help-26

[hoary burrow]

What is a positive injective proof

[stone rapids]

Positive case is when the function actually is injective

For this case, we have x^2 = injective for f(x) =x^2 with domain {0}

[hoary burrow]

Yea but

Your current statement lacks rigor then

[stone rapids]

Which one

[hoary burrow]

The positive bounded one

[stone rapids]

What

I don't mean positive as in positive real numbers. I mean positive as in 'actually true,' or 'correct'

Anyway I'll wait for OP to respond

[toxic aspen]

I'm gonna take a break

thanks for helping tho

.close

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[ruby tree]

Just so you get something out of this (strange) last part:
to prove f(x) = x^2 is not injective you only need to state a counterexample.
f(1) = f(-1) but 1 =/= -1
That's literally it

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[odd juniper]

Why does this imply that the sequence is not Cauchy?

ok wait

this might be very dumb

have i inverted the condition properly

$\exists \varepsilon>0 ;\forall n_0\in\mathbb N[\exists n,m>n_0,; |x_m-x_n|\geq\varepsilon]$

[thorny flame]

normalAtmosphericPa=101,325

[bold urchin]

yeah that seems right

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[boreal mulch]

how is this = -1

[boreal mulch]

LHS

when i do cot(pi/2)^(n+1) on my calculator it says math error

[cedar wagon]

Cuz n+1 means nothing to him

[boreal mulch]

???

no

i did cot(pi/2)

[cedar wagon]

Oh ok

[rigid cloak]

Bro

[boreal mulch]

which is 1/tan(pi/2)

[rigid cloak]

Cot(π/2)

Is

0

Fr

[cedar wagon]

Tan pi/2 is ?

[boreal mulch]

why my calc says undefined

[cedar wagon]

Cuz tan pi/2 is ?

[boreal mulch]

undefined

[rigid cloak]

That's the answer

[boreal mulch]

asymptote

[rigid cloak]

U dividing 1/undefined

That's why

[restive inlet]

cot(t) = 1/tan(t) where defined

[cedar wagon]

Ok so improprer integral hit and limits kicks in

[rigid cloak]

Yes

[boreal mulch]

ok why does the bound give -1

[rigid cloak]

See

U put the limits and write expression here

[cedar wagon]

Pi/4 gives what ?

[boreal mulch]

1

[cedar wagon]

And since its F(b) - F(a) in the formula

So -1

[boreal mulch]

what if F(b)

[rigid cloak]

Fr

Bro listen

[restive inlet]

depending on which definition you use, like
cot(pi/2) = cos(pi/2)/sin(pi/2)
which is 0

[rigid cloak]

Do u know indefinite integral

[boreal mulch]

yes

[rigid cloak]

Use it

Put limits bro

[cedar wagon]

No need

See what ramonov write

[boreal mulch]

ohh

[rigid cloak]

Cosmos put the limits in the expression

U will get it

[boreal mulch]

i cant do 1/tan(pi/2) since its undefined?

[rigid cloak]

Bro

[cedar wagon]

Well it requires limit

[boreal mulch]

i didnt go this deep into limits

[rigid cloak]

No bro

[boreal mulch]

tan(pi/2) is undefined so i cant do 1/tan(pi/2)?

[rigid cloak]

I mean in the original question

Where u have

The expression put limits

See tan(π/2) is undefined

Bcz tan(π/2)=sin(π/2)/cos(π/2)

[cedar wagon]

No you cant like an image calculation

[rigid cloak]

And cos (π/2)=0

[boreal mulch]

i see

[rigid cloak]

Fr

[cedar wagon]

But you can write write lim b->pi/2 of 1/tan(b)

And get 0 too

But its limit

[rigid cloak]

Fundamental theorem of calculus

B tends to -π/2

[cedar wagon]

So you have F(pi/2) - F(pi/4) = 0-1 = -1

[boreal mulch]

i get it now

ty

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[rigid cloak]

Fr

[sudden temple]

@cedar wagon sorry to disturb, but you're in uni right?

[cedar wagon]

Soon

[sudden temple]

oh okok

[rigid cloak]

Yo

[sudden temple]

🤟

[sudden temple]

why is it named Love YOU Gesture?

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[fading bobcat]

i need math help

[fading bobcat]

with tree diagrams

[cedar wagon]

Post up a picture of the problem and what you have tried (if you have) and we will look up

[fading bobcat]

ty

[cedar wagon]

18. ?

