help-33

[dense epoch]

Yes

[fleet urchin]

substitute

[pliant siren]

lmao

[fleet urchin]

simple as that .

[dense epoch]

Substitute what with what

[fleet urchin]

this

in

(x+y)/(x-y)

[pliant siren]

you might be onto something

haven’t thought of that lmao

[pliant siren]

$\frac{x+y}{x-y}$

[dense epoch]

6y-5 + 5x+6 over the same thing?

[elfin berry]

DerTheo

[fleet urchin]

take y as common from numerator and denominator

[fleet urchin]

no .

it'll be

(6y/5+y) / (6y/5-y)

now take y common and solve

simple .

[pliant siren]

$\frac{6y/5+y}{6y/5-y}$

[elfin berry]

DerTheo

[fleet urchin]

yup

take y common

and cancle it

[dense epoch]

y over -y?

[pliant siren]

$\frac{6/5}{6/5}$

[elfin berry]

DerTheo

[pliant siren]

$1$

[elfin berry]

DerTheo

[dense epoch]

O

[fleet urchin]

$\frac{y(6/5+1)}{y(6/5-1)}$

[elfin berry]

Jawad

[fleet urchin]

CANCLE Y DAMM IT

IT'S SIMPLE

[pliant siren]

oh youre right my mistake

[fleet urchin]

i learned this in 8th standard

[pliant siren]

7th you mean

[fleet urchin]

anyhow i remember and u dont

[dense epoch]

Learning this in 9th

[pliant siren]

sorry, my mistake, I canceled the ys out but I forgot to leave the 1 there, so that it was +-0 and you can ignore that

[fleet urchin]

how many times did u make that mistake

[pliant siren]

rarely

when you try to be quick, you loose accuracy

also you dont have to be mad, its not your mistake

[dense epoch]

Take a chill pill

So what next

[pliant siren]

youre done

2.2 / 0.2

11

[dense epoch]

Imma be honest

I don't get it

[fleet urchin]

yea ok i apologize
we all have our own pace ..

[pliant siren]

ill try writing it all down

[pliant siren]

hope you can read it

[dense epoch]

I get it now thank you so much

[pliant siren]

youre welcome

use .close now

[dense epoch]

.close

[marsh citrus]

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[marsh citrus]

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[robust notch]

In the first picture I need to do something like in the 2nd picture

With the 105pi/2

Pls help

Please

.close

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@placid geyser Has your question been resolved?

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[marsh citrus]

@green badge Has your question been resolved?

[green badge]

<@&286206848099549185>

[marsh citrus]

@green badge Has your question been resolved?

[gaunt pagoda]

Did you cover actio = reactio, newton's third law?

@green badge you need to apply newton's laws here

F=ma and actio=reactio

from the parcel, we can tell the acceleration of the lift: The man can feel it in his arm

F=ma where m=3kg and F is 22N

then you have the acceleration

Then we can use the acceleration together with the 760N to get the mass of the man+parcel

from the mass of man+parcel, get the mass of the man by subtracting the parcel's mass

[marsh citrus]

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[marsh citrus]

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[willow wind]

can somebody help me understand this? if that isn't what that limit means, then what does it mean? is it moreso just what f(x) approaches generally as x -> infinity?

[willow wind]

this is the function graphed, so i think that's what it means

[sharp vessel]

The limit means, that f(x) gets "as close as you want to" to 0

[sharp vessel]

even if f(x) = 0 always, the limit is 0

there is no requirement of NOT being equal to 0

it can be equal to 0

[willow wind]

ohhhhhhh

[sharp vessel]

"getting as close as you want to" as in, if you want to get within 0.0001 distance of 0

(on the y)

I can tell you an x, such that from that point forward, f(x) will be that close to 0

or less

[willow wind]

ok i think i understand

thank you

[sharp vessel]

np

[marsh citrus]

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[marsh citrus]

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[spiral fable]

Basically on question 9 it looks easy but you have to give your answer through calculus, just struggling and would appreciate a hand

[novel juniper]

k=725, right?

[spiral fable]

yh it does

[novel juniper]

In that case could you formulate an expression for perimeter

[spiral fable]

wdym?

[novel juniper]

Write an expression for perimeter in terms of two variables first

let the length be l and bredth be b

[spiral fable]

like 2l + 2b = the perimeter

idk

[novel juniper]

yes

now what's the area

in terms of those same variables

[spiral fable]

?

[novel juniper]

what's the area in terms of l and b

[spiral fable]

l x b = area

[novel juniper]

yes, now can you express area purely in terms of l

[spiral fable]

area/b=l

[novel juniper]

remember that$l+b= \frac{725}{2}$

[elfin berry]

Why am. I here

[spiral fable]

l=(725/2)/b

[novel juniper]

l=725/2 -b not by b

[spiral fable]

ohh yeah

my bad

so l=(725/2)-b

[novel juniper]

Now what's the area only in terms of b?

