help-10

[candid yarrow]

[glossy yew]

olkkkkk

[ruby path]

There's a high chance that you'll find someone with good probability skills here

Laugh goddamnit

It's a good joke

[glossy yew]

very funny

question 2

[candid yarrow]

you could have waited longer than 30 seconds before sending that message, maybe 15 minutes before pinging helpers

[candid yarrow]

!show

[obtuse pebble]

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

[glossy yew]

me?

[candid yarrow]

no, just a joke. Can you explain where you’re stuck

[glossy yew]

i dont know how to simplify this

[candid yarrow]

should be 3/4 in the second line

[glossy yew]

yah

misetake

[candid yarrow]

after that you can simplify it

[glossy yew]

okey

so

6x9/4^4

=54/256

=27/128

correct?

[candid yarrow]

looks correct

[glossy yew]

lets go

on last thing

question 13

[candid yarrow]

do you know what the 68-95-99.7 rule says? and state it here

[glossy yew]

nooo

i dontr

[candid yarrow]

have you been taught it?

[glossy yew]

yah

i know the formula

but first can you explane to me what( 0.42, 0.66) is

@candid yarrow

[candid yarrow]

if you don’t mind my correction, the way to spell it is “explain”

[candid yarrow]

that’s the value that the population will give

[glossy yew]

but why is it in x,y term

[candid yarrow]

it represents “open interval”, I think

it means “from 0.42 to 0.66” probably

[glossy yew]

so is the 68% interval

[candid yarrow]

blue is (0.42, 0.66)

[glossy yew]

yyes

that 2 devation

so that means

to find the mean

we can

do 0.42+0.66/2

q = .54

bro i hate probablity

<@&286206848099549185>

i think bro left

[violet kindle]

Yea i think this is the right mean

But you wanna find the green interval up there

Can you find the standard deviation from (0.42, 0.66)

[glossy yew]

yah i dont know how

[violet kindle]

So the interval (0.42, 0.66) is the interval (mean - 2*stdev, mean + 2*stdev)

By the 68-95-99.7 rule

Right?

[glossy yew]

yah

[violet kindle]

So find stdev

[glossy yew]

okey

i know the formula for the stdev

but whats p

[violet kindle]

Wdym p

What formula

You dont need any formula

[glossy yew]

p in the stdev formula

[violet kindle]

You do not need the stdev formula

Youre overcomplicating it

[violet kindle]

Try this

Just use this

[glossy yew]

so a dev apart is

66-0.54

/2

[violet kindle]

Which is?

[glossy yew]

.12

[violet kindle]

,calc (0.66-0.54)/2

[warm shale]

Result:

0.06

[glossy yew]

.6

[violet kindle]

0.06 is your standard deviation

So you have mean, and you have stdev

[glossy yew]

0.6

[violet kindle]

No

Its 0.06

[glossy yew]

yah

true

[violet kindle]

To find a 99.7% confidence interval you want the interval (mean - 3*stdev, mean + 3*stdev)

Find that and you have your answer

[glossy yew]

lets gogogogoogog

i see it

it connects

[violet kindle]

Nice

[glossy yew]

0.54 - 3x0.06, 0.54 + 3x0.06

[violet kindle]

Yeah

[glossy yew]

0.54 - 0.18

is 0.36, 0.72

[violet kindle]

Nice

Thats your intervla

Good job

[glossy yew]

wow

is int in [] or ()

[violet kindle]

I'd put (), same as the problem description

[glossy yew]

okey

are you up to help me more in probablities

[violet kindle]

Might not be able to

Someone else will

[glossy yew]

oh okey

um

[candid yarrow]

can you say where you’re stuck

[glossy yew]

okey

so

we learn 3 confidence intervals

but not other percentages other then

[candid yarrow]

and on which question?

[glossy yew]

68%, 95%, and 99.7

question 1

where we need to find the kangroo poplution thats tail thats biger then 92cm

[candid yarrow]

100 - 8 = 92

how many standard deviations is the kangaroo with a 92 cm long tail away from the mean?

[glossy yew]

wiat

something is connectimng

92 is the first devation back

so that means (x > mean) + (92 => mean)

so 0.5 + 0.34

84% of kangroos have a tail longer then 92cm

[candid yarrow]

this is right

[glossy yew]

so for the second question is the same thing

100 x 2x8

• 2.5

because we only want the negavtibe

[candid yarrow]

so can you state the answer you got?

[glossy yew]

wait i think this is wrong

100 - 2x8

100 - 16

84cm

[candid yarrow]

this is right

[glossy yew]

my answer is

2.5% of kangroo have 84cm

tai;

for the third question nothing is connecting

[candid yarrow]

111 is 11 away from 100

the distribution is symmetrical about 100

so 89 should look the same as 111

and what do you know about the number of kangaroos with tail length under 108 cm?

