help-28

[steady flicker]

gracias

I got it 😭

I got the right answer finally

that took longer than it should’ve but thanks for being patient

[tender nacelle]

No problem, prob and stats has a learning curve involved with it, everything will be way easier once you get the hang of it

[steady flicker]

also I just realized the 59.9 part should be the # of inches to be below the shaded region

Like the red part

[tender nacelle]

yeah

[steady flicker]

that’s why I went in circles

[tender nacelle]

its not exacctly 60 because we rouinded 1.66 to 1.7

[steady flicker]

I see

anyways thanks! have a good rest of your evening :)

[tender nacelle]

You too 🙂

[steady flicker]

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[runic trench]

Hello, is the ratio in gradient alyways begin with 1: ?

[torn gust]

might need to give more context

[runic trench]

it’s ok i just realised i was doing the question completely wrong

sorry for wasting your time

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[muted plaza]

I can't do 26.(iii)

Someone please explain how to break 6t10

and what you were thinking when you were breaking it

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@muted plaza Has your question been resolved?

[raven quail]

Add 1 to both sides, factorise the cos terms together and the sin terms together

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[grizzled tinsel]

hello

[grizzled tinsel]

how would i approach this integral?

[full forum]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[grizzled tinsel]

the sqrt is the problem

[sharp flame]

try looking at it a little differently

[grizzled tinsel]

i can only think of

sqrt(1-x^2)

for the arcsin

but

[sharp flame]

alright good we're on the right track

[grizzled tinsel]

its not 1 - x^2 its x - x^2

[sharp flame]

what would you have to substitute to make that possible

[glossy valve]

uoǝu

[sharp flame]

is the sub clearer now?

[grizzled tinsel]

so i get

1/u * arcsin(u)

right?

well i mean

[sharp flame]

uh

[grizzled tinsel]

nevermind

i meant

alright so

i guess

U = 1/sqrt(x)

and then the rest is dv

[sharp flame]

why not just sqrt(x)

[grizzled tinsel]

yeah first step is u = sqrt(x)

and thenm

then

intergral by parts?

[sharp flame]

yup then you'll have the standard arcsin integral that you thought of at the start

[sharp flame]

you'll that that won't be needed

[grizzled tinsel]

thats true

all i need to do

is add 1/2 inside

and 2 outside

and then i can do the intergral

so its

2 arcsin(u) right?

yeah thats it!

thanks

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[slate night]

f: R -> R

and n belongs to N

I need to study the continuity of the function

I presume I start by calculating the limit

but to do so there are 3 different cases that I need to consider, right?

[wise wyvern]

Can't

[wise wyvern]

Yes.

[slate night]

you can't?

ok

[wise wyvern]

Can't directly evaluate the limit

When you don't know x

[slate night]

for x<0, x=0 and and x>0

right?

[wise wyvern]

You'll need those cases like you said.

[wise wyvern]

x < 0

[slate night]

oh yeh thats what i meant

mb

[wise wyvern]

yeah that

[slate night]

do you think there is anything else i should know before attempting this

it seems quite difficult

[wise wyvern]

Not really, that's about all.

[slate night]

Alright, Ill give it a go then, Ill be back soon ig

[slate night]

when doing (e^nx)', you get e^nx * (nx)'

here should i treat n or x as a constant?

or are they both treated as variables?

[wise wyvern]

n is the constant, but I'm not sure why you even need the derivative.

[slate night]

L'hopital

no?

[wise wyvern]

Not really needed.

[slate night]

for the case x>0

[wise wyvern]

Like

Sure

But

[slate night]

oh ig u can also factor

the e^nx right?

[wise wyvern]

yes.

[slate night]

alright

[wise wyvern]

With Lhopital you'll find the derivative is always that.

[slate night]

ok, so for x>0 i got x, for x < 0 i got 1 and for x = 0 i got (1+x)/2

is this right?

[wise wyvern]

For x = 0 (when you already know what x is) why not sub that in (1+x)/2

[slate night]

wdym

[wise wyvern]

what's (1+x)/2 when x = 0

[slate night]

1/2

oh

yeh

lol

so are these correct then?

[wise wyvern]

yes

[slate night]

alright

[slate night]

and now i do f(x) equals each of these whith their own respective domains and see if it's continous or not, right?

like a piece wise function

[wise wyvern]

yes.

[slate night]

alright

ill do that quick too then

[slate night]

i get something weird

for x0 = 0

the limit from the left is 1, right?

and from the right it is x, right?

but f(x) = 1/2

which means its not continuous

ig the question doesnt specfy if it does have to be or not

so it might be fine

[wise wyvern]

At x = 0 only

[slate night]

yewh thats what i meant

[wise wyvern]

That was exactly the question, whether or not it is.

