help-23

[cedar stream]

where theta is the angle from the positive x axis

so its quite simple, just plug in the angles from the diagrams into theta

[spark leaf]

sin 26 and cos 26?

[cedar stream]

other way round

[spark leaf]

no diff

[cedar stream]

(cos26, sin26)

rinse and repeat for the other angles

[spark leaf]

ok

thanks

[cedar stream]

okayy

should i close now?

[median sand]

Close what?

[cedar stream]

okay thank you for being the first person that i helped, i will close now

.close

[spark leaf]

hi mercury

[spark leaf]

np, btw only the original poster and helpful can close posts

[median sand]

Hi Moey!

[median sand]

Mods too.

[spark leaf]

what am i supposed to do here

[median sand]

Eh, check ASTC rule.

[spark leaf]

whats that

[median sand]

Then draw out an unit circle.

You would figure that one out easily.

[spark leaf]

isnt what i expected

[mild plaza]

It is ASTC rule not Acts

[spark leaf]

uhh i dont get this for example when cos is negative in s it syas its negative in A it says its positive

[mild plaza]

Please clarify

ASTC NOT ACTS

[spark leaf]

see how a negative gives both a negative and positive

that makes no sense

[mild plaza]

It stands for All, Sin, Tangent, Cosine

[spark leaf]

ok but why is it when cos is negative the answer is both positive and negative

[mild plaza]

The answer of what

[spark leaf]

look at what i circled

[mild plaza]

I think that trigonometry works that way . Sometimes Cos is negative and sometimes positive. Its value is positive or negative

[spark leaf]

then why is there a rule that states the obvious

[mild plaza]

To allow us to remember

[median sand]

Ehh... why?

[mild plaza]

It is like metaphor helps us remember stuff

[spark leaf]

i dont get the rule

[mild plaza]

The rule states that all trigonometric functions are positive in 1st quadrant

[median sand]

Well cos of theta here is positive, right?

[spark leaf]

ohhhhh

[median sand]

Which means the adjacent and hypotenuse are positive.

[spark leaf]

now i get the rule

[median sand]

If you mirror it, then adjacent negative.

Which gives cos negative.

[mild plaza]

Every point on the circle is cosx, sinx

[spark leaf]

i get it

[mild plaza]

That's why they say , a picture is worth a thousand words

[spark leaf]

whose they

[mild plaza]

Everyone mean to say not specifically

Its (cosx, sinx) . That's the form of each piece on circle in coordinate plane

It obvious

As hypotenuse is 1

[median sand]

Eh... ?

[mild plaza]

What its true

[median sand]

Sounds confusing.

[mild plaza]

At x= 0 the point is 1,0 as radius is 1

And at x= π/2 point is 0,1

Find a single counter example if possible

[median sand]

Counter example for?

[spark leaf]

@median sand idk im stupid

[median sand]

Minus 2pi.

You rotate it anticlockwise and still give the same coordinate.

It is like a clock.

If it points at 4, you rotate the clock 12 more hours, still give you number 4.

[spark leaf]

360-400?

that =40

[burnt notch]

The other way round, I'd say

[median sand]

Yea it is.

[burnt notch]

But you'd get the same result

[median sand]

Ehh still the same.

[burnt notch]

(luckily cos is even)

[median sand]

Cos(-40)=cos(40).

[spark leaf]

i dont get the reason tho

[median sand]

But for cos only.

[median sand]

Read this please Moey.

[burnt notch]

Thanks to the concept of PERIODICITY

[median sand]

Lucky him

[spark leaf]

ohh so it loops

[median sand]

It does.

[spark leaf]

makes sense

thanks

[median sand]

I think it is kinda like mods ?

Also works like a clock maybe.

[burnt notch]

Yeah that's the same concept

[spark leaf]

does it work with tan?

[median sand]

Similarly.

But note that sin and tan are not even.

[spark leaf]

yes

[median sand]

Such that sin(-x)=-sin(x).

And tan(-x)=-tan(x).

[spark leaf]

this is the question

please dont answer it

[median sand]

Yeah apply same concept.

[spark leaf]

thanks

[median sand]

Well also note that tan also the same value if quadrants are opposite diagonally.

Such that tan x = tan pi+x.

[spark leaf]

well c and d cant be the same a and b cant be the same

is it any?

[median sand]

There is.

[median sand]

Check this.

[spark leaf]

so i have to times it by pi?

[median sand]

Plus.

[spark leaf]

ok

[median sand]

For example, tan 15 = tan 195.

There is one pair here.

[spark leaf]

how should i put it in a calc

[median sand]

What do you mean?

You can just count logically.

[spark leaf]

cause i did tan(15)+pi it gave me me decimals

[median sand]

First, identify which pairs are in opposite quadrants that are diagonal.

[median sand]

Tan (15+180)

If you use pi, then change 15 to radiants lol.

[spark leaf]

b-c,d,e and a-c,d,e

[median sand]

Great.

Now check if any angle in first quadrant satisfies the other = 180 + itself?

A.k.a. If you have an angle x in quadrant 1, is other angle = 180 +x?

[median sand]

Which in third quadrant.

[median sand]

It is just an example, now try tan 15 if it is the same.

[spark leaf]

its B and D

just by looking at it no calculations

[median sand]

Yes.

There we go.

[spark leaf]

that was simple

[median sand]

I know right.

[spark leaf]

how do i do b

[burnt notch]

What was question (a)?

