help-10

[jolly ginkgo]

Pi/2 swiping a line to get 1/4th circle then swiping 1/4th circle by pi/2 to get First quadrant of sphere

Hard to explain without visuals

[eternal raft]

no no i understood

[jolly ginkgo]

Ok

[eternal raft]

i was strugging with how to identify if its only a quadrant or a full sphere

but i got it

[jolly ginkgo]

Alright

[eternal raft]

thenks for the help!

[jolly ginkgo]

Try the questions I gave they are good ones

[eternal raft]

i will surely!

[jolly ginkgo]

.close

[obtuse pebble]

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[dapper ferry]

how does that turn into that

[jolly ginkgo]

(4-1) sqrt(3)

[dapper ferry]

where does the -1 come from

[high lily]

the - in front of the first sqrt(3)

[dapper ferry]

oh ok I get it

next question

why is it sqrt(12)*3 and not sqrt(12)+3

[jolly ginkgo]

Why would you bring a + when it's not there?

[dapper ferry]

because when 1/2+1/2 happens it's 1+1/2

[obtuse pebble]

@dapper ferry Has your question been resolved?

[tardy epoch]

no?

,calc 1/2 + 1/2

[warm shale]

Result:

1

[tardy epoch]

,calc 1 + 1/2

[warm shale]

Result:

1.5

[dapper ferry]

I meant (1+1)/2

my bad

[tardy epoch]

that addition property has nothing to do with multiplication

[dapper ferry]

there's an addition in the middle

[tardy epoch]

simplify each of those fractions before adding

it doesn't matter how you simplify, but you're getting confused so try another way

[dapper ferry]

ok I understood now thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[spark ridge]

Hi, I’m confused on what it’s asking for and how to about solving it

[severe reef]

If the right hand side is equal to left hand side, what should that mean for coefficients of x ?

[spark ridge]

Dumb it down 1 more step please

[severe reef]

Nvm forget this

On the right side we have a perfect square expression in x + some constant b

[spark ridge]

This

So how would we go about solving it?

I learn from seeing solved problems

[severe reef]

Try expanding the (x-a)²

Then you would get a squared term of x in the right hand side of the equation as well

And also a linear term of x

And why is that beneficial to us? Because we can compare coefficients of x that way

[spark ridge]

Can you solve it on paper or send a worked example through texit

[severe reef]

Wait let's think about it this way

$ax²=bx²$

[warm shale]

Radiation 𝕏

[severe reef]

What can you say about a and b

If they are some numbers

Forget the original problem for now

[spark ridge]

What can I say? Wdym?

[severe reef]

a times some number is equal to b times the same number

What can you say about b and a? Are they equal?

[spark ridge]

Yes I think so

[severe reef]

Great now let's add another term

$ax²+qx=bx²+rx$

[warm shale]

Radiation 𝕏

[severe reef]

Now is it reasonable to conclude if these two polynomials are equal,
a=b and q=r?

[spark ridge]

Polynomials?

I’m soo far behind I will never pass at this rate

[severe reef]

It's alright dw

So a polynomial is basically an expression of algebraic terms

[idle thunder]

yes you will

you got this

[spark ridge]

❤️😭

This question was in a completing a square work sheet

[severe reef]

We can go by that method as well

[spark ridge]

Where am I completing a square

Yes please do

[fathom flicker]

polynomial - means basically "many terms" it is just a name used to call those equations you have up there. Anything involving x's or x^2s (or even x^100s, etc...) added together or subtracted is called a polynomial. don't get caught up on the names

[fathom flicker]

https://youtu.be/McDdEw_Fb5E I highly recommend watching this short video to understand the method of completing the square

[spark ridge]

Austin you a legend that was a good explanation

[spark ridge]

Please do

[spark ridge]

Thank you I will watch that

[fathom flicker]

definitely would be more than worth your time

[severe reef]

I recommend trying it yourself after watching the video suggested by Austin

If after that you still have any problems, feel free to ask

[spark ridge]

I practically know how to complete the square but the only thing confusing me is the way the question is worded

[severe reef]

After completing the square in the left hand side, we'll get an expression of the sort (x±a)²+b right?

[spark ridge]

Am I completing the square for the 2 questions?

[spark ridge]

Yes

[severe reef]

Then all that's left is to equate the corresponding values

[warm shale]

AustinU

[fathom flicker]

you need help with this? ^

I recommend expanding (x-a)^2 and then grouping like terms on the RHS

[severe reef]

Yes that is what I was trying to explain earlier, but it seems they are supposed to do it using completing the square method

[fathom flicker]

seems unnecessary

[severe reef]

True, but we can't help it if the homework instructions are like that lmao

[fathom flicker]

yeah then you need to complete the square for the LHS

[fathom flicker]

then it should be clear to you what values "a" and "b" must be for the equations to be the same on both sidess

[spark ridge]

I got to here

Is this righ so far?

[fathom flicker]

no

you are completing the square incorrectly

and lack the appropriate parenthesis

[warm shale]

AustinU

[fathom flicker]

if you are struggling with completing the square then I would go back to the video

[spark ridge]

Okay just a sec

Is it -12 in the end?