[slate night]

f(0) not f(x)

[wise wyvern]

And you just figured, it isn't at x = 0

[slate night]

alright

makes sense

thanks alot

it was easier than expected

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[formal osprey]

hey guys

i need help with this discrete mathematics question

sorry for the low quality image

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[paper wedge]

Is anyone able to help with this

Nvm

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[storm steeple]

Hello

[tulip oriole]

they give you two coordinates so you can find the slope and the y intercept

slope is y2-y1/x2-x1

[storm steeple]

And a,b and c?

[tulip oriole]

well you need to find the equation that relates P to Q

so you need the slope and y intercept to find it

[storm steeple]

Ok

Which will be 230-30/50-10 right?

[storm steeple]

The slope

[tulip oriole]

yes

[storm steeple]

Ok

And then?

[tulip oriole]

then you can find the y intercept

using the slope and one of the coordinate

plug them in y = ax+b

then solve for b

[storm steeple]

That will be the answer for 8 d right?

[tulip oriole]

well that'll help get you all of the answers

for now we need the equationsthat relates P to Q

[storm steeple]

How can we find that?

[tulip oriole]

do you know what linear equations are

y = ax+b

you'll need something like that

where the slope is a and the y intercept is b

[storm steeple]

Yup

[tulip oriole]

alright so you found the slope right

now if you plug the slope and one of the coordinate given into y = ax+b, you should be able to solve for b

[storm steeple]

For y and x I can just plug any of the 2 values right?

[tulip oriole]

yeah

like 30 and 10

[storm steeple]

Yup got it

Then?

b= -20

[tulip oriole]

now you have an equation for the price and supply

which is Q(P) = slope * P + y intercept

plug what you found and that should be answer a

and for b and c, you only need to plug in the value they give you and solve for the missing variable

for b) they give you Q, find P

for c) they give you P, find Q

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[storm steeple]

@tulip oriole thank you!

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[dense acorn]

so i was trying to find d/dx arctan(x) and I came across to pretty much the final step that it equals 1/(tan^2(arctan x) + 1) and i was confused on this step

[dense acorn]

I got stuck here because i thought tan^2 x means tan(tan(x))

so how does tan^2(arctan x) = x^2

because from what i see now tan^2 x = tan x * tan x = tan(arctan x) * tan(arctan x) = x^2

So i'm just trying to figure out what f^n(x) actually means

[hot herald]

in this context tan^2(k) = (tan(k))^2

[dense acorn]

Is it always this?

for trig functins

[hot herald]

for trig yes

[dense acorn]

Ok thank you

[hot herald]

unless the power is -1, in which case function inverse is represented

[dense acorn]

yea

thank you

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[spare nacelle]

hello, am I doing this right? for which $z \in \mathbb{C}$ do these series converge $\sum_{k=1}^{\infty} z^k / k, k z^k$ and $k^2 z^k$ ? I calculated the limsup (convergence radius) and got $R = 1$ for all three, is then $z \in (-1, 1)$ correct for all three?

[glossy valve]

based

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[spare nacelle]

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[cyan wedge]

which one

[timid wasp]

(b), (c), (d)
or you could solve (b) and give tips

[granite torrent]

solving for y in terms of x is a pretty simple way to solve this problem

[timid wasp]

i tried to, but idk if it's right

could you do it?

[granite torrent]

i could do it, but then what?

the point is you need to post your own work for someone to properly help

[timid wasp]

also, i only did the (a) and (b)

[granite torrent]

that's fine as well

[timid wasp]

but im not sure if my solution to (b) is correct

[timid wasp]

ohh

its because i wrote that in portuguese

im from brazil

[steel cove]

If you move the y’s to the other side, you get (some polynomial)=xy+5y

Figure out where to go from here

[timid wasp]

alr

[timid wasp]

look

i will write what i did

(b) $y=\frac{8x^{3}-x-13}{x+5}$\
$8x^{3}-\frac{x(8x^{3}-x-13)}{x+5}-x-\frac{5(8x^{3}-x-13)}{x+5}-13=0\implies x^{2}+17x+52=0\implies x_{1}=-4\text{, }x_{2}=-13$\
that means (-4, -521) and (-13, 2197) are also solutions

i found these two solutions

and the (1, -1)

is it correct?

[glossy valve]

punisherblade

[granite torrent]

this top part here is fine

what you want to do here is polynomial division to rewrite that expression

[steel cove]

What happened in step 2

[timid wasp]

i did the math in my notebook

and wrote the result i got in latex

[timid wasp]

okk

i got $8x^{2}-40x+199+\frac{-1008}{x+5}$

[glossy valve]

punisherblade

[timid wasp]

its not an exact division

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@timid wasp Has your question been resolved?

