help-26

[fading sparrow]

There's another that says "the preimage of -3 is 19"

It means when x is 19

Is 2x^2 + 1 = -3

[hybrid raptor]

So 2(19)²+1?

[fading sparrow]

yes

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[subtle viper]

Is the length of the latus rectum 4 or 8

[subtle viper]

Pls 🥹🥹🥹

[fierce ember]

8

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[quiet cape]

Two circles with different radii have chords AB and CD, such that AB is congruent to CD. Are the arcs intersected by these chords also congruent? Explain.Consider the type of triangle that may be drawn by connecting the endpoints of a chord to the center of a circle.Compare the triangles made by two circles with different radii.
.

[quiet cape]

I found that the arcs arent congruent

I'm just double checking

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[quiet cape]

.reopen

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✅

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[dense anvil]

find dy/dx in terms of x and y by differnetiating xsiny + ysinx = 1 implicitly

[dense anvil]

d/dx(xsiny+ysinx) = d/dx(1)

idk what to do after this

[tall wolf]

do you know implict differentiation?

[dense anvil]

not rlly

?

[tall wolf]

https://youtu.be/xbviQHhU1rA

[dense anvil]

aight ill give it a watcj

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[leaden wing]

Do I (4x^3-16x^2+4x+24):(x-2)?

[worthy storm]

you have two unknowns, a and b

try plugging in two values of x, so you will have two equations involving those unknowns

[leaden wing]

Plugging in?

[worthy storm]

evaluating?

[leaden wing]

Sorry english isnt my strongest languange 🥲

Like do i need to simplify it?

Oh is it like a(x^2-2x+1x-2)?

[worthy storm]

example, if f(x) = x^2, then "plugging in x=5" means evaluating f(5) = 5^2 = 25

try doing that with two numbers

[leaden wing]

Is it like
F(x)= a(x+1)(x-2)(x-b)?

[worthy storm]

example of what i mean

if i let x = 1 in the equation, then the left hand side becomes 4(1)^3 - 16(1)^2 + 4(1) + 24 = 4 - 16 + 4 + 24 = 16

do the same with the right hand side

[leaden wing]

So do i need to try every number? Like im suppose to guess..?

[worthy storm]

no, just two numbers

any two

that will give you two equations involving the two unknowns a and b

then you can solve those equations

[junior dock]

you dont need to do so much

just open the brackets

you will get a polynomial in the third degree

compare with LHS

you will get the answers

[leaden wing]

So i just need to factorize the left side..?

[junior dock]

that or open the right side

[leaden wing]

How to open the right side?

[junior dock]

you multiply each term with the other

[leaden wing]

Something like this?

[junior dock]

https://www.youtube.com/watch?v=3s_lroR5_1U&ab_channel=maths520

[junior dock]

yes

but factorizing works also 👍

[leaden wing]

So like
a(x^2-1x-2)(x-b)

And do i just multiply it again?

[junior dock]

yup

[leaden wing]

🥲

Hm..

Is it suppose to be like this?

[junior dock]

$a(x^3-x^2-2x)-b(x^2-x-2)$=$a((1x^3-(b+1)x^2-()x+2b))$
so b comes out as 3 and a=4 it think

[thorny flame]

dextowen

[junior dock]

$a(x^3-x^2-2x)-b(x^2-x-2)$=$a((1x^3-(b+1)x^2+(b-2)x+2b))$

[thorny flame]

dextowen

[junior dock]

b=3, a=4 on comparing

@leaden wing check if you undersatnd

[leaden wing]

Where did a go?

[junior dock]

a is common

just like i took 4 common

i took 'a' as common

that's why a=4

[leaden wing]

So like you get it out and put it infront?

[junior dock]

yes and you divide each term in the bracket by that same amount

[leaden wing]

I think i understand how to do that type of question but i dont understand the upper part

Should like a be a^2 or..?

Is it like
a-a
And x get in, b get in?

[topaz sinew]

@leaden wing Has your question been resolved?

[leaden wing]

<@&286206848099549185>

[unkempt umbra]

which part is the upper part

the (x + 1)(x - 2) part?

[leaden wing]

This part

Where does the a on ab go?

[unkempt umbra]

it is factored out

there's two terms there
ax[x² - x - 2] and -ab[x² - x - 2]

actually we only need that both terms here have an a

if you'd like those lines go
ax[x² - x - 2] - ab[x² - x - 2]
= a ( x[x² - x - 2] - b[x² - x - 2] )
= a(x³ - x² - 2x - bx² - bx - 2b)
is this clearer?

[leaden wing]

So like

a-a and it becomes..a?

[unkempt umbra]

there's no subtraction

[leaden wing]

a:a..?

[unkempt umbra]

does
ab + ac = a(b + c)
make sense?

we aren't actually doing anything to a, just using the fact that multiplication is distributive

[leaden wing]

So

A is like kicked out right?

