help-26

[barren patrol]

wait but how do i find the range of k after that

i got 24k > 0

[celest shuttle]

so basically k>0

so that's the range right

[barren patrol]

oh that's the range?

[celest shuttle]

yea [0,infinty)

[barren patrol]

ohhh ok ty

.close

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[orchid void]

how do i do this?

[dim plover]

areas of triangles and rectangles and trapeziums

[orchid void]

Im sorry i don't understand, that's actually my brother's but he asked me to get help.

I couldn't find anything in the internet, thats why i went here.

Maybe can you provide me with the answer?

[dim plover]

eg for the first part, you have a trapezium whose parallel sides have lengths 1 and 3, the distance between them is 3, so the area is 1/2 * (1 + 3) * 3

and its below the x axis, so it should have a negative sign

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[timber valve]

still need help?

[timber valve]

.close

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[quaint bronze]

Solve this equation giving solutions in the range 0 to 2π. cos^2x -sin^2x = 1

[quaint bronze]

I am correct in thinking the only solutions are 0 2π

[idle fiber]

probably

I'd had to check

[quaint bronze]

well is it? I think there's a typo in my answer booklet as it also gives π as a solution but that seems wrong

[fluid phoenix]

no

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[mild rivet]

Hi

[topaz sinew]

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[forest cedar]

how do i know while plotting a linear programming sum that a region is bounded or unbounded

[forest cedar]

can provide questions and stuff if someone doesn't understand my question

like in this question

the lines are made, how do i decide if its unbounded region or bounded region to be calculated

[neon iron]

200 people stand in a circle in order to 1 to 200 . number 1 has a sword he kills the next person (i.e. No. 2) and gives the sword to the next living person (i.e. No. 3). all people do the same until only 1 survives. wich number surivives to the end?

[forest cedar]

dawg

😭

<@&286206848099549185>

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[whole quiver]

I understand what is being done in the solution, but I don't understand how the pinching theorem is being used

[vast tangle]

u should know that it is 0

just memorize that its zero

its a property of limits

u can use hopital to prove that it equals 0 but thats it

[vast tangle]

its not being used lol. 1-cosx/x is the pinching theorem

thats just proving that it equals 0

but 1-cosx/x is part of that theorum as it is a fundamental property of trig lims

[whole quiver]

ohh okay

thank you 🙂

[whole quiver]

wdym it is the pinching theorem?

pls help lol I'm revising for a test

[vast tangle]

if u ever see limx->0 of 1-cosx/x u say 0

just memorize

they wont ask u to prove it

just memorize

[whole quiver]

alright

do you know other things (limits) worth memorising?

[vast tangle]

uhh what test is this

limits only?

or trig limits

only

like whats on the test

and ill tell u what u need to memorize

[whole quiver]

limits, derivatives, continuity, inequalities, and linear algebra (Gaussian elimination, simultaneous equations, basic vectors)

[vast tangle]

ok i dont know linear

[whole quiver]

that's okay 😄

any help is much appreciated

[vast tangle]

for derivatives
memorize all the derivatives of trig, logs, ln, e^x,

and derivatives of a^x

[whole quiver]

derivatives with first principles

probably nothing beyond first principles

[vast tangle]

so no chain rule?

[whole quiver]

oh yeah

chain, quotient, product rules

[vast tangle]

did u learn trig derivatives?

like d/dx (sinx)

which is cosx

[whole quiver]

yeah

[vast tangle]

ok so memorize that

memorize d/dx of sinx

cosx

tanx

u should know those 3

[whole quiver]

ok

[vast tangle]

and know how to prove them

memorize limit definition of derivative

know the chain rule version of trig dervatives

do u know what the derivative of lnx is?

u should know that

and its chain rule version

logx as well

and e^x

[whole quiver]

okay

[vast tangle]

also, know the 3 conditions of continutiy

memorize that

[whole quiver]

ok

[vast tangle]

memorize how to do limits at infinity

lim as x approaches infinity

[whole quiver]

ok

[vast tangle]

u need to know how to do that

[whole quiver]

thank you so much

[vast tangle]

np

[whole quiver]

.close

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[whole quiver]

.reopen

[topaz sinew]

✅

[vast tangle]

uhh also, know absolute value limits

remember, when there is an absolute value in a limit always do lim as x approaches the right side and left side (2 seperate limit statements)

[whole quiver]

ok

[vast tangle]

if they dont equal to each other, limit does not exist

[whole quiver]

yup

i understand

thank you very much

[vast tangle]

same for piecewise functions

[whole quiver]

yup

[vast tangle]

and thats pretty much it

[whole quiver]

practiced that

ok

[vast tangle]

in terms of memorization. for more stuff to memorize, ur textbook usually has a memory aid

that u can't use but u can look over it before ur test

i would look at the memory aid in the textbook before u do ur test just to refresh ur mind

[whole quiver]

alright

thank you

[vast tangle]

i wish u the best of luck

u will do great

wishing u all the best in ur test

dont stress too much

if u studied well (which u have) u will do very well

now last part of studying is sleeping

extremely important to sleep before test

i know too many people who pull all nighters and then fuck up their test

its rlly not worth it studying too much.