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[steady flicker]

hi! i have no idea how to find the standard deviation from these graphs. i have no idea where to start :(

[steady flicker]

we're supposed to sort the sets to the standard deviation

[torn gust]

you'd use that standard deviation is about how far things are from the middle

the middle gray bar adds 0 to the standard deviation because it's right there

[steady flicker]

heyyyy plurmorant

[torn gust]

this left graph has a huge standard deviation, while the other is really small

[steady flicker]

long time no see 😭

specifically for graph A, i have no idea how to start

so i took the mean

2.5

and i looked at the x-axis numbers and their frequencies

lemme edit it

i did something like that with all the other values

but when i try dividing everything by the sample size, it's just zero

i wasn't there for standard deviation practice either

[torn gust]

you'd get rid of all the negatives since it's just distance to the center

I don't think the question wants you to calculate anything though

[steady flicker]

oh

wait

ohhhh

i think i see

so i'm supposed to square after multiplying it out?

like square -2

[torn gust]

you'd do before

for that 2nd gray column I'd do 4*(1.5-2.5)^2

since you're adding the distance squared and there's four of them there

but I'd just look at the shapes to get the answer

this sort of formula where f is like the column height 4

you just need to look at what shapes are more spread out though

[steady flicker]

i understand the majority of this, but why multiply by 4?

oh wait

[torn gust]

it's as if there are 4 data points there

each one with a (1.5-2.5)^2

[steady flicker]

i think i'm starting to understand this now

so (1.5-2.5)^2 models the distance from the median to the nearest columns?

aka the deviation

[torn gust]

it's from the 1.5 points to the mean

[steady flicker]

wait

oh

got a lil confused by the location of the red points

lemme try calculating this tho i'll get back to u

okay i defo got the wrong standard deviation for A @torn gust

i'm a little lost on how to read the graph

[torn gust]

idk if you can get the numbers they give you, since they don't give you the actual data, just certain ranges they can be in

like it's a qualitative question, you just have to rank the graphs by sight

[steady flicker]

ohh

[torn gust]

oh I guess it does work with the n-1 equation

[steady flicker]

it does?

[torn gust]

this one

like I get 1.124 this way https://www.wolframalpha.com/input?i=sqrt((8(1.5-2.5)^2%2B4(0.5-2.5)^2)%2F19) edited

[steady flicker]

ohhh ya i was trying to use that one but im pretty sure i got it wrong...

[torn gust]

but you just have to say that A looks like a squished together graph so the standard deviation is small

[steady flicker]

got it

if we were to use the n-1 equation, why would we put 1.5 in as x

[torn gust]

you do it for all the columns, 0.5 1.5 2.5 3.5 4.5

[steady flicker]

ohhhh so it's just linear or something liek that

okay i get it i think

omg

i think i finally get it now

tysm @torn gust

i'm no longer as cooked as before for my makeup math test

good night! thanks for helping me :)

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[steady flicker]

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✅

[steady flicker]

wait @torn gust last question, how to find f in the equation you gave me?

is that the frequency of the nearest deviation?

[torn gust]

height of the bar

since frequency is the y axis

[steady flicker]

height of the nearest bar?

like the one next to the mean?

@torn gust height of which bar?

[torn gust]

you add everything

for the 1st bar it's 2*(0.5-2.5)^2, and there's stuff for every other bar

[steady flicker]

just for understanding imma break this down for confirmation
so the first 2 would be the frequency of the first bar ("y-val"), the 0.5 is the "x-val," the 2.5 is the mean, and the square is just kinda there (as part of the formula)

and f=2 in this instance

so f is the value of the bar being focused on? @torn gust

[torn gust]

right

[steady flicker]

omg

yayyyy

thanks!

no more questions lol

goodnight! :)

[torn gust]

gnight

[steady flicker]

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[quaint prawn]

I'm not sure how the result follows from this fact

I follow until the green underline bit just fine

I just don't see how it becomes what the question asks for

@slow canopy

[slow canopy]

Sir I might be confused. Isn’t that last integral just exactly what the question wants after you rewrite that real part as P_r(theta - phi)

[quaint prawn]

I don’t see why we can write it as that

is it letting z=re^(itheta)

ah

that's it

bruh

[quaint prawn]

no well, if we let z=re^(theta) then we get the first part in yellow, but how do we get the second equality? factoring the e will give us (1+r) no?

[slow canopy]

Are you confused on why the yellow equation is true? Like independent of the question?

