help-17

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[rain hollow]

a pool get filled up with 2 pipes. Using the first pipe only the pool get filled up in 6 hours less time then using the second pipe only. They started filling up the empty pool with both pipes. But after 5 hours they stopped the first pipe. Because of this filling up the pool took 3 hours more. How many hours dose it take to fill up the pool fully with both pipes

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[bitter pilot]

It is known that each of 150 musicians plays one or more of the instruments violin, viola,
Cello plays. 94 play violin, 35 viola and 55 cello. 12 play violin and viola, 17 play violin
and cello and 4 play all three. How many musicians play viola and cello?

$\$

I started with $$\abs{V \cap C} = \abs{V} + \abs{C} - \abs{V \cup C} = 35 + 55 - \abs{V \cup C}$$

I don't know how to get $\abs{V \cup C}$, any ideas?

[twin meteor]

𝔸dωn𝓲²s

[regal bane]

You're ignoring violin information

Let A be the set of viola players. You want to start breaking down |A U C U V|

[vast shale]

yeah this is principle of inclusion exclusion on three sets

you used it on 2 sets

|A U C U V| = (all singletons) - (all pair intersections) + (all triple intersections)

[dark kiln]

this is elementary you;re given the middle part

just fill in from inside

ok nvm there's no more elementary things left

[bitter pilot]

mine looks different

[dark kiln]

*minus x

[dark kiln]

i mixed up the instruments

[bitter pilot]

I will try that thanks

this should be then solvable with the diagram i made

.solved

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[dark kiln]

i think it's usually like mine, the 4 is included in 17

[vast shale]

or just use the formula

[dark kiln]

but maybe it's like 30/70 so neither is too rare

[vast shale]

,w 150 = 94 + 35 + 55 - (12 + 17 + x) + 4

[twin meteor]

Results provided by WolframAlpha

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Requested by kizzyyyyy

[dark kiln]

this is obviously wrong though, you would need negative X to add up to 150

[bitter pilot]

wdym

[bitter pilot]

yea that's what's in the sol. too

[dark kiln]

these 7 numbers sum to 150

94 + 55 +...

you need negative x

maybe you're not intending the numbers to add up to 150 it's like a diagram that's not supposed to be taken literally?

[bitter pilot]

i dont think that's how it works A + B + C = musicians

[dark kiln]

there's no interpretation where you get 150 from 3 numbers

[bitter pilot]

ok

[bitter pilot]

you btw

seem to know alot

for someone who pretends that doesnt

pre uni😆

[vast shale]

this is pre uni stuffs, is it not?

[bitter pilot]

hahah maybe maybe not

[vast shale]

it's jsut drawing this

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[prime estuary]

Hello may I get some assistance with my work in call

[vast shale]

yes

do not ask to ask

just tell me the damn question

[vast shale]

draw all the triangles you can make onto the circle

and label the angles you already know

[rocky valley]

how you solve this 💀

[vast shale]

w8

[vast shale]

just triangles

[rocky valley]

show

how

[prime estuary]

I currently dont have the supplies avalible for that to be possible

[plucky trout]

There's a formula

CDF is related to x

U prolly have learned it

Think about all the info u are given

[prime estuary]

can't lie I ain't pay attention much when it can to circles it confused me too much

[noble heath]

Remember, the angle between two tan lines is equal to half the different of the intercepted arcs

[vast shale]

you can make x arbitrarily far away from the circle

[plucky trout]

How would you make an angle far away from the circle

[noble heath]

Use this

This is what your looking for

[vast shale]

there are no tangent lines

[noble heath]

One sec

The same rules still apply though

It’s the same intercepted arcs

[vast shale]

what thes the m mean

i've never seen that before

is it just notation for an angle

[vast shale]

are there any previous questions related to this

[rocky valley]

gary got muted

for what

💀

[vast shale]

no idea lol

as far as i can tell, there's no way you can answer this question

because there's nothing fixing the position h so k can be arbitrarily far away. as k goes away, the angle will decrease

[vocal sleet]

@prime estuary Has your question been resolved?

[vast shale]

this looks so wrong

im just assuming cg divides the circle evenly

if u have a right angle within a triangle and you extend it to touch the circle at 3 points, it will make a line that cuts the circle into 2 evenly

i think i've messed up somewher

ohhhhh im dumb

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[rocky compass]

In remainder theorem, how is x=a?

[rocky compass]

Please help me someone

[subtle helm]

!original

[vocal sleet]

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

[rocky compass]

Case 1 x=a

[heavy yoke]

,rotate

[rocky compass]

For x=a, x-a should be equal to 0, but how is (x-a)=0

That’s my doubt

[subtle helm]

is what you're asking: why is x - a = 0 when x = a?

[brisk moss]

@thin vale

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@rocky compass Has your question been resolved?

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[still mica]

10

[still mica]

No idea how to approach this

[subtle summit]

can you figure out when cos x / sin ax is periodic?

[still mica]

I know as information that it's periodic when a is rational, I don't know the reason

[subtle summit]

try to figure out what its period is for different fractions

then you can show that

• it's periodic when a = s/t for integers s and t
• it's not periodic when a is irrational

[still mica]

I'll try

[still mica]

How do i find the period
Say cos(x) /sin(2x)
Do i assume cos(x) /sin(2x) = cos(x+h) /sin(2x+h) and solve for h?

