help-0

[night path]

but how do I do it algebaraeicaly or whatever you spell it

[fierce prairie]

uh that is algebraic lol

f(0) = 250 - 25(0)

[night path]

it says show work

[fierce prairie]

it's just math from there

[winter tangle]

calculations*

[night path]

how do I do the math

[fierce prairie]

can you do 25 times 0?

[night path]

oh

what is f(0) tho

[fierce prairie]

that's what you're solving for here

[night path]

so f(0) = 250 what else

[winter tangle]

that's it

[night path]

oh ok thanks

[fierce prairie]

ye

[night path]

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[sick pilot]

question is asking about solutiosn to the system of equations

i already deduce that the matrix A can not have one solution by showing that the inverse cannot be computed

If A is a square matrix, then if A is invertible every equation Ax = b
has one and only one solution. Namely, x = A’b.

and applying the above

would using,

If Ax = 0 has only the zero solution. THen Ax=b has at most one solution

justify that there are no infinite solutions

hence the matrix A is inconsistent

or are there betters way to approach this question

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.close

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[bleak root]

prove that (12)(34) is a normal subgroup of A4

[bleak root]

I think that I can brute force it by using the aHa^{-1} \in H as a normal subgroup test, but that would take a while

any one know any better solutions?

[mortal trellis]

well first off (12)(34) is an element of A4, not a subgroup. but ok

instead of bruteforcing it's enough to just calculate (abcd)H(abcd)^-1 (where that's not cycle notation but instead pi(1)=a, pi(2)=b etc)

and in general it might be useful to know the formula for pi tau pi^-1 for a cycle tau and any permutation pi

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[bleak root]

I don't think that i have it in my notes. Can you refer me to some online/book resources that might answer my question?

[mortal trellis]

$\pi (a_1 a_2 \cdots a_k) \pi^{-1} = (\pi(a_1) \pi(a_2) \cdots \pi(a_k))$

[ocean seal]

Denascite

[mortal trellis]

in cycle notation

try proving that

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[glossy geyser]

Hello, I have a normal vector of a plane and I now want to calculate a family of vectors (dependent on a parameter) which are orthogonal to the normal vector.

My first idea was to say that x,y,z are the cords of my vectorfamily. Then say that normal vector * {x;y;z} = 0, but then I have 3 variables, how should I do this? Let’s say for example I have a normal vector {3;5;0} then I would do 3x+5y+0z=0 but how do I continue?

[limpid turret]

You should read up on vector equation of a plane

[keen pasture]

So you have a plane in the normal form and want to get it into the vector form?

[glossy geyser]

I have a plane in normal form and want to get a family of lines which are in the plane.

[keen pasture]

Normal form is n1x1+ n2x2+n3x3 + a is = 0

You can just plug in random values for x1, x2, x3 and find points in the plane

Each of these points is connected by a line in the plane

[glossy geyser]

like this:

where g_a is in E and n is the normal vector of E

[glossy geyser]

Yeah, this works, but I want to get every possible line (g_a) dependent on a parameter (a)

in german this is called „Geradenschar“, I guess its family of lines in english?

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@glossy geyser Has your question been resolved?

[mortal trellis]

Planes are given 2 parameters

That said, n*(x,y,z) is just a linear system of equations with one equation and three unknowns

Solve it like any other linear system

i.e. here introduce two parameters and then calculate the third coordinate from them

After that you can set one of the parameters arbitrarily and then get a family of vectors only depending on one parameter

[glossy geyser]

sorry, I dont get it, like this? But how do I continue?

what should I do with the 0z?

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[inner kayak]

The solar radiation power per unit area depends on the distance from the sun. The Earth's distance from the Sun is very large relative to its distance from the Moon. Find with the help of the data appearing in the body of the question, what is the solar radiation power per unit area that reaches the surface of the moon on the side facing the sun?
You were assigned to design the first solar power plant on the moon. The power plant needs to generate electrical energy with a capacity of 25 megawatts (megawatts = million watts) for the first lunar settlement. You have at your disposal panels of photovoltaic cells with an efficiency of 22%, that is, 22% of the light energy that strikes them is converted into electrical energy. The average power per unit area of ​​the solar radiation that reaches the Earth's atmosphere during the day is 1360 watts per square meter (square meter). The solar radiation power on the surface of the moon is unknown. The lunar system is built so that the panels are directed towards the sun during the daytime

[inner kayak]

How on earth shall I even start approaching this question

?

