help-10

[timid silo]

I was told to ????

[fickle turret]

Somebody professionell 😀

[fickle turret]

Not to make a mixture. It's either dx or du

So first of all: Where does the /8x come from?

[timid silo]

Was told to do this ?? I have no clue what i am doing

[fickle turret]

u = 2x/5

So far it is correct

[wary vigil]

in this picture you've got it into the shape of

$\frac{1}{1 + a^2}$

[warm shale]

Katharine

[wary vigil]

which is a known integral

,w integral 1/(1 + a^2) da

[warm shale]

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Requested by Katharine#1478

[wary vigil]

does that make sense?

[timid silo]

Yes but i don’t know how to do the next step

[wary vigil]

well you have

$\frac{3}{25} \int \frac{1}{1 + (\frac{2x}{5})^2} dx$

[warm shale]

Katharine

[wary vigil]

you understand that you have this?

[timid silo]

100%

[wary vigil]

cool

now you can perform a substitution

to make it look more like

[wary vigil]

this

by saying

$a = \frac{2x}{5}$

[warm shale]

Katharine

[timid silo]

X = sqrt U-25/4

[wary vigil]

either way is fine

i think i understand your thing

although you forgot that it's u^2

the point is you can substitute a or u

for 2x/5

but to do that we have to change the dx

we want da

[wary vigil]

because we now have this

still good so far?

[timid silo]

Think so

[wary vigil]

$\frac{3}{25} \int \frac{1}{1 + a^2} dx$

[warm shale]

Katharine

[wary vigil]

we now have this

but we have dx

we want da

so to change from one to the other

have you learned that?

[timid silo]

DA/DX = 2/5?

[wary vigil]

yes

and so to find dx in terms of da

[timid silo]

2da/5

= dx

[wary vigil]

are you sure?

[timid silo]

5da/2

[wary vigil]

exactly 😄

right so

we can replace our dx with our da

$\frac{3}{10} \int \frac{1}{1 + a^2} da$

[warm shale]

Katharine

[wary vigil]

now we have the perfect integral

arctan

[timid silo]

No

Actually yes

Your right

[wary vigil]

so if we do the integration to get arctan

what does it look like then?

[timid silo]

This

[wary vigil]

you still have an integral

[timid silo]

Tan-1

[wary vigil]

arctan

[timid silo]

What’s arctan

[wary vigil]

or tan^-1

having this

$\frac{3}{10} \int \frac{1}{1 + a^2} da$

[warm shale]

Katharine

[wary vigil]

we can now integrate it

to get

[timid silo]

Forget the a as the denominator

That right

[wary vigil]

$\frac{3}{10} \arctan(a)$

[warm shale]

Katharine

[wary vigil]

this is exactly why i hate tan^-1 notation for arctan

$\tan^{-1}(x) \neq \frac{1}{\tan(x)}$

[warm shale]

Katharine

[timid silo]

Oh i get it

[wary vigil]

the integral goes away

$\int \frac{1}{1 + a^2} da = \arctan(a)$

[timid silo]

How would you integrate arc tan would it he arcsec^2

[warm shale]

Katharine

[wary vigil]

i don't actually know what

$\int \arctan(a) da$

is

[warm shale]

Katharine

[wary vigil]

,w integral arctan(a) da

[warm shale]

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Requested by Katharine#1478

[wary vigil]

it

it's a lot more difficult 😄

[timid silo]

Alright thanks for your help anyways understand it now

[wary vigil]

$\int \frac{1}{1 + a^2} da = \arctan(a) \neq \int \arctan(a) da$

[warm shale]

Katharine

[wary vigil]

this is important 🙂

[tardy epoch]

how did you get tan in the last line?

[wary vigil]

misunderstanding as to what tan^-1 (x) means

[timid silo]

.

[wary vigil]

because tan^-1 is horrible and i hope it dies

[timid silo]

What is tan^-2

[wary vigil]

idk

[tardy epoch]

RIP a million calculators

[wary vigil]

,w tan^-2(x)

[warm shale]

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Requested by Katharine#1478

[timid silo]

Its technically right

[wary vigil]

whenever you have arctan

just write arctan

don't use tan^-1

[timid silo]

Tan^-1 is different to arctan

[wary vigil]

no they mean the same thing

but exactly because of that confusion

i think tan^-1 is bad

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[uncut robin]

what have you tried?

also, which of the questions do you need help with

[obtuse pebble]

@proper raptor Has your question been resolved?

[proper raptor]

question 1 and quest 2 part a

i dont really know how to prove that it is a rectangle and i just dont understand question 1

[uncut robin]

well for question one let's look at the first quadrant

we have two lines right

y=ax
y=bx

[proper raptor]

ok

[past fog]

what are the properties of a rectangle

simplistic

I wanna walk you through this

Not do it for you

[proper raptor]

equal opposite sides right angles

[uncut robin]

bruh, we were doing question 1

[past fog]

I’m going to do q2 for him first

Draw all 3 points and connect those that form the rectangle, do not do the connection that makes a weird looking side

[proper raptor]

ok

[past fog]

I promise this will help you actually understand what’s going on

[proper raptor]

ive drawn i

it

[past fog]

Send a picture

So you did part B right? For q2

[proper raptor]

uh no I haven't yet but I can solve it

[past fog]

okay so, let’s say we draw in the fourth point, could you assume where that might go?