[quaint prawn]

now I am yes

[slow canopy]

That would just be multiplying top and bottom by e^itheta and doing algebra

[quaint prawn]

oh

yes that would be why

by e^(-itheta)

[slow canopy]

Yes

[quaint prawn]

ABCD

Helpful soon mashallah sir @slow canopy

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🔠

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[loud haven]

is there any way to find a non numerical solution to sin(x) = x cos(x)?
I dont think there is but i need to find at least the first non zero solution

[kind jay]

When is sin(x)=cos(x)?

[loud haven]

thats pi/4 or well, more comonly pi *n + pi/4

with n in Z

[frozen plover]

IS there only one solution which IS 0

[loud haven]

nope, there are many solutions. I have already put it into wolframalpha and got numericals out but preferably would find a way to get it analytically

[frozen plover]

Idk what to do exactly but maybe try to study the function sin x - xcos x

Like with mean value theorem maybe WE Can get sth like there IS multiple solution on an interval

And see when the function IS strictly increasing or not

[loud haven]

basically its about nuclear form factors. The result F(q²) = 3/a³ * (sin(a) - a cos(a)) is set to 0 and you then need to obtain a to solve for nuclear radi in the a = 2p *sin(θ/2) * R /hbar

with R being the radi, everything else in that equation is known

[frozen plover]

K

[loud haven]

so yeah, if i need to find it numerically anyway since there is no other way i will just do that. Doesnt really matter much

just making sure that i am not missing a trick

[fast peak]

for most equations you can write down you cant solve them analytically

thats pretty much expected

[loud haven]

Yeah, its expected and kinda used to it but havent had them occur with sin and cos yet. Welp, i am just gonna use the solutions of wolframalpha then and get to calculating relativistic impulse and be done. Thanks

[fast peak]

well you are combining trig with polynomials

its not just sin and cos

[loud haven]

yeah true

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[thorny kettle]

how does that become the answer using laws of log

pls help

[dull seal]

Did u try doing anything?

[thorny kettle]

idk what to do after this

i tried distributing but after a few more steps it just led me back to the original equation

distributing*

[dull seal]

Show it

[thorny kettle]

one sec

[thorny kettle]

here

im so lost

pls help

[dull seal]

Go here

Divide by log3 both sides

[thorny kettle]

\what after

[dull seal]

Sorry i missread ur solution

In your work

Dont put the x inside the log

Let it like

xlog3+log3

Same in the other side

After that, gather x together

And factor x outside

Are u following me?

[thorny kettle]

how do i gather the x's together

oh wiat

one sec

\this right??

[dull seal]

Looks fine

Wait

I think u missed signs

Log5 - log3

And xlog3 +2xlog5

[thorny kettle]

why minus though\

id ont

get it

[dull seal]

U have the equality

xlog3+log3=-2xlog5+log5

[thorny kettle]

yes but u said gather the x's together

so i put em on the left and the rest on the right

[dull seal]

That is a way to tell u use the rules u know from equations to make the x’s kn the same side

U have to add (2xlog5-log3) to both sides of the equation to gather th x’s

[thorny kettle]

whats after that

[dull seal]

Show what u got

[thorny kettle]

im still confused canu just show the whole equation

cus this is the only rpbolem im having a hard time at

[dull seal]

U have this right?

[thorny kettle]

ye

[dull seal]

Ok, subtract log3 to both sides of the equation

After that, add 2xlog5 to both sides of the equation

Then factor out the x and show what i got so i can check

[dull seal]

Ok now is correct

Now divide both sides by (log3+2log5)

And like that u isolate the x already

The only work left is usong log properties to make the final answer better looking

[thorny kettle]

ohhhh

okay

i got it now

3 times 25

got it

thank

you

!

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[glossy terrace]

How can prove that for isometry T : R^n -> R^n, if dist(X, T(X)) is constant for all X in R^n then T is translation?

[glossy terrace]

I defined displacement vector f(X) = T(X) - X and assumed that there exists two points p, q such that f(p) is not f(q)

And trying to show contradiction but I have no idea how to make one

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[lofty wigeon]

Conjecture a simple formula for an if the first 10 terms of the sequence {an} are 1, 7, 25, 79, 241, 727, 2185, 6559, 19681, 59047.

[lofty wigeon]

I recognized a pattern but I don't know how should I write a formula

How should I go in these types of question?

6, 18, 54 , 162. The amount of increase is the previous amount times 3

[fast peak]

what type of thing corresponds to multiplying by 3 again and again

[lofty wigeon]

3^n

[paper kelp]

search up geometric sequence

[lofty wigeon]

Found it. ty

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[craggy plinth]

dont get this

[craggy plinth]

i really need help figuring this out

i was able to get a which is just 055*

b is 095* because it's just 55 + 40

[lime jackal]

what is your question @craggy plinth ?

