]

ya

subtract 24ab - 10b-18a from 30ab +12b +14a

[hard gull]

rewrite that as an expression and simplify
30ab +12b +14a - (24ab - 10b-18a)

[still temple]

ok

thkna

.close

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[severe island]

hi, bit of a dumb question

[severe island]

would $A_0$ be $f\qty(\frac {x_a + x_b}{2})$

[elfin berry]

kanna

[past maple]

Not necessarily

$A_0$ is $f(x_0)$

[elfin berry]

d

[severe island]

and B = (f(x_a) + f(x_b))/2

[severe island]

no let's asssume it's the midpoint

x_0 is the midpoint

[past maple]

But what are x_a and x_b?

[severe island]

i was looking for a graphical representation of this

this is a property of a convex function

[past maple]

Are they x_1 and x_2?

[severe island]

yes

[severe island]

this is the context

[past maple]

In that case, yes

[severe island]

would the left hand side be A_0 and right hand side be B?

[past maple]

Yes

[severe island]

okay fair enough, thanks

[severe island]

oh lol i didn't see the that the diagram already labeled a and b

probably why you were confused

[past maple]

Yes

[marsh citrus]

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[mortal badger]

okay so it is very probably elementary but i cannot seem to find the obvious proof as to why the row space of A and gA are the same (where g is an invertible matrix )

[unborn condor]

g being invertible means that it's equivalent to a product of elementary matrices, which means that A and gA are equivalent by rows

[marsh citrus]

@mortal badger Has your question been resolved?

[mortal badger]

oh, so every invertible matrix is the product of the 3 elementary matrices?

ah wait now that i think about it, does that follow directly from the algorithm of gaussian elimination

[marsh citrus]

@mortal badger Has your question been resolved?

[mortal badger]

anyone answer so i can stop procrastinating

[past maple]

It can be written as a product of elementary matrices

Not necessarily 3, it can be another number

[past maple]

Yes

[past maple]

So this follows from that fact. Elementary matrices do not change the row space (or the column space, if you do Ag instead of gA)

[marsh citrus]

@mortal badger Has your question been resolved?

[mortal badger]

ab thanks a lot

[marsh citrus]

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[chilly plume]

How do I find the limit of the sequence an=nsin(3/n)

[lucid zenith]

just find the limit of the expression inside as n tends to infinity

[chilly plume]

Why is there no difference between that and a regular limit?

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[tulip marsh]

can someone verify my work?

[tulip marsh]

<@&286206848099549185>

<@&286206848099549185>

[flat raft]

F(-5) is not what you claim it to be; it is negative, so it cannot be the maximum (since you know F(0) = 0)

also 9pi/4 + 3 is definitely not 21 pi/4

[tulip marsh]

how would i add the 3 then?

wait, it can not be the maximum?

[flat raft]

no, because maximum is the highest F(x) goes, and F(-5) is less than zero, and you know that F(0) = 0 is higher by virtue of that

and as for adding the 3, try writing out the individual steps for adding fractions

[tulip marsh]

got it

[tulip marsh]

but im not sure i follow this

because we are finding the area under the curve, and in the interval 0 to -5 the area seems the largest

are you saying becasue the area is a negative # it is not the largest?

[flat raft]

I'm saying that F(-5) cannot be the maximum because we know that F(0) is bigger than it

[tulip marsh]

if we are talking bigger than F(0) and still largest then it would be F(2)

[flat raft]

F(2) is indeed the maximum

[tulip marsh]

i supppose i was fooled when it gave the hint

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[idle hinge]

is this correct

[idle hinge]

is this correct?

[desert dirge]

how did you get 105?

[idle hinge]

i did 14/8 * 60

[desert dirge]

nearly

just be careful

[idle hinge]

was it correct?

[desert dirge]

depends how youre bracketting that

actually no

the period would be total time/ number of cycles right

[idle hinge]

yes

i want to convert to seconds

[desert dirge]

thats all well and good

but why 14/8

[idle hinge]

is it .02?

8*60

480

then 14/480

[desert dirge]

no

[desert dirge]

youre doing it the wrong way

[desert dirge]

youre doing the opposite

[idle hinge]

what is the formula called

i thought it was cycles/time

[desert dirge]

time/cycles

cycles/time is frequency

[idle hinge]

hm i see

so how do i do this

[desert dirge]

just flip your fraction

480/14

[idle hinge]

oh i see alright

so 34?

what does 34 represent?

