help-26

[agile harness]

work energy theorem

W=deltaKE

calculate the work done over the two second interval

[potent blade]

oh

wd= f*d

[agile harness]

then a set it equal to m/2(vf^2-vo^2)

[agile harness]

work=fd for constant force

which applies here

the only forces doing work on the block are mg and F

which are really the forces in the x direction that you need to consider

thus (F-mgsintheta)d=m/2(vf^2-vo^2)

[potent blade]

I have a question, how is mg doing work, when it is becoming balanced by normal reaction force?

[agile harness]

because it has an x component

in the direction of motion

[potent blade]

oh i see

[agile harness]

or rather opposite the direction of motion

[potent blade]

every force can be split, right?

[agile harness]

sure

[potent blade]

I see, thanks

[agile harness]

but i mean not all forces have non zero components

in whatever coordinate system you define

[potent blade]

yeah true

[agile harness]

F doesn’t have a y component

N doesn’t have an x component

[potent blade]

ahh

okay, but it will always have either one

[agile harness]

you’ll need to find d from d=vot+1/2at^2

where a=F-mgsintheta

[potent blade]

ok sure

thanks

[agile harness]

just a bit of algebra

[potent blade]

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[clever citrus]

If something like acacac is passed

it shouldnt be accepted right?

but the DFA accepts it

we have 3 a's 0 b's and 3 c's

3 + 0 = 3 mod 3 = 0 ?

even if I use the corrected DFA

it still is accepted by the DFA

[meager dawn]

3=3

[clever citrus]

how

isnt 3 mod 3 = 0

[meager dawn]

yeah, but there are 3 a’s and 3 c’s

so both are 0 mod 3

so we’re good

[clever citrus]

3 = 3 mod 3 then no?

wait whattt

Im confused

[meager dawn]

that’s also true

[clever citrus]

how come

[meager dawn]

okay so mod stuff is you just want to reduce everything to its remainder when dividing by 3 and THEN check equality

so like if i want to ask if 1=4 (mod 3) i check to see if both remainders are equal

[clever citrus]

this would be truw

4 mod 3 = 1

but in our case

we have 3 + 0 = 3 mod 3

3 = 3 mod 3
3 = 0

[meager dawn]

the (mod 3) is meant to be for the whole equivalence relation

[cunning snow]

a is congruent modulo r mod b
Then a is also congruent modulo r-b mod b it's a property

[meager dawn]

like “mod 3” isn’t really an operator that you’d apply to one side in this case, it’s meant to be a modifier for the conditions in which you’d consider two numbers “equal” (they’re equal mod 3 if and only if they have the same remainder mod 3)

[cunning snow]

bruh that isn't allowed in modular arithmetic

[clever citrus]

wait so over here its just saying amount of a's + amount of b's = amount of c's?

[meager dawn]

it’s saying that if you take number of a’s + number of b’s, and also take the number of c’s, then take both of their remainders after dividing by 3, those will be equal

[clever citrus]

ohhhhhh

so the mod 3

is for both sides

not just rhs

[meager dawn]

the mod 3 applies to the = sign, and yeah you can think of it as equality after “applying the operation” to both sides

[clever citrus]

I seee

alr ty

alsoo

acc one more q

say I had something like

a + b = c mod 4

then the DFA would be similiar but instead have the 1st and 4th state as the accepting state and there would be another transition from 3rd state to 4th state for a, b and 4th state to 3rd state for c

and then I remove the 2 transitions from 1st to 3rd state and instead have them from 1st to 4th state ?

[meager dawn]

probably the fourth state would not be an accept state

similar to why the third state is not an accept state in the current DFA

[clever citrus]

oh im slow

I thought it was im tripping

ignore that

other thn that its correct

[meager dawn]

yep

[clever citrus]

alr I see now

ty

[meager dawn]

also, seems as though you’re no longer a calc III victim

more like a based DFA enjoyer now

[clever citrus]

nah im forever a calc III victim. Prof fucked us up on the exam

praying he curves but I doubt it

again ty

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[severe lichen]

(x-2)/(x^2-4)

[severe lichen]

I have a removable discontinuity at x = -2 right

[pseudo sonnet]

yes

[severe lichen]

And the limit is 1/4

Because my derivative is 1/(x+2)

And when x approaches 2

The limit approachs 1/4

[neon iron]

y

e

s

[severe lichen]

1/2+2

okay so what about when x = 2

when x = 2 I have 0/0 so does that mean there's a non removable discontinuity at x = 2?

This is what I have so far

[topaz sinew]

@severe lichen Has your question been resolved?

[reef fjord]

what is (x-2)/(x^2-4) if you simplify it

[severe lichen]

(x-2)/(x-2)(x+2)

[reef fjord]

and so (x-2)s 'cancel' and you get what?

[severe lichen]

1/(x+2)

[reef fjord]

now note you can plug x=2 just fine

this is what you would typically call a removeable discontinuity

[severe lichen]

But if I did it before the cancellation I would get 0/0

[reef fjord]

yes, but 0/0 is what we call in calculus an 'indeterminate form'

it can be anything limit-wise depending on the function

[severe lichen]

but the function is 0/0 as well

So it's just undefined at x = 2

So it's a removable discontinuity

[reef fjord]

but note what happens at x=-2

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@severe lichen Has your question been resolved?

