help-23

[long shore]

yes

so we can take 3 out and what does that make it?

[rigid path]

Ok

Uhh

A^2 b^3 + 2ab^4

?

[long shore]

yes!

ok so

,tex $a^m \times a^n = a^{m+n}$

[flat frigate]

suds

[long shore]

right?

[rigid path]

Yes

[long shore]

so we can also take out the a and b

[rigid path]

Uhhh ok

[long shore]

so how many a's and b's can we take out from both of them without getting a negative number of a's and b's

remember $a^4 = a \times a \times a \times a$

[flat frigate]

suds

[rigid path]

Uhhhh 1 a and 3 bs

[long shore]

yep

so we have taken out 3, a, b^3

so we can put that at the front of the equation

[rigid path]

Kk

[long shore]

,tex $3ab^3(a + 2b)$

[flat frigate]

suds

[long shore]

does that make sense?

[rigid path]

Ohhh

Ye

[long shore]

ok do you want to try it with the original equation?

its the same thing

[rigid path]

Uhh ok

??

Idk I feel dumb

[long shore]

give me a second to check

[rigid path]

Ok

[long shore]

ok so thats basically it but you just might have put something in wrong

its 24 out the front

ill send the original equation again

[rigid path]

Kk

[long shore]

,tex $ = \frac{18r^5t^5 - 36r^4t^6}{48r^3t^6 - 24r^4t^5}$

[flat frigate]

suds

[long shore]

so that was right with the number of r's and the number of t's

but 24 is out the front

and there are 2t's in the brackets

so

,tex $24r^3t^5(-r + 2t)$

[rigid path]

Oh

Ok

[long shore]

yup?

ok so we can also change it to take a negative out the front

[rigid path]

Huh?

[long shore]

and all that does is reverse the signs in the brackets

[rigid path]

😭

[long shore]

,tex $-24r^3t^5(r - 2t)$

[flat frigate]

suds

[long shore]

see its still the same equation?

but you are just changing around the negatives to positives?

[rigid path]

Uhh

Ok

[long shore]

,tex $\frac{18r^4t^5(r - 2t)}{-24r^3t^5(r - 2t)}$

[flat frigate]

suds

[long shore]

so thats the full equation

the reason that we want to take the - out the front is so that the things in the brackets are the same

because then we can cancel them out

[rigid path]

Ohh

[long shore]

,tex $\frac{18r^4t^5}{-24r^3t^5}$

[flat frigate]

suds

[long shore]

yes?

[rigid path]

Cancel out the t?

[long shore]

yup we can cancel out the t as well

[rigid path]

Kk

[long shore]

,tex $\frac{18r^4}{-24r^3}$

[flat frigate]

suds

[rigid path]

So that should be uhm

Divided by 3?

[long shore]

we gotta handle the rs still

so we have the index law of

[rigid path]

Oh ok

[long shore]

,tex $\frac{a^m}{a^n} = a^{m-n}$

[flat frigate]

suds

[rigid path]

Kk

[long shore]

so what does that make r?

[rigid path]

42 r?

😭

[long shore]

ok so

,tex $\frac{r^4}{r^3}$

[flat frigate]

suds

[long shore]

yes?

if we forget about the other stuff for a second

[rigid path]

Ok

[long shore]

so it will be 4-3

so its just r

[rigid path]

Yeah

[long shore]

,tex $\frac{18r}{-24}$

[flat frigate]

suds

[long shore]

and because 24 and 18 are divisible by 6

[rigid path]

Ye

[long shore]

,tex $ = \frac{3r}{4}$

[flat frigate]

suds

[long shore]

and thats it

[rigid path]

Oh

Kk

[long shore]

does all that make sense to you?

[rigid path]

Yes

I think so

[long shore]

if you need any more help you can dm me if you want

[rigid path]

Okkk

I'll friend you then?

If that's OK with you

Accidental caps

Ye

[long shore]

yes thats fine

[rigid path]

Kk tyyy

.close

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[safe radish]

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[rare hornet]

Hello

[rare hornet]

Hello help me

Answer this.

Finding the roots of the pelyoomial function, f(x) = 2x⁴+x³ - 16x² + 3x + 18

[silent narwhal]

Ok so using factors of 18

[hazy elbow]

rational roots theorem

[silent narwhal]

Find when f(x)=0

Then u know ifs a factor as if u divide by x=(the factor) u get 0 remainder

Then u can just do polynomial division

Should be simple from their on tag if u need help

[rare hornet]

Ok

How about this?

[silent narwhal]

Does it equal zero?

Then h know x=-3 is a factor

Hence u can use x+3

And divide the polynomial by it

[rare hornet]

Thank you.

