help-10

[timid silo]

Speed is the rate of change of position, acceleration is the rate of change of speed

[honest flicker]

Idk

[timid silo]

I don't know why I mentionned that I just thought it was a neat idea

Props to them

[honest flicker]

Ur so cool

[timid silo]

Don't say that, I'm gonna blush

@honest flicker So... Got any more questions?

[ruby path]

I think he's cooler than you think he is

[honest flicker]

Yes

Did I get it right frfr

I forgot to send the question

Stuff

Idk Abt the second one

[timid silo]

Boys, one at a time please

[honest flicker]

Plz help

[timid silo]

Yes, I was looking at it

You got it right

[honest flicker]

Oh

OMG

I got it wrong😭

[timid silo]

They're both ascending at the same rate but at 4 seconds in the hot air balloon is 4/2+4=6 meters high, which is higher than the blue baloon at 4 seconds in

[timid silo]

what the fuck

[honest flicker]

It's like 3 screenshots worth an explanation

Will send

Brb

Its so slow

SEND

BRO

PLZ

@timid silo

[timid silo]

??? It says you got it right

[honest flicker]

No

I picked red balloon for the second one

[timid silo]

Second one?

You only sent one question

[honest flicker]

Second question

At the bottom

Which balloon ascended faster?

[timid silo]

Ah dangit, I missed that

Indeed they're both ascending at a speed of 1/2 meters per second

@honest flicker Is that it?

[honest flicker]

Yes I don't have enough time to redo the quiz rn😭

.close

[obtuse pebble]

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[heady shale]

Write the following as powers of 2, 3 or 5: 25 x 5^-4

[heady shale]

my problem is 25 x 5^-4

how do i go about answering it

if 5^2 is 25

[quaint wing]

Yes

[heady shale]

how do i solve the rest of it

whats my next step

[quaint wing]

If you have x^2*x^-4

How do you simplify that

Think about exponent rules

[heady shale]

a^m * a^n is a^m+n

so 2 - 4?

[quaint wing]

Yeah exactly

And what is x in this situation

[heady shale]

oh thats simple

5

so 5^-2?

[quaint wing]

Yeah

[heady shale]

thats all?

its actually simple

[quaint wing]

Yeah

[heady shale]

thanks man

[quaint wing]

Np

[heady shale]

❤️

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[flint egret]

i'm stuck at converting that imaginary part to an argument

[obtuse pebble]

@flint egret Has your question been resolved?

[brazen gorge]

recall euler's formula:
$e^{i \phi} = \cos(\phi) + i\sin(\phi)$

[warm shale]

artemetra

[brazen gorge]

use it to simplify your denominator ($e^i$)

[warm shale]

artemetra

[obtuse pebble]

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[hard minnow]

Could someone validate if my proof for this problem is correct?

[flint egret]

thank you, i will try working with that

[hard minnow]

This is what I got for part a and b:

[obtuse pebble]

@hard minnow Has your question been resolved?

[obtuse pebble]

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[steady lodge]

hi

[steady lodge]

how many ways there are to place 25 identical cookies into 7 distinct jars such that exactly 2 are empty

my method --

j1+j2.......+j7 = 25
j1 = j2 = 0

j3+...j7 = 25
ji >= 1 , i = 3,4,5,6,7

hence x1+x2+x3+x4+x5 = 20

but idk how to solve ahead.

<@&286206848099549185>

[obtuse pebble]

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.

[steady lodge]

oh , i am sorry

[daring sorrel]

For the first part you are only considering one case of 2 jars being empty

j2 and j3 could be empty while j1 and j4,j5,j6,j7 have cookies aswell

[steady lodge]

okay , so i multiply final answer by 7c2

[daring sorrel]

yeah

[steady lodge]

but how to get to final answer?

[daring sorrel]

Im not exactly sure what you are doing for the 2nd part. To start though you can 1. ignore the other 2 jars (seems like you did that) and 2. consider that each of the 5 remaining jars can have min 1 cookie max 21 cookies

[steady lodge]

yee

[obtuse pebble]

@steady lodge Has your question been resolved?

[obtuse pebble]

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[fluid meadow]

Is this dealing with derangements?

[wispy wadi]

no need for derangements (but yeah you're essentially couting number of derangements/n!)

[obtuse pebble]

@fluid meadow Has your question been resolved?

[obtuse pebble]

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[hybrid pagoda]

How is this wrong 💀

I need help

[coarse vault]

try dy/dx is 5

[hybrid pagoda]

Wdym

How

@coarse vault

[coarse vault]

dy/dx is 5 so y = 5x+1

solve by implicit differentiation

[hybrid pagoda]

Idk how to do that

Can you write down for me so ik

I’m so new to this

[coarse vault]

I mean it would be easier if u looked it up i am sure there are many videos explaining it.