[severe lichen]

its an asymptote

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[trail goblet]

hello! this is a pre test for our course and I have no idea how to answer them

[trail goblet]

topic is inequalities, i think

i might be doomed fr

[hexed quartz]

how much time do u have?

[trail goblet]

30 minutes

[hexed quartz]

well

[trail goblet]

but its ok ^^ ill try to learn

[hexed quartz]

to solve it u have to isolate the variable

[trail goblet]

how would i do that

[hexed quartz]

for example

take the first one for example

u put every term with a variable on one side of the equation

and every other term on the other

[trail goblet]

would the sign change?

[hexed quartz]

it depends

an easy way of seeing this

[trail goblet]

should i distribute

[hexed quartz]

yes

easy way of solving it is imagine it as 2 functions seeying where they meet and seeying where one is higher than the other

[trail goblet]

hmm

after distributing i got 5x + 8 < - 3 - 18z

is that correct?

[hexed quartz]

yes

put all the x on the left

[trail goblet]

okay, now how would one isolate the variables from the constants

[hexed quartz]

all the constants on the right

[trail goblet]

just put it there? no rules whatsoever?

5x - 18z < 8 - 3

like thatt?

[hexed quartz]

the minus become a plus

[trail goblet]

oh

does the inequality sign change too?

[hexed quartz]

and the plus from the 8 become a minus

inequality remains

[trail goblet]

i got 5x + 18z < - 11

is that correct?

[hexed quartz]

its not a z its an x

[trail goblet]

oh shoot my bad

23x < - 11

is that correct?

[hexed quartz]

yes yes

[trail goblet]

nowww whats nexttt

[hexed quartz]

divide both sides by 23

have u done it?

[trail goblet]

yes!

x < -11/23

[hexed quartz]

YES

that's the answer for 1

for question 1 that is the answer

[trail goblet]

ohh

its simple

[hexed quartz]

yes yes

[trail goblet]

but what if the inequality sign is different?

[hexed quartz]

what do u mean?

u mean reversed?

it would still apply

the answer would still have been -11/23

[trail goblet]

like if its not less than

what if its greater than or equal to

[hexed quartz]

if it was more than the answer would still have been the same

[trail goblet]

oh so the inequality sign does not matter

i will try answering the rest, and ill send them here. thank you for teaching me! i learned immediately

[hexed quartz]

the inequality sign sometimes changes when you are solving the problem

that's the tricky part of the inequality sign

[trail goblet]

oh, can u givr an example?

or is it in my problems

[hexed quartz]

yes

-x smaller than 1

[trail goblet]

nvm ill try answering each of them and send it here to make sure im doing it righr

for number 2 i got -13/46 greater than/equal to w

idk how to do the sign here

if i wanted to put the variable to the left, should i change the sign?

it would then be w less/equal to -13/46?

[hexed quartz]

u change the sign when u multiply or divide by negative numbers

[trail goblet]

did i do it correctly?

[hexed quartz]

i think there was a mistake

can u show me how u have done it quickly

[hexed quartz]

its not -13/46 its - 12/46

[trail goblet]

oh sorry

[hexed quartz]

u made a small error when calculating 30-18

[trail goblet]

yeah lol my fault

but the rest okay?

[hexed quartz]

also u divided by a negative number so the sign of the inequality changes

[trail goblet]

so it is less than now?

[hexed quartz]

its not smaller or equal to but greater and eqaul to

[hexed quartz]

i dont understand what u mean

[hexed quartz]

yes yes

that is correct

u can simplify -12/46 but idk if it's obligatory in ur case

[trail goblet]

alright

i got 3

[hexed quartz]

no no

[trail goblet]

i meant number 3

xd

[hexed quartz]

oh ok

[trail goblet]

i made another error of simplifying 14/20 to 8/10

i should work on my basic math at this point xD

[hexed quartz]

yes

did u write 7/10 at the end?

[trail goblet]

yes

for number 4 im not sure if i simplified it right

i got 2/3 -1/3z on the left

and 1/9z + 8 - 14/9z to the right

havent simplified that one yet

[hexed quartz]

yes

[trail goblet]

is it -13/9z + 8?

[hexed quartz]

on the right yes

u still have time for others?

or is it finished?

it's done😔

30 mins has run off

[trail goblet]

we have 10 more

not all of us are finished

im stuck in number 4 because of the fraction, i dont think i got 13/9 + 1/3 right

i got 14/9

[hexed quartz]

let's continue from here

[trail goblet]

now what if i have 2 inequality sign in my problem

[hexed quartz]

take the +8 and put it on the left

[hexed quartz]

what

[trail goblet]

done

[hexed quartz]

the the -1/3*z from the left and put it on the right

[trail goblet]

yes

[hexed quartz]

calculate for each side

[trail goblet]

2/3 - 8 > -13/9z + 1/3z

[hexed quartz]

yes

2/3 -8 what does that give u after calculation

[trail goblet]

-22/3

[hexed quartz]

yes

now the tight side

[trail goblet]

-14/9?