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[jade nexus]

Hi

[jade nexus]

How do I do question 4

And how do I reopen a message

[loud dawn]

,rotate

[loud dawn]

@jade nexus wdym reopen a message

[neon iron]

what is the circumference of a circle and what is that area?

[sweet shard]

.reopen. but only works if no one else asked a question after yours was closed in the same channel

[jade nexus]

2 pi r

38 pi and 193 pi

But it’s for two circles though

[topaz sinew]

@jade nexus Has your question been resolved?

[loud dawn]

@jade nexus let's say the first circle has radius r1, and the second circle has radius r2

they tell us that the sum of the circumferences of the two circles is 38pi

[thorny flame]

abs_0

[loud dawn]

are you following

[jade nexus]

Yes

[loud dawn]

ok

they also tell us the sum of the areas is 193pi

so we have two equations

\begin{align*}
2\pi r_1 + 2 \pi r_2 &= 38\pi\\
\pi r_1^2 + \pi r_2^2 &= 193\pi
\end{align*}

[thorny flame]

abs_0

[jade nexus]

Yes

[loud dawn]

from the first one, we can deduce that r1 + r2 = 19

and the second one says r1^2 + r2^2 = 193

hmm

[jade nexus]

Ok

I sub r1 or r 2 into the equation 2

?

[loud dawn]

if we square the first equation we would get $$r_1^2 + 2r_1r_2 + r_2^2 = 361,$$ so $$2r_1r_2 + 193 = 361$$

[thorny flame]

abs_0

[loud dawn]

which means r1r2 = 84

hmm

[jade nexus]

Is it possible to use the substitution method

[loud dawn]

yes nice

I didn't realize that

try substitution lmk what you get

[jade nexus]

Sure give me a min

Hi

I would get

R1= 6.541 or r1= 12.458

[loud dawn]

r1(19 - r1) = 84
r1^2 - 19r1 + 84 = 0
(r1 - 7)(r1 - 12) = 0
r1 = 7 or r1 = 12

oh huh

this is a cool problem

I think I'm right but idk

would they give you decimal answers

[loud dawn]

should be 168, not 163

[loud dawn]

I substituted into r1r2 = 84

but they come out the same

[jade nexus]

Ah yes then I get 7 and 12

Thanks a lot

[loud dawn]

np

but what are you putting as the answer

[jade nexus]

I sub both the numbers into the other equation

Wait but

Yeah

[loud dawn]

no I mean

they said calculate "the" radii of the circles

[jade nexus]

Then they have two possible answers I guess

[loud dawn]

but there's two possibilities

yeah

[jade nexus]

Yeah then I find r2

It would be the same answers 7 and 12

[topaz sinew]

@jade nexus Has your question been resolved?

[loud dawn]

ah that's interesting

so it really is one answer

"if the first circle is 7, then the second circle is 12"
"if the first circle is 12, then the second circle is 7"
the circles aren't in any particular order, so it's just "7 and 12" and that's it

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[gentle valley]

Use the Pythagoras theorem to calculate x

Calculate the angles

Use law of cosines

Or just trigonometric functions

[junior widget]

Cannot be correct

[gentle valley]

It is a right angled triangle so yh never mind the law of cosines

Should be 24

[junior widget]

Why are you handing out answers?

And telling people to use law of cosines on a right triangle...

[gentle valley]

Ah mb xd

Idk braindead

[junior widget]

Set up the ratios

sin = opp/hyp

cos = adj/hyp

tan = opp/adj

csc = hyp/opp

sec = hyp/adj

cot = adj/opp

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[frosty cape]

Hey, I'm confused on smth

[frosty cape]

In which quadrant does cos(180°) lies??

[swift cave]

if i'm not mistaken, cos(180) is just a value

180 degrees is quadrant II and III, so for cos, that means it's a negative value, since it's left of the y-axis

[topaz sinew]

@frosty cape Has your question been resolved?

[frosty cape]

So how do I find the value of cos(180)

Like it was in II quadrant I'd just use 180-t, and if ut was in III I'd use t-180 but idk in which quadrant cos(180) is

??

[swift cave]

not sure what this "t" is, but if you look at a unit circle, 180 degrees is pi radians. and at pi radians, the x-value distance from the origin to a point at pi radians is gonna be -1

cos of anything is just measuring the x-value distance from the origin to that point

see that point at (-1,0)? that's at 180 degrees. the y-value is 0, but the x-value is -1

[frosty cape]

Oh

Ok

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[humble nova]

I'm confused on what to do here

[humble nova]

not sure if this would work

can someone please confirm with me?

[shadow lodge]

I don't think it does

It only works when multiplying the 2 exponents

[humble nova]

so how should I start?