Can you give the correct answer?

[silent narwhal]

Have a go and and tell me I’ll correct

[rare hornet]

I got it. Thank you very much

[safe radish]

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[feral beacon]

how would i set up this question

[hazy elbow]

well, f(x)=25-x^2

so, f(c)=25-c^2 which is also equal to 71/3

solve for c

and you'll get two values out of which one would be rejected due to restriction on the domain

[safe radish]

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[lean coyote]

Sup

[lean coyote]

My problem is 1.2A

I need to find the sum of possible a s that

Has x y z have only one solution

I can t solve it with a matrix

Because it s not square

If i knew how to make it square

Maybe i would find a solution

As a matrix it looks like this

But idk what to do with it

[safe radish]

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[safe radish]

@lean coyote Has your question been resolved?

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@lean coyote Has your question been resolved?

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[warm crest]

im kinda confused on how an infinite series converges, and i dont mean in terms of solving a problem i mean in general

[warm crest]

if there are infinitely many terms how do they converge because i took a limit

Ive been doing the problems but any time i think any deeper than "Taking the limit of the summation proves its convergent or divergent" i get really confused

[peak estuary]

do you just mean that you are confused about how adding something infinitely often can still only give you a finite number?

[warm crest]

yes

[peak estuary]

imagine a circle

[warm crest]

mmmm circle

[peak estuary]

at each step shade half of the remaining part of the circle

so after the first step half the circle is shaded, then 3/4, then 7/8 and so on

yes?

[warm crest]

when you say step do you mean on the line of the circle or on the circle

[peak estuary]

the area

step by step you shade in some part of the area

[peak smelt]

do you understand how an infinite sequence converges?

[warm crest]

oh

yea

[peak estuary]

well a series is just a sequence of partial sums. if that helps

[warm crest]

oh, thx

[peak smelt]

its funny, i thought it was easier to see how an infinite series of fractions does converge than doesnt

suppose you draw a line

then you draw a point thats x distance above the line

then if u go to the opposite side of the line an equal distance, then move a tiny bit closer

then u go back to the top, but a bit closer again

then u keep switching sides

this is a type of infinite series that converges eventually to what?

the line

[peak estuary]

well ok but that only works for alternating series. not every converging series is alternating

[peak smelt]

just giving an example of another easy to visualise infinite series

another way you can think of it

take a piece of paper

rip it in half

then take one of the halves, and rip that in half

and add it to the first piece

then rip the new piece in half again

keep ripping in half and adding it to the other pieces

u are never going to get the full piece of paper right?

you keep adding a bit more and more, but never reach the full piece of paper

you keep adding more pieces, but they just get smalller and smaller

[safe radish]

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[earnest flume]

<@&286206848099549185>

Couldn't find the min area

[slow summit]

My guess is the minimum area would be achieved when the folded over section is half of a square

[earnest flume]

thats the max area

28^2/2.

shd be using differentiation optimization

<@&286206848099549185>

[safe radish]

@earnest flume Has your question been resolved?

[safe radish]

@earnest flume Has your question been resolved?

[earnest flume]

<@&286206848099549185>

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@earnest flume Has your question been resolved?

[safe radish]

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[fossil vine]

Im not sure how to start the proof

[peak estuary]

what does it mean to be odd?

[fossil vine]

it can be written as 2n+-1

right

[peak estuary]

yes

[fossil vine]

so how could i use that

[safe radish]

@fossil vine Has your question been resolved?

[peak estuary]

plug a=2n+1 and b=2m+1 in

[safe radish]

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[lean otter]

How do u solve this one? ( 6)

[plucky elk]

,rotate

[hollow pagoda]

,rotate

[lean otter]

Oops

Let me take a pic unrotated

Here

[hollow pagoda]

pick 1

oof its french?

[lean otter]

Italiam

Uh

Idk how to do 6

[hollow pagoda]

pick 1 and maybe try to translate to english?

tell what we should do in 6

[lean otter]

Ok hold on

"calculate the values of the following expressions, applying the properties of exponents whenever possible"

[hollow pagoda]

k i can do 4 with you and let u try to solve 6 alone?

[lean otter]

Ik how to do 4 and 5, im just confused on 6

[hollow pagoda]

can you show what you have done in 6?

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[lean otter]

Ye

Idk what to do after this @hollow pagoda

Specially the fractions divided by fractions that are already divided by fractions

😭

[hollow pagoda]

can you calc:
(1/2) / (1/5)

[lean otter]

Wdym?