[fading bobcat]

yeah 17) and 18) i’m fine with 19)

[cedar wagon]

Ok so the last question of 17) is using total probability formula

[fading bobcat]

yeah

[cedar wagon]

They ask you to find P(P) so its equal to all the issues leading to P

So 0.75 for the first try of the test

[fading bobcat]

yes

[cedar wagon]

+0.15 to the second try

[fading bobcat]

yes

[cedar wagon]

So its 0.9

[fading bobcat]

ohhh

okay i got it now

what about question 18), idk if i have to take away one from the denominator

or if it’s the same

[cedar wagon]

Supposing the two spine are independant and identicaly defined, then its the same numbers in the second spine

[fading bobcat]

okay

[cedar wagon]

Ok so they ask you to find the probability of odd knowing even happen

So its right or left one ?

In first spine

<@&268886789983436800>

Wth happen

Ty

[fading bobcat]

oh ok, i got it now i know how to answer the questions its just drawing the diagrams thats a bit tricky for me

[fading bobcat]

first spine, then the 2nd from the first... 1/3 x 2/3

2/9

[cedar wagon]

Wait

Yeah no that should be this

Nvm

[fading bobcat]

yeah

[cedar wagon]

I dont see whats question c

The picture ate it

[fading bobcat]

there is no 18) c)

[cedar wagon]

Oh ok

[fading bobcat]

yep

i finsihed the sheet

now i have more...

wait ill show

[cedar wagon]

Jumping from b to d, maybe they should go back to school too

[fading bobcat]

and idk if i did this correctly

[cedar wagon]

Ok many mistakes

In calculations

[fading bobcat]

oh

[cedar wagon]

Diagrams are great

[fading bobcat]

tysm

[cedar wagon]

0.8*0.8 = 0.64 not 1.6

And there is no need to need to add something

6b)

What did you write at 5b as result ?

Im reading 1.7

[fading bobcat]

1.17

[cedar wagon]

The result is 0.42

[fading bobcat]

oh

[cedar wagon]

Well actually a probability will ALWAYS be
0 < probability < 1

So if its more

[fading bobcat]

i think i mistakingly added instead of multiply

[cedar wagon]

Otherwise diagrams are well done

[fading bobcat]

thanks

[cedar wagon]

And its the same for the one lasting

[fading bobcat]

okay i understand my mistakes now

[fading bobcat]

what about this?

[cedar wagon]

The same for all

No conditional probability all identicaly defined

[fading bobcat]

oh

[cedar wagon]

Both questions are the same as done before so

[cedar wagon]

Gj

[fading bobcat]

tysm

what about this

[cedar wagon]

So far, its all good

[fading bobcat]

yes

[cedar wagon]

Well for the last questions you have to make another step for the third chocolate

With reducing one on numerator and denominator

Same as for choco 2 but for choco 3

[cedar wagon]

Indeed

[fading bobcat]

tysm, what about 21)

[cedar wagon]

So a card games is /52

[fading bobcat]

i still dont get it

[cedar wagon]

He has 1/52 of probability to pick an ace

And if its red then it depends on if its Diamond or heart

[fading bobcat]

okay

[cedar wagon]

The flip is 50-50

So 1/2 both sides

And now just wrote the gain of each ending

[fading bobcat]

wait

[cedar wagon]

Its not 0$ its -2$

Since it cost

Its not given back if he looses

[fading bobcat]

oh ok

[cedar wagon]

Ok

Answers are not hard from this

[fading bobcat]

yep, im working on them

[cedar wagon]

I gtg

[fading bobcat]

oh, okay. thank u so much for your help, i finally finished my hoemwrok that iv been procastinating on doing so your help really saved me. take care

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[topaz sinew]

@keen totem Has your question been resolved?

[icy sky]

looks right

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[strong sable]

I asked this before, but i really couldn't figure it out, what's the obvious u sub here ?!

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@strong sable Has your question been resolved?

[strong sable]

<@&286206848099549185> \

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@strong sable Has your question been resolved?

[bold urchin]

If $I$ is the integral you want consider $-I + 1 - 1$. Use one of the hypotheses then set $u = xe^{1-x}$.

[thorny flame]

pola_touche

[strong sable]

If I is the integral you want consider -I + 1 - 1
Why?

[bold urchin]

to use that $1 = \int_0^1e^{1-x}f(xe^{1-x})dx$

[thorny flame]

pola_touche

[bold urchin]

and make the u sub on the whole I + 1 -1 nice

[strong sable]

so you mean I is this?

