help-0

[strong flame]

ok but i think i have a small issue

chatgbt isnt working in my country as far as i know

[worn fox]

@strong flame expand Rsin(x+α) and set equal to your original expression to find R, this will tell you your max and min

[lone heart]

@strong flame Has your question been resolved?

[lone heart]

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[chilly dove]

idk where to start

this question confuses me completely

[half stream]

Try setting up equations for each of the statements

So an equation for "the average of the first two marks is 50", "the average of the second and third marks is 75", "the average of the third and fourth marks is 70", and "the average of the first and fourth marks"

Tell me which equations you wrote when you have them @chilly dove

[lone heart]

@chilly dove Has your question been resolved?

[chilly dove]

ye idk how to do that

[hazy ferry]

How about the first mark = A, the second one = B , ....

[chilly dove]

k

yea still idk how to do it

[half stream]

lets say each mark is a, b, c, d

[chilly dove]

yea

[half stream]

(a+b)/2 would be the average of the first 2 marks

the problem says (a+b)/2 = 50, so this is our first equation

[hazy ferry]

$50 = \frac{A + B}{2}$

[half stream]

now, there are 3 other equations that you must form in a similar way

[chilly dove]

ok so a and b should be 50 right

[ocean seal]

DieterBowlin

[half stream]

not quite

[chilly dove]

it could be 50

they both could be 50

[half stream]

a+b has to be 100, so 50+50 works, but so does 49+51, 48+52, ...

[chilly dove]

yea

[half stream]

instead of trying to find what each mark is individually, we can take some shortcuts

[chilly dove]

mhm

[half stream]

but first, try setting up the other 3 equations

[chilly dove]

(b+c)/2 and (c+d)/2

[half stream]

yes, and there is another one. but first, what do each of those equations equal

what are the average marks that those two equations represent

[chilly dove]

i dont really understand what u mean

[half stream]

so in our first equation, we have (a+b)/2 = 50. we need to finish the other equations by putting what they equal to

so (b+c)/2 equals what according to the problem?

[chilly dove]

(b+c)/2=75

[half stream]

yes, now what about (c+d)/2 ?

[chilly dove]

(c+d)/2=70

[half stream]

good, so we have the first 3 equations, but there is one more that relates the average of the 1st and 4th marks to any unknown average. lets just call the unknown average x. can you set up an equation that relates the average of a and d to x?

[chilly dove]

idk how to do that

[half stream]

well, we don't know the average of a and d, so we just call it x. x in this problem is just an unknown value, just like a,b,c,d are unknown values

[chilly dove]

alright

[half stream]

x is what the average of a and d equals

so what is the average of a and d?

[chilly dove]

x

[half stream]

well yes, but what is the full equation?

[chilly dove]

a+d=x

[half stream]

so close, a+d is the sum of the two values, what is the average of the two values?

recall that the average of a group of variables is all the numbers added up divided by the number of variables

[chilly dove]

yea im sorry but ur just confusing me even more

so a+d/x

[half stream]

not quite

(a+d)/2 is the average of a and d

[chilly dove]

yea

[half stream]

we add up a and d and then divide by 2 because there are 2 variables (a,b)

[chilly dove]

mhm

[half stream]

now we need to set (a+d)/2 equal to the variable that we called x earlier, can you set up an equation for this?

[chilly dove]

(a+d)/2=x

[half stream]

good job!

okay, so now we have (a+b)/2 = 50 (b+c)/2 = 75 (c+d)/2 = 70 (a+d)/2 = x

[chilly dove]

yep

[half stream]

the first thing we should do is move the 2 that divides each average

[chilly dove]

mhm

[half stream]

do you know how to do this?

[chilly dove]

not rlly

no

[half stream]

do you have any ideas?

[chilly dove]

do we just divide the sum by 2 if we remove the /2?

idk

[half stream]

very close, i like your thought process

[chilly dove]

thanks

[half stream]

we are already dividing each sum by 2, so we need to do the opposite of division

$\frac{A+B}{2} = 50$

[ocean seal]

JWCfive

[chilly dove]

multiply

[half stream]

good job!

what do we need to multiply by?

[chilly dove]

2

[half stream]

excellent!

now lets do that to each equation. can you convert each equation by multiplying both sides by 2?

(a+b)/2 = 50
(b+c)/2 = 75
(c+d)/2 = 70
(a+d)/2 = x```

here are the equations for reference

[chilly dove]

(a+b)2 = 100
(b+c)2 = 150
(c+d)2 = 140
(a+d)2 = x^2

[half stream]

close

let me put the equation in a different format

$\frac{1}{2}(a+b) = 50$

[ocean seal]

JWCfive

[chilly dove]

i dont see the difference

[half stream]

it is jsut a little easier to see the fraction

okay so say we have an example average 1/2 * (m+n) = 100 (this isn't part of the problem but just for reference). multiplying by 2 on each side gives us 2*1/2*(m+n) = 2*100

[chilly dove]

oh ok

[half stream]

$\frac{1}{2}(m+n)=100\2\times\frac{1}{2}(m+n) = 2\times100$

woops

wait

[ocean seal]

JWCfive

[half stream]

there we go

[chilly dove]

yep

[half stream]

we can simplify this further by writing $\frac{2}{2}(m+n) = 200$

[ocean seal]

JWCfive

[half stream]

now what is 2/2 equal to?

[chilly dove]

just times 2

[half stream]

not quite, remember that anything divided by the same thing is equal to 1

so 3/3 = 1, 5/5 = 1, 8/8 = 1, x^2 / x^2 = 1, so on

this means 2/2 is also equal to 1

[chilly dove]

so it kinda just cancels it self

[half stream]

so we get $m+n = 200$

[ocean seal]

JWCfive

[half stream]

exactly!

good job!