[hollow pagoda]

just do it

its an example

[lean otter]

Ok

It's 5/2 i think

Your lack of awnser makes me think i awnsered it wrong and now you are praying to the gods of math for my downfall

[hollow pagoda]

no u did it correctly lol

[lean otter]

Oh

Lol

[hollow pagoda]

if u could write what u got here would be better because i didnt rlly understand what is written

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

8/5 - 9/7 × ( 22/15)-3/43 x ( -43/60) : 9/4- (11/12): 2,5-2,1/2-0.99999

@hollow pagoda

[lean otter]

😭

[lean otter]

Its dynover

[hollow pagoda]

im back lol @lean otter

[lean otter]

We are so back

[hollow pagoda]

8/5 -66/35 + (1/20) : (9/4) -(11/12) : (5/2) - (1 + (1/20) + 0.99999)

you agree with this?

this is according to what u send here

[hollow pagoda]

.

[safe radish]

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[wanton shell]

How do I rationalize?

[scenic thunder]

Help!

[icy lance]

#❓how-to-get-help !

!occupied

[safe radish]

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).

[icy lance]

!

[safe radish]

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[safe radish]

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[lean otter]

a) Determine the equation of the line that passes through the points (-1,5) and (1,1 )
b) Determine the equation of the line that is parallel to the one in problem a and that passes through the origin

[lean otter]

I don’t know how to determine the equation

I tried making a table

Of x and y

[hollow pagoda]

its a line

[lean otter]

Oh

Uhhh why

[hollow pagoda]

you mean (1,1) ? or (11)

[lean otter]

Oh yeah sorry 1,1

[hollow pagoda]

good find the slope

you know how?

[lean otter]

It’s a slope?

Like the round one

And not a line?

So u have to use that complicated formula

?

Or is it y = kx + m

[hollow pagoda]

in order to find the slope of a line:
you need to use:
slope = a = delta(y) / delta(x)

[lean otter]

Yeah

I did 5-1/1-1

Is that incorrect

To find the slope

[hollow pagoda]

its 5 - 1 / -1 - 1

which is 4 / -2 = -2

[lean otter]

Oh

Okay

And then y = -2x + m

So I can replace y with any of the y values

???

[hollow pagoda]

then u need to find m

by subsituting 1 point

lets say (1,1)

[lean otter]

Mhm

1 = -2*1 + m

Right?

[hollow pagoda]

so m = ?

[lean otter]

3

?

😅

[hollow pagoda]

good so what is the equation?

[lean otter]

Y = -2x + 3

ohhhHhhH

Thank you

I got it now

[hollow pagoda]

good luck

[lean otter]

.close

[safe radish]

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[wind hemlock]

How do I transform this into the uv-plane?

original question is here

the graph is the region on the xy-plane

[quiet juniper]

<@&268886789983436800>

start from y <= x < = y+2 bounds

try to make that into a <= u <= b for some real numbers a and b

Second set of bounds is 0 <= y <= 1

I think u can figure that one out

[wind hemlock]

uh

[wind hemlock]

dont you need to consider the inequality with y in the first expression?

[quiet juniper]

i did

[quiet juniper]

.

[wind hemlock]

yes that

[quiet juniper]

yeah

[wind hemlock]

so to go to a constant 'a' and 'b' you would need to plug in values of y(?)

[quiet juniper]

no

u can be anything

Doesnt even have to be a constant

[wind hemlock]

well 'a' and 'b' have to be constants tho right

[quiet juniper]

Ye

[wind hemlock]

and since currently 'a' and 'b' are in terms of y (?)

then you would need a value of y to solve for 'a' or 'b' (?) doesnt sound right

[quiet juniper]

no

u can be an expression in terms of other variables

[wind hemlock]

yes

[quiet juniper]

including y

[wind hemlock]

yes

[quiet juniper]

Maybe take the integrand for some further inspiration

I mean loik at the integrand

Wishful thinking

[wind hemlock]

i mean i could take u = x-y but what would v=

[quiet juniper]

take a guess

[wind hemlock]

0?

[quiet juniper]

no

[wind hemlock]

u^0.5?