[spark leaf]

i was lazy so i just copy pasted the answers

[burnt notch]

What do you notice about, let's say, sin100° and sin80° ?

How are the two values? Are they equal or different?

[spark leaf]

they are the same

[burnt notch]

Awesome, and how are 100° and 80° related to each other?

They sum up to which angle?

[spark leaf]

180

i get it

the answe is just =theta

[burnt notch]

Good

[burnt notch]

sin(θ)

To be precise

[spark leaf]

umm

i feel like u guys are just tired of me

[cedar stream]

hi again

[spark leaf]

hi

are u tired

[cedar stream]

you need help with this?

[median sand]

Nah.

[spark leaf]

yes

[cedar stream]

||ofc not im doing this for fun, however lets not talk abt that too much cuz i dont wanna derail this||

[median sand]

C or d?

[spark leaf]

c

[cedar stream]

which answer do you need to justify

[median sand]

Yeah appeal to the fact that which quadrant sin and cos positive.

[cedar stream]

he needs to prove the quadrants in which sinx and cosx are positive?

[spark leaf]

so idk

[cedar stream]

uhhuh

right so moey what is the question exactly

what do tou need to justify

[spark leaf]

sin(180-theta)=sin theta

[cedar stream]

ah okay

[spark leaf]

so idk how

i have the answer and solution but im not looking at it till we solve

when u get back ping me

[cedar stream]

okay

okay so what is sin

@spark leaf

[spark leaf]

wassup

[spark leaf]

opp/hyp

[cedar stream]

on a unit circle how would u visualise it

[spark leaf]

its the y axis

[cedar stream]

yes correct

what would it take for sin theta to be the same withkut the same angle

[spark leaf]

cos is the x axis

[spark leaf]

a full loop or half

no just half

[cedar stream]

yea

just half

[spark leaf]

yep

[cedar stream]

so, theta minus 180

right

[spark leaf]

yes

[cedar stream]

so thats enough right

[spark leaf]

is it

[cedar stream]

i would say it is

[spark leaf]

thats the book answer

[cedar stream]

mhm

from how i understand it we did the same thing in basically a more laypersons way

[safe radish]

@spark leaf Has your question been resolved?

[spark leaf]

@ who ever is left

[burnt notch]

Always the same concept 😭

sin(180° - θ) = sin(θ)

[spark leaf]

180-137?

[burnt notch]

Yeah

[spark leaf]

ohh ok very simple

[burnt notch]

Yep

[spark leaf]

but i need to learn how to deduct large numbers again

lol

[mild plaza]

What you mean by deduct large numbers?

Subtract?

[burnt notch]

Or maybe they meant deduce the sine of large numbers

[mild plaza]

If the number is > than 360 it becomes x-360*x%360

where % is remanider

[median sand]

Sin 137 also give that decimal thing.

Since 137+43=180.

[spark leaf]

@ who ever is left

[median sand]

Which question ?

[spark leaf]

c

[median sand]

C?

[spark leaf]

yep

[karmic merlin]

Well we see that the angle for Q is -theta, so can you think about what it's coordinates would be

[spark leaf]

-theta, cos

i think u guys had enough

ill sleep and do this tmrw

its 1.30am

.close

[safe radish]

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[safe radish]

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[glass sparrow]

hello can someone give me an idea for the last question pls

[fathom jewel]

You basically wanna do an integral, but you need to get the bounds right

[muted sapphire]

You know how integrals work right

[alpine shell]

(look at the rest of the test)

[muted sapphire]

Well then this should be pretty simple

[glass sparrow]

ohh yes

i don’t kbow how to find the bound

[alpine shell]

well an idea: is to write the integral of the curve from 0 to k to get the area as a function of k, and try to find the k such that that area is half of the total

(you will therefore also need to calculate the total area, and you'll get basically an equation to solve for k)

[glass sparrow]

so first step

we will integrate 100/x+5 from 0 to k right

[alpine shell]

yeah

that way your result will be some function of k for that area

does that idea make sense to you? do you see where i am going with it?

because that's the important part, not just get the answer you know, but learn how to get there

[glass sparrow]

ohhh i think i get it

[alpine shell]

ok, and now that you have that, you must make it so that this is equal to half the total area for some k

so you can say this = total area/2, and that gives you an equation that you can solve and get your k

[glass sparrow]

u mean solve m this to get K right

[alpine shell]

who is m?

[glass sparrow]

solve this:)))

so i get

[alpine shell]

oh oops, yeah

[glass sparrow]

hehhehe

[alpine shell]

hmm

did you get the equation?

[glass sparrow]

i get this equation

[alpine shell]

why = 0?

[glass sparrow]

cuz I don’t know that is itb= to

I have to solve for k right

Ohh hahhaa

I see eit

[alpine shell]

yeah, but it is not = 0, think about what you just have, the 100ln((k+5)/5) what does that represent?

[glass sparrow]

;))) cuz I put x

It looks better now

[alpine shell]

your equation is still wrong

[glass sparrow]

It represent the total area

3072?

[alpine shell]

it is not the total area

what you have represents the area up to k

[glass sparrow]

Ohhh

[glass sparrow]

Ohhhh

[alpine shell]

what you have is the red area

and you want this to be half of the whole thing

does that make sense?

[glass sparrow]

Yesss

[alpine shell]

cool

[glass sparrow]

So then I will let this euqla to the half of the total area I have before

[alpine shell]

well you need to calculate the total area, and then make what you have equal to half of that

you havent yet calculated the total area normally

[glass sparrow]

oh i will calculate the total area by integrate y from 0 to 40?