[unkempt umbra]

yeah but it's still there

[leaden wing]

To the front?

[unkempt umbra]

it's now outside, so that means the same thing as each term inside having an a

[leaden wing]

Yeah

But there was a previous a right

So

[unkempt umbra]

this is the same a

we are doing the same thing above it
a(x - b) = ax - ab

[leaden wing]

Ohhh

[unkempt umbra]

we then multiply (ax - ab)(x² - x - 2)

really you could have just left the a outside

just done (x - b)(x² - x - 2)

and then you'd have a(x³ - x² ...)

[leaden wing]

I thought the a on the front was already outside

[unkempt umbra]

it was but he brought it in

make sense now?

[leaden wing]

Yeah

[unkempt umbra]

👍

[leaden wing]

God 😭

Thank you sm

🙏

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[ocean fjord]

how would we factorize an equation of the form kmn = x(m+n) + c for some known constants k, x and c

[thorny flame]

sirgareth1

[topaz sinew]

@ocean fjord Has your question been resolved?

[ocean fjord]

<@&286206848099549185>

[craggy haven]

uh you can't really factorize that

[peak heart]

what question is it

[craggy haven]

you can solve for x though, is that what you mean?

[ocean fjord]

nah i was just wondering if there was a general factorizing algorithm for that

apparently not, its case based

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[lunar gust]

can someone help with this problem

[topaz sinew]

Please don't occupy multiple help channels.

[lunar gust]

i feel like i am messing up somewhere

[noble laurel]

Don't occupy multiple channels

do you want to keep this or #help-8 open?

[lunar gust]

sorry i didn't mean to

i didn;t know how to close the other one

keep this

[noble laurel]

Okay

use (.close) when you are done with a channel

I will close the other one for you

[lunar gust]

thank you

[topaz sinew]

@lunar gust Has your question been resolved?

[noble laurel]

What is the actual question here?

[lunar gust]

i feel like i am going towards a negative y value

but if i am splitting the parabola to find an equal value on either side it should be positive

let me send the question

[noble laurel]

No like what is the question you are trying to solve

[lunar gust]

this is the question

[noble laurel]

Okay

so your work is a little hard for me to see, so I am just going to give a general idea here

and if you've already tried it then forget I said it

but

think about what the area of the region bounded by y=25x^2 and y=b will be

and what the area of the region bounded by y=25 and y=b will be

now you should be able to come up with integral formulas for both of these regions

and you should be able to evaluate both integrals

and then since you want the areas to be equal, set the two evaluated integrals equal to eachother

and then you have an equation which you will then use to solve for b

[lunar gust]

idk if that is any clearer

but i will redo it with what you told me

ty

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[violet umbra]

this was in my multiplication/chain rule calculus homework and i'm not too sure how to go about it

[valid marsh]

Do you know how to find the minimum of a function

[violet umbra]

besides graphing it, no

[valid marsh]

How about finding critical points?

[violet umbra]

i'm not sure

[tall wolf]

,w graph x^2

[thorny flame]

Results provided by WolframAlpha

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Requested by Arctic#1647

[tall wolf]

whats the derivative at the vertex

[valid marsh]

just noticed that you can divide the top by the bottom

and get a quotient and remainder

[violet umbra]

oh

[violet umbra]

um

[valid marsh]

you can rewrite the top as (x+y)^2 - something

[violet umbra]

yep

[valid marsh]

what do you get when you do that?

[violet umbra]

(x+y)^2 - xy i think

[valid marsh]

then you can separate it

into two fractions

[violet umbra]

ok

1 - (xy)/(x+y)^2?

[woeful stone]

You would need to find the Hessian matrix. I can't remember the precise mathematical theorems. I recommend you refer to James Stewart Calculus. Solution to textbook are on quizlet

[violet umbra]

i have no idea what that is sorry

[distant nymph]

This can be solved easily using A.M. >= G.M. inequality. However, if using calculus, then you'll probably have to differentiate.

[violet umbra]

how would i differentiate it if there are two variables

[woeful stone]

You would use partial differentiation.

[violet umbra]

i dont think i've learnt that yet

[valid marsh]

are you allowed to use am gm inequality

[violet umbra]

i mean i've learnt it but its not the topic right now

so i'm not sure

but i guess not

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[violet umbra]

.close

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[topaz sinew]

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

[fringe owl]

3

[whole slate]

show your work?

[fringe owl]

ok

let me type it out

[whole slate]

Can you write and post?

if possible

or your TeX

[fringe owl]

i did

[whole slate]

you have two equations here

[fringe owl]

yes

ones postitve other negative

i didnt type it down tho

[whole slate]

first apply log 10 to first equation as well as log 10 to second equation

[fringe owl]

i just did x=asqrt10^b minus or plus 2

[whole slate]

$x+2= 10^{a+b}$,
$x-2=10^{a-b}$

[thorny flame]

dotdoc.