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@still mica Has your question been resolved?

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@still mica Has your question been resolved?

[subtle summit]

sounds about right

2(x + h) though

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[fresh tide]

tan^-1(sqrt3)= pi/3
tan^-1(1)= pi/4

[fresh tide]

how?

I honestly can't do without plugging it in the calc and getting a decimal point

[sharp frost]

Do you know trignormetries?

From this

[fresh tide]

a lil bit

[sharp frost]

Since tan60=root 3

Tan^-1 of root 3 is pi/3

[fresh tide]

Ah I see

.close

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[jagged cargo]

A survey is hosted for 1000 students about choosing at least 1 in 3 subjects they feel easiest. 320 students chose philosophy, 480 students chose law, 360 chose probability theory. What is the maximum number of people that chose all 3 subjects?

[jagged cargo]

so i made venn diagram

let A be 320, B be 480 and C be 360

then id have 1000 = A + B + C - ( A ∩ B ) - ( A ∩ C ) - ( B ∩ C) - ( A ∩ B ∩ C)

therefore 160 = ( A ∩ B ) + ( A ∩ C ) + ( B ∩ C) + (A ∩ B ∩ C )

from here not sure what to do

[urban edge]

that should be + A n B n C

you pied wrong

[jagged cargo]

oh + ?

[urban edge]

yeah

[jagged cargo]

ok so id have

( A ∩ B ) + ( A ∩ C ) + ( B ∩ C) - (A ∩ B ∩ C ) = 160

[urban edge]

we want to maximize the number of people with all three, how can we do that

[jagged cargo]

( A ∩ B ) + ( A ∩ C ) + ( B ∩ C) = 0?

[urban edge]

remember that anything in AnBnC is in AnB etc

[jagged cargo]

im not sure

( A ∩ B ) + ( A ∩ C ) + ( B ∩ C) = 160?

oh wait no

[urban edge]

would it be efficient to have a voter vote for two classes?

[jagged cargo]

not sure im following, wdym "efficient"?

[urban edge]

will it maximize AnBnC

[jagged cargo]

well no

oh wait so

( A ∩ B ) + ( A ∩ C ) + ( B ∩ C) is minimum?

[urban edge]

so lets make the naive assumption that people either voted 1 or 3 (it works out here, you might need to deal with edge cases later)

[jagged cargo]

alright

but wouldnt that make (A ∩ B ∩ C ) = -160?

[urban edge]

no

remember that AnBnC is a subset of AnB etc

[jagged cargo]

ah

but if AnB = 0

then (A ∩ B ∩ C ) = 0 as well?

[urban edge]

lets say the number of people who voted for all three is X

a person who voted for all three will also count towards voting for A and B or A and C or B and C

[jagged cargo]

but no one votes for 2

which can only mean there are no people who vote for 3

which means A, B and C are independent sets

[urban edge]

AnB is voting for at least A and B

AnB = (voted A and B and did not vote C) + (voted A and B and did vote for C)

[jagged cargo]

oh so

AnBnC + AnB\C = 0?

[jagged cargo]

i believe if we are assuming something like this, the venn diagram would look somewhat like this?

[urban edge]

yeah

[jagged cargo]

right so

here, 1000 = A + B + C - 2(AnBnC)?

since AnBnC is counted thrice when we are referring to A, B or C

[urban edge]

yeah thats one way to do it

[jagged cargo]

but this is just an assumption is it?

does it maximize AnBnC?

[urban edge]

yes

the students have to voye for at least one class

[jagged cargo]

oh alright

[jagged cargo]

but what do you mean when you refer to "edge cases" here?

[urban edge]

by maximizing the number of people who vote for all 3, we minimize the other cases

lets say we had 361 votes for C

[jagged cargo]

well i dont see the problem

just tweak C = 361?

[urban edge]

what happens to the equation 1000=A+B+C-2(AnBnC)

[jagged cargo]

oh wait

it outputs a non integer number

[urban edge]

we allow one person to vote for two classes

[jagged cargo]

yes, but the maximum would still be 80 isnt it?

[urban edge]

yes

[jagged cargo]

say if C = 363, then it would be 81

[urban edge]

yup

[jagged cargo]

oh alright, i can visualize it now

thanks!

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[raw flax]

Help

For a function

[jagged cargo]

no lol

[raw flax]

😡😡😡

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[vast shale]

,rccw

[vast shale]

<@&286206848099549185>

...

.close

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[chilly garden]

can i get help on # 10

[weak edge]

Graph goes down 1 unit

Put x tending infinity to get y asympote

And put y tending infinity to x asympote

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@chilly garden Has your question been resolved?

[rocky compass]

How is x=a

[pliant socket]

You still need a hand with this?

[chilly garden]

so for 10

tranformations are

shift right 1 and vertical stretch by 4

and vertical = 0 and horizontal =-1?

[pliant socket]

Dilate yes, shift down by 1 not right as the entire graph moves down by 1 value of y

[chilly garden]

it doesnt translate down 1 unit?