[quiet juniper]

Look at the second set of bounda

[wind hemlock]

0<=v<=1

[quiet juniper]

yeah

[wind hemlock]

wait then does this become a unit square

in the uv-plane

[quiet juniper]

yeah thats a lot easier to inregrate over

[wind hemlock]

oohhhhhh

[quiet juniper]

Well

Not unit square actually just a rectangle

[wind hemlock]

oh yeah cuz the bounds of u would be 0 to 2 i think

[quiet juniper]

yup

[wind hemlock]

oh this makes sense now

is there a fast way to look at it or is it just intuition/trial

[quiet juniper]

intuition/trial i guess

ideally you end up inregrating over a rectangle or a rectangular prism or whatever

[wind hemlock]

yeah that ould be the best

ok thx for the help

[quiet juniper]

np

[wind hemlock]

.close

[safe radish]

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[wind hemlock]

.reopen

[safe radish]

✅

[wind hemlock]

one more thing. Since u = x-y what would be the Jacobian? v is kinda doing its own thing so it isnt related to x or y

[quiet juniper]

oh right yeah

v = y

[wind hemlock]

oh is it because they both have same range on their respective planes

[quiet juniper]

yeah

[wind hemlock]

oh ok got it

[wind hemlock]

so then we can set x = u+v and take partials of that for x?

since y = v

[quiet juniper]

yuuuup

[wind hemlock]

yayyyy

ok thanks

that was the last question

.close

[safe radish]

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[hard forum]

help

[gritty glacier]

red travels along the length of a semicircle which has a radius of 10 metres

[hard forum]

object 2

so blue path

[gritty glacier]

yes

[hard forum]

you said red

object one is just 10 pi meters right?

[gritty glacier]

yes

which is approximately 31.4159 meters

[safe radish]

@hard forum Has your question been resolved?

[hard forum]

ok but for the blue path?

is the blue path 28.28 m?

[safe radish]

@hard forum Has your question been resolved?

[safe radish]

@hard forum Has your question been resolved?

[wheat condor]

Nope - try again

@hard forum Blue goes up & back down - two parts

$a^2 + b^2 = c^2$

[flat frigate]

G. Spark

[hard forum]

I’m getting the same answer tho?

-10^2 + 10^2 = c^2

sqrt(200) = sqrt c

which is 14.14

then same on the other side which is 28.28?

[wheat condor]

YEP - Oh, I see, the whole trip, yes - twice that

[hard forum]

ok thank you how about that yellow line?

[wheat condor]

Can you see how to do it?

[hard forum]

just use this?

[wheat condor]

Yep. five times.

[hard forum]

ok thank you

.close

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[loud bronze]

hi my prof is just talking about cyclic subgroups

nvm

i understand it lol

.close

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[magic tusk]

is a position vector literally just the 'i' and 'j' thing?
so to find it
i would do
sally = [30, 40]

x = 30cos50
y = 30sin50 + 10 (sally's friend is on a platform so it increase's y-axis right?)

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[true estuary]

for this question do i just need to investigate b>0, b=0, b<0?

[drowsy karma]

it is too less, you should play with Jacobi's matrix and its interepreation of stability of such origin and its kind

you also need to identify eigen values

[true estuary]

i dont think so

surely u can just use the trace

and determinant

to identity sink nodes spirals etc

let trace = T and detA = D, u can use T^2 - 4D

[drowsy karma]

i know it but didactically

[true estuary]

whether it is less than, equal to, or more than 0 is a different story

[drowsy karma]

you need to su eJoacobis matrix

for example, if both eigenvalues are positive, then the node is unstable

[true estuary]

ye

ok ill have a go

thanks

[drowsy karma]

yw)

[true estuary]

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[wicked tide]

Jason considered two similar televisions at a local electronics store. The generic version was based on the brand name and was three eighths the size of the brand name. If the generic television set is 12 inches by 24 inches, what are the dimensions of the brand name television?

How do I solve this?

[safe radish]

@wicked tide Has your question been resolved?

[fickle pendant]

let the width of the brand name television be W. how can you write an equation in terms of three eighths, W, and 24?

I'm assuming this isn't some weird TV that is twice as high as it is wide

but the same thing applies

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[jade gust]

how?

[lean otter]

Thats the general form of writing answers

[jade gust]

ye how’d they solve for x

[lean otter]

You got
cos2x = -2
x = {arccos(-2)} / 2

[frigid spruce]

eragon

[lean otter]

Then using that and the general form

That answer came

[lean otter]

Yajat

[jade gust]

?

[lean otter]

What?

[jade gust]

what does any of what you’ve said mean

[lean otter]

This?

[jade gust]

ye mate

i mean like anything youve said

[lean otter]

Oh wait okay

[jade gust]

general form of writing answers?

[lean otter]

Your original questions was
2cos²x + 1 = 0?

[jade gust]

where’s this comin from

[jade gust]

right

[lean otter]

I thought this was your original question, my bad

[jade gust]

I’m just tryna figure out how they solved for x

[lean otter]

Do you know any formula for cos2x ?