[fringe owl]

yes thats what i did

yep

[whole slate]

right, before we do anything

[whole slate]

we clearly know this system and the original system in question are equivalent right?

[fringe owl]

yes

log_10(x+2)

=a+b

[whole slate]

and you agree,if we multiply the same thing in both sides in any of the equations, the solution will not change.

[fringe owl]

like both equations

[whole slate]

I mean take one equation, multiply both sides of the equality with same thing

[fringe owl]

sorry whats the thingg

[whole slate]

Let’s take $x+2= 10^{a+b}$

[thorny flame]

dotdoc.

[fringe owl]

yes

[whole slate]

Will the solution change, if I multiply both sides with same quantity?

say, If multiply both sides by 2?

[fringe owl]

yes

[whole slate]

will it change?

[fringe owl]

yes

but just will be multiplied with 2 right?

[whole slate]

Why do you think so?

[fringe owl]

they will always be the same

[whole slate]

exactly, the won’t change, as you can always cancel

now multiply the left side with (x-2) and right side with 10^{a-b}

[fringe owl]

both equations or one

you just square the equation no?

[whole slate]

The other one

[whole slate]

I took this equation

multiply both sides with other equation

[fringe owl]

ok

[whole slate]

because the solution won’t change if you multiply same thing with both side

Even though it looks different, we are multiplying the same thing but disguised

[fringe owl]

yes

so (x+2=10^(a+b))x-2=10^(a-b)

[whole slate]

$(x+2)(x-2)=10^{a+b}10^{a-b}$

[thorny flame]

dotdoc.

[whole slate]

this is what i mean

[fringe owl]

oh yeh

sorry i though it differently

[whole slate]

because (x-2) and 10^{a-b} are same. It’s like multiply 2 in both sides

what i said is the reasoning of solving system of equations

[fringe owl]

yep

[whole slate]

usually you add or substrate two equation to solve it, it’s because of the above reasoning

[fringe owl]

It says in the = sign

[whole slate]

Now can you proceed?

clearly you can get rid of b

[fringe owl]

so solve in terms of x only

if u getting rid of b

[whole slate]

can you see b is getting removed?

[fringe owl]

with the plus and minus of b

[whole slate]

exactly

[fringe owl]

but cant you do that only in division

where you subract the powers

[whole slate]

you can do without ever operations you want provided the nothing mishap happens

ie you don’t get to divide by zero or so

[fringe owl]

oh k

cause -b++b=0

[whole slate]

in the exponents, yes

[fringe owl]

so then u get 10^2a

[whole slate]

if you divide you cancel off a right?

not b

[fringe owl]

yes

cause u subrtracing powers in division

[whole slate]

so that’s not case where

also i cannot guarantee x cannot be equal to -2 being an division by zero

[fringe owl]

should i check the answer sheet?

[whole slate]

proceed from here

solve quadratic

[fringe owl]

ok

[whole slate]

you multiply both equations because you want to get rid of b

so clearly that’s what we did

[fringe owl]

ima use the quadratic formula

[whole slate]

what’s (a+b)(a-b)?

[fringe owl]

a-b^2

all squared

[whole slate]

you are done, we don’t want to run for quadratic equation

[whole slate]

It’s a^2-b^2

[fringe owl]

yes

i said a-b all squred

[whole slate]

that’s wrong

[fringe owl]

i might be saying it wrong

(a-b)^2

[whole slate]

$(a-b)^2= a^2+ 2ab+b^2$

[thorny flame]

dotdoc.

[whole slate]

don’t worry about formula, can you multiply them normally?

[fringe owl]

so u wanted me to expand

[whole slate]

how do you multiply (x+2)(x-2)?

[fringe owl]

u mean expand

[whole slate]

yes

[fringe owl]

using foil

[whole slate]

do it

[fringe owl]

ok!!

[whole slate]

i’ll prefer the word distributive property than foil

but all good

[fringe owl]

never heard of word distributive

so its x^2-4

[whole slate]

notice, it’s x^2-2^2

[fringe owl]

oh sorry

x^2+4x-4

[whole slate]

or $(x+2)(x-2)= x^2-2^2$

[thorny flame]

dotdoc.

[whole slate]

you are right

[fringe owl]

so x^2+4x-4=10^a

[whole slate]

notice $(x-a)(x+a)= x^2-a^2$

[thorny flame]

dotdoc.

[fringe owl]

in terms of a?

[whole slate]

where do you get the other x from?

[fringe owl]

double the product?

[whole slate]

use this to evaluate (x-2)(x+2)

what do you get?

if you multiply $(x+2)(x-2)= x^2+2x+-2x+2^2= x^2-4$

[thorny flame]

dotdoc.

[fringe owl]

yes

so x^2-4=10^a

[whole slate]

proceed from here to solve x

[fringe owl]

thers two variables

x^2=10^a +4

[whole slate]

take root you are done.