[pliant socket]

It does

If the -1 was in the denominator then it would translate right

You can test it by subbing values btw, it's good for double checking

[chilly garden]

oh my bad i meant shift down 1

[pliant socket]

Yh ok cool
Btw never heard of stretch vertically, I'd say dilate away from the y axis

[chilly garden]

im loosing sleeps prob why lol

yeah there so many ways to call it

but in my class our teacher calls the transformation like that

[pliant socket]

But that's up to you I guess

[pliant socket]

Fair enough, I've taught myself to use dilations with the numerator as the dilation factor

Asymptotes are right

[chilly garden]

do mind helping with the last two steps

im done the first two and need help on the rest

[pliant socket]

Ok, so first mark the asymptotes on the graph & label them (y=-1 etc)

[chilly garden]

got it

[pliant socket]

Then what I'd do is plot a few points by subbing in, 2 points for the left and 2 for the right parts (4 points they ask for)

(Unless the 4 points are something else that the question is referring to)

Preferably any x/y intercepts as points

Then sketch the general shape like 1/x but by joining the dots so you have the right dilation

(Use the marked dots and asymptotes)

Part 4 is fine, since it extends to ∞ & -∞ for both x & y you just do the domain and range excluding the asymptotes

Idk what notation you'd want use, that's up to whatever you're comfortable with

[chilly garden]

cool. thanks for the guideline

😯

[pliant socket]

Sweet, hope that helps

[chilly garden]

.close

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[real barn]

1/8+ -2/8

[real barn]

How to do this

[fickle arrow]

Hello

1/8 + -2/8

They have the same denominator

So you can add the numerator together

(1 + (-2)) / 8

[real barn]

Yes

And then

[fickle arrow]

1 + (-2) = -1

So the answer is -1/8

[real barn]

Okay understand

[fickle arrow]

Alright

[real barn]

Thank you

I close now!!

[fickle arrow]

Np

[real barn]

. close

. close

.close

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[crimson lily]

Translation: Find minimum distance a between origin and the points on the curve:

[crimson lily]

i use lagrange multiplier and get to these equations but i get stuck on how to solve it

[lucid bane]

are you required to use lagrange multipler?

[crimson lily]

no, but all the previous tasks were using it so i would expect it to atleast be possible

[lucid bane]

have you expressed the distance from the origin as a function of x and y?

[crimson lily]

yes i use the square distance as f function, so f=d^2=x^2+y^2

since a=sqrt(x^2+y^2)

i meant a not d

i think i have found out how to do it myself

.close

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[misty dragon]

hello, i found the inverse of the function but i dont understand y my ans is wrong

[tough valve]

Show your work

[fervent wasp]

!status

[vocal sleet]

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

[misty dragon]

i got

f^-1(x) = √(x + 11) + 4

[misty dragon]

i got an answer but i was told that it's wrong

[winter thistle]

Please show your entire work and not just answer.

[misty dragon]

alr

so

x = y^2 - 8y + 5

x = (y^2 - 2*4y - 4^2) - 4^2 + 5

x = (y - 4)^2 - 11

x + 11 = (y - 4)^2

√(x + 11) = y - 4

√(x + 11) + 4 = y

therefore

f^-1(x) = √(x + 11) + 4

thats my working

[winter thistle]

when taking sqrt, you need to consider both positive and negative sqrt

[misty dragon]

i see

lemme make sum changes

wait a sec

[winter thistle]

[misty dragon]

wait hold up

so x ≤ 4

since its inverse

the range of y is y ≤ 4

so i can sub it back in

[winter thistle]

the range of y is <= 4?

oh I see what you mean

[misty dragon]

yup

[winter thistle]

the range of f^-1

[misty dragon]

mhm

oh i get it now

it shld be negative sqrt

[winter thistle]

yup

[misty dragon]

alr cool

ty for the help

[winter thistle]

np

[misty dragon]

just a further question

tho

whenever i come across questions like this,

i gotta plug the range/domain bac in the inverse

to know whether its positive or negative like u mentioned earlier

right

[winter thistle]

yes, or you can sketch the graph

[misty dragon]

by completing the square?

like if i dont use a calc i mean

[winter thistle]

the inverse of a function is always the reflection of the function in y=x

[misty dragon]

oh right yea

[winter thistle]

so here the red is your quadratic, and reflected in y=x, then from the graph you can determine positive/negative

[misty dragon]

gotcha

thanks again man

[winter thistle]

you're welcome

[misty dragon]

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[slow atlas]

Hi quick question can someone explain why the inside bounds are 2costheta

and 0

I am bit confused by that

[last osprey]

I need math section

[slow atlas]

can you use a different thing please

I am using this one

just delete your messages and go to one of the availbe ones

better question can someone just explain to me how to find the bounds for the integral that is all I need

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@slow atlas Has your question been resolved?

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[vast shale]

im not getting 2

and i shud be getting 2

i got 1

instead

.close

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[vast shale]

.close

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[elder forum]

I have function ax^b + cx and I can get any points from this function. Is it possible to get a, b and c by having points? If yes then how?

[elder forum]

b < 1

[hoary cypress]

u need 3 set of points from the function

and then solve the 3 eqns for 3 variables

[elder forum]

but how to do it?