[glossy yew]

is the first devation

[candid yarrow]

put all this together and you can get the answer

[glossy yew]

how is 89 look the same as 111

[candid yarrow]

same distance from 100

[glossy yew]

so -11 and 11

84% have a tail lower then 108

[candid yarrow]

yes

[glossy yew]

so 89 = 0.8 as well

[candid yarrow]

how did you get that

[glossy yew]

well

i thought that ]

scene 111 = 0.8

and 89 is the same 111

[candid yarrow]

it’s actually 0.08

[glossy yew]

yah

missing the 0

yeaks

so we have 108

which is 0.34

then we got

0.5 - 0.08

which is 0.48

n

no

0.42

= 76

[obtuse pebble]

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[modern ravine]

Is this maths statement correct?

[modern ravine]

I wanted to express: "the numbers 5 and 7 are divisors of both 35 and 70"

Or is there a more correct way of expressing it?

[glossy basalt]

yes there is

[modern ravine]

What would it be?

[timber island]

You could just write two separate statements

oh waits its 35 and 70 💀

so 4 in that case

[modern ravine]

Oh god

Why are mathematical expressions so dumb

Like why can't we simply have ands and ors like in programming

[timber island]

Good question

ask the guy who set these expression paterns up

[modern ravine]

probably a high arab hundreds of years ago

[timber island]

bruh why asian

[glossy basalt]

in case you wanna write it in a single sentence, you can say something like
A={5,7}, B={35,70}, for all x in A and y in B, x|y

[timber island]

if i were to say x belongs to A/y belongs to B, would that change the meaning

[glossy basalt]

oh, i just wrote "in" because of the sign

$\in$

[warm shale]

biscuityxd

[timber island]

I call that belongs to, same thing but longer

[glossy basalt]

this is of course a better way of writing it

because it clarifies the situation

[modern ravine]

So this?

Are the commas ok?

oh forgot 1 part

[timber island]

mm u missed y

[glossy basalt]

forall x in A, forall y in B

moreover, you can change B into something like
{35n:n in N}, it will still work

[modern ravine]

can you explain that?

[glossy basalt]

since 35 is the lcm of 5,7

if you wanna generalize it, you can just say any multiple of 35 is divisible by 5 or 7

[modern ravine]

Thanks

.close

[obtuse pebble]

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[compact root]

Second time asking for help but...

I get what to do with 5) and 6), but i don't know where to start (what vector base should I use?)

[obtuse pebble]

@compact root Has your question been resolved?

[obtuse pebble]

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[karmic hedge]

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

[karmic hedge]

I got the right answer but I still have a question

It says I can use any of those symbols like <, =, => etc, but I can only see a use for the not equal sign, when would I use any of the other ones?

[sage geode]

E.g. When the right limit is |a| and the left limit is a

Then the continuity/discontinuity would be dictated by the sign of a

[karmic hedge]

Ah ok I think I get you

Thanks!

❤️

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[daring kestrel]

can someone explain where the 3 went at the end?

i got 3 - 3cos²(x) - 3cos(x)

[whole dock]

You had $3(1 - cos^2(x))cosx$

[warm shale]

coldtee

[daring kestrel]

yeah

[whole dock]

$cosx(3 - 3cos^2x)$

[daring kestrel]

yeah

who takes the loose cos(x)

oh right both

[whole dock]

So what seems to be the problem

Oh ik maybe

[daring kestrel]

i misread the ^3 as ^2 at the end

now ik what i did wrong

[warm shale]

coldtee

[whole dock]

Maybe this is better

[daring kestrel]

yeahh

you're really good with this bot lol

thanks a lot

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[eager star]

help

[eager star]

how do derive this

[fathom flicker]

what are the marks for?

[eager star]

wdym lmao

[eager star]

they are obviously going to be marked

[fathom flicker]

for what?

[eager star]

but its multiple questions

so there is a. and b.

each worth 2 marks

[fathom flicker]

What type of assignment is this? Quiz? Test?

[eager star]

test

but its a practice

what?

ook...

[fathom flicker]

https://tenor.com/view/side-eye-dog-suspicious-look-suspicious-doubt-dog-doubt-gif-23680990

[eager star]

alright bro

[eager star]

i have the answers

i just need to find out how they got that derivative

[fathom flicker]

oh cool

ty for showing that

[tardy epoch]

then just show them

save us all a lot of time

[eager star]

so how did they get that derivative?