[fathom flicker]

yes but again you are writing it incorrectly

(x-3)^2 is not equal to (x^2-3)

[spark ridge]

I’m doomed...

[fathom flicker]

and on the right hand side

for no reason

you are rewriting (x-a)^2

as (x-a^2)

which isn't true?

so

[spark ridge]

Oh bruh

Like that?

Where do I go from here?

[fathom flicker]

is it not obvious from that final line what a and b must be?

[spark ridge]

3 and 12??

Please tell me I'm right

[ancient jacinth]

yessir

[spark ridge]

YESSSS

.close

[obtuse pebble]

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[spark ridge]

Thank you @severe reef @fathom flicker absolute legends

[obtuse pebble]

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[sharp pecan]

!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

[crimson flax]

1

[gloomy pine]

Are you familar with the sum of 2 vectors ?

[crimson flax]

yeah

this is just asking to draw the position not calculate it hence we got no values

[gloomy pine]

I get you but there is a way to draw a Vector as the sum of 2

Do you know it ?

[crimson flax]

no

can you show me

[gloomy pine]

Its very visual

I recommand you watch a video

[crimson flax]

do you have any

[gloomy pine]

Let me check Real quick

[crimson flax]

would it just be this for part (i)

[gloomy pine]

https://youtu.be/VFRW0f0XUU8

[crimson flax]

so how would you visualise part ii then

its division by 2

ok

[gloomy pine]

Hey

Well to divide a Vector by 2

Draw the Vector with half the magnitude in the same direction

[obtuse pebble]

@crimson flax Has your question been resolved?

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[unreal basin]

Can anyone help me check a calculation for second moment of area here?

[unreal basin]

I feel that this is the wrong way to approach it and I should perhaps try a table with 3 triangles?

I can find the second moment with respect to the Y axis just fine since it is not a conflict when the integration bar dA reaches that weird flat part where y=a

[obtuse pebble]

@unreal basin Has your question been resolved?

[obtuse pebble]

@unreal basin Has your question been resolved?

[obtuse pebble]

@unreal basin Has your question been resolved?

[obtuse pebble]

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[smoky onyx]

\begin{document}

\textbf{Problem 2:}
$\int _0^{\infty }:x\left(0.07e^{\left(-0.07x\right)}\right)dx$

let u = x , $dv=0.07e^{-0.07x}dx$, $du=dx$ , and $v=-e^{-0.07x}$

[smoky onyx]

$\int _0^{\infty }:x\left(0.07e^{\left(-0.07x\right)}\right)dx$

Was working on this problem here

my approach was IBP

but kinda confused on how id go about it from here

can i get some help with this please

[rugged barn]

You are almost there

The only mistake you made is that you don’t want to have du, you want to have du/dx to use the formula

Same actually, it’s not dv=0,73etc.. it’s dv/dx=0,73etc..

[smoky onyx]

im trying to catch your drift here

[smoky onyx]

but this threw me off 😭

am i missing something here

this better ?

[obtuse pebble]

@smoky onyx Has your question been resolved?

[obtuse pebble]

@smoky onyx Has your question been resolved?

[obtuse pebble]

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[scenic elbow]

ive been trying to find the answer to this for so long and i cant find it.

Why does the area increase until the length becomes greater than the width?

This was a question my teacher gave to me to research on my own, but I cant find any websites or sources even coming close to this. Please help

[obtuse pebble]

@scenic elbow Has your question been resolved?

[obtuse pebble]

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[lavish drum]

im rly stupid

[lavish drum]

how do u do

integral from 3 to 6, (2(x-4)^2)dx

i tried expanding it n stuff then doing rev power rule

i got rly big numbers and stopped

[warm schooner]

set u = x-4

[lavish drum]

OHH

[urban patrol]

should be big numbers

[lavish drum]

then solve for x?

[urban patrol]

Thats smarr

smart

[lavish drum]

x = u+4

wait no

[urban patrol]

no, solve for u

I mean

Integrate u

[lavish drum]

nvm im confusing this w something else

[warm schooner]

do you know how to do u substitution?

[lavish drum]

then u distribute right

yeah

2(u^2)

[urban patrol]

You can set the bounds equal to when x is that

so upper bound is now 10

[lavish drum]

yeah

okay

mind blanked

ty

[warm schooner]

👍

[obtuse pebble]

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[grave ore]

.close

[obtuse pebble]

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[fading sluice]

pleas help me

[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

[fading sluice]

4. I got an answer and would like my work checked

Segment Addition Postulate bc it said that its on the same line meaning that its collinear

sorry if i took long to answer i was feeding my cat

thank you

.close

[obtuse pebble]

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[warm siren]

can someone explain how this is wrong?

i though its just the x and y values but divided by the mass added up

nvm its flipped

.close

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[ruby elm]

Problem 9 is a tough one. I have tried to think of ways to use what is given, but could not proceed. Please help with this one.