[formal yew]

just sub in the x value

for the x cord

and use system of equations

what points do u have

[proven spire]

You do realise these aren't linear equations?

[formal yew]

you can still use system of equations

[proven spire]

Well, I mean, you might still be able to

But a part of me feels like three points might not uniquely define the equation

[elder forum]

(705.293,300) (221.046,120) (83.021,60)

[proven spire]

holy cow

Wait, is the "." there a thousands-separator?

[elder forum]

decimals separator

"," separates x and y

[proven spire]

👀

...Why were those the values you have

[elder forum]

I get these from some complicated game simulations

[proven spire]

Oh now I see why this was a question lol

[elder forum]

and I need to get needed velocity to jump to specific height

[proven spire]

So is the b there a dampening factor or something?

Cos otherwise I'd have assumed it was a 2 (though you'd said b<1)

[elder forum]

dampening factor? I dont undertand?

[proven spire]

What's the b there for?

[elder forum]

to get specific curve

[proven spire]

So this is a curve that represents a jump, right?

i.e. it's a parabolic curve?

So the input is the forward distance travelled, and the output is the height of the jump at that distance?

[atomic flax]

hi can yall help me with math its gonna be in the exam it counts as 80 marks

[elder forum]

x are wanted max jump height and y are needed start upward velocity

[proven spire]

You're asking in an occupied help-channel (note the category this channel is under). Ask in a channel marked MATH HELP (AVAILABLE)

[proven spire]

Started upward velocity? of a jump?

Like, how long the jump button is held?

[elder forum]

character is standing with 0 valocity and when jumped there is added Impulse velocity once

[proven spire]

And as they're running, that velocity affects how high they can jump?

[elder forum]

yes

[proven spire]

Right I see

(Why would you want that though? Physics doesn't work like that usually)

[elder forum]

Im programming something and need to know needed impulse by height

[proven spire]

The direction you run in is perpendicular to the upward force from jumping, so the jump would not be affected by how fast you were

btw, I've asked ChatGPT -and asked it for the python code to calculate this- to get a≈5.075,b≈0.439,c≈0.297

import numpy as np
from scipy.optimize import least_squares

# Define the function based on the given system of equations
def equations(params, x, y):
    a, b, c = params
    return a * (x ** b) + c * x - y

# Input-output pairs
x_data = np.array([705.293, 221.046, 83.021])
y_data = np.array([300, 120, 60])

# Initial guess for a, b, c
initial_guess = [1, 1, 1]

# Solve the system using least squares method
result = least_squares(equations, initial_guess, args=(x_data, y_data))

# Extracting the solution
a, b, c = result.x
a, b, c

[elder forum]

thats exactly what I needed

[proven spire]

(why do you know b<1, anyways?)

[elder forum]

because its going more to horizontal not vertical

[proven spire]

Oh right, I see

So how did you come to the values you did?

Cos I get you handpicked them, but what made you pick the values you did?

[elder forum]

some simulations with setted start velocity and gettet highest point (blue)

its height per time

[proven spire]

Formulae?

Just a s=ut+1/2 at^2 -style graph?

Where you've varied the a value...?

[elder forum]

(v(previous) + v (delta)t) * 0.9936 v = v - gt and it adds point every tick which has 0.02s

[proven spire]

So this is just a negative quadratic function

[elder forum]

red yes

but red is without *0.9936

[proven spire]

Yh, the blue is a simulation-approximation of another negative quadratic, is what I meant

[elder forum]

actually it looks like quadratic but i thinks that its not

but maybe its possible

[proven spire]

It likely is

Because it's a physics simulation of gravity over time

You're plotting this on Desmos, aren't you?

[elder forum]

yes

[proven spire]

Try some regression calculations (you can find some examples)

You can get it to give you an approximate equation that fits your data, though you need to specify how it looks

I can't recall how it works exactly, but you want it to give you an output like f(x) = ax^2 + bx + c (and if done right, a should be negative)

[elder forum]

thanks

I think that its all

[proven spire]

Ah, you got your data as a table right?

like, x1 in one column and y1 in another?

[elder forum]

which data?

[proven spire]

the blue data points

[elder forum]

yes

[proven spire]

Add this after it (with the ~ sign)

That'll give you coefficients to approximate a quadratic regression through those points

to check it, plot f(x) = ax^2 + bx + c

So you can get something like this

[elder forum]

looks helpful

[proven spire]

(the residuals represent the error of approximation AKA the offness of the approximation)

Similarly the R^2 is a measure of how well the data fits the graph (the closer to 1 the better, where 1 is perfect)

[elder forum]

Ok tested, maaany thx for time

works perfectly

.close

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[still mica]

Whts that?

[mild flower]

,tex .plane geom

[twin meteor]

hayley

[still mica]

Are we assuming its a square?

Can you solve for x?

The sides of the square are equal

One side is 2x + 4 other is x + 5

What is B? inn the eq

You cant multiply a point huh

Do you understand why 2x + 4 = x + 5

Theres some miscommunication

You agree BC = DC

?

BC = DC

x + 5 = 2x + 4

You can solve for x now?

Yup

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[vital mountain]

i have a question about the high lighted part
why so we take the 42xy and break it into two parts(42x)(42y) rather than just deriving it into 42 (calc 1 implicit differentiation)

[regal bane]

xy is a product of two variables. You need product rule for it

[vocal sleet]

@vital mountain Has your question been resolved?