[spiral fable]

b=(725/2)-l

[novel juniper]

That's b

what's the area

[spiral fable]

not sure

[novel juniper]

l((725/2)-l)=A

Can you minimise this using calculus now?

[spiral fable]

dont worry mate, think i just figured out

appreciate your help tho

sorry wasting your time

.close

[marsh citrus]

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[solemn plank]

how do i do b?

[solemn plank]

i got part a correct with the area

how do i use that to find the volume rotated across y

this is the graph for reference

[stark trail]

Shell method, washer method, disk method, …

[solemn plank]

disk method

i think

im not too advanced

[stark trail]

Have you tried it?

[solemn plank]

my book doesnt mention these methods
it just says like to find volume rotated around y you do integral of x^2 respect dy

[stark trail]

I’m sure that is not all it says

[solemn plank]

ill send ss

@stark trail

[stark trail]

Okay so what’s stopping you from using that

[solemn plank]

i dont get what bounds im supposed to pick

[stark trail]

It shows in the image

[solemn plank]

for this example i found the area between 0 and 0.5
now i need to rotate that around y
should i use the bounds f(0) and f(0.5) ?

[stark trail]

Reread the question if you’re confused about the bounds

[solemn plank]

i have been reading it for an hour bro LMAO

its ok

thanks for help

[solemn plank]

ye the corresponding area is pi/6 u2

[stark trail]

..

You’re not rotating pi/6 u2

You’re rotating a region

When it says the corresponding area

It doesn’t mean the numerical value

It means like that region , that area

[solemn plank]

OH kk so this region is gonna be somewhere like between y = 1 and y =2 (suppose)

which can then be easily rotated

[marsh citrus]

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[surreal shoal]

Hi I need some help with conic sections here is the assignment 3 questions only

[marsh citrus]

Please don't occupy multiple help channels.

[surreal shoal]

Here it is

[main idol]

!1c

[marsh citrus]

Please stick to your channel.

@surreal shoal Has your question been resolved?

[surreal shoal]

No

[main idol]

Read this

[surreal shoal]

Oh sorry

.close

[marsh citrus]

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[dreamy shadow]

yo i need help

[dreamy shadow]

yo

@short latch perhaps you can help me?

[short latch]

no

[dreamy shadow]

:(

<@&286206848099549185>

[short latch]

<@&286206848099549185>

[sand fable]

lmao

[dreamy shadow]

?

[winged pier]

@dreamy shadow dont ping helpers until after 15 minutes

[winged pier]

btw you can just state your question you don't need to ask anyone for help before your question

[dreamy shadow]

aight bro, i jus need help this shit is due in 10 minutes.

[short latch]

who wanna play fortnite

[dreamy shadow]

aight, someone kick this "amor" fella.

[winged pier]

how many times does 1 show up?

[dreamy shadow]

bro

aight never mind

LMAO

i know how to do it

good lookin out.

[short latch]

idiot.

[dreamy shadow]

?

i forgot

[winged pier]

👎

[dreamy shadow]

thank you moon caik

[short latch]

🤦🏻‍♀️

[dreamy shadow]

do you mind helping on the rest of my questions buddy?

[short latch]

[dreamy shadow]

[short latch]

[sand fable]

do you two know each other outside of this server lol

[short latch]

i have no idea who he is

[dreamy shadow]

we go to college together

[short latch]

???

[sand fable]

insane

[short latch]

hes l;ying

[dreamy shadow]

yeah i lied

i think they want me

[short latch]

[main idol]

! redir

[marsh citrus]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[dreamy shadow]

<@&286206848099549185>

.close

[marsh citrus]

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[still temple]

can the median formula be applied to more than type cumulative frequency distribution?

[still temple]

or should i just convert this to the 'less than' type data to find its median

[marsh citrus]

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[hushed whale]

i am stuck on how to approach this problem as the about line is offset so it is confusing me

ok so i got the answer 11pi/6 but i would like to have it explained as well

[marsh citrus]

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[marsh citrus]

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[remote barn]

let's say APD = a and APB = b, and AB=x
10-6cosa=x^2
8-2sqrt7cosb=x^2
16-6sqrt7cos(180-a-b)=16+6sqrt7cos(a+b)=2x^2

aside from that I can't figure much out

cos(a+b)=cosacosb-sinasinb but idk how that's supposed to help when I can't find out sin

I suppose there's also a graphical way to approach this

let B=0,0 and D=s, s
x^2+y^2=9
x^2+(y-s)^2=1
(x-s)^2+(y-s)^2=7

s^2-2sy=-8
s^2-2sx=6
and we could
s^2-2sy+8=0
s^2-2sx-6=0
since s is positive
s=y+-sqrt(y^2-8)=x+sqrt(x^2+6)

Yeah I have 0 idea

Don't know any theorems behind this

Please send help

<@&286206848099549185> (it's been 15 minutes)

[marsh citrus]

@remote barn Has your question been resolved?