[ruby elm]

If it is not readable, pls tell, I will type the question.

[obtuse pebble]

@ruby elm Has your question been resolved?

[spice rune]

would be great

[ruby elm]

9. If sum of the series sigma n=1 to infinity, [ { arc sin (√(mod x)) + arc cosec(√(mod x)) }/ π a ] ^n is finite, where mod x > 1 and a > 0 then find interval of a.

[obtuse pebble]

@ruby elm Has your question been resolved?

[rugged barn]

It is

[obtuse pebble]

@ruby elm Has your question been resolved?

[abstract breach]

that's tooo~ easy

[obtuse pebble]

@ruby elm Has your question been resolved?

[ruby elm]

Anyone reach anywhere?

[smoky onyx]

how would i solve it from there tho

it has me soo confused

[ruby elm]

Well, maybe its a little too tough. Need to ask my teacher. Shall I end this?

[ruby elm]

You nailed it my friend. The small observation about mod of common ration being less than 1 was fantastic. Can you give me some tips for solving even harder maths problems?

Understood. Tnx.

Shall I close?

.close

[obtuse pebble]

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[brave oxide]

How is this integration wrong?

[timid silo]

The C_2 term said byebye?

U forgot to integrate that too

[brave oxide]

It’s a constant right so I have to add a V 1 and V2 to it

Is it correct like this?

@timid silo

[timid silo]

No

[brave oxide]

Wdymmm why?

The last V2 has to be V1

Is it correct then?

[obtuse pebble]

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[little mango]

can someone explain this to me?

[wary vigil]

Which?

[obtuse pebble]

@little mango Has your question been resolved?

[little mango]

not yet

number 18

@wary vigil number 18

[obtuse pebble]

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[languid rune]

So I managed to get an answer for question 9 based off of the second-order differential equation for a damped oscillator. This is what I have so far:

[languid rune]

One of the things I'm unsure of is whether my answer for the equilibrium point is correct (10.b). And furthermore, I'm not sure how to move forward for part c

This is the third time I've opened a help channel for this question 😅

[obtuse pebble]

@languid rune Has your question been resolved?

[languid rune]

<@&286206848099549185>

[timid silo]

huh

[languid rune]

This is the problem

I'm mostly getting stuck on part 10b and c but I'm not sure if it's just me or if I screwed up at an earlier state

[obtuse pebble]

@languid rune Has your question been resolved?

[timid silo]

this channel is already occupied

[obtuse pebble]

@languid rune Has your question been resolved?

[languid rune]

Somebody help pls

[obtuse pebble]

@languid rune Has your question been resolved?

[obtuse pebble]

@languid rune Has your question been resolved?

[obtuse pebble]

@languid rune Has your question been resolved?

[languid rune]

This is a bit ridiculous

[slim leaf]

#odes-and-pdes might be helpful

[timid silo]

then yall should search up "Xing Buildz Youtube"

[obtuse pebble]

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[tough hawk]

Can I get help?

[tough hawk]

With this equation

?

@ruby fulcrum

.close

[obtuse pebble]

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[crisp sparrow]

Hello! I wam wanting help with this Maths question, I'm in Year 12 Methods. :) I just don't understand how to do part d I think I did the others right... Thankyou!!

[midnight spear]

do u have a screenshot

[crisp sparrow]

yess!

sorry the other image must not have worked

[midnight spear]

@crisp sparrow part D is what ur stuck on?

[crisp sparrow]

yes!

[obtuse pebble]

@crisp sparrow Has your question been resolved?

[timid silo]

have you tried finding the derivative of (4x^3 + px^2 + qx + 3) e^{-2x} first? You can then just equate the coefficients

,, kx^3 e^{-2x} = e^{-2x} (-8x^3 -2px^2 -2qx -6 + 12x^2 + 2px + q)

[warm shale]

!Kiz__

[timid silo]

you want the right hand side to only contain x^3 terms because otherwise you'd have something like:
ke^{-2x} (x^3 + O(x^2)) which doesn't simplify to the form on the left side

[obtuse pebble]

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[twin edge]

I cannot understand why this limit is zero

[twin edge]

I understand the substitution and all the steps leading to the final expression. What I don't understand is how the final limit equals to zero

[obtuse pebble]

@twin edge Has your question been resolved?

[fathom flicker]

okay well it looks like inside the square root if r=0 then r^4=0 and the inside of the square root in the denominator goes to sqrt(1)

and the r^4 divided by r^2 on the bottom will just go to r^2 on the top

so you are left with r^2 times something on the top

and just sqrt(1)+1 on the bottom

if r -> 0 then you get 0^2 * something on the top

divided by 2

and 0/2 is 0

[twin edge]

oh I see

also another question: when introducing a trigonometric substitution do we consider the angle $\theta$ constant?

[warm shale]

ναζμπαμπα

[fathom flicker]

a trigonometric substitution to what?