[proper raptor]

0,3

[past fog]

Draw it in with an open circle

Just to separate it

[proper raptor]

wdym by with a open circle

[past fog]

like don’t fill the dot

For the point

Make it an open circle

[proper raptor]

ok

[past fog]

So now, can you prove the three points form three of the vertices?

Knowing that phantom, let’s call it, point is there

[proper raptor]

well then dont i just prove that 2 angles are right angles with gradients

[past fog]

try it

[proper raptor]

does this work?

[uncut robin]

yeah, you just had to prove that ABC is a 90 degree angle

leave alone the vertex D for now, you don't even know that the coordinates are (0,3)

(spoiler: they are)

[proper raptor]

also to find the fourth vertex do i just find the gradient of AB then use y=mx+b?

just making sure

[uncut robin]

well no

[proper raptor]

does this not work?

[uncut robin]

yes but why did you say b = 3

[proper raptor]

well cause 1= -2 +b so you + 2 on both sides

and 3=b

whats the other method for solving if not?

im hot garbage at coordinate geoemtry

[uncut robin]

well by finding that b = 3
you just got y=-2x+3
the equation of the line parallel to AB which goes through point C

but you can't just find D saying that x = 0, or that y = 3 and substituting in that equation

those are assumptions

so now, you have two methods of doing this

midpoint or intersection of lines

probably the second one is easier

[uncut robin]

now you do the same thing but for the line parallel to CB which passes through A

and then find the intersection

[proper raptor]

sorry i had to go do something

so i just find both lines

and use intersection of lines to find point d

well the unknown point

[uncut robin]

yes

[proper raptor]

ok

can we go back to question 1 lol

i still dont know how to solve that

[uncut robin]

yeah sure

so, first quadrant

you have two lines

which one has a bigger slope

[proper raptor]

y=bx?

[uncut robin]

(which one makes a bigger angle with the x axis)

[proper raptor]

the 2 lines as in y=bx and y=ax right

[uncut robin]

yes exactly

so b>a

now

the 3rd line is in the second quadrant

so the slope is actually negative because the angle the line forms with the x axis (or y=0) is greater than 90 degrees

so that means that both a and b are bigger than c

[proper raptor]

oh

so if the angle formed is greater than 90 its always negative?

[uncut robin]

yes

[proper raptor]

tysm for your help

[uncut robin]

np

[proper raptor]

.close

[obtuse pebble]

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[lucid smelt]

I'm not getting anywhere with a math/physics problem involving orbits

[devout solar]

Cool

[lucid smelt]

hold on I just dropped something

[devout solar]

an apple?

[lucid smelt]

Ha, no, a piece of adhesive I play with between my fingers

[devout solar]

lol

relevant piece of information

[lucid smelt]

Ok, I'm trying to determine the optimal launch trajectory for a rocket going from surface to orbit, assuming a circular destination orbit, and ignoring air resistance.

Optimal in the sense of minimal delta-v

[turbid sail]

Up

[lucid smelt]

That's... not how orbits work

[turbid sail]

🤣🤣 I’m playing I shall leave now

[lucid smelt]

😑

[devout solar]

so we have to get to a set point or what

[lucid smelt]

The requirement is that the periapsis is above a certain height

Beyond that, I don't actually care much about the specific orbit

[devout solar]

so just blow alot of shit up and we should get there

[lucid smelt]

Well Kerbal gives me a fair amount of shit to blow up, but not an infinite amount

[devout solar]

So we find other suppliers

[lucid smelt]

...Kerbal is a video game

[devout solar]

fuck him then

[lucid smelt]

😑

Let's start with the basics

The vis-viva equation says that for all orbits: $$ v^2 = \mu \qty(\frac{2}{r} - \frac{1}{a}) $$

[warm shale]

derp_commander

[lucid smelt]

$a$ is the semi-major axis of the ellipse the orbit traces, $r$ is the distance from the center of the body, and $v$ is the orbital speed.

[warm shale]

derp_commander

[lucid smelt]

So we want to end up there

But we're starting with a projectile under constant acceleration

$$ x''(t) = F\cos(\theta) $$
$$ y''(t) = F\sin(\theta) - g $$
$$ x(0) = y(0) = x'(0) = y'(0) = 0 $$

[warm shale]

derp_commander

[lucid smelt]

$$ r = y + h_0 $$

$h_0$ is the radius of the body being orbited

[warm shale]

derp_commander

derp_commander

[lucid smelt]

I think?

I don't know, I'm lost trying to go from the ballistic equations to the orbital ones

[obtuse pebble]

@lucid smelt Has your question been resolved?

[lucid smelt]

Technically "at rest on the surface" is another orbit, so one could treat it as a Hohmann transfer

[obtuse pebble]

@lucid smelt Has your question been resolved?

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[noble kindle]

Is this correct?

So there’s only 5 Cosets of {1,10} in U(11)

[obtuse pebble]

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[dim rock]

https://media.discordapp.net/attachments/928138725991534672/994460169377419364/20220707_095917.jpg

[tardy epoch]

What's your question

[obtuse pebble]

@dim rock Has your question been resolved?

[obtuse pebble]

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[queen canopy]

help

[tardy epoch]

https://www.cuemath.com/algebra/exponent-rules/

[obtuse pebble]

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[tardy epoch]

what formula did u use

[kindred oasis]

Yeah it's correct 👍

You said 75W but used 125W in the problem

Except that, what you did is correct

Yeah

You probably sent the wrong photo, but it's correct

Don't worry

👍

Yeah it's correct

$$kWh = \frac{watts \times time(hours)}{1000}$$

[warm shale]

Andrea276

[kindred oasis]

Start with this formula and try to think what you can do next

You don't have Watts, but kilowatt-hour

Watts? Amperes? Joules?