[alpine sable]

Bruv 💀

[tribal haven]

first thing is to summarise useful figures

and you might've not pasted the full question, with information given you can only get an answer with some assumptions

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[wary hound]

help

[wary hound]

i can't find a

my work

<@&286206848099549185>

[strong furnace]

@wary hound think about the limit of sin(1/h^2) and whether it converges

[wary hound]

i know limitx x to 0 hsin(1/h^2)=0

but i don't know how to deal with sin(1/h^2)

[strong furnace]

you can't use l'hopitals since the limit of derivative does not exist

[wary hound]

But the question stat that f(x) is differentiable everywhere

so the limit should be exist

i am stuck

<@&286206848099549185>

[strong furnace]

in this case the derivative will still have an existing limit but that is only because you have the choice on a ( which you can choose such that it works ) however you can have functions whose derivatives exist but the limit does not exist an example would be limit of sin(x) at infinity where clearly the derivative exists but the limit of the derivative does not exist

for this case you have to adjust the value of 'a' such that you get a differentiable function, you don't need to use l'hopital's, look at the initial limit, and get rid of all the 'h' and you'll notice the problematic a*sin(1/h^2) in the limit (which will diverge for certain values of a), which you can remove by choosing 'a'

[wary hound]

so if i choose a=0,how to explain ?

[strong furnace]

for any other choice of a you can show the limit diverges

[wary hound]

like that?

.close

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[dire sage]

Would this be correct? I am trying to proof Collorary 4 of Spivak's Differential Geometry Vol.2 but I have very weak Linear Algebra.

[long nexus]

holy

[dire sage]

this are the definitios used

[long nexus]

idk how to do this sorry🤔

im in 7th grade in swiss

[wanton tusk]

hahaha this is uni lvl math

[long nexus]

i will learn this in 9th or 10th grade

[dire sage]

Oh My Bad lol

[wanton tusk]

huhhh

[long nexus]

ik XD

[dire sage]

what age is 7th grade?

[long nexus]

13-14

im 14

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[west girder]

impossible

not in usual school curriculum

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[alpine sable]

Why is it that when finding an angle in the third quadrant

[alpine sable]

You’d do 270 - theta

Or 180 + theta

[serene flame]

because that way you can roll it to the first quadrant

and its trig functions are equal in magnitude

as for why specifically those angles, well you're taking the difference between the extreme of the 3rd quadrant and the 3rd quad angle so you get the 1st quad one

[ornate condor]

🛞 roll

[serene flame]

roll indeed

also I can't.see that first symbol

[ornate condor]

:c

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[tall coyote]

Hello can someone explain to me this pls ?

[tall coyote]

There's someone ?

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[hard patio]

it is generally frown upon to have to download some file here

it would be better

if you send in screenshot

[alpine sable]

send a normal picture

[tall coyote]

ok

but there 12 pages

I dont understand

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@tall coyote Has your question been resolved?

[alpine sable]

Lmao me neither

Fuck theory of computation

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[tall coyote]

.close

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[steady basin]

how do they get that to be the cumulative density function

[noble sinew]

You don’t how to get CDF from PDF?

[steady basin]

integrate

but wheres the -16/9 come from

[noble sinew]

From integrating?

[steady basin]

how

when u integrate 2t

u get t^2

[noble sinew]

You are not just finding antiderivative

There are bounds in your integral

[steady basin]

5/3 and 4/3?

but that gives u 1

[noble sinew]

No

Check again how to find CDF?

[steady basin]

CDF is the probability P(X<=x)

[noble sinew]

Clearly you aren’t sure how to find CDF from PDF

So check that since you tried twice and was wrong both times

[steady basin]

integrate between 5/3 and -infinity?

if thats not correct than idk

cuz CDF is P(X<=x)

so u integrate between x and - infinity

of the PDF

[noble sinew]

So -inf to x?

And not 5/3?

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[wheat crystal]

@vale wigeon help i just understood what u meant when cancelling the 6!s lmaoo

i forgot u can rewrite a/b divided by c/d as a/b * d/c

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[alpine sable]

is : hypotenuse/2 = height of the triangle

[alpine sable]

always correct?

[tired kettle]

This is not generally true no

[alpine sable]

ah okay

shit

how do i solve this shit then

[tired kettle]

depends on the triangle

[alpine sable]

for the A1 triangle

in my teachers sketch she found the hypothenuse with this theorem

[tired kettle]

is h here the length you are calculating?

[alpine sable]

no i just need to find how tf did she get 2h

this is her sketch but she skips like 5 steps

[tired kettle]

so you know all three angles in the triangle

[alpine sable]

yes

[tired kettle]

do you know sohcahtoa?

i.e sin(x) = opposite/hypotenuse

if you call the height of the triangle h

as she has

then you have sin(30) = h/hypotenuse

sin(30) = 1/2

so 1/2 = h/hypotenuse

or hypotenuse = 2h after rearranging

[alpine sable]

sorry to interrupt

we cannot use that unfortunately

oh hold on

yeah we can’t

[alpine sable]

but for example for random triangles

is this theorem true?

like 2•height=hypotenuse

i know using trigonometry is better it’s just that we get points removed if we do..