[tardy epoch]

chain rule

have you learned that yet

[eager star]

yes i have

[tardy epoch]

use $a^x = e^{\log(a^x)}$ and log rules

[warm shale]

rie.mann

[eager star]

.close

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[distant seal]

log rules again?

but whats it like with substraction

[tardy epoch]

Subtraction is adding the negative

,tex .log rules

[warm shale]

rie.mann

[distant seal]

so 4 (log (x-7)/5 log(x))

and what was that other rule

where the base went to the exponent or something

i would assume d bc it doesnt make sense to have 4/5 anywhere since its not part of the log like that and would make sense if it went to the exponent based off the rules

correct?

... gulp

[high lily]

where's that coming from

[distant seal]

i didn't mean that as my answer

i means (log(x−7))^4(log(x))^5

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[sage dagger]

I don’t know how or where to start, I could only draw a diagram up so far

[smoky dove]

Look for any properties of triangles

Tjat can help to find certain angles

And work from.there

[high lily]

seems like this is supposed to be a 3d problem

where those are angles of elevation to the top of Uluru

[obtuse pebble]

@sage dagger Has your question been resolved?

[obtuse pebble]

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[long hedge]

shouldn't the points by pi/2 and 3pi/2

half of pi is pi/2

not pi/4

[plain stag]

3pi/4 is my favourite number that falls in between pi and 2pi

definitely just a mistake(s)

[native inlet]

yea better wording then whatever the hell I was typing

[plain stag]

you're good

[long hedge]

I love the organic chem tutor so much but when he makes typos like this it drives me crazy

[plain stag]

i bet

[obtuse pebble]

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[timid silo]

I know I still have to find the height of the tree

But did I do something wrong for the other steps?

[lethal sand]

everything is correct, just plug in your inital value and find A

[timid silo]

@lethal sand this is what the answers got tho

[lethal sand]

can you show me the steps above that you have cropped out?

[lethal sand]

this is definitely incorect

[timid silo]

This is part I

i

[lethal sand]

,w partial fraction 1/(x(45-x)

[warm shale]

Results provided by WolframAlpha

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

Requested by just_waler#0

[lethal sand]

hmmm nvm, apparently the question made a typo

[timid silo]

I’m confused

[lethal sand]

right then, so part i should have been showing that 1/(H(45-H)) = 1/45 * (1/H - 1/(45-H))

the question missed a factor

[timid silo]

Ahhh ok

So technically my solution is correct?

[lethal sand]

well no, because your solution used the result from part i, your solution is incorrect

it's also missing a factor of 1/45

[timid silo]

Ohhhh ok

So it should be like this?

[lethal sand]

no, the lhs is still 1/(H(45-H))

the rhs is what you wrote

$\frac{1}{H(45-H)} \neq \frac{1}{H} + \frac{1}{45-H}$

[warm shale]

just_waler

[timid silo]

Oh okay

Got it

Thank you so much

.close

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[astral aurora]

My answer is 4

But why can't it be 7? If I took separately

[rocky goblet]

...well for which x is 4sin^2(x) + 3cos^2(x) = 7?

[candid yarrow]

they are not at maximum at the same time

[astral aurora]

Yes

I want it clear more

How it works

[rocky goblet]

well consider this simpler example

what's the maximum value of x - x?

[royal basin]

^

[astral aurora]

0

[rocky goblet]

yep

but if we take the maximum of each term individually, x can be arbitrarily large, and -x can also be arbitrarily large
but they're not both large at the same time, x - x is always 0

[royal basin]

just because one function's maximum value is M, and another function's maximum value is N, does not necessarily mean the maximum value of their sum is M+N.
you can say that the max value of the sum is at most M+N. but it often will be smaller.

[astral aurora]

How it can be 7

If angles are different

4sin^2x+3cos^2y =7

If one angle is x and other y

[royal basin]

that's not what bee asked you.

and that's also got nothing to do with your question.

[astral aurora]

What?

This is the question

[royal basin]

no it isn't.

the angles under the sine and cosine aren't x and y.

they're x and x.

they are the same angle.

[astral aurora]

And i want to make it clear the confusion

I don't know what's bee asking

[royal basin]

for which x is 4sin^2(x) + 3cos^2(x) = 7?

[astral aurora]

I am trying to make other confusion clear

What if angles are not same

Can it be 7?

[royal basin]

then it is a different question entirely

[astral aurora]

Yes different question

Related to this question

[royal basin]

yes, the maximum value of 4cos^2(x) + 3sin^2(y), when x and y both range over R, is 7.

no, it is not related to this question.

the relation is very distant, at best.

[astral aurora]

This is the relation to make it clear for students why it is not 7

So there is a relation

If angles are different then it will be 7

If angles are the same then it will be 4

[royal basin]

i don't think that's a helpful way to frame it.

