help-27

[hasty crane]

Yeah?

[tame matrix]

When differentiating

[hasty crane]

Wait you mean

[tame matrix]

Yes

[hasty crane]

This is

What happened

[tame matrix]

Yes

[hasty crane]

OHHHHHHH

https://tenor.com/view/drake-computer-notebook-explaning-gif-21152267

I appreciate it thank you

My brain farted on that one

Are you okay b?😭

[rigid pewter]

i am thanks for asking😄👍

[hasty crane]

Alright bet

[hasty crane]

I’m gonna close now

[hasty crane]

🗣️

.close

[devout snow]

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[lament schooner]

how did this become this

s_n = a-ar+rs

[upper schooner]

Factor the left hand side

[lusty sapphire]

factor this, specifically

god that's blurry

Factor $s_n-rs_n$

[woven radish]

SWR

[devout snow]

@lament schooner Has your question been resolved?

[lament schooner]

im duymb asf

preschool math is too hard

ty

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[viscid goblet]

I know how to solve this but I am not sure about my answer

[safe fractal]

Well what's your answer?

[viscid goblet]

199.5 , 340.5

degrees

Also

Is there a faster way of doing this other than guessing it

[lusty sapphire]

!show

[devout snow]

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

[safe fractal]

Can you use a calculator?

[viscid goblet]

yes

I just use random numbers

[safe fractal]

Does your calculator have a sin button?

[viscid goblet]

and try to get closer and closer

[viscid goblet]

yea it's Casio 991ex

[safe fractal]

Does your calculator have a shift button?

[viscid goblet]

yes

[safe fractal]

shift then sin then type the value you have (-1/3)

This is called the inverse sin function

[viscid goblet]

cool

let me try

[safe fractal]

It'll only give you one solution

you have to figure out the other one on your own

though it's not difficult if you know how to get to it

[viscid goblet]

I just guess a number

that equals -0.33

[safe fractal]

No guessing involved

[viscid goblet]

is there a way?

[safe fractal]

Yes

[viscid goblet]

show pls

[safe fractal]

I will show you da wae

But first, what did you get in your calculator?

[viscid goblet]

-19.471

for Sin^-1(-1/3)

[safe fractal]

So that's your first answer then

[viscid goblet]

yep

how to find the other

[safe fractal]

Sketch

There are technically infinitely many answers here, but let's focus on the ones in the first period

Since your answer is negative add 360° to it

[viscid goblet]

yea

[safe fractal]

To put it into the first period

Don't mind my artistic skills

So you now have the blue angle

That's the first solution

how to get the green one?

[viscid goblet]

no don't worry g

[viscid goblet]

is it on the same angle from the x line

or the y line

like angle of Blue = angle of green line?

[safe fractal]

No it's not

[viscid goblet]

ohh

[safe fractal]

Blue angle is clearly larger

[viscid goblet]

yeah you mean if we start from postive side x

ok got it

[viscid goblet]

these 2 points lie on a circle and they both touch the line y=-1/3 right?

[safe fractal]

ye

[viscid goblet]

if we use a protractor we can find the angle

is there a mathmatical way to find the angle?

[safe fractal]

Well you have to calculate it

Yes

One thing to note here is that these 2 angles are the same

Oh is that what you meant before? Sorry if I missunderstood you

[viscid goblet]

yea that

but I understood the idea

it's like circle theorem

like in nature

[safe fractal]

Anyway, the green angle would be 180° + yellow right?

[viscid goblet]

yes

[safe fractal]

and the blue angle is 360° - yellow right?

so since you know the blue angle, you can calculate yellow

[viscid goblet]

yes for sure

[safe fractal]

and from yellow get the green angle

[viscid goblet]

ohhhhhh

that's genius

no more guessing

yeah my answers were correct before

but now I can prove them

@safe fractal thanks g

[safe fractal]

ye, what you did before was a numeric method

congrats

you invented computer maths

[viscid goblet]

I found this harder than differentiation

that image you drew is what clicked for me in the end

@safe fractal

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[wooden cairn]

can u get some help plz with a markov chain question

[wooden cairn]

im rly stuck

[hot pike]

show

[wooden cairn]

have u done markov chains before?

[hot pike]

kinda

[wooden cairn]

lmk if u understand it then ill ask u my dilemma

[hot pike]

yea sorry this is too complex

i just know the basics

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[pale orbit]

how do u do part a

[lost laurel]

I swear someone asked this just a while back

use the identity $(2a+b)^2$

I think

[woven radish]

ƒ(Why am. I here)=I don't Know

[faint hearth]

Also can use: AM≥GM

[faint hearth]

And for the two variable case, the AM-GM inequality can be proved using this fact

[balmy slate]

$(2a+b)^2 >= 0$

[woven radish]

Z3ITRA

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[obsidian drum]

What percentage of the distribution falls above D7?
a. 7%
b. 30%
c. 70%
d. 90%

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[ionic swift]

Hey everyone, on the internet I didnt find any clear answer. What is $log^2 n$ normally ? Is it $log(n) \cdot log(n)$ or rather log(log(n))?

[woven radish]

barış

[ionic swift]

It should be the first

But I suck at logs

So I am not sure :/

[upper schooner]

"usually" (i.e. from my experience) it's used to refer to (log(n))^2, though of course there may be some places where it's the latter that's meant

[ionic swift]

Alright. Thank you very much!