[twin edge]

x = rcos($\theta$) for example

[warm shale]

ναζμπαμπα

[fathom flicker]

no, so that is swapping from cartesian (x-y) coordinates to polar (r-theta) coordinates

so instead of having functions involving x's and y's

you have functions involving r's and theta's

as variables

and like you would normally write a cartesian equation as y=f(x)

you normally write a polar equation as r=f(theta)

but theta is definitely not kept constant as a rule of thumb

[obtuse pebble]

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[twin edge]

what I don't understand is why when substituting and x and y for rcos($\theta$) and rsin($\theta$) the variable in the limit is only r and not $\theta$ as well

[warm shale]

ναζμπαμπα

[obtuse pebble]

@twin edge Has your question been resolved?

[twin edge]

what I mean is why is it $\lim_{r\to0}f(r, \theta)$ and not $\lim_{(r, \theta)\to(0, \theta_0)}f(r, \theta)$

[warm shale]

ναζμπαμπα

[twin edge]

nevermind, I think I get it now. It has to do with the fact that $x^2+y^2=r^2$, there's no $\theta$ dependence

[warm shale]

ναζμπαμπα

[sonic anchor]

intuitively you are looking at a circle with infinitely small radius

so letting r->0 is enough

[twin edge]

yeah, thanks

.close

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[crimson gorge]

what is the determinant of A?

[grizzled shore]

the first one?

what properties of determinants do you know

[obtuse pebble]

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[obtuse pebble]

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[weak meadow]

if I have a line

[weak meadow]

L1=(9,3)+(1,6)p

and I want to find a line that overlaps L1 on (9,3)

how would I do that

.close

[obtuse pebble]

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[covert sundial]

how do I check the convergence of the following infinite sum?

[timid silo]

what have you tried?

what do you know about the order of growth of polynomials

versus logarithmic functions

[covert sundial]

the integration method, but idk how to calcule the integral, also tried the limit of an+1/an which is 1, so no help

[timid silo]

you mean the integral test?

[covert sundial]

yes

also

[timid silo]

this isn't an elementary function and it doesn't have an elementary antiderivative, so I wouldn't recommend that

[covert sundial]

I tried comparing it to 1/n, 1/lnn is greater or equal than 1/n and 1/n is divergent

[timid silo]

yes

[covert sundial]

am I allowed to do that?

like.. does it result that 1/ln(n) is divergent

[timid silo]

yes it does

[covert sundial]

I see, I didn't think it was correct, thank you

[obtuse pebble]

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[obtuse pebble]

@blissful bane Has your question been resolved?

[obtuse pebble]

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[inland patio]

hey guys
this is a rotation question
when we have two bodies and we want to find the kinetic energy of their centre of mass(in terms of I and omega), how do we do it?

[inland patio]

i understood for I, but how do we find omega of centre of mass

[obtuse pebble]

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[obtuse pebble]

@inland patio Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[scarlet oriole]

Hey

I need help on this one

For c2 from 1,1 to 1,0

How does the computation work?

[obtuse pebble]

@scarlet oriole Has your question been resolved?

[obtuse pebble]

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[tame narwhal]

!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

[deep brook]

i know the pythagoerean identities and stuff

im just not sure how to apply it here

2

[tame narwhal]

!show

[obtuse pebble]

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

[deep brook]

I tried raising the whole eqquation to the power of two to see if I could get 9cos^2x - sin^2x +9 = 0 but then I realized sin would be raised to the power too so that doesnt work

idk

ig i could set sin^2x = to cos^2x +1

wait no isnt it 1- co^2x

idk what to do after that though

if I have cos^2x +3cosx +2 = 0 idk

[tame narwhal]

Do that, then see if the quadratic formula will get you to the solution

[deep brook]

oh ok

[tame narwhal]

letting, say, $u = \cos{x}$

[warm shale]

cwatson

[deep brook]

a= cos^2x b=3cosx) c = 2

[tame narwhal]

then $au^2 + bu + c = 0$

[warm shale]

cwatson

[deep brook]

ohhh

actually that makes more sense ok

is there another easier way to solve as well other than plugging it in?

[tame narwhal]

$a = 1, b = 3, c = 2$, without the cosines

[warm shale]

cwatson

[tame narwhal]

Got it?

[deep brook]

-1

and

-2

[tame narwhal]

so $u = -1$ or $u = -2$. Then what?

[warm shale]

cwatson

[deep brook]

not sure

[tame narwhal]

Guess

[deep brook]

plug in?

[tame narwhal]

Plug what in, where?

[deep brook]

in the original equation idk

[tame narwhal]

What is $u$?

[deep brook]

u is cosx

[tame narwhal]

Right... then?

[deep brook]

-cosx and -2cosx?

-3cosx?

[tame narwhal]

Not quite, you're getting closer I think

if $u = \cos{x}$ and you've already found that $u = -1$ or $u = -2$, then...

[warm shale]

cwatson

[deep brook]

cosx = -1

[tame narwhal]

Yes!

[deep brook]

or -2

[tame narwhal]

And so now you solve for x

[deep brook]

arccos(-2)?

or -1

[tame narwhal]

Which of those do you think is correct/valid?