[chilly dove]

do i just apply what u did to the 4 equations

[half stream]

yes

[chilly dove]

2/2(a+b) = 100
2/2(b+c) = 150
2/2(c+d) = 140
2/2(a+d) = x^2

[half stream]

good except for 2 things: first you should go ahead and simplify the 2/2 to a 1

second: x^2 means you multiply x by x, so x*x

[chilly dove]

alright

[half stream]

in this case, we want 2*x

so we get a+b = 100 b+c = 150 c+d = 140 a+d = 2x

[chilly dove]

yep

[half stream]

now for the next step, lets set aside the last equation and focus on the first 3

[chilly dove]

alright

[half stream]

remember that for our solution we need to find a+d and then divide it by 2, so first we need to find a+d

which of the first three equations has either an a or a d?

[chilly dove]

c+d = 140 and a+b = 100

[half stream]

good!

now here we can do a very useful trick

for our solution, we first need to find a+d, so how can we get a and d into one equation from those two equations?

[chilly dove]

im not sure

[viscid pawn]

[half stream]

okay, something special you can do with equations is add them together. to do this, you must create a new equation where each side is the sum of each side of each first equation

[chilly dove]

that confuses me even more

[half stream]

thats alright, it is sort of a sneaky tactic

for this step, we would get (a+b) + [c+d] = (100) + [140]

I put each part from either equation in different brackets so you can see

[chilly dove]

alright

[half stream]

but without the brackets it would be a+b+c+d = 100 + 140 or a+b+c+d = 240

[chilly dove]

alright

[half stream]

now from here, we need to make the equation only have a+d. to do this, we need to eliminate the b+c. in order to do this, we need an equation for b+c. we have an equation like this in the 3 that we got earlier

do you recall an equation that has b+c?

[chilly dove]

yea its just b+c = 150

[half stream]

good!

now, for the previous step, we added two equations together. we can also subtract two equations by subtracting one side from the other side. I'll explain it here with sample equations:

2m+n = 11
m+n = 7
``` to subtract these, we just subtract each side

(2m + n) - [m + n] = (11) - [7]

2m + n - m - n = 11 - 7

2m - m + n - n = 4

m = 4

this is just a sample equation for explaining

do you see what i did?

[chilly dove]

yea

[half stream]

okay, so now please apply this to ```
a+b+c+d = 240
b+c = 150

[chilly dove]

(a+b+c+d = 240) - (b+c = 150) = 240 - 150 = 90

[half stream]

yes, but I wouldn't have the other side of the equals in the parentheses, just write
(a+b+c+d) - (b+c) = 240 - 150 = 90

otherwise excellent

[chilly dove]

alright

[half stream]

from here, we can distribute the - to (b+c) to get ```
a + b + c + d - b - c = 90

which variables can we eliminate?

[chilly dove]

b and c

so we're left with a+d

[half stream]

good! can you write the equation that eliminates this?

[chilly dove]

a+b+c+d-b-c = a+d

[half stream]

yes, and what is the a+d equal to?

[chilly dove]

90

that should be the answer right

[half stream]

good! now remember, the question asked for the average of a+d, which means (a+d)/2

[chilly dove]

yea

[half stream]

right now, we have a+d = 90, so what do we need to do to get our answer?

[chilly dove]

divide it by 2

[half stream]

good, can you write an equation for this?

[chilly dove]

(a+d)=90/2

[half stream]

almost, whenever you do an operation on an equation, you must do it to both sides

[chilly dove]

(a+d)/2=90/2

[half stream]

excellent! now what is 90/2?

[chilly dove]

45

[half stream]

Perfect! that is the answer, but before u go, can I show you another methhod that you might like a little more?

[chilly dove]

i think this one is good enough

[half stream]

Trust me

[chilly dove]

it's 1:30 for me i gotta go anyways

[half stream]

The one that we used is called elimination, and there are a few more, but another one is substitution, where you change the b+c in a+b+c+d=240 to 150 (because b+c=150). doing this, we get a+150+d = 240 and then subtract 150 from both sides to get a+d = 90

But anyways I hope I could help, sorry if some of my explanations were bad

[chilly dove]

no its ok i could understand, thank u very much for helping me

[half stream]

Of course!

[chilly dove]

have a great day 🙂

[half stream]

You too!

[chilly dove]

thanks

[half stream]

I just know u are gonna be a master at math

[chilly dove]

great, you too

[half stream]

thanks! well see you later

[chilly dove]

yep bye

[half stream]

bye

[lone heart]

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[chilly dove]

Can anyone help me on this question?

[hazy ferry]

would be a step to find out the amount of integers first

[chilly dove]

yea

[hazy ferry]

you know how to do that

you've got the sum and the average

[alpine cape]

the sum of all integers (329) divided by the number of integers (X) = average (47)

therefore 47*X=329

which is 7

[chilly dove]

mhm

[hazy ferry]

since one integer is 97 ---> 329 - 97 = 232 (sum of the remaining 6 integers)

[chilly dove]

oh ye i was gonna type the same

[hazy ferry]

and if you want to make one integer as large as possible the other ones must be as small as possible

[hazy ferry]

considering that, the value of 5 of the remaining 6 integers should be straight forward, if you want to make them as small as possible

[chilly dove]

i got the answer

232

[hazy ferry]

wait

what

[hazy ferry]

the other 5 numbers cant be 0

[hazy ferry]

In the question it says "A set of positive integers, each of which is different"

[chilly dove]

oh

yea i've done so much math today my brain is just kinda screwed

[hazy ferry]

its alright

you want to make 5 integers as small as possible

[chilly dove]

mhm

[hazy ferry]

they are all different and positive

so what are the numbers

[chilly dove]

idk tho

[hazy ferry]

you are looking for the smallest positive integer, the second smallest, ...

[hazy ferry]

wait, you know what a positive integer is, right?

[chilly dove]

ye

[chilly dove]

2?

[hazy ferry]

thats the second smallest, the smallest is 1

[chilly dove]

ye but it has to be positive i thought

[hazy ferry]

1 is not positive?

[chilly dove]

wait nvm i got it confused with even

[hazy ferry]

np

so the first two are 1,2,..,..,..,

what are the other 3 ?

[chilly dove]

3,4,5

[hazy ferry]

exactly

so you subtract those from 232

and you have it

[chilly dove]

217

[hazy ferry]

yep

[chilly dove]

alright thanks

[hazy ferry]

np

[lone heart]

@chilly dove Has your question been resolved?

[lone heart]

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[alpine sable]

given 6 jugs, three of which contain 10 black balls and 20 whites, another contains all black, and the last two each contain 85% white balls and the rest are black, we randomly pick a jug and remove a ball, what is the probability that its white?

[alpine sable]

[
\frac{1}{6} \cdot \left(3 \cdot \frac{20}{30} + 1 + \frac{85}{100}\right)
]

[ocean seal]

metnal

[alpine sable]

is my answer correct?

[vale wigeon]

appears so

er

wait

the monochromatic jug is all black balls but you want a white ball

so the 1 in the middle should be 0

[alpine sable]

ohh

so @vale wigeon another question

an investor buys 2 shares today, the probability of the first stock going up at the end of the day is 0.4 and the second 0.7, the probability of neither going up is 0.2

the probabbility of neither going up, does it imply $P(\overline{A \cup B}) = 0.2$?

[ocean seal]

metnal

[vale wigeon]

it doesn't imply that so much as it means that

[alpine sable]

ok so it means that

but

using de morgan

i get $P(A \cap B) = 0.2$

[ocean seal]

metnal

[alpine sable]

but this confuses me, is this correct at all?

how is the probability of neither going up equal to the probbabiblity of both going up

ohhh i know whats wrong

it should be $P(\bar A \cap \bar B) = 0.2$ lol ofc

[ocean seal]

metnal

[alpine sable]

thanks ❤️ .close

.close

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

an investor buys 2 shares today, the probability of the first stock going up at the end of the day is 0.4 and the second 0.7, the probability of neither going up is 0.2

[alpine sable]

\begin{align*}
P(\overline{A \cup B}) &= P(\overline{A} \cap \overline{B})\
&= 0.2\
\end{align*}

[ocean seal]

metnal

[alpine sable]

not sure how to continue

trying to get to $P(A \cap B)$

[ocean seal]

metnal

[alpine sable]

the answer to this is probabbly that $P(A \cap B) = P(A) \cdot P(B) \iff P(\overline{A} \cap \overline{B}) = P(\overline{A}) \cdot P(\overline{B})$

[ocean seal]

metnal

[alpine sable]

but im not sure if this is correct or how to prove it

[viscid pawn]

so the probability of either going up is 1-0.2

i assume it's 0.8

[alpine sable]

yes

how does that help

$P(A \cup B) = 0.8$

[ocean seal]

metnal

[viscid pawn]

isn't that what you're looking for ?

okay no sorry i'll look into it

[alpine sable]

[
P(A \cap B) = P(A) \cdot P(B)
]

[ocean seal]

metnal

[alpine sable]

i need this to prove independence

oh i missed this formula

[ P(A_1 \cup A_2) = P(A_1) - P(A_2) - P(A_1 \cap A_2) ]