[astral aurora]

Wdym?

-1<= sinx<=1
0<= sin^2x<= 1

[royal basin]

when you just say it like that, you hide the reason why the max value drops to 4 when you make the two angles the same.

[royal basin]

beside the point.

[astral aurora]

4 comes by this or sin^2x+cos^2x=1

What??

I guess you are not understanding what i am trying to clear

[royal basin]

and you are not understanding what i am saying.

[astral aurora]

I am asking two questions simultaneously

Okay your turn what is your point?

I am saying there is a relation and you are saying there is not

[royal basin]

i am not saying there is no relation

[astral aurora]

Even questions aren't the same but there is a relation

To point out the angle so that students can make it clear

[royal basin]

i am saying the relation is distant.

i am also saying that if you try to explain the relation as "If the angles arent the same the answer is 7 but if they are the same it's 4" then that is a poor explanation.

[astral aurora]

Then what will be the proper explanation?

I am simply pointing out why it differs 4 to 7 because of angle that's it

[royal basin]

the max value of 4cos^2(x) + 3sin^2(y) is 7 because:

• 4 cos^2(x) ≤ 4 and 3 sin^2(y) ≤ 3 for all x and y, therefore 4cos^2(x) + 3sin^2(y) ≤ 7 for all x and y
• the value 7 is achieved at (x, y) = (0, pi/2)

[astral aurora]

I will explain them mathematically. You didn't ask me to write it like this

[royal basin]

4cos^2(x) + 3sin^2(x) = cos^2(x) + 3cos^2(x) + 3sin^2(x) = cos^2(x) + 3
and the max value of cos^2(x) + 3 is 4 because:

• cos^2(x) ≤ 1 for all x, therefore cos^2(x) + 3 ≤ 4 for all x
• the value 4 is achieved at x = 0

[astral aurora]

I explained them mathematically why it is 4

So I'll explain them why it is 7 same way

[royal basin]

you should understand the difference between "f(x) ≤ 4" and "the max value of f(x) is 4"

[astral aurora]

Difference with what?

[royal basin]

... i don't understand what you just said.

[astral aurora]

I didn't understand what you mean

I meant difference to what?

[royal basin]

the two statements

A. f(x) ≤ 4 for all x
B. the maximum value of f(x) is 4

are different
and you should understand the difference between them

[astral aurora]

Ohh wait a minute

B consisted in A

[royal basin]

if you say "B is consisted in A" nobody will understand you.

[astral aurora]

Then remove= sign

A includes B

A has maximum value 4 and less than values so it consists B in it

[royal basin]

you are using the word "consists" incorrectly even from a non-mathematical viewpoint

also "A has maximum value 4" is nonsense -- A is a statement, statements don't have such a thing as a maximum value

i think you may be trying to say this:
"Any function which satisfies B also satisfies A"

which is correct

[astral aurora]

So if someone asks what is the maximum value of the Statement a what will you say?

[royal basin]

i will say "your question does not make sense"

assuming i'm not being held at gunpoint, etc.

[astral aurora]

It is clearly seen f(x)<= 4

So it can't exceed 4

So you meant the maximum value of it is not 4?

[royal basin]

...

[ionic summit]

pap pap skeeter is here

[astral aurora]

??

[ionic summit]

i shall help with my overly simple ideas

[astral aurora]

My question is what is the maximum value of statement A what it can get

[timber island]

this..?

[ionic summit]

its 7

cus i mean

cos max is 1

so is sin

so like 4 + 3 is 7

[timber island]

Bruh ofc not

[astral aurora]

Nope. We are moving far far away

[timber island]

.

[astral aurora]

Statements vs statements in English

[astral aurora]

No

[timber island]

What is statement A?

[astral aurora]

The angle is the same so it can't be 7 it will be 4

If the angles are different then surely 7

[royal basin]

congrats on completely missing the point

[royal basin]

@timber island here are statements A and B

[ionic summit]

sadge

[timber island]

Thanks

[astral aurora]

Should I close?

If anyone hasn't something to write

[timber island]

I rlly don't get the difference myself tbh, except that B doesn't have a x range specified but I assume that isn't necessary

[astral aurora]

B has max value 4
A has (-infinity to 4]

Here max value of both is 4
Which means B consists in A

[timber island]

Unsure, better ask Ann herself

[astral aurora]

??

.close

[obtuse pebble]

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[timid silo]

How can I solve this?