[deep brook]

this is pi

[tame narwhal]

Good, that's right

[deep brook]

yeah

arccosines cant be too big right?

why is that?

[tame narwhal]

Well what is the range of cos and sin? i.e., the possible values they can take?

[deep brook]

ohh sin(pi) that makes sense

[deep brook]

only up to 2pi

[tame narwhal]

No that'd be the domain

[deep brook]

ohh

1 and -1

[tame narwhal]

yes

so you know that $\cos{x} = -2$ is impossible, so you can "throw that out"

[warm shale]

cwatson

[deep brook]

makes sense

this is a angle sum and difference question

i dont remember the formulas for them or how they're derived

i know how they work i think

[deep brook]

student A is correct no?

[deep brook]

idk what student B did in step 2 but it def seems wrong

and for question 12

would it be:
cos(180+120) = cos(180)*cos(120) - sin(180)*sin(120)

would there be a better way to split it up?

sin(360-60) = sin(360)cos(60) - cos(360)sin(60)

dunno if I fully understand this one either

[lethal falcon]

you guys needs to look at the unit circle

[obtuse pebble]

@deep brook Has your question been resolved?

[obtuse pebble]

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[frosty valve]

i don't have the physical copy of the paper but if i were to draw it, how do i draw it?

[sick jewel]

pull the compass out until 5 cm , put the pointed part at P and then draw a curved surface above the line PQ

then pull the compass again until 4 cm, put the pointed part at Q and then draw a curved surface above the line PQ

The point where the curved surface intersect is the point R
join P and Q to R

[lethal blade]

damn thats some typing speed

[timid silo]

how do i

get a

help channel

[sick jewel]

take anyone of the help channels under the category math help (available)

just post your question and that channel will be yours

[timid silo]

thanks

[frosty valve]

@sick jewel i dont get the 5cm and 4cm as well

is it something like this?

[obtuse pebble]

@frosty valve Has your question been resolved?

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[kindred shuttle]

how can i get the integral of a function like (x^2 + 5)/(x^2 + 1)?

[pine sail]

(x^2+5)/(x^2+1) = 1 + 4/(x^2+1)

And then trigsub

On an unrelated note, that's Kurt Cobain isn't it?

[kindred shuttle]

thx

[kindred shuttle]

yes

[pine sail]

keep up the good work.

[kindred shuttle]

thx

.close

[obtuse pebble]

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[kindred shuttle]

um..ur username

[kindred shuttle]

.close

[obtuse pebble]

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[pine sail]

Radiohead reference.

[kindred shuttle]

creep

[pine sail]

weirdo

[obtuse pebble]

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[kindred musk]

Hello guys, so I am trying to figure out why the correct answer to this problem is 1/2, I feel like it shouldn't be, especially after I tried doing some integrals.

[kindred musk]

I feel like it should be one of these other answers with n in it.

[median dome]

so what do you think it should be

[kindred musk]

1/2n?

But it's not an answer.

1 honestly makes sense too.

It's a cube, the volume should be 1 * 1 * 1.

[median dome]

what if n=1

[kindred musk]

If n is one, we integrate and get 1/2.

Hm.

[median dome]

what if n=2

[kindred musk]

We integrate and get 1/4.

The coefficient from n1 remains, 1/2 * (n^2 / 2)

Evaluating it as 1 gets us 1/4.

And the patter seems to persist.

[median dome]

wait what does it mean by "it's over a cube"

it won't be a cube for higher dimensions

but anyway

[median dome]

why

[kindred musk]

Because this, no?

n will always be 1.

[median dome]

(\int_{0}^{1}\int_{0}^{1} x_1 ,dx_2\ ,dx_1)

[warm shale]

kheerii

[median dome]

finally

so you have this

[kindred musk]

Haha, yes.

[median dome]

evaluating the first integral

what would you be left with

[kindred musk]

x^2/2

= 1/2

[median dome]

why

you're integrating wrt x2, not x1.

[kindred musk]

Oh.

[median dome]

the x1 is just a constant

in the x2 world

[kindred musk]

Ohh.

Ok.

so it's x1 * x2

then.

[median dome]

yes

and

[kindred musk]

x2 = 1

[median dome]

applying the limits of integration

[kindred musk]

so just x1

[median dome]

you'd be left with x1 again

yes

this would go on up until the final integral

[kindred musk]

Ok, and I do that again, the same thing happens?

[median dome]

at which point you'd get x1^2/2, which would evaluate to 1/2

[kindred musk]

OHH

[median dome]

the order of performing the integrals is

important

[kindred musk]

Only the last variable

[median dome]

indeed

[kindred musk]

is dx1

I see, that makes so much sense.

Thank you!!

[median dome]

no worries

[kindred musk]

.close

[obtuse pebble]

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[obtuse pebble]

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[echo notch]

What did i do wrong?

[median dome]

any context of the problem?

[echo notch]

Imagine the sun is infinite distance away, so the sun rays are paralel to the planet. I want to calculate the time an orbiting satelite is inside of the shade

[inland mulch]

just redo the 3 calculations ?