how stupid of me

[ocean seal]

metnal

[alpine sable]

.close

[lone heart]

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[stray swallow]

so here the circle is rolling towards the y axis

and it is blocked by the function

[limpid turret]

Oh like the circle is going to collide with it?

[stray swallow]

and i want to find the coordinates of the center

[stray swallow]

yeah right

[limpid turret]

Interesting problem

Okay

Well the two shapes will definitely have to share a tangent when they touch

That's for starters

[stray swallow]

yeah thats for sure

[viscid pawn]

well the problem means that you need to find when f(x) = 2*r

[stray swallow]

but i cant find the x0 where the tangent is defined

[limpid turret]

You can get closer than this

[viscid pawn]

💀 this has to be the exact point of contact tho

or

oh yeah wait sorry

yeah mb

[limpid turret]

What you want to do is create an equation L(x) that describes the distance from the point (x, f(x)) to the line y=4, but traveling orthogonal to the tangent of f(x)

Then find where the equation is 4

This is a first guess anyway

[stray swallow]

yeah but the tangent of f(x) is defined at x0 the point of contact

and i dont know x0

[lone heart]

@stray swallow Has your question been resolved?

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[ruby lion]

I am desperate

I got no idea how to solve this

x and y are known

c b w and r are all unknown

It has to be solvable, it's based on a solvable problem (which cone or cylinder approximates a 3 dimentional curve at a given point)

this feels unsolvable

the issue is that it has to be solvable

sorry I Had forgot about a 9 somewhere, here is 100% the most simple way I can put it

[lone heart]

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[ruby lion]

oops

[sullen meteor]

Maybe isolate the variables like b=1–c-r

Then plug in all those points of b

[ruby lion]

hmmmmmmmm

alright thanks I'm gonna try

[lone heart]

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[gloomy citrus]

question 13 a b c

they are asking for similar things but i got no clue how to do it

part A makes total sense though

[pseudo ice]

What’s the top limit of the integral, sorry? Looks cut off in the pic

[gloomy citrus]

mb

1

[pseudo ice]

(also you’re supposed to wait 15mins before pinging helpers btw)

[gloomy citrus]

soz

[pseudo ice]

Part b looks pretty much similar to part a, but you split it into two bits

[gloomy citrus]

so just half it?

[pseudo ice]

The smaller value looks like that’s from taking the “lower” of both of the rectangles, and the larger one with the taller one

[gloomy citrus]

i dont understand

[pseudo ice]

More work out the area of this blue and purple (sorry for the poor drawing, on phone)

Add them up and you should get something

Do similar for the larger ones if you get me

[gloomy citrus]

righttt

[pseudo ice]

And then c is basically the same idea but just longer and more pain haha

Rah they ask you to do it even more in d as well, they’re really working you

[gloomy citrus]

hold on whats the lenght/height for the lower rectangle

should i just plug the x coordinate into the formula

?

[pseudo ice]

Yeah that’s what you do, the x coordinate where it meets the graph (if that’s explained clear?)

[gloomy citrus]

are you sure these graphs are the same

[gloomy citrus]

how do u do the third one 😭

[pseudo ice]

They look to be the same (bearing in mind that they’re not to scale and all)

[pseudo ice]

Create three pairs of rectangles with width 1/3 between 0 and 1

[gloomy citrus]

in this graph?

[pseudo ice]

I think so, that’s the bottom diagram, no?

[gloomy citrus]

it is

[pseudo ice]

Let me try at least one of them and see (edit: yes it should!)

[gloomy citrus]

i think i got it

thanks

.close

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[bronze geyser]

Hello!

[alpine sable]

hi

[bronze geyser]

Im not sure on how to further simplify this : (2n^3 - 2n^2) / (4n^5 - 4n^4)

[alpine sable]

$\frac{2n³-2n²}{4n^5-4n^4}$

[ocean seal]

rbit ✨

[alpine sable]

anything u can divide both from top and bottom?

[bronze geyser]

the n ^ 3

and n^ 5

to get n^-2

I thnk

[alpine sable]

better not get any negative exponents

[bronze geyser]

oh

wait

[alpine sable]

stay positive

[bronze geyser]

u can do 2n^3 - 2n^2

to get 2n

and same woith bottem

to get 4n

2n/4n = 0.5

[alpine sable]

no

[bronze geyser]

oh

[alpine sable]

if you have n*n*n - n*n

thats not n

(n+n+n) - (n+n) would be n

what happens if you just divide n from top and bottom?

[bronze geyser]

that will be negative?

[bronze geyser]

Oh yeah

you minus one n

each

I think

[alpine sable]

whats (2n³-2n²)/n?

[bronze geyser]

2n^2-2n

[alpine sable]

so dividing n from top and bottom gives

$\frac{2n²-2n}{4n^4-4n^3}$

[ocean seal]

rbit ✨

[alpine sable]

and can you divide n again?

[bronze geyser]

yess

2n-2/ 4n^3 - 4n^2

[alpine sable]

yes, then, anything else you can divide?

or cancel i should probably say

[bronze geyser]

umm

lemme tihnk

[alpine sable]

$\frac{2n-2}{4n^3-4n^2}$

[ocean seal]

rbit ✨

[bronze geyser]

multibly by -2?

[alpine sable]

why multiply?

[bronze geyser]

because you are dividing

then u oppisite

oh

wait

[alpine sable]

we are doing fractions, we want to cancel things out

[bronze geyser]

actually yeah

[alpine sable]

like, 2/6 = 1/3

there i divided 2 from top and bottom

[bronze geyser]

so divide by 2

yeah

n-1 / 2n^3 - 2n^2

[alpine sable]

$\frac{n-1}{2n^3-2n^2}$

[ocean seal]

rbit ✨

[alpine sable]

anything else?

[bronze geyser]

no

[alpine sable]

ok, then i want you to factor 2n³-2n²

[bronze geyser]

but wait is 2n^2/4n^4 * 2n^2-1 / 4n^4 - 1 = 2n^3 - 2n^2 / 4n^5 - 4n^4?

[bronze geyser]

k

2(n^3-n^2)

[alpine sable]

yeah, now factor out the n² too

[bronze geyser]

2(n^2 - n ) ^2

[alpine sable]

not quite

[bronze geyser]

oh

[alpine sable]

you have 2(n²*n-n²*1) right?

[bronze geyser]

yes

[alpine sable]

so if you put the n² to the front

[bronze geyser]

how

2n^2 ( n-1 )?

[alpine sable]

yes

[bronze geyser]

oh

what about the *'s

okay I gues

[alpine sable]

do you see how it equals 2n^3-2n^2?

[bronze geyser]

yes

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

[alpine sable]

so then the fraction becomes

[bronze geyser]

1 / 2n^2

[alpine sable]

$\frac{n-1}{2n^2(n-1)}$

[ocean seal]

rbit ✨

[alpine sable]

right

[bronze geyser]

yessir

then the n-1 cancels out

and the top is left wiht 1

thakns

.close

[lone heart]

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[stoic folio]

https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles/sine-and-cosine-of-complementary-angles/v/example-relating-trig-function-to-side-ratios

[stoic folio]

Can you guys see this video?

Why would anybody use 3 sides of a triangle to describe an angle? That fills me with much anger and rage!!!

[last ether]

Because not all triangles are right triangles

[stoic folio]

wrong...one of the examples is triangle: DEC. It is a right triangle. But why describe using 3 legs?

