help-28

[worldly drum]

if have doubt mark me plz

so, the idea is

[gloomy warren]

I will get the intersection point

First

[worldly drum]

ok

bit

but

do you understand

the idea?

[gloomy warren]

Not really

[worldly drum]

ok

[gloomy warren]

Ik how to get the intersection point

But not the equation of straight line

[worldly drum]

oh

you have 2 points

the intersection and origin

put this on the general equation of a line

(y - yo) = (x - xo)*m

oh, its on the space

i will make this in a paper for you

[gloomy warren]

I got the intersection point for -6 -5 14

[worldly drum]

ye

its correct

and the equation?

[gloomy warren]

Hm

[worldly drum]

this the vectorial equation of a line

on space

[gloomy warren]

Why did u do 0-Ip

[worldly drum]

i need an director vector

of the line

[gloomy warren]

Loke

Like

The distance is the intersection point

[worldly drum]

now i will teach you how the vectorial equation works

do you know what is an vector

?

if yes, an vector in some cases represents movement

so, if we take an point

and a vector

we can move this point

with this vector

"move"

its like we can create other points using summing a vector and a point

so, the vectorial equation

do it

(x, y, z) = (a point) + k{its an number that will define the "size" of the vector} * (the director vector)

what is the use for the director vector?

this will say the direction of the "movement"

are you there? @gloomy warren

so, the infinity possibilities of "k" value, will represent an line

if k = 1

you can have the exactly point when k = 1

the vector will "move" the base point for someplace in the space, because of the sum of the director vector and this base point

its exactly what youve done

O - Ip is the director vector

[gloomy warren]

I'm reading

[worldly drum]

ok

and the point is the base point

the base point could be the origin too

[gloomy warren]

I refer to Ip as D and k as t

That's what I missed

[worldly drum]

okay

so

do you understand it all?

Ip = Intersection Point

if no,

https://www.nagwa.com/en/explainers/105130403094/

[gloomy warren]

Hss

Yes

I understand now

Thanks

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[deep lantern]

Can someone help me type this into wolfram alpha

I want to double check my answer

[woeful inlet]

I don't know about wormalpha

But I put it in:
https://www.integral-calculator.com/

The answer is

Steps:

[deep lantern]

Thank you

How would you type a base in log using the integral calculator?

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[heady vine]

How do i solve b

[severe basin]

!status

[full forum]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[heady vine]

1

[severe basin]

Start by applying this rule ^

[heady vine]

Bro u still here

?

[woeful inlet]

Wait a lil

People may go to different channels and help multiple people

But alr you've written it correctly

Now simplify the powers

[heady vine]

How

[woeful inlet]

What is $8 * \frac{-3}{4}$?

[glossy valve]

rubicminer

[heady vine]

-6

[severe basin]

Simplify the exponent in denominator as well

[heady vine]

-15

Is that the final answer?

And then i flip it to make it postive?

[heady vine]

Yh

[severe basin]

Simplify (16)^(-3/4)

[heady vine]

0.125

[woeful inlet]

Very well

Now you have:
$0.125 * \frac{w^-6}{y^-15}$

[glossy valve]

rubicminer

[woeful inlet]

How can you further simplify it? (hint: negatives)

@heady vine

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[heady vine]

.close

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[fathom arrow]

how do i deal with the nominator?

[round gust]

Nominator?

You mean top or bottom

[warm abyss]

Multiply the numerator out and use the linearity of integrals

[warm abyss]

It's as confusing as writing miximum

[round gust]

Lol

[fathom arrow]

numerator *

top

[round gust]

Use the formula

For (a-b)³

[fathom arrow]

can it be x^3 - 1?

[round gust]

Sure

[fathom arrow]

1^3 is 1 anyways

so if that's the case

i can split it into

x^3/x^2 - 1/x^2

right?

[round gust]

Hm

Whats the -1/x² part

That doesnt =1

Or what are you trying to do?

[fathom arrow]

is that wrong?

[fathom arrow]

what do you mean?

i am just trying to split them into to different terms

so that i can flip the x^2

x^3/x^2 is going to be x

[round gust]

Yes

[fathom arrow]

and then 1/x^2 can be x^-2

now i can integrate

right?

[shut cobalt]

Looking at the given integral, since the highest power of x in the numerator is 3 and in the denominator is 2, we should just expand (x-1)^3 and then just break it up into fractions and integrate them all separately. Or use synthetic division.

The former is better and quicker, imo.
Synthetic division is just overkill.

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@fathom arrow Has your question been resolved?

[fathom arrow]

how to expand it?

someone already told me that before but i don't know what is meant by expand it

i didn't study math in english

[shut cobalt]

(a-b)^3=a^3-3a^2b+3ab^2-b^3
Once, you do this you can break it up into fractions. And then you can integrate.