[jade gust]

do i have to touch this or sum

[lean otter]

No need

[jade gust]

in what way would that help

[lean otter]

Cus otherwise

Using 2cos²x + 1 = 0

You go and try to find for x

There is no answer

As

You get

cosx = √(-½)

And you cant have a negative under root

[lean otter]

{Note: btw if youre in lower grades answer is x doesnt exist
If youre in higher grades, you need to show how x doesnt exist}

[lean otter]

What grade youre in

Like just so i know you know stuff like complex numberz domain on trigo functions etc

Atleast

[jade gust]

first

[lean otter]

Then

Just tell your teacher
cosx = √(-½)

From the given question

Therefore, it is not possible

Therefore, x doesnt exist

{for 1st graders}

[jade gust]

nah but i wanna figure out how WA got that

[lean otter]

(or you can use cos2x formula and then show
arccos(-2) which again doesnt exist)

[lean otter]

Do you know cos2x = 2cos²x - 1 ?

Just use that

[frigid spruce]

Eragon

[lean otter]

2cos²x + 1 = 0
2cos²x - 1 + 2 = 0
cos2x + 2 = 0
cos2x = -2
2x = arccos(-2)
x = {(arccos(-2))} / 2

@jade gust youre welcome

[lean otter]

Yajat

[frigid spruce]

dont ping me

lmao sorry

[jade gust]

aigh cheers was overcomplicating it

.close

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[lean otter]

When they take away 4^2 from both sides, why does it go to the back of x^2 = 8.5^2 as -4 ? I thought it goes away no? Cos you took it away. Can someone explain please

[real ocean]

could you rephrase your question

[lean otter]

the guy said he wants X^2 by itself so he subtracts 4^2 from both sides, im confused on how subtracting 4^2 makes it reappear behind x^2 = 8.5^2 in the image

[real ocean]

ok

so when you subtract 4^2 from both sides

On the left hand side

you get

x^2+4^2 -4^2

which is just equal to x^2

and on the right side of the equation

you';re left with 8.5^2-4^2

[lean otter]

thanks

.close

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[frosty granite]

For underlined but

[frosty granite]

Don’t understand how they’ve manipulated

[raven heart]

what's 15.1.2?

[junior smelt]

Manipulated in which step?

[frosty granite]

First bit

[junior smelt]

Getting the dy/dx? How do you define a general utility function that's homogenous of degree h?

[frosty granite]

Think he’s using this?

[junior smelt]

thinking then, if you had...

$U(x,y)$, then differentiating it with respect to $x$ would give you $U_1(x,y) + \dv{y}{x} U_2(x,y)$, wouldn't it?

[flat frigate]

@junior smelt

[junior smelt]

If you had something like $U(x,y) = c$ for some constant $c$ then you'd get $U_1(x,y) + \dv{y}{x}U_2(x,y) = 0$ and so that would rearrange to $\dv{y}{x}U_2(x,y) = -U_1(x,y)$ and what they ended up getting

[flat frigate]

@junior smelt

[junior smelt]

The final rearrangement bit is that they used effectively this but as $U_1(x,y) = U_1( x \cdot 1, x \cdot \frac{y}{x})$

[flat frigate]

@junior smelt

[frosty granite]

Why is there u1 and u2

[junior smelt]

Multivariate chain rule

[frosty granite]

Wait is that this

[lament garnet]

Yo

[junior smelt]

yep pretty much!

[lament garnet]

I have question

What is four hundred and fifty six written as a numerical expresion

[junior smelt]

!occupied, thank you

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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).

[frosty granite]

Ohhh

But then there’s the degree thing

[junior smelt]

what about the degree thing?

[frosty granite]

What is the point of writing this bit out

Because in the next bit he doesn’t really use it?

Oh wait

Yeah he does

But even then I dont get that manipulation

I can see why he’s done that now tho, to cancel the commonalities

[junior smelt]

Because you know that $U_1$ and $U_2$ are homogenous of degree $h-1$, as per here, you'd know that, for example taking $U_1$,
[
U_1( {\color{green} \lambda} {\color{red} x}, {\color{green} \lambda} {\color{orange} y}) = {\color{green} \lambda^{h-1}} U_1({\color{red} x}, {\color{orange} y})
]
so because we can write
[
U_1(x, y) = U_1({\color{green} x} \cdot {\color{red} 1}, {\color{green} x} \cdot {\color{orange}\frac{y}{x} } )
]
we get that it becomes ${\color{green} x^{h-1}} U_1({\color{red} 1}, {\color{orange} \frac{y}{x}})$

[flat frigate]

@junior smelt

[frosty granite]

Ohhh so the x is taken out from the inside because that becomes the scalar?

[junior smelt]

Yep pretty much

[frosty granite]

Ohhhhhh

Thanks so much for writing that all out

[junior smelt]

Thought that would make it a bit clearer

[frosty granite]

Yh lmao this q was bugging me

Makes sense now

[junior smelt]

perfect that it does!