[last ether]

I'm not watching the video

You can describe it with trig ratios

But not all 3 are needed

[stoic folio]

If you haven't looked at the video then how can you answer my question?

[last ether]

I ain't gonna watch the whole thing. I am surmise you're asking about trigonometric ratios

If you have any other triangle that's not a right triangle, you'd still have to use trig (law of cosines, usually, or vectors) to describe the angle

We typically use the 3 sides if we need to find the length of something

Or similar triangles

If I wanted to find the length of c, I can use the relationship each side and the angle has with each other

[stoic folio]

You don't even have to watch the whole video to answer my question. You literally have to watch up to the first 20 seconds to see what I'm asking. But you won't. face palm

[last ether]

And I answered it

[proven rune]

J4w4 umbraleviathan is right

[last ether]

Literally calm down lol you're getting stressed and projecting it

[stoic folio]

Bruh, I'm not stressed. My jaw is on the floor how somebody can think they can answer a question about a video they won't watch the first 20 seconds of. 🙂

[last ether]

Because I can guess what the topic is based on your question and the title of the video

I don't need to watch it lol if it's literally basic trig

[stoic folio]

Right, that's all your doing is...guessing.

[last ether]

Well no because I know how to do this stuff already

And based on your question on "describing an angle with 3 sides" it's most likely trigonometric ratios for geometry

[stoic folio]

Right, I guess what keeps going over your head is I actually had a question about the video question, not trigonometry itself.

[last ether]

Perhaps be more specific then

[stoic folio]

Ah, now you want specifics. 🙂

[last ether]

Your frustrations were general to all triangles

[stoic folio]

That's why I posted a link to the video from the get-go.

[last ether]

I don't need to help you if you're gonna be a bitch about this

Like seriously chill

[stoic folio]

To provide context for my question, why you just completely ignored because " Hey I know trig already". 🙂

[last ether]

You asked a general question and a link to a video that describes that general phenomenon, and I answered it with a general answer

And frankly, barely any helpers here are gonna watch a whole video

If you're gonna waste my time, I'll just go

[stoic folio]

Right, general answers that don't explain the video. Thanks for the effort though bro. Come back and help when giving up 20 seconds of your life isn't too time consuming, okay. 🙂

[last ether]

But if you read this, you'll also realize it can be used to prove things

[alpine sable]

i guess you know these are not 3 sides but 3 points right?

[stoic folio]

Agreed, that's what I meant to say. My mistake. After reading some of the comments on the video I see that other students had the exact same concerns as me and somebody answered them because they understood the students question. There is a standard being followed that I wasn't aware of because the teacher doing the videos used it here for the first time.

Up until this video the teacher would label the angles as theta. Now he is labelling them with 3 points and that was new to me.

[alpine sable]

yeah, if it were 3 sides then ofc it would be nonsense, but 3 points can describe 2 sides with one shared point

[stoic folio]

Even still one must know the standard. It's explained as being the middle value that is the angle in reference.

[alpine sable]

right, i dont even know if it has to be clockwise or counterclockwise rn

[stoic folio]

From the comments: "Yes, the middle letter in that notation is the referenced angle by convention :).
<ABC and <CBA both reference the same angle."

[alpine sable]

i would disagree ABC and CBA refering to the same angle if you allow for angles from the outside

[stoic folio]

Which kinda makes sense now that I know the convetion. If you trace 3 points on a triangle the middle point will be where the angle is.

.close

[lone heart]

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[lone heart]

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[wooden jewel]

I understand that I need to use the fundamental th. of calculus with chainrule but I dont understand how the upper bound can be sinx

how does that make sense 💀

[sweet burrow]

when you have numbers they're implicitly functions of any variable as well

the integration is w. r. t. variable t, therefore after finding the antiderivative and plugging in bounds you'll get something that depends on x

[wooden jewel]

im not sure i understand.. but sinx is between -1 and 1 so why can the upperbound be sinx when the lowerbound is 1?

[last ether]

They can be functions

Numbers by themselves can also be functions

[sweet burrow]

the integral from 10 to -5 makes total sense if you think about it

[wary stream]

Basically, the function is in terms of t, hence the current variable. That integral will put it in terms of x

[last ether]

But bounds don't always have to be constants

[wary stream]

They can be shapes if you wanted it to be

[wooden jewel]

wait thats possible?

[sweet burrow]

recall the properties of definite integral

[tacit arch]

If you can integrate 2t-1, then the right side is just a function of x without any integrals

[wooden jewel]

okok

[wooden jewel]

wdym by right side

[alpine sable]

Of the equals sign

[wooden jewel]

Oh alrrr

Ok thanks guys i understand now

.close

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[viral pine]

good mornin USAAA

[viral pine]

how should I have approached this question

i know some of my workings were incorrect

since a is a constant, some of the rules that I’ve applied here wouldn’t actually work

but then idk how to go about doing this

[wary stream]

Technically it's night time in the USA

[viral pine]

[wary stream]

What's the full problem?

[alpine sable]

You should be plugging in x = 2 after differentiating

And you should be differentiating with respect to x instead of a

[viral pine]

Ohh alr

as for the 2nd part of the question

how did they get 2

-2*

[sweet burrow]

just do what the question is saying

find x at which the derivative is equal to 1

[viral pine]

ohh alr

by the way

$\frac{d}{dx}e^(x)^{\frac{1}{3}$

$\frac{d}{dx} {2e^3}$

[ocean seal]

...1

[viral pine]

ik ill need to shift 2 to the front but

do I apply the power rule afterwards to the e^3

[alpine sable]

e^3 is just a constant

so the derivative would just be 0

[viral pine]

O

[sweet burrow]

should you apply power rule to 5^2

[viral pine]

no

[tribal haven]

you can if you're differentiating with respect to the variable 5

[viral pine]

something went wrong

9a

$\frac{d}{dx} {e^[(x)]^{\frac{1}{3}$

[ocean seal]

...1

$\frac{d}{dx} {e^[(x)]^{\frac{1}{3}$
```Compilation error:```! Missing } inserted.
<inserted text> 
                }
l.57 $\frac{d}{dx} {e^[(x)]^{\frac{1}{3}$
                                         