[vital mountain]

when i do the product rule it is coming out wrong

[vocal sleet]

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

[vital mountain]

should the product rule be being done with just xy? or with 42xy?

im confused as to what we are finding the product of

[strange crater]

it can be done just with xy. but you'll still have to multiply everything by 42

[vital mountain]

so x is a function of y which allows me to use the product rule on it?

but

im not sure when to know if i can remove the constant

and not have to derive it

[strange crater]

x is not a function of y. y is a function of x

[vital mountain]

right sorry

but the fact that there is one function of another is the reason for the product rule?

[strange crater]

it's like if y were sin(x), for example. you would use the product rule then, correct?

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[shut storm]

How do i show this in a diagram

for a i got = -2

[fiery elbow]

cosecx=1/sinx
sin(5pi/4) is the same as sin(225°) and it is the same as -sin(45°)=-sqrt(2)/2
1/that is -2/sqrt(2)

[shut storm]

i do not understand lol

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[fiery elbow]

which part

[shut storm]

all the / =

im just trynna see what it would like as a diagram

i dont understand what the teacher wants

i was thinking maybe something like this

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[bitter pilot]

Show the validity of the sieve formula in the case that all $n$ sets $M_1, M_2, ..., M_n$ are identical.
$\$
My work:

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

oops

the minus oopsie

.solved

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[crude harbor]

im struggling with part d atm, and i dont know if i did part c correctly either

heres my solutions though

[brisk moss]

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[crude harbor]

<@&286206848099549185> pretty please 🙏

[snow hare]

Hm, Interesting.

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[unreal cargo]

I was trying to evaluate (\sqrt{-i}) using the fact that (e^{i\theta}=\cos(\theta)+i\sin(\theta)) but I ran into a problem. The steps I followed are the following:

At the angle (\frac{3\pi}{2}), it is known that Cosine will be 0 and Sine will be -1, and so it follows that (e^{\frac{i3\pi}{2}}=-i), and so (\sqrt{-i}=\sqrt{e^{\frac{i3\pi}{2}}}).

We can also say that (\sqrt{e^{\frac{i3\pi}{2}}}=\left(e^{\frac{i3\pi}{2}}\right)^{\frac{1}{2}}=e^{\frac{i3\pi}{4}})

That would seem to mean that (\sqrt{-i}=\cos\left(\frac{3\pi}{4}\right)+i\sin\left(\frac{3\pi}{4}\right)=\frac{-1}{\sqrt{2}}+\frac{i}{\sqrt{2}})

This would then be simplified into (\frac{i-1}{\sqrt{2}}) to which, once we rationalize the denominator, we get (\frac{\sqrt{-2}-\sqrt{2}}{2})

The problem is that my calculator tells me that, in truth, (\sqrt{-i}=\frac{\sqrt{2}-\sqrt{-2}}{2})

[unreal cargo]

Okay I was kind of hoping the bot would kick in uh

Gimme a second

[calm light]

you need dollar signs to trigger the bot

just put one at the beginning and end of the post

[unreal cargo]

Oh y'know what the Discord copy thing fucked it

Here lemme try

I was trying to evaluate (\sqrt{-i}) using the fact that (e^{i\theta}=\cos(\theta)+i\sin(\theta)) but I ran into a problem. The steps I followed are the following:

At the angle (\frac{3\pi}{2}), it is known that Cosine will be 0 and Sine will be -1, and so it follows that (e^{\frac{i3\pi}{2}}=-i), and so (\sqrt{-i}=\sqrt{e^{\frac{i3\pi}{2}}}).

We can also say that (\sqrt{e^{\frac{i3\pi}{2}}}=\left(e^{\frac{i3\pi}{2}}\right)^{\frac{1}{2}}=e^{\frac{i3\pi}{4}})

That would seem to mean that (\sqrt{-i}=\cos\left(\frac{3\pi}{4}\right)+i\sin\left(\frac{3\pi}{4}\right)=\frac{-1}{\sqrt{2}}+\frac{i}{\sqrt{2}})

This would then be simplified into (\frac{i-1}{\sqrt{2}}) to which, once we rationalize the denominator, we get (\frac{\sqrt{-2}-\sqrt{2}}{2})

The problem is that my calculator tells me that, in truth, (\sqrt{-i}=\frac{\sqrt{2}-\sqrt{-2}}{2})

[twin meteor]

Lober

[unreal cargo]

There sorry

[loud walrus]

You made a mistake when u say sqrt(-1)=cos(3pi/4) etcc

Sqrt(-1) = -(cos(3pi/4)+isin(3pi/4))

There is a minus that should multiply the whole rhs

[unreal cargo]

Wait where does that come from?

[loud walrus]

That is up to you to find. I prefer you think about that more

[unreal cargo]

Are you referring to the plus-minus trait that square roots have?

Cause I think based on how roots work, (\sqrt{-i}=\pm e^{\frac{i3\pi}{4}}) correct?

[twin meteor]

Lober

[loud walrus]

What is -i?

It is what u wrote squared right?

[unreal cargo]

Yes

[loud walrus]

Ok now take square root in both sides

[unreal cargo]

What?