[tired kettle]

no that is not true in general

[alpine sable]

ah man

do you know how i could potentially find 2h without trigonometry?

[tired kettle]

what is d?

c+h?

when what is c

[alpine sable]

d is the basis of the whole triangle

[tired kettle]

yes

[alpine sable]

the base of both triangles

[tired kettle]

she has used this

[alpine sable]

c is the base of the square triangle we created

[tired kettle]

area is a1

[alpine sable]

she didnt use trigonometry

she only used certain theorems

and i cannot find them, she mentionned this one once and i wrote it down but it’s wrong

or i misheard something

[tired kettle]

she is using trig implicitly

for example

on the 45 degree 45 degree 90 degree triangle

she can label the two lengths h

because of the angles

[alpine sable]

yes right

[tired kettle]

just like on the 60,30,90 triangle

she can label them h and 2h

[alpine sable]

how doe?

is this theorem true for 60 30 90 triangles then?

we call it isosceles in french but idk if it exists in english

so is it true for isosceles triangles?

no its not isosceles im dumb

[tired kettle]

the 60 30 90 is not an isoceles

yeah the 45 one is

[alpine sable]

yes

[tired kettle]

its true for 60,30,90 triangles yes

because sin(30) = height/hypotenuse

[alpine sable]

finally 🥲

[alpine sable]

yes but shes stupid

thanks a lot doe

i can finally move on

[tired kettle]

all good

[alpine sable]

❤️ 💗💯💯💗💯

.close

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[swift monolith]

The Baltic Sea has an area of ​​approx. 413,000 km² and holds approx. 20,000 km³ of water. The salinity varies between 2‰ and 20‰ depending on the areas and depth. How much does all the salt in the Baltic Sea weigh? Assume that the average salinity is 9‰ and that 1 liter of seawater weighs 1 kg. Answer in basic power form.

[swift monolith]

i need help answering this question

what i did was 0.009 x 20,000

but i think im missing a step

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[swift monolith]

!close

.close

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[hot marsh]

Can anyone help with physics

[viral pagoda]

Dude

There's a separate server for physics

[hot marsh]

It asks me to verify via phone

I dont wanna

[viral pagoda]

Try receive sms cc

[viral pagoda]

Or come to jee pro discord server

https://discord.gg/7vfuZKpm

[hot marsh]

The what

[viral pagoda]

There's this section for physics doubts here

[viral pagoda]

Or
Google
Receive sms cc

[hot marsh]

FML

[viral pagoda]

?

[hot marsh]

@100 server limit

[viral pagoda]

Wdym

[hot marsh]

.close

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[fading aurora]

Given
The Laplace transformation of a function f: $$\frac{4-z^2}{(z^2+4)^2} \cdot e^{-z}$$

[fading aurora]

ah crap

texit bot down everywhere?

nvm then ill come back later

.close

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[mellow linden]

how did they get 1.5 and 4

this is precalc btw

[alpine sable]

'seems' probably means read off the graph

[solemn juniper]

The derivative is 0 at x=1 and x=2, so they guessed it was highest in the middle

Not a bad estimation tbh

(Or means of estimating, rather)

[mellow linden]

what about the 4

.close

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[severe mural]

can someone explain how I would find rate of change for b

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@severe mural Has your question been resolved?

[severe mural]

<@&286206848099549185>

[deep fable]

im not a helper, but i think the rate of change would just be delta y over delta x

so change in y over change in x

might want to double check that

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[rustic zinc]

<@&286206848099549185>

[manic wind]

!15m

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

[manic wind]

do you know how to find the slope?

[rustic zinc]

i forgot

my teachers dont teach me enough

[manic wind]

slope can be thought as how steep the line is

have you seen rise / run?

[rustic zinc]

ok so I can use this example you showed me on the current question I am on?

[manic wind]

yeah

[rustic zinc]

ok

okay, so what would the y-intercept be?

the slope is 3

y=mx+b ?

[alpine sable]

plug 0 into x, and you'll get the y-int

you have to find b right? and you have y = mx

you said slope is 3 so y = 3x + b

you just said the slope is 3?

or i read it wrong

so m = 3

ok I will write it down for you, one second

so y would be your output for x right?

so you want the y-intercept, when x is 0

Holy shit I'm bad at math I need to write this down

slope should be -3/2?

your points are (-2,6) and (4, -3)

so rise/run like the other guy said is change in y divided by change in x

you should have:
(-3 - 6) -9 -3
--------- = ------ = ------
(4 - [-2]) 6 2

So now your equation is
y = (-3/2)x + b

Yeah, it's -1.5

Exactly

So here is how you find b: you already have two points to choose from to solve for b

Yes

I am dearly sorry if I barely helped but do you understand how you're supposed to solve for b?