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.```

[viral pine]

sorrow

[green folio]

Ya didn't apply chain rule in d/dx(3e^x^1/3)

[viral pine]

oh I can’t apply the 9th rule to it right away?

[sweet burrow]

it's not e^x

it's e^f(x)

[alpine sable]

Also is this your final answer for a?

[viral pine]

it was

gonna fix it em

rn*

[alpine sable]

U can apply the 9th rule if the variable you are differentiating with respect to is the exponent of e

The general case for this goes n^x differentiated with respect to x is; n^x × ln(n) but since ln(e)= 1 e^x is a special case

[viral pine]

I see

so I should’ve applied rule 10 instead,

?

[alpine sable]

That's when the exponent of e is a function of variable you are differentiating with

[viral pine]

$\frac{d}{dx} {e^[(x)]^{\frac{1}{3}$

[ocean seal]

...1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[viral pine]

[alpine sable]

What ?

[viral pine]

hold on

$\frac{d}{dx} {e^[(x)]^{\frac{1}{3}}$

[ocean seal]

...1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[viral pine]

$\frac{d}{dx} {e^(x)^{\frac{1}{3}$

[alpine sable]

$\sta}\frac{d}{dx} {e^{x \frac{1}{3}}}$

[ocean seal]

...1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[viral pine]

$\frac{d}{dx} {e^[(x)^{\frac{1}{3}]$

[ocean seal]

...1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[viral pine]

ok I’ll just send a pic of it idk

irl 📻

[alpine sable]

Ok wait

[ocean seal]

Vanilla's Social

[alpine sable]

Do u mean this ?

[viral pine]

x to the power of 1/3

not x *1/3

ye

[alpine sable]

Ah! Ok

[viral pine]

x^1/3 is a function right

[ocean seal]

Vanilla's Social

[alpine sable]

Yes it is in the form x^n where n is constant

In this case n = 1/3

[viral pine]

which rule is that

[alpine sable]

So u have function e^f(x) where f(x) = x ^ 1/3

[viral pine]

i forgor to print out the quotient & product rule

[alpine sable]

U don't seem to have it ig, I can't see it their

[viral pine]

o

are there any other rules that I’m missing

aside from these 3

[alpine sable]

Nah dw

There aren't many and u can manipulate a basic function to make it easier

[viral pine]

alright I’ll brb 5 mins

[lone heart]

@viral pine Has your question been resolved?

[lone heart]

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[lone heart]

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[dapper veldt]

given triangle ABC A(5,18) C(1,6) points E is on AC and D is on AB such that DE || BC area of ADE = 18 area of trapezoid DECB = 14. Find E

[raven haven]

show work

[dapper veldt]

I found area of abc is 32 AC is y=3x+3 I tried drawing heights from both B and A to try to see if that'll work so far it hasn't and AC = 4*sqrt(10)

[raven haven]

ok do you have a diagram?

[dapper veldt]

not an accurate one yet but imma draw one

[raven haven]

doesnt have to be accurate

[dapper veldt]

i mean rn I have a random triangle with all the information ik written on it

[raven haven]

sure that works

[dapper veldt]

oh I think you need to use similar triangles

yep

.close

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[bright remnant]

I thought A' meant not in A? Isnt the union part of A? Why is it being included?

[pseudo ice]

“In B, or not in A”

[lone heart]

@bright remnant Has your question been resolved?

[bright remnant]

so if it says in B, the union part is included regardless of it says 'not in A'

[pseudo ice]

Basically you just need to meet one of those. “Not A” is basically everything but that intersection, which the “B” part takes care of

[bright remnant]

oh ok

[bright remnant]

can u help me with 1 more question

[pseudo ice]

Maybe maybe, but in any case,

[bright remnant]

i need help with b

ik (a) is 9 metres

essentially, I know t = 0-4 is 24 metres, but idk how to find the distance of t = 4-5

[pseudo ice]

Guess that the “signed” area between t=4 and t=5 is something you’d work out in integral form and use that s(5)=s(2)

Then part a helps you with that

[bright remnant]

if s(2) = 15, s(5) = 15 right?

then 15 + 9 + 15 = 39?

[pseudo ice]

That is a true statement yes

[bright remnant]

the answer isnt 39 tho

[pseudo ice]

My idea was more that you get that $\int_{4}^{5} v dt$ and find that it’ll look like part a

[ocean seal]

chartbit

[pseudo ice]

Then take account that it’s signed area

Hmmm hang on a moment

[bright remnant]

yeah but i dont know how to get that 😢

[pseudo ice]

Had a brain fart moment, but yeah, this shouldn’t be 15

[pseudo ice]

You can, if you know the relationship between velocity and displacement…

Which I’m sure you do

[bright remnant]

you integrate velocity to get displacement

[pseudo ice]

Yep, and s is our displacement here, do you agree?

[bright remnant]

yep

[pseudo ice]

Shall leave you to think about that in the context of this then…

[bright remnant]

ok i figured it out

ty

[pseudo ice]

Managed to get the right answer in the end then?

And hey @alpine sable, merry Christmas to you ☺️

[keen plinth]

[alpine sable]

Istg I hate mobile discord it shows me typing when there is text already in the chat box

I'm just lurking

[keen plinth]

[alpine sable]

But merry Christmas to you as well chartbit!

[pseudo ice]

Hahaha exposed 😂 was thinking you’d have a better explanation than I did tbh 😂

[alpine sable]

Merry Christmas btw @keen plinth

[ocean seal]

♡LexQa♡

[alpine sable]

[alpine sable]

😵‍💫😵‍💫

[keen plinth]

happy christmas

[pseudo ice]

Happy Christmas to you as well snow ☺️

[pseudo ice]

Haha a lot of the time I’m expecting that I’ve either ended up mixing up explanations or am explaining something in a longer way than it needs to be 😂

[alpine sable]

No man I'm like wasted rn

[pseudo ice]

“And now we know that 2-5=3”

[alpine sable]

Head free, returned to monke

[keen plinth]

[pseudo ice]

I’ve just been feeling fried, and for like no reason

[alpine sable]

The Christmas vibes raaaah 😎

[keen plinth]

deep fried

[pseudo ice]

Faxxxx 😂 tis mad outchea 😂😂

[alpine sable]

[pseudo ice]

Autocorrect I hate you

Some weird choice to change my words to

[alpine sable]

we love autocorrect 🥰

[pseudo ice]

Actually nah gotta take that back, it’s saved my ass more than it’s caused me pain(!)