[hexed quartz]

,p

no

[trail goblet]

uh

[hexed quartz]

it should give u -10/9

[trail goblet]

oh it is 10

i forgot the sign from 13

-10/9 yes

[hexed quartz]

ok great

no we multiply each side by 9

and divide by -10 and u will get an answer

[trail goblet]

33/5 > z

[hexed quartz]

u forgot to change the sign

[trail goblet]

whaat which sign

[hexed quartz]

inequailty

[trail goblet]

oh right i divided with a negative

now in 5, theres two inequalities

[hexed quartz]

yes yes

now u u must calculate twice

u must first solve it without the second inequation

then solve it without the first inequation

[trail goblet]

howw

use the middle side on both inequalities?

[hexed quartz]

ur first inequation is gonna be this

[trail goblet]

okay

[hexed quartz]

and ur second inequaiont is gonna be this

and the way u will write ur answer will be different

[trail goblet]

that makes sense

ill solveb it first

[hexed quartz]

yes yes

[trail goblet]

3 >/_ x

and 2 </_ x

[hexed quartz]

are u wrting minux -x?

[trail goblet]

no

thats the ineq

[hexed quartz]

ok

yes that is the answers

NOW

[trail goblet]

greater/equal

[hexed quartz]

i recommend u draw a line like this with ur answers

[trail goblet]

and replace 2 and 3 with my answers?

how do i know which is where

[trail goblet]

i dont get it

[hexed quartz]

we see for the first inequation 3 >/_ x

for the second inequation and 2 </_ x

we must find all values of x where these two solutions apply

so everything between 2 and 3 are solutions

if we took something smaller than 2 the solution from the second inequation would not apply ( 2 </_ x)

if we took something greater than 3 the solution from the first inequation would not apply (3 >/_ x)

[trail goblet]

hmm

what do i write in the line

[hexed quartz]

idk how ur teacher tough u to write this but this is how it's been agreed to write from i come from

i would have wrote S=[2,3]

[trail goblet]

the line doesnt matter?

[hexed quartz]

the brackets?

u mean []

[trail goblet]

the one u draw

[hexed quartz]

u dont have to draw the lines

it's only for visualisation

[trail goblet]

okay

for number 6 i got -16/7 < x AND x <_ -1

same with S=[]?

[hexed quartz]

no

when -16/

16/-7 isnt included

and -1 is included

so u write S= [-1;-16/7[

open brackets on the right because -16/7 isnt included

[trail goblet]

okayy

okay last one

for the item 7

i got 0 < 1/4t

why is there a zero

[hexed quartz]

then t is bigger than 0

that can happen

[trail goblet]

what do i write?

[hexed quartz]

S=]0;1[

[trail goblet]

on the other i got -1/4 < -1/4t

do i divide

[hexed quartz]

u should multiply both sides by -4

[trail goblet]

oh

-1 < -1t

how do i write for both inequalities

with the both sides

[hexed quartz]

u write with both side of the brackets open

S=]...;...[

[trail goblet]

S=] -1;-1 [

??

[hexed quartz]

no no

[hexed quartz]

u havent finished

[trail goblet]

oh right

[hexed quartz]

u still have to get rid of the negative

[trail goblet]

1<t

[hexed quartz]

u forgot sometinn

[trail goblet]

1>t

[hexed quartz]

yes

now that's the solution of ur second inequation

what was the solution of ur first inequation?

[trail goblet]

how do write them in a single S=

[trail goblet]

this

[hexed quartz]

yes

i gave u the answer thinking u already had found somehting

that was my bad

[trail goblet]

now how do i write it with the second ineq

[hexed quartz]

no no that is the final answer

[trail goblet]

oh okay

thank you

[hexed quartz]

ur first solution was 0<t and ur second solution was 1>t. u combine both and u have S=]0;1[

[trail goblet]

OMGMGMG

i had to pass it

i wasnt finished lol

but ill finish here for the review

for the item 8

i got 2 fractions

[hexed quartz]

for the item 8, what have you foud

you should have one fraction

[trail goblet]

16/11 >/_ m and 1/11 </_ m

omg

bro my _

[hexed quartz]

ur first is right

16/11 >/. m this is correct

1/11 </ m how did u end up here?

[trail goblet]

what about the second one

[hexed quartz]

nvm it's also correct

[trail goblet]

okay

[hexed quartz]

both are correct

[trail goblet]

so how do i write it

[hexed quartz]

u write from the smallest to the highest

so in the inequality signs it says smaller or EQUAL to and greater or EQUAL to

u will have S =[...;...]

[trail goblet]

do i put the fractions in there?

[hexed quartz]

yes

fractions are still real numbers

Pls excuse me

im really tired

i wont be able to help u with 9 and 10

[trail goblet]

oh okay, thats fine!!

ill look it up from here

you helped me a lot !!

thank you so much for your patience and your time

[hexed quartz]

just the solution for 8 is S= [1/11 ; 16/11]

gl for 9 and 10

[trail goblet]

okayy, thank you!!

rest well

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@trail goblet Has your question been resolved?

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[tacit turret]

One message removed from a suspended account.