[alpine shell]

yeah exactly

[glass sparrow]

okok then i let the previous equatiin = to this /2 to find the k

[alpine shell]

exactly yes

[glass sparrow]

wowwww niceee

tyty

[alpine shell]

it may be helpful to have a drawing

[glass sparrow]

hehehehe yahhh:)))

[alpine shell]

i find its very often useful to do drawings to not forget what you are doing and where you are going

just like this, its enough

this basically sums up your whole problem

[glass sparrow]

woww true true

oh yeah, but normally for this type of question

is it non cas or cas active?

[alpine shell]

what does that mean 😅 im french so there is some vocabulary i dont have

[glass sparrow]

ohhhd:)))

like ican we use cas

for this type of question

the CAS calculator

[alpine shell]

this should be doable without a calculator

try to write your equation that you get down

[glass sparrow]

oh no i’m cooked

[alpine shell]

dont worry, once you have written it you will see its not that bad

[glass sparrow]

yeahh hehehe

are u doing vce math right now?

[alpine shell]

what is vce?

[glass sparrow]

it’s the victoria program

for y12 and y11

[alpine shell]

no because i am french, we dont have that i think

[glass sparrow]

ohhhh

wait what year are u currently in

[alpine shell]

im 20, im studying engineering rn, im down with school

[glass sparrow]

ohhh wowww

engineering is so good

[alpine shell]

i find it interesting yeah xD

[glass sparrow]

:))) woww xd

can u help me for one more pls

:))) i set it equal to 0

but the cas is broken

[alpine shell]

what equal to zero?

[alpine shell]

we need to finish your first problem

[glass sparrow]

the differentiate of h(x)

[alpine shell]

no?

[glass sparrow]

ohh yes

[alpine shell]

did you have an answer?

[glass sparrow]

i have k=-20 and k=10

but i don’t have the answer sheet

[alpine shell]

well do you think k =-20 makes sense?

and what about k = 10?

[glass sparrow]

hmm

i thibk k=10

is the half of 20

[alpine shell]

i didnt understand your reasoning

[glass sparrow]

hmmm

i’m trying to think

[alpine shell]

what would k = -20 mean?

try to think about the drawing

[glass sparrow]

it will be the reflect

[alpine shell]

it would be on the left of the y axis, that wouldnt be too logical now would it?

[glass sparrow]

ohhhh

it should be positive

[alpine shell]

we want to divide the area in half and you have your k outside of the area, that would be pretty weird

[alpine shell]

better, it should be between 0 and 40

or else it makes no sense even

[glass sparrow]

yess so only k=10 make sense

[alpine shell]

yup and that is the answer indeed :)

now for your other problem, i am going to sleep because it is 1:40 am for me, but you can create a new channel and ask someone else for help

[glass sparrow]

ohh okeyyy

wow it’s too late

[alpine shell]

it kinda is yes where i am

[glass sparrow]

good night good night thank youuu

[alpine shell]

my pleasure :) nice work from you too

cya

(dont forget to close this if you create new help channel)

[glass sparrow]

.close

[safe radish]

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[safe radish]

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[minor ember]

hey guys i’ve been staring at this projectile motion problem for a while and it’s driving me crazy i’m trying to derive the general formula for the maximum horizontal range r but here’s the catch the projectile is launched from a cliff of height h not from the ground can someone walk me through the math i’m specifically looking for the breakdown using quadratic solutions and how trigonometric identities come into play here i really want to understand how the starting height h changes the optimal angle compared to the usual 45 degrees we learn in class down for a step by step challenge if anyone is free to help thanks a ton

[tardy mango]

discord.gg/physics

[delicate bobcat]

I don't see why this couldn't be handled here it's a beyond common trope for math problems

What would be nice is some more specificity to the question being asked

[median sand]

Show us the picture of your question?

[primal bone]

Or more explicitly:

!xy

[safe radish]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

@minor ember Has your question been resolved?

[minor ember]

There’s no textbook photo, I actually came up with this myself. I'm self-studying and wanted to challenge myself by deriving the general formula for a launch from height h instead of the ground. I’m really stuck on the quadratic breakdown and the trig part. I made the diagram to show exactly what I'm trying to derive. Any help with the steps would be huge!

[median sand]

Then a graph or something would be great!

[safe radish]

@minor ember Has your question been resolved?

[median sand]

Well, what have you done already?

I mean, you can just first define v0, angle theta, h, and g though. And perhaps make some vertical and horizontal components for initial velocity.

[minor ember]

Thanks for the tips! I’m currently calculating the components. So v_{0x} would be v_0 \cos(\theta) and v_{0y} is v_0 \sin(\theta), right? I'll let you know what I get for the final range

I’m working on the time of flight now. Since it’s launched from a height h, should I use the quadratic formula for -h = v_{0y}t - 0.5gt^2? It looks a bit tricky

[plucky elk]

why -h

[safe radish]

@minor ember Has your question been resolved?