1 cycle every 34 sec?

[desert dirge]

the 34.3 is the time it takes for one cycle to be completed

ie the time period

[desert dirge]

yeah

[idle hinge]

i also need to know when to see if something is periodic

i am not understanding it

for 3 and 4

.flip

(╯°□°)╯︵ ┻━┻

[desert dirge]

,rccw

[idle hinge]

Accidental

[desert dirge]

a function is periodic if it repeats itself after some time

[idle hinge]

i see

[desert dirge]

do you think 3 is?

[idle hinge]

no

[desert dirge]

id agree

what about 4

[idle hinge]

yes

can i put yes for 4 and say it is following a trend)

?

[desert dirge]

from the little thats shown i think its fair enough to assume 4 is periodic yeah

[desert dirge]

might wanna be more specific

[idle hinge]

what else can i add?

[marsh citrus]

@idle hinge Has your question been resolved?

[marsh citrus]

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[glossy flint]

B

Because that 1+x² is in the denominator

I've never done such things, looks like economics

[marsh citrus]

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[sharp aurora]

Help please?

[marsh citrus]

@sharp aurora Has your question been resolved?

[main idol]

scroll to projections
https://tutorial.math.lamar.edu/classes/calcii/DotProduct.aspx

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[zinc scroll]

if i ran a game with 3 possible outcomes and I have the data/outcomes of 55 trials, how do i calculate what the probability was for the specific outcomes

[zinc scroll]

i know I use binomial distribution for the individual cases but how do i calculate it when i want the probability of the whole 55 trials outcomes?

do i use hypergeometric?

Outcome A: 6 ppl with a theoretical of 1/13
outcome B: 24 ppl with a theoretical of 6/13

outcome C: 25 ppl with a theoretical of 6/13

[marsh citrus]

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[zinc scroll]

<@&286206848099549185>

<@&286206848099549185>

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[bleak quartz]

Why would the reliability not be t^4?

[quaint elm]

well because imagine the top left one broke

the system still works because it can use the ones on the bottom instead

[bleak quartz]

but is the total reliability not dependant on both branches working?

[quaint elm]

no

[bleak quartz]

oh ok

ty

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[cunning heath]

how would u start

[marsh citrus]

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

[crystal wraith]

Don't you think "how would you start" is pretty self explanatory?

You sure aren't lazy talking unnecessarily.

[crystal wraith]

Have you ever done limits which deal with 1^infty before?

[cunning heath]

yep

so u do e^y , you argument that e^y is continous

and you rewrite it

i will send a screenshot

@crystal wraith

@still temple

[crystal wraith]

$\sqrt[x]{2} = 2^{\f{1}{x}}$

[elfin berry]

! What the hell am I doing here?

[crystal wraith]

Use series expansion.

[cunning heath]

what is series expansion?

[crystal wraith]

Ah.

[cunning heath]

taylor?

[crystal wraith]

Yes.

[cunning heath]

we should not use that, do you see a way other than taylor?

[serene basalt]

You tried put it in power of e it is 1^infinity form

[cunning heath]

thats what i did

[cunning heath]

this is the e^y form

this is 0 . infinity

[crystal wraith]

Fate.

You can probably use lhopital other than series expansion?
It's 0 * infty. Rewrite that x in the numerator as 1/(1/x) so that it becomes zero?

0/0

[cunning heath]

Yea, that is possible

do you think, splitting it into 2 fractions and using a^n - b^n formula will help?

@crystal wraith

[crystal wraith]

What formula?

[cunning heath]

a^n - b^n = ......

a^n - b^n = (a-b) (a^n-1 + a^n-2 . b + ....)