[sharp dew]

Differentiate it

The keyboard looks so fire

So clean

So neat

[tacit turret]

One message removed from a suspended account.

One message removed from a suspended account.

[sharp dew]

Have you learned about differentiation

[tacit turret]

One message removed from a suspended account.

One message removed from a suspended account.

[sharp dew]

Go and get it

Go and get it! Get your revenge

[tacit turret]

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One message removed from a suspended account.

One message removed from a suspended account.

[sharp dew]

Cool

Cool water

[tacit turret]

One message removed from a suspended account.

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[fleet basalt]

How do I turn these into standard equations?

[deft holly]

Q2?

[fleet basalt]

1b and 1c’s kinda unfamiliar for me

I could graph but I can’t really figure out these questions myself

[deft holly]

so you wanna do 1b and c first?

[fleet basalt]

yeah. I got 1a figured last night but I didn’t understand what to do at 1b and 1c

Anything with fractions basically

[deft holly]

these fractions are still square numbers

the right hand side (radius) is already to be a fraction

so for b, we can write 4/9 as (?)^2, and the ? will be your radius

[fleet basalt]

Then I can find my radius after I graph it correctly with h&k right?

[deft holly]

not real sure what you saying, im guessing thats the centre so yes

[fleet basalt]

yeah it is the center

And about 1c?

Do I divide the given numbers by 16?

[deft holly]

multiply 16 to the other side

[fleet basalt]

how do I do that??

[deft holly]

just

times each term by 16

[fleet basalt]

Can you dumb it down for me? I’m real weak on this one

[deft holly]

so

x/3 +y/3 = 1

is the same as

x + y = 3

right?

[fleet basalt]

yes

[deft holly]

same thing here

[fleet basalt]

Ah so I multiply both two and five

By 16

[deft holly]

multiply the whole term

(x+3)/4 = 5

is same as x+3=20

kind pf thing

[fleet basalt]

i still cant understand 💔

[deft holly]

:(

[fleet basalt]

im genuinely struggling w/ it im just tryna pass

[deft holly]

dwdw

uhhh

ill do an eg

[fleet basalt]

omg an example

I get the last part but what do I do with the multiply part

[deft holly]

cause they both on a fraction of 9

i can times all pf the terms by 9

and the left hand side is a nice circle formula

with no fractions

[fleet basalt]

Also I have to make it a standard equation like this

[deft holly]

ye

thats what i thought

[fleet basalt]

So should the answer be (x-2)2 + (y-5)2 = 4 squared

since the square root of 16 is 4???

[deft holly]

the - should be positive but yep

[fleet basalt]

The first parenthesis should be a positive then?

[deft holly]

positive

[fleet basalt]

oh all of them

[deft holly]

if you wanna write them as -k, -r

you do -(-2)

-(-5)

[fleet basalt]

I got it all correct all I gotta do is just turn them all to positives then

[deft holly]

yea your radius is fine

[fleet basalt]

yep, now all I gotta do is 2a-f

Endpoints are where the circles end up right??

[deft holly]

where the line ends

cause diameter arc, the points where diameter hits the circle

[fleet basalt]

Ah alright

what about C??

these things

[deft holly]

C is centre

[fleet basalt]

oh okay

Well i suppose all I gotta do is just turn them into standard equations like the ones on top

[deft holly]

yep

[fleet basalt]

What about 2b tho

[deft holly]

what bout it?

[fleet basalt]

its another fraction 💔

[deft holly]

ye

[fleet basalt]

wat do I do with it?

[deft holly]

you can just leave it as a fraction

[fleet basalt]

oh okay

[deft holly]

it can do that

[fleet basalt]

wat about these

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@fleet basalt Has your question been resolved?

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[gaunt cosmos]

How does one determine the coefficients using pascal's triangle

[ionic root]

hello

do you know the pascal triangles??

[gaunt cosmos]

no i am not sure what that is

[ionic root]

ok

so a pascal triangle looks like this

1
1 1
1 2 1
1 3 3 1

[ionic root]

yes

[gaunt cosmos]

but then how do i know what the coeffcients are

[ionic root]

so look at the triangle

1 + 1 = 2

6 + 4 = 10

[gaunt cosmos]

yes

[ionic root]

then keep doing it until the second term is 7

[ionic root]

then write it in the order like that

[gaunt cosmos]

So the coefficents would be in order: 1, 7, 21, 35, 35, etc?

[ionic root]

a^7 + 7 * a^6 * b + 21 * a^5 * b^2 +...+b^7

[ionic root]

you got it bro

[gaunt cosmos]

lol thank you

.close

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[ionic root]

got a little issue 🙂

[gaunt cosmos]

oh wait do u have more info sorry

[ionic root]

info what??

[gaunt cosmos]

did I close too early?

[ionic root]

no

close it:)

[gaunt cosmos]

ok thank you! :))

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[ionic root]

can somebody help me pls

[topaz sinew]

Please don't occupy multiple help channels.

[ionic root]

Find positive integers (x, y, z) so that $xyz = x^2 - 2z + 2$

[thorny flame]

Nguyễn Tuấn Minh

[ionic root]

<@&286206848099549185>

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@ionic root Has your question been resolved?