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[devout snow]

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[dusty island]

l

sdl

lg

ldll

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[stone thorn]

help pls

[pastel pasture]

what have you tried

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[waxen grotto]

i need help with something simple

[waxen grotto]

300 x 0.25

im not getting the correct numbers

I understand 300 times 0.25 = 75 but i need help manually doing it correctly

[regal moat]

0.25 = 1 / 4

300 * 0.25 = 300 / 4

300 /4 = 75

[devout snow]

@waxen grotto Has your question been resolved?

[waxen grotto]

i was taught to frame the question like this:

300
0.25

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[shell heron]

$\vec{\nabla} \dot \vec{\nabla}\cross f = 0$

[woven radish]

Mephisto

[shell heron]

no idea how to fix this... it's basically div(rot(f)) = 0

can this identity by interpreted as
the rotation around a point, cannot change instantaintiously, bu changes gradually instead?

[devout snow]

@shell heron Has your question been resolved?

[hot steeple]

$\vec{\nabla} \cdot \vec{\nabla} \cross f = 0$

[woven radish]

Katharine

[shell heron]

thanks

[hot steeple]

although i don't think this is correct

as i don't know what the curl of a function is

i mean a function $\rightarrow \mathbb{R}$

[woven radish]

Katharine

[hot steeple]

aka scalar

Do you mean $\vec{\nabla} \cross \vec{\nabla} f = 0$?

[woven radish]

Katharine

[hot steeple]

or maybe $\vec{\nabla} \cdot \vec{\nabla} \cross \vec{A} = 0$?

[woven radish]

Katharine

[shell heron]

@hot steeple my bad

It should be that the rotor of the gradient is always 0

no woops

wrong again

divergence of the curl of a vector field is 0

that's the identity

[shell heron]

so this

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[shell heron]

.close

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[frosty gyro]

A cyclic quadrilateral has length 25, 39, 52, 60 in order, what is the circumdiameter?

[frosty gyro]

So I tried to use ptolemy's and got AC*BD = 3640 but don't think there is a next step

I found the area with Brahmagupta's and got 36*49 = 1764

I tried to then use the center to form 4 isoceles triangles but I don't think that works because the angles are unknown

Not sure what to try next

[low coyote]

Do you know these formulās?

[frosty gyro]

I know it but I don't want to bash it and we didn't learn it in class so I can't use it

only things I can use are Ptolemy, Brahmagupta, and the quadrilateral with the largest area with a certain sequence of side lengths is always cyclic

[low coyote]

I'm afraid the formula of the radius from the data you have is the above so I can't see a simpler way. It could only be a simpler way if there were some symetries in your special case, but I didn't check

You could also perform the proofs of the formula yourself, it takes time though

[frosty gyro]

I mean the side lengths look kinda purposeful

[low coyote]

Yes maybe, I don't have a piece of paper atm sadly

[frosty gyro]

it's alright, thank you

[dense lynx]

it's good you noticed

consider the significance of those numbers and you will probably see your way forward

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[full wind]

I need help with some Desmos restrictions.
The equation for the circle right now is: $(x-2.56)^2 + (y-2.96)^2 = 0.7^2 { x \ge 1.934$ } { y \ge 2.384 } $.
Without the restrictions it is: $(x-2.56)^2 + (y-2.96)^2 = 0.7^2$

[woven radish]