[marsh citrus]

@remote barn Has your question been resolved?

[sand fable]

it’s good you got here

[sand fable]

solve for x and y in terms of s, and plug those into the third equation you have here

[marsh citrus]

@remote barn Has your question been resolved?

[cunning fiber]

Alternatively, find angles PAB and PAD by the law of cosines and then use the Pythagorean identity

[marsh citrus]

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[torpid wing]

hi

[torpid wing]

Can someone explain the calculation of y to me ?

what type of an equation is this ? and why was it added to 1.78 ?

[marsh citrus]

@torpid wing Has your question been resolved?

[marsh citrus]

@torpid wing Has your question been resolved?

[tacit fractal]

tangent calculating

Since the angle is fixed, the tangents are equal to each other

y is opposite

[torpid wing]

oh ok

but why the difference between 1.94 and 1.78?

[tacit fractal]

for the other opposite

[torpid wing]

and why the summation to 1.78 ?

[tacit fractal]

on the far side

what is P exactly

[torpid wing]

50

[tacit fractal]

he said vertical movement of P

50

[torpid wing]

oh

[tacit fractal]

what is 50

[torpid wing]

wait

p =B

so calculating for p

I think I got it

add 1.78 to y

[tacit fractal]

ok there is np then

[torpid wing]

to get full B

full displacement of B

right ?

P is the point of 50KN force

which is the same as B

yh it's solved

thanks for the hints

How do I close this ?

@tacit fractal

<@&268886789983436800> how to close this channel ?

[quaint hill]

.close

[marsh citrus]

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[torpid wing]

.reopen

[marsh citrus]

✅

[torpid wing]

lol

.close

[marsh citrus]

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[marsh citrus]

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[sudden mauve]

Can someone tell me where I went wrong

[sand fable]

everything you did is right but there's an xe^(-x) term you didn't plug in bounds from 1 to 0 for

🤔 you can actually just do that in your final step rn and still get the right answer

[marsh citrus]

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[static ether]

Regarding gradient of summation

[static ether]

How would I find the gradient of f(a)?
a is an 1xn vector,
x^(i) is a nx1 vector,
and y^(i) is a scalar

[marsh citrus]

@static ether Has your question been resolved?

[main idol]

f(a) is a constant

Oh no it's not

[static ether]

it's a value in R, so I guess?

it's a real number

f: R^d -> R

d* not n actually. Either way idk how to take the gradient of this ;-;

it should be possible cuz, given a vector a, it makes sense it has a gradient in the plane (described by the a,x - y whatever part)

<a,x> is the dot product of a and x btw

[static ether]

There's this hint if it helps

[main idol]

Try doing the individual partial derivatives for just n=3

Once you figure out n=3, the pattern will be clear

[static ether]

How would you do this for n=1? 😓

That's all I would like to know

[main idol]

...

[static ether]

;-;

[main idol]

Do you know the definition of gradient

And f_i

[static ether]

I do know f_i, it's [<a,x^(i)> - y^(i) ]^2

[static ether]

The right side of the second screenshot without the summation pretty much

I think I know the def of gradient too

even then...

we're taking the gradient of a dot product. how does this even make sense? 😭

(a_0 x^(i)_0+...+a_d x^(i)_d -y)^2

the issue is the entries in the vector a are not variables, just real numbers

(maybe this doesn't matter?)

[main idol]

a are constant

Again, just figure it out for small dimensions first

d=3

[static ether]

do you know a isn't a function of anything?

I think you know

[main idol]

It's your problem you should know

[main idol]

.

[static ether]

yeah...

[static ether]

the gradient of f would involve derivatives of f with respect to a. I guess, I'm confused because a is a constant (oh maybe this is my problem)

I should try to think of vector a as a variable

ok, I'll try to make sense of it in lower dimensions

[main idol]

Conventionally, gradient is derivative with respect to x

[static ether]

I thought the argument vector a was the "x" here

ok maybe that is also a problem

[static ether]

I'll think of this first

[marsh citrus]

@static ether Has your question been resolved?

[marsh citrus]

@static ether Has your question been resolved?

[marsh citrus]

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[still temple]

can someone explain to me how I can get the image of this matrix ?

[still temple]

ImF = span((1, 2))?

isn't it just that?