[bold urchin]

yeah

[strong sable]

but then what's -I+1-1, isn't it just $-\int_0^1 xe^{1-x}f(xe^{1-x})dx$+1-1

[thorny flame]

riyob

[strong sable]

$-\int_0^1 xe^{1-x}f(xe^{1-x})dx+ \int_0^1e^{1-x}f(xe^{1-x})dx-1$

[thorny flame]

riyob

[bold urchin]

now put the integrals together and compute the u sub

[strong sable]

$$\int_0^1 [e^{1-x}f(xe^{1-x})- xe^{1-x}f(xe^{1-x})]dx-1=\int_0^1 [(1-x)e^{1-x}f(xe^{1-x})]dx-1$$
$$=\int_0^1 f(xe^{1-x})dx-1$$

[bold urchin]

no i did not use integration by part, compute the u sub

[strong sable]

ooh so it's
$$ u = xe^{1-x}. $$

$$ \boxed{\int_0^1 f(u) , du - 1.} $$

[thorny flame]

riyob

[bold urchin]

yessir

so -I = -1 gg

[strong sable]

Tysm -I +1-1 is a fancy idea

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[bold urchin]

a trick of the trade nothing more

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[noble fable]

Anyone send notes for jee

[noble fable]

For trigonometry

Contains a lot of problems with solutions

Like solved examples

Like around 60-100 solved examples in a book or notes

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[harsh shell]

how do u know if p^ is given or if u have to do p^=x/n

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Please don't occupy multiple help channels.

[noble laurel]

It depends on the context in the question, do you have a specific question you are working on?

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[woven pagoda]

i have vectors a, b and c. if a x b = n and c x b is also n, does that mean a, b and c lie on the same plane?

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@woven pagoda Has your question been resolved?

[gray mulch]

Yes. One way to prove it: subtract the equations: axb - cxb = 0 => (a-c)xb =0 => (a-c) and b are collinear => a, b c are linearly dependent and lie on the same plane.

[woven pagoda]

ty 🫡

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[desert dagger]

how has the mark scheme rewritten the top equation as the bottom one?

[thorny remnant]

$n \log a=\log a^n$

[thorny flame]

convergence

[desert dagger]

and rhs?

[desert dagger]

lhs i got using this but fourth root of 9 wont give square root 3 right

nvm got it

[thorny remnant]

it does

np

[desert dagger]

and one last thing

how does the ln x turn into just x

are we just cancelling ln on both sides?

[gray mulch]

Use e^x on both sides.

[thorny remnant]

kinda yeah

[desert dagger]

okay thnaks

thanks*

.closew

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[placid mango]

can anyone help me with this formula

[placid mango]

i cant seem to find the pattern

[cedar wagon]

In the three first, the first number of the multiplication is the answer and in the three next, its the second number of the multiplication which is the answer. Ig the 7th come back like the start, so we take the first number of 7*0 which is 7

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[mystic siren]

for f(x) prove the limit is true

[mystic siren]

why can you take delta = epsilon?

i dont understand what can we decide delta to be, given obviously it depends on epsilon

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[toxic aspen]

How many ways can the letters PARALLELLOGRAMbe arranged such that no L are neighbours.

My initial thought was this:

---

I have 14 total spots. If i want to L's to be separated i would have 7 different spots for my Ls such that it would be 7*6*5*4 total positions to place the L's. Then i have 10 letters PARAEOGRAMto place on the remaining 10 spots. This would be 10!/(3!2!) because of the repeat A and repeating R.

I'm a little stuck now tho and would love a tiny hint

[toxic aspen]

!occupied

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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).

[toxic aspen]

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[topaz sinew]

Please don't occupy multiple help channels.

[chilly walrus]

the derivative of a CDF is the PDF

[harsh shell]

do u just derive each line

nothing else

cuz u know how sometimes when u go from pdf to cdf it adds like an extra line

will i not have to remove a line or smth

in the piecewise function

[chilly walrus]

line?

[harsh shell]

for example

[chilly walrus]

yeah

[harsh shell]

this is the pdf

this is the cdf

u see how theres an extra line

[chilly walrus]

eh

[harsh shell]

if i derive each line theres still gonna be that extra line

it doesnt remove it

[chilly walrus]

don't worry about that

[harsh shell]

its fine if i leave it?

[chilly walrus]

the "elsewhere" is doing the work here, what they really mean is 0, x < 0 and 0, x > pi

[chilly walrus]

just differentiate your function and i'll review your answer

[harsh shell]

i already checked the answerts

its correc

t

but will i ever get a cdf with more than 3 lines

[chilly walrus]

not impossible

just look at what is sensible lol

also helps if you visualize/graph it

[harsh shell]

oh okay ty

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[strong sable]

why

[brave coral]

Sum of Geometric sequence

$$ e \cdot e^{2i/n} $$

[thorny flame]

StrangeQuarkAL

[cedar wagon]

And you can put the e in front of the sum

So you have (e^(2/n))^i

Which is a geometric with a reason/ratio e^2/n

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[strong sable]

aah the messages delayed, i just saw it now

i get it, tysm

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[terse bronze]

https://www.desmos.com/calculator/cik091pzjx

[terse bronze]

hi what type of graph would this be?

if I had to linearize it then what x variable would I plot with what y variable

e.g. x^2 graphed against y^3 or something

i would like to simplify the equation as much as possible

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[toxic aspen]

Are there any points on the straight line 26x +15y = 131 that have positive integer coordinates x and y? If applicable, state all such points.