HelloLife

I need help with some Desmos restrictions.
The equation for the circle right now is:  $(x-2.56)^2 + (y-2.96)^2 = 0.7^2 { x \ge 1.934$ } { y \ge 2.384 } $.
Without the restrictions it is: $(x-2.56)^2 + (y-2.96)^2 = 0.7^2$
```Compilation error:```! Missing } inserted.
<inserted text> 
                }
l.50 ....56)^2 + (y-2.96)^2 = 0.7^2 { x \ge 1.934$
                                                   } { y \ge 2.384 } $.
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```

[rustic jetty]

so what is the problem?

[full wind]

i want the circle to be cut above the line but i cant get all of it

also laTeX does not like { }

[rustic jetty]

what is the equation of the other curve?

[full wind]

$(x+0.1)^2+(y+0.1)^2=3.94^2$

[woven radish]

HelloLife

[full wind]

i also have this as a restriction: y >= 0.37

[rustic jetty]

try this for the restriction instead

0.869141x + y \ge 4.95492

[full wind]

thank you

[rustic jetty]

np

[full wind]

.close

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[bitter halo]

The equation for the tangent plane to the graph of the function

$\f(x, y) = 3 + \frac{x^2}{16} + \frac{y^2}{9}\$

is given as $3x - 4y + 6z = 6$ at the point $(a,b,c)$ where $c = f(a,b)$.$\$

1. find the point $(a,b,c)\$
2. calculate the linearization of $f$ at $(-4,3)$.$\$

[woven radish]

milanesa de pollo

[low coyote]

Ask your question. We are not solvers.

We are feeling people

[bitter halo]

how do I do 1

,w D[3 + (x^2)/16 + (y^2)/9 , x]

[jaunty mantle]

If the point is on the graph of the function and on the plane at the same time it must satisfy both equations at the same time

[bitter halo]

yes

[woven radish]

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[bitter halo]

,w D[3 + (x^2)/16 + (y^2)/9 , y]

[woven radish]

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[bitter halo]

help

I need to clear z

$3x -4y +6z = 6 \
6z = 6 -3x +4y \
z = 1 -\frac{x}{2} + \frac{2y}{3}$

[woven radish]

milanesa de pollo

[bitter halo]

,w solve x/8 = -1/2

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[bitter halo]

,w solve 2y/9 = 2/3

[woven radish]

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[bitter halo]

if $c = f(-4,3)$

[woven radish]

milanesa de pollo

[bitter halo]

if $c = f(-4,3)$ then $\\implies f(-4,3) = \left(3 + \frac{(-4)^2}{16} + \frac{3^2}{9} \right) = \left(3 + 1 + 1\right) = 5$

its something like this for the linearization

,, L(-4,3) = f(-4,3) + f_x(-4,3)(x - (-4)) + f_y(-4,3)(y - 3)

[woven radish]

milanesa de pollo

milanesa de pollo

[bitter halo]

hopefully I am not just yapping

,, L(-4,3) = f(-4,3) + f_x(-4,3)(x - (-4)) + f_y(-4,3)(y - 3) \
f(-4,3) = 5 \
f_x(-4,3) = \frac{-4}{8} \
f_y(-4,3) = \frac{2 \cdot 3}{9} \
L(-4,3) = 5 + \left( \frac{-4}{8}\right)(x - (-4)) +\left( \frac{2 \cdot 3}{9}\right)(y - 3) \
L(-4,3) = 5 + \left( \frac{-1}{2}\right)(x +4) +\left( \frac{2}{3}\right)(y - 3) \

[woven radish]

milanesa de pollo

[bitter halo]

,w simplify 5 + (-1/2)(x+4) + (2/3)(y-3)

[woven radish]

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[bitter halo]

<@&286206848099549185>

[devout snow]

@bitter halo Has your question been resolved?

[devout snow]

@bitter halo Has your question been resolved?

[devout snow]

@bitter halo Has your question been resolved?

[devout snow]

@bitter halo Has your question been resolved?

[nova glen]

whats your question

your work for linearization looks alright to me

[bitter halo]

find the point a b c

[nova glen]

okay well you have a system of equations

try to find expressions for x, y, and z given the constraints

youll quickly see only one set of values makes sense

[bitter halo]

how to do that

[nova glen]

just simple algebra

rearrange the eqns

then substitute

[bitter halo]

how to do that can you show me

[nova glen]

consider only the eqn for f(x,y)

set that equal to z

and try to obtain an eqn for x and y

@bitter halo

[bitter halo]

can you draw it

[nova glen]

hold on

$z = 3 + \frac{x^2}{16} + \frac{y^2}{9}$

$z - 3 - \frac{y^2}{9} = \frac{x^2}{16}$

etc etc

[woven radish]

esca | এস্কা

[nova glen]

actually i think its easier if you solve for y first

but do you have a better idea of the process now

[bitter halo]

I dont get it

wdym

[nova glen]

okay lemme give you a hint

first rearrange the eqn of the plane

so that you get z = ...

then, plug in z for the equation z = f(x,y)

[bitter halo]

.

[nova glen]

if you expand and do some simple algebra, you can complete the square

[nova glen]

yep you got there

now eliminate z by substitution

and youll get a problem that involves completing the square

now normally when you have 3 unknowns but only two equations, it has infinte solutions

however, in this case it just so works out that there is only one real solution that makes sense

that was a pretty big hint

[bitter halo]

can you draw it

[nova glen]

wdym draw it

i dont think theres any good visual intuition here other than literally "a plane tangent to a bowl looking thing"

[bitter halo]

but in latex what you mean

completing the square or something

[nova glen]

by completing the square all i mean is youll get some equation involving $(x+a)^2$ and $(y+b)^2$ where a and b are just numbers. you can figure it out with some work

[woven radish]

esca | এস্কা

[bitter halo]

how to do that

[nova glen]

work it out lol

just play around with the numbers

youll get it

[devout snow]

@bitter halo Has your question been resolved?

[devout snow]

@bitter halo Has your question been resolved?

[devout snow]

@bitter halo Has your question been resolved?

[edgy dirge]

rewrite the plane formula as z=g(x,y). Then solve g(x,y)=f(x,y) for x and y, then plug them into the plane formula to get (a,b,c)

[bitter halo]

I rewrote the plane as a function of x and y

ahh you mean like this

first equation

this

,w solve 3 + (x^2)/16 + (y^2)/9 = 1 - x/2 + (2y)/3

[woven radish]

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[bitter halo]

ah I messed up something but dunno where

,w solve 3 + (x^2)/16 + (y^2)/9 = 1 - (-4)/2 + (2*3)/3

[bitter halo]

.close

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[hard cosmos]