[static quarry]

in this case yes

since (2,4) and (3,6) are in that span

[still temple]

ok

but I'm confused

I'm supposed to find a vector u

such that F(u) = w

[static quarry]

what's w

(1, 2)?

[still temple]

yes

yes

[static quarry]

it's the first column of the matrix

so you can take u to be the column vector (1, 0, 0)

in general the column vector with zeros everywhere except a 1 in the k'th row will give you the k'th column of the matrix

[still temple]

ok

but what confuses is that usually when I solve something like F(u) = w, I take the matrix, simplify it (so in this case, i just get rid of the second row) and the do the dot product with a vector u with the compnents u_1, u_2 and u_3

and then set the equations that I get from the product equal to vector w

[static quarry]

i'm not sure i follow, can you give an example of what you mean?

[still temple]

one sec

im gonna draw itr

here

I did what I usually do and got a contradiction 0 = 2

[static quarry]

wait, why did you zero out the second row of the matrix

you need that row too

[still temple]

Can I not do that?

[static quarry]

that's what is going to build the 2 in w

[static quarry]

not in this situation you can't

[still temple]

shouldn't it still give me a solution since it's the same matrix simplified?

[static quarry]

you can do it with the augmented matrix

namely $$\begin{pmatrix}1 & 2 & 3 & 1 \ 2 & 4 & 6 & 2 \end{pmatrix}$$

[elfin berry]

Bungo

[static quarry]

then after simplifying, your last row will be 0 = 0, no contradiction

[still temple]

here I haven't simplified the second row i get

yeah 0 = 0

[static quarry]

yea

[still temple]

but this doesn't help

[static quarry]

your second equation there is just 2 times your first equation

it doesn't tell you anything new

[still temple]

exactly

[static quarry]

row reduction gets rid of the redundant equation

but if you only clear the zeros in the original matrix, not the augmented one, then your last row will give you the imnpossible equation 0 = 2

so just be careful about the context

[still temple]

so you just made an obsevation and came to the conclusion that the vector u has to be 1, 0, 0

since we want the first column to be our answer

but is there a way to do it that works in every situation?

[static quarry]

yea that was a shortcut for when you want your output to be one of the columns of the original matrix, then it's easy

but that's not a unique solution, there are infinitely many u vectors that work

so that leads to the question of how to do it in general

you have the equation
u1 + 2u2 + 3u3 = 1

three variables, one equation

set two of the variables to any values you like and solve for the third

the (1, 0, 0) solution can be obtained for example by setting u2 = u3 = 0 and solving for u1 = 1

but you could do any crazy thing like say:
u1 = 14 and u3 = 2
then your equation becomes:
14 + 2u2 + 6 = 1
or
2u2 = -19
u2 = -19/2

so another solution is (14, -19/2, 2)

btw, if your second equation hadn't reduced to 0 = 0 then you would have two equations and 3 variables
then you have one degree of freedom, set one variable to anything you like and solve the two equations for the other two variables

[still temple]

Oh ok

got it

@static quarry can I show the question I'm working on? I have no clue how to do the second part

[static quarry]

yea sure

[still temple]

what we just did was part of part a

[static quarry]

ok what are you calling the second part? (b) or some of the other stuff in (a)?

[still temple]

no I meant part b

Actually, I think it might be a better idea to not this part for now.

I have now idea what any of this stuff means

I missed the lessons for them

[static quarry]

fiber just means inverse image, the set of all vectors that get mapped to x by F

[still temple]

is x a point?

[static quarry]

the inverse image will be of the form z + N where z is some vector such that F(z) = x, and N is the null space of F

[static quarry]

yea, a point in the image

a point in the image in this case will be a scalar multiple of (1, 2)

[still temple]

I didn't quite understand fiber

[static quarry]

maybe you have heard the term "inverse image" or "preimage"? they all mean the same thing

[still temple]

so it's a set off all the vectors from which space? that get mapped to a point x in which space?

[static quarry]

the set of all vectors from the input space (R^3 in this case) that get mapped to x, which is an element of the image of F, which is a subspace of R^2. So x lives in R^2.

for example (1,2) is in the image of F

and so are all scalar multiples of (1,2)

in fact that's all that is in the image here

scalar multiples of (1,2)

[still temple]

that'S what we are trying to show mathamatically right?

[static quarry]

well it's kind of clear from the matrix

the columns are all scalar multiples of each other

so any one of them suffices as a basis for the image

in general, the image of F would be the set of all linear combinations of the columns of F
image is the same thing as column space btw

[still temple]

I'm not understanding it

I'm going to go read the lecture notes

and I'll be back

[static quarry]

ok

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[velvet elk]

I don't understand question a

[velvet elk]

I guess equation of a circle?