[shadow lodge]

2^(x+1) = 2 * 2^x

So left side equals 3 * 2^x

[humble nova]

how does it equal 3?

[shadow lodge]

2 * 2^x + 2^x = 3 * 2^x

[humble nova]

I'm sorry I don't understand why we break the first part into 22^x+2^x = 32^x

[shadow lodge]

It'll make sense

later in the problem

You understand how it can be broken up right?

[humble nova]

not really

[shadow lodge]

2^(x+1) is really just 2^1 * 2^x

[humble nova]

oh ok

[shadow lodge]

So left side = 3 * 2^x

On the right

3^(y+2) = 3^2 * 3^y

= 9 * 3^y

• 3^y

= 8 * 3^y

Wait, wth

[humble nova]

wait so this?

[shadow lodge]

You don't have multiple equations?

[humble nova]

just 1 equation

[shadow lodge]

3^y

You can't solve it with 1 equation

Also, I'm not actually a helper

So not sure if I should be answering your q's

[humble nova]

it's fine

I'm not sure if I should ping someone else to help me

[shadow lodge]

Probably

But just saying though

It's not possible to solve for 2 unknowns with 1 equation

[humble nova]

wait are any of these questions solvable?

2 I think should be solvable

for the first one

I got to here

Not sure what to do next

<@&286206848099549185>

here's what i keep getting

[topaz sinew]

@humble nova Has your question been resolved?

[mellow venture]

$$2^2\cdot 2^x - 2^x=48$$
$$2^x(4-1)=48$$
$$2^x3=48$$
$$2^x=16$$
$$x=4$$

[thorny flame]

Wew "Fractalogist" Tbh 🐧

[mellow venture]

factoring helps very much

[neon iron]

u could write it as log2(48) i think

[topaz sinew]

@humble nova Has your question been resolved?

[humble nova]

we haven't learnt logrithms though in my class

[timber valve]

Dm me @humble nova

[abstract wadi]

Isn't that kinda incorrect

[timber valve]

Yah but it's almost there

[abstract wadi]

Well yes.

[timber valve]

nice

[mellow venture]

how

AHAHAHAH

[abstract wadi]

$2^2 \neq 2$

[thorny flame]

What the hell am I doing here?

[abstract wadi]

[mellow venture]

$$2^2\cdot 2^x - 2^x=48$$
$$2^x(4-1)=48$$
$$2^x3=48$$
$$2^x=16$$
$$x=4$$

[thorny flame]

Wew "Fractalogist" Tbh 🐧

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[neon iron]

what happened here

[pulsar falcon]

Where is here?

[topaz sinew]

@neon iron Has your question been resolved?

[patent oxide]

Substitution and manipulation

[topaz sinew]

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[carmine oyster]

How is beta+alpha=theta?

[carmine oyster]

,rotate

[vale jacinth]

That is not necessarily true

[dim crag]

There is some information missing, i guess its that the angle at B is also $\beta$. If thats the case, then ABC and DAC are similar, which yields the desired conclusion

[thorny flame]

Alexander42

[topaz sinew]

@carmine oyster Has your question been resolved?

[carmine oyster]

No, only this much is given(and also it's given that BD:DC = m:n, which I don't think has any relevance to this)

[worthy storm]

well it's not true in general that alpha + beta = theta. suppose for example that D is the midpoint of BC, and the lengths of AB and AC are equal. then that forces theta = 90 degrees, but alpha + beta need not be 90 in general

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[ruby mist]

how do i shift -9x - 9/x

[tough nest]

shift?

[ruby mist]

in order to find the value of x³ - 27/x³
we will have to get rid of -9x -9/x

[ruby mist]

how do i get rid of it

[vale jacinth]

?

Your original equation is (x-3/x)^3=5^3?

I don’t see why you don’t directly use this

x—>x^3 is bijective

[ruby mist]

wdym

i dont get it

[vale jacinth]

?

x-3/x=5

a^3=b^3 iff a=b

And that is quadratic

x=(5+/-sqrt(37))/2

[ruby mist]

what does quadratic mean

[vale jacinth]

What?

You don’t know how to solve y for y^2+py+q=0?

[ruby mist]

umm

[tough nest]

in your original question you have to find x right?

[ruby mist]

no

i have to find the value x³ - 27/x³

sorry for not specfing the question

[tough nest]

so using a computer i found that your expansion is not correct?

[vale jacinth]

?

[vale jacinth]

He solved x then he will have that expression

[vale jacinth]

And this is x

[ruby mist]

where did +27/x³ come from

i am so confused

[tough nest]

i expanded the cube

it gave me that

[ruby mist]

where did i go wrong here

[vale jacinth]

I still don’t get why you are so reluctant to solve x

[topaz sinew]

@ruby mist Has your question been resolved?

[topaz sinew]

@ruby mist Has your question been resolved?