[hybrid pagoda]

Can you write down

I work better with one example

Then I can figure out rest

@coarse vault

[coarse vault]

this problem is very long. https://www.emathhelp.net/en/calculators/calculus-1/implicit-differentiation-calculator/?f=x^2%2By^2%3D(x^2%2By^2-5x)^2&type=x&px=0&py=1

[hybrid pagoda]

Thank you

@coarse vault

How do you get y^n

[obtuse pebble]

@hybrid pagoda Has your question been resolved?

[high lily]

why do you wnat y^n

[hybrid pagoda]

Wdym

[coarse vault]

once u get y' u must differentiate again using implicit

y' = -9x/8y so use quotient rule to find y''

[hybrid pagoda]

Alright

Thank you

[hybrid pagoda]

Why twice

[high lily]

you're begin asked for y''

(the second derivative)

(which is the derivative of the first derivative)

[hybrid pagoda]

Hmm

It says it’s wrong

[high lily]

what are you getting? show your work

[hybrid pagoda]

Wait

Is this right

@high lily

[high lily]

no

[hybrid pagoda]

What i do then

[high lily]

you're not differentiating -7x/y properly

differentiate it properly apply appropriate rules

like quotient rule and chain rule

[hybrid pagoda]

Yes

But idk what to do

If there’s a y value

Because I don’t think you can just simply substitute it in like you’d do with an x value

[high lily]

the same thing you did when finding the first derivative

chain rule

or the derivative of y is literally the derivative of y

[hybrid pagoda]

-7x/y is erm

-1/7(xy)

?

@high lily

[high lily]

no

do you know the quotient rule?

[hybrid pagoda]

No

I think I do

But idk what to do when y value

f’(g)-g’(f)/g^2

[high lily]

y' = y'
d/dx y = dy/dx

[hybrid pagoda]

Ok

So

-7(y)-dy/dx(-7x)/-7x^2

@high lily

[high lily]

looks off

how are you geting that

show work

[hybrid pagoda]

Ok

Idk

@high lily

[high lily]

you're not applying quotient rule properly

[hybrid pagoda]

How

[high lily]

list your
f
f'
g
g'

[hybrid pagoda]

f = -7x

g= y

f’ = -7

g’ = dy/dx

[high lily]

yes
and use those in

[hybrid pagoda]

I did

[high lily]

you did not

check what you did/ do it again

[hybrid pagoda]

But the answer doesn’t want a dy/dx in it

@high lily

[high lily]

we're getting there

one step at a time

[hybrid pagoda]

Wait

[high lily]

its still wrong

[hybrid pagoda]

How

I give up at this point

I have no clue what to do

[high lily]

pay close attention to your expressions

[hybrid pagoda]

I did

[high lily]

what is f
what is g'
and from that what is fg'

[hybrid pagoda]

^

[hybrid pagoda]

^

[high lily]

no

i'm asking for g'

[hybrid pagoda]

dy/dx

Or 1

[high lily]

no

dy/dx is dy/dx

NOT 1

[hybrid pagoda]

oh

than I have no clue

[high lily]

so you have
f = -7x
g' = dy/dx (or y')

now what's fg'

[hybrid pagoda]

-7(dy/dx)

[high lily]

no

how are you getting that

[hybrid pagoda]

no way it’s y’

[high lily]

no, that's not the issue

[hybrid pagoda]

What’s the issue

[high lily]

idc if you use dy/dx ior y'

what's f

[hybrid pagoda]

-7x = f

[high lily]

yes

and what do you get when you multiply that by y' or dy/dx

[hybrid pagoda]

That multiplied it beside it right

[high lily]

yes

and can you write that down

[hybrid pagoda]

Ok I did

[high lily]

and what did you end up writing

[hybrid pagoda]

Hm

-7y-y’(dy/dx)

[high lily]

no

[hybrid pagoda]

f

[high lily]

don't skip ahead or overthink

[hybrid pagoda]

okay

[high lily]

you'er way overthinking this

[hybrid pagoda]

It’s y’(dy/dx)

[high lily]

no

STOP rushing

you have

f = -7x
and g' = y'
right?

[hybrid pagoda]

Yes

[high lily]

literally just write those two things being multiplied together

[hybrid pagoda]

7y’

[high lily]

no

[hybrid pagoda]

wait

Bro what

f = -7x

g’ = y’

[high lily]

yes

[hybrid pagoda]

f(g’) = -7(y’)

[high lily]

no

[hybrid pagoda]

HOW

[high lily]

f isn't -7

[hybrid pagoda]

😵‍💫

[high lily]

you've said this many times now

that f=-7x

[hybrid pagoda]

?

[high lily]

yes

where is the x wandering off to?