Can you send your work? I think there's something wrong

The conversion from kWh to Wh wasn't needed if you keep 1000 in the denominator. And it would be in Wh, not in W. Also in this case you found the amount of watts, not the work, so you should use the formula $Watts = Amps \times Volts$

[warm shale]

Andrea276

[kindred oasis]

And not P=W/t, since W is not work and you don't need power anyway

So it would be $$1.35kWh = \frac{watts \times 0.75h}{1000}$$
$$watts = \frac{1.35kWh \times 1000}{0.75h} = 1800 W$$

[warm shale]

Andrea276

[kindred oasis]

,calc 1800/110

[warm shale]

Result:

16.363636363636

[kindred oasis]

Yeah

You're welcome. Pay attention to the units of measurement, they're a good way to know whether your work is correct or you used some wrong formulas

95% efficiency means that only 95% of what goes in is actually used

$\frac{95}{100}P_i = P_f$

[warm shale]

Andrea276

[kindred oasis]

You have the energy required to boil the water. Is it the same as the energy you need to input?

[obtuse pebble]

@dire copper Has your question been resolved?

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[fringe panther]

I do not understand how the first sentence and second sentence of this question relates. It seems that the first sentence of this question is useless, and the answer will simply just be p(n) = 48n/80. However, I’m pretty sure that is not the answer. Did I misunderstood the question, or can someone guide me through this question?

[grizzled shore]

first sentence, 120 donuts for 60$

how much did each donut cost?

[fringe panther]

0.5$

[grizzled shore]

ok second sentence

she profited 48 dollars from 80 donuts

how much did she profit per donut?

[fringe panther]

48/80 = 0.6

[grizzled shore]

ok well the first part of the question doesn't have anything to do with the answer

[fringe panther]

That was exactly what I was thinking

But there must be a reason why it’s there

[grizzled shore]

so your answer will just be p(n) = 0.6n

idk maybe there's a part 2

or maybe they removed it from the question and forgot to delete that first part

[fringe panther]

There isn’t

[grizzled shore]

could be other reasons

[fringe panther]

I’m guessing they wanted to say she sold 80 donuts for 48 bucks but instead said they got a profit of 48 bucks from 80 donuts

Could that be a possibility?

[grizzled shore]

no because they used profit for both parts of the question

so if u answered anything else you'd just be wrong

[royal basin]

there is a difference between profit and revenue lol

[grizzled shore]

no but they used the word profit for both parts

[fringe panther]

Like, maybe they wanted to let us to do something like price - cost = profit, but they messed up on the wording

[royal basin]

i don't think the question has any mistakes

her profit before any sales is -60 and her profit after selling eighty donuts is +48

it is fairly transparent from here to find the price for which each of her donuts was sold

[fringe panther]

Oh, I think I might’ve understood the question

Let me try

So the profit is n(0.6-0.5)?

and the function would be p(n) = n(0.6-0.5) or 0.1n?

Since the profit of buying a donut is -0.5, and the profit of selling one is 0.6

Is that correct?

@royal basin

[royal basin]

bad

[fringe panther]

Oh, which step did I do wrong?

[royal basin]

basically all of them

according to your formula the profit before any sales is 0*(0.6-0.5) = 0, when it should be -60

the girl paid a lump sum of 60 bucks to get her donuts after all

[fringe panther]

That's true

[royal basin]

so before any sales are made she is at a monetary loss of 60 dollars

and then she made a total of 48 - (-60) = 108 dollars in revenue from the sale of 80 donuts

do you understand this so far?

[fringe panther]

Hold on, I think I misunderstood the whole question

Let me redo it

So from the start, camille loses $60 by buying 120 donuts

and after she sold 80, she got a profit of $48 dollars

Meaning that she earned 108 dollars by filling in the $60 dollars that she used and earned an additional $48

Therefore, she sells each donut for $108/80, which is $1.35 dollars

Thus, the function should be: p(n) = 1.35n - 60

[royal basin]

yes exactly

[fringe panther]

Thank you for helping

I'm pretty bad at word problems

Thank you for helping me understand this question, I really appreciated.

.close

[obtuse pebble]

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[fringe panther]

.reopen

[obtuse pebble]

✅

[fringe panther]

Hold on, I think I might've made a mistake

Because since -60 is just 120 donuts, what if the value of n is larger than 120?

Wouldn't she need to get more donuts?

If she sells one donut for $1.35, and each donut costs her $60/120 = $0.5, she only gets a profit of $1.35-$0.5 = $0.85

There's something I'm not understanding

@royal basin Sorry to ping and annoy you again, but I have a question

Nevermind, I'll just ask my parents.

.close

[obtuse pebble]

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[red snow]

how do u solve this?

[tranquil arch]

you can solve CD first

$CD = \frac{1}{2} BC$

[warm shale]

秋水

[red snow]

then do I use Pythagoras?

[tranquil arch]

yes

[obtuse pebble]

@red snow Has your question been resolved?

[obtuse pebble]

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[shut island]

How do you write sum with $$...$$?