[craggy plinth]

I don't understand how to solve them.

I dont gethow to get C through to F

[lime jackal]

F?

[craggy plinth]

i dont get how to get these

im not sure how to get any of these

i''ve tried lots of stuff i simply do not posses the ability to understand

[wet patrol]

are u sure b is 095?

[craggy plinth]

nope, my bad

i got a and C

[wet patrol]

ah k

[craggy plinth]

So for example, the answer to b is 235*

i have absolutely 0 clue how they got that

[wet patrol]

ah ok

[craggy plinth]

same goes for D - F

[craggy plinth]

do yk?

[wet patrol]

yeah

when we find the bearing of A from B

it is asking us to find the bearing of A if we were to go around B

[craggy plinth]

we're turning from B clockwise tog et to A

i get that

[wet patrol]

ah k

[craggy plinth]

but i dont get how they found 235 is what im saying

[wet patrol]

well visual another north line from B

[craggy plinth]

oh wait it's just 180 + 55

cuz alternate interior angles

[wet patrol]

yeah

[craggy plinth]

ok what about D then

A from C

??

[craggy plinth]

oh that's easy

180 - 95

wrong answer

off by 10*

so

55 + 40 + theta = 180 right

95 + theta = 180

theta = 180 - 95

theta = 85

180 + 85 = 265

the answer being 275

@wet patrol what do you think

[craggy plinth]

correct

my question is, how did we get that 5 *

like how would someone know "hey that's 5 degrees" just by looking at it

[wet patrol]

look at the two angles from before

if we draw a 90 degree line to above C

and 55+40 add to 95

then it must over exceed the line by 5 degrees

[craggy plinth]

O

OH

OK

thanks

bye thanks a lot

that's it

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[wet patrol]

ok

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[hoary aspen]

why cant i just write it as e^3x+1

since integral of e power anything is just the same? idk 😭

[plain fox]

differentiate that with respect to x

[hoary aspen]

wdym

[plain fox]

$$\int \frac{1}{x+1} = \ln |x+1|$$
differentiating both sides with respect to $x$
$$\frac{1}{x+1} = \frac{d}{dx} \ln |x+1| = \frac{1}{x+1}$$
u can verify ur answer by differentiating

[glossy valve]

JustToPro

[plain fox]

cuz indefinite integral is the anti derivative

[hoary aspen]

i mean why is this invalid

[plain fox]

according to u $$\int e^{3x+1} = e^{3x+1}$$
differentiating both sides we get
$$e^{3x+1} = \frac{d}{dx} e^{3x+1}$$
$$\implies e^{3x+1} = e^{3x+1} \cdot 3$$
which is not equal to each other

[glossy valve]

JustToPro

[plain fox]

$$\cancel {\frac{d}{dx}} \cancel{ \int dx } = 1$$ btw

[glossy valve]

JustToPro

[hoary aspen]

shouldnt it be 2*e^3x+1

[plain fox]

review ur notes on chain rule

[hoary aspen]

alr

[plain fox]

also to solve this u can use u sub

[hoary aspen]

got it

[hoary aspen]

so i substitute 3x+1

but thats not helping me holup

[plain fox]

why not

u = 3x+1
du = 3 dx
dx = du / 3

[hoary aspen]

after that?

[plain fox]

$$\int e^{3x+1} dx = \int e^u \frac{du}{3}$$

[glossy valve]

JustToPro

[hoary aspen]

ahh

ty

[plain fox]

np

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[void crystal]

I don't get this

[void crystal]

In the vector field representation, what is my axis going to be?
w=z is supposed to look like a straight line
Then should w=iz be hyperbolic?

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[full forum]

Please don't occupy multiple help channels.

[pulsar kernel]

Closed the other one

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[icy pendant]

wondering if answered correctly

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@icy pendant Has your question been resolved?

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@icy pendant Has your question been resolved?

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@icy pendant Has your question been resolved?