[fringe owl]

x=sqrt(10^a +2)

but the question asks for terms of a only

[whole slate]

x in terms of a.

that’s what you got

your expression for x in terms of a

[fringe owl]

oh yeh

thanks

do u have time for one more question

[whole slate]

before that

how many solutions you get for last problem?

[fringe owl]

2

[whole slate]

great

Can you open a new channel

i might have to go now

[fringe owl]

ok sure

bye

[whole slate]

ill help if possible

Have a great day

[fringe owl]

you too

.close

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[sharp plume]

10^(2a)

[topaz sinew]

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[fringe owl]

yes thats wt i meant

[fringe owl]

.close

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[neon iron]

How do you get an equation in the form of

[neon iron]

$$ L * M$$ into $$ \frac{L}{M}$$

[thorny flame]

Brandon H

[neon iron]

Goal: Use L'Hopital rule:

[flat kindle]

$L*M = \frac{L}{M^{-1}}$

[thorny flame]

WhereWolf

[neon iron]

[flat kindle]

?

[topaz sinew]

@neon iron Has your question been resolved?

[molten raven]

dimensional analysis!

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[rose idol]

So I'm on the inductive step right now and I'm a bit stuck

[rose idol]

It's the definition part I'm having trouble on

[topaz sinew]

@rose idol Has your question been resolved?

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@rose idol Has your question been resolved?

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[rose idol]

.reopen

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✅

[rose idol]

.close

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[neon iron]

how do i write up the most generel solution to this

[grim jacinth]

Just solve the ODE, pretty much

f'' = f, f'' = -f

[neon iron]

is that how you write it up?

[grim jacinth]

The solution?

[neon iron]

yh

[grim jacinth]

Well you have to solve the ODE first

I never solved for f

[neon iron]

try to solve the first one

then ill solve the 2nd

[grim jacinth]

That's up to you my friend

[neon iron]

if i understand what u did in th1 1st

[grim jacinth]

Ik the solutions to both

[neon iron]

what

yes that is why i am here asking you to help me solve it

[grim jacinth]

I recognize them

So you don't know how to start?

[neon iron]

no

i do not

[grim jacinth]

Are you taking a course in ODEs rn?

Your prof should go over how to do these types of problems

[neon iron]

yh

i dont think this exactly was ever covered

we did like "check if this is the solution to the ode" and where it intersects and such

[grim jacinth]

In that case idk if your prof is expecting you to know the approach to do these

[neon iron]

i have finished school

i am merely doing this for fun but u refuse to tell me how to procede

idk what the problem is

[grim jacinth]

Ah ok

This may help:

https://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx

The idea is to guess f(x) = e^(rx) for some scalar r, and then solve for r

[neon iron]

any specif reason why is is e?

so i guess f(x) = e^rx

and then it becomes r^2=1 or

[grim jacinth]

Bc taking derivatives of e^(rx) gives you scalar multiples of e^(rx) out

Yes, if r^2 = 1, then r = ?

[neon iron]

1,-1 i guess

so f1(x)=e^x and f2(x)=e^-x

[grim jacinth]

👍

[neon iron]

f(x)=ke^x+k1e^-x

never heard of this before

is this supposed to be trivial

[grim jacinth]

Nope, did this for basically 4-5 weeks in my ODEs course lol

And slightly diff variations of it

[neon iron]

if it was the exact same

but with -f(x)

would it then just be the same solution but switched - and +

[grim jacinth]

Well, how does the r^2 eqn change?

This is the main idea, because r is what determines your roots

[neon iron]

anyways

thanks for the help

[grim jacinth]

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[neon iron]

Help with this

[neon iron]

I have a limit that tend to 0

My problem is that I don´t know what to do now

send the problem

I will put it with latex

Sorry I will give it a try more.

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[distant nymph]

Hello

[distant nymph]

Never mind.

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[radiant temple]

How do I equal the denominator of the second term with the first term, so how do I make (x-3)ˆ2 equal that of the denominator of the first one?

[gleaming thunder]

Just like you would with numbers

Take the product of both, or to make it smaller, the lcm (which I won't attempt to define but I hope you see what I mean by that)

[radiant temple]

Yeah but im not trying to get the lcm really, im supposed to simplify it till I get something that has the denominator of the first term

[radiant temple]

This could be wrong but ill just show you the calculations of the computer and its simplification

[gleaming thunder]

Write each fraction as properly factored and it should be much clearer

[gleaming thunder]

Computer is disgusting

[radiant temple]

I forgot a +1

[radiant temple]

Yeah it is lol but the derivative im taking is also quite disgusting, ill just rewrite my stuff and see if it gets any clearer

[gleaming thunder]

Notice they both have a (x-3)^3/2 in the denominator. What's the difference between the two ?

[radiant temple]

Ok I redid it more clearly

[radiant temple]

I dont see how the second one has 3/2

[gleaming thunder]

"properly factored"
Did you simplify the fractions first?

