help-27

[faint hinge]

I got $\frac{\frac{17}{3} + \frac{4x}{3}}{4} \rightarrow \frac{17}{8}$

which was 2.125

[woven radish]

OceanBro

[faint hinge]

This is a good question tho ngl, it's got me puzzled

@sonic kelp what about the edge cases

like for example 1

[sonic kelp]

Then that doesnt work

I'm stumped

[devout snow]

@sonic kelp Has your question been resolved?

[sonic kelp]

<@&286206848099549185>

[daring solar]

middle row left column and middle row right column should be 15/6 and that should be right

[devout snow]

@sonic kelp Has your question been resolved?

[devout snow]

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[alpine forge]

not sure how to draw the innercylinder shown – what's the reasoning on how to draw it? i can do the regular one

[gusty sparrow]

You'd like to know how to draw a hollow cylinder with thickness? @alpine forge

[alpine forge]

um

for this volume question

i usually draw the cylinder on the outside

the normal thing

i know for washers how to do the inner and outer circles

but im not sure how to do it here

[alpine forge]

so hope that helps

[gusty sparrow]

basically, you want help to draw this?

I suggest drawing the inner cylinder first and draw the outer cylinder next

think of it like scaling the inner cylinder on all the axis (except z-axis) to get the outer cylinder

[alpine forge]

what's the point of drawin an inner one?

[gusty sparrow]

wait, what do you need help with?

i'm not sure

on how to draw the diagram right?

[alpine forge]

no why is there an inner ccyclinder there

i dont understand

[gusty sparrow]

for this?

[alpine forge]

yep

[gusty sparrow]

hmm, sorry... I'm not sure of the actual question. I thought you only needed help on drawing the diagram...

[alpine forge]

alright no problem thanks

[devout snow]

@alpine forge Has your question been resolved?

[devout snow]

@alpine forge Has your question been resolved?

[devout snow]

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[restive wigeon]

confused as to why a = 1 here, just need some clarification on this one thing

[stone stump]

it's not saying a=1

$\frac{a}{\abs{a}} = \frac{1}{\abs{a}} \cdot a$

[woven radish]

Denascite

[stone stump]

and |a|=sqrt(65) by the calculation before that

[devout snow]

@restive wigeon Has your question been resolved?

[restive wigeon]

kk

ty

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[karmic seal]

Hello, I have this problem on the left and the answer, allegedly, on the right. I've been trying to figure out how to factor it into such a simple form, I've plugged it into Wolfram Alpha, but I have no answers. I included the factored form, but I'm sure it doesn't simplify to the alleged solution. My life is in shambles. Please help.

[devout snow]

@karmic seal Has your question been resolved?

[karmic seal]

<@&286206848099549185>

[limber pumice]

Let's leave the term t²-1/t²-4 alone for now

Focus on 9t+9/4t+12

[karmic seal]

okay

[limber pumice]

Is there any Common factor?

[karmic seal]

t+3 you mean

[limber pumice]

The numerator, take nine as common factor

[karmic seal]

whoops

sorry i did that already

[limber pumice]

Take 4 as common factor in the denominator

[karmic seal]

yeah i did that

my notes are a bit of a mess but i'll try to show you what i did

[limber pumice]

Ok

[karmic seal]

i factored it correctly to that point

[limber pumice]

Factored the quadratic too?

[karmic seal]

yeah

[limber pumice]

What did you factor it into?

[karmic seal]

this

the two multiplied fractions i factored into this

so then i would just add them normally the way you do with fractions right

[limber pumice]

Wait

[karmic seal]

I end up with this

Ok I'm w8

[devout snow]

@karmic seal Has your question been resolved?

[limber pumice]

Are you sure the textbook is correct? I mean it could be possible ble that you saw the answer to some different excercise

[karmic seal]

that's what i'm saying, i think the book has a misprint

i'm guessing they changed the question from the previous edition but forgot to change the answer

[limber pumice]

Maybe

[limber pumice]

Cause this is as far as you can factorise

[karmic seal]

yeah at best you can factor (t+1) from the numerator but that's not making every single power and coefficient of T disappear from the equation

anyway thanks for confirming i'm not crazy

i think that settles the matter

[limber pumice]

This qn made me think I'm crazy lol

[karmic seal]

same!

😩

this question really ruined my self-esteem

yeesh

anyway thanks a lot for your help

.close

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[jaunty bane]

Hi all! I need help on how to continue solving this problem:

[jaunty bane]

My strategy:

Use the area of a sector formula (A = theta/360(pi*r^2))

Solve for r:

Area_circle = pi*r^2

1 = pi*r^2

r = 1/sqrt(pi)

A_sector = theta/360(pi*(1/sqrt(pi))

A_sector = theta/360

so now all i need to do is solve for theta (angle DOC)

[chrome girder]

true

[jaunty bane]

otherwise known as arc DEC

yeah

but i dont know how to do that

[restive river]

Hmmmmm

I tried to find other arc to put property Lol

[chrome girder]

you can figure out distance from E to DC, then use arccos to get angle of D above EO

[jaunty bane]

yeah i tried to solve for the major arc to do 360- major arc

but couldnt do that

[restive river]

One moment

[chrome girder]

nvm this doesn't work

[jaunty bane]

hm

[chrome girder]

or that may be part of it but not the whole thing

[restive river]

What if u join D and C

And complete the square?