[toxic aspen]

I solved the dio equation such that

-524(26) + 917(15) = 131

But how do i find the point?

iirc a point is found from

(x,y,z) = (x,y,z) + t(x,y,z)

[opal vault]

?

you're in 2D

a point is (x,y)

[toxic aspen]

i see

[opal vault]

points on the line are (x,y) such that 26x + 15y = 131

didn't you just find such (x,y)?

[toxic aspen]

I don't know

I'm just able to solve dio equations

[opal vault]

?

you were tasked to solve 26x + 15y = 131

so what are the solutions (x,y)?

[toxic aspen]

Hm

x = x_0 + bn
y = y_0 - an

So my solutions should be..

x = -524 + 15n
y = 917 - 26n

I think?

[opal vault]

not +

on both

ok

so (x,y) = (-524 + 15n, 917 - 26n)

[toxic aspen]

makes sense

so now i have to find where both x and y are positive

[opal vault]

well right now the x0 you took is negative

so you gotta do +15, -26 until x > 0

[toxic aspen]

Yeah i figured that

but that seems very hard to do in head

[opal vault]

it's arithmetic

euclidian division

-524 = ...*15 + (remainder mod 15)

[toxic aspen]

oh

until it's positive?

[opal vault]

the first guy that's gonna be positive is the remainder

[toxic aspen]

That's what i mean like

I can't do 15 * ... in my head to get 524 hahaha

[opal vault]

try 150* first

and then do + 15 to get closer

[toxic aspen]

150*15?

that doesn't seem right

[opal vault]

no

I suggested 150

[toxic aspen]

I know that 15*10 is 150

[opal vault]

as it's 15*10

so

[ruby tree]

What he means is that, to do it in your head, you can just start from 150+150+... and then do +15+15+...

[opal vault]

try 150*1

[toxic aspen]

o.o

[ruby tree]

3*10*15 = 450 but 4*10*15 = 600 so 524 is going to be between 30 and 40 times 15

[toxic aspen]

Well yes

i could manage to find that out but

jesus

[ruby tree]

But what? You prefer doing long division 524 by 15?

[toxic aspen]

Nah i can't do it hahah

I can figure out that it's between 30 and 40 times

but then what haha

[ruby tree]

Then you take 30*15 and start adding 15

What's 31*15?

[toxic aspen]

i literally can't do it

I'm not used to this high multiplication

15*30 = 450

but i'm missing

524-450

74?=

15*2 = 30 15*3 = 45 15*4 = 60 15*5 = 75

[ruby tree]

But it's just repeated addition

[toxic aspen]

so my guess isanother 3 times?

15*33?

no 4 times?

[ruby tree]

15*34 = 450 + 60 = 510 and you want 524 (or just above)

[toxic aspen]

i unironically can't do the multiplication

so i wont get full poitns on the question

lol

[ruby tree]

Again it's just repeated addition, there isn't much easier than that

[toxic aspen]

15*35 = 525

+1 remainder

[ruby tree]

15*35 = (15+15+15)*10 + (15+15+15+15+15) = 525

Correct

[toxic aspen]

so x = 1?

yes

and then find y

[ruby tree]

n = 35 and x = 1, yes

[toxic aspen]

917 = -26 * ... + remainder

917 = -26 * 35 + 7

(1,7)

I wonder why we can't just have normal questions like the book gives "solve the Diophantine equation" and nothing else

It's clear that they want us to fail

[ruby tree]

Because solving a Diophantine equation is not usually a goal by itself

Unless you're inventing a way to solve them

[toxic aspen]

Do you have a reminder of how i converted one base to another

$x = (BA0BAB)_{16}$

[thorny flame]

Merineth

[toxic aspen]

Assume that i wanted to write this in binary and base 8

Would i have to convert B, A, B, A and B into numbers first?

$(1001,1000,0,1001,1000,1001)_{16}$

[thorny flame]

Merineth

[ruby tree]

In this case, because 8 and 16 are powers of 2, the easy way is to convert to binary first

[toxic aspen]

This was not mentioned in the book or anywhere

I assumed i divide the numbers

and then the remainders

[ruby tree]

Yeah that's the general way

[toxic aspen]

https://youtu.be/FFDMzbrEXaE?t=656

like so

So once again i'm supposed to know something new and it's not just a simple (convert)

I have to know that they are powers of 2 and use it somehow?

[ruby tree]

You don't have to, you can use the general way

[ruby tree]

But clearly you know how to convert a base-16 digit into binary already

[toxic aspen]

That was just luck

I remember it from the first time i took the exam.