[hybrid pagoda]

uh

When you fill the inside

[high lily]

wdym fill the inside

[hybrid pagoda]

Don’t you replace the x with a bracket

[high lily]

no?

[hybrid pagoda]

And then evaluate through that

oh

[high lily]

x is x

x is x

-7x is the product of -7 and x

[hybrid pagoda]

ok

[high lily]

-7x * y'
is the product of -7x and y'

[hybrid pagoda]

Alright

[high lily]

yes

[hybrid pagoda]

ok

[high lily]

now note that you have an expression for y' or dy/dx from earlier

[hybrid pagoda]

Noted

Wait

Y(-7)+7x(y’)/y^2

Doesn’t it simplify to -7+7xy’/y

@high lily

[high lily]

no

[hybrid pagoda]

ok I gave up now

[high lily]

you need to review your algebra

[hybrid pagoda]

yeah I’m dropping calculus

Because I’m sure it’s -7+7xy’/y

if it’s not right than prove it

[high lily]

$\frac{a + b}{c} \redneq \frac{\frac ad + b}{\br{\frac cd}}$

[warm shale]

ℝam()n()v

[hybrid pagoda]

Ok

-7y+7xy’/y^2

?

@high lily

[high lily]

$\frac{-7y + 7xy'}{y^2}$

[warm shale]

ℝam()n()v

[high lily]

yes

[hybrid pagoda]

Ok

[obtuse pebble]

@hybrid pagoda Has your question been resolved?

[obtuse pebble]

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[sly geode]

Help Me

[sly geode]

HelLo

I appreciate helps

[viscid gull]

probs easiest to write all in polar form first, then calculate the fraction

Exponential Form (same thing)

[sly geode]

How do i do the exponents s**t

[viscid gull]

first find the magnitude

Then divide the complex number by the magnitude and use the unit circle to find what the argument is

[sly geode]

Ight the exponent will be. i and a degree so idk how to add the expentn to that

[viscid gull]

all of these lie on nice values on the unit circle

exponent rules r the same

[sly geode]

So just multiply

[viscid gull]

e^(a + b) = e^a*e^b still

[sly geode]

for example like i can get an expenent of 4(i25degreee)

[viscid gull]

yeah

[sly geode]

OK i think i got is

Thx

[viscid gull]

(e^a)^b = e^(ab) also

[sly geode]

Give me a sec to understand that

I didnt mean to do a special react

[viscid gull]

lol

[sly geode]

but yea makes sense

Thx

[viscid gull]

np

[sly geode]

.close

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[dull quartz]

Hi i need help

[dull quartz]

@forest sinew Here are my questions 🙂

Needing help with 3i first

[forest sinew]

okay, whats a critical value?

how can you find the critical values of f given f'?

[dull quartz]

you make the function equal to 0 @forest sinew

[forest sinew]

yea

so wheres f' = 0?

[dull quartz]

yes

i got 4 for the numerator

but for the denominator.. would i get the same answer by solving the bottom as a whole comapred to doing it seperately?

@forest sinew

[obtuse pebble]

@dull quartz Has your question been resolved?

[dull quartz]

hello?

needing help with 3ii

[forest sinew]

sorry gary 1s

[obtuse pebble]

@dull quartz Has your question been resolved?

[forest sinew]

@dull quartz sorry, i fucked up my hand on a screw on my desk, are you still around?

[dull quartz]

yes im on 3ii now

[forest sinew]

click the ❌

^

or the bot will close the channel

[forest sinew]

right, so if you check the sign of f' in between critical points it will tell you the answer to this Q

[dull quartz]

so i pretty much put 0, 4 and 12 on a number line?

wouldnt it be always increasing?

[forest sinew]

what, f(x) should only be 0 at 4?

[dull quartz]

how did you do it?

[forest sinew]

sorry, youre right, my bad

0 4 12

it shouldnt always be decrease

https://www.desmos.com/calculator/dlzlfrwxmq

in should be decreasing between 4 and 12, i think

you just plug in any value

if you wanna check the region between 0 and 4

try 2

or 3

[dull quartz]

so do i draw a number line?

and plot 0, 4 and 12 on the line

[forest sinew]

well you know where the roots are

theres 4 regions then

negative, between 0 and 4, between 4 and 12, and bigger than 12

you gotta check one value from each region

gary im goin to bed

i posted to let other ppl know to please check in here

do you have any last minute questions

@dull quartz

[dull quartz]

k

ima try it out and lyk what i get

[dull quartz]

i get math error....

[obtuse pebble]

@dull quartz Has your question been resolved?

[dull quartz]

Hello, anyone here? i need help with 3iii

<@&286206848099549185>

[dull quartz]

@keen garnet

[keen garnet]

Why ?