[warm shale]

A random bird

[shut island]

So that TeXit writes it like in the Image

[long peak]

$\sum_{bottom}^{top} stuff$

[warm shale]

Tom

[shut island]

Ok thanks

[knotty crow]

$$\sum^{\infty}_{k=0} \Big(\frac{1}{2}\Big)^{k}$$

[warm shale]

Modus

[shut island]

Thx

.close

[obtuse pebble]

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[compact gate]

I need to find a value for a such that the system has zero solution, one solution and infinite solution

[knotty crow]

Cramer's formula, u know?

[compact gate]

I was able to get the third row (-R1+R3) as 0 0 2-a^2 | a-2 Either I made a silly arithmetic mistake or don't know how to move forward

[compact gate]

I'm not supposed to use that to solve this sytem.

[knotty crow]

ah ok

so then Gauss-Jordan elimination to row-echelon form and Kronecker Capelli theorem

[compact gate]

yes I am farmiliar with that idea (though never heard of the name til now).

I know to have infinite solutions, I would want the bottom row to be all zeros, but what value of a will give that?

[knotty crow]

wait, you're trying some a's?

[compact gate]

guess and check? no

I'm trying to find it analytically

[knotty crow]

I'll do row echelon form on the paper

[compact gate]

thanks :)\

[knotty crow]

ah

Just two operations

[long peak]

you could also have the bottom row equal to another row to get this

[knotty crow]

so I got on the main diagonal 1, -6, a^2 - 2

so for a = +/- sqrt(2) there are no solutions

[compact gate]

I follow so far, the part that is confusing me is the infinite solutions.

[knotty crow]

generally I'd say for a^2 - 2 != 0 we have one solution, for a = +/- sqrt(2) we have no solutions and there's no "a" for which we have infinitely many solutions, cuz ranks are equal to 3 and can't both equal to 2 (or less)

[compact gate]

Okay, I thought as much, but the way the question is worded it seems as though there should be a way to get a value of "a" such that there are infinite solutions.

I was going crazy.

just to confirm, you believe that there is no way we can choose a value of a such that there are infinite solutions?

[knotty crow]

by Kronecker Capelli to have infinitely many solutions we should have rank of matrices equal to each other and less than number of variables

here it's impossible (from what I've got) because then 2 - a = 0 and a^2 - 2 = 0

[compact gate]

okay thanks a lot for your time.

.close

[obtuse pebble]

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[tight geyser]

I asked this question like two days ago in #calculus but I haven't really got a reply that clarifies my question so Im opening a help channel. My qns is in the latex image I created.

[tight geyser]

Links:

1. Original integrals I became confused about in Examples 1-3 of:
   https://tutorial.math.lamar.edu/classes/calcII/TrigSubstitutions.aspx

2. Wolfram Alpha showing somehow, for negative x bounds (from -4/5 to -2/5), you just still have 2(tan theta - theta) but you put |x| instead of x in the functions:
   https://www.wolframalpha.com/input?i=2(Sqrt[25(-2%2F5)^2-4]%2F2-arccos(2%2F(5(|-2%2F5|))))-(2(Sqrt[25(-4%2F5)^2-4]%2F2-arccos(2%2F(5(|-4%2F5|)))))%3D2pi%2F3-2sqrt3

Desmos plotting showing the visualised 3 functions
3. https://www.desmos.com/calculator/azpiw2gghg

[knotty crow]

btw

I've checked this line:

and wolfram says it's ok

[tight geyser]

wait wolfram says they're equivalent?

Lemme check

Wait I'll check it again, when I plugged it in the other day it didn't work

Wait why did you delete your msg

[knotty crow]

because it's spam, haven't you seen that?

[tight geyser]

Oh yeah oops, haven't opened a help channel for a lonnng time lol

[knotty crow]

generally

only difference is the sign

it seems that the minus before the whole thing is unnecessary

[tight geyser]

Hmm

[knotty crow]

what was the substitution?

[tight geyser]

But

[tight geyser]

$x=\frac{2}{5}\sec(\theta)$

[warm shale]

grass

[knotty crow]

where's the abs value

result is valid

[knotty crow]

ok I'll try

[tight geyser]

Im trying to show to you that taking away the negative doesn't make it equal. But yeah if we replace x with |x| it works

For visualisation ^

[tight geyser]

Thanks! : )

[tight geyser]

Yeah?

[knotty crow]

f' = g

wolfram draws same sketch also

[tight geyser]

Hm yeah but the thing is in

In the example in Pauls' Online Math notes,

This was the indefinite integral calculated for nonnegative x

[knotty crow]

for x > 0 it still works

[tight geyser]

Why can't we take the blue part and substitute it into the green one? Since it's just a constant multiple difference of -1, it seems to me like it should work but it doesn't

[knotty crow]

but what, blue is in terms of "x", you mean after x = 2/5sec(theta)?

[tight geyser]

Yeah but in the first one, 2(tan theta - theta) = 2 blue (for positive x), so why can't we sub this into green to give -2(tan theta - theta) = -2 blue (for negative x, where -2 was the derived coeff.)

[knotty crow]

well

blue part is for non-negative x's

this is why the redundant minus I was talking about was unnecessary

you've used that "-2" coeff

instead of 2

or vice versa for negative "x" I guess

[tight geyser]

But -2 was supposed to be the correct coefficient though

And this coefficient works when we're deal with the expression in terms of theta, but I don't get why this coefficient suddenly doesn't work when we try to sub in the blue part to get an expression in terms of x (even though this blue part works for coeff +2)

Additionally, how would one derive $\sqrt{25x^2-4}- \arccos \left( \frac{2}{5|x|} \right)$, because it isn't immediately clear to me how you would get the $|x|$ part when integrating

[warm shale]

grass

[obtuse pebble]

@tight geyser Has your question been resolved?