ABCDEF 11 12
1000 1001 1010 1011 1100 1101 1110 1111

i think it was something like that?

[ruby tree]

Yes

Well not quite

There's no G and H in base 16

$0_{16} = 0000_2 \
1_{16} = 0001_2 \
... \
8_{16} = 1000_2 \
9_{16} = 1001_2 \
... \
F_{16} = 1111_2$

[thorny flame]

Nel

[toxic aspen]

ok

So i have to remember this?

[ruby tree]

It just makes it much easier to go between bases that are powers of 2

[toxic aspen]

0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111
0 1 2 3 4 5 6 7 8 9 A B C D E F

Yeah i remember now that 1010 was A

but how do i go between the powers?

[ruby tree]

Use binary as an intermediate

[ruby tree]

This is correct, now you concatenate every group using 4 digits for each

That gives you your binary, and to get the octal, you make groups of 3 and convert each one into an octal digit

[toxic aspen]

uh okay so binary has 2 digits between each line, 8 has 3 and 16 has 4?

[ruby tree]

No

Each binary digit is made up of 1 binary digit

Each octal digit is made up of 3 binary digits

Each hexadecimal digit is made up of 4

[toxic aspen]

Yes that's what i meant

[ruby tree]

You said 2 for binary

[toxic aspen]

$AB = (1010|1011){16} = (101|010|011){8} = (10|10|10|11)_2$

[thorny flame]

Merineth

[toxic aspen]

Like so?

If i want to get AB in base 5
Do i take 10101011 and divide it by 5?

[ruby tree]

No, that last one is nonsense

[toxic aspen]

Then i dont get it

Completely lost

[ruby tree]

The first four numbers (0 to 3) in binary are written 0, 1, 10, and 11

Right?

[toxic aspen]

huh

I have no idea

[ruby tree]

You are feigning ignorance, it's not helping

How do you count to 10 in decimal?

[toxic aspen]

I'm trying to find in the book where i read it

i sincerely dont remember

[ruby tree]

What, how to count to 10?

You need a book for that?

[toxic aspen]

I dont remember bases

[ruby tree]

That's not what I'm asking

[toxic aspen]

1 2 3 4 5 6 7 8 9 10

[ruby tree]

I see a big number here, not a count to 10

Ok, now why is the tenth number represented by two symbols? Why not write it with yet another single symbol?

[toxic aspen]

as a single digit?

Oh

A?

[ruby tree]

Sure, why don't we count like 1 2 3 4 5 6 7 8 9 A ... in daily life?

Unfortunately I'm gonna have to go but the point is that we chose to only have 10 symbols for writing numbers (for slightly more complex historical reasons, but that's basically just a choice)

We call that the decimal system, or base 10

In binary, or base 2, we choose to only use 2 symbols, namely 0 and 1

So when you run out of symbols for the first digit, you go to the left and add another digit

In decimal, after 9 you go back to 0 and add a 1 in front, to make 10

In binary that happens after 1

0, 1, 10, ...

In base 4, that happens after 3

0, 1, 2, 3, 10, ...

[toxic aspen]

I see

[ruby tree]

So to take your notation:

$AB_{16} = (1010|1011){16} = (101|010|011){8} = (10|10|10|11)_4 = 10101011_2$

[thorny flame]

Nel

[ruby tree]

Alright I'm off now, good luck

[toxic aspen]

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[scenic girder]

Hello guys.

[scenic girder]

So im in khan academy right now, and I dont understand why they say that the cos(x) graph is the pink one

It clearly isnt the cos(x) graph

[naive quail]

yes the pink one is the cos graph

[scenic girder]

Huh

[naive quail]

yep

Starts at 1 and goes all the way down to -1 and back up to 1

y = cos x

[scenic girder]

But the period isnt 4 aka 2pi

[naive quail]

Well, They have just adjusted the x-axis to their likings

[scenic girder]

Huh

Dont quite get what you mean

[naive quail]

It doesn't really matter what the x-axis is. There is nothing set for the x-axis

They just say x

But what is x, we don't know. It can be anything

[scenic girder]

But then what are the 2 yellow points?

[naive quail]

oh

it is 2pi

cos look

2 x Pi is 6.28...

right

[scenic girder]

Right

[scenic girder]

and so what?

uh

[naive quail]

That is 6.28

[scenic girder]

Oh

[naive quail]

So 2pi

lol

[scenic girder]

OHH

Okat

well then another question

How was I supposed to get the answer they wanted from me?

[naive quail]

That is just very weird in terms of representation

[naive quail]

What is the question?

oh

found it

[scenic girder]

1 moment

f is a trigonometric function of the form f(x)=a*cos(bx+c)+d

Below is the graph of
[f(x)]. The function has a maximum point at
[(\pi,6)] and a minimum point at

[\left(-\dfrac{3\pi}{4},2\right)].