[visual linden]

Because the displacement is technically negative, considering the body is moving down

[safe radish]

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[azure harbor]

Congrats on helpful

[safe radish]

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[lost cliff]

can anyone explain to me for the question a)ii)

[small sandal]

if i recall the number of arrangements in a circle is (n-1)! so if were separating the vowels and what not. so in the first 12 letters we have {a, b, c, d, e, f, g, h, i, j, k, l} and from that set we have the vowels being {a, e, i}

provided i have understood the question correctly

[lost cliff]

theres 7 consonants and 5 vowels

[small sandal]

ah k (didnt know the 12 letters they were referencing)

[lost cliff]

its ok

i asked ai, it gave a different answer than the marking scheme

[small sandal]

so instead we got something like this {V, C, V, C, V, C ...}

[lost cliff]

yea

[small sandal]

where V and C are vowels and consonants respectively

and we know that #C > #V

so we can also have arrangements such as {C, C, V, C, V, C, ...}

[lost cliff]

yes

[small sandal]

hm ok

well, the maximum # of ways they can be arranged is 11!

but thats provided that {V, V} is a legal input

[lost cliff]

there will be permutation

[small sandal]

yea

[lost cliff]

its not 11

[small sandal]

they are in a circle no?

[lost cliff]

the letters can be arranged

Theyre not identical

[small sandal]

they are arranged in a circle, so the maximum # of ways they can be arranged (without any extra rules) is (n-1)! so, (12-1)! or 11!

then we work down from that number

so we can do this process for the 7 consonants and 5 vowels

so, for C we get (7-1)! or 6! = 720. And to make sure no two Vowels can touch they need to be placed between each consonant. so we get get 7C5 for this arrangement, and then how many ways can our Vowels be arranged in general? 5!

do you think you can go from here?

-# and as predicted, our value is significantly less than 11! since thats the maximum ways in general

[small sandal]

so, you should get 6! * 7C5 * 5! yea. Do you know how to turn this into a more simplified version which includes a permutation?

[safe radish]

@lost cliff Has your question been resolved?

[safe radish]

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[proven smelt]

Can someone check this please?

[pallid bone]

line 2

x^3 * -1 / y

instead of x^3 * y

[proven smelt]

But when u diff log_e(1/y) u get 1/(1/y) = y?

[proven smelt]

why is there a Negative too btw

[pallid bone]

1/(1/y) d/dx(1/y)

y* -1/y^2 * dy/dx

-1/y * dy/dx

is there something wrong?

[idle jacinth]

is that ln (1/y) in numerator??

[proven smelt]

where

[idle jacinth]

there wont be negative in the first term of numerator

[idle jacinth]

and the denominator is kinda wrong from mine
hold on

[agile scaffold]

(apologies for butting in muffin et al), the second line when you apply the chain rule to the second term
$x^3 \log_e(\frac{1}{y})$

$3x^2 \log_e(y^{-1})$ is correct but on $\log_e(y^{-1})$ you forgot to take the derivative of the innermost function, i.e. $1/y$

so when you take $\frac{d}{dx} \left( \log_e(y^{-1}) \right) = \frac{1}{y^{-1}} \cdot \frac{d}{dy}(y^{-1}) \cdot \frac{dy}{dx}$ this should be $= y \cdot (-y^{-2}) \frac{dy}{dx}$ by applying the chain rule

[flat frigate]

reze ♡

[flat frigate]

reze ♡

[proven smelt]

ohh so it’s just easier to take a negative power out of the log

i see

[idle jacinth]

yea keep it minimalistic

[proven smelt]

Is this better?

[idle jacinth]

yess sir

[safe radish]

@proven smelt Has your question been resolved?

[safe radish]

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[worn geyser]

Hi, i need help with the task below. I have already done (a), and i got R = 7 for the first sphere, and R = 10 for the second one. For b, do i need to find some kind of plane in order to solve the problem? I marked the circle of intersection on the picture below, but i am not sure if thats correct.

**The task:
**A sphere is given by its equation x^2+y^2+z^2-2x+6y-6z-30=0. Another sphere is given by its equation x^2+y^2+z^2-10x-18y+6=0.
(a): Find the radius of the spheres.
**(b): **The intersection curve between the two sphere surfaces is a circle. Find the radius and the center of the circle of intersection.

[astral glacier]

You can equate the two expressions

[worn geyser]

Ohhh yess

Then i would get the equation for the circle of intersection?

Or noo

[cunning pasture]

Think.

[worn geyser]

I don know, the plane?

[brave wolf]

Unfortunately not necessarily, because you're still gonna have 3 variables in there.

[brave wolf]

Yeah, if you equate the non-zero sides, you're gonna get the plane in which the circle lies

[fathom jewel]

It's the same approach like last time, you'd set up a line that goes through the centers respectively and find the normal vector of the plane

[worn geyser]

Oh okai

[worn geyser]

But first i need to find the plane

And theen the line

When i know the normal vector

[cunning pasture]

What if we do (x^2+y^2+z^2-2x+6y-6z-30)^2 + (x^2+y^2+z^2-10x-18y+6)^2 = 0?

[worn geyser]

Right?

[fathom jewel]

Just equate the spheres as already mentioned

[brave wolf]

-# this'd not be very helpful, you'd get some sphere which contains the circle, but it'd be difficult to extract the circle from it

[worn geyser]

Okei

[worn geyser]

8x + 24y -6z -36 = 0

And soo the normal vector would be (8, 24, -6)

Okai got it

Thank youuu

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[sick timber]

i got a question to derviatives so ik that ln(e^x) = x but why is log(10^x) = x or 10^log(x)=x i dont understand

[karmic merlin]

that's the definition of a logarithm

[agile scaffold]

the logarithm undos an exponential function with the same base

ln is the logarithm of base e, log is usually representative of the logarithm of base 10

[cunning pasture]

The base of the log in log(10^x) = x must be 10 for it to be true.