Nah I have too much time

Ask away

Now I got a notebook so I can solve it myself if my brain can't comprehend it :p

Sure

Idk what to say to prove I am not an npc

Yes, I would first convert each of the answer choices to y = mx + b format so you should do that

I think you can plug in the point and slope better if you have it written in the format already

Not quite

So when there's parentheses, you have to multiply the 4 to both of the terms in the parentheses

For 4(x+2 ) you have to multiply x by 4 and 2 by 4

Yes

No, that's y - 5 not 5y

you're subtracting 5 from y

So you can add that 5 to both sides

So you want to find y alone

You don't want y - 5 because that's not the proper y = mx + b format

Yeah so you add 5 to 4x + 8

so yeah it equals 4x + 13

that's a

now try b on your own

like answer choice b

and it's the y intercept

where will the point be on the y axis when x is 0

Do you think it's correct?

Try plugging in the point you have

(-2, 5)

Well I can't just tell you the answer

Yeah but I'm pretty sure that's against the rules and generally this is a really important topic to grasp

Yeah but understanding is not the same as giving you the answers

Anyways we can work it out together

Plug in (-2, 5) into y = 4x + 13

So you should do it

Also I feel like you should ask your teacher for help on this if you still need some help

yeah but that's not how life works lol

I cannot emphasize enough how important this is for higher math, even if you're not planning on taking calc or anything

Either way we can do it together but genuinely slopes are incredibly important

That's awesome!

Either way if you want to work on this I'm happy to help but if you don't want to do it then you should probs close the ticket

So, we should get back to work lol

Dang really I wish I stopped high school in sophomore year

also you're my little brother's age which is tripping me up

Anyways back to algebra

can you plug in (-2,5) to y = 4x + 13 and see what you get

if 4(-2) + 13 equals 5 that means a is one of the answers

yes

if you see that the points plug into the equation where the right side equals y and the slope of the equation is 4, then that is an answer

Idk about b but I agree with a and d

wait what's your equation for b

How did you get that

I got y = 4x - 13

so the original is y + 5 = 4(x-2)

it can be both

We've already established 4(x-2)

so that is y + 5 = 4x - 8

yes I'm trying to show you how b looks in y = mx + b form

so y = 4x - 13 for b

I didn't see that (-2,5) plugged into that

I got A and D

I said I didn't know if b was right

So I showed you how I was doing it and why I don't think it's right

if you plug (-2,5) into y = 4x - 13 you get:
5 = 4(-2) - 13
so 5 = -8 - 13
5 = -21

but 5 ≠ -21 so it can't be b

I just showed you why it's not b

I mean I am not you so if you think it's a and d go for it

3 tries, if you mess up we can try again

I gtg after this though so

[young crow]

Hey

I think its A and D

[alpine sable]

I swear that's not an alt

[young crow]

I ve been reading

But in B you get 10=-16

When you plug in the point

Alright

I d start by calculatong the slope

Do you know how to do it?

Or well you can try plugging in the points and checking if the equality holds

Yeah

Thats right

Yes

Haha

Correct

Now i d suggest you check with the points if the equality holds

You should try both of them

The points you are given represent (x,y)

So you replace the x with the first number of the pair

And y with the second

[tacit arch]

oh they got banned

.close

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[grave niche]

i dont understand why we take half of your A value

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[summer merlin]

how would you solve this problem?

[severe mural]

so it would be y over x when finding slope?

[gray isle]

by substitution

[summer merlin]

but like how

[gray isle]

do you know what substitution is, have you ever done it before?

[summer merlin]

yeah but not with much practice

[gray isle]

the variable y is already isolated in your first equation

substitute that into your second equation

[elfin moat]

or u can multiply the second equation by -3 :>

[gray isle]

that'll give you an equation with a single variable x, which upon solving will give you its value

or u can multiply the second equation by -3 :>
the question wants you to do it by sub

[elfin moat]

oh

[summer merlin]

ok

[gray isle]

that'll give you an equation with a single variable x, which upon solving will give you its value
then sub that value of x back into either equation to get the value of y

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[pliant yew]

what is this question even asking

like what does it mean for a vector to take a column vector as an input

oh nvm

.close

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[orchid shore]

It's asking for the chances that it's negative, but I would assume its when it passes the Y-axis, but would it be the same as the chance a variable is drawn? I have no clue

[orchid shore]

I think maybe the writing should be in the second section

[lone heart]

@orchid shore Has your question been resolved?

[orchid shore]

no

[opaque lynx]

https://i.imgur.com/rMhHsbg.png
is it possible that a line at any angle other than 0 degrees in the x/z plane is infinately large in the x dimension (occupies the whole X dimension)?

[crude rose]

wdym?

like the line hits the plane z = 0 at all points?

[opaque lynx]

at 0 degrees the line is pointed directly at you, and at any other angle it occupies the entire x dimension

[crude rose]

well tilt it 90 degrees

so i guess not?