[thorny flame]

Sayo

Sayo

[naive quail]

So, as they have already made a start to it on the images you have sent me, I would start by visualising it. So I would put these values on a graph to see how it will be like

As they have done it here

Do you get this bit?

[scenic girder]

Mhm

Well

Mostlyu

THEY just put a graph of cos(x)

[naive quail]

This bit is the one that has the values from the question

MAximum point and the minimum point

[scenic girder]

uhm

right

But thats not the cos(x) graph

Isnt it?

I mean they are different

[naive quail]

It isn't exactly it, no. But it is a transformed graph of cos(x)

That's what you are trying to work out

[scenic girder]

mhm

[naive quail]

How has the cos x graph been transformed, what is the equation that describes this change

[scenic girder]

Ohh ok

I got it

what next

[naive quail]

okay cool

So, normally the difference between the maximum point and the minimum point of a cos graph is only 2 units with respect to the y-axis

But for this new graph what is the difference in the y direction?

so from (pi, 6) to (-3pi/4, 2)

[scenic girder]

Uh

its like

7pi/4

[naive quail]

Let me word it differently.

Here, this difference here is different than the normal cos x graph, right?

normally it is from 1 to -1

but now it is from 6 to 2

[scenic girder]

Ohh

yeah

So 2

4*

[naive quail]

yes

4

nice

normally, the difference was 2

from 1 to -1

but now, it is 4

So we doubled the whole thing! The y-values have doubled

So we can adjusted the equation as to become y = 2 (cos x)

Right? Makes sense?

[scenic girder]

Oh right

Yeah

[naive quail]

Cool, right. Next.

So the graph has been translated up.

Instead of being on the x-axis, it isn't even touching it anymore

[scenic girder]

What do you mean not touching i

t

Like not being close?

[naive quail]

the x-axis

This graph is not touching the x-axis

not crossing it

[naive quail]

.

It has been moved up

yea?

[scenic girder]

Ohh right

[naive quail]

Okay, so we have to come up with something to alter our equation now (which is y = 2 (cos x) ) to move this graph up. Can you have a go?

[scenic girder]

Hm

So like

move the graph up?

[naive quail]

YES

exactly

when we move up a graph, what changes in the equation?

y?

x?

both?

[scenic girder]

Y

[naive quail]

yup

[scenic girder]

We could probably move up the midpoint

[naive quail]

We could i think. What do you propose?

[scenic girder]

Hm

yeah move the midpoint

that way we go up

[naive quail]

Yea, that would work but for this question, you can't really know where the midpoint is. Can you see where it is?

[scenic girder]

Hm

2+6

=

8

and 8/2

[naive quail]

yep

[scenic girder]

=4

[naive quail]

Exactly, so right now, we are at 0 for y = 2 (cos x). What would i do to y = 2 (cos x) to move ALL of the y-values up by 4?

[scenic girder]

uh

move rthe midline by 4?

[naive quail]

Yes, when we did this to the equation we would get y = 2(cos x) + 4

moves all of y values up by 4

got it?

[scenic girder]

Mhm

[naive quail]

yea?

[scenic girder]

What next?

[naive quail]

Okay cool

We are done with everything in the Vertical direction/y-axis

Now we need to work on the x-axis

These are what we have done so far

[scenic girder]

Alrighty

[naive quail]

It is so hard to do stuff on here, sorry if im taking too long :P

What is the difference here?

Can you identify it?

[scenic girder]

All good

uh

10?

[naive quail]

Well, nearly but we need it in terms of pi because that's what we are working with

Let's remember what it was for a normal cos graph

How many pi is it from a point on a wave to another point on the same place but on the next wave

Like, from a minimum point to another minimum point

[scenic girder]

2 pi

[naive quail]

Exactly, but instead we have different value now and we can work this out by doubling the difference between the x-values of the minimum point and the maximum point.

So (-3/4pi, ...) and (pi, ...), what is the difference?

hold on

[scenic girder]

Hm

7/4pi

[naive quail]

yes

So, now double it

[scenic girder]

Uh

14/4 pi

[naive quail]

yep, we can write that as 7/2 pi

yes

[scenic girder]

Yeah

[naive quail]

So, we come to this stage now

7/2 pi is the same as 3.5pi

[scenic girder]

Ohh right

[naive quail]

Yea, so 4/7 multiplied by x gives us 3.5pi

Now, we are still not there yet

[scenic girder]

Alright

[naive quail]

hold on

before that

Now our equation has got to y = 2 (cos (4/7x)) + 4

[scenic girder]

But how did we get 4/7

oh wait

yeah how

[naive quail]

let me do it on my phone to send it.