[frosty granite]

.close

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[bleak hinge]

Just purely out of interest, is there a formula (one formula, not plotted lines) to plot a graph like this (in desmos for example)?

[bleak hinge]

(Please ping when replying to me, thanks)

[fleet condor]

probably since you can do floor, but I don't see an obvious way

you could surely do it in a 2 part piecewise function

if they don't have to be perfectly straight, sin(x)+x looks vaguely like that

[safe radish]

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[glacial cairn]

Like this? https://www.desmos.com/calculator/hjnyowijrn

[final halo]

https://www.desmos.com/calculator/cu82uyzs3i
Same thing with one equation (took me more time to figure out than I'd care to admit )

Oh it's i guess it's just the same as yours nel

Ah mine just has no mod

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[glacial cairn]

Yes because you can make the mod function from the floor function

[glacial cairn]

But yours is nice

Anyway sorry for the channel

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[cobalt tapir]

can anyone help me what step the took to transform this ?

[fleet condor]

looks like like a lot of fraction simplification and combination

or is this V() an operator? if it is then i have no idea where the R went

[cobalt tapir]

how did the simplify and comboine it?

where does omega^2 come from and 2omega

[fleet condor]

did you try combining all the fractions on the bottom?

[cobalt tapir]

how would i do that @fleet condor ?

[fleet condor]

common denominator then add them

the same way one would add any two fractions

[cobalt tapir]

and if i would to that, why does the nominator go from 1 to so many symbols

[fleet condor]

well, then you combine it with the j\omega C, probably

i havent actually done it, this is just the standard way one would simplify this

although i happen to think I dont know the context to this and they are actually using properties of either some V() function, or like j is the imaginary number or something

since it does not actually appear to simplify into that very well

[cobalt tapir]

indeed j is the imaginary number

[fleet condor]

Alright, you should disregard what I said, I was unable to simplify it to that by combining fractions. Hopefully another helper knows:

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@cobalt tapir Has your question been resolved?

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[lean otter]

How to solve rational expressions undefined

[frigid spruce]

if im not wrong i think its already simplified

also why would you even do m=3?

[halcyon carbon]

idk whats going on either

[frigid spruce]

show original question ig?

[halcyon carbon]

do you mean you want to find the values of m such that your expression is undefined?

[lean otter]

Yes I am having a hard time finding the “let”

[halcyon carbon]

!xy

[safe radish]

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

[lean otter]

How to solve thus

[fickle pendant]

what's the question

[lean otter]

It’s there

The picture

[fickle pendant]

solve in what way

[frigid spruce]

r u serious rn😭 this looks very different to what you sent earlier

[lean otter]

Rational expression undefined

[lean otter]

I solved it now

But I don’t know how to answer number 1

10*

PLS

[fickle pendant]

we cannot help you because you have not given us a question

[frigid spruce]

its just the same thing instead you have q in place of x

[fickle pendant]

is this solve for q=5? is this simplify? is this???

[lean otter]

HUH

[halcyon carbon]

the rational expression is undefined when the denomiator is 0 so thats what you have to solve

i think thats tthe question

[frigid spruce]

oh mb im just sleepy

gn

[halcyon carbon]

gn

[fickle pendant]

oh so the question is "if this fraction is undefined, what is the value of q"?

[halcyon carbon]

i guess so

[lean otter]

Where trying to find the value to make the answer undefined

[fickle pendant]

so yes, the denominator = 0

[lean otter]

Pls help

[halcyon carbon]

have you done this?

[lean otter]

I am confuse right now

😓

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@lean otter Has your question been resolved?

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[raven hawk]

I dont understand how they constructed w2 = u-w1

for the orthogonal projections proof thingy

[acoustic goblet]

sorry

[raven hawk]

.c,osep

].CLSEO

.CLOSE

/ASDKJ

DSGK

.C.OLSE

.colse

.close

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[gritty glacier]

most calmest guy while doing math

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[slender timber]

Why are these not equal?

[empty gyro]

Have you tried graphing both equations in the given intervals?

[halcyon carbon]

sqrt{x^2} = |x|

[narrow grove]

Hi

how the actual heck do u do that

Get the app symbolab

[halcyon carbon]

[slender timber]

How would I solve this integral then?

[halcyon carbon]

the second one?

[slender timber]

The first one in my message

[halcyon carbon]

split into two intervals

when cos(theta/2) is positive

and negative

oh just double the arae

[slender timber]

Was it a bad idea to make this into a cos^2 then?

Oh does changing to theta/2 require you to change the bounds?