.close

[marsh citrus]

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[iron acorn]

may I please have help on 1 c and d, it is year 11 math

[marsh citrus]

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@iron acorn Has your question been resolved?

[iron acorn]

close

.close

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[teal geyser]

Help with trigonometr

[teal geyser]

Give me hint on this one

[torpid mountain]

cos 2x = 1 - 2sin²x

its one of many double angle identities

[marsh citrus]

@teal geyser Has your question been resolved?

[restive jacinth]

yea and 0.28 would be the answer

[marsh citrus]

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[analog plover]

why does line 2 = line 3?

[devout mauve]

(1-a)^3=-(a-1)^3

[analog plover]

i mean the highlighted part

[devout mauve]

fuck me I cant read

factor out the 13(a-1)

note that 13*4=52

its easier to multiply out line 3 and see that you get line 2 back

[analog plover]

ty

.close

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[placid anchor]

i

[placid anchor]

need help with this

the outer line of a trophy is described with the following functions in the picture

i first and foremost dont understand s(x) function
it says the s(x) is between (2;1) & (8: 0.5)

am i supposed to use the two points to determine the functions value or how else youd call it

because it seems the s(x) is a linear function or am i wrong

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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[fluid kiln]

What's the name of this graph?

[marsh citrus]

@fluid kiln Has your question been resolved?

[frigid patio]

It seems to be a mix of two things called, subbotin distributions.
Seen here: https://www.desmos.com/calculator/hhbwes3ff1

And here, https://stats.stackexchange.com/questions/201038/does-this-distribution-have-a-name-fx-propto-exp-x-mup-beta

And here, https://stats.stackexchange.com/questions/203629/is-there-a-plateau-shaped-distribution

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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[stuck solar]

im getting sqrt 34/5

but theres no option for it

[lunar light]

what brought you to that conclusion?

[stuck solar]

i did the pythagorean theorom to find opp side

since cos = adj/hyp

[lunar light]

correct

so what did you get the opposite side as?

[stuck solar]

sqrt 34

[lunar light]

howcome? the adjacent squared + the base squared = the hypotenuse squared

square root 34 would be what you would get if you considered the 3 as the height and 5 as the base, or vice versa. 5 is the hypotenuse

[stuck solar]

so i did

sqrt(5^2-(-3^2))

5^2+3^2

and i got 34

considering 5 as hyp

[lunar light]

wrong

5^2 = x^2 = 3^2

[drowsy crag]

You square -3 aka (-3)^2 = 9

[lunar light]

how does that matter? -3 squared is the same as 3 squared

[drowsy crag]

To me it seems like -3 squared is -9 for him

[stuck solar]

no

squaring or raise to any even number is positive

[drowsy crag]

Yes i know that im saying i think he got -9

[drowsy crag]

By this message

[stuck solar]

i dont get it

[lunar light]

look, hypotenuse squared = base squared + height squared

[drowsy crag]

Idk why he did 5^2 - (-3²)
Instead of 5² - 3²

[stuck solar]

right

[lunar light]

you have the base as 3 and hypotenuse as 5

so you get 5^2 = 3^2 + height squared

so height squared is 5^2 - 3^2 which is 4^2

so height is 4

[stuck solar]

i see

i over complicated it

but

hyp= b + h right

to get h

hyp - b

if i substitue values

wont it become

5^2 - -3^2

since 3 is negative

[inland hemlock]

i’d draw a lil unit circle on the side ngl

and draw y = -3/5

check intersections

it gets dead easy

[marsh citrus]

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[dawn vale]

I have a question: I'm simplifying from a previous post that didn't get solved.

[marsh citrus]

Please don't occupy multiple help channels.

[marsh citrus]

@dawn vale Has your question been resolved?

[late hazel]

Do we know either of the distances between the known 16.21 length and Y1 or Y2? or the distance between the left end of the 4.1 segment and Y1?

[marsh citrus]

@dawn vale Has your question been resolved?

[late hazel]

Atleast using trigonometry this is the best i could do with the given information

There may be some analytical things one could do so determine the exact length that would force the 1.1 length line to be constrained between the two legs of the triangle but this is what I had time for at the moment

Having any combination of the lengths that I labeled d1, d2, or d3 and d4 (both of these together) you could solve this problem

[marsh citrus]

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[random reef]

want to make sure i have this correct, superscript is column and subscript is row

[marsh citrus]

@random reef Has your question been resolved?

[marsh citrus]

@random reef Has your question been resolved?

[random reef]

<@&286206848099549185>

[marsh citrus]

@random reef Has your question been resolved?