Given that a=x+(1/x) solve for x in terms of a

[dawn lintel]

Multiply X on both sides

[hard cosmos]

i got ax-1=x

[dawn lintel]

You'll get quadratic equation

[dawn lintel]

ax-1=x^2

[hard cosmos]

Oh

Then what from there?

[dawn lintel]

Do you know to proceed?

Hmm quadratic formula

[hard cosmos]

Ok

[hard cosmos]

Can u subtract x^2 from both sides

But there’s no 0

[dawn lintel]

There is

[dawn lintel]

ax-1=x^2

Means x^2-ax+1=0

Just substitute the coefficients

[hard cosmos]

Ty!

.close

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[open badge]

can send rules of intregation

[devout snow]

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[golden hollow]

Can someone explain the final part of this question. I’m new to definite integration so I’m not really getting the same answer as solution when I apply f(b)-f(a)

[devout snow]

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[sacred wren]

can someone help me to understand this problem ?

[thin tapir]

Let a be the total white area and b the total black area

And assume that the total area, a+b, is 1

Then 1 is divided in 9 parts

4 are white, 4 are black and 1 has the same proportion as the whole figure, as the pattern is infinite

[green kelp]

it's a sum of infinite GP

[thin tapir]

So the total white area equals 4/9 (because of the white squares) plus the total white area fivided by 9, a/9, because of the square at the centre

So a = 4/9 + a/9

[sacred wren]

4 white and 4 black, so when we calculating the total white area, isn't it be just 4/8 ? how is it 4/9 ?

[thin tapir]

The total white area is the four big squares, plus the four smaller squares, plus another four until infinity

[sacred wren]

are we considering that middle square also ?

[thin tapir]

Yes

[native mica]

the total shaded area is 4/9 +1/9(4/9 + 1/9(4/9 + ......))

[sacred wren]

Now that I understand it, we are going to calculate the shaded area as 4/9 for the first cube. Now for the inside cube, it becomes 4/9^2, and if we even go further, we get 4/9^3 +..... 
But I have a doubt: once we are done with the first outer cube and go inside to calculate the shaded area, are we ignoring the shaded squars of the outer cube and only considering the inner cube shaded region? because the sequence is like 4/9 + 4/9^2 + 4/9^3 + ...

isin't it be like 4/9 + 8/9^2 + 12/9^3 + ...

[thin tapir]

In this case, you are considering on each term only four squares

And at each step the squares are 9 times smaller than the previous time

So it must be 4/9 + 4/9^2 +...

[sacred wren]

If we consider each square at a time, does getting smaller effect the shaded region because its just another square that like we calculate the shaded region of the the first one 4/9?

okay i got it now. its square of equal area

.close

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[sacred wren]

can someone explain me this problem? i don't understand what it trying me to calculate.

[frozen aurora]

the airport had <that many> departures throughout the entirety of 2001

what's the mean number of seconds between one departure and the next one?

[sacred wren]

so should i just add 738+114, to get the total departure and then find out the mean number of second between each departure ?

[frozen aurora]

i think it's a typo

they meant that comma as a decimal separator

they had 738114 departures

although that's quite a lot lol

[sacred wren]

oh oki

ty for the help

.close

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[cedar echo]

So ive got a pythageors theoreom question and ill just show you

[cedar echo]

I need to find a side length but honestly I'm just stuck

I'm going for dinner now ill come back to this in 20 mins

[hybrid snow]

The little lines means they're congruent sides. There's two ways to go about this. Either use the 45-45-90 theorem or let both sides be x

[cedar echo]

Ah alright thanks

[devout snow]

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[limber wren]

hi could someone please help me with this😭 the markscheme requires it to be solved with R=cos(x-a)

[devout snow]

@limber wren Has your question been resolved?

[limber wren]

No 😭

[limber wren]

<@&286206848099549185>

[lunar harbor]

I assume you meant $R\cos(x-\alpha)$

[woven radish]

Civil Service Pigeon

[lunar harbor]

note the maximum of cos x is 1, so the maximum her eis just R

and to find where the maximum occurs, recall that $\cos x$ is maximized at $x=2\pi n$, where $n$ is an integer

[woven radish]

Civil Service Pigeon

[lunar harbor]

then just do your standard transformations fare

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@limber wren Has your question been resolved?

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[misty root]

Hi! I'm in AP Precalculus and I'm currently doing homework for Polar Function graphs, and I'm very stuck on how to deal with finding radius for a polar function. If I could get the example and help on how to find it, I'd be very grateful.

I understand to replace the theta (ex. 8cos(4 • pi/6)), but I've forgotten what to do after that.

(Ignore the 8, j don't know if it's correct.)

[quick peak]

what exactly is your problem

[misty root]

I don't know how to exactly go past this point.

[quick peak]

what does the question say

[devout snow]

@misty root Has your question been resolved?

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[solar geode]

How to solve for x? (geometry)

[soft nest]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[solar geode]

I was gone this day and only have pictures of notes, and this was cutoff

[soft nest]

Do you know the Pythagorean theorem?

[solar geode]

gonna continue in the geo channel

.close

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[devout snow]

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[restive river]

heres my attempt

(its wrong obiously)

the answer is 1/64

but i got 1/16

can someone explain why?

[stone locust]

looks correct to me

[restive river]

oh so the solution (1/64) is wrong

[stone locust]

yeah that's for strictly more than two

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[hexed ocean]