[solid swift]

Will a picture of the answer do?
maybe lmk if you can't understand a part :)

[topaz sinew]

@ionic root Has your question been resolved?

[cunning snow]

It's a diaphantine eq find one sol and then use the formula to find the gen form

[topaz sinew]

@ionic root Has your question been resolved?

[inner wren]

There are infinite solutions i suppose

x = -10 and y = -10 and z = 1
x = -9 and y = -9 and z = 1
x = -8 and y = -8 and z = 1
x = -8 and y = 3 and z = -3
x = -7 and y = -7 and z = 1
x = -6 and y = -6 and z = 1
x = -5 and y = -5 and z = 1
x = -5 and y = 1 and z = -9
x = -4 and y = -4 and z = 1
x = -4 and y = -1 and z = 3
x = -4 and y = 0 and z = 9
x = -4 and y = 1 and z = -9
x = -4 and y = 2 and z = -3
x = -4 and y = 5 and z = -1
x = -3 and y = -3 and z = 1
x = -2 and y = -2 and z = 1
x = -2 and y = 0 and z = 3
x = -2 and y = 2 and z = -3
x = -2 and y = 4 and z = -1
x = -1 and y = -1 and z = 1
x = -1 and y = 1 and z = 3
x = -1 and y = 3 and z = -3
x = -1 and y = 5 and z = -1
x = 0 and y = -10 and z = 1
x = 0 and y = -9 and z = 1
x = 0 and y = -8 and z = 1
x = 0 and y = -7 and z = 1
x = 0 and y = -6 and z = 1
x = 0 and y = -5 and z = 1
x = 0 and y = -4 and z = 1
x = 0 and y = -3 and z = 1
x = 0 and y = -2 and z = 1
x = 0 and y = -1 and z = 1
x = 0 and y = 0 and z = 1
x = 0 and y = 1 and z = 1
x = 0 and y = 2 and z = 1
x = 0 and y = 3 and z = 1
x = 0 and y = 4 and z = 1
x = 0 and y = 5 and z = 1
x = 0 and y = 6 and z = 1
x = 0 and y = 7 and z = 1
x = 0 and y = 8 and z = 1
x = 0 and y = 9 and z = 1
x = 0 and y = 10 and z = 1
x = 1 and y = -5 and z = -1
x = 1 and y = -3 and z = -3
x = 1 and y = -1 and z = 3
x = 1 and y = 1 and z = 1
x = 2 and y = -4 and z = -1
x = 2 and y = -2 and z = -3
x = 2 and y = 0 and z = 3
x = 2 and y = 2 and z = 1
x = 3 and y = 3 and z = 1
x = 4 and y = -5 and z = -1
x = 4 and y = -2 and z = -3
x = 4 and y = -1 and z = -9
x = 4 and y = 0 and z = 9
x = 4 and y = 1 and z = 3
x = 4 and y = 4 and z = 1
x = 5 and y = -1 and z = -9
x = 5 and y = 5 and z = 1
x = 6 and y = 6 and z = 1
x = 7 and y = 7 and z = 1
x = 8 and y = -3 and z = -3
x = 8 and y = 8 and z = 1
x = 9 and y = 9 and z = 1
x = 10 and y = 10 and z = 1

[cursive patrol]

it appears you have neglected the part of the question which specified "positive" integers

[wary tulip]

that also does not show there are infinitely many. but yes, there are. take (x,x,1) for any positive integer x

[neon iron]

Lol

why is that many infinite?

[wary tulip]

(1,1,1) is a solution, (2,2,1) is a solution, (3,3,1) is a solution, etc

[inner wren]

there are a lot thought

though*

[wary tulip]

it may be a pell-like equation but i am both bad at factoring and not motivated to try

[neon iron]

what would make you motivated to try

[inner wren]

wait

let z = `

z = 1

then there are infinite?

[wary tulip]

i just said that lol

[inner wren]

oh

[inner wren]

yes

[ionic root]

i found a solution when x = y and z = 1

[inner wren]

yes

x,y -> N
where z= 1
is a general soltuion

[wary tulip]

interesting use of the word general

[wooden osprey]

can you provide a picture of the problem?

[ionic root]

i found it lol

x = 4, y = 1, z = 2

[neon iron]

?

[ionic root]

i'm vietnamese so the symbols r different

[wooden osprey]

just send it

[ionic root]

no phone btw

🙂

[neon iron]

i found it too!!! (0,1,1)

[wooden osprey]

okay, can you restate the question verbatim, in vietnamese?

[ionic root]

what is vertabim

[neon iron]

can't you transfer the picture to your pc in some way or the other

[wooden osprey]

0 isnt a positive integer

[neon iron]

verbatim = word for word

[wooden osprey]

gửi câu hỏi nguyên văn

[neon iron]

was a typio for (1,1,1)

*typo

[ionic root]

no

[wooden osprey]

....

[ionic root]

x = y, z = 1 is always true

[neon iron]

(4,1,3)

[ionic root]

I can prove it

[wooden osprey]

how would you prove it?