[agile scaffold]

in any case, where does this involve derivatives if i may ask?

[sick timber]

idk im new to maths writing an exam tmrw

[sick timber]

ah

[sick timber]

makes sense

[agile scaffold]

would that be all?

[sick timber]

uh no

uh

what does d/dx mean

[median sand]

Derivative?

[sick timber]

ah

so derivative of e^x is still the same

because ln(e)= 1

?!

[karmic merlin]

No it's because of how e is defined

Tho the proof would depend on which definition you're using

[sick timber]

but isnt that the special thing abt the constant e the function valueu equals the gradient

or am i understanding it wrong

[karmic merlin]

Oh ye that's one of the definitions

Or something similar

[opaque fern]

yea, id say so

[agile scaffold]

sure, yes

[opaque fern]

because [
\dv x(a^x) = a^x \2\ln(a)
]

[opaque fern]

although i feel like that's more of a consequence of change of base rather than directly from derivatives

[agile scaffold]

though that is more to do with how the number e is constructed

i.e., this becomes a tautological truth that the gradient of e^x is proportional to e^x itself

[karmic merlin]

Though do note e^x isn't the only function with that property, any function of the form k×e^x for real values of k also satisfy that fact

[sick timber]

thats what i wanted to ask

so in the exam theres prob an question with like N(t)= initial valueu times e^kt

buut what is even k expressing

[agile scaffold]

the growth rate

[opaque fern]

relative growth rate

[sick timber]

so k>0 is growth

[agile scaffold]

well, the decay rate too

yes

[opaque fern]

it is analogous to the differential equation [
\dv[N]t = k\2N
]

[timid obsidian]

Organic math? 😢

[opaque fern]

The growth (the derivative) is proportional to its current size (N)

[agile scaffold]

pardon?

[sick timber]

so negative numbers is decay or what

[agile scaffold]

yep

[sick timber]

and fractions?

[agile scaffold]

depends

if it is negative it is a decay rate too

[timid obsidian]

Decay is smth wen any organic substance rots or smth-

[agile scaffold]

if it is positive, the converse

[agile scaffold]

oh, not talking about that here

[timid obsidian]

Alr

[agile scaffold]

fractions only imply that the growth/decay rate is relatively slow

[sick timber]

ah

[agile scaffold]

for illustration sake

[sick timber]

oh now i get it

and it always goes against y=1 bcs e^0=1 ?!

[agile scaffold]

im presuming youre asking whether or not the graph will ever approach y = 1, that is not necessarily the case, and only is true iff you have e^kt + 1

that is, the graph has an asymptote at y = 1

wait im confused i might be not interpreting what youre saying correctly

[sick timber]

no problem

i meant that the main function of e cuts thorugh y=1 thats

[opaque fern]

they are saying the y-intercept is y = 1 because e^0 = 1

[agile scaffold]

ah

[sick timber]

and it doesnt matter what value k has because for exapmle e^2 times 0 is still e^0

[agile scaffold]

well, yep

the vertical intercept is indeed = 1 for e^kt

[sick timber]

okay well thats it with my qeustions

thank u reze

[agile scaffold]

np!

if youre done, type .close

[sick timber]

.close

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[graceful lichen]

how to continue further?

[ionic blaze]

I'm prolly wrong on this but you can consider

subtract [x] and {x} to both sides and add 1 and factor LHS ||to ([x]-1)({x}-1)=1||

since {x} is fractional, you can make assumptions such as: ||it must be less than 1 so {x}-1 is less than 0 or negative, since the RHS is positive, [x]-1 must be less than 0 and [x] must also be less than 1||

now we have numbers ||0 to -inf|| to consider, although I'm not entirely sure on that last part

[graceful lichen]

here is solution tho

i dont fucking understand it

after 2nd step

[ionic blaze]

-# I sold 🥀

is 0 not a solution

wouldn't 4 also give

0

since it's 4+0/whatever

[brave wolf]

this is weird

how did they get here?

were there some other conditions in the problem?

[graceful lichen]

i think this book is cooked

[brave wolf]

are you sure you are reading the correct pages

page 1 starts well, but page 2 talks about unrelated stuff

[graceful lichen]

mb it was my mistake

im cooked

[brave wolf]

oh lol

[brave wolf]

this looks better

[graceful lichen]

im tired will solve it later

closing this

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[weary latch]

Okay, so I think I get the 5 km/h part, and I get the 35 km/h part, but after that, it gets a little shaky. How should I set up this problem?

[craggy tree]

you pretty much have it, you're just missing the northward 7km/h component of the current

then use pythagoras

[weary latch]

Or wait, do I start the vector at the end of the (12, 0) vector?

[crystal wigeon]

just plus the coordinates

[crystal wigeon]

you can add them in any order

[weary latch]

Does that include the (0, 7) coordinate or is it just the (0, 35) + (12, 0) ones?

[crystal wigeon]

plus them all

since 7km and 35km are in same direction, so let it be y and 12km is x positive

so (12,42)

let north be y and east be x

[weary latch]

Got that part right. Now I assume I have to do the Pythagorean Theorem on those values.

[crystal wigeon]

yes

[weary latch]

So 12^2 + 42^2 = c^2.

[visual sorrel]

isnt it (7, 42) cuz theres 5 km/h on the west asw

[crystal wigeon]

you are forgetting something

ariah is walking 5km in west direction, which is opposite to east

so you gotta minus that

[visual sorrel]

oh mb i thought u missed that 😭

[weary latch]

So that's (2, 42).