[gaunt pagoda]

what is the question

@random reef

[random reef]

is my thinking correct about the notation

just havent seen it stacked like that

[gaunt pagoda]

Depends on what you want to write? that notation means something yes

it's a set of vectors where the elements squared are the same

[random reef]

oh thats just saying (x_1)^2

[marsh citrus]

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[marsh citrus]

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[tender plover]

Would be extremely greatful if I could get any help on question 6

[marble flame]

Try to graph each single equation

Then for domain and range

If inequality with equal sign then use solid line

If not dashed

@tender plover

[tender plover]

Okok

Lemme try

So would I just put each equation in my calculator and graph them first?

[marble flame]

I will sketch the first equation

Hand on there

Hang*

[tender plover]

Kk

[marble flame]

@tender plover

[tender plover]

Oh I see

U started from 5

[marble flame]

Because of it being y intercept

[tender plover]

If it’s negative

Don’t u go down two

And the to the right there

Three*

[marble flame]

Since negative is next to the fraction

You can choose whether it down two

Or 3 left

[tender plover]

Oh

Is that like a math rule

Never knew that

[marble flame]

Because of the condition I went 3 left

[tender plover]

I see

[marble flame]

X is less than or equal to zero

[tender plover]

Does this look correct for the first line?

Can’t really see it well my bad

X is equal to or less then to zero

[marble flame]

Yes

[tender plover]

I forgot like what u do with information

[marble flame]

Wait I'll give you a great source

[tender plover]

kk

[marble flame]

https://algebra.flippedmath.com/85-solving-systems-of-linear-inequalities.html

Check this flipped math

[tender plover]

Kk

Damn

Ngl I don’t understand a lot in the video

Like the shading graphing dotted line all that

[marble flame]

Ngl the video

[tender plover]

Should I watch it fully

[marble flame]

Second grap is like this

Wait

[tender plover]

Kk

[tender plover]

Ooof okay

So by the second graph we are talking about -x if 0<x<4

Right?

[marble flame]

Yes

Are you able to graph the 3rd

[tender plover]

Not really but hold on

For the second graph could u explain the thought process

Like the open circle

And why you started from 0

I’m trying to piece it together but like idk

[marble flame]

Open circle because the inequality sign does not have equal

Started from 0 because of the condition given

@tender plover

.....

[tender plover]

Oh man I’m lost sorry

Give me a second

So open circles are for inequality signs with no line under it

I wrote the wrong thing in my notes fuck

So if -x is smaller then x

Wait so if 0 is smaller then x

And -x is smaller then 0

Is that what u mean by condition

[marble flame]

Don't overcomplicate things

If you are confused

Draw a numberline

[tender plover]

Kk hold on

[tender plover]

I c

[marble flame]

Yes

[tender plover]

So for this line it ends in a closed circle cause of the <4 part

Less then or equal

To 4

That part right

[marble flame]

Yes

[tender plover]

So how come for like that first line we did

It was like -2/3

It was like plotted differently yk what I mean

-2/3x+5 we did like rise over run to plot it

[marble flame]

Yes

The 2nd equation is x

[tender plover]

And then for second equation I’m like confused

Yea

[marble flame]

-x**

[tender plover]

Ye

[marble flame]

So linear

But opposite side

B/c -

[tender plover]

So you cant do what we did in the first graph right

[marble flame]

Open desmos to see graph

[tender plover]

-x/1 won’t work

[marble flame]

@tender plover we did same logic

[tender plover]

Kk

[tender plover]

I’m so sorry I’m like so bad at math

Just give me a second

This is new to me

This second line is confusing me idk why

Like what to do

[marble flame]

If you are confused do it like this

And then apply the domain on the graph

[tender plover]

Kk

Something like this?

[marble flame]

Because 5 in and intercept meaning (0,5)

@tender plover no

I am explaining

[tender plover]

Oh my god im dumb

My bad

[marble flame]

We will start from 0

And make close hoke

[tender plover]

Kk

[marble flame]

Got is

Got it?

[tender plover]

Like this ?

[marble flame]

Wait

[tender plover]

Closed hole

My bad

[marble flame]

Don't do anything now wait

[tender plover]

Kk

[marble flame]

It will become like this if we applied the logic we just used

[tender plover]

Mhm

That’s the first graph

Right

-2/3x+5

But then the if x is less then or equal to 0 part

[marble flame]

2nd equation

[tender plover]

Oh

Wait how’s that the second one

I’m so lost lmao

[marble flame]

I meant now I am going to explain 2nd

[tender plover]

Kk

[marble flame]

Ignore the graph

Above it

Now got it???

@tender plover

[tender plover]

Kinda

I see

What u mean by like

The number line

[marble flame]

This

[tender plover]

Yea

So now how would u graph this though?