[obtuse pebble]

@tight geyser Has your question been resolved?

[obtuse pebble]

@tight geyser Has your question been resolved?

[obtuse pebble]

@tight geyser Has your question been resolved?

[obtuse pebble]

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[final girder]

How would I simplify 8 root 243?

[timid silo]

factor 243

[final girder]

121.5

Oh wait

1 and 243

Wait so it would just be 8 root 243

Because

243 is a PRIME number

[dapper bloom]

No

243 is not prime

[final girder]

How do I factorise it then

[timid silo]

2+4+3=9, so 3 divides 243

[final girder]

81 and 3

[cloud dove]

nine does too

because 2+4+3=9

[dapper bloom]

Now factor 81

[final girder]

9 and 3

3 and 3?

So 243 should be 3 root 3?

[dapper bloom]

Keep going until you have 243 fully factored into primes

[final girder]

Okay so

3 x 8 Is 24

24 root 3

Wait

What

[pine sail]

Divide 243, by 3. Keep doing that until you get 1.

Or some other prime number.

[dapper bloom]

You want to write 243 in its prime factored form

[final girder]

Got 3

[pine sail]

Yes, but not just one time right?

Because 3*1 is 3 not 243.

[final girder]

Okay

243 / 3 is Root81 Root3

Root81 is Root9 Root3

[dapper bloom]

I think your arithmetic is wrong

[pine sail]

An example,
24 = 3 * 8
8 = 2 * 4
4 = 2 * 2
Therefore, 24 = 2 * 2 * 2 * 3
24 = 2^3 * 3

What I did is essentially called prime factorisation, that's what you have to do.

[final girder]

Yeah I do that

[dapper bloom]

Here remember 81=9^2 so 3sqrt(81)=9(3)=27 not 243

[final girder]

243 = 81 x 3

Don't I have to root until I get two prime numbers?

Or factorise?

[dapper bloom]

Mmm not really if you want to factor it

[final girder]

Simplify

I should've specified that huh

[dapper bloom]

You saw 81 is divisible by 3 earlier right?

[final girder]

Yes

81 is 27*3

[dapper bloom]

So now find a way to factor a 3 from 81

Yes

[final girder]

27*3 is 9

[dapper bloom]

So 243=27(3)(3)

[final girder]

9 is3*3

[dapper bloom]

So 243=9(3)(3)(3)

So 243=3(3)(3)(3)(3)

Or just 3^5 ofc

[final girder]

How even

[dapper bloom]

What?

[final girder]

I forgot something

Had to find highest square

Which is 81

[dapper bloom]

Yeah, or you can just do 3^5=3(3^4)

[final girder]

243 is root 81 x root 3

[dapper bloom]

Same thing

[final girder]

So 9 root 3

[dapper bloom]

Mmm

Why the root

243 is NOT sqrt(81)sqrt(3)

Oops

[final girder]

It is

[dapper bloom]

No it isn't

[final girder]

Root 81 x root 3 = root 243

[dapper bloom]

Okay but that is different from saying 243=sqrt(81)sqrt(3)

Which is what I was disagreeing with

[final girder]

Oh yeah

My bad lool

[dapper bloom]

I agree with this tho

You see how 81 is a square too?

[final girder]

Yes

Forgot

Thanks guys .close

.close

[obtuse pebble]

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[dapper bloom]

Ah

I was gonna say

You can think of square roots as 1/2 powers if it helps

[obtuse pebble]

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[timid silo]

I’d appreciate if someone could help for these 2 questions.

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

Apparently no one likes to take the challenge :/

[obtuse pebble]

@timid silo Has your question been resolved?

[novel knoll]

Because you haven’t shown any effort on your own

Or what you don’t understand

[timid silo]

It’s my final exam due to tomorrow and I was so busy with the others. So, to be honest, I have no idea at all. I can’t solve it on my own I just need someone to solve on paper and explains the steps. I’m sorry if I sound selfish but it is what it is..

[tardy epoch]

no one's gonna do your final exam since it's against the rules

maybe hire a tutor

also no one's gonna do a whole problem for you

[timid silo]

She’ll ask something similar. It’s not my actual final exam.

She said she would just change the numbers.

.close

[obtuse pebble]

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[frail karma]

Hi i have a question about trig

How do we write cos(2 cos^-1(X)) in algebraic expression in x

[frail karma]

Is that like simplify?

[gilded needle]

yeah i guess they want you to eliminate the trig functions

hint: do you know a double angle identity for cosine?

[frail karma]

Yea

cos 2x= 1-2sin^2x

cos2x=2cos^2x-1

[errant zinc]

Assume cos^-1(x) = theta or in other words cos theta=x
And you need to find cos (2theta) which can be done using the formula you stated above

[frail karma]

oh

cos(2 cos^-1(X))= 1-2sin^2(cos^-1(X))

[frail karma]

Like that?

[errant zinc]

Use the other one

[frail karma]

cos(2 cos^-1(X))= 2*cos^2(cos^-1(X))-1

[solar trellis]

Yes because you can easily simplify cos(cos^{-1}x)

[frail karma]

I am still confused

[solar trellis]

How would you simplify that?