[scenic girder]

alr

[naive quail]

Here

so the x in cos x now represents 3.5pi, so the 4/7 kind of accounts for that

we did 2pi/3.5pi to get 4/7pi

[scenic girder]

Mhm

ohh got it

[naive quail]

you sure?

I can find a textbook example if you like? or are you sure ?

[scenic girder]

Well

If you would it would be helpful

[naive quail]

Sure

Here is a diagram you will find useful

[scenic girder]

alright

Uhm

what next

[naive quail]

You sure, im still looking for an example. But okay, imma get that for you when i can

[scenic girder]

Alright

Yeah, just continue with your explanation

[naive quail]

okay, now the equation got to y = 2 (cos 4/7x) + 4

so the maximum turning point of his new graph is (pi, 6) but right now, we haven't adjusted the x-axis movement yet (translation).

With the equation y = 2 (cos 4/7x) + 4 our maximum point is (0, 6) rather than (pi, 6). How do you think we should adjust it to make it (pi, 6)?

[scenic girder]

Right

[naive quail]

So in this case we need to turn the equation into y = 2 [cos4/7(x-pi)] + 4 to account for this translation

[scenic girder]

Right

Wait

What do you mean by the turning point

[naive quail]

I meant Maximum point on the graph

[scenic girder]

Ohh

okay

[naive quail]

so here

[scenic girder]

Mhm

Uhh

[naive quail]

This point here, is at (pi, 6) so we need to do (x-pi) adjust for this

[scenic girder]

But why x-pi?

I dont quite get it

wont that give us less than pi?

[naive quail]

i would write the whole proof for it

[scenic girder]

Or wait

2pi-pi

= pi?

\

OHH like that?

[naive quail]

hold on

No

So here

[scenic girder]

mhm

right

[naive quail]

If you look at the 4th one down, it says f(x-c) makes the (x+c, y)

so right now our equation was y = 2 cos(4/7x) +4 which means that our MAXIMUM POINT for this equation is at (0, 6) and we need it to be at (pi, 6)

To achieve that we need to subtract pi from x to shift the whole graph by pi amount on the x axis

[scenic girder]

Right..

so 2pi-pi?

[naive quail]

Well, if you take 2pi as your x, then yes

[scenic girder]

And thats our case here

ah nvm

OH

OHHHH

I get it, @naive quail Now I get it

at least i think so

Anyways, I gotta head to sleep now

Im really glad for the help you gave me today

I think I understood everything I needed

If I need help, ill ask you again

Farewell!

[naive quail]

Good, im glad i could help. dont hesitate to ask me questions on dms as well brother

[naive quail]

You're welcome

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[ashen elk]

this is the question being referred too in the image

[acoustic pecan]

r=sqrt(103)
hypotenuse of that triangle must then be sqrt(2*103)
why did you assume the 90 degree angle would mean the other point is at h-1?
that would work if the radius was sqrt(2)

but thats it

[ashen elk]

because

if

the point is one value postivie

the other value should be 1 value negative because its on the same y axis

if the triangle was split in the middle sohuldnt

the two triangles be congruent

[acoustic pecan]

yes, but im still not following why that demands a separation of 2

[ashen elk]

becaues the x value is 1

like

if H,K was at 0,0

that means the point A would be at 1, sqrt(102)

that means point b would be at -1, sqrt(102)

meaning those two points are seperated at a distance of 2

[acoustic pecan]

the slope of the line from the center to A would be sqrt(102)
the slope of the line from the center to B would be -sqrt(102)
those lines arent perpendicular, so the angle isnt 90

[ashen elk]

its stated that the

oh

wait but the central angle is 90

[acoustic pecan]

yes, but your reasoning contradicts that

so its wrong

[ashen elk]

oh i see

oh i get it

alright thanks

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[delicate dragon]

I need help with understanding cuantity of limits, and would love to see examples of it.

[acoustic pecan]

#❓how-to-get-help

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[gloomy pine]

could someone please help walk me through the process of solving this first order ODE using bernoulli? Ive attached the question and the key, ive gotten how to go from the question to the initial form of the solution, but the substitution is confusing me a bit. I am not allowed a calculator so please keep that in mind when talking about solutions, thanks!

[topaz sinew]

@gloomy pine Has your question been resolved?

[gloomy pine]

<@&286206848099549185>

im especially confused with the section after the Integrating Factor step

is it part of the IF process to add a dx to each component? why is the first dv? what is happening in the step going from x^2dv+2xvdx=6x^2dx to d[vx^2]=6x^2dx

<@&286206848099549185>

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[lean pawn]

“What is the measure of an angle if twice the measure of its supplement is 30 more than 5 times the measure of its complement?”

[lean pawn]

heres wht i have so far

y = 5z/2 + 15

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@lean pawn Has your question been resolved?