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[fathom arrow]

thank you, i'll try that rn

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[pulsar cradle]

Where did i go wrong in my sigma notation because the answer is wrong

This is literally the answer

Why is it marked wrong

So i found that the answer is (33cbrt(11))/4 but like where did the 4 or 33 come from? When was the 11 cubed

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[torn jolt]

need help

[torn jolt]

i need to decompose

1-x²

its a function compose

[narrow granite]

f°f(x) = 1-x²?

[torn jolt]

no

if f(x)=1-x²

from g°h° etc idk

i need to decompose 1-x²

like x---->x² idk

@narrow graniteso

[gritty rose]

can you just take a picture of the question

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[sleek quest]

could i get a hint for this problem? just a hint, i would like to solve it mainly myself:
Winnie the Pooh has 8 clay pots with honey; they weigh 1, 2, 3, …, 8 pounds. Each one is labeled with its weight in pounds. Baby Roo played a joke on Winnie, placing a piece of cheese weighing 1 pound into one of the pots. How does Winnie the Pooh determine which clay pot has cheese in it in two weighings, using a two-sided balance scale with no weights? (The scale only shows which side has larger weight or if the two sides have the same weight.)

[frosty geyser]

try to combine weights in the first weighting

[sleek quest]

ive tried 4 and 4 and 3 and 3 but the one with cheese can be on either side

should i try 2 and 2?

@frosty geyser

[frosty geyser]

you need to weight all 8 weights in the first weighting

then formulate a follow-up for both possible outcomes

[sleek quest]

so 4 and 4 then?

[fierce basin]

@frosty geyser Have you worked out such a solution that uses that as a first step?

[frosty geyser]

can't quite make it work

with 2

but 3 weightings would

no glue how you would do that then, since I believe you can only remove half of the remaining possibilities with each weighting

[fierce basin]

@sleek quest My hints for you are as follows:
Use the fact that the balance has a trichotomous output.
In deciding what you will do for a second weighing, consider all objects at your disposal.

[fleet cedar]

does your solution work?

i can only think of binary search but that does it in 3 like jiggly's

[fierce basin]

Would you like me to message it to you?

[fleet cedar]

sure...

[frosty geyser]

right, I guess 3 and 3 is the way to go for the first weighting

[fleet cedar]

ah i got it

yeah thats quite clever

[sleek quest]

ok ill try with 3

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@sleek quest Has your question been resolved?

[sleek quest]

what if the ones you chose dont have the cheese?

thats what confuses me

anyone?

[frosty geyser]

then you know that one of the remaining ones has it lol

[sleek quest]

but how will you know if the ones you chose doesnt have it

[sleek quest]

@frosty geyser ?

[frosty geyser]

have what

[sleek quest]

the pot with chesse

[ember gulch]

i need help pls

[sleek quest]

cheese i meant

[sleek quest]

#❓how-to-get-help

[sleek quest]

and then how would you figure out which remaining one has it?

its all very ambiguous

[sleek quest]

@frosty geyser ?

anyone?

[sleek quest]

^^^

???

im so confused

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@sleek quest Has your question been resolved?

[sleek quest]

pls

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[torn jolt]

how do i solve for k in system of linear equations?

[torn jolt]

lets say 1/2x +5/3y=2

and kx+5/2y=3

the question is for what values of k does the linear system have:

infinite solutions

and one solution

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@torn jolt Has your question been resolved?

[torn jolt]

<@&286206848099549185>

[unkempt shoal]

@torn jolt hey there

[torn jolt]

hello

[unkempt shoal]

Have you studied linear algebra?

[torn jolt]

yea

the teacher didnt teach us the k variable so i dont know how to solve it

[unkempt shoal]

OK

[torn jolt]

could you teach me

[unkempt shoal]

Did you go over expressing linear systems in terms of matrix vector multiplication?

[torn jolt]

what is that

probably not

[unkempt shoal]

You translate a system like you have now as a matrix multiplying a vector

[torn jolt]

i didnt do that yet

[unkempt shoal]

OK let's try it with an easy example

[torn jolt]

the chapter im on is finding solutions

[torn jolt]

ok

[unkempt shoal]

Make an easy 2x2 matrix and let's call it A

[torn jolt]

what is a matrix

[unkempt shoal]

Oh

Well

I thought you studied them when you told me you're studying linear algebra

[torn jolt]

nope

[unkempt shoal]

OK I'll try to explain it without using matrices but give me 5min

[torn jolt]

what are matrices

https://tenor.com/view/crying-meme-black-guy-cries-sad-man-thank-god-for-my-reefer-hood-news-gif-24902056

maybe its cause i dont know the words

[unkempt shoal]

You'll study them don't worry about it

[torn jolt]

is it ok if you show me?

[unkempt shoal]

A matrix is something like this

[torn jolt]

what the heck is that

[unkempt shoal]

Don't worry about it you can kinda solve the question without using matrices

[torn jolt]

:0

oki

[unkempt shoal]

OK so if you want to determine 2 variables, you need 2 equations right?