[dull quartz]

wym

[keen garnet]

Why did you ping me

[dull quartz]

i need help

[keen garnet]

Read the rules

[dull quartz]

Ok

[dull quartz]

I need help

[chilly fjord]

isn' t the critical point for the function just 4?

[dull quartz]

yes and 0 and 12

[chilly fjord]

how did u get those?

[dull quartz]

set denom to 0

im stuck on 3iii

[chilly fjord]

don't u have to solve f' as a whole?!?!?

[dull quartz]

wym

set num and denom to 0 and solve for 0

[chilly fjord]

like just multiply both sides by the denom and you'll be left with x-4=0

[dull quartz]

exactly

[chilly fjord]

like i don't understand why you would set them both to zero seperately

[dull quartz]

but you got to do it both to num and denom

[chilly fjord]

oh

[dull quartz]

idk either

[chilly fjord]

okk

[dull quartz]

but i think ive done it

can you help me with 3iii?

like idk what its asking @chilly fjord

[chilly fjord]

i think you check whether the function is increasing or decreasing

[dull quartz]

llike exactly the same as for First deriv test?

[chilly fjord]

which i think is second

[dull quartz]

second is about concavity

[chilly fjord]

so if f''(x)<0 then i think it is minima

[chilly fjord]

yeah

man i've forgotten single variable calc

[dull quartz]

haha

[chilly fjord]

i've got a midsem in 2 days and a final in like 3 weeks

[dull quartz]

damn you joined today too

[chilly fjord]

yeah

i needed help with my assignment

[dull quartz]

ive just got 4 exams in 2 weeks

fml

[chilly fjord]

calc questions are cursed

[dull quartz]

and now your helped ppl with assignemnts?

[chilly fjord]

i didn't help anyone lmao

[dull quartz]

your trying to help me *

[chilly fjord]

trying yeah lol

[dull quartz]

thought you were a helper at first

so what do i do again?

[chilly fjord]

my brain has been scrambled

[dull quartz]

bruh XD

[chilly fjord]

how do you create a help channel ?

[dull quartz]

#❓how-to-get-help

Help

[chilly fjord]

thanks

[dull quartz]

np

Help

[dull quartz]

.close

[obtuse pebble]

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[clever oasis]

help a) please idk how to start

[warm cloak]

do you know the equation of a circle ?

[obtuse pebble]

@clever oasis Has your question been resolved?

[clever oasis]

yes

(x-h)²+(y-k)²=r²

[clever oasis]

<@&286206848099549185>

[obtuse pebble]

@clever oasis Has your question been resolved?

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[timid silo]

S semigroup

[timid silo]

show that if ab=ba implies a=b

then a^2=a and abc=ac for every a,b,c in S

[obtuse pebble]

@timid silo Has your question been resolved?

[sage geode]

Associativity suggests that a * (a * a) = (a * a) * a, use the property of your semigroup to that equation

[timid silo]

yes i did that

but how to prove abc=ac

[timid silo]

any idea>

?

[sage geode]

Still can't think of anything

[wispy wadi]

abc = abc
so acabc = abcac
ie (ac)abc = abc(ac) which implies abc = ac

[kind hawk]

from where are you getting the second line

[wispy wadi]

a = aca and c = cac because in this semi group, a² = a and c² = c

so from aba you deduce aaba = abaa so a = aba

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[onyx dust]

Probability question: we had a question about something game-related. Assuming projectile has a chance of p% (in the particular case it's 24%) of causing a particular effect. Then what is the chance of 3 projectiles causing the effect? I was sure that because the question would be p1 causes effect or p2 or p3. So because it's an OR we add the probabilities, I suggested the chance of the effect being caused would be 3*p%. Somebody else though, suggested it is (1 - (1-p%)^3) because the chance of no projectile causing the effect would be (1-p)% (chance of projectile not causing effect) and 3 times in a row of that happening would be that to the power of 3 (since it's an AND, so the probabilities are multiplied). Which solution is correct now? The problem is that both variants yield completely different results (something around 75% vs something around 55%). I'll try to make a quick write-up of the events.

what happens?                              predicate          chance
proj. causes effect                        A                  p%

proj. doesn't cause effect                 ~A                 1-p%

at least 1 of 3 proj. causes effect        A | B | C          3p%

3 proj. don't cause effect                 ~A & ~B & ~C       (1-p%)^3

~(3 proj. don't cause effect)              ~(~A & ~B & ~C)    1-(1-p%)^3
(equivalent to at least causes it???)      <=> A | B | C
                                           (DeMorgan)

[obtuse pebble]

@onyx dust Has your question been resolved?