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[bold kernel]

Are my answers correct here and I’m really unsure between the transpose and the inverse rule

[bold kernel]

-12 should be wrong actually hmm

Oh wait the transpose determinant is the same as A determinant

But the scalar 3A I’m unsure of really

[dusk adder]

What is 3 in matrix form?

@bold kernel

[bold kernel]

It’s -324 I think I got it

It’s the scalar to the power of n

N being the number of rows or columns here

Which is 4

Times the determinant of A

[dusk adder]

Yes, do you see why?

[bold kernel]

Hmm 3 in matrix form

So the scalar 3 as a matrix

I’m unsure

3 0 0
0 3 0
0 0 3

Maybe

I’m not sure haha

[dusk adder]

3=3(I)

[bold kernel]

Ohhhhhh

Right

[dusk adder]

Here I is 4x4

[bold kernel]

Which I sort of spelled out right?

[bold kernel]

Here

[dusk adder]

You had the right idea yeah

Just missing one dimension

[bold kernel]

Yes

So essentially you’re saying we’re multiplying that 3I by the A matrix

And finding the determinant of that product

But as an algorithm let’s say

I should think of it as scalar to the power of n times det(A)

Where n is the rows

Or columns whatever since it’s square

[dusk adder]

Yes

[bold kernel]

Got you

I appreciate it

[dusk adder]

Np

[bold kernel]

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[grave berry]

why is 1/3 used first in the interval notation?

instead of 8

[finite moth]

cuz 1/3 is before 8 on the number line

and intervals are in order of least to greatest

[grave berry]

thanks

will intervals always be in the order of least to greatest if it was for range?

[finite moth]

yea, that's just how intervals work

[grave berry]

ok

thanks

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[trim wharf]

the question asks

wheter or not a residual is big or small

there is a pattern so i know the second part of part is a pattern

but i dont know how im supposed to know whether or not a residual is small or large

<@&286206848099549185>

<@&286206848099549185>

bruhhhh

<@&286206848099549185>

<@&286206848099549185>

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[violet sundial]

How can i simplify this?

By steps?

<@&286206848099549185>

[thorny remnant]

!15mins

[topaz sinew]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[violet sundial]

Okay thanks

[thorny remnant]

take (t+2) common in the numerator

[violet sundial]

How sir?

The other (t+2) has squared

[thorny remnant]

$\frac{(t+2)(2t^2-2t(t+2))}{t^4}$

[violet sundial]

It's derivatives and this is what ive got, i want simply it.

[thorny flame]

convergence

[thorny remnant]

like this

[violet sundial]

Ohh isee

And then sir?

[thorny remnant]

simplify $2t^2-2t(t+2)$

[thorny flame]

convergence

[violet sundial]

Wait it has squared sir.

[thorny remnant]

i know

i took (t+2) common

[violet sundial]

Ok so there is only one (t+2)

[thorny remnant]

(t+2)^2=(t+2)(t+2)

[violet sundial]

How can i simplify this sir?

[thorny remnant]

2t^2 (t+2)-2t(t+2)(t+2)

[thorny remnant]

expand it

[violet sundial]

Yes

[thorny remnant]

now you take (t+2) common

what do you get?

[violet sundial]

I need to expand this right?

Do i need to subract 2t^2 -2t?

[thorny remnant]

no think carefully

you have 2t^2 -2t(t+2)

[violet sundial]

Yeah

2t ^ 2 is same as 2t x 2t

[thorny remnant]

now if you had to multiply a(b+c) how would you do it?

[violet sundial]

Chain rule?

[thorny remnant]

yes

[violet sundial]

(ab + ac)

[thorny remnant]

now think a=2t ,b=t,c=2

[violet sundial]

(2t +4 )

[thorny remnant]

no

[thorny flame]

convergence

[violet sundial]

2t^2

[thorny remnant]

and what will be 2t(2)?

[violet sundial]

Ohh okay

[thorny remnant]

so what will be 2t(t+2)

[violet sundial]

(2t^2 + 4t)

[thorny remnant]

now in this there is a minus so what will happen?

to (2t^2+4t)?

[violet sundial]

(2t^2 - 4t)

[thorny remnant]

no

[violet sundial]

Change 2t^2 to negative

[thorny remnant]

and?

[violet sundial]

(- 2t^2 - 4t)

[thorny remnant]

yes

now you have 2t^2-2t^2 -4t

how will you simplify this

[violet sundial]

-4t

Combine like terms

soo

my final answer to my book is -4(t+2) / t^3

[thorny remnant]

yes

there you go

[violet sundial]

where do we get t^3 sir

i'm confused

i'm thinking power rule

but i think its not

[thorny remnant]

$\frac{-4t(t+2)}{t^4}$