[echo notch]

The radius of the planet is 200.000 km and the orbital period is listen above

I did them like 10x and i dont understand why the answer is wrong

[median dome]

where did the measurements 100,000 and 450,000 come from

[echo notch]

They were given

omg

450.000 was given

[inland mulch]

they look correct to me

[echo notch]

for 100.000 i divided the radius of the planet by two but i think i shouldnt have done that...

[median dome]

450.000 was given for what

[echo notch]

the distance of the satelite from the center of the planet

[echo notch]

Yep oops this was my mistake

anyway thanks for taking a look

.close

[obtuse pebble]

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[obtuse pebble]

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[woven mantle]

why does it want me to factor this

[royal basin]

we need to know the times at which the ball's height is zero

in other words, we want to solve the equation -5t^2 + 30t = 0

[timid silo]

it’s easier to solve for t

[woven mantle]

why

[timid silo]

$0 = -5t(t - 5)$

[warm shale]

pulse

[timid silo]

now do you see why it’s easy to find t?

[woven mantle]

yes

[royal basin]

you've solved equations by factorization before, yes?

[woven mantle]

how can i tell when a problem would be easier in factored form

[royal basin]

no one-size-fits-all there

[woven mantle]

wdym

[timid silo]

factoring always makes it easier to find the values where it’s equal to 0

otherwise, you’d be using the quadratic formula or other methods

[royal basin]

i am asking you whether you have used the idea/concept/technique/whatever of factorization or factoring in order to solve equations

most likely quadratic ones

[woven mantle]

haven't worked with quadratic stuff yet

or it hasn't been called that in my classes

[royal basin]

this problem you've posted involves a quadratic front and center lmfao

[woven mantle]

so it hasnt been called it in my class

[royal basin]

dunno what else it'd be called if not quadratic

is this problem from your class or from somewhere else?

[woven mantle]

i missed 4 weeks so they sent me this website

saying it had the same material

so idk

[royal basin]

well, i have to say these missed weeks explain it then.

perhaps it is during those missed weeks that the missing material was covered

[woven mantle]

thewy said this covered it

[royal basin]

we are at an impasse

[obtuse pebble]

@woven mantle Has your question been resolved?

[obtuse pebble]

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

[timid silo]

so you are looking for a series $(\sum_i^na_i)n$ such that $a_i>\left(\frac27\right)^n$ (for a positive proportion of i in N) and $0<\lim{n\to\infty}\frac{a_n}{\left(\frac27\right)^n} <\infty$

[warm shale]

Toby

[timid silo]

(just latexing your question)

[timid silo]

(bad notation sorry)

this doesnt work because 4/7 < 2/7

perhaps you could try something like 1/7 instead?

um

lol

nvm

hmm

what about ^(x-1)?

?

$\left(\frac27\right)^n<\left(\frac27\right)^{n-1}$

[warm shale]

Toby

[timid silo]

which is what we want, right?

we want a_n to not be bigger than (2/7)^n

[timid silo]

which it isnt

it should be what you are looking for

2/7 < 1

so thats not true

,w (2/7)^(n+1)> (2/7)^n in reals

[warm shale]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Toby#2275

[timid silo]

np

[obtuse pebble]

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[tepid eagle]

Find the parameter a that solves this identity

[tepid eagle]

with a ≠ 0

I'm stuck with this exercise, can u help me?

[obtuse pebble]

@tepid eagle Has your question been resolved?

[tepid eagle]

<@&286206848099549185>

[timid silo]

Yo.

How can I help?

[valid canopy]

Yes?

[hexed agate]

You could differentiate 1/4x^2 and equate that either the original function

[tepid eagle]

thank u very much

it was easier than I though

[obtuse pebble]

@tepid eagle Has your question been resolved?

[obtuse pebble]

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[fresh kayak]

okay so ive been watching the lesson like 3x and idek how t figure anything out

[obtuse pebble]

@fresh kayak Has your question been resolved?

[fresh kayak]

.close

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[timid silo]

Need the answer please

[nocturne minnow]

We're not here to give out answers

[fierce lagoon]

[rough talon]

lol try the prob lol dont ask for answers

[obtuse pebble]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[quiet drum]

I’m having trouble finding AE

[limber quartz]

,rotate

[quiet drum]

<@&286206848099549185>

[rough talon]

ok

first look at triangle CED

sorry AED

[quiet drum]

yeah

[rough talon]

so

[quiet drum]

oh

wait

[rough talon]

sinceAE is straight up

its a right triangle

so yea

[quiet drum]

9.2^2-7.82^2 = AE

[rough talon]

yes

[quiet drum]

AE^2**

since ur here

can you tell me how to reset my caculators memory

[frosty spoke]

try looking it up on Google

[obtuse pebble]

@quiet drum Has your question been resolved?