[radiant temple]

Maybe? To the best of my abilities, where im certain at least, I think

Ah wait

[gleaming thunder]

sqrt(x-3)/(x-3)^2, can't you simplify that ?

[radiant temple]

Oh shit I see it

[radiant temple]

I see both 3/2 now, Thank you ill just work it out and come back if I get it or not

[radiant temple]

I got it, thank you 🙏

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[timber mirage]

stupid question but how are they even congruent

i know 27/9 = 30/10 ofc

the problem is i can't really visualize which side in the big triangle corresponds to the side in the smaller triangle

any tips for that?

[deep tendon]

they are not congruent, they are similar

[timber mirage]

yes okay yeah

i meant similar

anyway i don't see how i can orient the two triangles to easily see that

[deep tendon]

CD corresponds to TU, CE corresponds to TV, DE corresponds to UV

[timber mirage]

okay and how do you know?

[deep tendon]

you need a rotation to visualise it

[deep tendon]

SAS

side angle side

[timber mirage]

i don't get it

yes i know sas

but aren't they suppsoed to be in same orientation before you can see which side corresponds to which

[deep tendon]

no

not necessarily

the orientation doesn't matter at all, just that if it is in the same orientation it is easier to see

[timber mirage]

okay then how do you know that EC corresponds to VT

and why isn't VT corresponding to CD instead

[deep tendon]

because CDE is similar to TUV in that order , by SAS

[deep tendon]

the ratio is 1 is to 3..

[timber mirage]

bruh i know the ratio 😭

i just don't get how you get that order

[deep tendon]

yeah like EC is 9, VT is 27, ratio is 3

can't be the other way

[deep tendon]

SAS

SAS property...

[timber mirage]

bro

[deep tendon]

do you really understand SAS

[timber mirage]

SAS says two sides and one sandwiched angle

you keep saying SAS

i know sas

i just wnat to know how you know which side corresponds to which side in the triangle

how did you figure that out

which doesn't 27 correspond to 10

and 9 correspond to 30

[deep tendon]

it means:
given that angle ABC=angle DEF
BA: ED=BC:EF
Then ABC and DEF are similar IN THAT ORDER

that's sas

[deep tendon]

because 27/10 is not 9/30

similar triangles mean you can scale one to get the other

[timber mirage]

Okay here

TVU = CDE

in that order

[deep tendon]

yes

wait no

[deep tendon]

wrong

[timber mirage]

💀

[deep tendon]

it refers to triangles

[deep tendon]

the angles are the same

indeed

[timber mirage]

that's what i'm saying

where did you get the orientation

[deep tendon]

but TV:CD is NOT the same as VU to DE

[timber mirage]

T and C are at the same place

[deep tendon]

so you can't get that

bruh

bruh

read what i sent again

[timber mirage]

bro

[deep tendon]

try to use this to show TVU and CDE are similar

you will fail

[timber mirage]

i know 😭 it'll fail because

10/27 =/= 9/30

[deep tendon]

yeah, so it's the OTHER WAY ROUND

[timber mirage]

it's easy in this example but what if we have some super complicated shit

how would i just know

without trying each combination

that's what i've been asking all this time lol

[deep tendon]

there are only two combinations

what i meant by that was like if you had two angles only

you know?

[timber mirage]

there's 6 combinations

what

maybe i'm missing some conventions

[deep tendon]

so if angle ABC=angle DEF then either ABC is similar to DEF or CBA is similar to DEF, IN THOSE ORDERS

[deep tendon]

like if you have two angles then the vertex of that angle is in the middle for both

not necessarily since you can permute vertices but the idea is there

there aren't 6 combinations

[timber mirage]

okay hold on so you're saying i have to try between VTU or UTV

[deep tendon]

yes

since the angle is fixed

if you get what i mean

[timber mirage]

literally don't lmao

maybe it's too late

i'll check later ig

thanks anyway

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[tough viper]

o

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@tough viper Has your question been resolved?

[neon iron]

maybe, you know, ask your question instead of saying 'o'?

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@tough viper Has your question been resolved?

[restive inlet]

this ain't going anywhere
if she has a proper question, she can open a new channel later

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[slim wadi]

Number 14

How am I supposed to find the Fourier series ?

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[slim wadi]

<@&286206848099549185>

I'm trying to find $b_n$

[thorny flame]

Eggman's

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[slim wadi]

All right thx...

But I got it haha

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[undone flicker]

|(z-3)/(z+3)| =2 this is equation of circle how?

[undone flicker]

Or it is the equation when diameter end points are z1,z2

[drifting swift]

this rewrites as $|z - 3| = 2|z+3|$

[thorny flame]

Ann

[drifting swift]

square both sides, write $z = x + iy$ and you get $(x-3)^2 + y^2 = 4[(x+3)^2 + y^2]$

[thorny flame]

Ann

[undone flicker]

Yes. I know this represents circle equation

But my question is different i guess

[undone flicker]

This one

[drifting swift]

no, the 3 and -3 don't have anything to do with the diameter...