[crystal wraith]

That works for integer n.

you have 1/x which is like 0

[serene basalt]

You just have to write it in power of e then l'hospital

[stoic slate]

This answer is correct, but I would have left a bit more time to him to work on this

\text{If } $\lim_{{x \to A}} f(x)^{g(x)} = 1^\infty$ \text{, then this limit will be equal to } $\lim_{{x \to A}} e^{g(x) \cdot (f(x) - 1)}$

[elfin berry]

Samuel

[gleaming pecan]

this limit can be calculated using the relations between the arithmetic and geometric mean, from the monotonicity of the logarithmic function, and from the standard inequality for logarithms, namely:

this, of course, can also be generalized

[cunning heath]

like

,w limit as n approaches infinity of ((sqrt(2) + sqrt(3))/2)^n

[elfin berry]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by fate8967

[gleaming pecan]

and next other way of solvign the problem is:

also elementary solution, based on basic limits, not derivatives

[cunning heath]

Oh yeah

This is the one

[gleaming pecan]

but the most elementary is the first one i have shown above

[marsh citrus]

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[crystal wraith]

One at a time, which one are you doing first?

@dense epoch

[stuck solar]

its the same question

[crystal wraith]

lol is it? My bad, too ignorant of me if that's the case.

Then,

!status

[marsh citrus]

What step are you on?
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2. I have begun but got stuck midway.
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6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@dense epoch Has your question been resolved?

[dense epoch]

1

1st image is b second is a

[marsh citrus]

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[wise oxide]

Bacteria starts at 900. After one hour it’s 1000. How much hours till 5000? I did 5000=900(9/10)^t. Ln50/9/ln(9/10)=t

[wise oxide]

Did I do it right

[lost bane]

HELP PLS

draw the parallelogram ABCD in the figure; O is the intersection point of its diagonals. The point K is outside the plane of the parallelogram. KA=KC and KB=KD. Prove that KO is perpendicular to (ABCD).

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[rapid hound]

The vector field 𝐏(𝑥, 𝑦, 𝑧) = (4𝒚 + 𝐴𝒛) exp[−𝑗(𝐵𝒚+ − 5𝒛) ⋅ 𝒓] satisfies the differential
equations: 𝛁 ⋅ 𝐏 = 0 and ∇ × ∇ × 𝐏 = 9𝐏 (𝒓 = 𝒙𝑥 + 𝒚𝑦 + 𝒛*𝑧) is the position vector in the
rectangular coordinates). Determine the constants 𝐴 and 𝐵.

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[green geyser]

Hey people, I am still sort of struggling with combinatorics. Things have gone better lately, but it's just this one thing.

Now, the thing I am struggling with is, probabilities.

I understand that you should calculate the mentioned possibilities divided by all possible possibilities.

The issue is, I am sometimes getting the "all possible possibilities" part wrong, I am locked in a mindset where I clump together everything and then take that number of things over something.

Sometimes, the number of possible possibilities is something like this (a over b)^n and sometimes you don't exponentiate it.

Is there a... Strategy to get that right? When to exponentiate and when not to?

[white willow]

Yes!
I understand what you mean, I also had that kind of issue.

However, there are 4 main cases that you have to memorize.

1. Is repetition allowed?

2. is (123) treated as the same case as (321)?

Like if you're making a smoothie, it doesnt matter if you put the bananas before the blueberries.

[green geyser]

Yeah no, order is alright

I understand 1 and 2

It is just the repitition part that fucks me over sometimes.

That's where I go: Oh, I'll clump it together

and then realize after I'm wrong, "AH IM STUPID! There is repitition."

[green geyser]

and sometimes, you take an element from A and then from B.

Multiply them by treating A and B as sets.

[white willow]

hm I havent encountered that, but yeah the repetition can frick u sometimes

[green geyser]

I mean:

If you have 4 books and then 3 apples.

How many ways can you pair a book with an apple?

Order doesn't matter.

(4 over 1) * (3 over 1)

That kind of stuff

[white willow]

Yeah same scenario as if you have a play, and you need to pair 5 women with 6 men

5 * 6 = 30

[green geyser]

And then, there's probabilities.

It fricks me over, because the number of actual possibilities can be very differently set up.

For example:

4 women, 3 men

Pick 3 people:

Wouldn't that be:

(7 over 3) right?

[white willow]

Like 3 random, no repetition and order doesnt matter?

[green geyser]

yes

vs

I have 4 bottles. Each bottle contains a set amount of things. (Let's say 5 apples and 2 oranges)

How many ways can you pick up an apple and an orange?
If you pick 2 objects out of each.

(7 over 2)^4

[white willow]

So in how many ways can you pair 3 people out of 7?

[green geyser]

Realized there were 7 people

oof

💀

[green geyser]

7x6x5/(3!)

[white willow]

Yeah you'll just have to study the theory.