Gimme a second I gotta like try to understand this

[marble flame]

I see what you don't understand

[tender plover]

Ok so I understand how the number line is like representing the x axis

Axis

[marble flame]

Yes

So fine till now?

[tender plover]

Yea

[marble flame]

For the last equation there is a shortcut that saves a lot of time

It is called 1-3-5 shortcut to graph quadratics

[tender plover]

Kk

[marble flame]

I'll show it

Wait

I still did not apply condition

So 1-3-5 shortcut says

Multiply coefficient of a with each of them

Then at the vertex

Go right 1 step then go up or down depending on what you get when you multiply a coefficient with 1 3 5

@tender plover

[tender plover]

Yea

[marble flame]

And apply same left

[tender plover]

That formula

-b/2a

[marble flame]

Yes

Get vertex first

Then apply the shortcut

Getting the idea????

@tender plover

[tender plover]

So find the vertex of the third graph

First ?

[marble flame]

Yes

Then what will you do?

[tender plover]

Tiny problem

So wait

How would u graph that third thing

That line looks complicated

[marble flame]

Its not

I'll simplify the concept

[tender plover]

Idk how to graph using the equation

That’s like my main problem right now

[marble flame]

Ok

Wiat

Wait*

[tender plover]

Kk

[marble flame]

I introduced the shortcut because I know you will graph it using a table

[tender plover]

Oh wait

Hold on a sec

Is this what you mean ?

[marble flame]

Yes

[tender plover]

And then this table right

This is what I did in class

But the thing is then for that first graph could u also just put it in the calculator

And the second graph also?

[marble flame]

This explains the shortcut

[tender plover]

Kk ty

Give me a second

[marble flame]

https://m.youtube.com/watch?v=tFg_uJPZSg0&pp=ygUdMS0zLTUgc2hvcnRjdXQgZm9yIHF1YWRyYXRpY3M%3D

If you need more help

[tender plover]

Oh

Wait so this second line right

I mean third line sorry

[marble flame]

No we did not apply domain

Which what I am doing now

[tender plover]

Is this the formula u used for the graphing of the third equation?

This equation wouldn’t work on the first two graphs right?

That’s confusing me

[marble flame]

And get final answer as shown below

[tender plover]

I see

[marble flame]

@tender plover understand fully?

[tender plover]

A lot more I just need some time to try like do it on my own

Thank you so much for the help

[marble flame]

Practice my friend and you will master this skill

[tender plover]

Yes sir

[marble flame]

Understand the concept you are good at math but you need to practice

Always believe that you are capable of acing such simple things

[tender plover]

Yeah your right sorry this took a while a lot of this isn’t stuff I’ve ever seen before

.close

[marsh citrus]

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[short pagoda]

How do you solve this

[stark trail]

it's an equation

[short pagoda]

oh

i know

[sinful thistle]

dyk how to convert log bases?

[marsh citrus]

@short pagoda Has your question been resolved?

[sinful thistle]

@short pagoda do you know how to convert log bases? use that here.

[marsh citrus]

@short pagoda Has your question been resolved?

[marsh citrus]

@short pagoda Has your question been resolved?

[short pagoda]

Is there a problem somewhere?

[merry bay]

;[ Lah, lemme check.

[short pagoda]

yeah, no problem hehe

yes

[merry bay]

I used the wrong equation, nvm.

[short pagoda]

No need to rush, my friend

[marsh citrus]

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[mortal hinge]

Find the rate at which Rs.4000 will give Rs.630.50 as compound interest in 9 months , interest being compounded quarterly

[marsh citrus]

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[marsh citrus]

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[sharp vessel]

(x+1)^2 is never negative

What are you even trying to do

(x+1)^2 > 0
and then you wrote
x + 1 > 0
which isn't necessarily true

you can't just take the square root

sqrt((x+1)^2) is not x + 1

its |x+1|

just like sqrt((-2)^2) is 2 and not -2

[marsh citrus]

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[marsh citrus]

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[restive anvil]

Hi, I'm doing the integral of a rational function of sin and cos and I don't understand how to do the steps I've marked in red.

I don't understand how you can delete the whole second part of the integral in one go, I can't see where it cancels out.

[whole sleet]

I assume they just multiplied the denominator out

Huh, then again, the degree seems wrong

[restive anvil]

I still don't understand. Can you show me? Sorry, but I'm really lost.

[whole sleet]

.

I'm looking at it. It's a bit suspicious though.