[torn jolt]

yep

[unkempt shoal]

But it doesn't count if the second equation is a multiple of the first

For example

If you havr x+y=1

And 2x+2y=2

It's basically the same equation

Just multiplied by a number

[torn jolt]

wut

doesnt that mean it has infinite solutions

wait

[unkempt shoal]

Yeah, because the second eqaution is "dependant" on the first

We need 2 "linearly independent" equations

[torn jolt]

so how do we solve an equation with 3

well my question has one equation with only two

and the second with 3

[unkempt shoal]

Let's go step by step

[torn jolt]

ok

[unkempt shoal]

Forget about the question for now, we'll build some notions firat and then get back to it

First*

[torn jolt]

oki

btw what do you mean by build notions

[unkempt shoal]

So do you agree that in order to fully determine 2 variables we need two equations, and that the sexond one must not be a multiple of the first?

[torn jolt]

yes i do agree

[unkempt shoal]

Perfect

And if the second one does come out to be a multiple, what happens?

[torn jolt]

you get infinite solutions

[unkempt shoal]

Perfect

OK so now, the 1st question is implicitly asking you to determine k so that the second equation doesn't come out as a multiple of the first

And the 1nd question is asking you the opposite

2nd*

Sorry I flippes them

The 1st ask for infinite and 2nd asks for 1 solution

[torn jolt]

i did not know that

wait what are yo

u talking about

[unkempt shoal]

The first question is: determine k as to have infinite solutions, right?

[torn jolt]

nop its determine k to have one solution

maybe i typed wrong

sorry

[unkempt shoal]

OK because I just checked your comment and it was reversed 😛

[torn jolt]

:o

my bad

wait how did u find out that they were multiples of each other?

[unkempt shoal]

I didn't, k is the thing that determines whether they are multiples or not

1st, you are going to determine k to have 1 solution (AKA they are not multiples)

Then you are going to determine k so you hqve infinitely many

[torn jolt]

yea dont we just change k so that it isnt the same as the slope for the first equation

[unkempt shoal]

You got the idea, but it isn't quite that simple. Oh and can you please recheck the numbers for me because it looks like there is a mistake

[torn jolt]

$ 1/2x +5/3y=2$

why isnt it working

[unkempt shoal]

Two dollar signs are needed

[torn jolt]

$$ 1/2x +5/3y=2$$

[glossy valve]

ok.

[torn jolt]

thanks

$$kx+5/2y=3$$

[glossy valve]

ok.

[torn jolt]

these are the two equations

[unkempt shoal]

Ok

[torn jolt]

wait should i use and easier example

or is this okay

[unkempt shoal]

OK try multiplying the first equation with 3/4

And try to notice something

Did you notice anything?

[torn jolt]

nope

i got

$$3/8x+15/4y=3/2$$

[glossy valve]

ok.

[torn jolt]

or is this wrong

[unkempt shoal]

Gimme a sec

Oh I got a better idea

Devide the first by 2 and the 2nd by 3

[torn jolt]

shoot

it was wrpmg

give me a sec

[unkempt shoal]

It's OK this one is better

[torn jolt]

ok

i will trythat

[unkempt shoal]

Tell me qhat you get when you're done

[torn jolt]

$$1/4x+5/6y=1$$

[glossy valve]

ok.

[unkempt shoal]

Yeah and the second?

[torn jolt]

$$kx/3+5/6y=1$$

[glossy valve]

ok.

[unkempt shoal]

Don't they look similar?

[torn jolt]

yes

except for the 3 and the 4

[unkempt shoal]

So what should k be to not have the same equation?

[torn jolt]

1

cant it be any number

since the denominator is different

[unkempt shoal]

There is exactly one number that it can't be

K should be different from 3/4

Else, you would end up with the same equation

[torn jolt]

yep

i just got that

i think i now understand how to do this

[unkempt shoal]

Nice

[torn jolt]

is that all there is? nothing else

or am i wr ong

[unkempt shoal]

Yup that is pretty much it

[torn jolt]

oh wow

thank you so much

[unkempt shoal]

No probs

[torn jolt]

https://tenor.com/view/thank-you-sticker-thanks-sticker-line-sticker-bear-sticker-bear-gif-26476682

.close

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[nova island]

How to check if two vectors emanating from the origin in R^3 are parallel?

[nova island]

I thought it was just if they are scalar multiples of each other but my book seems to think differently?

[old hamlet]

I didn't get what you meant by R^3

[severe basin]

In three dimensions

[old hamlet]

Oh alright

[nova island]

Right in 3 dimensions

[old hamlet]

What's the book saying

[nova island]

$\mathbb{R}^3$

[glossy valve]

goobybalooby

[nova island]

I have two vectors, (5,-6,7) and (-5,6,-7) but it says they are not parallel

Do they have to be in the same octant?