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[fluid phoenix]

does this just happen because of arcsin's range?
$$arcsin(cosx) = arcsin(sin(x+\pi/2)) = x +\pi/2$$
but
$$arcsin(cos \pi) = arcsin(-1) = -\pi/2 $$
which isn't equal to $$\pi + \pi/2$$

[thorny flame]

Swas.py

[tough nest]

3pi/2 = -pi/2 + 2pi

[fluid phoenix]

mhm

but arcsin's range is [-pi/2, pi/2]

[fluid phoenix]

that's the only reason why this doesn't work for x > 0 right?

[tough nest]

arcsin is funky like that

all of them mean you need to be careful

[fluid phoenix]

alright

.close

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@neon iron Has your question been resolved?

[topaz sinew]

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[mellow shadow]

im not sure how to do the questions from 43-48

[topaz sinew]

@mellow shadow Has your question been resolved?

[mellow shadow]

<@&286206848099549185>

[leaden tusk]

use derivative rules, f and g are functions of x

product, quotient, chain, etc

[mellow shadow]

well ye but how would u do 43

[leaden tusk]

that notation just means (g'-f')(2)

[mellow shadow]

oh ok so 3

[leaden tusk]

yeah

[mellow shadow]

44 is 11

45 -1/25

46 1

how would u do 47

[leaden tusk]

$(f\circ f)'(2) = f(f)'(2)$

[thorny flame]

a disappointing son

[leaden tusk]

composite functions call for the chain rule

[mellow shadow]

oh, so 47 is -7

and is 48 -1/4?

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[signal granite]

hey

[signal granite]

Q -> P == ¬Q v P (implication elimination)

correct?

does this mean
(P v ¬Q)
== (¬Q v P)
== (Q -> P)

therefore,
(P v ¬Q) == (Q -> P)

[topaz sinew]

@signal granite Has your question been resolved?

[long stirrup]

yes

[signal granite]

to confirm

(P -> Q) == (-P v Q) == (Q v -P)

@long stirrup

[long stirrup]

yeah

[signal granite]

thanks! i rmb u helped me before

.close

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[gritty sluice]

Given a 2D Vector A and a 2D Vector B , find the center and radius of the circle that intersects both of the vectors, has the shortest arc length between the two vectors, and is tangent to both vectors.

This is a problem that I'm working out for a group pathfinding/AI algorithm I'm making for a game. Essentially, I'm trying to determine the radius and center of rotation of a 'formation' or 'group' of AI's in order to generate a set of points along the circle's circumference to pathfind each AI by, thereby rotating each member of the 'formation' accordingly. If you know any better methods for this, then that would also help!

[gritty sluice]

Please DM me if you respond. 🙏

I just realised this has a lot of impossible cases

.close

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[livid pelican]

please can someone help see where I went wrong?

how would I approach this then?

oh yes, I see

like this?

thank you

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[livid pelican]

.reopen

[topaz sinew]

✅

[livid pelican]

so you'd split something up like this because 1/T-Tr is not a constant but it can be differentiated with dT?

yep sorry, that's what I meant

thanks

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[tribal cypress]

Can someone help me with this?

[leaden tusk]

integrate and use your conditions to find C

then do it again

[tribal cypress]

I did but when I get my first integration I struggle to find C

[leaden tusk]

show work

[tribal cypress]

@leaden tusk

[sweet shard]

you set f'(x) = 18 instead of f'(0) = 18

[tribal cypress]

Ohhhhhhhhh

Omg

C=18

Ok

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[eager dragon]

Hello guys, just a quick question. What does this notation mean? I haven’t attempted anything because I don’t understand what it’s asking.

[sweet shard]

what notation

[eager dragon]

Not really the notation. More with what f: (0,2) —> (0,π/2) means

[sweet shard]

https://mathinsight.org/function_notation

[eager dragon]

I don’t think it means that’s it’s differentiable on that vertical line segment

Wouldn’t make sense

[sweet shard]

what vertical line

[eager dragon]

Ok never mind I see

It’s not a vertical line because those are not points. But what does the arrow represent moving from 2 to pi/2

[sweet shard]

.

[eager dragon]

Ok thank you

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[devout halo]

No.

[warm finch]

What are i and j supposed to be?

[devout halo]

You need to find the derivative of cos(t), sin(t), and 1/2 * t.

What you have is not it.

You need to find the derivative with respect to t.

Yep.

No you don't for this case.

if t was equal to something else then you will most likely apply the chain rule. Bad wording but whatever.

np.

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[storm monolith]

What do you know, geometrically, about the roots of a complex number?

[shadow cloak]

they are sepaarated by pi/2

[storm monolith]

That's important

[shadow cloak]

they sometimes come in conjugate pairs if and only if their product is real? Is that true

i mean they always come in conjgate pairs if and only if their products are real

whereas they wouldn't come in conjugate paris iff their product is imaginary

@storm monolith

[storm monolith]

You don't need to use qualifiers in these sorts of phrases

Like you don't need to say 'always'

But what do you mean?