Tho it doesnt seem to be a square

[jaunty bane]

yeah and angles D and C would need to be tangent to the circle

to be 90 degs

[restive river]

You will get an isosceles triangle tho

[chrome girder]

is there a justification that AEB is congruent to DOC
if there is I can't think of it

[jaunty bane]

i wont be able to use triangle sum theorem though

because BCE and OCE are not supplementary

[jaunty bane]

hmm i dont know, but that would help a lot

[restive river]

(Brainstorming)

[jaunty bane]

well if we wanted to prove that they are congruent, we could do it by CPCTC after proving that triangles DOC and AEB are congruent

[chrome girder]

like if the problem said "isosceles triangle (AB is the non-same side)"
would DOC = AEB

[jaunty bane]

hm maybe we could make a square like hakken said by drawing a segment DC, right?

[chrome girder]

DC is a good idea

[jaunty bane]

then DC would be congruent to AB

[restive river]

Similarity?

Ab = Dc

[chrome girder]

if you call the closest point on DC to E, "F"
you can find the ratio of DC to EF
and that's like 2sintheta/(1-costheta)
if you're in trig idk if this helps

[jaunty bane]

yea im not in trig so idk D:

[chrome girder]

alr

do you know about equations of circles in cartesian coords

[jaunty bane]

yes

[chrome girder]

so you can coordinate bash this

there's probablyyyy a more elegant solution

[jaunty bane]

how D:

i'd atleast need 1 coordinate to start it off maybe

[chrome girder]

set the middle of the circle as the origin
side length of square is variable s

the answer here is making itself seem obvious but making itself hard to justify ;-;

[devout snow]

@jaunty bane Has your question been resolved?

[chrome girder]

ah
AE = AB = AD
DO = EO
so ADOE is a kite
AD is parallel to EO

that might be too close to an answer for the liking of the people here but uh eheh

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[restive river]

here P is a power set then what is the meaning of this?

[pseudo basin]

the power set of {1, 2}.

[restive river]

ok

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[restive river]

why iam getting the answer {2,3,5}

[restive river]

A = {1,3,5}

B = {2,3,5,7}

C is (-infinity,8]

my calculations

not infinity it is negetive even numbers

The answer given is (c)

[finite briar]

C is missing 2, 4, 6 and 0

[restive river]

ok even after that i am getting {2,3,5} as answer

[finite briar]

Huh? How

[restive river]

(A union C)= {0,1,2,3,4,5,6,8}

(A union C) intersection B = {0,1,2,3,4,5,6,8} intersection {2,3,5,7}

= {2,3,5}

@finite briar

<@&286206848099549185>

[devout snow]

@restive river Has your question been resolved?

[sage burrow]

If you google this example you will find two different versions which differ in answer (c). one version has {2}, the other {2,3,5}. i think you get the wrong version of the example. the second interpretation is, that the definition of B and C has mistakenly swapped.

[restive river]

ok, thx that really helped ....... ah ! i really should have googled it

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[restive river]

How would I find the height of this cone?

[pseudo basin]

the total height of the shape is 9 cm

and that's made of the radius of the hemisphere and the height of the cone

[restive river]

how do u know that tho

[pseudo basin]

...from looking at the picture?

[restive river]

i dont understand 😭

[restive river]

but what does the radius have to do with the height

[pseudo basin]

do you see that the red line i drew is the height?

[restive river]

yes i see that

[pseudo basin]

do you see that it is made of the radius of the hemisphere (above the marked point) and the height of the cone (below)?

[restive river]

oh yeaa

i see it now

thank youuuu

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[quaint citrus]

So I'm just learning about double integrals, and I got this question wrong. It's because Fubini's Theorem is contradicted in some way, right? What is the reason why this is wrong?

[quaint citrus]

Here’s the theorem I learned about this

[topaz beacon]

f(x,y) has to be continuous

[wooden zodiac]

indeed

[quaint citrus]

Yea thats what i was thinking, but then what would it mean for f to be discontinuous only on a finite number of continuous curves?

i thought that was what was going on here

[topaz beacon]

the whole line x=2y is discontinuous

[quaint citrus]

ohhhhhhhhhhhhhhhhhh

thanks man, got it

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[tulip latch]

How is it (0,0)?

Cos(0)=1

Oh

Hold on

Its radius

[vivid kite]

Yeah

If you are trying to graph it

You don’t need r

=

I mean it’s just cos graph with amplitude 3

[devout snow]

@tulip latch Has your question been resolved?

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[hollow stirrup]

What should I write for this question

I need help really quickly cuz I got to go soon

[restive river]

$\overrightarrow{D} = \overrightarrow{A} + \overrightarrow{B}$

[hollow stirrup]

It says error

[woven radish]

sppetsnaz

[hollow stirrup]

That’s the answer?