[wispy wadi]

the somebody else is right

"it's a OR so we sum probabilities" doesn't work

the proba that none of the projectiles touches is (1-p)^3
so the proba that at least one touches is 1-(1-p)^3

if you could just sum probabilities like you did, you would find probabilities over 100% almost all the time

you would flip a coin twice and have 100% of having 2 heads but also 100% of having two tails

it wouldn't have any meaning

[onyx dust]

but wasnt there a addition rule for probabilities when they're connected by OR?

[wispy wadi]

P(A or B) = P(A)+P(B)-P(A and B)

and similar formulas for more that 2 events

[onyx dust]

oh i see 🤔

so the 72% (= 3 * 24%) do include the fact that all projectiles may have caused it or only 2 of them

so you said that 1-(1-p)^3 is correct

which should mean that 1-(1-p)^2 (lets consider only 2 for now to make formula easier) should be equal to p + p - p^2 right?

ah yeah, which seems to be the case

forgot to reverse the plus/minus of p²

alright, thanks

.close

[obtuse pebble]

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[timid silo]

I’m stumped on the last question, could someone’s help me please?

Two diagrams I have drawn so far.

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[eager niche]

Try solving it first in terms of theta, what expression do you get for a?

[timid silo]

Is this where i equate both to eachother(once I have fully resolved the right side) and sub in the identity to try get the answer?

[eager niche]

First can you do the trig identity for sin(theta+60) ?

Then 4 left hand side = right hand side

And solve for thetas on one side

[timid silo]

Oh ok I'll do that

[timid silo]

I got to here but can’t seem to see what I’ve done wrong

Ohhhhh nvm

I found out

Still ain't getting it

Ohhhhh wait

I know

I forgot the g

Got it, thanks. @eager niche

.close

[obtuse pebble]

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[tired nacelle]

So how I can do that

[tired nacelle]

Hmm so :

1. not ( P and Q) or (r)
2. not P or (P or Q)

[brazen gorge]

yes and you can simplify those

what's "not (P and Q)"?

hint: ||de morgan rules||

[tired nacelle]

Aaa we didn't learn Morgan yet

[brazen gorge]

oh okay

[tired nacelle]

Logic is First chapter and Morgan is the Third

[tired nacelle]

So it's will still like this or I can simplify it

[brazen gorge]

idk, what have you studied?

de morgan rules are very simple tho

[brazen gorge]

also for 2) because logic operators are associative, not P or (P or Q) = (not P or P) or Q, and (not P or P) is always true, so we get True or Q, which is just True

[tired nacelle]

Alright, I understand, thanks you

[tired nacelle]

Just about this law , what is it, name?

[brazen gorge]

idk i just googled definition of implication

you can see that it's true by implication's truth table

[tired nacelle]

Alright, tysm

[brazen gorge]

if you are done then write .close here

[obtuse pebble]

@tired nacelle Has your question been resolved?

[obtuse pebble]

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[slate agate]

Hey yall. Got a question about a particular equality:

[slate agate]

so regarding this

is equal to ---> The derivative of the function phi with respect to the magnitude of the unit vector

so d(phi)/dn'

This is really tripping me up

.close

[obtuse pebble]

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[silver plover]

for part a i got this as my answer:

for part b, am i able to just plug in -1 for d?

or do i need to do the whole process of the eigenvectors and values again?

[full flume]

guys i need some help im in grade 6 pls come to clone server

pls

[silver plover]

https://discord.com/channels/268882317391429632/488120190538743810

<@&286206848099549185>

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@silver plover Has your question been resolved?

[obtuse pebble]

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[woven marsh]

In the second equation of the proof, why isn't it =(z1,6z1,0,0,0)+(0,0,-2z4,z4,0)+(0,0,-z5,0,z5)

Is there a typo?

[woven marsh]

Course: linear algebra

[obtuse pebble]

@woven marsh Has your question been resolved?

[wispy wadi]

that's a mistake yeah

[woven marsh]

Thank you

[obtuse pebble]

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[azure anchor]

Intro to Linear Algebra. Need a check on if this is enough. This is not a proof-based class.

[kind hawk]

no

why couldnt it be 3* 1/3

[azure anchor]

because if U^T is the inverse of U, then det U^T = 1/detU

[viscid gull]

I think u just need 1 more line of reasoning and u probs already assumed it

[kind hawk]

well and that works for 3 and 1/3

[azure anchor]

okay, just gotta add that property of inverse then, specifically det U^T = 1/detU ?

[kind hawk]

no

thats not the property missing here

if det U^T = 3 and detU=1/3, then that equality works

[azure anchor]

the fact that U^T * U = I means U^T = U^-1 => det U^T = 1 / det U

anything multiplied by the inverse is 1. and if the det of U is -1, then det U^T = 1/-1 = -1, so det U = +-1

[kind hawk]

but who says det U=-1

[azure anchor]

thats just a special case

[kind hawk]

what did you write in the last line

look closely

[azure anchor]

det U = det U^T = +-1 ?

oh

i see, yeah i need to lay out the general case

[kind hawk]

?

the first equality is the important one

det U = det U^T

thats what you are missing

because then det U = 1/det U

equivalently (det U)^2=1

and then you have det U = +-1

[azure anchor]

i did some rewriting, how does that look?