[severe basin]

When you scalar multiply a vector say (a,b,c) in R³ with, it's either gonna squish or stretch the same vector depending on the sign of the constant.

[old hamlet]

I don't think so

[nova island]

Right they are collinear, but doesn’t collinear mean they are parallel?? Not all parallel vectors are collinear, but all collinear vectors are parallel?

Is there an error in the solutions of the book?

[unkempt shoal]

(Parallel vectors) is equivalent to (colinear vectors)

There might be a typo in the book

[severe basin]

Which book is it?

[old hamlet]

What's the solution in the book

[severe basin]

The wording seems like the one I'm using

[nova island]

The solution in the book is that they are not parallel

[unkempt shoal]

Let's imagine that the minus sign is on 6 instead of 7

[nova island]

Its a linear algebra book, let me see what it’s called

[unkempt shoal]

In that case they definitely are not parallel

[severe basin]

From the authors Friedberg Insel and Spence?

[nova island]

Yes, 4th edition

[severe basin]

Which exercise?

[old hamlet]

1st point should be in the 4th octant and the second one should be in the 6th

[nova island]

1.1.1 part c

[severe basin]

Are you referring to the a solution provided by a third party in the internet?

[nova island]

No, instructor provided the actual solutions manual so we can check our work

[severe basin]

They are actually parallel

[old hamlet]

Their cross product also comes up to be zero, so yes they are

[nova island]

Right?? Okay, just wanted to double check cuz I thought I was going crazy

[severe basin]

I checked my work on this exercise

[nova island]

Ohhhh that’s the other way to check! The cross product! Haven’t taken calc 3 in a few semesters

[old hamlet]

Just checked it

Determinant gives a zero vector

[severe basin]

I'm using 5th edition, but it turns I too have found out a lot mistakes in the solution

[nova island]

Wait, cross product being zero doesn’t mean orthogonal?

[unkempt shoal]

Yeah always gets me 😂

[old hamlet]

Dot product will be zero for orthogonal

[unkempt shoal]

That is inner product

[old hamlet]

Since it has cos

[nova island]

Right right right, thank you all for the help and reassurance!!

[unkempt shoal]

Yw 😊

[severe basin]

You are gonna find more errors in the solution

[old hamlet]

It's no problem

[severe basin]

(in upcoming exercises)

[nova island]

I believe it lol and I’ll be wary, but thanks again!

.close

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[spark pulsar]

help pls

[spark pulsar]

b)

i tried making a quadratic expression but for some reason its wrong even though i didnt it properly

[deft zodiac]

so blur

:c

have u done the scatter plot?

take a pic of it

[spark pulsar]

its like a rough sketch

but how can a scatter plot help me with this?

[deft zodiac]

um

[spark pulsar]

i have my vertex and 2 points on the grid

[deft zodiac]

they asked for a scatter plot?

[spark pulsar]

yeye that part is fine

but the second part of question b

[deft zodiac]

hm

well

wtv ur graph is like

just take the 2 points it hits the x-axis

those are the zeroes of the graph

so say the zeroes are a and b

[spark pulsar]

but since its a manual line of best fit the zeros are never gonna be 100% accurate right??

[deft zodiac]

the graph will be

k(x-a)(x-b)

[deft zodiac]

yea

meh just approximation

draw so it looks fine

[deft zodiac]

then pcik a point on graph sub in val to find k

[spark pulsar]

but would it work if i use the vertex form

like

4, 88 is vertex

and i take a random point like

2, 57

and i plug in the values

to find a

does that work

[deft zodiac]

um no

a and b are the values u read off the graph

u just calculate k

[spark pulsar]

ok hang on

i took the equation on the answer sheet

and put it in desmos

and the points on the desmos arent accurate to the table of data

@deft zodiac

[deft zodiac]

ya its an approximation..

[spark pulsar]

but

but

it has to be accurate

its a quadratic expression

[deft zodiac]

a quadratic can at most fit 3 points (with different x values) when arbitrarily chosen

perfectly anyway

after that it depends on if the points line up

chances are they dont

[spark pulsar]

oh

ic

tyyy

[deft zodiac]

np

[spark pulsar]

.close

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[deft zodiac]

[spark pulsar]

[full forum]

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[carmine needle]

need help with number 19

[round gust]

Pythagoras

Draw the way he went

Dont tell her the solution

[safe herald]

sorry

[round gust]

Delete

[safe herald]

done

done

[round gust]

👍

[safe herald]

in college?

[round gust]

Im not in college

[carmine needle]

i think our teacher drew us something like this

[round gust]

Good

[safe herald]

use pythagoras

[round gust]

Now you need to find the missing lenght

[safe herald]

hypotenuse

[carmine needle]

is this like the right formula?