[loud walrus]

Sqrt(-i)= sqrt((cos(3π/4) + isin(3π/4))^2)

And be careful with parenthesis

Sqrt is the most outside

$\sqrt{(-i)^2} \neq (\sqrt{-i})^2$

[twin meteor]

Samuel

[unreal cargo]

Wait okay let me try evaluating this because now I am kind of confused

Your equation is right but how does this show that the rhs must be multiplied by -1

Wait yeah so then this works

[loud walrus]

Ok i almost say something

So what do you get

[unreal cargo]

I mean this evaluation proves my initial result correct, so I'm gathering that (\pm\left(\cos\left(\frac{3\pi}{4}\right)+i\sin\left(\frac{3\pi}{4}\right)\right)) is a solution to (\sqrt{-i})

[twin meteor]

Lober

[unreal cargo]

I do not know how I didn't see that now that it's laid out in front of me

Thank you

[loud walrus]

Only the minus is correct

[unreal cargo]

How so? If you square the positive version, you also get -i

[loud walrus]

It is very late here and maybe i am not using my brain but

This says the opposite

[unreal cargo]

No I think it must be tiredness

whether you multiply (\cos\left(\frac{3\pi}{4}\right)+i\sin\left(\frac{3\pi}{4}\right)) by -1 or not, when squared, it will give -i

[twin meteor]

Lober

[unreal cargo]

I just did not catch that before because I did not think to try squaring what my answer was to check it's validity

[loud walrus]

Right, it is tiredness lol

[unreal cargo]

thank you man, you should probably go sleep hahah

[loud walrus]

Well, at least we got to the point

[unreal cargo]

that had me stumped for a few days ngl

.close

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[outer stream]

"Find the tension in the attachment of the cable to point D and C. The 220-lb uniform tank is suspended by means of a 7.0-ft-long cable, which is attached to the sides of the tank and passes over the small pulley located at O."

I've tried, and what I try doesn't seem to work. How should I go about finding the tension within the attachment point D?

[outer stream]

I tried placing force vectors and using cosine to project the force onto points C and D in the form of 110lb downward, for the weight of the tank

And then projecting OC and OD onto the vertical vector shown F

(was wrong)

@spice sun

<@&286206848099549185>

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[outer stream]

.reopen

[vocal sleet]

✅

[outer stream]

I would greatly appreciate help

[snow hare]

Well a pulley is frictionless, the tension in the new cable section will be the same as the tension in the original cable section

We conduct a equilibrium problem,
where the forces acting upwards (the tension in the cable on the left side of the cut) must be equal to the weight of the tank (220 lbs) acting downwards.

[vocal sleet]

@outer stream Has your question been resolved?

[outer stream]

F(z)sin(theta)=220

ends with 230, which was wrong

where hypotenuse=3.5, adjacent = 1, and opposite side = (3.5^2-1)^1/2

so F*(3.35/3.5)=220

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[outer stream]

seems no one can help 😦

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[stuck moss]

someone can help?

[summer lichen]

what have you tried

[stuck moss]

I can only get either no max or no min

not both

[summer lichen]

you have to have f continuos on (0,1] but the domain of f doesn’t have to be restricted to (0,1]

[rugged orchid]

What did you do to get no max

[stuck moss]

just anything with the open side would work

like 1/x

[rugged orchid]

Perfect

[stuck moss]

or just x

[rugged orchid]

We’ll build on that

[rugged orchid]

f(x) = x has a maximum on (0, 1]

[rugged orchid]

We can use this

[stuck moss]

sure

[rugged orchid]

Now we sort of need this to go to - infinity near 0 too right?

[stuck moss]

yea

but how could we do that

[rugged orchid]

Perhaps we could make it alternate

Do you know a way to make a bunch of alternating sort of wavey function near 0?

[stuck moss]

sin(1/x)

[rugged orchid]

Aha

So sin(1/x) kinda zigzags between 1 and -1 near 0

Now how can we make it zigzag between -inf and inf instead?

[stuck moss]

sin(1/x)/x

[rugged orchid]

Yep

That has no min no max and is continuous on (0, 1]

Tada

[stuck moss]

ooooh i see

I didnt think about those functions

[slow brook]

what's up people

[rugged orchid]

There may be other constructions but that’s how I thought about it

That’s the first one I thought about

[stuck moss]

thank you Frosst

[rugged orchid]

👍

[stuck moss]

.close

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[hard citrus]

i need some help

fr

@past thicket

u tryna help?

or

<@&286206848099549185>

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[dusty terrace]

What I doodled out so far

,rotate -90

[inner osprey]

did you see my previous message

ah you did

yeah that pretty much ends it

it's important that DE is half of BE

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[summer vector]

Write an equation for the nth term of the geometric sequence then find a6.

[summer vector]

Write an equation for the nth term of the geometric sequence then find a6.

[silk osprey]

what do you think the ratio is

[summer vector]

-1/4

[silk osprey]

mhm

so what do you know about geometric sequences

what is the formula

for the nth term

[summer vector]

Bruh to be honest I don’t know shit I got like last minute work to do for online school and it’s just been stacking up cuz I don’t get it at all

[silk osprey]

an=a1(r^(n-1))

essentially it’s saying to get to the nth term you multiply by the ratio n-1 times

from the first term

so for the second term

you’d multiply by it once

for the fifth term you’d multiply by it 4 times

do you see the pattern

[summer vector]

So an=a1(r^(n-5) ?