[restive river]

Yes

[hollow stirrup]

Thankyou so much

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[restive river]

Yw

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[cloud stone]

let f be a function defined by the expression

[cloud stone]

where φ is an arbitrary differentiable function and a, b, c are real constants.
Show that you have:

I'm kinda lost

Idk if I have to use the implicit function theorem here

[stone stump]

well the function is defined implicitly. so probably

[cloud stone]

how can I do so then?

<@&286206848099549185>

[analog trellis]

Since you have one equation you consider z to be the singular dependent variable and x and y to be independent.

Differentiate implicitly with respect to x and y separately and solve for the common φ'.

[cloud stone]

so, I have to differentiate the whole expression?

with respect to x and y

[analog trellis]

Can you differentiate half of the expression? 🧐

[cloud stone]

nope

I think

ohh, I think I already see what you mean, ok thx for helping :)

[analog trellis]

Glad to help. 🙂

[cloud stone]

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[slender wigeon]

Hi I'm doing this integral and my result is wrong the book says Ln 3x-1/(3x+1)^2
I don't understand how he got the (3x+1)^2

[slender wigeon]

My result is Ln 3x-1/3x+1 i didn't write it correctly in the photo

[chrome girder]

you can find integral of those easier by cancelling out a 3 so the 2nd integral will be -2/(x - 1/3) which integrates to -2ln(x - 1/3) + C
and you get a -2ln(3) out of the +C

[slender wigeon]

And the ^2?

[slender wigeon]

It shouldn't be - 2ln(3x-1)?

[chrome girder]

you get a factor of 3 out of the +C

so it’s the same

[slender wigeon]

Oh ok

[chrome girder]

the ^2 is due to logarithm rules

[slender wigeon]

Oh I don't know well the logarithm rules

[chrome girder]

aln(x) = ln(x^a)

if x > 0

[slender wigeon]

Mh

In the exercise how it should be?

[chrome girder]

wdym

[slender wigeon]

Like for me who is aln(x) if I have 3ln(3x+1)-ln(3x-1)/3

[chrome girder]

that’s not quite right

[slender wigeon]

I don't understand what I should do here

[chrome girder]

here

[slender wigeon]

Something like these?

[frosty cradle]

that is very faint

[slender wigeon]

?

[chrome girder]

like this

forgot the dx

oh well

[slender wigeon]

But why u get x +1/3

[frosty cradle]

divide numerator and denominator by 3

[slender wigeon]

Ohhh ok

So after that how u get the ^2?

?

Someone?

Ohh if I get the integral of 2/x+1/3 it's 2 ln x +1/3 = Ln(x+1/3)^2?

[slender wigeon]

Is this right?

[frosty cradle]

are you asking if $2 \ln(x + 1/3) = \ln(x + 1/3)^2$? if so, then yes

[woven radish]

cwatson

[slender wigeon]

But if i have - 2ln(3x+1) the result is the same?

[frosty cradle]

with the minus sign it would then go in the denominator

[slender wigeon]

Because now I have 3ln(3x-1) - 2ln(3x-1)

I tried again the integration

[slender wigeon]

Oh ok

[slender wigeon]

But the 3ln?

How I get rid of the 3?

[frosty cradle]

I have to go but if you are here in a bit I can try help

[slender wigeon]

Oh i don't think that I'll be there

But thanks for your help and your time

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[restive river]

i need help

pls

<@&286206848099549185>

only for question 21

i need an example for this

to know how to do these problems

please help me

[scenic ether]

For question 21 its a rectangle.

@restive river

[restive river]

wait

what

wait actually?

thank you mane

lol

i meant to send this one

[scenic ether]

the question says to give the specific name of the quadrilateral

[scenic ether]

oh

[restive river]

yeah ik

ty though

[scenic ether]

np

[restive river]

so can you help me with this one?

[scenic ether]

i can try

[restive river]

s

thanks

[scenic ether]

which problem specifically?

[restive river]

ig #6 for now

[scenic ether]

just break the whole figure up

so for example

find the volume of the rectangular prism

and then find the volume of the pyramidal top

[restive river]

so do you just multiply it?

[scenic ether]

for the rectangular prism

[restive river]

4x4x3

[scenic ether]

you just multiply the length, width, and then the height

[restive river]

okay

but how do i do the top one then

[scenic ether]

you would need to know the formula of a pyramid

[restive river]

1/3bh

[scenic ether]

so since the rectangular prism has a square base

that means that the base is 4

[restive river]

so the base would be

yuh

[scenic ether]

yes

[restive river]

would the hieght be 3?

or 4

[scenic ether]

so far we have 1/3(4)(h)

[scenic ether]

no the height has to be perpendicular to the entire square base

did you guys learn the Pythagorean Theorem?

[restive river]

yes\

but im kinda braindead

[scenic ether]

ok

try to use it

just think about it

[restive river]

wouldn't the answer be 69 1/3

?