I got 1768 after the Pythagorean Theorem.

Oh wait, that's wrong.

[crystal wigeon]

(12,42) + (-5,0)

[weary latch]

It's actually (7, 42).

[crystal wigeon]

yes

[weary latch]

So that's 1813.

Now we just have to square that.

[crystal wigeon]

yep

sqrt(1813)

[weary latch]

Rounded to the nearest tenth, that's 42.6.

[crystal wigeon]

yes

[weary latch]

Got it! Thank you so much!

[rain adder]

a+b whole square?

[weary latch]

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[static harness]

How many more equations to finish this

[whole hill]

That looks really sick so far bro

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@static harness Has your question been resolved?

[safe radish]

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[shy osprey]

How did they multiple this? I don’t see it

[plucky elk]

$\frac{4}{12} = \frac{1}{3}$

[flat frigate]

riemann

[shy osprey]

thank you

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[glass sparrow]

hello can someone help me w this

[swift fjord]

do yk how to find average value

for a function

[candid wraith]

first i would plug into the average value thing

[glass sparrow]

y-y1/x-x1

[swift fjord]

lol that seems to be something for linear functions maybe

[light shoal]

probably you want to use an integral?

[swift fjord]

in calculus you should learn a general form to find the average value of any function over any interval

[glass sparrow]

oh so integral of Ia-x)^2 right

[swift fjord]

yea you intergrate over the interval

and divide by the length

so integral from -1 to 1 of (a-x)^2

and the interval has length 2 so u divide at the end

[glass sparrow]

is it?

[candid wraith]

i would expand (a-x)^2 but it doesn't really matter

[light shoal]

you should pick up a minus sign when you integrate, because of the -x

[swift fjord]

yea its good except for the minus sign from chain rule

[light shoal]

(or just change it to (x-a)^2)

[glass sparrow]

can i do ((a-x)^3/-3))

for the integrate or (a-x)^2

[swift fjord]

yea

[glass sparrow]

but after i done it

[swift fjord]

all you do is expand and solve

should be easy

i think the cubic terms cancel as well so its just a quadratic equation

[glass sparrow]

after i done it

i got a=3/2

[swift fjord]

you should get two values

since its quadratic

[glass sparrow]

3/2 or -3/2

[candid wraith]

ye

[swift fjord]

nice

you got it

[glass sparrow]

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[ionic blaze]

I don't think that top part, (1+cos(2x+pi))/2 is equal to cos^2(2x+pi)

cos^2(x)=(1+cos2x)/2 not (1+cosx)/2

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@proven smelt Has your question been resolved?

[proven smelt]

oh isit not

okay thx

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[glass sparrow]

hello

[glass sparrow]

can soemone help me w this pls

[light shoal]

which part, and where are you stuck?

[glass sparrow]

the part b

[light shoal]

are you sure about your answer for (a)? this looks questionable

[glass sparrow]

this is the coordinate of my B

i sub 10pi

to find y

[light shoal]

shouldn't it just be 20

not 21.whateverthatis

anyway, for part (b), yes you want to calculate h'(x) and set it equal to zero
but where did the pi's come from?

the derivative of sin(x/20) should just be (1/20)cos(x/20)

[glass sparrow]

ohhh

i see you

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[exotic crystal]

How can I prove this?

[exotic crystal]

What I currently have:

[plucky elk]

Is f continuous?

[exotic crystal]

It doesn't say

Let's suppose that $\lim_{x\to \infty} f(x)=L.$ That means that $\forall \varepsilon>0, \exists N>0 \text{such that if } x>N \Rightarrow |f(x)-L|<\varepsilon. $ Now take $\delta=\frac{1}{N}$, then and let $t=\frac{1}{x}$. Then if $0<t<\delta \Rightarrow \frac{1}{t}>\frac{1}{\delta} \Rightarrow |f(\frac{1}{t})-L|<\varepsilon$

[flat frigate]

YuyuHenry

[exotic crystal]

which then shows that $ \lim_{x \to 0^{+}} f(1/x)=L$

is that correct?

[flat frigate]

YuyuHenry

[exotic crystal]

I asked chatgpt to translate what I had in an image

Like, is this correct?

<@&286206848099549185>

[severe pond]

what do you think?

[exotic crystal]

I think it is, but I need double confirmation

because of my insecurities

[severe pond]

it is correct

[severe pond]

did you not come up with it

maybe that's why

[exotic crystal]

Nah, I got an assignment about the topic marked wrong

Thanks tho

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[split fulcrum]

probably the easiest way is to use the identity for 1-cosx but if you don’t know it then might not be obvious

[proven smelt]

oh i see. thanks.

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[proven smelt]

.reopen

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✅ Original question: #help-23 message

[proven smelt]

I’m finding my bounds are both 1, but this doesn’t make sense?