[silk osprey]

no

where’d you get 5

[summer vector]

Damn

[silk osprey]

so for a6

which is the 6th term

[summer vector]

You said it would go and multiply

[silk osprey]

that means you let n=6

[summer vector]

So n is 6

[silk osprey]

mhm

r=-1/4

a1=-192

[summer vector]

Wait what so how I do make an equation

[silk osprey]

a6=-192(-1/4)^(6-1)

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[gleaming glacier]

help?

[inner osprey]

help!

[gleaming glacier]

!help

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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[molten remnant]

please can someone help me solve this question

[zinc quail]

sry LightWorkery2 claimed it first :)

[molten remnant]

what

i'm new to server

[zinc quail]

he sent a message a few seconds before, you did everything correct, he was just a little faster, but you can claim another open channel

[molten remnant]

ok how

[zinc quail]

just send a message in one of these

available ones

[gleaming glacier]

i need help with 9th grade math, it is the end of semester and im almost at a B, if i get 10/10 then i can change my C to a B, please help me🙏

[inner osprey]

can you identify what kinds of congruences the problem statement gives you?

[thin vale]

post your question

[inner osprey]

they did

[gleaming glacier]

i missed school because my auntie was in the hospital so i missed this whole lesson. thats why im on here, idk how to do it

[molten remnant]

@zinc quail help no one is answering my question

[zinc quail]

yes it may take a few min depending on who comes along for guidance, everybody on this server helps voluntarily :)

[molten remnant]

ok thank you

[vocal sleet]

@gleaming glacier Has your question been resolved?

[gleaming glacier]

?

[zinc quail]

well you're given two angles and a side

so SAS can't be true

[gleaming glacier]

i already guessed it since no one was helping

thanks though

[zinc quail]

k but the reasoning is that the side provided is not between the two angles

[gleaming glacier]

i got an essay due in an hour, had no choice to do it asap

[zinc quail]

thereby requiring AAS

kk

[zinc quail]

yh depending on demand and supply help has delay

there's always more who seek than provide :3

[gleaming glacier]

18/20 GOD IS ON MY SIDE FUCKERS

FUCK MATH THE SHIT WE LEARN AFTER MIDDLE SCHOOL IS STUPID ASF AND WE WILL NEVER USE IT IN THE REAL WORLD LETS BE BRUTALLY FUKEN HONEST

[zinc quail]

k?

that statement is untrue

more truthful would be: if you attempt to avoid math after middle school THEN you'll most likely never use it.

@gleaming glacier at least argue for your position if you rant about it

nywys if you need guidance open a new channel

.close

[vocal sleet]

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[past prairie]

Hello i need somehelp to make this equation runable into a program i don't know where to start

[past prairie]

by program i mean like python etc....

[zinc quail]

it's unclear what you intend to do

do you want to just construct a function which takes a, b, c, t, alpha, beta, gamma and calculates r?

or is the program supposed to graph it?

[past prairie]

it an equation the unknown is t

[zinc quail]

okay but you know that the above equation can be simplified right

since the a's, b's and c's cancel out

[past prairie]

it different value but yes i can add them

[zinc quail]

k, well then you get:

which is

which is

giving you

with two possible values for t (that's why +- in front of the root)

[past prairie]

i put value in so that can be more readable

[zinc quail]

but there you inserted two different values for a,b,c each

became

[past prairie]

yeah that was a mistake sorry

[zinc quail]

what I meant before is that it's irrelevant what a,b,c have as values

since they just cancel out in the equation

the value of t doesn't change for different values of a,b,c

[zinc quail]

& are the steps clear with how I got to the end equation?

[past prairie]

what i need to do is to develop the expression and with the result do an quadratic equation with

[zinc quail]

develop the exact expression?

even though a,b,c are useless?

[past prairie]

they aren't useless

[zinc quail]

why

[zinc quail]

for calculating t they are useless

[past prairie]

maybe i will explain more of what i need to do

[zinc quail]

yh context is fairly unclear

[past prairie]

so i have an sphere and i need to know if a line goes in the sphere using the parametric line and sphere equation so i include the parametric line into the sphere equation wich give me the equation (...)²... = ... then using quatric equation determine if the line goes in or not

[zinc quail]

ok but isn't x = a + alpha*t then

instead of x = a + alpha*t - a

you used
x²+y²+z² = r²

for the sphere

and the line has form

(a b c) + (alpha beta gamma) * t

[zinc quail]

so if you insert x = a + alpha * t

[zinc quail]

yeah exactly

[zinc quail]

ah sry sphere isn't centered

yeah the error you made is that you used a,b,c

for the sphere center

and the line start

but they are different points

[past prairie]

yes

differents starting points

[zinc quail]

so you need to rename it differently

[past prairie]

wich one the sphere ?