Thank you autocorrect

Anyways y’all enjoy your day and take it easy 🙋‍♀️

[alpine sable]

You tooo

Also did like

OP die

Should I close this

Oh okay they solved it right

[keen plinth]

shrodingers OP

[alpine sable]

.close

[lone heart]

Channel closed

Closed by @sullen badge

Use .reopen if this was a mistake.

[pseudo ice]

Yeah now I’m wondering if they actually got the right answer(!)

[alpine sable]

Well they ran with it sooo

[alpine sable]

[keen plinth]

both alive and dead at the same time

[pseudo ice]

I’m just gonna take their last post as a yes and enjoy the rest of my day

[lone heart]

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[alpine sable]

magnitude of the average horizontal displacement of the tip, over the half hour, from its position at 4: 00?

[alpine sable]

Ive done all the other parts of the question such finding the average speed average acceleration.. total displacement and vertical displacement

but average horizontal displacement is confusing me

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[livid narwhal]

what have you tried?

[alpine sable]

well for a moment I thought it would be 0 since its the displacement but its not. Then I thought since I know the vertical and total displacement I can use Pythagoras but that wasnt it either.

[livid narwhal]

do you know about integrals?

[alpine sable]

a bit

what would I have to do?

[livid narwhal]

well im not sure if this is entirely correct

but my approach would be to define an angle, call it theta, and then find the magnitude of the hoz displacement vector in terms of theta

then integrate over the values of theta that would get from 0 to 30 mins

keep in mind, that it's not zero because you're considering the magnitude of the disp vector, not the values of it

if it were the values, then you'd be right

[alpine sable]

oh I see, I'll try that

thank you

.close

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[sand star]

H

[sand star]

.close

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[livid jewel]

why will the variance of a lineair regression model be high when p> n (p # predictors, n= #observations)

[pseudo ice]

,rccw

[vital chasm]

froof L.h.s=R.h.s

[vital chasm]

froof L.h.s=R.h.s

any one can solve it ?

[vital chasm]

thnx

[keen plinth]

apply half angle formula like 4 times ig

[vital chasm]

i need the steps of the solution

[keen plinth]

ive just rewritten your LHS

[vital chasm]

do u know R.h.s?

[keen plinth]

well the RHS is the RHS

theres not much to be done about that

just a huge nested radical

[vital chasm]

i can't solve it

[keen plinth]

you arent supposed to evaluate the RHS

you're supposed to evaluate the LHS

[vale wigeon]

do you know the identity cos^2(x/2) = (1+cos(x))/2

[vital chasm]

how can i make them identical?

[vital chasm]

no

[keen plinth]

you're kinda gonna need it

look it up, learn about it from somewhere

its the half angle identity

[vital chasm]

ok thnx

[vale wigeon]

once you do that, your first few steps will look like this-ish

[vital chasm]

can u comlete the solution?

[vale wigeon]

no. we do not give you complete solutions here.

[vital chasm]

it's hard qwestion and i need him tomorow

[limpid turret]

https://tenor.com/view/you-best-get-a-move-on-hurry-up-rush-quickly-get-started-gif-15315162

[vale wigeon]

not our problem, unfortunately.

#info and #❓how-to-get-help

[vital chasm]

at all thank you

[lone heart]

@vital chasm Has your question been resolved?

[vital chasm]

yes

sorry my bad

how can i move?

[vale wigeon]

if you have nothing else to say or ask about this problem, you can close this channel

to do this, type .close

[vital chasm]

.close

[lone heart]

Channel closed

Closed by @vital chasm

Use .reopen if this was a mistake.

[vital chasm]

.close

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[raven copper]

how do you solve such type of questions?

[raven copper]

plz help

[pliant cedar]

u put 2x instead of x

f(x) = 2^x + 2, f(2x) is when u put 2x in place of x

[raven copper]

@pliant cedar ik but look at the awensers

[pliant cedar]

its clearly not A or B

try putting it in the various options

[pseudo ice]

Is that $f^2$ composition ($f \circ f$, $f(f(x))$) or squaring ($(f(x))^2$)?

[raven copper]

.close

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[ocean seal]

chartbit

[lone heart]

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[vital chasm]

proof L.H.S=R.H.S

[pseudo ice]

Are you really not wanting to try finishing off the steps that you got in one of the previous threads you made at all?

[sharp scroll]

help

[grave island]

#❓how-to-get-help

[vital chasm]

i don't know the other steps i tride

can u complete the steps ?

pls

[pseudo ice]

You definitely should be able to… do you have the hint from last time?

[vital chasm]

no i am so dumb

[vital chasm]

i have am exam tomorw

[pseudo ice]

Ann did this for you last time

Think about that cos(\pi/8) and the identity at the top

[vital chasm]

i tried it's too hard

my brain stop moving

[vital chasm]

pls just help me on this

Is this correct ?

any help '

?

[pseudo ice]

Nice effort! let me read through that and check it for you

[vital chasm]

ok thnx

[pseudo ice]

Up to here I agree with (also note that you probably should know cos(\pi/4) off by heart

Below that I don’t get how you got there though?

[vital chasm]

is to correct?

is it+

[pseudo ice]

Anyways, another hint (which is pretty messy but will work) is to (once you’ve put in what cos(\pi/4) is here, or equivalently evaluated the line below that) is to get the “innermost” root with a common denominator and keep factoring it out of the roots

[vital chasm]

i don't relly undetand

i am not very good at engilish

i don't know evry think

can u do it in psge

??

[vital chasm]

???

[pseudo ice]

,w cos(pi/4)

[ocean seal]

Results provided by WolframAlpha

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

Requested by chartbit#3847

[vital chasm]

what about it

[pseudo ice]

Put this

[vital chasm]

what can i do else?

[pseudo ice]

Here

And rationalise the denominator (in other words, make $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$)

[ocean seal]

chartbit

[pseudo ice]

Then what does that give you?