[undone flicker]

Example 6

So it means |(z-3)/(z+3)| =2 is the equation of orthogonal circle?

So it has nothing to do with diameter

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[tropic gull]

i know this is a very simple question

but what is the difference with using 1/4(area of a circle)

and area of a sector

[wild brook]

hm

[wild brook]

Um, it's not 1/4th of a circle though

[wild brook]

that is how you get the area of a sector

[tropic gull]

how is it not 1/4th of circle?

[wild brook]

I don't know how to explain

[old geyser]

because it’s 1/8 no?

[tropic gull]

my brain isnt braining

how is there 8 segments

[wild brook]

Um

So the 1 1 thing

[old geyser]

a quarter would be that whole top right section

[wild brook]

it's a 45 degree equilateral triangle

[old geyser]

but that’s cut in half

[wild brook]

what this guy said

[tropic gull]

but yk for a full circle

its would be pirr^2

[old geyser]

yeah

[tropic gull]

a semi would be pir^2/2

would that not be semi/2

?

[old geyser]

yeah

no

because

if you did half of a semi that’s a quarter

the shaded part is from an eighth

half is all above X axis

quarter is all above and infront of x and y respectively

then that bit is halved

[tropic gull]

but im doing the whole 1/4 part-the triangle

would that not be the shaded?

[tropic gull]

ye ik

ok

im silly

if i do that

theres still a extra part

im such a silly billy

thanks gamers

[old geyser]

yeah you need to subtract the area of a triangle after

[tropic gull]

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[undone flicker]

Is there any short method idea?

I don't know how to solve this or search?

[frail reef]

you could just rewrite some of these

if you spot that some of these numbers are 90 - other numbers here

[mellow arrow]

Use these
tan(x)=cot(90-x)
cosec(x)=1/sin(x)
sin(x)=cos(90-x)

These should be more than enough

[frail reef]

try not to blatently give the answers

[mellow arrow]

did I?

[rich fossil]

i mean you've suggested the cofunction identities yourself

[frail reef]

suggested

not gave

[undone flicker]

I got it i am solving wait

[mellow arrow]

seems like the OP was unaware of these

so it was better to give them off. They, in no way do give the answer

[undone flicker]

Answer 1

D option

[mellow arrow]

correct

[undone flicker]

Thank you all

One more question

[frail reef]

D?

[undone flicker]

Yes

[frail reef]

where does the tan60 go?

[undone flicker]

Hang on

[mellow arrow]

wait by answer 1 You mean option one right?

[undone flicker]

1/√3

This is the answer

[frail reef]

good

[undone flicker]

By mistake

273

I know 366 days

52 weeks

But 2/7 is the answer which they are focusing on week only

[mellow arrow]

364 days=52*7 days
So, in 364 days, all days have passed exactly 52 times

in a leap year there are 366 days

so how many days extra?

[undone flicker]

2 days so in this week only 2 possibilities

Why are we ignoring 52 weeks in probablity

[mellow arrow]

out of 53, 52 tuesdays have already passed in 364 days

they are a must

all leap/non leap years have a minimum of 52 tuesdays

[undone flicker]

Yes they do

So now total possibility?

54/366

Right?

Or 2/7?

[mellow arrow]

no

how do you think 54/366 is right?

your reasoning behind it?

[undone flicker]

52 Tuesday in each week +2 our maximum possibility

@mellow arrow

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@undone flicker Has your question been resolved?

[undone flicker]

52/366 +2/7 ?

52 is fixed so there is no possibility?

[undone flicker]

209

My answer is B

(x-2)^2
(X+2)^2

<@&286206848099549185>

[undone flicker]

211 my answer by substitution method alpha+beta =-2

Only option 3 satisfied

Is there any other method?

[fallow heart]

I would use the fact that the polynomial is divisible by (x + √2)(x - √2) which is also x² - 2. From here I would factorise the polynomial as (x² - 2)(x² + 2x - 15) and calculate the roots of the second factor, which turn out to be 3 and -5

[undone flicker]

Hmm interesting but a little bit long

[fallow heart]

But I didn't understand yours

What do you mean by substitution ?

Your a and b are the solutions?

[undone flicker]

I mean if we add roots no option satisfied

[fallow heart]

I don't get it, sorry

[undone flicker]

Alpha+beta+√2-√2 =-2

@fallow heart

[fallow heart]

Yes, and what is the problem?

[undone flicker]

I was looking for other methods

[fallow heart]

Ok but it is your method that I don't understand. Didn't say that it was wrong, but I've never seen it. Can you explain a bit more?

[undone flicker]

The sum of roots is equal to -b/a

[knotty ledge]

α+β = -2 which is satisfied by (3)

[fallow heart]

Ahnn the Vieta's formulas okok I got it thx

[undone flicker]

208

I checked it random number

[fallow heart]

Well, that method you used I believe it's the fastest yeah

[undone flicker]

I see thanks

[fallow heart]

Yw

[undone flicker]

208 <@&286206848099549185>

[distant nymph]

Do you know how to write a number in expanded form using the place value?