[timid silo]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[timid silo]

So far I have that
area ABC = area(ABD) + area(BDC) + area(ADC)
e=f=g (radius of the incircle)
area(ABC) = 1/2 *e (AB + BC + CA)

How can I proceed

Almost looks like there isn’t enough information

i just have to find 30e from that info somehow

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.close

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[wild root]

Find the size of angle ABC

wait heres a more simplified

x+y=105

2x+75-y=180

2x+2x=210

2x-y=105

y=105
x=0

[royal basin]

what do these messages with a single minus sign mean

[wild root]

i used a

system of linear equation

i just put the - so it looks less confusing

sorry english not my language.

[royal basin]

x+y=105
2x+75-y=180

these equations look correct

how did you get 2x + 2x = 210?

[wild root]

wait i wrote it wrong sorry

how do i do that line

at the text

[warped aspen]

Can you do simultaneous eq?

[royal basin]

that is what he's doing...

[royal basin]

what line?

[warped aspen]

oh

[wild root]

nvm found the line

x+y=105
2x+75-y=180

2x+2y=210
2x-y=105

and we subtract

and uhh

[royal basin]

ok, but 2y - (-y) = 3y, not y.

[wild root]

OHHHHH

[pure geode]

yehp

[wild root]

OH MY GOD I JUST REALISED

WHILE WRITING THAT

YEAH THANK YOU ANN

[warped aspen]

💀

Nice

[wild root]

.close

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[timid silo]

If a lineartransformation T(x) = Ax has a n x n standard matrix A which is diagonlizable A = PDP^{-1} and the columns of P is a basis of R^n (because it contains N linearly independent eigenvectors?), then [T]_B = D. My questions is if i want to take a vector x and calculate T(x) = Ax in the other basis B, why does P^{-1} = [x]_B*?

[kind hawk]

sry I had to go away for a bit

let me be more precise

we have $P=(b_1, \ldots, b_n)$ and we know that $(e_1, \ldots, e_n) = I=P^{-1} P = P^{-1} (b_1, \ldots, b_n)$

[warm shale]

denascite

[kind hawk]

looking at this column by column, we see that $P^{-1} b_i = e_i$

[warm shale]

denascite

[kind hawk]

so, now lets write $x=\sum c_i b_i$ in basis $B$. Then $P^{-1} x = \sum c_i P^{-1} b_i = \sum c_i e_i = (c_1, \ldots, c_n) = [x]_B$

[warm shale]

denascite

[kind hawk]

or as a response to your DM, when you write down the system $[b_1, \ldots, b_n | x]$, that's really just $Py=x$ where $y=[x]_B$ are the coordinates. But this system you can just solve by multiplying it by $P^{-1}$ from the left

[warm shale]

denascite

[kind hawk]

@timid silo

[timid silo]

I get it

Many thx

So

Just to make sure

So when we have [x]_B it contains scalars to the eigenvectors

Then

[kind hawk]

this isnt really about eigenvalues and eigenvectors

this is just about basis change in general

[timid silo]

I know

[kind hawk]

ok

[timid silo]

But in the case when the standard matrix is similar to a diagonal matrix

We are using eigenvectors and eigenvalues til calculate Ax

By using a basis consisting of the eigenvectors

Right?

[kind hawk]

essentially, yes

we are translating into the basis of the eigenvectors

[timid silo]

Goodie

[kind hawk]

where applying the transformation is super easy

and then we translate back

[timid silo]

In general if the basis is invertible, Can i just take the inverse matrix and multiply it on a vector to obtain the coordinatevector?

Instead of row reducing the augmented matrix?

[kind hawk]

row reducing is equivalent to finding the inverse matrix in the first place

the inverse matrix is found by solving all the systems [b1, ..., bn | ei]

[timid silo]

Goodie

So in the future if i know the inverse of a matrix i Can just right multiply a vector on it to obtain the coordinate vector with respect to the original matrix?

[kind hawk]

yes

[timid silo]

My mind has been blown

Thx for the help!

[kind hawk]

youre welcome

[obtuse pebble]

@timid silo Has your question been resolved?

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[elfin burrow]

?

do you have a question

@onyx sinew

[civic socket]

He keeps spamming it lol

[elfin burrow]

yeah in help- 5, 6, 9

[limpid widget]

,w plot x^2 + (y - (x^2)^0.25)^2 = 1

[warm shale]

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[zenith ibex]

this integral

I can't do it with U substitution right ? it doesnt make sense

what method should I use

parts ?

[normal oracle]

yes, you can use u-sub

[violet sentinel]

why doesn't it make sense?

[zenith ibex]

1 sec, I'll write it

this is what I tried

[high lily]

poor choice of substitution

[zenith ibex]

[normal oracle]

u = lnt

right

[zenith ibex]

let me try again with lnt

it worked i just did dumb

thanks a lot]

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[sage geode]

Are you taking a test/exam rn?