[opaque lynx]

the z dimension is the one orthagonal to the plane we're looking at the line from

if it rotates in X/Z dimension (towards or away from you), does the line occupy the entire X dimension for any angle other than 0 degrees (where it's pointed directly at you)

[crude rose]

rotating in the xz dimention is up and down

right?

[opaque lynx]

no

[crude rose]

how

[opaque lynx]

rotating around the y axis

[crude rose]

yes

[opaque lynx]

where it goes in/out

[crude rose]

so 90 degrees

[opaque lynx]

not up down or left right

the z dimension is the one pointed at you

if the x/z plane is at any angle other than 0 degrees does the line occupy the entire X dimension?

[crude rose]

Yes

90 degrees

u cant

or any degree

to span the x axis

[opaque lynx]

what does this mean?

[crude rose]

if u rotate the line by anything in the x z plane

it comes up

off the x axis

[opaque lynx]

https://i.imgur.com/OKFFDVv.png

[crude rose]

or towards u

i suppose

[opaque lynx]

it rotates around the red pole

[crude rose]

the way u drew it

yea

if u lift it

its no longer on the x axis

[opaque lynx]

if you rotate it around the red pole going up and down

does it occupy the entire x dimension for any angle other than % 180 degrees?

[crude rose]

no

[opaque lynx]

you sure?

let's say the slope of the line is infinity to 1 in the z,x dimensions

so for every one unit in the x dimension it is infinity units on the z dimension

[orchid shore]

yall in my channel dawg

[tired echo]

i was about to say that you hijacked this poor man's channel

[orchid shore]

im just tryna get my thing done LMAO 😭

[orchid shore]

anyone?

[lone heart]

@orchid shore Has your question been resolved?

[lone heart]

@orchid shore Has your question been resolved?

[orchid shore]

close

!close

.close

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[keen seal]

So
Reimann’s sum and area under a line

[keen seal]

How did they get the 2 and the 10 for the part c?

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[keen seal]

.reopen

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✅

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[wind bloom]

i dont get why theres a zero cuz -12 and 9 is not on the interval 0,7

[alpine sable]

-12 and 9 describe the y values, the interval from 0 to 7 describes the x values

It changes signs as seen with the turn from negative 12 to positive 9,so there must be a root in between

[wind bloom]

so dont need to find the y values?

just factor and find the zeros?

[alpine sable]

You have to find the y values just to verify that there is a sign switch on that interval

If both y values were positive or negative, you cannot say that there will be a root on that interval with IVT

[wind bloom]

wdym by a sign switch?

oh wait i think i kinda get it

[alpine sable]

By the intermediate value theorem, if there is a continuous function with an interval of [a, b], then that function needs to take all values in between f(a) and f(b) too

[wind bloom]

so first i have to plug in 0 and 7 which gets me -12 and 9, then find the zeros and see if any of the zeros are in between -12 and 9 since thats the interval?

[alpine sable]

It basically means, if you have an interval running from 0 to 7, and you figure out that f(0) = -12 and f(7) = 9 then that means you can find values for numbers in between -12 and 9 on that interval too

[wind bloom]

yeah

[alpine sable]

Which includes 0 because 0 is in between -12 and 9

[wind bloom]

ok got it

thanks!!

.close

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[alpine sable]

Yep

This is for continuous functions on closed intervals, though

So keep that in mind

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[wind bloom]

wait

[wind bloom]

why is 6 a guaranteed zero

why not -2

[alpine sable]

X = -2 is not in the interval from 0 to 7

It may be a root, but you can't say anything about it with IVT, because it's outside of that interval

[wind bloom]

ohh

what about the interval -12 and 9?

[alpine sable]

That just tells u that the thing takes any value in between -12 and 9 on the interval from 0 to 7

Which is why we were able to say that there was a guaranteed root there by IVT

Which is x = 6 with f(6) = 0

[wind bloom]

ok got it lol

[alpine sable]

Aye aye

[wind bloom]

thx

.close

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[alpine sable]

Ofc

Good luck with studies

[wind bloom]

thansk

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[tulip summit]

how would I solve this without using l'hospital's rule?

[vale wigeon]

your last step is incorrect

sin^2(x)/cos^2(x) - 1 does not mean sin^2(x)/cos^2(x) * (-1)

[tulip summit]

Ah ok ok

[vale wigeon]

recommend substituting t := x - pi/4 to turn this into a limit at zero

may be easier to see how to proceed from there

[tulip summit]

I will try that

Also nice pfp

[vale wigeon]

thank you

[alpine sable]

thank you

[tulip summit]

I changed it to this, how would I go about turning it into a limit at zero?

[vale wigeon]

well, the first step would be to actually format your work properly and not fragmented

[tulip summit]

So just leave it like the step before?