[shadow cloak]

so in roots of unity, we are taught that they come in conjugate pairs

and they are equally spaced

but roots of unity only explores their product to be 1

but what about when you have other real numbers and also imaginary numbers included?

would they still be in conjugate pairs?

[storm monolith]

Oh yeah that's true for roots of unity

Not necessarily for other roots

[shadow cloak]

so for the case real numbers?

what would happen, do they still in conjugate pairs?

[storm monolith]

Yeah they should do

In general we always get an equally-spaced n-gon of roots, but they could be 'rotated' if you will

I'm not sure how accurate they want the sketch

[shadow cloak]

what about for the case of imaginary numbers:

[storm monolith]

That's not necessarily the case for imaginary numbers

I guess you could find a general formula for a 4th root of the complex number a + bi and plot that

[shadow cloak]

is it always not the case they are not in conjugate pairs?

[storm monolith]

It's not always the case

In fact just consider z^4 = i

[shadow cloak]

yeah

[storm monolith]

In general they form a regular n-gon about the origin and that's pretty much it

So you want to have them form a square basically

[shadow cloak]

what about irregular n gon?

[storm monolith]

Wdym

[shadow cloak]

when does an irregualr n gon occur?

[storm monolith]

It doesn't

Not with this

[shadow cloak]

okay

so how would we do that question?

we know that they are separated by pi/4 , but where does it begin?

[storm monolith]

Well that depends on a+bi

You could come up with a general formula if you knew what quadrant a + bi is in, or if we had some general formula for the argument of a complex number

[shadow cloak]

how did they know where to egin the rotation

[storm monolith]

I think they just picked it arbitrarily

Nothing suggests otherwise

[shadow cloak]

but we do know the arg

we know where the a+bi is placed

it is pi/2 * 4/5

[storm monolith]

Oh

That's an actual point

Okay nvm then

[shadow cloak]

so how can we figure out that arg

we construct a principle formula

[storm monolith]

You know the general approach for finding nth roots?

[shadow cloak]

so 5(theta) = 2pi/5 + 2pi * k

and then we divide the 5 to ther otherside and say for k = 0, +- 1, -2

[storm monolith]

I think that should be 4(theta)

[shadow cloak]

why 4pi/5

that would mean it is in the second quad

[storm monolith]

🤦‍♂️

Yeah

Well anyway 4(theta), I was imagining the wrong thing

[shadow cloak]

so we divide it by 4

[storm monolith]

Yes

[shadow cloak]

and then we know that a+bi lies on the first

and when divide that arg by 4 or make k = 0

it will be at the first line of the first quad

[storm monolith]

Yep

Then you can just rotate

I think the modulus is 2 there

So new modulus is 1.189 and so on

[shadow cloak]

how did you figure out the mod?

[storm monolith]

Just looked at the diagram

[shadow cloak]

yeah

[storm monolith]

Looks about twice the distance as 1, so take the 4th root

[shadow cloak]

so you are saying the distance from the origin to the point a+bi

is twice the point from the second circular line

[storm monolith]

First one

The first one has radius 1

I'm saying it's roughly twice that

[shadow cloak]

yeah

I would of just divide the radius of the first one by 4

[storm monolith]

Could be ever so slightly more

Yeah that's not what we do

Consider z^2 = 16

Worst example I could have picked

Is z = 8 a solution to that?

Going by the logic of 'divide' by the power

[shadow cloak]

yeah

I'm saying divide the length of the a+bi by 4

[storm monolith]

Which is basically same thing as my counter example

Let b = 0, a = 16

[shadow cloak]

what is b?

[storm monolith]

Would we just 'divide by 4' in this case

It doesn't really matter

[shadow cloak]

why?

[storm monolith]

I'm saying you can disprove that intuition if you consider the case where a+bi is real

No reason why it would for this purpose

[shadow cloak]

yeah

so how would you find the mod?

[mint raft]

when you do z^4 you basically multiply the angle by 4 in the complex plane

so if z is a 4th root then multiplying its angle by 4 should give the angle of a+bi

[shadow cloak]

yeah, but what about the mod?

[storm monolith]

In general consider $z = r(\cos{\theta}+i\sin{\theta})$, then $z^n$ has modulus $r^n$, and $a+bi$ has modulus $\sqrt{a^2 + b^2}$, the modulus of $z$ is then the nth root of the modulus of $a+bi$

Pain, why doesn't latex support cis

[thorny flame]

dk.dkn

[shadow cloak]

yeah

[mint raft]

hmm I assumed that the modulus (in this context) would be the spacing between the values of z, which is pi/2

[shadow cloak]

yeah

[storm monolith]

The modulus of a complex number is the magnitude of the distance of a complex number from the origin

[mint raft]

for angle you are just solving
4z=theta (mod 2pi)

[storm monolith]

You could call is the magnitude I guess

Not to be confused with modulo - some number

[mint raft]

yeah you are right, so that indicates that you should plot the roots closer to a “radius” of 1