[deep mauve]

4πr³/3 = 4π(1/8)/3 = (π/2)/3 = π/6, but i read something like 1/6π there, maybe you mistakenly wrote it like that (i'm talking about the volume of small copper sphere)

and yes i also found 5880 for the number of small copper spheres

because i think even though the base radius of the cylinder is 8cm, there is an obstacle of 1cm thickness

so we will calculate the volume of the inside as 7cm

π(7)²×20 = 980π

980π/(π/6) = 5880

and for cross sectional area

we will calculate the lateral area

and the surface area of ​​the base

and the surface area of the top

i guess it is like that

2π(8)×20+π(8)²+π(8)²-π(7)² = 320π+64π+(64π-49π) = 384π+15π = 399π

2π(8)×20 is lateral area

π(8)² is base area

i assumed there were no holes in the base

π(8)²-π(7)² is area of ​​the ring at the top

if we also calculate the area of ​​the interior

for lateral: 2π(7)×20 = 280π

for base: π(7)² = 49π

if we add them too

it is 399π+329π = 728π but i am not that sure for this answer

i mean i don't really understand what they wanted us to do in the 2nd question

maybe it is about my english

but 1st one and 3rd one look correct

i don't know if you are here still but i wanted to answer

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[sleek pecan]

hey

[sleek pecan]

Im in highschool and taking differential equations, I am having trouble figuring out how to solve this problem, (lapase transformations). I'm guessing the bottom is supposed go get into the form for e^atsinwt but I'm not really sure

so its not sideways

<@&286206848099549185> 👉🥺👈

ok i just figured it out of my own, for anyone wondering: guess a factor, then synetic division

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[sleek pecan]

lmk if u find a better way (:

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[red garden]

Not sure where to start with the first part

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[red garden]

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[azure token]

If anyone at all is available to call about integration I have a couple more than a few questions

[hot charm]

First define a partition of the interval

To get Delta x and the xi

[azure token]

xi=(3i)/n ? Delta x = 3/n?

[hot charm]

Then fill out those values in the sum

[azure token]

((3i/n)^2/3)+5)*3/n

[hot charm]

That looks about right

But you need to apply the sum over i

[azure token]

how?

[hot charm]

I'll do it on paper

[azure token]

would you be available to call btw?

[hot charm]

Sure

One sec

[azure token]

np

[hot charm]

Okay I'm free to call

[azure token]

dms

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[azure token]

no

[storm steeple]

Hello

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[crisp terrace]

need help

[crisp terrace]

dont really know where to start

[compact nest]

Since the graph isnt labelled you'll have to note each square as just 1 unit. The left line on the triangle is 8 units long, and for the other two lines we'll have to use the pythagorean theorem to find out their lengths

[crisp terrace]

can u show me how?

[compact nest]

Then with all sides we can find the perimeter by adding up all the sides. For the area we'll have to use the law of cosines to get one of the angles

I'll do the bottom one for u but try to get the top line urself after. Might take me a min. Im at work using my phone so just give me one sec :).

[compact nest]

Hope its legible lol

[crisp terrace]

bro, you just helped me so much

thank you so much❤️

imma try the other one now

[compact nest]

Let me know what u get 😄

[crisp terrace]

i got c=10

[compact nest]

Nice thats what I got

[crisp terrace]

so now we add them together

[compact nest]

For the perimeter yup

[crisp terrace]

8+10+2sqrt17

[compact nest]

Do u need an exact answer or r u allowed to round?

[crisp terrace]

im allowed to round

[compact nest]

To what decimal

[crisp terrace]

thousanthb

8+10+8.246

[compact nest]

Yup

[crisp terrace]

26.246

ok good

[compact nest]

Now are u familiar with the law of cosines?

[crisp terrace]

nope

they make us use the box method

[compact nest]

Ooo I got no clue what that is lol

[crisp terrace]

to find area we use 1/2bh or box method

1/2 base x height

[compact nest]

Unfortunately we do not have the height on this triangle

[crisp terrace]

what should we do

[compact nest]

Use the law of cosines XD

[crisp terrace]

ok

[compact nest]

Its not that scary I promise

[crisp terrace]

bet

[compact nest]

Thats the formula. Where big C denotes one of the angles opposite to side c

We are given the liberty of choosing whatever letter we want for whichever side we want

As long as we remain consistent with how we name them it doesnt matter

[crisp terrace]

so c^2=8^2+10^2-2(80) cos C

[compact nest]

I honestly think a yt vid on the law of cosines will do a better job of explaining than I could but I can try my best

[compact nest]

We actually already solved for all three sides so we can plug in a side for that c^2 on the left

And then we solve for big C

[crisp terrace]

i dont have a calc:(

[compact nest]

Daaaaaaayuum

[crisp terrace]

its ok

[compact nest]

R u on ur phone?

[crisp terrace]

youve helped enough i can get it from ehre

[compact nest]

GLGL 🙌

[crisp terrace]

thanks so much man great help

🤝

[compact nest]

Glad to hear that, yw

[crisp terrace]

peace

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[minor needle]

Can someone confirm to me that the answers they gave is just wrong

i get for the first one that dz/dx = yz/(e^z-xy)

[hot herald]

apply e^z = xyz from the original equation

[minor needle]

oh what we are allowed to do that

but would my other answer be right and get points too

[shrewd hamlet]

Ask ur teacher

[minor needle]

ok

[shrewd hamlet]

When u do logarithmic differentiation u do the same thing so

[minor needle]

thdx

thx

[hot herald]

depending on the route you take, the form of your expression may be different

[minor needle]

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[icy pendant]

4/3 x pi x 0.5^3 is 1/6pi for me

[full forum]

Please don't occupy multiple help channels.

