help-42

[trail jolt]

i am referencing off this question btw

if y’all get what I mean

otherwise pls lmk

[calm coral]

@trail jolt Has your question been resolved?

[calm coral]

@trail jolt Has your question been resolved?

[calm coral]

@trail jolt Has your question been resolved?

[proper marlin]

arg(z-1-2i) means the angle from the point 1+2i

what they did is correct

[trail jolt]

what about arg(-1-2i) aka the argument of Z = -1-2i?

are they the same thing?

[proper marlin]

no

arg(z-(1+2i)) means the angle of z from the point 1+2i

because its z-(1+2i)

same as |z-1| is the distance from 1

because its z-1

[trail jolt]

Ohh

So I’d need some other prompts to use the arg(Z-(1+2i)) and there’s nothing I can really do with that argument itself right?

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[calm coral]

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[orchid shoal]

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[lavish haven]

is this right? should i explain more?

[calm coral]

@lavish haven Has your question been resolved?

[lavish haven]

<@&286206848099549185>

[odd acorn]

I don’t see why you would

[lavish haven]

fair enough

i just feel like i always give too little info

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[ionic lake]

How many distinguishable permutations are there of the letters in ANSWER that end in a vowel

[ionic lake]

I got 240 and i feel like it’s wrong

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[hearty mountain]

$\sin2x = \root3/2$

[potent lotus]

j
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hearty mountain]

$\2sin2x = \root3/2$

[potent lotus]

j
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hearty mountain]

$\sin2x = \sqrt3/2$

[potent lotus]

j

[odd moon]

$\sin2x = \frac{\sqrt{3}}{2}$

[hearty mountain]

i need to solve for sin(x) and (cos)xfor the domain 0-360

[potent lotus]

k.tten

[hearty mountain]

but how do i solve for cos

after solving for sin

[odd moon]

cos x = sin (x + pi/2)

[wet finch]

if you know the values of x, plug them into cos

[hearty mountain]

no

this is a no calculator question

[odd moon]

... ok

[hearty mountain]

wait

no i mean

the ratio

for

[winter elbow]

You don’t need a calculator for that

[hearty mountain]

cosx

the RATIO

[odd moon]

what ratio

[hearty mountain]

rn sinx = 30,60,210,240

@odd moon

[odd moon]

so plug those into cos x?

[winter elbow]

Why would you need a calculator for this?

[hearty mountain]

holy shit

[odd moon]

can you explain why that isnt what you're trying to find

we dont even know what you want

[winter elbow]

lmao

[hearty mountain]

okay

let me give you

the full question

a) Prove that you can rewrite 4sinxcosx = sqrt(3) , as sin2x = sqrt(3)/2.

b) given this, find sinx and cosx for 4sinxcosx = sqrt(3) (for 0-360 degrees)

i already solved a

4sinxcosx = 2sin2x

2sin2x = root3/2

sin2x = root3

[daring gale]

it s sin2x = sqrt(3)/2

[hearty mountain]

woops

yeah

sorry

sin2x = sqrt(3)/2

yes

so sin(x) = 30,60,210,240 right

[winter elbow]

no

[odd moon]

you should probably write those as radians

[winter elbow]

If x=240, you’ll get 2x=480

Obviously, it’s not in the interval

[daring gale]

just find a sinx that s equal to sqrt(3)/2
then make sin2x = sin(pi/3)
2x = pi/3 + k2pi or 2 x = pi - pi/3 + k2pi, k belongs to 0 to 360

[hearty mountain]

bro gave me a chatgpt answer

[daring gale]

man I am a uni student I have done this before multiple times....

[hearty mountain]

listen

dude

ok

the answers in degrees for sin(x) is

30,60,210,240

am i correct

?

[winter elbow]

?

Snowy gave you similar explanation as well

[thin gulch]

cuz it's simple enough that even chatgpt can do it

[daring gale]

30 is not a solution btw

just take my answer and change the radians to degrees, then change k to any integer value (I said wrong before) and you ll get the answers in 0 to 360

maybe someone check if its right but the problem is not hard your book probably has it all explained

[hearty mountain]

i literally

just

graphed it

[daring gale]

srry man you re correct
I think you have it right

[hearty mountain]

you literally used chatgpt thats why u cant solve it

[daring gale]

bruh

[hearty mountain]

ur acc

no help

dont even try

by using chatgpt

[winter elbow]

<@&268886789983436800>

[hearty mountain]

?

why u pinging them

anywyas

[winter elbow]

We are not here to help as mandatory

Don’t take it for granted

[hearty mountain]

after solving for sin(x), how do i solve for cos(x)

[hearty mountain]

he shouldnt be using gpt

[winter elbow]

He didn’t

[hearty mountain]

yes he did i put it in and got the same answer

anyway

[winter elbow]

ChatGPT wouldn’t give such a brief solution

[daring gale]

that s why I missed...