[storm monolith]

Yes, but of course not simply 'divide' them

Because in reality the modulus of the new complex number is the nth root of the previous, not that divided by n

[shadow cloak]

so how do you think we should fine them?

divide the mod of the a+bi by half

and then another half

[storm monolith]

That's the same as dividing by 4

[mint raft]

a+bi looks to be around 2 so take the 4th root of 2

[shadow cloak]

yeah

[storm monolith]

You take the 4th root of the previous modulus

[shadow cloak]

I'm confused

[storm monolith]

They've basically given you a line

[mint raft]

that’s why they plotted z in the inner circle

[storm monolith]

(Although I think it's a little off)

[shadow cloak]

how would we know the 4th root of the previous miodulus

[mint raft]

the inner circle is supposed to be the 4th root in magnitude

you just guesstimate the value

[storm monolith]

I think you're trying to make this way too rigorous

[mint raft]

in the question they literally give a circle to plot the value so

[storm monolith]

We don't have clearly defined values here

[shadow cloak]

yeah

[storm monolith]

Well

We have heavily implied visual values I suppose

[shadow cloak]

yeah

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@shadow cloak Has your question been resolved?

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@shadow cloak Has your question been resolved?

[topaz sinew]

@shadow cloak Has your question been resolved?

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[empty sail]

The concept you should be applying is inscribed angles

Do you know how to apply inscribed angles?

Yes

So you know that the angle is 1/2 the arc. How could you make the arc the subject of the equation?

Did you find x?

What arc is needed to find angle x?

Yes

How would you set up the equation to find x?

Yes

So the 76 relates to the arc, the x relates to the angle, correct?

Is b an arc or angle?

What angle relates to b?

No

Still no

Where would the endpoints of arc b be?

You understand this, correct?

Can you determine which lines connect to the arc, in your question?

You can use paint or something to color the lines, and send that screenshot

Are you not on a Windows machine?

That's not an inscribed angle

Yes

[topaz sinew]

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[warped zinc]

not sure how they got that

[vale jacinth]

There is a result which you can prove it yourself:
$\frac{d}{dx}(\int_{g(x)}^{h(x)}f(t)dt)=f(h(x))h’(x)-f(g(x))g’(x)$

[thorny flame]

Cogwheels of the mind

[warped zinc]

what do u mean?

[vale jacinth]

?

I just gave a result, not my personal thought or explanation. There is nothing that I mean.

[warped zinc]

ok

ill have a look at that then

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[neon iron]

not sure what im doing wrong i get the answers but it is supposed to be setup different

[slate valve]

hi

you need to know the ratio

[neon iron]

hi

[slate valve]

30 degrees angle is 1:sqrt3:2

I can give you the photo one moment

[neon iron]

ok

[slate valve]

use this ratio

[grim jacinth]

Brawl Stars name?

[slate valve]

so c will be 5/sqrt3=5sqrt3/3

[slate valve]

dm

[neon iron]

ok

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[crystal forge]

.open

[crystal forge]

hey so here, i have all the answers but I'm trying to find the focus points

could you please go back to your help channel?

it says you are asking in #help-18

here I have all the a and b variables solves

solved*

a=3

b=sqrt3

[vale jacinth]

Just (-sqrt(6),0) and (sqrt(6),0) shift (-3,5)

[crystal forge]

I have the shift part down

but what does the Just (-sqrt(6),0) and (sqrt(6),0 mean?

[vale jacinth]

Those two points plus (-3,5)

So (-sqrt(6)-3,5) and (sqrt(6)-3,5) I mean

[crystal forge]

this is what i have right now, I'm trying to solve for c

a^2=b^2+ c^2

[vale jacinth]

a^2-b^2=c^2

Yeah

[crystal forge]

I'm just trying to make sure the answer is sqrt 6?

[vale jacinth]

That’s why I said sqrt(6)

[crystal forge]

ohh

sorry i didnt understand

tysm

[vale jacinth]

Np

[crystal forge]

.close

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[silver valve]

can someone explain this?

what did they take n as?

[vale jacinth]

3

[silver valve]

so k = -3,-2,-1,0,1,2,3?

[brittle tartan]

yeah and cos is an even function

which also accounts for -1 since cos 0

oh nvm

[silver valve]

im getting -2(cosx - cos2x + cos3x) = -1(cos(2x))/cos(x/2)

now what?

cosx - cos2x + cos3x = cos(2x)/(2 × cos(x/2))?
how did they get cos(7x/2)/cos(x/2)?

[brittle tartan]

what about cos 0

[silver valve]

right..

-cos(-3x) = -cos(3x)?

i can simply do that and simplify right?

i did it and got the expression on the right..

[brittle tartan]

bet

[silver valve]

ohk everything checks out

but what is that formula?

and how did they just equate that sum into the (-1)^n cos(n + 1/2)... expression?

that sum seems to be tailored specifically for the question in hand?

how tf does someone do that?