[undone flicker]

What??

I simply took random number

[distant nymph]

That works as well.

I was just trying to tell you the method.

[undone flicker]

But the random number is divisible by 6,9

[distant nymph]

Although only if you work through enough numbers to rule out wrong answers.

[undone flicker]

I didn't get it

[distant nymph]

For example, let's take the number 325.
According to place value, we can write

$$325 = 3\times 100 + 2 \times 10 + 5\times1$$

[thorny flame]

Enemagneto

[distant nymph]

Like that. Write the general form for a 5-digit number.

[undone flicker]

Ohh that method

12345
1×10000+2×1000+3×100+4×10+5

5×1000+4×1000+3×100+2×10+1

[distant nymph]

Write for a general number.

Take number to be pqrst.

[junior anvil]

whats the question

[undone flicker]

It will be like 9999+ 999+ 99+ 9

So 9

[distant nymph]

Answer is correct. Reasoning is not.

[undone flicker]

Could you correct it?

[distant nymph]

Correct what?

[distant nymph]

Difference will just not be like this always.

Each term will be divisible by 9 instead.

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[ripe pecan]

im stuck on question 4

[ripe pecan]

i know i use the formula x+-2;004(s/sqrt(n))

<@&286206848099549185>

the answers are 46 and 42?

oh

well

its

95 percent confidence level

i just wanna make sure im using the right formula and i want my work checked

@neon iron

so did i get it right

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[young sundial]

im trying to figure out what formula there using to get the monthly payment here . GoodLeap 25 year 4.99%
total loan amount $28,652.44
With 30 percent tax credit $122.09/mo
Without 30 percent tax credit $173.90/mo
TOTAL MONTHLY PAYMENTS $36,626.00
2. GoodLeap 25 year 5.99%
total loan amount $26,297.77
With 30 percent tax credit $124.28/mo
Without 30 percent tax credit $176.42/mo
TOTAL MONTHLY PAYMENTS $37,285.00
3. GoodLeap 20 year 5.99%
total loan amount $25,768.35
With 30 percent tax credit $135.69/mo
Without 30 percent tax credit $193.37/mo
TOTAL MONTHLY PAYMENTS $32,565.00

[young sundial]

its usefull to know that this is for a solar loan and the fed gov is paying 30 percent of the cost which is applied to the first monthly payment and the second monthly payment is if the 30 percent is not applied

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[young sundial]

<@&286206848099549185>

if its not possible to reverse engineer the formula let me know

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[fierce zealot]

would the line integral over a closed but non simple curve be 0?

[cyan mountain]

can anyone help me with this question?

[fierce zealot]

please use another channel

this one is taken

[cyan mountain]

oh mb

[sweet shard]

Depends on the curve
https://tutorial.math.lamar.edu/classes/calciii/fundthmlineintegrals.aspx

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[undone flicker]

what if m=1 then?

[vivid flame]

A base of a log cannot be 1

It would be undefined

[whole slate]

It can be evident by seeing anything raised to 1 is always 1

[spare smelt]

here's the deal right: 1^x is not invertible. so you obviously have that 1^2 = 1^3 but that doesn't mean 2 = 3

[undone flicker]

That's why in exponential and logritm we use a>1

Or 0<a<1 but never use equal sign to 1

[whole slate]

fractions can serve as base for logarithms

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[wild brook]

send formula '

[wild brook]

forthis

the image was 40 kilobytes, how LONG COULD IT HAVE POSSIBLY taken

[wild brook]

2 / (2x3x4) + 2 / (3x4x5) + ... + 2 / (88x89x90)
Bear with me guys, I'm too tired to type *

[wild brook]

Currently I have 0 progress

2
10 + 4
60 + 24 + 12
I see

a pattern

I gotta extend

2
10 + 4
60 + 24 + 12
420 + 168 + 84 + 48

Interesting

2
14
96
720

2
2x7
2x2x2x2x2x3
2x2x2x2x3x3x5

That seems

Odd

wait

it's over

divided by 1! and such

Oh wat

no

Nvm I have no idea where this is going someone help

[molten raven]

use partial fraction

[wild brook]

?

[molten raven]

Partial reaction decomposition

The sum probably telescopes

[wild brook]

I hear chemistry

wait is it something like

2/3 * 1/4 * 1/5 + etc etc

[molten raven]

No

[wild brook]

can you like

show example

[molten raven]

Partial fractions are like 1/n(n+1)=1/n-1/(n+1)

[wild brook]

huh

doesn't that not make sense

[molten raven]

Yeah I’ll latex it

[wild brook]

Oh wait

[molten raven]

$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

[wild brook]

ohh

[thorny flame]

kappa

[wild brook]

oh

what if it's 1/(n)(n+1)(n+2)

oh wait

Wait let me do it

[molten raven]

Ok there’s a method though

[wild brook]

1/n - 1/(n+1)(n+2) = 1/n - 1/(n+1) + 1/(n+2) right?