can someone please help me with this

[restive river]

ok thanks

[restive river]

let acceleration be A

you know that Vi = V

and it travels length C+L

[hexed ocean]

what's Vi

oh nvm

[restive river]

V initial

oh and you need time

thats how long it takes the truck to travel L metres

use eqution t = d/v

distance over velocity

the substitue that into

$\Delta d = v_i t + 0.5at^2$

and isolate a

and thats your answer

[woven radish]

Derivative

[hexed ocean]

is there a way to do it without that equation

[restive river]

maybe there is but thats the only way i know

[hexed ocean]

alr

[restive river]

is this physics or math?

[hexed ocean]

maths

[restive river]

ah maybe u need to use derivative

is this calculus?

[hexed ocean]

yeah this section uses chain rule alot

[restive river]

ah

well you know $a = \dfrac{dv}{dt}$

[woven radish]

Derivative

[restive river]

and i told you time is $\dfrac{L}{V}$

[woven radish]

Derivative

[restive river]

for the car it will travel $C+L = V dt$

[woven radish]

Derivative

[restive river]

also you know $dt = \dfrac{dV}{a}$

[woven radish]

Derivative

[hexed ocean]

right

[restive river]

but $t = \dfrac{L}{V}$

[woven radish]

Derivative

[restive river]

i must have made a mistake elsewhere

[hexed ocean]

o

[restive river]

because i need to subtitude all these things

like i dont want $dt$ in my equation

[woven radish]

Derivative

[restive river]

$C+L = \dfrac{V dV}{a}$

ah

look at that

nvm

[woven radish]

Derivative

[restive river]

$a = \dfrac{V dV}{C+L}$

[woven radish]

Derivative

[restive river]

well

whats V dV

hmm

i think this is good bsides the dV

because we have all the elements

i need to get rid of dV

thats the only thing

not sure

[devout snow]

@hexed ocean Has your question been resolved?

[devout snow]

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[restive river]

Hi can someone help me

[dawn lintel]

Which option

[restive river]

All except b if possible please

[dawn lintel]

You can find f(-6) from graph

Hey?

@restive river

[restive river]

hey

[dawn lintel]

.

[restive river]

sorry i didnt think u would respond that quick

[restive river]

not necessarily

[dense lynx]

are you saying f(-6) = 2

[dawn lintel]

Yes

[dense lynx]

that is false

[restive river]

bruh

[dense lynx]

the graph provided is the graph of f', not f

[restive river]

did u even bother reading it

[dawn lintel]

Oh mb

I just saw the graph

Wait a few mins

[radiant dune]

we can write the eqn of that straight line which passes through (-6,2) and (-2,0)

thats your f'x for that interval

integrate with proper bounds/conditions to get f(x)

[dense lynx]

to solve part a, you wish to make use of the second fundamental theorem of calculus

that is,

[restive river]

without an a and b or a f(x) it'll be difficult

[dense lynx]

yep that's why the problem gives you f(-2) = 7

[restive river]

How would that help?

[dense lynx]

since you can replace one of a and b with -2, and replace one of a and b with -6

(you're given f(-2) and you want to find f(-6))

see what happens when you do this and see if you can make progress on the problem then

[restive river]

But it wouldn't necessarily give f(-6)

It would give the area between those two bounds

[dense lynx]

uhh

idk what to say except just try it and see what you get

[dense lynx]

hm

so you plugged in the numbers correctly

though that integrand should be f'(x) not f(-6)

[restive river]

7-f(-6)

What

[dense lynx]

by this equation, on the LHS you are integrating f'(x) from -6 to -2

you replaced the f'(x) with something else but you can't do that since there's no a or b there

[restive river]

So what would I do

[dense lynx]

you integrate f'(x) from -6 to -2

[restive river]

Oh I get it, thanks.

.close

[devout snow]

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[dense lynx]

fascinating

yw though

[devout snow]

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[fair bolt]

Hello, can someone explain me this

[steel patio]

what is the context

[fair bolt]

I need to find the height of the wall

[steel patio]

divide 45 by 15

i think

[fair bolt]

Then?

[rigid pewter]

huh

do you know how to solve a right triangle?

[steel patio]

im in sixth grade

i dont know much

im having trouble with the circumfrance of a circle

i dont understand how to

[rigid pewter]

feel free to open a help channel

[steel patio]

how

[fair bolt]

#❓how-to-get-help

[rigid pewter]

#❓how-to-get-help

wow

[fair bolt]

🙏

[rigid pewter]

this question was for both of u lol

[fair bolt]

I dont

[rigid pewter]

this should be a refresher https://www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-solve-for-a-side/v/example-trig-to-solve-the-sides-and-angles-of-a-right-triangle

[fair bolt]

Oh OK, I remembered

but how do I do it with those two degrees?

[rigid pewter]

the green line is also 40

[brazen meadow]

is the ground flat💀

[fair bolt]

So it have 3 degrees?

[rigid pewter]

uhh can you rephrase that

[fair bolt]

45,40,15

Sorry if I don't explain myself well, English is not my main language.

[rigid pewter]

it's okay

you should think about this as two triangles

pick one and focus on it

[fair bolt]

If I pick the 15° one, what I have to do?

[rigid pewter]

do you know which ratio to use

[fair bolt]

No,sorry this is a new topic for me

[rigid pewter]

ok that's fine (that said i actually do have to go right now😅if nobody else comes to help maybe ping @ helpers)

[fair bolt]

.close

[devout snow]

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[devout snow]

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[visual hazel]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[atomic saffron]

1

[visual hazel]

you have two sides of right triangle

how to get third side

[atomic saffron]

pythag

sqrt5

[visual hazel]

what is cos(2θ) in terms of sin

[atomic saffron]

do not know what you mean

[visual hazel]

you know the double angle identities for cos?

[atomic saffron]

yeah

2cos^2(θ) - 1

[visual hazel]

another one?

[atomic saffron]

ohhhh

cos2x = cos^2x - sin^2x ?

[visual hazel]

another one?

[atomic saffron]