[brittle tartan]

its a pretty well known series if you've worked with lots of series but, yeah its kinda ridiculous

[silver valve]

never worked with series...what is it called?

[brittle tartan]

not sure, but it shows up a lot in my assignments

[silver valve]

is there any way to evaluate cos(x) - cos(2x) + cos(3x) using basic trig?

like sum to prod, prod to sum, multiple and sub-multiple angle formulae?

[brittle tartan]

maybe with some angle addition formula

thats what i was thinking

[silver valve]

how?

[brittle tartan]

well cos 2x is well known and cos 3x is cos(2x + x)

[silver valve]

you mean i should express in terms of cos(x)?

[brittle tartan]

well, its a good strategy in general, but if you were looking for whats on the rhs, it comes from sequence analysis

[silver valve]

4cos^3(x) - 2cos^2(x) - 2cos(x) + 1

[silver valve]

no im looking to prove that for x = π/7
-1 + 2(cosx - cos2x + cos3x) = 0

i.e π/7 is a zero of this expression

[brittle tartan]

oohh

i see

and im guessing you dont want to have to memorize the rhs for like a test or smthg

[silver valve]

if i get that RHS expression using sequences (which ive never studied before) i can easily do 7x/2 = π .. but yeah...

[silver valve]

no wdym?

[brittle tartan]

oh i thought you wanted to find some intuition behind the expression instead of using the rhs to evaluate

[silver valve]

nah

[silver valve]

want to prove this

its a "prove that" type ques

[brittle tartan]

oh, did you try plugging in already

[silver valve]

yeah... but π/7 isnt so neat

dunno how to calculate it

[brittle tartan]

specifically, the denominator right cos(pi/14)

[silver valve]

splitting, multiplying, adding, subtracting .. nothing seems to give me a standard angle

[silver valve]

what?

[brittle tartan]

you plugged x = pi/7, and got cos(pi/2)/cos(pi/14) im assuming

[silver valve]

no

lemme try

well even if i did cos(pi/14)? how do we calculate this?

[brittle tartan]

this is already zero cuz of numerator

[silver valve]

ohh k

so what did you simplify?

cos(π/7) - cos(2π/7) + cos(3π/7)??

[brittle tartan]

nahh i just used the formula

rhs

[silver valve]

whats that?

[brittle tartan]

we have that -1 + 2(cosx -cos 2x + cos 3x) = cos(7x/2)/cos(x/2)

so if LHS is zero then so is RHS

[silver valve]

yeah but i didnt understand that sum formula so how will I use it on such questions?

i need to find a different approach

https://math.stackexchange.com/questions/2139075/prove-that-cos-pi-7-is-root-of-equation-8x3-4x2-4x1-0 its from here btw

[brittle tartan]

what did you think about the given solutions

[silver valve]

i didnt understand any

the one whose ss i sent... i partly understood

except ofc that last bit

[brittle tartan]

which one? im not sure

[silver valve]

the one with 3 upvotes

[silver valve]

this is the tb sol... but i dont think anyone can possibly just "know" to do these steps

so i want a general approach

found this online

but idk wth is happening

.close

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[candid dove]

is scalar product two dimensional?

[candid dove]

kindly ping me while replying

[vale jacinth]

I never saw someone define the dimension of a binary operation

[candid dove]

i mean cross product is basically three dimensional right

at minimum three dimensional

[vale jacinth]

Again, how did you define the dimension of all this

[candid dove]

well take two vectors in a single plane then the cross product is in the perpendicular plane

thus x y z three dimensions?

[vale jacinth]

Or are you saying for what kind of n we can define dot product and cross product on R^n?

[candid dove]

am i misunderstanding smthn?

[candid dove]

i dont even know wht tht meant

[vale jacinth]

If this is what you are asking, then the answer is any positive n/ n=2,3 respectively

[candid dove]

m learning i mean and teachers just swoop it away pretty quickly

[vale jacinth]

You can define dot product on any n dimensional spaces,you can define cross product on 2 or 3 dimensional spaces

[candid dove]

ahhh

so it doesnt matter

like u need no think abt the dimensions when it comes to dot product

[vale jacinth]

Yeah

Any dot product is equivalent to (x,y)—>x^TAy for some symmetric/hermitian matrix on R /C

Doesn’t matter, just it always exists for any dimension

[candid dove]

ooo alright

thanks a lot mate

.close

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[rapid minnow]

(x+y)12=t
or
(1/2+2/3)12=t

considered as linear equation?

[topaz sinew]

@rapid minnow Has your question been resolved?

[rapid minnow]

.close

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• Show your work, and if possible, explain where you are stuck.
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[neon iron]

Hi, please could someone check if my answer is right?

[cinder raven]

,,turn

u can skip a step at the second last step

do u see base = 12
and 12^2

u can auto cancel and take power as the answer

but yes its correct

[neon iron]

Okay! yep I will change that, thanks for your help!