[scenic ether]

finally

[restive river]

oh nvm

[scenic ether]

so by the Pythagorean theorem, h^2+2^2=4^2

and then just solve for h

h^2=12

$h=\sqrt{12}$

[woven radish]

m64sky

[scenic ether]

so thats the height

[restive river]

oh

[scenic ether]

once u find the volume of the pyramidal top, you need to add the volume of the pyramid and the volume of the rectangular prism

[restive river]

so thats what i got

as the final

[scenic ether]

interesting volume

wait let me check

ok so the volume of the pyramid is $\frac{4*sqrt{12}}{3}$

and the volume of the rectangular prism is 48 cm^3

[restive river]

yup

[scenic ether]

so 48+(4sqrt12)/3

so (144+4sqrt12)/3

[scenic ether]

yes ik

[restive river]

it was simplfied to that

[scenic ether]

ur right

[restive river]

so

so would i just put my answer as that?

[scenic ether]

sure

[restive river]

as this

[scenic ether]

some people prefer simplified answers

[restive river]

oh okay

[scenic ether]

use the simplified answer

[restive river]

yeah i will thank you

[scenic ether]

you are welcome

[restive river]

how to end session

[scenic ether]

.close

[devout snow]

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[scenic ether]

i ended it

[restive river]

oh okay

thanks

[scenic ether]

you're welcome

have a good day

[devout snow]

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[vapid fractal]

I need some help regarding a simple stats/prob question

[clever mist]

[devout snow]

@vapid fractal Has your question been resolved?

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[vapid fractal]

I’m sorry

[vapid fractal]

My pc crashes

Crashed

@clever mist hope you can still answer my simple question

[clever mist]

ok..?

why me

[restive river]

You're HIM

[clever mist]

send

[vapid fractal]

Aight so my first try was using the formula

(X - Mean )/ standard deviation

That’s (670-660)/35, from that I got .29

To get the probability of getting more than 670, I’d have to subtract the z score from 0.5 and then subtract the answer from 1

Doing that gets -38.85% of getting more than 670?

[vapid fractal]

I mean u did say I can just ask

[clever mist]

whats the question

[vapid fractal]

Sorry I thought I already posted it

[devout snow]

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[manic citrus]

Hi, can anyone help me understand part b please

[somber shore]

hi

,tex Lets say $\sin{x}=\sin{\alpha}$

[woven radish]

NotDyatlov

[somber shore]

what can we say about x and alpha?

in the question $\sin{\frac{\pi-3x}{12}} = \sin{\frac{\pi}{3}}$

[woven radish]

NotDyatlov

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[wooden axle]

How to solve this question

[surreal frigate]

u substitution

[sonic smelt]

The function is periodic with period one

So that would be the same as $2000\int_0^1e^{\frac{x}2 - \floor{\frac{x}2}}\dd{x}$

[woven radish]

A Lonely Bean

[finite briar]

Isn't the anti-derivative e^0

Since the exponents subtract

[sonic smelt]

Oh wait

I made a mistake

The period of 2

Should have been $1000\int_0^2e^{\frac{x}2 - \floor{\frac{x}2}}\dd{x}$

[woven radish]

A Lonely Bean

[sonic smelt]

For $0 < x < 2$ you should have $e^{\frac{x}2 - \floor{\frac{x}2}} = e^{\frac{x}2}$

[woven radish]

A Lonely Bean

[finite briar]

Oh..

[devout snow]

@wooden axle Has your question been resolved?

[devout snow]

@wooden axle Has your question been resolved?

[sonic smelt]

What have you got so far?

[wooden axle]

I don't know how to solve this with geogebra

[sonic smelt]

You don't need geogebra

[wooden axle]

I don't know how to solve

[sonic smelt]

Rewrite the integral like that

[wooden axle]

But what can we say?

[sonic smelt]

Okay if why that step is possible is unclear to you

Then

Try graphing e^(x/2 - floor(x/2)) in the geogebra

Since you have that open

What you will notice is that the graph repeats every 2 units

Here we are solving for the area under that curve from 0 to 2000

And, thanks to the repetition, we can say that's just the same as 1000 * the area under the curve from 0 to 2

[coral locust]

@wooden axle

Does this make sense?

I think the part that's hard to comprehend here is what $$x - \floor{x}$$ means

[woven radish]

1pkame(一筒龜)

[coral locust]

But in the end, you'll get that this becomes the part under the floating point, and all you have to do is "move the graph" to where you actually know how to integrate the function (in this case, [0,2) so that you can state all the messy part above as a simple elementary function)

[devout snow]

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[charred bolt]

Don’t know where to start, I was thinking maybe show f’ = a + g’

But don’t know how to continue

[wooden veldt]

General strategy: suppose f(x)=f(y) and show that x=y
Probably but differentiating f(x)=f(y) and using the info given

[devout snow]

@charred bolt Has your question been resolved?

[charred bolt]

Ok

Suppose f(x) = f(y)

Then we also have that ax+g(x) = ay+g(y)

We can also have a+g’(x) = a+g’(y) since f’(x) = f’(y)

So g’(x) = g’(y)

[charred bolt]

Don’t know where to go from here

[devout snow]

@charred bolt Has your question been resolved?

[charred bolt]

<@&286206848099549185>

[thin tapir]

You can't use f'(x) = f'(y)

x and y are two fixed numbers so you can't differentiate

[charred bolt]

Ok

[charred bolt]

But f is differentiable, so f’ = a + g’

And then you just plug in x and y into that equation no?