That's probably your best shot.

But you'll get it.

[green geyser]

The thing is, I understand you have to pick 2 out of each bottle.

however, for some reason I make the mistake of writing

(4x7 over 2)

[white willow]

Ok so how many ways are there to pick 1 apple and 1 orange in a bottle?

5*2 = 10

10 ways to get 1 apple and 1 orange

[green geyser]

The way questions are asked here fucks me over

[white willow]

Since you have 4 bottles and the choices are independent across bottles, you multiply the number of ways for each bottle

10 ways per bottle * 4 bottles = 40

Or thats at least how I think

[green geyser]

I'm intepreting it as:

(7 over 2)^4

Because that's the total ways you can pick 2 objects from a group of 7. <-- repeatedly

[white willow]

But now you're grouping apples and oranges into the same group

[green geyser]

But that wouldn't be an apple and an orange anymore right? That would just be disregarding the identity of the object

[white willow]

Exactly.

So thats why we multiply to get the total amount of ways per a bottl

Use that method for cases where you have 2 alternatives

Now you took 2 choices and you had 2 fruits

[green geyser]

So when it comes to the probabilities:

You ask:

How many ways can I get this object?
Through
How many ways can I combine the objects here?

[white willow]

Yes!

And since each bottle have their own fruits, we simply multiply the bottles together

[green geyser]

Because there is a repitition, technically

[white willow]

If it was for example for a common box of fruits, we would treat it differently.

[green geyser]

when you make your choice the next time, the amount of fruits restore <-- not actually, but mathemathically

Hence, exponentiation is needed.

[green geyser]

So if the things are seperated into groups

vs It's not

[white willow]

Yes, and if the selection from this group affects the other "bottles"

[green geyser]

So, if the selection from this group doesn't affect the other bottles, then repitition is in order

because you're ending up with 5 elements or whatever it was

again

[white willow]

Repetition is allowed, no matter if the group affects other ones.

I'm talking about the last step, multiplying the ways with the amount of bottles

(10 x 4)

[green geyser]

im not sure...

By repition you're treating it as if you're re-adding the elements

[white willow]

Yes that's what we're doing right? Once we have a a unique case we put them back and take 2 random new ones.

Or are you refering that we never put them back?

[green geyser]

well we never did that ye

but

The next bottle will have the same elements again

And again

Until the multiplication is finished, then it ends

[white willow]

Wait a second.

[green geyser]

I'm not sure if I can explain it properly

[white willow]

So you're saying that there are 5 apples and 2 oranges per bottle

case 1: You take out 1 orange and 1 apple

case 2: You take out another orange and another apple

[green geyser]

Yeah

i think?

[white willow]

case 3: You can't do anything because there are no more oranges in the bottles

[green geyser]

wait no

I mean just picking them out, one by one

like, once you picked that one, go to the next one

[green geyser]

Agh, I think I overcomplicated it

Refer back to the (7 over 2)^4 example

[marsh citrus]

@green geyser Has your question been resolved?

[marsh citrus]

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[spring oracle]

,rotate

[spring oracle]

I know i need to turn inf - inf to inf/inf but I don't know how to go on about this one

[whole sleet]

You don't need to do that

Well, I suppose you could by factoring out ln(x)

[static quarry]

you could see what happens to e^(8x - 7ln x)

[spring oracle]

are you guys referring to assigning it to y and then taking natural log of both sides?

!quit

.close

[marsh citrus]

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[ancient acorn]

can

I get help

[thorny kestrel]

does this mean i just have to set the matrix equal to [1,2,0]

and solve for that

[marsh citrus]

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).

[ancient acorn]

ohh

[waxen remnant]

try doing this first

[marsh citrus]

@thorny kestrel Has your question been resolved?

[waxen remnant]

didnt see this but yes

[marsh citrus]

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[hybrid hill]

Studing PMCC dont get how to use the formula. how do i find r from this

[night mica]

what is PMCC?

[hybrid hill]

product moment correlation coefficient

this is the answer slide

i dont get how to get to the answer R

[night mica]

i cannot help you with this ..sorry

[runic temple]

omg stats

so x bar and y bar are the mean values of the first and second variable types

in this case x bar is the mean of the math scores

it would help to label the column titles

the titles of the empty columns are
x-xbar
y-ybar
(x-xbar)(y-ybar)
(x-xbar)^2
and (y-ybar)^2

[hybrid hill]

how were the numbers found in the fourth and fifth column i get the rest.