[idle jacinth]

oh yea see in periodic functions

such as in cos and sin

the graph goes up x axis and below and up and below and so on

[light shoal]

something went wrong here

the integrand on the first line is nonnegative, but on the third line is is not

probably a sqrt(x^2) = x type of blunder

[idle jacinth]

so if u find area (integration) or bounds on the given limits then basically u will be cancelling everything and will make it zero

[burnt notch]

And also the angle should be θ/2 @proven smelt

[tacit prawn]

^

[burnt notch]

Because you used double angle formula

[tacit prawn]

$(1 - \cos \theta)^{3/2} = \left( 2\sin^2\left(\frac{\theta}{2}\right) \right)^{3/2}$ is what it should look like if i’m not mistaken

wait

yeah

wait why’d id post the first part

[idle jacinth]

u have to break the limits
from 0 to pie and from pie to 2pie
or just 2 times 0 to pie
in order to not get same bounds

[tacit prawn]

$\left( 2\sin^2\left(\frac{\theta}{2}\right) \right)^{3/2}$

[flat frigate]

Yatto

[tacit prawn]

yeah this

[light shoal]

OP hasn't replied since the initial post, maybe let's not do the problem for him while he snoozes 😁

[split fulcrum]

You snooze you lose

[light shoal]

facts are stubborn things

[proven smelt]

ohh. this helps.

hm lemme try again

[tacit prawn]

well if we’re showing him what he SHOULD be getting then maybe it can help him see how to get said answer 😴

[safe radish]

@proven smelt Has your question been resolved?

[proven smelt]

I got it, thanks guys!!

[safe radish]

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[sleek current]

Is my understanding correct?

[sleek current]

Cause the cube is symmetrical b CB has the same magnitude of b, but goes the opposite direction, thus -b

So doing this CF = (-b) + c

[light shoal]

looks fine

[sleek current]

Thanks!

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[gilded star]

how to construct it in geogebra?

have done this so far

i know that D lies on this circle

which has AC as diameter

[rancid lion]

Hint: there's a pair of similar triangles

[errant ravine]

https://cdn.discordapp.com/attachments/908078029778083872/1502275307128750081/Animated_illustration_of_thales_theorem.gif

-# sorry to intterupt jus thought this might be helpfull

-# for refrence angle B here is always 90 degrees!

[gilded star]

i think i see now

since BC and AB intersect in an angle of 90 we can rotate the square in 90 degrees and then BC gonna be parallel to AB

and then CBH gonna be equal to BAG

but how can i use this to construct tho

[golden nebula]

A solution that you could try is related to this. You know the lengths of BC,BA and CA. Also, AD is the side length of the square. All these points lie on a circle

[idle jacinth]

can we not use vectors and coordinate geometry to do this?

[golden nebula]

A hint for you to try is to treat AD as a chord and find the length of it

[gilded star]

i know how to solve it, i just want to construct it perfectly from here

using synthetic methods

[gilded star]

this is an approximation btw

lets say we have this, where AB is 4 and BC is 3

how can we have ABC inscribed in a square, construct this square

[safe radish]

@gilded star Has your question been resolved?

[brave wolf]

maybe if we start like this

A (origin) will be the common vertex of the square and the triangle

we're basically trying to circumscribe a square now

so we draw the 2 semicircles - the 2 neighboring corners of the square must lie on those

and they must form 90° angle with A

[brave wolf]

so let's rotate one of the semicircles 90°

and where they intersect (the rotated with non-rotated), there is our corner

[brave wolf]

since you know what the triangle is, your problem is basically just circumscribing a square s.t. they share one vertex

so the solution works generally for any non-obtuse triangle

[gilded star]

nice solution, thank you

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[lavish ermine]

hi I need help

[lavish ermine]

not arriving at the correct answer

[safe radish]

@lavish ermine Has your question been resolved?

[tardy mango]

(In the future, please make the first message you send the actual question since that's what the bot pins. This saves us the effort of having to change the pin manually.)

[tardy mango]

I'd advise you to not write so small

seems like you changed from 0.65 to 0.62 mid-calculation?

[lavish ermine]

nah its 0.65 that's my handwriting sorry

[tardy mango]

I mean
$$v_n=200 \cdot 0.65^{n+1}+50 \cdot \frac{0.65^{n+1}-1}{0.65-1}$$
should be fine. Though you can simplify it tbf.

[flat frigate]

Civil Service Pigeon

[tardy mango]

,w 200+50(\frac{1}{\frac{13}{20}-1})

[flat frigate]

Results provided by WolframAlpha

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Requested by civil_service_pigeon

[lavish ermine]

the key says it should be raised to n

instead of n+1

[tardy mango]

$$v_n=200 \cdot 0.65^{n}+50 \cdot \frac{0.65^{n}-1}{0.65-1}$$

[flat frigate]

Civil Service Pigeon

[tardy mango]

as in this?

[lavish ermine]

ye

hmm

it should be 180 at n = 0

[tardy mango]

yeah

seems like they mis-shifted by 1 and/or didn't read what they wrote carefully

[lavish ermine]

oh okey

lol sounds good to me

thanks for the help

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[mortal marten]

Determine the coordinates of the intersection points between the asymptotes of the hyperbola and the circle passing by the focus points of the hyperbola

[mortal marten]

not sure how to start

[loud wren]

use the 2 points to make the hyperbola equation prolly

then make the circle equation and solve using algebra

that would be a rough sketch

[safe radish]

@mortal marten Has your question been resolved?

[mortal marten]

yeah (-42)^2/21^2 - (-20sqrt3)/b^2 = 1

b = 20

x^2/21^2 - y^2/20^2 = 1

is rthe equation ofthe hyperbola

and the focus points so c^2 = a^2 + b^2

c^2 = 21^2 + 20^2 c^2 = sqrt841

c = -29 and 29

which means the circles radius is 29

x^2 + y^2 = 29^2 is the circles equation

[safe radish]

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[weary latch]

I understand that I have to do 6cos(160) + 6sin(160) and stuff like that, but how do I put them together to find the length and angle of both of them combined?