[safe herald]

yep

[short siren]

Ye

[round gust]

You know a and b and you want c

[short siren]

Generally: the sum of the square of both legs equals the square of the hypotenuse.

[safe herald]

correct

[short siren]

Yep

[carmine needle]

wait ok i get it now

So can i say 6km away?

is that correct?

[safe herald]

approx.

but you should rather find out the square root of 34

[carmine needle]

so is it better to say 5.83km?

[short siren]

Yep

You see how many sig figs they want

If they don't specify, assume they want 2 decimal places

[carmine needle]

alrrrr thank youu

[short siren]

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@carmine needle Has your question been resolved?

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[frozen knoll]

I need to solve limit using LH

how to I deal with ^1/x

[hot wadi]

Take log both sides

[frozen knoll]

Ill try

something like this?

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[slate shell]

Not really sure what to do can someone help me please

[rough plaza]

i think your approach on
ax-by=2
bx+ay=1
is a good start

ax-by=2 ... (1)
bx+ay=1 ... (2)

try finding
(2)+(1)
and
(2)-(1)
first and simplify them and see what you can get

[full forum]

@slate shell Has your question been resolved?

[slate shell]

is that right

[rough plaza]

i would factor out the x and y instead of a and b

[slate shell]

Do I add them both together to get 2sinx+2cosy=2

[rough plaza]

hmmm

I'm thinking

my bad, i think i got it wrong lol,
lemme do it again

[slate shell]

it’s alright 👍

[rough plaza]

abx-b²y=2b
abx+a²y=a

(a²+b²)y=a-2b

this looks better

[rough plaza]

would this help?
@slate shell

[slate shell]

One second I’m just trying to think of what to do next

so I can make a^2+b^2 =1

[rough plaza]

take your time👍

[slate shell]

So then I have y=a-2b so y=cos-2sin ?

[full forum]

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[rough plaza]

you can keep your a,b first

and then solve for x

[slate shell]

.reopen

[full forum]

✅

[rough plaza]

hi

did you get x?

[rough plaza]

lemme check

a²x-aby=2a
b²y+aby=b

(a²+b²)x=2a+b

therefoŗe
x=2a+b
y=a-2b

[rough plaza]

yours correct, just not simplify enough

anyways,

2a+b
and
a-2b

is always a real number right?

[slate shell]

yes

[rough plaza]

so, that means, no matter what theta is, x and y are real, right?

[slate shell]

Yep

sorry for my slow replies I was just working on this

i think u can also substitute x and y back into the equation

and I got sin^2*cos^2=1 for both equations

[rough plaza]

yea, you can do that

[slate shell]

gosh that question was hard

[rough plaza]

it is!

[slate shell]

Thanks for the help👍

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[candid prairie]

what am i doing wrong

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[candid prairie]

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[full forum]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[undone axle]

i have begun but got stuck

i know that l = r(angle)

how would i find the angle insdie circle

[grave elm]

In degrees, full angle is 360°, what is it in radians?

[undone axle]

2pi

[grave elm]

great

So what would the red angle be?

[undone axle]

in decimal form thats 6.28 rad

so would i do 6.28 - 1.75

[grave elm]

Correct, but it's better to just keep it as (2pi-1.75) for now

because pi is irrational and you want to avoid rounding errors

[undone axle]

true

thanks bro

i will do 4 x (2pi - 1.75)

to find arc length

then plus 8

cause it wants perimeter

[grave elm]

Yep

[undone axle]

thanks man

[grave elm]

Do you know how to do area?

[undone axle]

yh

[grave elm]

Okay, great. Good luck

[undone axle]

thanks

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[dusty jewel]

it says convert to polar, isnt this already in polar?

[gritty rose]

it isn't

[dusty jewel]

a+bi is not polar?

[gritty rose]

it isn't

[dusty jewel]

how

[gritty rose]

https://math.libretexts.org/Bookshelves/Precalculus/Precalculus_(OpenStax)/08%3A_Further_Applications_of_Trigonometry/8.05%3A_Polar_Form_of_Complex_Numbers

[dusty jewel]

z = r(cosx + isinx) is cartesian no

[gritty rose]

no

[gritty rose]

this is in cartesian form

[dusty jewel]

u mean cartesian

[gritty rose]

yes

[dusty jewel]

oh

nice

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• Be polite and have a nice day!

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[undone bluff]

I need help with 9th grade maths

[thick minnow]

[undone bluff]

the question is evaluate (0.000064)^5/6

how on earth can you solve this

[thick minnow]

Do you want an approximate answer? Then use a calculator

[undone bluff]

and what do they exactly mean by evaluate

[undone bluff]

I need with work

my teacher gave this sum as hw and i was absent the previous day so i dont rlly know how to do it

[undone bluff]

formula?

[round gust]

Use this

[undone bluff]

hmm leme try

2^10x10^-10?

i got this right?

ahh this maths is too much

phy better

[thick minnow]

phy is maths + understanding. worse than maths

Just saying

[undone bluff]

also tho is it even possible to express a decimal (which has a bar) to be expressed in the form of p/q?