.close

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[cinder raven]

in another world u can take
(3^2 * 2^4) = 144
144 =12^2 and skip more steps

[neon iron]

Oh right, I'll have a go at doing that too. Thanks 😊

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[ruby meadow]

Hello can anyone help me with some home

[topaz sinew]

Please don't occupy multiple help channels.

[wheat grove]

please use only one channel

[ruby meadow]

Ok I’m really sorry I’m new

Well can u help me

It involves algebra

<@&286206848099549185>

[exotic radish]

@ruby meadow how about you define each term?
For example each nickel equaling a certain amount

[ruby meadow]

I got x = nickel

That’s all right now I’m really confused

[exotic radish]

listen

[ruby meadow]

Yes go ahead

[exotic radish]

How about we remove the names and try to simplify the question

Let's do this

Each nickel is 5 cents right?

[ruby meadow]

Ok yes let’s do this

[exotic radish]

Each dime is 10 cents

[ruby meadow]

Yes

[exotic radish]

Now if thomas had x-46 amount of 0.05$, he'd have 4 times as many 0.05$ as 0.1$

Right?

[ruby meadow]

Yes

[exotic radish]

Now he has 9.20$ in total

[ruby meadow]

So x - 46 = the amount of dimes

[exotic radish]

Yep

[ruby meadow]

Ok and x = nickels

[exotic radish]

y = dimes

4(y) = x

Cuz remember it said 4 times nickels as dimes or sumn like that

[ruby meadow]

So 4(y) = dimes

[exotic radish]

yeah apparently

[ruby meadow]

Ok

Why 4 (y) why not 46 x y

[exotic radish]

Why would it be 46y?

[ruby meadow]

But I have to solve the question with one variable

That’s what’s confusing me

[exotic radish]

How many of each coin does he have

So how many y's and x's are there

[ruby meadow]

Total 46

[exotic radish]

Jesus this question is worded really weirdly

[ruby meadow]

Loo

Lol

[exotic radish]

By "4 times as many nickels as dimes"

[ruby meadow]

What grade are u in

[exotic radish]

Does it mean that the amount of nickels are 4 times the amount of dimes

11th

Or is it backwards

[ruby meadow]

Umm
It says 4 x as many nickels as dimes

[exotic radish]

Well what does that even mean???

For each 4 nickels there's a dime?

[ruby meadow]

Lol I’m in grade 8

[ruby meadow]

Yes

No

[exotic radish]

Bruh

[ruby meadow]

2 nickels is 1 dime

5 + 5 = 10
Right

[exotic radish]

We're not converting em for now

[ruby meadow]

Ok

[exotic radish]

I just wanna simplify the question

[ruby meadow]

Ok

[exotic radish]

Because oh my god it's confusing

💀

[ruby meadow]

Ya fr

[exotic radish]

On god

[ruby meadow]

Ya

Can I show u another question

[exotic radish]

So 46-(All Nickels)=4(All Dimes)

[exotic radish]

Sure

[ruby meadow]

1 sec

These 2 are really confusing

@exotic radish

[exotic radish]

Bro what.

[ruby meadow]

I know right

[exotic radish]

B r u

[ruby meadow]

What

[exotic radish]

This some real shit

[ruby meadow]

Lol

[exotic radish]

These mf geese bruh

[ruby meadow]

Lol

[exotic radish]

Istg they talk to eachother in math questions

And they know math

Like ong they op asf

[ruby meadow]

Lol

So do u have any idea

[exotic radish]

Ok so

It's saying

original amount of flock : x
many more : y

x+y+y/2+y/4+1 = 100

[ruby meadow]

So x = amount of geese

[exotic radish]

$\underbrace{1}{goose}+\underbrace{x}{flock}+\underbrace {y}{many more}+\underbrace {\frac{y}{2}}{half of flock}+\underbrace {\frac {y}{4}}_{quarter of flock}$

I just got one question

By "as many more"

Does it mean the same original flock number?

Or smth else?

[ruby meadow]

No it says how many geese are in the flock

There only 1 flock

U know what skip those

I’ll send another because I need to understand it

[thorny flame]

Plumbum

[ruby meadow]

Is this understandable

@exotic radish

[exotic radish]

Sean's age : x
12-x = 36-2x

Ok this one is finally easy!

[ruby meadow]

Lol

What does 12-x =

Like who’s age

[ruby meadow]

Is that the equation

[exotic radish]

$12-x=36-2x\
\implies 2x-x = 36-12\
\implies x = 24$

Yep

[thorny flame]

Plumbum

[exotic radish]

Sean's age is 24

[ruby meadow]

Is that the equation

[exotic radish]

Yes

[ruby meadow]

Ok why did u do 36-2x

And who’s age is 12-x

[exotic radish]

Let's do it like this

Say sean's age is x, yep?

[ruby meadow]

Yes

[exotic radish]

It said 12 less than Sean's age = 36 less than twice his age