[vale wigeon]

$\lim_{x \to \pi/4} \frac{\tan^2(x) - 1}{x - \pi/4} = \lim_{t \to 0} \frac{\tan^2(t + \pi/4) - 1}{t}$

[ocean seal]

Ann

[vale wigeon]

this is what you should have

now apply difference of squares and/or tangent sum identity on the top

do some algebraic fuckery to it

[tulip summit]

Oh ok ok

[vale wigeon]

eventually, hopefully manage to pair up tan(t) on the top with t on the bottom

[tulip summit]

So they cancel into 1?

[alpine sable]

thank you

Whats the square root of 66?

[tulip summit]

.close

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[alpine sable]

How do I calculate this? aside from making the pi 3.14 and subtracting

Arg Z = -(π - tan^-1 (-4/-4))

[trim wagon]

[alpine sable]

tan ^-1(-4/-4) is 0.78

[trim wagon]

?

[tacit arch]

wot

[trim wagon]

,w -4/-4

[ocean seal]

Results provided by WolframAlpha

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

Requested by Deep.#7373

[trim wagon]

[alpine sable]

O

[tacit arch]

,calc -(pi - atan(1))

[ocean seal]

Result:

-2.3561944901923

[alpine sable]

.

Thanks-

[trim wagon]

[tacit arch]

,calc pi/4

[ocean seal]

Result:

0.78539816339745

[tacit arch]

oh i never thought about that

[trim wagon]

[alpine sable]

Jeez man

I appreciate it

.close

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[tacit arch]

,calc -3pi/4

[ocean seal]

Result:

-2.3561944901923

[tacit arch]

great, i can still add basic fractions in my head

[trim wagon]

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[lost pond]

how to i determine if it is side angle side or side side side?

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@lost pond Has your question been resolved?

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[trim wagon]

<@&268886789983436800>

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[sleek gyro]

.close

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[fair swallow]

Can you paint 9 pairwise non-adjacent squares of a 5x5 grid black such that, on the 16 remaining squares, you can draw 4 lines, each connecting 4 adjacent squares, each without lifting your pen or going over the same square more than once? (adjacent squares must share a side)

Here's an example with 8 squares painted black instead of 9.

[night geyser]

Does a t shape make a valid "line"? If so, simply colour the diagonals black

[fair swallow]

no, you can't draw a t shape without lifting your pen

[night geyser]

I mean I think you can

But if you mean without passing through the same cell twice

Then sure

[fair swallow]

sorry, i meant *each without lifting your pen or going over a square more than one.

*once

<@&286206848099549185> ?

[night geyser]

No

Any collection of 9 circles must divide the grid into at least 2 regions (why?)

Reason about the shapes of those regions.

[fair swallow]

hmm

[night geyser]

Hint: a lot of circles must share corners

If two squares share corners of the same line segment, it becomes very hard to draw a line "out of" that inlet

It's possible but quite forced

[fair swallow]

sorry? i didn't quite get that

[night geyser]

As an example, a construction like this is already impossible

Because both these inlets need lines leading out of them

Which is illegal

Sorry for the low quality sketch, I'm on phone

[fair swallow]

oh yeah
but how can i prove that we must make one of these constructions?

i get that lots of black squares must share corners, which makes it hard to draw lines

[nimble jackal]

Guys please help me with this h/w

I don't get the concept

[fair swallow]

but i cant really [[formalize?]] it for a proof

[nimble jackal]

What does 1/3 = to?? In equivalent fractions

[night geyser]

Get your own channel

[nimble jackal]

Wdym?

[night geyser]

#❓how-to-get-help

[nimble jackal]

K

[night geyser]

random-internet-guy here's perhaps a better way to look at it

Every valid line is composed of two "sublines" of length two

[fair swallow]

yes

[night geyser]

This rules out the T shape (why?)

[fair swallow]

so you can't make it by connecting 2 segments of length 2

[night geyser]

Colour the grid two different colours

Note that 13 are yellow, 12 are blue

But our 8 sublines must have equal parts yellow and blue

So there are 5 yellow dots, 4 blue dots

[fair swallow]

yes, i get that

(i actually already got this result before, but i just didn't know what to do with it)

[night geyser]

WLOG pick a blue circle on the edge (why do we know at least one blue circle is on the edge?)

Now proceed by cases

[fair swallow]

if4 blue circles are all in the center, we wouldnt be able to draw a line to the square in the middle

oh alright

thanks!

.close

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[versed crater]

why cant u divide both sides of a congrugence

[versed crater]

simply i mean

for example 3x = 12 mod (15)

sure x = 4 mod 15 is one soltuion but there are others

[ornate condor]

um

there are zero divisors

[echo socket]

3x = 12 = 27
x = 9 ?