[undone bluff]

i mean in phy the numericals are fun and also the concept is kinda fun ngl

[thick minnow]

yes, it is possible for all rationals

[undone bluff]

So it is a rational number?

[thick minnow]

Lemme give you a hint, 2^6 = 64

[thick minnow]

yes, it is

[undone bluff]

even when it is a bar?

[thick minnow]

wdym by a bar?

[undone bluff]

non-term decimal

[thick minnow]

There's 2 types of non-terminating decimal. (1) repeating one ie the pattern repeats like 1/3 = 0.33333

[undone bluff]

i mean the one that goes forever

[thick minnow]

(2) non-repeating. <- These are not rationals

[undone bluff]

and also circles concept is a pain

[undone bluff]

hmm intresting

[thick minnow]

Compare this with the value of root(2) = 1.4142135623730951....

[undone bluff]

i got my ans in powers its fine right i dont need to simplify right?

[thick minnow]

There is no bar cause the decimal values do not repeat

[undone bluff]

Yes

[thick minnow]

Most likely you need to

[undone bluff]

omg bru

[thick minnow]

cause the answer is simple enough

[undone bluff]

2^10x10^-10

gon take ages

[thick minnow]

That seems wrong

[undone bluff]

the question is evaluate (0.000064)^5/6

this was the question

[thick minnow]

yes, write it down as a fraction

removing the decimal

[undone bluff]

so first put it as a decimal then do the powers?

[thick minnow]

nah, write the decimal as a fraction then see about the powers

[undone bluff]

alr one min

64/1000000

now?

you here?

[thick minnow]

yes

what did I mention about 64

[undone bluff]

2^6 = 64

[thick minnow]

Can we say something similar about the denominator

[undone bluff]

wait..

nah i dont see anything

whats there in 64/1000000

[thick minnow]

count the number of zeroes maybe

[undone bluff]

6

[thick minnow]

Do you see it now?

[undone bluff]

Yes

we can cancel cant we

[thick minnow]

yes

[undone bluff]

ig?

wait les go

now ?

[thick minnow]

what do we have now

[undone bluff]

(64/1000000)^5/6

so now we have

we cancel the 6 zeros right?

[thick minnow]

${\left( \frac {2^6}{10^6} \right)}^\frac56$

[undone bluff]

what now to do with the powers?

this confusing man

[thick minnow]

${\left( \frac {2^6}{10^6} \right)}^\frac56 = {\left( \frac {2}{10} \right)}^{6 \times \frac56}$

[undone bluff]

what can i cancel?

[glossy valve]

numbpy

[thick minnow]

Do you see something you can cancel?

[undone bluff]

hmm

(2/10)^41

if we convert

or we can cancel?

[thick minnow]

How did you get 41???

[undone bluff]

6x6+5

isnt that how u convert mixed to imprper?

or do we even need to convert?

[thick minnow]

Nooooooooo, 6 times 5/6 = 5

The 6's cancel

[undone bluff]

OHHHHHHHHHH

[thick minnow]

It's a product

[undone bluff]

man..

so now its just power of 5

right

[thick minnow]

yeah

You can just evaluate the power

Then write it as decimal

[undone bluff]

0.00032

[thick minnow]

yep

[undone bluff]

LETS GO

I DID IT

soo all sums which involve power of fraction are like this?

[thick minnow]

Most likely yeah, you need to remember some powers then it would become easier

For harder questions hopefully they would give some kind of hint

[undone bluff]

So i dont need to use formulas for this right

i alr gotta learn 18 formulas cuz of this cuboid,cube,spher,hemisphere lateral sufrace and stuff bru

[undone bluff]

That woud be jolly

this is just powers and cancelling right?

[thick minnow]

Yeaaah, it kinda sucks ngl especially later you won't even need it

[undone bluff]

bro istg even volume of cone they wont leave me

[thick minnow]

Mostly that, yes

[undone bluff]

Thanks a ton for the help man rlly appreciate it

[thick minnow]

sure, no worries

.close to close the channel

[undone bluff]

.close

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[deft igloo]

Find the extremes of
$1-\sqrt{x^2+y^2}$

[glossy valve]

Git commit

[woeful bramble]

Where are you getting stuck?

[deft igloo]

partial derivatives equal to 0 are unsolvable

but there is a maximum in (0,0)

Take a look

@woeful bramble

[woeful bramble]

I guess it's a bit tricky since the partial derivatives aren't defined at (0,0), but you can see that df/dx = 0 when x = 0, and that's true for all y. So as you approach (0,0) from -y or from +y, the limit of df/dx will be 0. Does that make sense?