[west shard]

Anyways, I don’t think you deserve help.

[hearty mountain]

ok im not trying to argue

can you guys just

tell me how to solve for

cos(x)

[winter elbow]

We are not help to be your servant either

[west shard]

If this is the way that you are going to treat people who are trying to help you, then you do to deserve help.

[daring gale]

I corrected again man if you still think I can t help its ok

[hearty mountain]

yeah okay

but how would i solve for

cos(x)

im confused

[daring gale]

cos(x) = sqrt(3)/2?

[hearty mountain]

remember the original equation was

4sinxcosx = root3

we got sin(x) by converting it to 2sinx = root3/2

but how do i solve for cosx now

[candid crane]

You rang?

[daring gale]

you know sinx = cos(90-x)

[winter elbow]

yes

He kept name-calling to helpers

[hearty mountain]

i didnt, and we moved on

[hearty mountain]

yeah

and

?

[candid crane]

Was something deleted? I don't immediately see any namecalling (or at least none worse than "dude") in the scrollback.

[daring gale]

first of all you missed again its sin2x = sqrt(3)/2
if we are talking about 4sinxcosx..
knowing how to change sin to cos
just do cos(90-2x) = sqrt(3)/2

[hearty mountain]

no im pretty sure you're wrong

i think we have to use the

pythagorean identity

[daring gale]

the thing is ... cos(90-2x) = sqrt(3)/2 is the same
sin2x = sqrt(3)/2
if you wanna solve as "cosx" that s atleast how I interpret your question
can t see other way

[hearty mountain]

so do i do this

for

each

x value

i solved for

<@&286206848099549185>

[daring gale]

cause after that is just find once again a cosx that s equal to sqrt(3)/2
and make it
90 - 2x = 30 + 2k180 or 90-2x = -30 +2k180

[hearty mountain]

example

pls

[daring gale]

cos x = 1/2
cos x = cos 60
x = 60 + 2k180 or x = -60 + 2k180
if x belongs to 0 to 360
k = 0, x= 60
k = 1, x= 300
checking if solutions are right
x = 60
cos 60 = 1/2
x = 300
cos 300 = 1/2

[calm coral]

@hearty mountain Has your question been resolved?

[calm coral]

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[dusk ember]

Hi I need help figuring how to go about this question, So from what I know A union not b I believe is 12458?

but when it comes to intersect C I don't know what would that include? I assumed 4 and 5 but its incorrect

[surreal patio]

isnt 7 also a solution?

[dusk ember]

I don't understand how it could be since its intersect C I think that would be 4 5 and 6

[surreal patio]

b compliment means that 1478 are shaded
a union b compliment means 124578 are shaded
a union b compliment intersection C are 457

[dusk ember]

omg

that was correct

how does that work ? if you could go more in depth so i can understand how

Im struggling to understand sets with venn diagrams

[surreal patio]

i just like to shade

so B compliment means 2356 are not a solution

[dusk ember]

i see

ok wait so one thing i dont get is why is a union b complement intersection c wouldnt include the 8?

since its also outside b

[surreal patio]

A union b compliment adds 2 and 5

[dusk ember]

ohhh

[surreal patio]

it doesnt include 8 tho

[dusk ember]

hmm okay, i think i get it a little better

[surreal patio]

the black part is everything intersection C

and the black is 4 5 7

[dusk ember]

I see

hmm

would breaking it down to set notation be a good way to go about questions like these?

i think i get confused more when shading

[surreal patio]

breaking it down as in first making Bcompliment then A union Bcompliment then A union Bcompliment intersection C??