[frail karma]

X

It will be just x

[solar trellis]

Right so plug that in

[frail karma]

2*cos(x)-1

[frail karma]

Am i doing it right?

[solar trellis]

No

[frail karma]

I am not sure how

[solar trellis]

Understand this expression

[frail karma]

I still dont get it

I read it again

[frail karma]

Can you help me more

Been stuck on it for a while

[solar trellis]

Ask yourself if you really understand what the rhs of this equation means; if you can say it in words

Maybe think about how you would evaluate that for some numerical value of x

[frail karma]

Can you write the steps down on a paper

I am getting lost

[solar trellis]

Nah

[frail karma]

ok

This is not helping me

[obtuse pebble]

@frail karma Has your question been resolved?

[obtuse pebble]

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[flint silo]

OOO HOO AH HOO OOOO OH HOO OOOO AH HOO AH OOO OOH EEK HOO OOH OOOO OOH AHOOO HOO AH HOO OOOO OH HOO OOOO AH HOO AH OOO OOH EEK HOO OOH OOOO OOH AHOOO HOO AH HOO OOOO OH HOO OOOO AH HOO AH OOO OOH EEK HOO OOH OOOO OOH AHOOO HOO AH HOO OOOO OH HOO OOOO AH HOO AH OOO OOH EEK HOO OOH OOOO OOH AHOOO HOO AH HOO OOOO OH HOO OOOO AH HOO AH OOO OOH EEK HOO OOH OOOO OOH AHOOO HOO AH HOO OOOO OH HOO OOOO AH HOO AH OOO OOH EEK HOO OOH OOOO OOH AHOOO HOO AH HOO OOOO OH HOO OOOO AH HOO AH OOO OOH EEK HOO OOH OOOO OOH AHOOO HOO AH HOO OOOO OH HOO OOOO AH HOO AH OOO OOH EEK HOO OOH OOOO OOH AHOOO HOO AH HOO OOOO OH HOO OOOO AH HOO AH OOO OOH EEK HOO OOH OOOO OOH AHOOO HOO AH HOO OOOO OH HOO OOOO AH HOO AH OOO OOH EEK HOO OOH OOOO OOH AHOOO HOO AH HOO OOOO OH HOO OOOO AH HOO AH OOO OOH EEK HOO OOH OOOO OOH AH

[candid sigil]

Cool

[flint silo]

OOO HOO AH HOO OOOO OH HOO OOOO AH HOO AH OOO OOH EEK HOO OOH OOOO OOH AH

[candid sigil]

What is your question

❓

[versed cave]

@flint silo I'm not sure, have you tried with wolfram?

[flint silo]

OOH?

[versed cave]

wolframalpha

[flint silo]

AHH OHH EEK AHH?

Ah, yes, i am starting to understand. Thank you for your guidence

i shall be removing myself from this situation now

thank you

[versed cave]

If you have more questions feel free to ask, otherwise .close the channel

[obtuse pebble]

@flint silo Has your question been resolved?

[obtuse pebble]

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[shut island]

$$\sum{infinity}_{k=0}*(1/2)^k$$

[shut island]

...

[boreal condor]

sum of infinite geometric?

[shut island]

Yeah

[cerulean tulip]

Ur trying to evaluate it right

[boreal condor]

u remember the formula for that?

[solid sun]

$$\sum_{k=0}^\infty(1/2)^k$$

[warm shale]

BenJr

[shut island]

Exactly

thx

[solid sun]

Try writing the partial sum.

[shut island]

I know how it works but not with infinity...

[solid sun]

Let S = 1 + r + r^2 + … + r^(n-1) + r^n

[shut island]

so with ininity being n is easy and I understand it

So for example $$\sum_{k=0}^n(1/2)^k$$

[warm shale]

A random bird

[solid sun]

If you multiply S by r, we have
rS = r + r^2 + … + r^n + r^(n-1).

[shut island]

Yeah

[solid sun]

So you can solve for S then take the limit.

[cerulean tulip]

Thats the original sum without the 1

[solid sun]

S = 1 + r + r^2 + … + r^(n-1) + r^n
rS = r + r^2 + … + r^n + r^(n+1).

[shut island]

Yeah. Got it

[shut island]

But what is it infinite times? S= 1+r... what is the solution for that?

[cerulean tulip]

Well the rhs is the original sum without the 1

[solid sun]

$$\sum_{k = 0}^\infty ar^k = \lim_{n\to\infty}\sum_{k=0}^n ar^k$$

[cerulean tulip]

bruh

[warm shale]

BenJr

[solid sun]

The rhs without lim is the partial sum of the series. Which is S.

We are trying to find a formula for S so we can take the limit.

The … is just a notation for all the stuff between r^2 and r^(n-1).

[shut island]

Ok

Yeah

[solid sun]

For your series’s your dealing with, a = 1 and r = 1/2.

S = 1 + r + r^2 + … + r^(n-1) + r^n
rS = r + r^2 + … + r^n + r^(n+1).

Observe what happens when you take S - rS.

If you can do that then you can find a formula for S.

[shut island]

can you explain it again? What happens with 1 + r + r² + ... + r^n i do understand but what exactly is rS?

[solid sun]

Let take an example. For n = 5 we have S = 1 + r + r^2 + r^3 + r^4 + r^5. But I can write it as S = 1 + r + r^2 + … + r^5 and it is understood that it means what I previously had.

[shut island]

Yes

[solid sun]

As for S - rS, the reason for that is a lot of terms of S appears in rS.