[thin tapir]

No. You have two different numbers, x and y, and two different functions, f and f'. You know that f(x) = f(y), however that does not mean that f'(x) = f'(y) because f' is different from f

For example if f(x) = x^2 then f(1) = f(-1) but f'(1) is different from f'(-1)

[thin tapir]

This can be useful, what can you say about f' using it?

[charred bolt]

f’ < 2a

[wooden axle]

@sonic smelt

[sonic smelt]

Yeah that seems to be the same

[wooden axle]

But they have 1000

[sonic smelt]

Because of the substitution

Just like we did

[thin tapir]

And can you say f' > something?

[charred bolt]

2g’ < f’ < 2a

@thin tapir

[thin tapir]

Not necessarily

[thin tapir]

This is also false in general

It works if a > 0

But a might be negative

[charred bolt]

Ok idk what to do then

[thin tapir]

You can prove that either f' < 0 or f' > 0

[charred bolt]

Ok and how is that useful here

[thin tapir]

f is strictly increasing, or strictly decreasing

[charred bolt]

ok how can I use that here though lol

[thin tapir]

It proves directly what you want

[charred bolt]

i want to show that f(x) = f(y) ==> x=y

[thin tapir]

Which is equivalent to x≠y ==> f(x) ≠ f(y)

And remember that if a and b are numbers, then either a < b, or a = b, or a > b

[charred bolt]

ok?

[thin tapir]

So

If x ≠ y

Then x < y or x > y

Then f(x) < f(y) or f(x) > f(y)

And then f(x) ≠ f(y)

[charred bolt]

ah ok i see

[devout snow]

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[gray phoenix]

For which angles v in a right triangle does sin v < cos v apply?

[proud perch]

Well a bigger angle has a bigger sin

[gray phoenix]

okay which angles are bigger as cos than sin

[wooden veldt]

Use the unit circle, considering angles between 0 and 90°

[gray phoenix]

what

[gray phoenix]

what do you mean

[devout snow]

@gray phoenix Has your question been resolved?

[polar chasm]

since it's right triangle, 0<v<90

now just look at unit circle and see where sin v < cos v

Do you know where cos v = sin v?

[gray phoenix]

A little to the right

[polar chasm]

yep, and at what v or theta?

[gray phoenix]

I do not know

[polar chasm]

cant you just guess it?

If not, let's just call cos(v) and sin(v) a for now. Notice that the triangle is isosceles

since it's isosceles, the other angle must be also v

and since it's triangle, its angles sum up to 180. Which means v+v+90=180

so v=45, and cos(v)=sin(v) precisely when v=45

considering 0<v<90

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[devout snow]

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[lofty monolith]

can anyone explain part b

[somber shore]

hi
|t-a| <= b
it means -b <= (t - a) <= b

[lofty monolith]

oh

so youve changed it to that form so its comparable to the 5<=t<=13 part?

[somber shore]

yes

[lofty monolith]

oh i c

okay that helped a lot

i used simultaneous eqns and got a=9

thanks

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[golden frigate]

how do i find the circumference of a circle knowing its radius

[pseudo basin]

have you heard of the number pi

[golden frigate]

no

oh ye

yes

[pseudo basin]

okay then surely you know the formula C = 2πr, which in some sense defines pi?

[golden frigate]

not really

i know what pie equals

but i dont know pi

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[mild basin]

What am I doing wrong here? The last step of line 2.. is it not product rule?

I multiply 2ax and e^(2ax)

[scarlet sequoia]

2a is constant here

you can do product rule for xe^(2ax)

[mild basin]

But both the base and power have x

[mild basin]

Am I not allowed to do this?

[scarlet sequoia]

which gives e^(2ax) + 2axe^(2ax)

so I guess it should be

[mild basin]

So my issue is I chained too much

But I’m not sure how my answer is incorrect

[scarlet sequoia]

so wait, you did chain product rule and chain rule for whole expression, right?

[mild basin]

I believe that I messed up here.. I shouldn’t have added this to line 2?

[scarlet sequoia]

your answer is right

[mild basin]

Yes for the second derivative power and product rule

[mild basin]

Really?

[silent mantle]

It's correct

[mild basin]

Symbolab says it’s wrong

Lemme double check

[scarlet sequoia]

you only can simplify it further a bit

1st term + 3rd term = 4ae^(2ax)

[mild basin]

Oops

My mistake

I forgot to include x

[winter patrol]

are you using x as multiplication or forgot to enter the x into symbolab

[mild basin]

the last term, i wrote it as 2ae by mistake

now it's 2axe and I seem to have the same answer

https://tenor.com/view/woman-throws-axe-axe-throw-throw-axe-gif-23305344

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[restive river]

i can find a eq of line using point slop form and slop intercept form..right??

[pseudo basin]

sure can.

[restive river]

okie

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[dull otter]

why does a geometric series converge?