[dusk ember]

yes

[surreal patio]

ig everyone has thier own preference

yeah thats ok as well

[dusk ember]

because then i think its easier for me to find the intersection between c and a union not b

[surreal patio]

u could visualize that as well , Bcompliment means everything in U except B

so 1478

A union B compliment means everything in A and everything in B compliment

so 14578

A union B compliment intersection C means everything in A and everything in B compliment that is common in C

so what remains is 4 5 7

cuz C is 4576 and Aunion B compliment is 14578
so the common stuff is 457

[dusk ember]

yeah

i think thats easier for me to understand, thank you for your help i appreciate it !

[surreal patio]

yeah u can do it like that or shade , whatever u prefer

!done

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[quiet island]

Determine the volume of the tetrahedron $\Omega$ bounded by the planes $2x+y+z=2$, $y=0$, $y=x$, and $z=0$.

[potent lotus]

Halex

[quiet island]

Could someone please help me check that my answer is correct?

This is the integral I obtained from the problem: $$\int _0^{\frac{2}{3}}:\int _{\frac{y-2}{2}}^y:\left(2:-:2x:-:2y\right)dxdy$$

[potent lotus]

Halex

[calm coral]

@quiet island Has your question been resolved?

[quiet island]

were you meant to say it was correct? @mint scarab haha

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@quiet island Has your question been resolved?

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[distant remnant]

what is the solution for this
2 raised to modulus of x + 3 raised modulus of x + 4 raised to modulus of x = 9

[wanton spear]

Sorry, could you provide this in the original context?
Or add parentheses

It's a bit hard to understand what's being asked here

[distant remnant]

ok wait pls

[amber wedge]

2^|x| + 3^|x| + 4^|x| = 9

?

[distant remnant]

yes

[amber wedge]

i would suggest solving this by inspection

[distant remnant]

what is that

[amber wedge]

fancy way of saying use an inspired guess and check if it works

[distant remnant]

oh hit and trial method

[amber wedge]

yeah

[distant remnant]

btw how many methods of solving are there

?

thanks

@amber wedge thnaks

for help

[amber wedge]

i dont think there are any other methods

[distant remnant]

oh ok

thanks a lot

[amber wedge]

np

[distant remnant]

ok

thnx again

/close

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[slow delta]

Where did I do a mistake? This isn't the right answer

[slow delta]

It seems everything is fine..

[dull wagon]

you forgot to root D

when applying quadratic formula

$x = \frac{-b \pm \red{\sqrt{\black{\Delta}}}}{2a}$

[potent lotus]

ℝαμΩℕωⅤ

[slow delta]

Ah 😭 thank you

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[opaque trellis]

Hey, I got a very simple problem originating from an argument with a friend of mine:

In the game Russian roulette, how big is the chance you don't get hit?

My answer is 5/6, his is 1/6
His logic is that you don't add the odds up because you only do it once. To be clear, it's a full random game in which we only play once. Could any of you tell me who's right and maybe add the calculus behind it? Thanks before hand

[swift laurel]

the probability of landing on any particular barrel is 1/6, but you win if you land on any of the 5 empty barrels

so if barrel 6 is loaded, the probability of winning is the probability of hitting barrel 1 or barrel 2 or barrel 3 or barrel 4 or barrel 5, which is the sum of each of their probabilities (since they are mutually exclusive), which is
1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 5/6

alternatively, if the chance of losing is 1/6, then the probabilities of winning and losing must sum to 1, so the chances of winning must be 1 - 1/6 = 5/6

[opaque trellis]

Thanks a lot, I hope this will help us settle the debate!

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[indigo coyote]

.

how does the (8-x) chane to (x-8)?
how does the inequality sign change? is it because both the sides are multiplied by -1?

[vernal crest]

yes

[spiral dagger]

Yes

[verbal finch]

yes

[vernal portal]

Yes

[spiral dagger]

Lol

[vernal crest]

damn guys

[vernal portal]

Lol

[vernal portal]

I was here first grr

[spiral dagger]

Several people are typing...