So when you take the difference it tends up canceling leaving only a few terms left.

That is why I said to observe what happens when you take S - rS.

Let’s use our previous example.

S = 1 + r + r^2 + r^3 + r^4 + r^5
rS = r + r^2 + r^3 + r^4 + r^5 + r^6.

Hence S - rS = 1 - r^6.

So S(1-r) = 1-r^6.

[shut island]

yeah

[solid sun]

S = (1-r^6)/(1-r), which is what we want.

A formula for S.

[shut island]

So the solution would be S = (1-r^(n+1))/(1-r)?

[solid sun]

Yep.

But do you understand why that is.

[shut island]

Yeah

[solid sun]

Good.

Since we have have the formulae for the partial sum of the series, we can take the limit as n goes to infinity.

Compute $$\lim_{n\to\infty}S = \lim_{n\to\infty}\frac{1-r^{n+1}}{1-r}$$.

[warm shale]

BenJr

[solid sun]

Let’s use a fact that is not proved in calculus class, $\lim_{n\to\infty}b^n = 0$ iff $|b| < 1$.

[warm shale]

BenJr

[obtuse pebble]

@shut island Has your question been resolved?

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[bright lotus]

The graphs of y = x^2 and y = (x - 6)^2. Wouldn't only the blue one(y = x^2) open downwards or am I wrong?

[boreal condor]

why do u think it would open downwards?

[bright lotus]

wait

oh mb it's upwards sorry cause I saw the negative

So both are opening upwards

[boreal condor]

yes

[obtuse pebble]

@bright lotus Has your question been resolved?

[bright lotus]

THank you

.close

[obtuse pebble]

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[graceful quail]

I seem to have made a mistake in my unit circle

[graceful quail]

Here they get pi/3

In my circle I got pi/6

[boreal condor]

4p/3 -3pi/3

is pi/3

[boreal condor]

180 degrees is pi radians correct?

[graceful quail]

Yes

[boreal condor]

so then for your reference angle

you go from the x axis

[graceful quail]

That is 9pi/6 not 4pi/6

[boreal condor]

to your terminal side

are you looking for cos and sin of 4pi/3?

[graceful quail]

Yes

[boreal condor]

reference angle always starts from the x axis

[graceful quail]

Wait are you saying that I reversed the subtraction?

[graceful quail]

Wdym?

[boreal condor]

why are you subtracting 8pi/6?

ignore my fail of an attempt at writing

[graceful quail]

(3pi / 2) * 3 allows me to evaluate for theta by allowing me to use the common denominator of 6 to subtract the two values

[boreal condor]

but why are you trying to subtract 3π/2?

what are you trying to get from that

[graceful quail]

This was a successful attempt at one of these problems

[boreal condor]

ah now i see why

ok when u have stuff with π/4 or multiples of it

[graceful quail]

I think I confused the hell out of myself lol

[boreal condor]

yeah you kinda did

the proper way to find the reference angle is

5π/4 - 4π/4

=π/4

you only use the x axis

for reference angles

never the y axis

[graceful quail]

What did I do

[boreal condor]

3π/2 is the y axis

you found the distance to the y axis

when you need to find the distance to the x axis

see in my screen shot

how i used the x axis

even though the terminal side is closer to the y axis

[graceful quail]

That’s the leg right

[boreal condor]

yeah

[graceful quail]

I have it connected to the x axis

[boreal condor]

it's cause you did 3π/2 - 4π/3

when looking for reference angles

you are subtracting from your original

[graceful quail]

So the other way around

[boreal condor]

think of your starting side

sec let me draw smth

ok so your angle is 4π/3

you know that 180 degrees is pi

so you reach the blue line

make sense so far?

@graceful quail

[graceful quail]

I understand I think

[boreal condor]

do you agree that π radians takes you to the blue line from starting from initial side?

[graceful quail]

Of course

[boreal condor]

so then

how many radians

are left

of 4π/3

after traveling π radians

[graceful quail]

1pi/3

[boreal condor]

so then what is your reference angle?

[graceful quail]

Oh shit

Wait what lol

So it has to be subtracted in terms of the x axis

[boreal condor]

yes

exactly

[graceful quail]

The original angle has to be subtracted from the x axis

That is because we draw the triangle toward the x axis

[boreal condor]

yeah

so then what are the cos and sin vals

[graceful quail]

I’m going to retry it thank you

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

not sure how the last term changed throughout the problem

[high lily]

exponent laws were applied

[obtuse pebble]

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[obtuse pebble]

@polar mango Has your question been resolved?

[obtuse pebble]

@polar mango Has your question been resolved?

[tardy epoch]

Define what it means for a step to minimize the function

And what descent direction means

f_k also needs to be defined

What is the cross section of the function in that direction?

[obtuse pebble]

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[obtuse pebble]

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[frigid jasper]

first, i think this is better suited for the physics server from #old-network. second, is this a test?

[rough bough]

this is a maths questiion

[frigid jasper]

seems very much like physics.

[rough bough]

this is from a mathematics paper

[frigid jasper]

dont think a maths paper would ever ask you to apply F = ma to a "system of particles"

[rough bough]

ii my syllabus it does

[obtuse pebble]

@rough bough Has your question been resolved?

[gilded needle]

well what do you get when you write down the forces acting on each of the objects

[frigid jasper]

@rough bough you there?

[obtuse pebble]

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[obtuse pebble]

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[leaden shoal]

how?