We can't help you in that case

[obtuse pebble]

Channel closed

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[hard citrus]

bro deleted the message 💀

.close

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[graceful pollen]

If the average value of a discrete data is not a whole number, do we have to round it?

[kind hawk]

depends for what purpose you want it

[graceful pollen]

for (3)

[kind hawk]

I dont know the context, I dont know the problem statement

[graceful pollen]

here

it's 5.5 but is that supposed to be rounded up to 6

[kind hawk]

I would just leave it at 5.5 but I'm not in your course. maybe your teacher mentioned at some point what they want you to do

[graceful pollen]

right thanks

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[graceful pollen]

<@&268886789983436800>

[obtuse pebble]

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[lilac parcel]

"we consider a function f(x) = sin(x^3) and its taylor development around a = 0 to be: ... "

wihtout calculating its derivatives, find the value of the ninth derivative of 0

[lilac parcel]

not sure where to begin

[kind hawk]

well they already wrote down the taylor series for you

just have to extract the relevant term

[lilac parcel]

gotta write my answer as a factorial term

[tardy epoch]

Find the 9th power in the Taylor series

And equate it to your given series

[obtuse pebble]

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[mellow glacier]

Can someone tell me how we got to this?

[mellow glacier]

i got 3e^2 - e

[gilded needle]

what is $e^u u - e^u$ when you plug in $u=1$?

[warm shale]

Bungo

[mellow glacier]

0 ?

wait

e^u ( u-1)
Put u = 1 then it should give 0

am i wrong?

[gilded needle]

right 0

and what is $e^u u - e^u$ when you plug in $u=2$?

[warm shale]

Bungo

[mellow glacier]

e^2 (2-1)
e^2?

[gilded needle]

yep

so what's your overall answer?

[mellow glacier]

oh.

wait so what did i do wrong

i did this:

oh fckkk

man i am so dumb

i didnt multiply the 2 to all of em

got it thanx

[gilded needle]

haha it happens

[mellow glacier]

.close

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[timid silo]

Hi Bungo

[timid silo]

Sorry to bother

I just did this question on Khan Academy

Give me a min, i'll just sketch it out

there you go awkward sketch

anyway it asked me to find the area above the curve from 7 to 4

From what I understood, you told me that as long as it's above, it'll be positive

the x values don't matter

so I'm confused why here it becomes negative

here's another one

[violet kindle]

You’re confused why the answer is 3?

Is that it?

[gilded needle]

notice that it's asking for $\int_7^3$, not $\int_3^7$. When the integration bounds are in the wrong order like that, it means you change the sign. The integral from 3 to 7 is -3, the integral from 7 to 3 is 3. This is a convention, there's not really an obvious physical reason why it has to be so. It's mainly so that $\int_a^c = \int_a^b + \int_b^c$ remains true regardless of the order of $a,b,c$

[warm shale]

Bungo

[timid silo]

gotcha

just something to memorize then

[gilded needle]

pretty much yea

[timid silo]

many thanks for your help

.close

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[drowsy blaze]

Bro how do we do this

[obtuse pebble]

@drowsy blaze Has your question been resolved?

[rich silo]

Show us what you have tried!

[quaint sonnet]

!show

[obtuse pebble]

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

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[marsh spoke]

How do i know to put A out the front? Why is it not just B/(x-1) + C/(x+1)

[gilded needle]

because the degree of the numerator is the same as that of the denominator (2)

if you only had B/(x-1) and C/(x+1), then when you add those two together, the highest power of x in the numerator would be at most 1

[steel gust]

a=1, b=1, c=-1

[gilded needle]

how is that helpful?

[steel gust]

idk

[marsh spoke]

ohhh okay so when the degree is the same u add it to the front? but if it was a cubic up top then u dont need it?

[gilded needle]

OP wants to know why you need A at all

[steel gust]

cuz the question says A

[gilded needle]

if it was a cubic, you'd need something like A + Dx in front

basically, imagine doing the long division

[marsh spoke]

oh right

[marsh spoke]

ahh yeah that makes sense

thank you

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[dire fable]

hi

[dire fable]

pls helppp

pls

[high lily]

are you being asked to solve for / isolate M?

[dire fable]

yeha

[high lily]

what have you tried

[dire fable]

sub the subjet question

how come i the Ms minus out

[high lily]

wdym

[dire fable]

liek when i exnad the bracket m^2

i move it over and it just cance3ls

[high lily]

and?

what's the issue

[dire fable]

i cant get the value of M

[high lily]

can you show your work

[dire fable]

It becoimes zero

k

[high lily]

$(M-m)^2 \redneq M^2 - m^2$

[warm shale]

ℝamonov

[dire fable]

the answer is

i still dont know

[high lily]

you didn't expand (M-m)^2 correctly

[dire fable]

?