[pastel pasture]

doesn't always converge

[dull otter]

ik my g

but like when r is less than 1 and greater than -1, why does it converge

[pastel pasture]

If you think about it

If you square something that is in between -1 and 1

Or cube

Or in general

Multiply it by itself

It gets smaller and smaller

¯_(ツ)_/¯

[dull otter]

ye

but

[pastel pasture]

That's the poor man's reason

[dull otter]

if you look at the harmonic series

the terms do get smaller

[pastel pasture]

yes I knew you were gonna bring that up

[dull otter]

yet it diverges

[pastel pasture]

[dull otter]

so how do we know for sure it converges

[pastel pasture]

There are many tests to check if a series converges or not

https://en.wikipedia.org/wiki/Convergence_tests

[dull otter]

im talking about a geometric series with common ratio greater than -1 and less than 1

so how do we know a series like that converges?

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[slender wigeon]

Hi I'm trying to do this integral but I'm stuck because i don't know how to get A
I know that for B i need to assume X=3
And for C x=-2

[slender wigeon]

Also i saw some people using system for doing fractions it is a better system then this?

[signal crag]

By systems I’m assuming you’re talking about a system of equations

[slender wigeon]

Yes

[signal crag]

And if so, that is significantly longer

[crisp fossil]

it's the same process

[signal crag]

You might want to consider comparing coefficients

[crisp fossil]

just done differently (system of equations vs comparing coefficients)

it's actually faster to do a system of equations if it's complex and you know how to use matrices

[signal crag]

It feels like more work

And gets more confusing

With the amount of terms

[slender wigeon]

Oh because I don't remember systems of equations very well

[crisp fossil]

well we won't know until the equations are written out lol

do you know how to do that Kurama?

or well I should rephrase

[slender wigeon]

And that's why i wanted to know if using them is the proper way for solving integrals

[crisp fossil]

yes you need partial fractions to solve this

compare coefficients, which will give you a system of equations, which can be solved to give you the three fractions equivalent to your original, which you then integrate

[slender wigeon]

And this is the method that use a system of equation

? *

[crisp fossil]

rn you have
[4x^2-4x+6=A(x-3)(x+2)+Bx(x+2)+Cx(x-3)]

[woven radish]

Scythe

[slender wigeon]

Yes

[crisp fossil]

they're all the same thing really, sorry for any confusion

first you want to expand

[4x^2-4x+6=A(x^2-x-6)+B(x^2+2x)+C(x^2-3x)]

[woven radish]

Scythe

[crisp fossil]

here is where you get your system of equations

we compare coefficients

the x^2 coefficient on the left side must be the same as the x^2 term on the right side

otherwise, they are not the same function

there is no coefficient for x that makes it a replacement for a missing x^2 term

[slender wigeon]

Ohh ok so 4= terms with x^2

[crisp fossil]

yes

4=A+B+C

next, we compare the x coefficients

[slender wigeon]

And after then I make a system of equations

[crisp fossil]

again, there is no constant coefficient or x^2 coefficient that can replace an x coefficient

[slender wigeon]

For the 3 coefficient

[crisp fossil]

one equation for each power of x

x^2, x^1 and x^0

[slender wigeon]

Oh ok

[crisp fossil]

there will be three variables in your system, A B and C, which you will solve for

for instance, your first equation is A+B+C=4, as we found earlier

do you know how to get the rest?

[slender wigeon]

Mh yea i thinks yes

But i have one more question

[slender wigeon]

If I wanted to use this method how do i get A?

[crisp fossil]

OHHH I know what you're talking about

haven't used this method in years

so look

you have [4x^2-4x+6=A(x-3)(x+2)+Bx(x+2)+Cx(x-3)]

[woven radish]

Scythe

[slender wigeon]

Yes

[crisp fossil]

so these two equations have to be the same at every point

if they are not, then they are not the same f(x) functions

so, we check at x=0 for instance

[4(0)^2-4(0)+6=A((0)-3)((0)+2)+B(0)((0)+2)+C(0)((0)-3)]

[woven radish]

Scythe

[slender wigeon]

Ohh

[crisp fossil]

[6=-6A]

[woven radish]

Scythe

[crisp fossil]

personally I never use this because the system of equation method works for all my problems, not just partial fractions

but this works too, I'd completely forgotten

but yeah for each factor in your fraction there's usually a value of x to make them 0

[slender wigeon]

Because with this method sometimes i struggle to find the value for x

[crisp fossil]

so 0 for x, -2 for x+2, 3 for x-3

and you should end up with c1=A(c2), so A=some constant

[slender wigeon]

Yea thanks

So the method with system is better?

[crisp fossil]

it's more useful generally (in more cases)

[crisp fossil]

but this method is so much easier

I strongly recommend using it for partial fractions when you can

[slender wigeon]

So i can only use it if the exponent of the denominator is bigger than the numerator

[crisp fossil]

I mean yes, but that's always

if your numerator is the same or higher degree, long divide

[slender wigeon]

Oh ok thank you so much

By the way do you know some YouTube channel for system of equations?

[crisp fossil]

nope! :D

very sorry

khan academy is very good though iirc

look up khan academy substitution or elimination system of equations iirc

[slender wigeon]

Okk thank you so much

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[lofty monolith]

for a i dont really have it in that form

got -3kh/800pi

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@lofty monolith Has your question been resolved?