3x = 12 = -3
x = -1

I'd rather look at this congruence by referring to the definition of it

[versed crater]

ah Thanks that is a mcuh bvetter way of lookjing at it

[echo socket]

Looks like mod is getting divided too though

[versed crater]

i was watchign a vdieo where they had some wierd methods for doing this

[vale wigeon]

you can divide everything by 3 and get x = 4 (mod 5)

and that way it is legitimate

[echo socket]

Yeah, 3x = 12 (mod 15) means that 15 | 3(x - 4), 5 | (x - 4), x = 4 (mod 5)

[versed crater]

similalry i thought that if a=b mod m and if d divides me a=b mod d

is this true

for example lets say x= 4 mod 12 then does x = 4 mod 4

but also 4 mod 3

[vale wigeon]

yes

[versed crater]

ah thanks

.close

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[versed crater]

.reopen

[lone heart]

✅

[versed crater]

sorry one last thing

consider the congrugence 4x=4 mod 12

x = 1 mod 12 is a leigitmate solution

x= 1 mod 3 and thefore x = 4 mod 3 and x = 7 mod 3 so x = 1 mod 12 x= 4 mod 12 and x = 7 mod 12

are all legitimate solutions

[lost roost]

How should I factor x^4 + x^3 + 2x^2 -x + 3

thanks

[versed crater]

but surely x= 1 mod 4, x= 5 mod4 , x= 9 mod 4 so x= 5 mod 12 x=9 mod 12

but these dont satify 4x = 1 mod 12

how come ?

[versed crater]

use a differnt channel mate

[lost roost]

sorry

[versed crater]

sorry one last thing
consider the congrugence 4x=4 mod 12
x = 1 mod 12 is a leigitmate solution
x= 1 mod 3 and thefore x = 4 mod 3 and x = 7 mod 3 so x = 1 mod 12 x= 4 mod 12 and x = 7 mod 12
are all legitimate solutions

but surely x= 1 mod 4, x= 5 mod4 , x= 9 mod 4 so x= 5 mod 12 x=9 mod 12
but these dont satify 4x = 1 mod 12
how come ?

[lost roost]

you were thinking 4x=4 mod 12 but why did it end up being 4x = 1 mod 12

[versed crater]

ah yes

buyt strill

it dosent satisfy

[lost roost]

we know x = 1 mod 12

wait

(x-1) is a multiple of 12, so it is also a multiple of 4

so we know (x - 1) = 0 mod 4

and

x = 1 mod 4

[lone heart]

@versed crater Has your question been resolved?

[versed crater]

Not yet

[lost roost]

so what

[lone heart]

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[primal lintel]

need help with c

[vale wigeon]

let's call the picky girl and boy Alice and Bob respectively

then Alice's neighbors are both girls and Bob's neighbors are both boys

[primal lintel]

yep

it would kind of have to look like this

[vale wigeon]

not quite

number the positions around the circle 1 through 8, starting from Alice

(i.e. Alice stands in position 1)

then Bob can stand in positions 4, 5 or 6, but in no others

so he has 3 options

[primal lintel]

Okay, yeah that makes sense

bob could be any of these

[vale wigeon]

yeah

but whatever he is, there will be 6 ways to pick who gets to be next to him

[primal lintel]

3p2 right

okay

and there will be 6 ways to choose who's next to alice

[vale wigeon]

indeed

and after that there will be 2 people left and 2 seats for them

[primal lintel]

6 * 6 * 2?

@vale wigeon they also have a * 3 in there

6 * 6 * 2 * 3

[vale wigeon]

.

bob has 3 options for where he could be.

[primal lintel]

ah shit

thanks

any chance you could help me with one more little one?

@vale wigeon i'm not sure why they are multiplying by 12 in the second line

[vale wigeon]

the probability of exactly one voter voting for party B is 12C1 * 0.4^1 * 0.6^11

the number of voters for party B follows a binomial distribution

[primal lintel]

ah okay

little confusing but just need some sleep haha

thanks

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[broken wraith]

i^52

[broken wraith]

I need to understand this

[echo socket]

i^52 = (i^2)^26 = (-1)^26 = 1

[broken wraith]

How?

I need to learn how can I solve it by myself

[echo socket]

Just remember that i^4 = 1, that's it

[broken wraith]

Ok

[echo socket]

So i^4k = 1 is also true for integer k

[broken wraith]

K

[broken wraith]

Why did you put negative in the solution

[echo socket]

That's just a further explanation

i^2 = -1
So
(i^2)^26 = (-1)^26

[broken wraith]

If the power from multiples of four the the solution equals one?