[high lily]

?

[dire fable]

how do i expand'

then

ohhhhhhhhh

ohhhhhhh

ohhhhhh

[high lily]

binomial theorem, distributive property

[dire fable]

ty

.close

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[polar idol]

so i know we can get the parametric equations of each line provided

but what next?

from line one i get another equation B(-3,0,5) + t(1,2,3) and if i repeat the process for second line i can get another equation

but i dont know if those are useful

(x,y,z)= C(2,-1,-3) + t(0,-2,1)

i have to go but if someone can provide the answer and close it would be very appreciated.

[obtuse pebble]

@polar idol Has your question been resolved?

[steel gust]

@smoky dove

[obtuse pebble]

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[hearty summit]

Have been trying to solve c for this question

Not sure which of my values are wrong

[hearty summit]

<@&286206848099549185>

[obtuse pebble]

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[obtuse pebble]

@hearty summit Has your question been resolved?

[obtuse pebble]

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[sacred root]

Hi

[sacred root]

Where did I go wrong??

[royal basin]

3x^2 - 15x + 18 is definitely not (x-2)(x-3)??

[sacred root]

,w 3x^2-15x+18

[warm shale]

Results provided by WolframAlpha

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Requested by _basudev#0

[sacred root]

See

[bright quail]

it's 3 times that

[royal basin]

^

[sacred root]

3(x-2)(x-3)

??

[bright quail]

can't u also do a telescoping thingy

[sacred root]

What's that

[royal basin]

3(x-2)(x-3)
yes and it matters here lmfao

[heavy depot]

something happened

[royal basin]

you understand that 3(x-2)(x-3) and 1(x-2)(x-3) arent the same thing right??

[heavy depot]

the board says -28 but you're saying +18

[royal basin]

also lets just make sure...

[bright quail]

1/(x-1)(x-2) = 1/(x-2) - 1/(x-1)

[sacred root]

Ye

[royal basin]

,w expand (x-3)(x-4) + (x-1)(x-4) + (x-1)(x-2)

[warm shale]

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Requested by ann_dec#0

[royal basin]

aight that parts correct

[sacred root]

Yep

[bright quail]

for telescoping

you use the idea of 1/(x-1)(x-2) = 1/(x-2) - 1/(x-1)

[royal basin]

well you asked where you went wrong and you were told where you went wrong

[bright quail]

if u write all of them similarly some things cancel out and you are left wiht a nicer expression

[sacred root]

So basically 3x^2-15x+18 = 3(x-2)(x-3)

Alr

Thanks I got it

Its 9,-4

.close

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[slate kayak]

anyone know how to find volume of solid rotating the region under the graph of y= x^(2/3) over an interval

[royal basin]

no, nobody in the world knows this

it's an open problem

don't you know?

(/sarcasm)

[slate kayak]

lol sorry for always not phrasing my question right

but as for what i tried

i tried V = pi * integral from 2 to 5 of (f(x))^2

which is this wrong?

assuming the interval is [2,5]

i drew the graph and it makes sense to me

not sure where my logic is false lol

[royal basin]

integral from 1 to 4
the interval is [2,5]

which axis are you rotating around bwt

btw*

[slate kayak]

interval is 2 to 5

and x axis

[royal basin]

ok

[indigo blade]

what is the interval?

oh ok

[royal basin]

so you found $\pi \int_2^5 (x^{2/3})^2 \dd{x}$

[warm shale]

Ann (glomed)

[slate kayak]

yes

[royal basin]

and you are sure you took the integral correctly

[slate kayak]

indeed

[royal basin]

and you entered the answer into your homework system

[slate kayak]

i used a calculator

so i guess it should be right

[slate kayak]

to be exactly precise i have two questions of very similar nature

the actual question is

x^(7/8)

but i think the questions are exactly the same

except the f(x)

[royal basin]

i mean ok like

[slate kayak]

same specifications

[royal basin]

an answer that you thought should be right is getting rejected

is that whats happening

[indigo blade]

are you getting 50.77

[slate kayak]

yes so basically it wants me to find the volume of the solid that is obtained when rotating the region under the graph of x^(7/8) about x axis over an interval [1,4]

[slate kayak]

yes

[indigo blade]

and its not the answer?

[slate kayak]

no

[indigo blade]

bruh

[slate kayak]

this is the actual question

my answer isn't 50.77 but since our answers are close i guess you switched intervals

[indigo blade]

ok'

[slate kayak]

my answer i put was 4/11 (32 sqrt(2) - 1) * pi

@royal basin would this be the right answer according to you?

i put V = pi * (integral from 1 to 4 of (x^(7/8))^2)

[violet kindle]

I may be wrong, but i'm not sure if that integral is set up correctly

[indigo blade]

the answer of this question should be 510pi/11

[violet kindle]

What did you get for this one?