[deft igloo]

If I assume x = 0

then second df/dy becomes

$-\frac{y}{\sqrt{x^2+y^2}}=-\frac{y}{\sqrt{y^2}}=-\frac{y}{\left|y\right|}=\left(-+1\right)$

[glossy valve]

Git commit

[deft igloo]

And I need both df/d to be 0

[woeful bramble]

Yea, fair point. For this one I might do a qualitative argument as opposed to trying to find it via derivatives. You can say that sqrt(x^2 + y^2) is always >=0, and since it's being subtracted the function will have a maximum when its 0, which only happens when x=0 and y=0

I'm not sure if that's an acceptable answer for your situation though

[deft igloo]

I would do this on exam

But the problem remains tbh. How to do it mathematically

[woeful bramble]

Yea that's tricky, since as you noted the limit of the partial derivatives doesn't exist at 0. I'm not sure how you would do this, although I would note that the argument I made above is mathematically valid. Perhaps someone else might come along and be able to help, sorry I couldn't help more.

[deft igloo]

Np king ❤️

[full forum]

@deft igloo Has your question been resolved?

[deft igloo]

<@&286206848099549185>

[full forum]

@deft igloo Has your question been resolved?

[rough plaza]

would it be suffice to say
-x/√(x²+y²)=0 AND -y/√(x²+y²)=0
implies
(x=0 and x²+y²≠0) AND (y=0 and x²+y²≠0)
and hence lead to contradiction

[deft igloo]

but (0,0) is an maximum

[rough plaza]

yea, that's where df/d is 0 or undefined

hmmm

if it just needed to be undefined, we can just use the fact that the denominator is 0 at (0,0)

[deft igloo]

We need to find extremas, so df/d = 0

I cannot be undefind

[rough plaza]

why can't be undefined?

[deft igloo]

ok I didn't know that tbh

ok, lets assume that (0,0) is an extreme point

now

how do we proceed

[rough plaza]

then, just say that df/dx doesn't exists due to x²+y²=0

more formally, of course

since you'll be doing it in an exam

[deft igloo]

I mean, how do we show that this point is min, max or saddle

[rough plaza]

hmmm....

i can only do it when i realized its a max

like just check since √(x²+y²)>=0 for all x,y in R
then f(x,y)=1-√(x²+y²)<=1
also, f(0,0)=1, therefore its a max

[torn jolt]

hi\

so

oks

pls

help

me\

ad

[deft igloo]

xD

Bro

[torn jolt]

pls,

[deft igloo]

U can get an empty room

This one is occupied

ok, so

this works

But what if the same situation happens with some crazy function and this is not obvious

I mean, is it the only way, or we can do something with df/d that will give us the anwser?

[torn jolt]

how tf do I get an empty room

[rough plaza]

!help

[full forum]

Please read #❓how-to-get-help

[rough plaza]

i think it's the only way for now

[deft igloo]

Dam

[rough plaza]

let's just hope there won't be crazy functions in the exams

[deft igloo]

ok, math isn't always easy it seems xd

thanks bro

for the help

@torn jolt im closeing this room, so u can get it if u want

.close

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[torn jolt]

thanks man

[full forum]

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[torn jolt]

Can I have some help on this

[tulip beacon]

Love the photography

[torn jolt]

Thanks

[light sonnet]

What's that line about photomath, symbolab?

[torn jolt]

It says you are limited to using those to help you

[light sonnet]

So then that means you shouldn't be using discord

[torn jolt]

But this is a formative assignment so it doesn't count towards my mark

[tulip beacon]

Why does this discord have strict rules about quizzes and tests

[torn jolt]

Idk

But this isn't a test I swear

[light sonnet]

Because academic dishonesty

[tulip beacon]

I’m sorry but if someone just said ‘this is not a test’ surley you can’t really dissprove it

[torn jolt]

What teacher allows to bring a test home

[light sonnet]

Take home test lol

[torn jolt]

Never got one of those before

[tulip beacon]

Anyway first thing is to find where the first line crosses the y axis

[tulip beacon]

Never had one either

[torn jolt]

I wish I had one

[light sonnet]

I've had plenty

[tulip beacon]

All my exams in my degree and throughout high school are written

Same my cousin had one coz of covid

They seem fun

[torn jolt]

Do u know how I can solve this problem

[light sonnet]

I suggest getting the equation in slope intercept form first

y = mx + b

[torn jolt]

Which one

The 2x +7y -14 one

[light sonnet]

Both

[torn jolt]

Ok

Ok done

The first one is y=-2/7x + 2 and the 2nd one is y=1/3x-4

[light sonnet]

So the first part says that your equation you are creating has the same y intercept as the first one, what's the y intercept?

[torn jolt]

The fist number right

[light sonnet]

y = mx + b, do you know what m and b means?

[torn jolt]

M is the slope

B is y intercept

Oh so it's the number at the end

[light sonnet]

So your problem is asking you to find the equation of a line that passes through the same y intercept as teh first equation

What is that y intercept?