[molten raven]

yes

[wild brook]

what if the above is 2

so does this mean like

2/n + 2/(n+2) + 2/(n+4) etc etc?

[molten raven]

This isn’t your original fraction

[wild brook]

I thought the negative would be carried over

[molten raven]

You have to split 1/n(n+1)(n+2)

[wild brook]

Yes
which split into 1/n - 1/(n+1)(n+2) = 1/n - (1/(n+1) - 1/(n+2))

[molten raven]

Those aren’t equal

[wild brook]

Wat

[molten raven]

1/n(n+1)(n+2) isn’t equal to 1/n-1(n+1)(n+2)

[wild brook]

Oh

[molten raven]

Look up cover up method on Google, then employ that technique on this fraction

https://brilliant.org/wiki/partial-fractions-cover-up-rule/

[wild brook]

OHH

OHHH

1/(n+2) * (1/n - 1/(n+1))

1/(n^2+2n) - 1/(n+2)(n+1)

[molten raven]

Yeah now you use it again

[wild brook]

Is this a sponsorship for brilliant-

[molten raven]

I wish I was getting paid to sponsor brilliant lol

[molten raven]

Use it on 1/(n)(n+2)-1/(n-1)(n+2)

[wild brook]

1/(n^2+2n) - 1/(n+1)+1/(n+2)

[molten raven]

Yea now split the first fraction

[wild brook]

1/n - 1/(n+2)

which leaves

1/n - 1/(n+1)

It all makes

so much sense

[molten raven]

No those aren’t equal

[wild brook]

Oh come on * cries *

[molten raven]

Lmao

1/(n)(n+2)=A/(n)-B(n+2)

Solve for A and B so that the equation is always true

[wild brook]

• cries *

[molten raven]

It just happens that A=B=1 when splitting 1/n(n+1)

[wild brook]

Pell's equation?

[molten raven]

No it’s not in pells format

Expand it and match like terms

[wild brook]

Me when less than .9 bits per second when 5 cm away from the router

[molten raven]

So 1=An+2A-Bn=(A-B)n+2A. This means A-B=0 and 2A=1, so A=B=1/2

[wild brook]

A(n+2)-Bn = 1

[molten raven]

Yeah

There’s no n term on the RHS so we want A=B so the n’s cancel out

[wild brook]

Sorry

BRO

my wifi is

WAIT

NOOO

I COULDVE SOLVED IT ON MY OWN

I sent these messages 5 minutes ago

._.

[molten raven]

Damn

[wild brook]

I took 7 screenshots

[molten raven]

Lol

[wild brook]

you get that A=B=0.5

[molten raven]

Put the whole thing together now

Yes

[wild brook]

1/(n^2+2n) - 1/(n+1)+1/(n+2) = 1/2n - 1/(2n+4) - 1/(n+1)+1/(n+2)

Oh no

not again

NOOOOOOOO

nvm it sent

[wild brook]

so this simplifies to

[molten raven]

Ok lemme check on wolfram to see if everything is right so far

[wild brook]

1/2n-1/(n+1)+1/(2n+4)

[molten raven]

,w partial fraction 1/(n*(n+1)(n+2))

[thorny flame]

Results provided by WolframAlpha

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

Requested by kappa_07

[molten raven]

Ok now we have to multiply by 2

[wild brook]

that makes it a bit more complicated

what if we

[molten raven]

it makes it nicer

[wild brook]

2(1/n(n+1)(n+2) + ...

[molten raven]

You could do that

Hopefully it telescopes

[wild brook]

(1/2n - 1/(n+1) + 1/(2n+4)) + (1/(2n+2) - 1/(n+2) + 1/(2n+6)) +

[wild brook]

which simplifies to

1/2n - 1/(2n+2) - 1/(2n+4) + 1/(2n+6)

I see a pattern

[molten raven]

Can we generalize it?

[wild brook]

?

[molten raven]

How do we do it with more terms

[wild brook]

1/2n - 1/(2n+2) ... - something something

[wild brook]

the third terms disappear if expanded

[molten raven]

Yea

[wild brook]

except when final numbers

[wild brook]

let wolfram alpha do the work?

https://tenor.com/view/spinning-rory-charles-graham-rag-n-bone-man-turning-around-the-room-is-spinning-gif-18986961

[molten raven]

You could get the answer without partial fractions then lol

[wild brook]

I'm only human after all

[molten raven]

I gtg, if you add more terms you will see a pattern.

[wild brook]

so like if we multiply by 2 we get like

1/n-1/(n+1)-1/(n+88)+1/(n+89)

Which should be

1/2-1/3-1/89+1/90

and then appling the multiplication thing