[quiet drum]

@quiet drum

@formal vault

<@&286206848099549185>

[rough talon]

,rotate

[quiet drum]

@rough talon

[rough talon]

connect and solve

pythagorean theorem

right

because you can drop the vertex to the s=center of the square

[quiet drum]

oh

[rough talon]

and the center to the side should be 18/2=9

[quiet drum]

i see

[rough talon]

and you get a right triangle

yup 😄

[quiet drum]

so

60^2 - 9^2

[rough talon]

yup

[foggy idol]

It's the 9,60,61 triple

😉

[quiet drum]

61?

[rough talon]

no hypotenuse 60?

the hypotenuse is 60

so

[quiet drum]

59

.32

[rough talon]

no

[quiet drum]

is x

[rough talon]

yes

well technically its 3sqrt391

but yes to the nearest hundreth

[quiet drum]

whats the difference between a slopping edge and a slopping face

,rotat

,rotate

[quiet drum]

@rough talon

[rough talon]

ok

[quiet drum]

This one mb

,rotate

[quiet drum]

What’s the diff between a sloping edge and face

[rough talon]

ok so the sloping face is the literal face

the sloping edge is the altitude of the face

so for a: find the base of the face to the vertex(top) of the face

and b is find the length of the face

[quiet drum]

so a i where i wrote x correct>

?

is*

[rough talon]

ok

im confused waht the x is

srry i gtg now

[obtuse pebble]

@quiet drum Has your question been resolved?

[obtuse pebble]

@quiet drum Has your question been resolved?

[obtuse pebble]

@quiet drum Has your question been resolved?

[royal basin]

.close (op muted)

[obtuse pebble]

Channel closed

Closed by @royal basin

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sage geode]

Generally a tangent to a line is the line itself

[warped hearth]

Bruh

[plain pine]

whered you come from

[warped hearth]

I spent 2mins typing tht and u types it before me

[plain pine]

just copy and paste in another help channel

[warped hearth]

It's fine I'll just delete my msg 😐

[plain pine]

same goes for graph?

[sage geode]

Yeah

[plain pine]

my problem is how am i supposed to solve this

[plain pine]

^

[sage geode]

Oh, wait, I misread the question

My bad

[plain pine]

you got me trippin

[sage geode]

Can you send the full question?

[sage geode]

Ah, alright

So, the function shown is an even function, right?

[plain pine]

what do you define as an even function

because usually the terms i use are completely different

[sage geode]

Algebraicly it means that f(-x) = f(x) for all x in the domain and visually it means that the graphs is symmetrical around the y-axis

[plain pine]

yes

f(x) or f'(x)?

[sage geode]

Therefore the derivative is an odd function

[sage geode]

f(x) is even and f'(x) is odd

[plain pine]

alright

[sage geode]

So, we are given that f'(-1) = 6

Since the tangent to the graph at x = 1 is y = 6x + 7

[plain pine]

inst it f'(-1)

[sage geode]

Yes I made a typo

[plain pine]

okay

[sage geode]

And so, since f' is odd, we have -f'(1) = 6 and f'(1) = -6

Great, let's now evaluate f(1)

The tangent to the graph at x = -1 is y = 6x + 7 or y = 6(x + 1) + 1

So we must have f(-1) = 1

We also know that f is even, so f(1) must be 1 as well

And this is it, we just plug in these values into y = f'(1)(x - 1) + f(1)

[plain pine]

ah okay

ok makes sense ty

.close

[obtuse pebble]

Channel closed

Closed by @plain pine

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[chilly lark]

hi! this is the question im trying to do

[chilly lark]

and here's my working so far

i dont really know if what ive done so far is right.. nor what do next

[neon eagle]

try expand the thing u wrote

[chilly lark]

wdym?

[timid silo]

(y+2) (y-1) isn't correct

[jolly ginkgo]

yeah

[chilly lark]

oh 😭

[jolly ginkgo]

wrong factorization

[timid silo]

Use quadratic formula

[chilly lark]

okok

[timid silo]

Also

You will have to do some wild guessing/approximation to get the answer

Unless you're allowed to use a calculator

I guess they wrote it in decimal because writing it in its proper irrational form would give the answer away

[chilly lark]

okay i got 1 + root 2 and 1 - root 2

[timid silo]

YeH

Yeah*

That's right

[chilly lark]

its a calc question 😌

[timid silo]

Okay that's great. That was possibly the hardest thing if u weren't able to use a calculator

But anyways let's examine your solutions

Do both of your solutions fit the domain specified in the question?

[chilly lark]

yes i think?

[timid silo]

Are you sure

Okay let's approximate it

You know pi is 3.14

[chilly lark]

how do you know if it fits 😅

yep

[timid silo]

pi/2 is 1.5 something something

sqrt(2) by memory is about 1.4 something something

So 1 + 1.4 = 2.4

You have a calculator so you can always directly input those values

Okay so now, is 2.4 lesser than 1.5, the maximum value of x the function can have?

(Hello chartbit )

[chilly lark]

wait but isnt 2pi 6 something 😭

[unreal musk]

Just to clarify, those are the solutions for y=sin(x) yea?