[azure anchor]

i dont think thats right, considering that U^T = U^-1, so how can det U = det U^T? it should be det U^T = 1/det U

[kind hawk]

those can be true at the same time

and in this case they have to be true at the same time for the proof to work

det A = det A^T is a general property of the determinant

right now your last line says nothing about the value of det U

cause it cancels out anyway

[azure anchor]

the only case i can think of where det A = det A^T = det A^-1 is when det A = +-1, but for the general case of it being != 1, would that still hold?

[kind hawk]

well quite conveniently thats literally what you are supposed to show

[azure anchor]

det U    =  x
det U^-1 = 1/x
det U^T = det U (by IMT)
therefore,
det U = det U^-1

something like that?

[kind hawk]

what do you mean with IVT?

[azure anchor]

invertible matrix theorem, an invertible matrix's determinant is equal to the determinant of the transpose.

lmao

IMT

not intermediate value theorem

[kind hawk]

this has nothing to do with the matrix being invertible

det A = det A^T always holds

[azure anchor]

damn it, so how can i prove det U = det U^-1

wait what

well it must be invertible for det != 0

[kind hawk]

you are confusing things

here the matrix is invertible, yes

but the property that det A = det A^T always holds, whether A is invertible or not

its a general property of determinants

that you should know

[azure anchor]

ahh found it youre right

holds for any square matrix

det U^-1 = det U must mean by detXY = detX*detY that it can only be +-1 right? but how can i write that?

[kind hawk]

combining the equations you have you get det U = 1/det U

so (det U)^2=1

[azure anchor]

im not v familiar with the proof notations so let me know if it looks stupid anywhere

[kind hawk]

its fine

[azure anchor]

would that be a satisfactory answer you think?

[kind hawk]

yes

[azure anchor]

thank you. sry i was a bit dense, i haven't had a lot of practice with these types of problems

[kind hawk]

experience really goes a long way

[azure anchor]

yeah. i feel like i learned a lot just by this one problem. thanks again!

.close

[obtuse pebble]

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[sacred jetty]

can i haave somehelp

[ripe merlin]

What does it mean by in the range?

.close

[obtuse pebble]

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[jaunty goblet]

@wooden cipher sorry for the Ping but I still got a doubt about that vectors question

[wooden cipher]

Sure

Whats your question?

[jaunty goblet]

U see u said the swimmer has to swim at an angle

[wooden cipher]

Yes

[jaunty goblet]

But when u resolve the swimmer and river u would get 1

Why don't we use the resolved velocity

[wooden cipher]

What do you mean?

[jaunty goblet]

Umm hold on

So u see the square

[wooden cipher]

Yes

[jaunty goblet]

Lemme pull up the question again

https://media.discordapp.net/attachments/1162852092680077394/1162862266094465084/x.png?ex=653d7ae1&is=652b05e1&hm=627cb7b9b4acefb429e8dbed869ec2e0c09e36004226a7e0d679de2da79b3d40&

Does the hypotenus, 1 lie on the square?

Sorry for being so problematic

But I want to be sure I understand this

[wooden cipher]

The 1 is at an angle off the square

[jaunty goblet]

And the thing that's on the square is root3/2?

[wooden cipher]

Yes

[jaunty goblet]

So this right

So what does speed relative to the water actually mean

[wooden cipher]

The 1 vector should point down and left

So that when it adds to the current, it becomes a straigjt line down

[jaunty goblet]

I think this is the main thing I'm not understanding

[wooden cipher]

If you are in the water, floating with the current like the swimmer is, you will see him move at 1 m/s

Someone who is standing on the shore will see him move at different speeds depending on which part of the square he is swimming around

[jaunty goblet]

Makes sense

Lemme process this for a bit

I think it makes sense now. Thank you so much for being so patient

[wooden cipher]

Youre welcome

[jaunty goblet]

.close

[obtuse pebble]

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[keen heart]

so im supposed to fill in this proof

i've filled everythinb

im at the last line and i can't explain the contradiction

the 2nd last line is p * q | d

so as a whole p*q divides both n and d making them not in their simplest form?

but idk how that contradicts the fact its supposedly rational

need some help

[obtuse pebble]

@keen heart Has your question been resolved?

[obtuse pebble]

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[signal kite]

Suppose both x and y are integers, prove that if x^2· (y^2− 2y) is odd, then x and y are both odd.