[slate kayak]

This

@violet kindle

[slate kayak]

why

[royal basin]

i don't like how my question was not given a straight answer.

[violet kindle]

Hm

I got something different

[slate kayak]

Could you elaborate? I might've missed your question

[slate kayak]

Yes

[slate kayak]

I put this

[slate kayak]

for this question

[slate kayak]

which is this according to a calculator

[royal basin]

,w int[1,4] x^(7/4) dx

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by ann_dec#0

[royal basin]

ok, yeah...

@slate kayak do you have a screenshot of the problem & what happened when you entered your answer

[violet kindle]

Oh wait

It's x axis 💀

My bad

I dont have my glasses and i read y axis

[violet kindle]

Oh

U might have to type in the actual thing

[royal basin]

and the problem statement?

[slate kayak]

No

[indigo blade]

radius is (x^(7/8)), the area of each circular plate would be pi x (x^(7/4)) and then just integrate it over [1,4]

[violet kindle]

Weird

[royal basin]

7/6 no?

[slate kayak]

whoops

did i wrote something else

[violet kindle]

Oh noooo

[royal basin]

you treated the six as an eight

[slate kayak]

i'm mistyping alot when it counts ig lol

[violet kindle]

You solved it for 7/8

Not 7/6

[indigo blade]

broooooooo

[violet kindle]

At least you set up the integral correctly and did everything else right

[slate kayak]

🤦‍♂️

Omg

[slate kayak]

so everything i did is right except

it should've been a 7/6

[royal basin]

except for the input data.

[slate kayak]

not a 7/8 right?

[royal basin]

yes

[slate kayak]

V = integral from 1 to 4 of pi * (x^(7/6))^2

just confirming cuz i've been putting so many wrong numbers today

@royal basin this integral right?

[royal basin]

yes

[slate kayak]

Thank you

[obtuse pebble]

@slate kayak Has your question been resolved?

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[karmic hedge]

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

[karmic hedge]

Dont really know how to do this one either

[timid silo]

I suggest factoring x^2 inside of the square root first

[karmic hedge]

But the roots will be complex, no?

[timid silo]

What?

[karmic hedge]

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

[sage geode]

You were not asked to find the zeros of 5x^2 + 9

But instead factor x^2 out of it

[lyric ember]

you should only look at real numbers here

[karmic hedge]

So like 5 + 9/x^2 ?

[lyric ember]

$\sqrt{a^2+b} = \sqrt{a^2\left(1+\frac{b}{a^2}\right)} = a\sqrt{1+\frac{b}{a^2}}$

a not being equal to 0

[warm shale]

tobi.

[lyric ember]

now its correct sry

[karmic hedge]

so x √(5 + 9/x^2)

[lyric ember]

yes

[karmic hedge]

Ok but sry I dont really understand why where factorizing out this x^2 from the root

Like what are we trying to do with this

[lyric ember]

you are looking for horizontal asymptotes

a horizontal asymptote is a if a function always gets closer but never reaches the value if x goes to infinity or negative infinity

[karmic hedge]

Ok but why x^2? Why not x or x^3?

[lyric ember]

so that its easier to see what happens to the function if x goes to positive or negative infinity

[karmic hedge]

Ok I think I get you

Should I also multiply the rest of the fraction by 1/x^2 ?

[lyric ember]

you can factorize more

but you cant just multiply by 1/x^2 because then you would get a new function

only multiplying by 1 and adding 0 is okay

[karmic hedge]

so like this

https://i.imgur.com/t8dOD1U.png

Is this correct?

[lyric ember]

$\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$

[warm shale]

tobi.

[karmic hedge]

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

I forgot the squared

[karmic hedge]

Which 2 fractions should I multiply together?

[lyric ember]

$\frac{-x+4}{x+\sqrt{5x^2+9}} = \frac{-x+4}{x+x\sqrt{5+9/x^2}}$

[warm shale]

tobi.

[lyric ember]

multiplying by 1 here would help

you need to do the same as before

[karmic hedge]

So im not supposed to divide everything by x^2 like I did here?

[lyric ember]

if you do that then you get a new function

$a \cdot x^2 \neq a$

[warm shale]

tobi.

[karmic hedge]

Ahh I think I see what you mean 1 moment

[lyric ember]

$1 = \frac{a}{a} \quad a \neq 0$

[warm shale]

tobi.

[lyric ember]

maybe that helps

[karmic hedge]

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

You mean like this?

[lyric ember]

that was one to much

[karmic hedge]

Haha im so confused sry

[lyric ember]

right idea