[high lily]

how what

[leaden shoal]

how do i get rid of the x from the denominator?

[high lily]

so the above is what you attempted?

[leaden shoal]

hmm wait

[high lily]

not sure if you were asking about how to solve or what happened in the imnage

[high lily]

yes

[leaden shoal]

ok ty

.close

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[high lily]

thatworks

[leaden shoal]

tysm

[obtuse pebble]

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[cold edge]

S={1,2,3,4,5,6}
how many subsets of S with cardinality 2 have atleast 1 odd integer

[cold edge]

(6C2)=15
so there are 15 subsets with cardinality 2

what do I do now

[patent owl]

you can subtract the number of subsets that do not contain at least 1 odd integer

[cold edge]

alr let me try

is it 12?

Set of even integers = {2,4,6}
subsets with cardinality 2 = 3
15-3=12?

[drifting wraith]

correct

[cold edge]

ah,ty

.close

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[wanton dagger]

can I get help with this question, thanks

[wanton dagger]

So far I have this information

[knotty crow]

you see it's arithmetic sequence?

[wanton dagger]

Yep

[knotty crow]

it's good

now solve Tn > 0

and take the smallest natural n

[wanton dagger]

Hmm ok I’ll try

So this?

[knotty crow]

I meant:
$$-86+(n-1)3>0$$

[warm shale]

Modus

[wanton dagger]

Oh ok

Oh makes sene ok I’ll do that

Ok I got this

Meaning n=30

[knotty crow]

yes n > 89/3 and since n is a natural number we start from n = 30

so for n >= 30 terms of the sequence are positive

[wanton dagger]

I see

That makes sense now

Thanks!

.close

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[left hazel]

Can i get help with d?

[left hazel]

Im supposed to express it in gradient/slope intercept form. Afterwards im supposed to write down the gradient and y intercept

[hexed agate]

Ya

[left hazel]

Yeah but idk how to do that lol

[hexed agate]

Do you know the slope intercept form?

[left hazel]

Ik it has something to do with y=mx+b

[hexed agate]

Yes

[left hazel]

But other than that idk much else

[hexed agate]

Isolate y

[left hazel]

What do i do with the 6

[hexed agate]

If I ask you to make y the subject can you do that?

[left hazel]

Is it like 3y=-2y+6+0

Is it something like that

[hexed agate]

Yes

[left hazel]

OOOH

Okay tysm

[hexed agate]

2x tho

[left hazel]

Okok

I typed wrong

Tysm

[hexed agate]

And you have to move the 3 also

[left hazel]

For that

Oh yeahyeah okok

Ty

.close

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[wise ether]

A vendor has learned that, by pricing pretzels at $1.50, sales will reach 91 pretzels per day. Raising the price to $2.25 will cause the sales to fall to 58 pretzels per day. Let y be the number of pretzels the vendor sells at x dollars each. Write a linear equation that models the number of pretzels sold per day when the price is x dollars each.

[wise ether]

Ok ok

I tried to make this slope intercept form

Y=mx+b

I feel like when I do y2-y1/x2-x1

So 58-91/2.25-1.5

Which is -44

I feel like I’m doing something wrong if it’s negative

I get y=-44x as my answer

[fierce lagoon]

Use point slope form

Then convert to slope-intercept

[wise ether]

Oh ok

[obtuse pebble]

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[timid silo]

The 1.84

Sorry I'm sending better pictures

Is it a textbook mistake?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[pseudo swift]

It's not

Have you understood why in the meantime ?

[timid silo]

No..

I'm sending a picture

[pseudo swift]

we have $\theta + \frac{\pi}{2} = -\theta + 2k\pi$

[warm shale]

aPlatypus

[pseudo swift]

if we rearrange that we get $2\theta = -\frac{\pi}{2} + 2k\pi$

[warm shale]

aPlatypus

[pseudo swift]

yeah sounds like they have a typo indeed

[timid silo]

Yeah it misses a -

Thanks, at least I know I'm not losing it

.close

[obtuse pebble]

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[lean comet]

i am going nowhere with this

Any idea how to solve?

[pine sail]

Sure.

Just compare th first four.

First

Can you do that?

A good way to do that would be making their denominators same.

Denominators of the exponents.

[tardy epoch]

more rigorously, you'd have to show

$x^{1/x}$ is a decreasing function on $[n,\infty)$ for some $n$

[lean comet]

1, 1.414, 1.442, and I think fourth root of four is the same as sqrt2

[pine sail]

True.

So just compare sqrt 2 and cube root of 3.

[warm shale]

riemann

[lean comet]

So the answer will be cbrt3?

[pine sail]

No cube root of 3 is greater than sqrt 2.

[pine sail]

Now you'll have to do this.

Which you should have noticed by now,

Happens to be ||...||

You tell me

Lol.

[lean comet]

Okay I'll try that out! Thank you!

.close

[obtuse pebble]

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[sleek valley]

here we go again

[sleek valley]

@pseudo swift

[pseudo swift]

Re

Well I'm gonna start with what I suggested before the channel closed

[sleek valley]

"let's think of something a bit simpler w/o implicit differentiation
let's say I want you to find the tangents of y=x²+2 which pass through (0,0)
how would you do it ?"

[pseudo swift]

Yeah this

[sleek valley]

I would do the following:

,tex $y' = 2x + 2$

[warm shale]

Hairo

[sleek valley]

,tex $y' = \frac{dy}{dx}$

[warm shale]

Hairo