[runic temple]

xbar is the mean of the math score

since the total sum is 560, the mean is 56

65-56=9

thats the first entry

[hybrid hill]

yeah i realized. the rest makes sense now

thanks

[runic temple]

45-56=-11

youre welcome

[hybrid hill]

.close

[marsh citrus]

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[valid cape]

$\int \frac{x}{3x+\sqrt{9x^2-1}}dx$

[elfin berry]

FungusDesu

[valid cape]

i have no idea what to substitute

[static quarry]

maybe divide everything by x first

[valid cape]

that would further complicate the problem no?

[static quarry]

it would get rid of most of the x's

[valid cape]

since you have 1/x^2

[static quarry]

yea that might be problematic

multiply and divide by the conjugate?

of the denom

[thorny kestrel]

rationalize it

or say x=sec(u)/3

u=arcsec(3x) and dx =(sec(u)*tan(u))/3 du

[sand fable]

some very good things to point out indeed

[thorny kestrel]

it almost impossible to solve it with u sub

dont do that

[sand fable]

this problem reeks of trig sub with the sqrt(9x^2 - 1)

[valid cape]

hm, how to know when to use trig sub?

[thorny kestrel]

rationlize it

then split it into two

and solve it

[valid cape]

indeed i got $x(3x-\sqrt{9x^2-1})$

[elfin berry]

FungusDesu

[valid cape]

from here do i subtitute everything in the parentheses?

[thorny kestrel]

no

[valid cape]

or just the sqrt

[thorny kestrel]

dont use u sub here

until later

apply linearity

[static quarry]

just do it as two integrals

3x^2 and the other one

[thorny kestrel]

you will ahve x^3/3 *3 = x^3

https://www.integral-calculator.com/

just use this

this will sovle any intergral with steps for you for free

[valid cape]

hm alright, i think i got it now

thanks yall

.close

[marsh citrus]

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[rocky nova]

is 0 a natural number?

[static quarry]

[rocky nova]

Is there a set that includes positive ints and zero?

do I just say whole numbers in that case?

maybe non-negative ints

[static quarry]

i like $\mathbb Z^{\geq 0}$ as it's unambiguous

[elfin berry]

Bungo

[hushed egret]

non-negative integers is a good choice

i like $\Z_{\ge0}$ as it's not terrible unlike $\Z^{\ge0}$

[static quarry]

haha

[rocky nova]

thank you

[hushed egret]

,,\Z_{\ge0}^2 \textss{vs.} {\Z^{\ge0}}^2

[hushed egret]

there's a clear winner here

[rocky nova]

"A function f from the natural numbers to the non-negative real numbers..." what does this mean?

the function takes in natural numbers and spits out non-negative real numbers?

[static quarry]

depends on your author
N either includes 0 or it doesn't

[hushed egret]

whatever natural numbers means

[static quarry]

in very many cases it doesn't really matter

[hushed egret]

,,f : \N \to \R_{\ge0}

[hushed egret]

it's just a shift by 1 if you do/don't include 0 in N

[rocky nova]

okay, thanks again

.close

[marsh citrus]

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[still temple]

-2 1/9 - 7

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[formal raptor]

Why is this a virtual image? Also if the focal point is 4cm does that mean it's 4cm away from the lens?

[hazy hare]

Diverging lens always forms a virtual image

Focal length is the distance between the principal focus and the optical centre

[formal raptor]

In virtual images, is the focal point always behind the object?

[formal raptor]

Is focal point distance from the object?

[hazy hare]

No

[hazy hare]

check this

[formal raptor]

Is this true?

[marsh citrus]

@formal raptor Has your question been resolved?

[marsh citrus]

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[velvet surge]

In drawing a single card from a regular deck of 52 cards we have:
(Note that an Ace is considered a face card for this problem)
(b) P( black and a face card)
(d)P( black or a face card )
(e) P( face card or a number card )

[velvet surge]

okay so for b i did
black = 13/52 ace = 3/52 not 4 because one is included in the black so it doesn't make sense to include again
so i got 16/52 but thats not correct but neither is 17/52

wait it says AND

wtf 1/52 didnt work