[swift fjord]

just break into x and y components then add

[quiet plume]

What you have isn't $\vec{v} = 6\cos(160) + 6\sin(160)$, but rather $\vec{v} = (6\cos(160), 6\sin(160))$. When you add two vectors, you add them coordinate-wise.

[flat frigate]

Azyrashacorki

[quiet plume]

So if you have v = (a,b) and w = (c,d), then v+w = (a+c, b+d).

[weary latch]

Wow, that sounds much simpler than I was expecting.

So now I just have to add them up.

And do the Pythagorean Theorem.

[errant rapids]

yes pythag is the length

[weary latch]

Okay, I got 4.6 for the first one.

[quiet plume]

Exactly yes. Then for the angle the best thing to do is to make a quick sketch of which quadrant the resulting vector is in. From the sidelengths you can use trig to get an angle, and you can adjust it according to the quadrant you're in.

[weary latch]

Well, they're both negative, so that would be in Quadrant III.

Then, after that, I think I have to get the tangent, because I already have opposite and adjacent (the -4.27 and the -1.71).

[quiet plume]

Yep

You take the arctangent (tan^(-1)) of that ratio essentially.

[weary latch]

I got 201.8 after adding 180 to the arctan.

I'll try submitting that.

Oh wow! That was pretty simple.

Thank you so much!

I'll close this.

.close

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[safe radish]

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[signal wave]

Is 16.95 not get rounded to 16.9 when rounded to the nearest tenth?

[mortal sandal]

It does not

[signal wave]

Why?

[mortal sandal]

If the next digit is 5 or above it rounds up

[signal wave]

In the hundredths digits?

[mortal sandal]

Yes

[signal wave]

Tysm

.close

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[mortal sandal]

No problem

[safe radish]

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[spark leaf]

Please @ me I just need someone to explain question 4

[timid obsidian]

Which part?

[spark leaf]

I just dont know what im supposed to do except find tan, how tho

[timid obsidian]

Ur given a angle

Q

[spark leaf]

I dont get the example thats why im here

[timid obsidian]

Which is in between π/2<Q<π

[timid obsidian]

Or u can say Q belongs to 90<Q<180

In 2nd quadrant

[spark leaf]

I have to find the exact angle

[timid obsidian]

Oof-

[spark leaf]

Thats the answers

[timid obsidian]

Yk wat sin Q is?

[spark leaf]

What is it

[timid obsidian]

Perpendicular/hypotenuse

Yk dat?

[spark leaf]

Ohh the formula for din yes

[timid obsidian]

And wats tan?

[spark leaf]

Sin over cos

[timid obsidian]

Yes

Or u can say perpendicular/base

Right?

[spark leaf]

Ihh wait ik now sorry

Im wasting ur time

I gtg im at school and break is over

Thanks

Cya

[timid obsidian]

Np

[spark leaf]

Should I close

[timid obsidian]

Sure

[spark leaf]

.close

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[safe radish]

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[spark leaf]

Please @ me question 4

[spark leaf]

If there is a helper helping please dont interrupt unless they make a mistake, thanks

[golden nebula]

Which part are you stuck on?

[spark leaf]

Idk how to do it

[golden nebula]

Should we start with a?

[spark leaf]

Actually nvm

[golden nebula]

Kk

[spark leaf]

Do we find cos then do sin over cos?

[golden nebula]

Thats an option yes

[spark leaf]

Is that the easier way

[golden nebula]

You can use this table

If you know it

[spark leaf]

Ik that table

But I see no tan

Ohh wait its there

But whats sec

[golden nebula]

Sec = 1/cos

If you dont know its fine just finding cos and doing sin/cos

[spark leaf]

Im not gonna use it as i didn't learn it yet

[golden nebula]

Kk

But you do know that sin^2 + cos^2 =1 right?

[spark leaf]

Is it possible to find tan with no calc?

[spark leaf]

Yep

[golden nebula]

So use that to get cos

[spark leaf]

Then sin divide by cos

[golden nebula]

Some calculation is required

[golden nebula]

Yes

[spark leaf]

I can find sin and cos with out a calculator so why cant I find tan

[golden nebula]

Oh I thought you meant calculation

No you dont need a calculator

Let me know when you find tan for the first question

I think you can solve the rest too

Did you solve it @spark leaf ?

[spark leaf]

Ohh I can

Not yet

I was distracted

Sorry

[golden nebula]

Thats fine

Not an issue

Let me know when you find tan

[vague phoenix]

you know tan is sin/cos

[spark leaf]

Yes

[vague phoenix]

Well look at question 4a

we know we are in the second quadrant

[spark leaf]

Im doing the question rn

[golden nebula]

Damn i forgot about that

[vague phoenix]

then just follow ASTC rule -> Tan < 0

Given information for sin and cos there are just redundant

[spark leaf]

What did I do wrong

[vague phoenix]

Ehhh that way is bad

[spark leaf]

I found cos but its still wrong

😭

[golden nebula]

You got the sign wrong thats all

[spark leaf]

Cos sign?

[vague phoenix]

A better way is that sin theta = 1/3, then theta = sin^-1(1/3), then put -cos(theta)

[golden nebula]

Since cos was negative and sin was positive, what willthe sign if tan be?

[golden nebula]

I dont think op has learnt inverse trig

[golden nebula]

U wrote positive sqrt2/4 for tan

[spark leaf]

No i didn't