,w integral of cos(x) / (1 + cos(x)) between 0 and pi/2

[elfin berry]

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[whole sleet]

Well, they have the right answer

[restive anvil]

how would you do it without this example

oo ok i get it

[sand fable]

they multiplied the denominators (1 + (1 - x^2)/(1 + x^2)) and (1 + x^2) to get 2

[restive anvil]

.close

[marsh citrus]

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[restive anvil]

yep

ty

[sand fable]

ah ok

[marsh citrus]

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[spiral cedar]

would someone be able to explain how to do this question, I thought we just plug in each sample as a binomial distribution but i must be wrong

[spiral cedar]

wait they should be multiplied not added i believe

still incorrect though

.close

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[reef flare]

What is h'(y)?

[marsh citrus]

@reef flare Has your question been resolved?

[reef flare]

<@&286206848099549185>

[dim schooner]

h'(y) is a unknown function that depends on the variable independent y, if therfore when doing a partial derivation with respect to x then h(y) would be a constant and if we deriv a constant the result = 0

[marsh citrus]

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[primal ridge]

I am having trouble understanding what my professor did for this homogeneous equation example. Everything is fine up until near the bottom where the natural logs start to cancel. How do the natural logs cancel? And why does c+x become CX here?

[still temple]

must have considered C = lnc

[primal ridge]

was it then incorrect to write ln + c? or is that a change your allowed to make when working with the problem

[sinful thistle]

nah it's an arbitrary constant

[still temple]

you mean lnx +c ?

[still temple]

nope

[sinful thistle]

ln c is just converted to c here bc e^c is tedious

[still temple]

exactly

[sinful thistle]

why not just write "C" when it can be anything

[primal ridge]

I might be missing the concept here. But what your trying to tell me is that lnx + c = ln |u-3/u-2| is the same as lnC = ln |u-3/u-2|

[still temple]

try this

[primal ridge]

Oh. Well. I see the problem now... I didnt have the end of the example because I got a page mixed up.....

So if that's supposed to be the solution. Let me check if C = lnc works.

OK yea that all checks out. IG whats confusing me is how the addition changed to multiplication for ln and c. I guess its possible I missed some algebra he wrote down.

Or is that some sort of notation that im messing up on?

Oh wait I see now. Is it adding together the two logs and then being able to dvide it out?

That would get you frim lnx + lnc to lnxc. Which would then divide out to just xc and get the same solution.

I think I have the hang of this now. Thank you very much.

.close

[marsh citrus]

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[light hedge]

I got the answer I just don’t know what it means by calculating the magnitude

[marsh citrus]

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[bright rivet]

i need help its my last task for hw its easy probability

i dont understand here tried everything lol

[drowsy crag]

What is prob of red red and what is prob of green green

[bright rivet]

how do i work it out?

[drowsy crag]

Multiply

[bright rivet]

green with green and red with red?

its weird tho it asks me for one awnser

can i please get help with this?

i multiplied and it didnt work

[marsh citrus]

@bright rivet Has your question been resolved?

[marsh citrus]

@bright rivet Has your question been resolved?

[marsh citrus]

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[surreal glade]

hi

[surreal glade]

$$\frac{1}{(x+2)^2} \cdot e^{\frac{1}{x}} + \frac{x+1}{x+2} \cdot -\frac{1}{x^2} \cdot e^{\frac{1}{x}}$

[elfin berry]

alee

$$\frac{1}{(x+2)^2} \cdot e^{\frac{1}{x}} + \frac{x+1}{x+2} \cdot -\frac{1}{x^2} \cdot e^{\frac{1}{x}}$
```Compilation error:```! Display math should end with $$.
<to be read again> 
                    
l.49 ...cdot -\frac{1}{x^2} \cdot e^{\frac{1}{x}}$
                                                  
The `$' that I just saw supposedly matches a previous `$$'.
So I shall assume that you typed `$$' both times.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}{/usr/l
ocal/texlive/2023/texmf-dist/fonts/enc/dvips/lm/lm-rm.enc}{/usr/local/texlive/2```

[surreal glade]

$\frac{1}{(x+2)^2} \cdot e^{\frac{1}{x}} + \frac{x+1}{x+2} \cdot -\frac{1}{x^2} \cdot e^{\frac{1}{x}}$

[elfin berry]

alee

[surreal glade]

from this how can i get

[stoic topaz]

get what?

[surreal glade]

$-e^{1/x}\frac{3x+2}{x^2(x+2)^2}.$

[elfin berry]

alee

[surreal glade]

this

the idea is this one

but i dont get it

$\frac1{(x+2)^2} - \frac 1{x^2}\frac{x+1}{x+2} = \frac{x^2-(x+1)(x+2)}{x^2(x+2)^2} = \frac{-3x-2}{x^2(x+2)^2}.$

[elfin berry]

alee

[drowsy crag]

We are still at derivative right?

[stoic topaz]

just add the fractions?

your question isn't very clear to me