[deep mauve]

it is not actually

pi is in the numerator

not denominator

[icy pendant]

i used ur formula and still got same answer

[deep mauve]

i mean it looks like pi is in the denominator

when you type

1/6pi

so you mean

pi times 1/6

right?

[deep mauve]

yes

[torn jolt]

Bruh

[deep mauve]

it is true

[torn jolt]

That’s pi/6

Not 1/6pi

[deep mauve]

i thought you wrote it in the denominator

[icy pendant]

i write it like that

mb

[torn jolt]

Well you’re wrong

🫵

[deep mauve]

online messaging error, but the true way to explain this online is to write π/6 ig

[torn jolt]

🔥

Ye ik I’m just trollin

Imma dip out

[icy pendant]

i see now

[deep mauve]

okay

[icy pendant]

it does look like that mb

[deep mauve]

no problem

[minor needle]

🫵

[deep mauve]

btw this expression is also true

unless you write it in denominator

or you can write directly in the numerator

[deep mauve]

oh wait

maybe it is wrong you know

i guess you can't write to the left like this

i'm not sure tho

It is best to put a multiplication sign or parentheses in between

So we can conclude with certainty that this is π times 1/6

because as i know there is a concept called compound fraction, i don't know if it applies to pi

i'm talking about this

however, i don't want to say anything definitive about this

It is best to put a multiplication sign or parentheses in between as i said

[deep mauve]

If we are talking about multiplication

have a nice day

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[shrewd plover]

a, b, c are positive real numbers such that abc = 1.
Prove that the inequality
1/(1+a+b)+1/(1+b+c)+1/(1+a+c)<=1
always holds.

[shrewd plover]

a, b, c are positive real numbers such that $abc = 1$.
Prove that the inequality
$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+a+c}\le1$
always holds.

[glossy valve]

Areg Galstyan

[shrewd plover]

<@&286206848099549185>

[distant tundra]

Yes

[shrewd plover]

How do I go about proving this?

I tried using HM-GM-AM inequalities

in different ways

but can't seem to figure it out

[distant tundra]

Try a few numbers to prove this

[shrewd plover]

Wdym?

Like substituting numbers for a, b, c?

It does hold true, but I'm not sure how that could lead to a proof.

I managed to prove that
$\frac{1}{1+a+b}+\frac{1}{1+a+c}+\frac{1}{1+b+c}\le\frac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{3}$
But it seems this doesn't lead to a solution either

[glossy valve]

Areg Galstyan

[wicked stream]

ok $frac{(a+b+c)}{3} >= 1$

oops

i

[wicked stream]

noo noo noo

am gm works

abc = 1

a+b+c/3 >= abc

which is 1

a+b+c =3

i swear ive done this 1 before

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@shrewd plover Has your question been resolved?

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[dire fable]

find the position of the centroid(x,y) of the plane figure bounded by the curve y= 19 - 4x^2 and the x and y axis. Find the value of y. How do I go about this question im at a loss

[somber niche]

How do you find the centroid of a figure in general?

[dire fable]

yeah in the context of the question

[somber niche]

what formula do you use?

like, what is the centroid formula in general?

[dire fable]

im not sure on what formula to use

[somber niche]

hold on let me find it

https://www.intmath.com/applications-integration/5-centroid-area.php

here, use this as a reference

[dire fable]

i have got an answer of y = 1 , can you please check this for me ?

[somber niche]

what answer did you get for x?

oh it only asks for y

doesn't look right

I'd argue it has to be higher up

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@dire fable Has your question been resolved?

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[wispy gorge]

If $\lim_{x\rightarrow\infty}\cos(x)$ is indeterminate. Why is $\lim_{x\rightarrow\infty}\frac{\cos(\sqrt{x})}{2\sqrt{x}} = 0$?