That doesn't make sense to me though

I'm just using this

But backwards

[halcyon carbon]

It’s not bad you can just do the integral of 10cos(theta/2) from 0 to pi and double it

[slender timber]

That's strange

How do I know to be on the lookout for that sort of thing?

I get that it's because it went negative but I never had to do that before

[halcyon carbon]

Happens a lot with trigonometric functions due to their periodic/odd/even nature

Normally if you don’t take square roots you don’t have to worry about positive/negative unless you’re solving for the total area under the curve

[slender timber]

alright

thank you

[halcyon carbon]

Np

[slender timber]

.close

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[plucky elk]

no

oh no you're right

$\sin(-x) = -\sin(x)$

[flat frigate]

riemann

Snopes

[untold sky]

yes for even functions which cos is

y

and 19b immediately comes from the fact sin is odd

ye

ye

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[tough shale]

Find the Coterminal angle between 0 and 2pi

[tough shale]

Given: 15pi/18

My answer 5pi/6

Correct answer: 17pi/6

Can someone explain?

[fast gazelle]

that... isn't between 0 and 2pi...?

17/6 = 2 + 5/6, so it's more than 2pi

that's probably just a mistake...?

[tough shale]

Yes thats what I was confused about

Is my answer correct? or is there a different answer?

[fast gazelle]

well i don't actually know what "coterminal" means

...also wait hang on the given angle is the same as your answer

basically i have no idea

[safe bluff]

coterminal angles are "the same" angle as the original, so coterminal angles of x are of the form n2pi+x

though in this case, the given angle is between 0 and 2pi

so any coterminal angles will not be

[tough shale]

So how do I solve this when all I can do is + or - 2pi?

[safe bluff]

I’m… not sure. The given angle is already between 0 and 2pi, so it shouldn’t have a coterminal angle in that range other than itself. Worse yet, their answer isn’t even in the range.

[tough shale]

So this is an error for the teachers part

[safe bluff]

Assuming this is the exact wording of the question, it seems like the teacher has made an error.

[tough shale]

Yup

Still

Thank you for info!

.close

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[devout shale]

I was hoping someone could check my work for this

So equivalently we want to show that $$\sqrt{n}(\sqrt{n+1}-\sqrt{n})\leq C$$ for some $C>0$ for $n$ sufficiently large.

[flat frigate]

Austin

[devout shale]

Let C=1

then we want to show that for $n$ sufficiently large that $$\sqrt{n}(\sqrt{n+1}-\sqrt{n})\leq 1$$

[flat frigate]

Austin

[devout shale]

equivalently $$\sqrt{n(n+1)}-n\leq 1$$ ....
$$\sqrt{n^2+n}-n\leq 1$$.....

[flat frigate]

Austin

[devout shale]

Okay and now I compare the nth and n+1th term of this 'sequence'

$$(\sqrt{n^2+n}-n)-\sqrt{(n+1)^{2}-(n+1)}-(n+1))$$$$=(\sqrt{n^2+n}-n)-(\sqrt{n^2+2n+1-n-1}-(n+1))$$ $$=(\sqrt{n^2+n}-n)-(\sqrt{n^2+n}-(n+1))=\sqrt{n^2+n}-\sqrt{n^2+n}-n+n+1$$$$=1$$

[flat frigate]

Austin

[devout shale]

so the difference between consecutive terms is positive and hence it is decreasing

so it suffices to show 1 value of m where it is less than 1

and we just do like m=10

actually nvm I see an error

.close

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[lean otter]

Find the closest point on the plane 2x-3y+z = 1 to the point (1,1,-2)

[lean otter]

how do i do this problem?

[drowsy karma]

i take a line that goes through th egiven point (1,1,-2) and line must be perpendiculalr to the plane

so

direciton vector of line is the same

[lean otter]

as the normal

[drowsy karma]

as normal vector of the plane

and then

[lean otter]

ok

[drowsy karma]

you solve system

plane and line

[lean otter]

take point of intersection

[drowsy karma]

yes

[lean otter]

any other way to do this though?

or is that the only way

[drowsy karma]

that is simlolest thing i just though of now

[lean otter]

ok

it is very simple

but maybe i could use projection

[drowsy karma]

yes mahe you can fdn sem ready formal fo rthat

but who wud know them all haha

[lean otter]

yeah

[drowsy karma]

projeciton is also formula

[lean otter]

yes

[drowsy karma]

so first method is more lfe )

[lean otter]

yeah

anyway thanks

very much!

[drowsy karma]

yw

)

[lean otter]

.close

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[south linden]

could someone help me out with this question

this was what i did but i think im doing it wrong, i def have to integrate?

wait maybe i dont, but im not sure what i did wrong

[broken roost]

how did you get (1/e)^(x-1) in the second line?

i think that is where you messed up?