1-sin^2 x

[visual hazel]

you have sin = 2/3

so just put into that

[atomic saffron]

as theta

?

[visual hazel]

yes

[atomic saffron]

ok

[visual hazel]

and what is tan

[atomic saffron]

2/sqrt5?

[visual hazel]

opp/adj

opp is 2

adj is sqrt(5) you just found

so whats tan

[atomic saffron]

why is that needed

[visual hazel]

ok good tan is that

now you have everything you need

[atomic saffron]

okay

[visual hazel]

then just do basic arithmetic

[atomic saffron]

close

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[devout snow]

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[wispy cradle]

question 5 pleaseee

😦

i)

[covert root]

You have made no attempt

[upper schooner]

(do you have any idea of how to begin?)

[devout snow]

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[dusty falcon]

only own non-verified existence fronts the verified.

[devout snow]

Please don't occupy multiple help channels.

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[high ginkgo]

Why are these functions defined with domain X which is a subset of R instead of simply R itself

[ashen fossil]

because domain of every function need not be R

example root(x) → its domain is x>0

[high ginkgo]

but then what guarantees X is a subset of real numbers too

does the definition of addition only holds true for functions with a real domain?

[ashen fossil]

its just the definition of a funtion

[ashen fossil]

no it holds for complex functions too

its just the general representation

[high ginkgo]

well why would we imply unnecessary restrictions

[drifting anchor]

every set of real numbers that you can think of, will be a subset of real numbers
to ensure generality, we say that the domain of a real func. is a subset of R
if we went around making a theorem and say domain is the set of all real numbers, then that definition would break for functions who's domain is not all real numbers, say x^1/2n
and since it wouldn't hold for all types of functions, it's not much of a theorem is it?

so it isnt unnecessary restriction, 'cuz every function have all reals as a domain is not necessarily true

[high ginkgo]

I'm not saying that they should have the set of all real numbers as a domain but if the definition of addition is true for functions with domains which contain non-real numbers then why specify the domain to be just a subset of real numbers. Can we do something like X is a subset of R union (the set of complex numbers)?

[drifting anchor]

oh well they only tend to teach parts of real analysis in high school

you'll find all that in complex analysis

[high ginkgo]

oh

ok

.close

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[tawdry venture]

can someone help me on c

[devout snow]

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[past fern]

Hi, this may seem like a silly question but I'm trying to find the surface area of this 2d donut, the outer circle that's highlighted green. I have the diameter of both the inner circle and outer circle, but I'm not sure how to find the entire surface area of the outer circle in millimetres. Together they make 53mm if that helps

[past fern]

Can I please get some help

[radiant dune]

You know the big one and you know the small one

[past fern]

Yeah I do, I just don't know how to get the total surface area in mm of the shaded green part

Oh wait would the entire surface area just be 13mm

[tame palm]

Area of larger circle minus the area of the smaller circle.

[past fern]

So 53-40 which would be 13mm?

[tame palm]

pi*R^2 - pi*r^2

[past fern]

I'm confused, what would be the radius, would it be say pi53^2 - pi40^2?

[tame palm]

Yes.

[past fern]

The answer is 3789, but what that be in terms of mm? I'm really confused 😭

[tame palm]

When unsure about the units, just place the units inside of the variables and see what you get when evaluated.

[past fern]

I'm even more confused 😭 so pi * (53)^2 = area mm ^ 2? And that's for one?

[tame palm]

Yes. The units will be mm^2.

All you are doing is calculating the area of the entire larger circle which is given by pi * (53)^2 and subtracting the area of the smaller circle pi * (40)^2. The remaining area will be the area of the green donut

[past fern]

so the answer is still 3789, so it would be 3789^2, which is 14 mil...

[tame palm]

3789 mm^2

The numerical value you calculated is calculated separately from the units. At the end, the units should conform with our notion of area which is some unit squared. In this case, it is mm^2.

[past fern]

So 3789mm^2 would be the answer? I'm just not understanding, because that would equate to over 370 cm

[tame palm]

Try converting the radii given from mm to cm and see what you get.

40 mm = 4 cm
53 mm = 5.3 cm

[past fern]

That gives me 37.9

[tame palm]

Correct.

You use a unit that was 1/10th of the previous value. When that difference is squared, you get 1/100th of a difference.

3789/100 = 37.89

And the units change as well, pi(cm)^2 - pi(cm)^2 = Area cm^2

[past fern]

I think I understand, so the area of the green shade would be 37.89mm?

[tame palm]

3789 mm^2 or 37.89 cm^2

[past fern]

But that doesn't make sense, since the diameter of the green shaded circle is 13mm, and is 53mm from one end to the other from the entire circle (rather from the inner to outer circle edge), the hole in the middle is 40mm in diameter, so I don't understand how you'd get 37.89cm^2 from the area of the part that's shaded green

[tame palm]

Ahh, my bad. The diameter of the inner circle is 40mm so its radius is 20mm.

It's too early in the morning for me. 😛

The radius of the larger circle is (40+13)/2 mm.

[past fern]

dw it's too late at night for me i'm so tired and confused 😭

Oh so the area of the green shade is or would be 26.5mm?

[tame palm]

One moment, let me make a graph that will help you visualize the math.

[past fern]

Thank you so much

[tame palm]

https://www.geogebra.org/classic/wfqxn86d

To find the area of the green donut, you calculate the area of the green circle with a radius of 26.5 mm and subtract the area of the red circle which has a radius of 20 mm. The difference in the two areas will the area of the remaining green donut.