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[late pewter]

An automobile plant contracted to buy shock absorbers from two suppliers X and Y. X supplies 60% and Y supplies 40% of the shock absorbers. All shock absorbers are subjected to a quality test. The ones that pass the quality test are considered reliable. Of X's shock absorbers, 96% are reliable. Of Y's shock absorbers, 72% are reliable. what is the probability that one selected shock absorber is reliable

[clever mist]

!status

[devout snow]

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

[late pewter]

1

!1

.1

<@&286206848099549185>

will it be-
96/100×60/100 + 72/100×40/100

[elfin basalt]

yes

because

[late pewter]

this is-
prob(X reliable)×prob(coming from X) + prob(Y reliable)×prob(coming from Y)

[elfin basalt]

P(A) = P(A|B_1) * P(B_1) + P(A|B_2) * P(B_2)

P(B_1) = 0.6 (60% from supplier X)
P(B_2) = 0.4 (40% from supplier Y)
P(A|B_1) = 0.96 (96% of X's shock absorbers are reliable)
P(A|B_2) = 0.72 (72% of Y's shock absorbers are reliable)
P(A) = (0.96 * 0.6) + (0.72 * 0.4)

cuz law of total probability

[late pewter]

but I don't know how does it relate to one item of more than one

what if it asks selected 2 shock absorbers are reliable

[elfin basalt]

well

the calculation combines the probabilities from both suppliers, X and Y, considering their individual contributions (60% and 40%, respectively) to the total supply of shock absorbers. essentially this means that when you randomly pick one shock absorber from the total supply, regardless of whether it comes from supplier X or supplier Y, there is an 86.4% chance that it will be reliable.

IF you were to select multiple shock absorbers, the individual probabilities for each one would still be the same (86.4% chance of being reliable). HOWEVER, the overall probability of all shock absorbers being reliable would be different, as you would need to consider the probabilities of each of them being reliable simultaneously.

[late pewter]

okok so each shock absorber will have same probability of being reliable, so asking select 2 is a dumb question

[elfin basalt]

yeah

well its not a dumb question but the same

[tepid skiff]

hello

[late pewter]

.close

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[late pewter]

thanks @elfin basalt

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[idle kayak]

stupmed on this im on trig unit and this is a challenge question wondering how to go about solving

[hybrid isle]

i believe the median would help us

iirc, the median will pass the intersection and has length (AB+DC)/2

we can guarantee will pass the intersection because we are given AD=BC

make another perpendicular line to the median passing through the center, then we have split it all into right triangles and may easily use trig

[sage burrow]

draw the height at point A to the line CD. name this point P and name the part from DP as x. with ADP and PAC you have two right angled triangles.

[idle kayak]

ah alr i did not know this ty i will try this

How am I supposed to isolate the variable I am still kind of stumped since there is a missing length

[devout snow]

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[devout snow]

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[prisma frost]

How do I answer 6?

[devout snow]

@prisma frost Has your question been resolved?

[prisma frost]

Anyone?

[swift wolf]

i can't even see the question

[torn vessel]

if they're in geometric progression, that means (n+2)/(n-4) and (3n+1)/(n+2) are the same since they'd result in the common ratio. So set them equal and solve for n

[prisma frost]

Ty

. Close

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[formal haven]

A is chain rule
B is chain rule
C is just e^x and arctan derivative
D is chain rule

what am i missing

omg nevermind i see it theres a x next to the sin in B

.close

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[elfin hill]

What is your question?

1 is good

[hybrid snow]

Last page is asking you which of the 3 curves (the bolded, the dashed, or the gray one) is f, f' and f"

Do you know the relationship between f' and f" with f?

You'd start by finding two curves whose relationship is that one is the derivative of the other

And then analyze the third curve

For example if I have you this, could you tell me which one is f and which one is f'?

[devout snow]

@cosmic socket Has your question been resolved?

[hybrid snow]

Yeah

What's the significance of the x values when f' = 0?

Yeah but more importantly, they're x-locations for possible extrema of f

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✅

[hybrid snow]

I would say that

Yeah

Yeah

I'll check later

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[nova idol]

google what a normal distribution curve is

google empirical rule

[sharp sequoia]

maybe you can help yourself by reading 6 words

https://en.wikipedia.org/wiki/68–95–99.7_rule

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[junior flax]

why is the formula for a tangent line y-y1=m(x-x1)?

[half minnow]

well first off, it satisfies (x1,y1) being a point on the line (you get 0=0)

[half minnow]

since the derivative at x_1 is literally the slope of tangent line at x_1,f(x_1)

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[idle tinsel]

how do I do this integral?

[balmy carbon]

small hint

trigonometric substitution

[idle tinsel]

hmm how can you tell we will need trigonometry

[balmy carbon]

you have the square root of a number - x^2

[hybrid snow]

Use geometry

[balmy carbon]

first of all that is the equation of half a circle

[hybrid snow]

You don't have to do trig sub

[balmy carbon]

with radius 8

He can use geometry

but make sure the question allows it

like the instructor allows it

this could be an integration test

[idle tinsel]

the instructions let us graph the integrand

[balmy carbon]

well that probably just use circle area xD

it feels a bit scuffed though

[idle tinsel]

my first attempt i just tried using calc to solve it and got an answer of 0 lol

[woven radish]

Please give me something to compile, for example latex ,tex The solutions to \(x^2 = 1\) are \(x = \pm 1\).See ,help and ,help tex for detailed usage and further examples!