[vital chasm]

i don't no

Is this correct ؟

[pseudo ice]

The way you’ve done the cancelling doesn’t quite imply the line below it

Alternatively, why not notice that $2=\sqrt{4}$ and absorb that into the roots

[ocean seal]

chartbit

[pseudo ice]

Then do it again, and again…

Take that step nice and slowly…

[vital chasm]

ok thnx

bye

.close

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[wind cloak]

$\int \sqrt{\frac{u}{4g - (K - u)^2}}du$

[ocean seal]

NEONPerseus

[wind cloak]

g and K are constants

Weird integral because physics

Already made a substitution

Original integral was $\int \sqrt{C - 2g\cos (\theta_0x)}dx$

Where again C and theta-nought are constants

x is inside the cosine btw

[ocean seal]

NEONPerseus

[wind cloak]

Wolfram uses special functions but I just wanna know if it's possible with regular stuff

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@wind cloak Has your question been resolved?

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@wind cloak Has your question been resolved?

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[scarlet shell]

\documentclass{article}
\usepackage{amsmath}
\begin{document}
[
\lim_{x\to 0^-} [-\frac{1}{2t^2}] + \lim_{x\to 0^+} [\frac{1}{2t^2}]
]
\end{document}

[scarlet shell]

\documentclass{article}
\usepackage{amsmath}
\begin{document}
[
\lim_{x\to 0^-} [-\frac{1}{2t^2}] + \lim_{x\to 0^+} [\frac{1}{2t^2}]
]
\end{document}

[ocean seal]

Noctowl

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
    \lim_{x\to 0^-} [-\frac{1}{2t^2}] + \lim_{x\to 0^+} [\frac{1}{2t^2}]
\]
\end{document}
```Compilation error:```! LaTeX Error: Can be used only in preamble.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.57 \documentclass
                   {article}
Your command was ignored.
Type  I <command> <return>  to replace it with another command,
or  <return>  to continue without it.```

[scarlet shell]

Hi

Can anyone helpme with answer?

Sorry this channel is occupied

[worn fox]

Is this the original problem? Because as written its an indeterminant form

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@scarlet shell Has your question been resolved?

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[lone heart]

@lethal berry Has your question been resolved?

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[alpine sable]

can someone help me with this matrix

[vague coral]

find K^T and K^-1 separately and then multiply them

[alpine sable]

I found K'T but I cannot find K^-1

[vague coral]

its a 2x2 matrix so its ez to inverse if you know the formula

[alpine sable]

I know the formula but I dont know how to implement it and multiply with it

[vague coral]

$\begin{pmatrix} a & b \ c & d \end{pmatrix}^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}$

[vague coral]

wdym

[ocean seal]

Ｈｅｒｅｌｓ

[alpine sable]

I just dont know how to multiply k^t with 1/ad-bc (d -b -c a)

[vague coral]

???

you dont know how to multiply matrices ?

[alpine sable]

I do but I dont know how to add 1/ad-bc into equation

[vague coral]

its... just a scalar

$\alpha \begin{pmatrix} a & b \ c & d \end{pmatrix} = \begin{pmatrix} \alpha a & \alpha b \ \alpha c & \alpha d \end{pmatrix}$

[ocean seal]

Ｈｅｒｅｌｓ

[alpine sable]

you multiply the outside with everything inside ?

[vague coral]

thats what I did

[alpine sable]

alright thank you

.close

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[royal quest]

This is my diagram i drew diagonal bd so that i can apply the angle bisector theorem, and i tried law of cosines but i fo not think that is the right way to do it

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@royal quest Has your question been resolved?

[alpine sable]

You swapped x and √2x in the diagram

I have no clue tho

Is this from a math olympiad or something like that? It reminds me of those kinds of problems

[royal quest]

Yes it is, and thanks i did not notice that i swapped the two lime lengths

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@royal quest Has your question been resolved?

[royal quest]

<@&286206848099549185>

[pliant oak]

Change parameters
a=x²,b=cosy
Solve a,b then solve x,y

But In this problem you can just solve x because the problem just ask the value of x

[royal quest]

Ok thanks i got it

.close

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[rocky pulsar]

How do we solve for a and n if (1+x/a)^n first four terms are 1+5x+(45/5)x^+bx^3

[weary wyvern]

Log

[rocky pulsar]

Log??

How exactly?

This is the question

n/a = 5 🤔 but there's an inf values for 5

[near hollow]

ah ok this is not an equality

[near hollow]

this is wrong the equation

[rocky pulsar]

Yes 🥲

[rocky pulsar]

yes

Well I

[near hollow]

you need to use this theorem and "compare coefficients"

[rocky pulsar]

I mean I've tried

But 💀

[rocky pulsar]

Ended up with this equation

[near hollow]

So you need to find an n and a such that the coefficient of the constant is 1, the coefficient of the x term is 5,the coefficient of the x^2 term is 45/4

[rocky pulsar]

since NC2, we can't really do that

[near hollow]

yeah but you will get two equations

[rocky pulsar]

oohh

[near hollow]

this is right btw

simplify it to n=5a

and then you will be able to input that into the second equation to get rid of either a or n

[rocky pulsar]

alrr

So basically 5aC2/a^2 = 45/4

And we just have to do a trial-and-error process

[near hollow]

there's an easier way

[rocky pulsar]

???

[near hollow]

wait yeah trial and error

it's easier in this case

[rocky pulsar]

yes

Alright ty!

[near hollow]

np

[rocky pulsar]

.close

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[neat plaza]

Suppose that A1, A2, and A3 are distinct non-collinear points in a plane. Define an infinite
sequence of points as follows: let An for n ≥ 4 be the reflection of An−3 over An−1.

1. Show that there exists a point X such that the triangles XAkAk+1, for all positive integers
   k, have the same area.
2. Explain how you found X. Is the point you found the only point with this property?

[neat plaza]

hello can someone help?

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[devout spruce]

hello. i stuck at 14

[devout spruce]

hello, i stuck at 14. i must find the canonical representation of this quadratic form