[glossy valve]

Bennxy

[mighty bobcat]

What's the solution for this quesion

[wispy gorge]

!occupied

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[solemn gate]

first of all, the lim of one function is indeterminate doesnt mean the lim of other function containg the first is also indeterminate

[wispy gorge]

True, but I couldn't really figure out how it can be 0

[solemn gate]

lhopital maybe

hmm

[wispy gorge]

But isn't 0/0 or infty/infty tho

[solemn gate]

yeah num doesnt go

too inf

not sure bro

id sub t=sqrt(x) first

[wheat ruin]

well this isnt a direct proof but

[solemn gate]

cos(t)/2t

as t-> inf

[wheat ruin]

theres easier way

the max value of cos(x) = 1

as x -> infinty

the denominator will approach infinty

so we have lim x -> inf 1/x

which is 0

[wispy gorge]

Wait

You gave me an idea

Because cos(x) oscillate between 1 and -1

Which is always < infty

So if the denominator goes to infty

It will be something small divided by something enormous

So equals 0

[solemn gate]

so use squeeze thorem maybe?

[solemn gate]

using this

so -1<=cos(t)<=1

-1/2t<=cos(t)/2t<=1/2t

[wispy gorge]

I didn't knew there was a theorem for this

But that's what I meant too

But thanks for the help

[solemn gate]

so i recommend to learn this thorem, its sometimes helpful

[wispy gorge]

Yea looks interesting

I'll take a look

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[stone trench]

could you guys explain to me what grad is

[umbral dome]

if you have a function $f(x,y,z)$ then the gradient is the vector of its partial derivatives: [ \grad f = \ab (\pdv fx, \pdv fy, \pdv fz) ] (or of different dimension depending on the number of inputs)

[glossy valve]

pnoןɔ

[stone trench]

so here it's the vector of the partial derivatives of V

i understand but the problem is from where did this 1/r come from

[umbral dome]

https://youtube.com/watch?v=csF3EkH1hN4

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[regal orchid]

if a matrix has an eigenvalue of 0, does that mean it's not invertible ?

[rare dock]

yea, a matrix is invertible iff 0 is not an eiegenvalue

[regal orchid]

alr thx

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[left creek]

how the heck do I solve for this power series
\sum^\infty_n=1{1/2^n}

[left creek]

sorry, don't remember how to use the math latex formatting from the bot

[umbral dome]

,, \sum_{n = 1}^\infty \frac 1{2^n}

[glossy valve]

pnoןɔ

[left creek]

yep, thank you

[plush egret]

what does it mean 'solve' here

[left creek]

find the sum

[plush egret]

what have you tried?

sorry cloud

it wasnt nice to steal

[left creek]

not really anything, I don't know where to start. I know manually it just looks like 1/2+1/2^2+1/2^3...1/2^n, but I don't know how to actually get the sum without just... summing up until I get a few decimal places

[plush egret]

so this one is a pretty popular sum, theres a few ways to go

can you classify the kind of sum?

i mean, maybe it helps to say: it's geometric

theres a popular 'proof by picture' for this one but it sorta gives away the answer

[left creek]

is geometric a type of power series or is this not a power series?

[plush egret]

its a type of power series

[left creek]

ok

[plush egret]

here's the proof by picture

[left creek]

so. geometric has a specific sum formula though, right? it was like a/r-1 or something

[plush egret]

it does, yea, for certain r

$\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$

[glossy valve]

jan Niku

[plush egret]

as long as |r|<1

[plush egret]

this image can maybe help enforce why its 'geometric'

@left creek does that give you what you needed?

[left creek]

not really tbh? like, it's geometric, ok. a=1, r=1/2, which means that the sum is 1/(1/2)=2?

but that feels. wrong

but i can't place it

[plush egret]

it is wrong

your series starts at 1

so you need to do some adjustment

[left creek]

oh, right. oh, adjustment?

[plush egret]

yours is $\sum_{n=1}^\infty \frac{1}{2^n}$

[glossy valve]

jan Niku

[plush egret]

so, how can we get this to start at n=0?

say we wrote it as $\sum _{n=0}^\infty \frac{1}{2^n}$ naively

[glossy valve]

jan Niku

[plush egret]

how much are we off by? what error have we introduced?

[echo marsh]

wouldnt the sum of this just be 2

since 1/2^n when n=1 to infitnity is 1

u just add +1

or just use infinite gp sum cuz its still a gp, doesnt matter if u start from n=1 or n=0

[plush egret]

it matters where you start if you want to apply a formula

if your sum doesnt match the formula then you need to make it match

[echo marsh]

what formula

if u mean the sum of infinite GP

[echo marsh]

its valid whether n=0 or n=1 in this case

[left creek]

so. geometric. from the picture. looks like it's supposed to be one whole. which leads me to believe it's supposed to be 1. but. why

[plush egret]

the picture is not convincing?

you can look at some more involved derivations

it should seem....idk intuitive

[left creek]

is the picture only for this specific scenario?

[plush egret]

its only for this specific ratio

but it can be generalized without losing its geometric quality

its just not a nice square

you can also think of it like