[chilly lark]

is that the maximum value

[chilly lark]

yep

[timid silo]

Where are u getting 2pi from

The maximum is pi/2

Oh wIt

Wait

I misread

Well regardless

That solution shouldn't be true either way because it doesn't fit the domain of the arcsin function, does it not

[unreal musk]

(alt: think about what sin(x) must be between)

[timid silo]

Yeah I think I confused them a lot more now LOL

I trust you chartbit

[chilly lark]

yeah... 😭 its ok

[unreal musk]

Anyways I was just lurking carry on

[timid silo]

Oh lmaoo alright

So so @chilly lark

Sorry for earliee

But basically

Let's continue assuming that both solutions are correct

you have y = 1+ sqrt(2) and y = 1-sqrt(2) right

But what did you define y to be at the very beginning?

[chilly lark]

sinx..?

[timid silo]

Yep!

so you have sin(x) = 1+sqrt(2) and sin(x) = 1-sqrt(2) as solutions rn, right?

Our aim is to find x

[timid silo]

Can you solve for x for both in equations?

[chilly lark]

do i use the inverse of sin to solve?

[timid silo]

Yep!

You are completely right

BUT

Before we do that

Can you tell me what the domain and range of the arcsin(x) function (inverse of sin function) is

[chilly lark]

im... not sure how to find that 😅

[timid silo]

hmmm are u familiar with inverse functions in general?

And like one-to-one functions

[chilly lark]

kind of?

like the range of an inverse function is the domain of the original etc?

[timid silo]

Yes!

But there is one condition

That needs to be fulfilled for a function to be invertible

Do you remember what it is?

[chilly lark]

it can only have one value for x i think?

i mean like

there cant be multiple y values for the same x value

[timid silo]

That's what makes a function not a function actually!

You mixed them up haha

[chilly lark]

oh!

oops 😭

[timid silo]

There can't be multiple x values for the same y value

This is called injectivity

But here is the problem, we are trying to invert sin(x)

But does the sin(x) function have multiple x values for the same y value u think?

[chilly lark]

yes?

[timid silo]

Yeah!

Like

sin(pi/4) and sin(3pi/4) are the same

They both give u sqrt(2)/2

So that's a problem, because it means the function is not invertible

In this case we have to restrict the domain of the function to make it invertible

[chilly lark]

ok i get it!

how do we restrict it?

[timid silo]

I hope I'm not losing u btw pff

Oh nice!

[timid silo]

So we are trying to restrict the domain of the function so it doesn't have any multiple values of x for y

So we don't want overlap

Now I wont like make u guess what that is because it's kinda hard to visualise, but the restricted domain would be [-pi/2, pi/2]

Think about this interval for a second, are there any multiple values of x that would give u the same value of y?

[chilly lark]

nope i dont think so?

[timid silo]

Yeah exactly!

You might wonder why we chose that interval in specific

To explain it in the best way possible, imagine the xy plane in your mind

The way sin(x) function is defined is over the unit circle

Now we are trying to restrict its domain so there is no overlap, while still including all possible values for the domain

So you have to include both the positive and negative portions of the sin(x) function while still removing the 'overlap'

in the unit circle, y = sinx, so whenever the y axis is negative or positive, so is the sin function

So our choice of [-pi/2 , pi/2] would start making sense if you think of that interval as encompassing the fourth and first quadrant, because that's what the interval is defining

The same logic applies to the cosx function with its inverse, in that case we want to have both the positive and negative portions while removing any overlap, so we want to include only the first and second quadrants (where cosx is negative and positive)

so the restricted domain is [0, pi]

Anyways I know this might be a lot to take in but I'm hoping you are still with me? @chilly lark

[chilly lark]

hm i think i get it

mostly at least 😭

[timid silo]

Okay thats great!

So like back to what we are trying to do haha

So we restricted the domain of the sinx function to be [-pi/2, pi/2] and our range is still the same of [-1, 1]

Now we want to know what the domain and range of arcsin(x) is, what do you think? @chilly lark

(Remember what u said earlier about flipping the domain and range haha)

[chilly lark]

[-1, 1] and [-pi/2, pi/2] !

[timid silo]

Yep! You nailed it

That's the domain and range of the arcsin(x) function

So NOW the entire idea of why we did all of this

[timid silo]

For this, you said to take the inverse of sin to solve for x

So what do you have out of doing that?

like, just tell me what it would look like after you solve for x haha

[chilly lark]

2.99 and 0.16?

[timid silo]

No no I don't want numbers

I just want to know the form of the thing right now

Without you evaluating or calculating anything

[chilly lark]

ohh is it a cosine graph now?

[timid silo]

Why do u think so?

[chilly lark]

nevermind wait 😭 it would just be the original graph but flipped over y =x ?

[timid silo]

Like I'm just asking you to tell me what [
\map \sin x = 1 + \s 2 \textss{and} \map \sin x = 1 - \s 2
]
Would look like after u solve for $x$ haha

\vs{3 mm}
Like similar to [
\map \tan x = 1 \longrightarrow x = \map \arctan {1}
] just what the equation would look like after you solve for $x$

[chilly lark]

oh