[dark vector]

my instinct here is proof by contradiction

i.e. assume x=2n for some integer n and show that buggers smth up

and then do the same for y

[signal kite]

so this is what i came up so far, so my question should I do something else or is this the end for this problem?

[wispy wadi]

the contrapositive is x or y even => x²(y²-2y) even

not and

besides that, the idea of proof is right

but conclusion very wrong

[signal kite]

so when its a contrapositive and becomes or and vice versa or becomes and right?

[wispy wadi]

the negation of x and y odd is x or y even

and then the proof follows, if x is even, then x² is so x²(y²-2y) is
and if y even, then y² is, so y²-2y is, so x²(y²-2y)

I use that the square of an even number is even but obv you can show it too

[obtuse pebble]

@signal kite Has your question been resolved?

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[rough pewter]

how would i solve this: Find the approximate side length of a square game board with an area of 126 squared.

[zenith spade]

What’s the formula for the area of a square?

[obtuse pebble]

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[obtuse pebble]

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[full oar]

i need help with the last question

how do I know if it has a negative rate?

[high lily]

rate of change for a straight line is it's slope

[full oar]

huh

[full oar]

i need help w the last question

[native sorrel]

thats what he did

[signal kite]

negative slopes?

[native sorrel]

negative rate of change = negative slope

[full oar]

ok but how do I know if it’s negative

if y is negative?

[native sorrel]

it means, the y value is decreasing when x value is increasing

[full oar]

okk

[high lily]

if slope is negative

[full oar]

so all of them except for iii

[high lily]

slope of line 4 is wrong btw

[signal kite]

Line 4 has no slope tho

[high lily]

reasoning for c) is also wrong

[full oar]

oh

[full oar]

how do u write it then

idk how to do it for a straight line

[high lily]

same with reasoningfor b

[full oar]

what should i write

cuz idk

[high lily]

you mistakenly wrote x is the same

the lines are parallel because the slope is the same

[full oar]

ohh

[high lily]

as for perpendicular, the correct reasoning would be that the slopes are negative reciprocals of each other

your wording is way to vague

ALSO, not sure why you listed individual points on the side for part a)

[full oar]

wdym

[high lily]

the slope for ai) is -2/5
what's with writing (5,-2)

[full oar]

so i dont get confused

is it wrong?

[high lily]

yes,

[full oar]

oh rlly?

its not -2/5?

[high lily]

the answer is just -2/5

[full oar]

ok

[high lily]

(5,-2) is irrelevant
it's inclusion indicates misunderstanding

[full oar]

how do i write an expression to solve this

[high lily]

what's the question asking for

[full oar]

i have to fill in the numbers

ur adding 4 each time

[high lily]

from the first 4 columns, you can deduce there's a linear relationship and make an equation of a line from that

[signal kite]

y = rateOfChange(x) + y value when x is 0

[high lily]

that indicates the slope is 4,
you can also identify the y intercept from the table

[full oar]

still kinda confused

[signal kite]

to find slope look at y. x value 0,1,2,3 are consecutive, now look at y for 0,1,2,3 by how much is y increasing?

[full oar]

4

[signal kite]

okay 4 is your slope

now linear equation looks like?

y = mx + b

?

seen that before?

[full oar]

yes

y = 4x + b

whats b

[full oar]

oh wait

nvm i use that to find y

?

[signal kite]

b is y intercept so when x is 0 what is y?

[full oar]

wait huh

so 4*0 + y

= y?

what

[signal kite]

in your table when x = 0 y is what value?

[full oar]

3

[signal kite]

therefore b = 3

based on that how would build the formula?

m= 4, b=3

[full oar]

ok so x = 1y is 7?

[signal kite]

y=mx+b

[full oar]

so x is 0 right

[signal kite]

so replace variable m and b in the equation.

m = 4, b =3, formula: y =mx+b

you got this

[full oar]

what abt x

0?

[signal kite]

first right the formula based on what i said

[full oar]

so y = 4x+3

[signal kite]

perfect

[full oar]

ohh

hold on

[signal kite]

now input any number of x to get y

[full oar]

woahh

[signal kite]

you see your chart? you are trying to find y for when x = 57.

[full oar]

yup

its 231

[signal kite]

you got it. Good job

[full oar]

theres more

[signal kite]

okay whats your next question

[full oar]

its part of the same question

[signal kite]

okay so for question a) remember how I told you 0-3 are consecutive meaning they go from 0 1 2 3 throughtout that order by how much does y increase by

[full oar]

yes

i just write 4 right

[signal kite]

yes

[full oar]

for the second question

is it 3

[signal kite]

yes

[full oar]

oh wait

lol sorry i should have fully read it

[signal kite]

yeah no problem it okay

[full oar]

i have this one tho

its a straight line

[signal kite]

first we are going to plot point A and B and connect them.