[torn jolt]

2

[light sonnet]

Then it states that the equation you are finding is perpendicular to that second equation, do you know what that means?

[torn jolt]

Yea

Like a cross

[light sonnet]

Do you know how two equations could be perpendicular? Just given the equations?

[torn jolt]

No

[light sonnet]

It has something to do with the slopes of the two lines

[torn jolt]

-2/7 and 1/3?

[light sonnet]

First you don't need that first equation anymore, all you wanted from that was that it passed through the same y intercept

[torn jolt]

Oh ok

[light sonnet]

For your equation that you are finding and the second equation, you want them to be perpendicular

[torn jolt]

Yea

I don't know how to make them perpendicular

[light sonnet]

See if this helps https://www.youtube.com/watch?v=Uq8pSFaXPmQ&ab_channel=TheOrganicChemistryTutor

I would tell you what to do, but this gives you the opportunity to look things up and learn

[torn jolt]

Ok

Ok I watched the video

I know what it is now

[light sonnet]

What do you think you need to do?

[torn jolt]

I need to make an equation where it has the same y intercept as the first equation and is also perpendicular to the second equation

I think

[light sonnet]

I meant the part about making it perpendicular

You know what the y intercept is

Can you figure out how to make it perpendicular?

[torn jolt]

Isn't it you flip the top and bottom

[light sonnet]

And?

[torn jolt]

For -2/7 it would be 7/2

And you take away the negative

[light sonnet]

Well, you want it perpendicular to the other equation

[torn jolt]

So -3/1

[light sonnet]

Yes

[torn jolt]

Ok

So first I write y=-3/1

[light sonnet]

So you have the slope and y intercept, can you make the equation now?

[torn jolt]

Yea

Is the answer y=-3/1x + 2?

[light sonnet]

One thing, what is -3/1?

[torn jolt]

The slope

[light sonnet]

I mean math wise

What's 2/1?

[torn jolt]

-3/1 is M and 2 is the B

[light sonnet]

Or 4/1?

-6/1?

What's any number divided by 1 equal to?

[torn jolt]

Itself

It wouldn't change

[light sonnet]

So then what's -3/1?

[torn jolt]

Oh

It's just -3

[light sonnet]

So then this equals?

[torn jolt]

Y=-3x+2

[light sonnet]

Yes

[torn jolt]

Ok

Good

Thanks

[full forum]

@torn jolt Has your question been resolved?

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

find the surface area

[low grove]

.close

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[dull stump]

Three different numbers are selected at random from the set {1..10} Probability that product of two numbers is equal to third is??

[dull stump]

So I wanted to know how to do this without bruteforce or is bruteforce only method?

Do we have some bionomial or some method to do this?

[normal tree]

bruteforce isn't that bad

[dull stump]

cant apply it if they do some changes in question

[sharp flame]

Honestly it isn't even brute force you can do it mentally

[dull stump]

i m less on braincells

mentally outta question

btw how

[sharp flame]

Start multiply single digit numbers in your head

[dull stump]

yeet

so if bruteforce, no combination with one would be possible as we aint replacing numbers

we start with two and start making pairs in table?

is that the only way 😢

what if they make the set 1..100

[normal tree]

even for 1 to 100, counting up the outcomes isn't that hard

I can think of things like letting A, B, C be the sampled numbers and compute the distribution of log A + log B - log C via convolution or characteristic functions / Fourier transform

but I just can't imagine that they would expect you to do that

[dull stump]

okay they would keep it simple in exams tru

so bruteforce it is?

[normal tree]

yeah probably

[dull stump]

aight, also one more thingy

we cant use Combination formula in question saying items are picked consecutively but can when they say 2 items are picked or 3

isnt it kind of same like when 3 items are picked even then the quantity is being lessed from total yet answer differs

[wide sundial]

Let me see if this works

[normal tree]

I'm not sure what you're referring to

[wide sundial]

,w expand (x+2x² + 3x³ + 4x⁴ + 5x⁵ + 6x⁶ + 7x⁷ + 8x⁸ + 9x⁹ + 10x¹⁰ )^2

[dull stump]

umm how do i explain

[glossy valve]

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Requested by Frosst#1111

[wide sundial]

I wonder if this works

[dull stump]

how to select which term?

[wide sundial]

Give me a moment

,w expand (x+x² + x³ + x⁴ + x⁵ + x⁶ + x⁷ + x⁸ + x⁹ + x¹⁰ )^2

[glossy valve]

Results provided by WolframAlpha

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Requested by Frosst#1111

[dull stump]

there was another question, in dice roll probability of n appearing is proportional to n, can we apply:
,w expand (x+2x² + 3x³ + 4x⁴ + 5x⁵ + 6x⁶)

[wide sundial]

It needs to be on its own line

[normal tree]

is that not just doing convolution on the coefficients of the polynomial