[ionic root]

$xyz = x^2 - 2z + 2 \implies z = x^2 + 2 / xy + 2$

[neon iron]

i mean true with the restrictions on the denominator

[wooden osprey]

x^2

use $\implies$

[neon iron]

not sure what you're tryingt o do anyway

[fallow igloo]

🐐

[thorny flame]

Nguyễn Tuấn Minh

[wooden osprey]

$xyz = x^2 - 2z + 2 \Rightarrow z = x^2 + \frac2{xy} + 2$

[wooden osprey]

this is wrong

[neon iron]

what

he clearly meant (x^2 + 2)/(xy + 2)

[wooden osprey]

why did you only divide the 2 term by xy

[ionic root]

$If x = y \Rightarrow z = 1$

[thorny flame]

Nguyễn Tuấn Minh

[wooden osprey]

put the if outside of dollar signs

[ionic root]

if $x < y \Rightarrow z < 1 but z >= 1$

[thorny flame]

Nguyễn Tuấn Minh

[ionic root]

so z doesn't exist

[neon iron]

can you just type your problem in vietnamese

[ionic root]

if

[neon iron]

in your keyboard or whatever?

[ionic root]

ok

[wooden osprey]

i...i dont even know what you are trying to communicate

[ionic root]

Tìm tất cả các bộ ba số nguyên (x, y, z) thỏa mãn xyz = x^2 - 2z + 2

[wooden osprey]

hm

[ionic root]

solved

but can you guys check it for me

if x>y then x^2 + 2 > xy + 2 but z is an integer so x^2 is divisible by xy + 2

after playing with numbers and letters we have 2 * (x + y) is divisible by xy + 2

so there is a number k so that $2(x+y) = k(xy + 2)$

[thorny flame]

Nguyễn Tuấn Minh

[ionic root]

if k >= 2 then there is no x, y,z

-> k < 2 but k is an integer so k = 1

-> x= 4, y = 1, z = 3

[wary tulip]

i haven't read the channel in a bit but i believe the only solutions are of form (x,x,1) and (4,1,3)

[ionic root]

yes

my friends and i have proved it

but can sb help me in my second channel

[wary tulip]

didn't read what you wrote but yea i agree with the result

[ionic root]

it's help-33

helo

.close

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[ionic root]

.

.

.

.

.

.close

.reopen

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[severe marsh]

Hello

[severe marsh]

Between 7-8 is that speeding up on the negative direction

And for 10-11 is that slowing down?

[barren zodiac]

yes

the velocity is getting closer to 0, so the object is slowing down

[neon iron]

10-11 its slowing down from the "reverse direction". Imagine walking in a straight path. Velocity will be +ve. Now slow down your walking movement, your velocity slowly decreases but is still be +ve.

However, for your case 10-11, imagine yourself walking backwards, velocity will be -ve. Now slow down your velocity will likewise decrease.

As you slow down for both scenarios, your velocity will approach zero.

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[terse bronze]

I am trying to solve for part b

when I do the invBinom function on my calculator it gives a critical value of 14

but the critical value is 13

so I was wondering why the calculator outputs a value 1 away from the actual critical value

I heard it's because the function in the calculator might be continuous and we're doing discrete distribution but I don't really understand the concept well

could someone explain to me when I need to add 1 or subtract 1 from the value the calculator gives me

[topaz sinew]

@terse bronze Has your question been resolved?

[neon iron]

@terse bronze

[terse bronze]

hi!

[neon iron]

Could 13 be your

N-1

for DF

because it seems like

[terse bronze]

DF?

[neon iron]

This is a t-test

degree of freedom.

Inverse binom your

0.05 , 29, 0.6

where your number of trials will now be n-1

with the consideration of the degree of freedom

[terse bronze]

hmm ill try doing it with n-1

[neon iron]

LMK how it goes

also which graphic calculator is that!!

[terse bronze]

oh it works!

Tinspire

it gives me 13 now

[neon iron]

Anyways, i hope this explanation will help;

[terse bronze]

but how did you know to do n-1

[neon iron]

Im going to assuem that this

hypothesis test is a t-test as you have 30 samples (usually t-test is done with n<30).

So for t-test you usually have to find the degree of freedom which is n-1.

So since this is a one sample test; your
$H0 \geq .6$
$H1 < .6$

[terse bronze]

not sure

[thorny flame]

Sukiyaki

[neon iron]

Its a one tail test, then assuming that you were not given population stdev

its more difficult to do the z-test.

So, thats where i guessed u did t-test for.

[terse bronze]

hmm but I didnt use any t -test function on my calculator

[neon iron]

Ultimately without my yapping, it would be n-1 for t-test.

Oh what.

[terse bronze]

i just used inversebinom

?

[neon iron]

Hm, well...

Then im not really sure

but from what i learnt in statistics, i would use a t-test for this question.

So thats where i assumed the n-1 for degree of freedom.

[terse bronze]

I know we can use t-test function for normal distribution

but this is binomial

so I had to used invbionomial instea

maybe someone else will know

[neon iron]

well, yep.

[neon iron]

ill let someone have this question then.^

[terse bronze]

alright

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@terse bronze Has your question been resolved?

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@terse bronze Has your question been resolved?

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@terse bronze Has your question been resolved?