[past fern]

I did not understand that 😭 but I tried rooting it, would this be correct? Or the correct formula?

so the answer would be about 11.5mm

[tame palm]

Yes.

[past fern]

Okay thank you so much for all your help!

[tame palm]

No.

Why did you square root it?

[past fern]

no?

[tame palm]

The area of the large green circle is given by pi * (26.5)^2.

[past fern]

I thought bc you squared the number, you needed to root it again to get it back into mm

[tame palm]

The are of the inner circle that is removed from large green circle is given by pi * (20)^2.

[past fern]

Yeah I understood that part

[tame palm]

To find the area of the remaining green donut, you subtract the area of the inner circle that was removed from the larger green circle.

In the graph, the inner circle that was removed is represented by the red circle.

[past fern]

So it would be 2206-1257? And that would give 949

[tame palm]

Yes.

You calculate the area of each of these circles.

When they overlap, you have a remaining green donut whose area can be found by subtracting the area of the red circle from the area of the green circle.

[past fern]

OH yes I see that, but then how would you derive the mm unit from 949?

[tame palm]

Because the radius is measured in mm and it gets squared in the equation for the area of a circle, the unit is squared as well.

[past fern]

So then you'd square root it?

[tame palm]

No, you do not need to square root it.

That is the area of the green donut.

[past fern]

949mm^2 is is 900k I don't understand, that wouldn't be able to fit inside the green area

[tame palm]

The unit itself is squared, not the numerical value which has already been squared.

26.5 was squared in the first line and then multiplied by pi.

[past fern]

So it would just be 949mm, you'd drop the square since the value as already been squared?

[past fern]

I got the same thing, but drop the unit out of the equation. I just used the area equation which I showed before in the ss, got the 2206 and 1256 as well, but why do you need the mm unit apart of the equation?

[tame palm]

If the units are given, the units are required in the solution.

[past fern]

Because with what I did, you have the area in mm, you square it so it's no longer mm, then you root it so it goes back to the unit that it was before which was mm

[tame palm]

You do not square root it.

[past fern]

But the part I'm not understanding is how you could fit 949mm^ in that green area, the diameter from the top of the outer circle, to the bottom of the outer circle is 13mm, so that's 13mm all around. So I don't see how you could fit 949mm^2 into that

[tame palm]

Well ... that IS the area of the green donut.

It may not seem like it, but these two green areas are equal.

[past fern]

But that makes no logical sense, the equation must be wrong, because if that donut with those diameters existed irl, there wouldn't be 949mm^2 in that surface area, it wouldn't be physically possible

Because you're trying to get mm not mm^2

<@&286206848099549185> Would this be the correct equation for calculating the surface area of part shaded in green (in mm)?

[frigid zodiac]

Can u sum up the whole qn pls.

[past fern]

I'm trying to find surface area of the part shaded in green, would this equation be correct in finding it?

[ripe merlin]

why did you take the underroot?

[past fern]

I thought you'd root it so you could get it in mm again

[warm valve]

To convert metre to mm u multiply the length by 1000 but if u work with everything in mm units u will have ur answer in mm units

Or i should say

mm^2

[past fern]

Oh I understand, thank you for the help

.close

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[upper crescent]

I stuck at proving an upper bound for $\beta$ in the following context:

$\alpha,\beta,\gamma,\delta\in\mathbb{R}$ satisfy the following:

If $x\in[0,\frac{1}{4}]$, then $\alpha x\ge 0$.

If $x\in(\frac{1}{4},\frac{1}{2}]$, then $4\beta x+(\alpha-\beta)\ge 0$.

If $x\in(\frac{1}{2},\frac{3}{4}]$, then $4\gamma x+(\alpha+\beta-2\gamma)\ge 0$.

If $x\in(\frac{3}{4},1]$, then $4\delta x+(\alpha+\beta+\gamma-3\delta)\ge 0$.

$3\alpha+2\beta+\gamma=16$

$-2\alpha-\beta+\delta=-16$

Now I am able to show that $\alpha\ge 0$ and $\beta\ge -\alpha$.

For the upper bound, I found $\beta\le 16-2\alpha$ but I was told that it is not the least one. I want to show that $\beta\le \frac{16-4\alpha}{3}$.

Can anyone help please.

[woven radish]

Trenton

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@upper crescent Has your question been resolved?

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[elfin rampart]

How do i solve: For what angles in the interval 0 < v < 90 (degrees) is sin3v < 1/2 true

[covert root]

Let u = 3v

[elfin rampart]

yeah okay i got it

thanks!

made it easier

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@elfin rampart Has your question been resolved?

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[azure widget]

Hello

[azure widget]

Could someone explain?

@everyone.

@everyone

[vapid nest]

bruh

why did you ping?

[wooden veldt]

Explain what?

[azure widget]

How to find q?

[frigid jetty]

a2-a1

[azure widget]

Arithemtic sequence.

[frigid jetty]

Ye

b2/b1

[azure widget]

Becuase when I post, no one answers.

[azure widget]

Could you write it by hand and send me to DM, if not provlem?

[frigid jetty]

Nah

online

b2/b1 to find q

In geometric sequence

[azure widget]

Yes, but how much q is worth?

That is what is unclear to me.

@frigid jetty

[frigid jetty]

What

We don't have b2 and b1

So we can't find

[azure widget]

So the solution is b2/b1, and then we get q, cus we dont have value?

[frigid jetty]

It's just general formula

[azure widget]

So we found the quotient ?

@frigid jetty

[frigid jetty]

Ye...

If it's geometrical sequence

[azure widget]

It is a geometrical sequence.

@frigid jetty

[frigid jetty]

Ye

[azure widget]

So we found the ratio?

B2/b1=q?

@frigid jetty

[frigid jetty]

Ye

[azure widget]

Yes?

I dont know what is Ye.

@frigid jetty

[frigid jetty]

Ye is yes

[azure widget]