[full oar]

yup i did that

[signal kite]

you got sumthn like this?

[full oar]

yes

[full oar]

yupp

i did that

[signal kite]

okay so when x = 0 what is the value of y?

[full oar]

3?

[signal kite]

correct

[full oar]

so y intercetp is 3

[signal kite]

yes

[full oar]

okk

[signal kite]

so do you know what slope is

[full oar]

uhh

[signal kite]

explain it

[full oar]

y = mx+ b

3 = b

[signal kite]

that there is your y intercept

[full oar]

yea

[signal kite]

so slope is rise over run

[full oar]

yea

its -13, 0

so its 0/-13

this?

[signal kite]

slope can be positive or negative based on how the line runs so if line from left to right looks like its going down that a negative slope if line from left to right is rising then the slope is positive

[full oar]

oh shoot wiat

thats not right

its?

[signal kite]

example

[full oar]

is my answer correct?

[signal kite]

based on what i said does the black line have negative slope or positive?

[full oar]

negative

i think

no

postive

hang on

wait

😭

i cant think

postive

oh

[signal kite]

you are right the black line is decreasing while purple line is increasing

[full oar]

okk

is my equation right ?

[signal kite]

so now look at this line is it increaing or decreasing?

[signal kite]

unfortunately no its incorrect

[full oar]

OH

oop

how do u find the slope for a straight line

thats vertical and hortinzal

cant spell

oops

[signal kite]

now you are seeing it

[full oar]

ohh

[signal kite]

so based on how your line looks what is the slope?

[full oar]

zero

[signal kite]

correct

[full oar]

soo its y = 0x+3?

[signal kite]

okay so 0 times any number is what?

[full oar]

0

[signal kite]

to write a more efficient equation how about removing the 0x what would your equation be

[full oar]

3

y=3?

ok wait

[signal kite]

yes, y=3 is correct.

good job

[full oar]

this one is vertical

so its undefined

[signal kite]

yes, thats correct

[full oar]

but how do i find the y intercept

[foggy idol]

Y-intercept

Is when x is 0

[full oar]

but x is 5

so

[foggy idol]

No

[full oar]

huh

[foggy idol]

It wants the y-intercept

[full oar]

ye

[foggy idol]

Of rhe line created by the plotted points

5,8 and 5,-3

[signal kite]

since x is 5 there is no x = 0 therefore can you find the x intercept when there is no 0?

[full oar]

uhh

idk

[signal kite]

no so there is no value x = 0 therefor y-intercept does not exist. does the line ever cross the y axis?

[full oar]

no

i think

do i just write undefined?

orr 0

[signal kite]

y-axis is the line that crosses with the x axis therefor it never crosses and y intercept does not exist.

[full oar]

so undefined

?

[signal kite]

so for b you can write: y-intercept does not exist.

[signal kite]

then for part c that is what you are going to write.

[full oar]

so undefined in all the parts for question 10

[signal kite]

This is 10b response

[full oar]

oh ok

[signal kite]

part c: what is the slope of a verticle line?

[signal kite]

i'll help with a and b

so for A plot the point.

its tell you its a verticle line

[full oar]

so it’s like y = a number?

is it -5?

[signal kite]

Side note: You should save this for reference for the future

[full oar]

what abt the equation tho

[signal kite]

almost got it.

[signal kite]

so since it told you it would be a vertical line then it will always cross the x-axis therefore the format for vertical line is x = a number

got it?

you got the equation?

[full oar]

y = 0

[signal kite]

you only use y = some number for horizontal lines x= somenumber for verticle lines

so x = what number that would touch that purple dot? try desmos change that 0 to another number

[full oar]

3?

[signal kite]

yes thats right

x=3

equation of a horizontal line

what would be the equation?

[full oar]

thats the equation for a?

also sorry i went afk for a bit

what do i do for b

[signal kite]

plot b and find a horizontal equation for it

[signal kite]

use this to guide yourself

[full oar]

okk

[obtuse pebble]

@full oar Has your question been resolved?

[obtuse pebble]

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[honest flicker]

This is exactly one right

cuz like different slopes when put into slope intercept form

Y=6x+1
Y=-6x-1

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

What mathematical background does someone need to study abstract algebra? I've taken calculus I-III courses and feel fairly comfortable with linear algebra. Is that sufficient? If so, can someone recommend a good introductory abstract algebra textbook (preferably with a lot of exercises)?

[obtuse pebble]

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[hot wing]

turn this into mathematical symbols.

five less than x is less than five less x

for this it is

5-x < 5-x

[hot wing]

right?

is less than

[wanton hull]

no

5 less than x is x-5

[hot wing]

then x-5 < 5-x

?

god this is confusing

[sacred barn]

Looks right

[obtuse pebble]

@hot wing Has your question been resolved?