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[quaint bone]

For this question

I rewrote it as

pi/4 + pi/3

is that correct?

or am I on the right track?

[sudden temple]

erm, not quite

pi/3 + pi/4 = 7pi/12

[quaint bone]

what

should I rewrite it as

[sudden temple]

erm, compound angle formula is sin(a+b) thing right?

[quaint bone]

ye

[sudden temple]

you know what sin(2pi-x) is?

[sudden temple]

?

[quaint bone]

I forget

[sudden temple]

see, sin(2pi-x) = -sinx
since there is no change in function about x-axis and 2pi-x lies in 4th quadrant where sine function is negative

okay?

[quaint bone]

OHHH

I get it now

so what should I rewrite the equation as?

[sudden temple]

write 17pi as 24pi - 7pi

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[grand solar]

Can't x and y be close to 0 aswell

[grand solar]

if we had an asymptote at 0 or something

nvm I see

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[clever citrus]

How is this O(n^3)

I dont get it

for min weight edge it would go over the whole graph to get the min right and the loop runs O(n) times so wont O(n^2) make sense

second question: How were they able to write log_10(b) in terms of n

as n

would O(nlogb) be valid too

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[clever citrus]

<@&286206848099549185>

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[golden flint]

i dont understand why d) is D1

[finite moth]

do you know how to find expected value

[golden flint]

yeah

[finite moth]

so both D1 and D2 have the highest expected value

the question asks which one is most likely to show the highest number

since D1 has a number higher overall than D2, it's the right answer

[golden flint]

how did u get 1 and 2

in got 1 and 3 the same E(X)

[finite moth]

E(D1) = (1+1+4+4+7+7)/6 = 4

E(D2) = (2+2+5+5+5+5)/6 = 4

E(D3) = (3+3+3+3+3+6)/6 = 3.5

[golden flint]

333366

oh well it doesnt matter anyway

thankyou

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[deft holly]

how can i find the range

this is my reasoning for domain

[ember lodge]

you may want to draw graph that combines those two parameter x and y

[modern rock]

since you're after the range, you only really need to focus on y

[ember lodge]

hint:||It's a circle||

[deft holly]

i was scared of doing that cause theres some cases where the circle is restricted

[modern rock]

and if u have seen auxillary angles before you know that sin and cos will form a larger wave

[ember lodge]

don't be scared, lol

if u know it's a circle, then the domain of f = the range of f

[deft holly]

oh

fair enough

mmmm ok then

thx

[ember lodge]

whenever you see something with sin or cos with a parameter t, it's somehow related to circle or rounded-shape thing or even hyperbolic things

[deft holly]

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[deft holly]

if the parametric forms a circle, can i just say domain and range is that of the circle

and thats it?

[ember lodge]

ye

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[pulsar girder]

anyone know how to calculate the limit of this

[pulsar girder]

is it just 0?

im sorry how do we get 2/n?

ah thank you

bless

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[pulsar girder]

@potent ravine sorry bro, but how is harmonic series geometric or am i misunderstanding something? what is r here

the common ratio i mean

[pulsar girder]

🤔

[bold urchin]

the harmonic series diverges

but slowly

[pulsar girder]

bro im confused, i guess the nth partial sum of harmonic series is not the same as harmonic series

[potent ravine]

oh 1/n at the end

my bad, I should gts

[pulsar girder]

still the limit should be zero right? if we distribute then every summand has n in denominator

and 1 in the numerator

and then we get lim 1/n + lim 1/2n + ... + lim 1/n^2

right?

<@&286206848099549185>

[potent ravine]

should be diverging as @bold urchin said, forget whatever I said lol I didn't check the last term properly

[pulsar girder]

there is a 1/n in front

[pulsar girder]

.

[potent ravine]

note that 1/n converges but not the series, if you group the terms of the series as follows: 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16)...., it is clear that the sum of each group of terms in parentheses is greater than 1/2. keep grouping the terms of the infinite series to get a series which is greater than the infinite series of 1/2

[pulsar girder]

thanks bro i ve seen that somewhere. i know harmonic series diverges, but this is different

its 1/n * harmonic series

and i think i can distribute 1/n without problem

[potent ravine]

also yeah I apologize for what I said before, really made a huge mistake in seeing

[pulsar girder]

no problem, thanks anyway

[bold urchin]

np

[pulsar girder]

you guys still not sure whether the limit is zero, bcs im not either

[bold urchin]

eh like you will distribute an infinite number of time in the limit thats kinfa weird

anyways i think it converges

because

so the ratio will tends to 0

[pulsar girder]

i see

[bold urchin]

but that’s not formal

[potent ravine]

yes the overall series should converge, because summation 1/n is bounded and 1/n will be tending to zero at n tending to \infty so the overall series will be converging to zero

[pulsar girder]

i just got the info that i cant distribute

at least not in this case

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[frail thistle]

In trig if I am doing sine, cosine, or tangent do I put theta with or without the degree symbol. For example sin(55) or sin(55°)

[frail thistle]

Hope that makes sense

[agile harness]

without

[frail thistle]

Why

Khan academy says to do it with

[agile harness]

because the degree symbol is ugly

[frail thistle]