Apr 17, 2023, 2:24 AM

[hybrid snow]

Graph the integrand first

[idle tinsel]

i tried
(64 - x^2) ^ 1/2
then I integrated each part for
(64x -((x^3)/(3))) ^ x/2

do integrals not work like that

[hybrid snow]

No

[fathom thicket]

@idle tinsel thats cursed

[idle tinsel]

sad

[hybrid snow]

That's very cursed lol yeah

[hybrid snow]

Do this

[fathom thicket]

you need to just use integration by subsitution

in anycase, you can solve a general

$\int_{-R}^{R} \sqrt{R^2 - x^2} dx$

[hybrid snow]

why not graph the integrand

It's a semicircle

I hate trig subs and I'm sure everyone does

[woven radish]

doctor99268

[fathom thicket]

yh nothing wrong with that

but if they can see

[idle tinsel]

how does it help solve the integral?

[fathom thicket]

that you end up getting the area of a semi circle formula anyway

if you were to integrate by hand

it makes it more of a satisfactory answer

and also answers the question of why the formula of the area of a circle is the way it is

[idle tinsel]

can I ask, why does my cursed method work for solving this integral?

[fathom thicket]

because

(x/2 + 8) is linear

it can be split up into

x/2 + 8

well nothing changes

sqrt(x + 8)

cant be split up into

sqrt(x) + sqrt(8)

also you made an additional mistake of integrating sqrt(x) as if you just integrate the x only and keep the square root as normal

[idle tinsel]

hmm i think i gotta relearn integrals from scratch i have been doing them wrong this entire time lol

[fathom thicket]

@idle tinsel you should realise that

the dx

is multiplied

into the entire thing

as in, (x + 8)dx is xdx + 8dx

[idle tinsel]

oo

[fathom thicket]

the brackets is literally a multiplication bracket

alot of times, its not written

[idle tinsel]

wait i am a bit confused

xdx is the derivative of x right? then you take the integral of that shouldn't the operations cancel out?

[fathom thicket]

like, people just write $\int x + 8 dx$

[woven radish]

doctor99268

[fathom thicket]

but they are implicitly writing $\int( x + 8)dx$

[woven radish]

doctor99268

[fathom thicket]

no

hmm, it seems that you dont actually understand what is actually going on with integration

[idle tinsel]

i've understood it as the reversal of differentiation

[fathom thicket]

the process is the reversal of differentiation yes

but what integration is actually doing is

summing up small things

to make something that is, well not small

its in the name

integration

like literally

bringing things together

you have some infitessimal thing

that is

$f(x)dx$

[woven radish]

doctor99268

[fathom thicket]

like literally f(x) times dx

the sign $\int$

[idle tinsel]

does integration relate to sigma notation

[woven radish]

doctor99268

[fathom thicket]

yh

i was just about to say that

you know discrete vs continous

sigma is discrete

and that integration symbol is the continous version

[idle tinsel]

oo

[fathom thicket]

so

the function is not $\int ( ) dx$

its only $\int ( )$

if you get what i mean

[woven radish]

doctor99268

doctor99268

[idle tinsel]

how would each one differ?

like int(8) vs int(8dx)

[fathom thicket]

as in

its still int 8dx

but its

int(8dx)

not

int(8)dx

[idle tinsel]

oo

[fathom thicket]

its why you can split up integrals

because

$\int f(x) + g(x) dx = \int f(x)dx + g(x)dx$

[woven radish]

doctor99268

[fathom thicket]

and so sigma notation rules apply here

you can split it up to be

$\int f(x)dx + \int g(x) dx$

[woven radish]

doctor99268

[fathom thicket]

@idle tinsel im saying this, because when you get further in maths, youll see that the dx isnt some flat thing that just is stuck onto the integral

if you want the arc length of a curve. (like if the curve was a rope, how long would it be)

youd do

$\int \sqrt{dx^2 + dy^2}$

[woven radish]

doctor99268

[idle tinsel]

that example does help me understand integration actually

[fathom thicket]

i can walk you through it.

[idle tinsel]

one thing I am still confused on (doesn't really matter, as it works the same with or without dx as you said)
how is int(x) == int(xdx), like in my mind, int(x) is taking the integral of x, while int(xdx) is taking the integral of the derivative of x

[fathom thicket]

you cant really have int(x) on its own like that

$\int x$

[woven radish]

doctor99268

[fathom thicket]

that is cursed

as fuck

[idle tinsel]

am I supposed to read this as:
the integral of the derivative of x
$\int xdx$

[woven radish]

Price

[fathom thicket]

no

because what does xdx have to do with the derivative of x

[idle tinsel]

ohh they are different

i think i am confusing dx with d/dx

what is dx representing?