[echo socket]

Yes

[broken wraith]

Ok , thanks

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[echo socket]

It's correct

Show your work

Limit of f is not the same as limit of f'

I'd recommend rewriting x^x as e^(xlnx) and finding the limit of xlnx

[zealous drum]

i need helpp

about

[echo socket]

Post your question in an available help channel

[zealous drum]

ok

[echo socket]

$\lim_{x\to{0^-}}e^{x\ln{x}} = e^{\lim_{x\to{0^-}}x\ln{x}}$ since the function $e^x$ is continuous

[ocean seal]

A Lonely Bean

[echo socket]

It does

Rewrite xlnx as lnx/(1/x) and apply L'hopital's rule

Oh wait 0^-

x^x isn't even a real function for all x < 0 though

It's weird to approach it like that

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[upbeat hemlock]

Hellow ! I wanted to know how to get the derivative number... I figured out -h-2a+17 ? What should I "take" ? -2 ?

I toke Tf(a)

I probably know what to do next I use the equation of tangent and get the limit with

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[uneven burrow]

How to do this?

[upbeat hemlock]

do wut ?

[uneven burrow]

like prove it

[wanton tusk]

use identities

[wanton tusk]

no need so complex actually hahaha

[stark grotto]

i just want to be more detailed hahaha

[wanton tusk]

ohh ya I understand

bec I’m very careless HAHA

[uneven burrow]

Omlll tysmmmm i understand it nowwww

TYSSMMMM

[wanton tusk]

welccc!

[uneven burrow]

.close

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[bitter vessel]

Taylor series (I get the concept of estimating a function with polynomials i just don't know how to math it)

[vague coral]

do you know the formula of the taylor expansion ?

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@bitter vessel Has your question been resolved?

[bitter vessel]

yeah sum of f^(k)/k! (x-c)^k

Something like that??

[bitter vessel]

And k tends to infinity

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[sharp rock]

Playing with the actions, I've got things like

$g = x (x \star g) x^{-1}$

[ocean seal]

todadqa

[sharp rock]

and similar expression for x

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.close

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[lyric osprey]

Hello, I do not understand this question

Please help 🙏

<@&286206848099549185>

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[untold pond]

plz solve this equation 310(2700-2d)=1116d

[slate jolt]

you want to learn how to do it or just the answer?

[untold pond]

how to do it

[slate jolt]

so basically you have an equation

meaning that it states that both sides are equal

so you can keep construct a new equation from this one by appliying the same thing on both sides

since both sides are equal

applying the same thing on both sides will give equal results

so the equal results will be part of a new equation

and the goal with this kind of exercise

is to construct new equations that are implied by your first

using this rule

"apply the same thing on both sides"

ideally you want to end up with an equation

with

d = something

and you'll have it solved

so for now i think you should expand the bracket first

[untold pond]

i tried to solve it but cannot get d value

[slate jolt]

to simplify

i'll guide you

[untold pond]

thanks sir

[slate jolt]

so do this

[untold pond]

so do i have to multiply 310 with (2700-2d)

[slate jolt]

yeah

distribute

310 * 2700 + 310 *(-2d)

[untold pond]

837000-620d

[slate jolt]

okay so now lets isolate the term with d

by subtracting 837000 on both sides

do you see the logic?

[untold pond]

how to subtract 837000 on both side

can u show me how to do it

[slate jolt]

oh actually nvm you shouldnt do this there is also a d term on the right

actually we should add 620d on both sides

which would be like this

837000 - 620d +620d = 1116d +620d

and after simplifying

837000 = 1736d

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[lunar lintel]

how are we supposed to do this

part

dy/dx

for my dy/dx i got 2t^2

t= 0

but they didnt

[worn fox]

Show your work

[worn fox]

Why are you setting it equal to zero

[lunar lintel]

to find t?

[worn fox]

But why do you think dy/dx = 0

[lunar lintel]

idk

what am i supposed to do?

[worn fox]

Use the equations for x,y to find t for the point (1,3)

[lunar lintel]

how

[worn fox]

(x,y)=(1,3)

You can solve equations

[lunar lintel]

there is no where to plug that in

the equations involve t

not x and y

[tacit arch]

x = 1 + ln(t) = 1

[worn fox]

The equals sign is a wonderful thing

[lunar lintel]

??? how does x= 1 + ln(t) = 1

[tacit arch]

did you read this

and this

[faint estuary]

You need to find the t that makes the equation work.

[lunar lintel]

oh nvm

[tacit arch]

if x = a and x = b, then a = b

[lunar lintel]

ln(t) = 0

[faint estuary]

Yup.

Now put the same value of t into the y equation to check it.

[lunar lintel]

t = +-1

for the 2nd

but its not -1 since ln(-1) is not 0

[faint estuary]

t=1 is the solution for both the x and y equations, to get the point (1,3)

[lunar lintel]

so now i plug that into the dy/dx

to get slope

ok

[faint estuary]

Yep. Once you have your dy/dx, put that value in to get the slope.

Best of luck

[lunar lintel]

what if uget

2 values