And will mark wrong if you don’t

So it’s optional

[agile harness]

cap

[frail thistle]

Nah

[agile harness]

https://tenor.com/view/stop-the-cap-cap-laugh-video-reaction-youtuber-gif-18293251

[frail thistle]

I had to switch to degree on my calculator when doing it

I think it was khan academy that did that

Something did

[loud oasis]

i think the degree sign is better since otherwise you could get confused with radians

[agile harness]

🤷🏼‍♂️🤷🏼‍♂️

[frail thistle]

So if I did use degree

Would my answer have a degree sign

[agile harness]

if you’re putting it into
your calculator you need to change the mode based on whether you’re using radians or degrees

[frail thistle]

Ik

[agile harness]

i’ve never written the degree symbol when using trig

[frail thistle]

My beautiful TI-84 Plus CE python calculator 🙏

lol

[frail thistle]

I’ve never even taken trig

I’m in geo rn lol

[agile harness]

which has trig

[frail thistle]

True

Basic

But yeah

[agile harness]

my geo class did too

[loud oasis]

this is basically just a style thing, but $\sin(\ang{55})$ is the most unambiguous notation

[thorny flame]

cloud

[loud oasis]

particularly as radians become the "default" for trig functions in later classes

[frail thistle]

So does my answer have a degree sign tho

And length being a degree seems weird

[agile harness]

length?

[frail thistle]

Right?

Or am I not understanding something

[loud oasis]

the input of a trig function is an angle in degrees, the output of a trig function is not an angle so it should never be in degrees

[frail thistle]

Ok

Like an example problem is like “hypotenuse of a right triangle when theta = 55° and the opposite side is 35m”

I need to visualize this

So you need sin(55)=35/X

Right

Or sin(55°)=35/X

[agile harness]

it doesn’t matter

[loud oasis]

they are the same equation but typeset a bit different

[frail thistle]

Won’t it change the answer

So with degree it’s 42.727

[agile harness]

only if you used 55 radians instead

[loud oasis]

but 55 is understood to be an angle in degrees either way, you're just choosing whether to make that explicit or not

[frail thistle]

So if didn’t use it I’d get -35.009

Is that right

How can those be worth the same

Did I do something wrong

I’ve never gotten a negative length

Nah I give up

Idk

I clearly don’t know how to do this stuff

Because a negative length doesn’t make any sense

[loud oasis]

your calculator is in radians mode, so anything without a degree sign is interpreted as radians. you can change this by pressing "mode" and changing the setting

[frail thistle]

So if the answer is negative then you have to do degree

So it’s not optional all the time

Right?

[loud oasis]

just switch your calculator mode

[frail thistle]

But I thought you could use either or

I don’t understand

But you can’t use radians unless you convert the angle

So you would have to do 55 x pi/180

[loud oasis]

because sin(55 degrees) is not the same as sin(55 radians)

[frail thistle]

Ok but do some numbers have the same value?

For both

[loud oasis]

but if you switch your calculator to use degrees by default then it will use degrees if you don't have a symbol

[loud oasis]

0 degrees = 0 radians

there is a difference between "which is correct for writing down on paper" and "which one will my calculator interpret correctly"

[frail thistle]

Ok so the video I watched was wrong lol

He put in the angle

Without the degree sign

And didn’t convert it

[loud oasis]

you can do that too if you change the setting!

[frail thistle]

Brother

Please

Give me a moment

Ok

So I looked over the work I did last night

And I used the degree sign so I guess I just need to use that always

So I don’t ever need to convert

@loud oasis

So I guess the last thing I need to ask is why you would ever want to convert to radians

It’s just more work

[loud oasis]

so is your question "should i use the degrees symbol when writing down on paper?" or "should i use the degrees symbol when entering into my calculator?"

[loud oasis]

radians are more useful in calculus

[frail thistle]

In the calculator

Was my question

[loud oasis]

what kind of calculator do you have? can you send a picture of what happens when you do or don't?

[frail thistle]

TI-84 Plus CE Python graphing calculator

[loud oasis]

so you should press "mode" and change the setting from "radian" to "degree"

[frail thistle]

I would do that when finding theta

When using like sin^-1

I’d switch to degree because I can’t just add a degree sign to that one

[loud oasis]

if you are in degree mode, then it should give the same result in both cases

[frail thistle]

Yes but I was only using degree for that one because I didn’t know you needed to convert radians

So I would put in sin(55) for a 55° angle

Because I thought it was the same thing

I didn’t know radians had to be converted to

Now I know

[loud oasis]

why is your calculator in radians mode? do you use radians in any calculations?

[topaz sinew]

@frail thistle Has your question been resolved?

Channel closed

Closed by @frail thistle

Use .reopen if this was a mistake.

[frail thistle]

It’s the default

[topaz sinew]

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[misty sand]

i think i might be a little slow

how do i find a

[mint crescent]

You have $N_0$, so substitute in either of the population data points to solve for a

[thorny flame]

Civil Service Pigeon

[mint crescent]

Alternatively, note that $$\frac{3.0}{2.65}=\frac{N_0 a^8}{N_0 a^0}$$

[thorny flame]

Civil Service Pigeon

[misty sand]

wait just making sure