[south linden]

i dont eveen think i ended up using that

i was just trying to figure out how to attempt tthe question there

but i dont think i used anything up there

the M(t) is where my solution should start

[broken roost]

oh. ive actually never heard the terms like moment gen func before. wut is that? maybe i could help then

[south linden]

im not sure exactly how to describe it

this is what it is defined as

which is thius

however

i believe i used the method for discrete variables

which is now why im sayting i think i have to integrate

[broken roost]

not sure about that, but i think i can help you simplify the func.
the given func is e^(-x-1); -1<x<inf
this is the same as e^(-x); 0<x<inf

that'll make the function simpler

no?

[south linden]

oh

true

mm

howeever

i dont think i can change the bounds here

since i have another e^tx

so i think ill just tank the longer calculation to be sure

[broken roost]

hmm. that's there

[obtuse plover]

If u have this, what r u having trouble with

[south linden]

i didnt use it at first

i think i realized after i sent it that i atttempted it wrong lol

[obtuse plover]

U know how to do it now?

[south linden]

id assume so

give me one second

should look something like this yeah?

[obtuse plover]

Ye

[south linden]

waiit

wont this just diverge though?

since i have an e^ x

that e should be t-1

but shouldnt change anything

[obtuse plover]

Hm

Lemme see online

Ah ok

We gotta restrict t

If t-1<0 it’ll be a neg exponent so converge

[south linden]

i had a friend do it this way

logically it makes 0 sense?

not sure how infinity just dissapeared

[obtuse plover]

They just ignored the e^inf lol

Yea

[south linden]

yeah

lol

[obtuse plover]

.

[south linden]

well

for the question

i only need t=0

since im finding expected values

wait is that why you can ignore it possibly?

[obtuse plover]

U only need t = 0?

Why

[south linden]

im only ever going to need t=0

[obtuse plover]

Am I missing smth lol

[south linden]

maybe i am

[obtuse plover]

Where’s the t= 0

[south linden]

E[X]=M'(0)

[obtuse plover]

I don’t think that’s the case

[south linden]

so like you were saying

it would converge there?

[obtuse plover]

If t<1

Yes it converges

[south linden]

it would also converge for t=0 no?

since it just ends up exp(-inf)

[obtuse plover]

Well that’s included in t<1

[south linden]

oh

right

...

lol

so maybe thats reasonable logic to just drop it

idek

[obtuse plover]

Wdym reasonable logic

[south linden]

as in

why he did it

or why it works

[obtuse plover]

1/e(t-1) * e^x(t-1) from -1 to inf

1/e(t-1) *(0-e^(1-t))

e^(1-t) / e(1-t)

It’s like that

[south linden]

ah

ok

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[cobalt tapir]

can anyone help me how they did this part?

[safe radish]

@cobalt tapir Has your question been resolved?

[patent kindle]

u can write the expression as [((u^1/2)2)-(2/(u^1/2))]du

[cobalt tapir]

What?

[patent kindle]

after dividing the terms in numerator individually by rootu , u get this

[safe radish]

@cobalt tapir Has your question been resolved?

[sinful zodiac]

THEY SPLIT THIS INTO 2 FRACTIONS AND INTEGRATED EACH

[cobalt tapir]

then what about du/2? is it left alone?

[sinful zodiac]

DU/2 IS JUST 1/2 * DU

1/2 IS JUST A CONSTANT SO U CAN BRING IT OUTSIDE THE INTEGRATION SIGN

YOU COULD ALSO PUT THE 2 HERE AND THEN SPLIT INTO 2 FRACTIONS

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[manic shadow]

Regarding Combined Variations

[manic shadow]

No idea how the order of multiplying works here

need help on advancing

cant understand how u get s=16/1

[safe radish]

@manic shadow Has your question been resolved?

[mortal sandal]

that's supposed to be a divided by sign on the right

$s = \f{16}1\div \f49$

[flat frigate]

thewizardofOU

[manic shadow]

but where did the 16/"1" come from

[mortal sandal]

From the 16 in the previous step

[manic shadow]

previous step wdym? I believe only way you can get 1 is by dividing it by the same number? But after 16/4/9

you just get 16/4/9

[mortal sandal]

Yes

$16\div\f49$

[flat frigate]

thewizardofOU

[mortal sandal]

In other words

[manic shadow]

is this where we get 1

[mortal sandal]

then we just say 16 = 16/1

so we replace 16 with 16/1 since they're the same thing

[manic shadow]

oh

oh tyvm

.close

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