Alright thank you very much!

@everyone how do I close?

[lunar harbor]

.close

[devout snow]

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[lunar harbor]

Also the everyone ping doesn’t work

[azure widget]

Thabks.

[devout snow]

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[waxen steeple]

im struggling to negate the cauchy sequence

[waxen steeple]

would it go something

like

$\exists \epsilon > 0$ s.t. $\forall N \in \mathbb{N}, \exists a_n, a_m \in \mathbb{R}$ where $|a_n - a_m| \geq \epsilon$ $\forall n, m \geq N$ ?

[woven radish]

Ashwyn

[stone stump]

not quite

write down the definition for cauchy sequence

[waxen steeple]

$\forall \epsilon > 0, \exists N \in \mathbb{N}$ s.t. $|a_n - a_m| < \epsilon$ $\forall n, m \geq N$

[woven radish]

Ashwyn

[sand dove]

yes, maybe write "forall n,m" before the inequality statement

[waxen steeple]

oh rly

ive always put it after and no one has ever said anything 😭

[stone stump]

so and now take this, turn every forall into exists and vice versa and make < into >=

and thats your negation

[waxen steeple]

$\exists \epsilon > 0$ s.t. $\forall N \in \mathbb{N}, \exists n, m \geq N$ where $|a_n - a_m| \geq \epsilon$

[woven radish]

Ashwyn

[sand dove]

yep

[waxen steeple]

yay

thxthx

.close

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[restive river]

How do I approach further

[devout snow]

@restive river Has your question been resolved?

[dense lynx]

you may consider a substitution

in this case, the substitution is fairly identifiable simply by observing the differential equation you received after performing the reduction transformation

||but as a general note, you may recognize your reduced differential equation as a Bernoulli equation||

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[trail void]

Can someone help me with this? The first part is to determine the function on the left graph (I already got this one, picture below). The second part says to find the function of g(x) which is a transformation of f(x). You have to find a,b,c,d real numbers that satisfy g(x) = af(bx + c) + d.

[trail void]

oh and the function g(x) is shown on the right graph

<@&286206848099549185>

[devout snow]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

@trail void Has your question been resolved?

[trail void]

<@&286206848099549185>

[undone relic]

First find the function g(x)

[devout snow]

@trail void Has your question been resolved?

[trail void]

yeah.. how?

[azure widget]

When I post question, no one answers. I don't mean to disturb your time, but I would appreciate your help.

[trail void]

.close

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[lucid umbra]

Hi! I have a linear algebra related question - how do I find the angle between a plane (ax+by+cz=d) and the y-axis?

[acoustic leaf]

that could be many different angles. it might make more sense to ask for the angle between the normal vector to that plane and the y-axis

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[sonic canyon]

can anyone tell me what this formula is called in english? I know it's related to Bernoulli's distribution, but I can't find what it's called in English.

[graceful cosmos]

That's the binomial distribution

[sonic canyon]

one moment let me confirm

yup that seems right but it annoys me that the english Wikipedia page for it doesn't have this form of it written

either that or it's hard to find on said page

anyways, thanks

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[coral pendant]

does this have a differential at x = 2, my book says it does and accepts it as a correct answer but if 2+ and 2- isnt the same

[wooden veldt]

can you show how theyre not the same?

[coral pendant]

X = 1.9 (2-) 3.8-2 = 1.8 X = 2.1 4.2 - 2+

oh yeah nvm its the same

but im still confused though;

so if left differential and right differential isnt the same how does it have a differential

or am i doing it wrong

[devout snow]

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

[wooden veldt]

what is this supposed to be?

[coral pendant]

differential from the left

2 -

the function is discontinus isnt it

[wooden veldt]

thats just plugging in values close to 2 into x^2 - 2x, not the definition of the derivative

[wooden veldt]

but you also cannot conclude this from only checking 1.9 and 2.1

[coral pendant]

well its asking me if its differentiable at x=2

[wooden veldt]

okay sure you could find out it wasnt differentiable by showing it isnt continuous at that point

[wooden veldt]

but this is still the case

[coral pendant]

so wait the book says its differentiable and i say its not which 1 is correct lol

[wooden veldt]

why do you say its not differentiable?

[coral pendant]

cause its not continuous?

[wooden veldt]

you cannot conclude it isnt continuous just from checking 1.9 and 2.1

for most functions f(1.9) and f(2.1) arent gonna be the same

you need to take values closer and closer to 2 from both sides and see if f(x) is approaching the same value

[coral pendant]

ok lets see 1,99 and 2,01

cuz i dont think its continuous

[frosty cradle]

why not?

[coral pendant]

left and right isnt the same or i am too tired and i dont trust the book cuz it has lots of mistakes and wrong questions

[wooden veldt]

how have you decided that the left and right limit isnt the same?

[coral pendant]

nvm its a turning point

then it does

i forgot its a turning point

[wooden veldt]

2 is not a turning point of x^2 - 2x

[frosty cradle]

it's a turning point?

how did you conclude that

[coral pendant]

2x-2 = 0 at x=2 so inclination is 0 and x square - 2x is a parabolic equation

wait no

fuck

yeah nvm im too tired

im doing everything wrong

[wooden veldt]

you;re trying to determine whether x^2 - 2x is continuous at x =2 or not, this is good

you seem to think it isnt continuous at x = 2 but you're not telling us why

[coral pendant]

hold on

ok nvm i solved it

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@coral pendant Has your question been resolved?

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[vernal solar]

I need help i need to know if the answers on first pic is correct