[zinc quail]

either of them

so that a,b,c don't collide

for instance like this, which makes it more clear:

where c is the center of the sphere

and c_x, c_y, c_z its coordinates

Line start s and direction v

[past prairie]

yeah i see know

[zinc quail]

then we can insert as you did before

so far clear right

let's look at one of those terms:

can you expand it? @past prairie

[past prairie]

sx - cx

[zinc quail]

I mean if you factor it out

[past prairie]

oh

[zinc quail]

then you get this ^

try this first by hand

you can use the binomial rule:

(a+b)² = a²+2ab+b²

[past prairie]

but we still don't know t ?

so in a program that don't work

[zinc quail]

yeah I'll get to that

[past prairie]

okok

[zinc quail]

now we've expanded one of the terms, but the others can be expanded the same way

which turns

into

(warning: ugly equation)

[past prairie]

x)

[zinc quail]

but it's just the three terms expanded like above

ok and now we can group all of those terms by whether they include:

t²

or t

or neither t², t

and factor them out:

to make it visually a bit clearer:

so far understandable @past prairie ? :D

[past prairie]

yeah it okay for now x)

[zinc quail]

ok but now we'll condense this ugly blob into a beautiful flower again by naming these long terms:

and that's your quadratic equation! :o

[past prairie]

oh your right

[zinc quail]

which you can solve with the quadratic formula

so for it to be solved in python for instance:

def solve(sx,sy,sz,vx,vy,vz,cx,cy,cz,r):
    A = ...
    B = ...
    C = ...
    (quadratic equation for t, CAREFUL: there can be up to two values for t)
    ...

[zinc quail]

for A, B and C we just need to dump those long statements in from above

and from then onward we just work with A, B, C to solve the quadratic

process mostly clear?

@past prairie note what we've done, we took the sphere equation and the line equation

and set them equal

which means each value of t that we get

implies an intersection of the sphere and the line

therefore, if the quadratic formula gives you any value for t, then the sphere and the line intersect! 🦇

and.. ..if you want to go a step further, you can also explicitly say where the line hits the sphere using the values of t!

because the line has the formula s + t * v

so if you get two values for t for example

then you can insert those t values into the above formula

and you get the two points where the sphere and line meet :)

[past prairie]

nice

thx

i know the way to do it know

[zinc quail]

wish thee gl

[past prairie]

but wait how does i include t in A B C

[zinc quail]

t isn't involved in A,B,C

you can calculate them purely with the given values

[past prairie]

oh yeah

i didn't see it

[zinc quail]

remember we want to find t, so we can't use it in the calculations

[past prairie]

and what mean szz ?

[zinc quail]

oopsie

sz

[past prairie]

i was sure

[zinc quail]

I know it's just a task and you're likely only given some formulas and then told to compute or solve it in a way

but to paint a little bit bigger picture

the above approach is not limited to spheres and lines

you could also take spheres and spheres

and determine their intersection

or sphere and planes

or spheres and hyperspheres :O

various object intersections

a sample where this becomes very quickly useful is for example consider collision detection in games

say you want to build a little 2D or 3D physics game

and each object has a sphere or box hitbox

then for each calculation step

[past prairie]

i use that to determine if a object is aiming toward a sphere

[zinc quail]

hm was the above a task or did you come up with it yourself for a freetime project?

[past prairie]

the above tasl

task

[zinc quail]

kk

[zinc quail]

yea that works too

[zinc quail]

but if in that physics game each object is a sphere or box then you could calculate the intersections

and let objects bounce of each other!

it only gets a bit more complicated than the calculations we did above

or say in space something would fly in a straight line (unrealistic but whatever)

then you can determine if it hits some other planet

and exactly where

[past prairie]

yeah you can do a lot of thing with that

[zinc quail]

what I mainly want to state is that with a bit experimentation you aren't far off to already pretty neat things ;]

[past prairie]

uh i found -186 for discriminant : (

i forgot the sqr() for a but still with -156

[past prairie]

?

[zinc quail]

at the end

sz-cz

and here the blue brackets are redundant

can be written as C -= r btw :)

[zinc quail]

meaning don't hesitate to experiment with it

[past prairie]

wdym

[past prairie]

i dunnot x)

[zinc quail]

Sqr((sx-cx)) = Sqr(sx-cx)

[past prairie]

oh i see

but now the discrimant output 36

[zinc quail]

because the discriminant is B² - 4AC :-)

[past prairie]

i already correct this

[zinc quail]

ah and spotted another thing

why C -= r

instead of C -= r²

then everything should be fine

[past prairie]

so i have like a exercice an the rayon of the sphere is

sqrt(14)

so it 14

[zinc quail]

rayon?

[past prairie]

radius

sorry x)

[zinc quail]

k but in code

you'd still use C -= r²

just let the input values be float instead of int

and set r = sqrt(14)

[past prairie]

yeah but that don't affect the result

i can juste left 14 instead of sqr(sqrt(14))

[zinc quail]

yeah but the formulas look all correct

what's the desired value?

[past prairie]

1

[zinc quail]

and what are A,B,C

[past prairie]

wdym

[zinc quail]

when you run the code

[zinc quail]

hm and what are the center and line coords?

[zinc quail]

hm little confused as well, a sec

[zinc quail]

are you sure it's meant to be 1?

[past prairie]

uh yes

[zinc quail]

sry for delay, am back