[indigo coyote]

so, wouldnt (3x-5) also change its signs

[vernal crest]

my bad broskis 😔

[vernal crest]

no

[vernal portal]

Prolly divide by (8-x) and multiply by (x-8)

Assuming x isn't 8

[vernal crest]

it's intially
(3x-5)(8-x)
when i multiply this with -1 i only distribute it along one of these

not both

like you wouldn't say (-1)(xy) = (-x)(-y)

[vernal portal]

Yes but at least one of the two
(X-8) or (8-x) Is negative

No fromcthe first inquality

[vernal crest]

that's not what akina is asking ithink

[vernal portal]

Divide (8-x)

[vernal crest]

i think he awas asking why (3x-5) doesn't change to (5-3x)

[vernal portal]

Then multiply by (x-8)

Because u aren't multiplying by -1

[vernal crest]

no they are

[indigo coyote]

yes

[vernal crest]

the inequality sign is changing

[vernal portal]

Omg

[indigo coyote]

yea this

[vernal portal]

Inequality sign changes when you multiply by a NEGATIVE

Doesn't have to be-1

[vernal crest]

(-1)(xy) is either (-x)(y) or x(-y), not (-x)(-y)

substitue x and y with 3x-5 and 8-x

[vernal crest]

true

[vernal portal]

Ig

[vernal crest]

i was speaking in the context of this question specifically though

[vernal portal]

🫢

[indigo coyote]

yeah this is the answer i was looking for got confused here

thank u guys

[vernal crest]

np

[indigo coyote]

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[midnight sedge]

can someone help me explain where 59.5 came from

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[orchid summit]

I need a bit of help, I have found expressions for PE, NF and MG as well as 2 expressions for MR but im not to sure where to go from there.

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[tall moon]

not sure how to do this

you can simplify the 10^-3^0 to a 1

and make it so that it becomes $$3\log_{x}[1000}-2\log_{10}{x}+1=0$$

[potent lotus]

Skill_Issue
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[verbal finch]

do u know this formula?

loga base b

is 1/logb base a

[tall moon]

uhh not really

lemme google

[verbal finch]

ok

[tall moon]

$$\log_{b}{a}=\frac{1}{\log_{a}{b}}$$

this?

[potent lotus]

Skill_Issue

[verbal finch]

yes

[tall moon]

nope

but if its true i see how to use it

[verbal finch]

for the first thing

[tall moon]

i got $log_{10}{x}=-1$

[potent lotus]

Skill_Issue

[tall moon]

1/10?

[high topaz]

According to this yes

[tall moon]

alr ty

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[tall moon]

.reopen

[calm coral]

✅

[tall moon]

how about this?

[flat sleet]

since there’s two absolute values, there’s 4 solutions

3x+5+2x-3=25
3x+5-2x+3=25
-3x-5+2x-3=25
-3x-5-2x+3=25

solve x for each equation and just pick the highest and lowest one and sum them

[winter elbow]

No, there are only three

[reef nimbus]

there are 2 according to Wolfram

[winter elbow]

When 2x-3>0, then 3x+5 must >0

[tall moon]

uh i got -27/5, 17, -33, 23/5

17 and -33 dont work

[dull wagon]

make sure you account for the conditions for each case

[winter elbow]

I’m referring to his equations

[dull wagon]

don't blindly solve all combinations of +-

[tall moon]

alright, how about this, this would be trivial if we can make all of them in log_3(x) but im not sure how to do that with the log_3(x/9)

[flat sleet]

oop sorry

[winter elbow]

log_3(x/9) =

log_3 (x)-

log_3 (9)

[tall moon]

is it an integer? i got log_3(x)=7/2

where did i go wrong?

ugh wtf why does my camera always send it in the wierd .heic format

[ember ledge]

it can be disabled in the settings

[tall moon]

wait this is a downward curve

actually like basically everything is wrong :(

ok i got it

if i have like some sort of $log_{a}x \cdot log_{a}y$ can i simplify it some way

or like combine em

[potent lotus]

Skill_Issue

[tall moon]

how bout this

i have no clue what complex notation means

[verbal finch]

complex number is generally written as x+iy

where i is iota

x and y are real numbers

familiar with this? @tall moon

[winter elbow]

Just set Z = a+bi

And you’ll be all good

[verbal finch]

yep

[tall moon]

yes

alr so apparently it would become a+bi+isqrt(a^2-b^2)=12+9i?

so a=12 right?

[verbal finch]

if z = a+ib then zbar will be a-ib

[tall moon]

oh yea

[verbal finch]

and also |z| is root(x^2+y^2)

[tall moon]

sqrt right?

[verbal finch]

yes

square root only

[tall moon]

its i|z|

[verbal finch]

yes

so i(root(x^2+y^2))

[tall moon]

a=12 right?

[verbal finch]

yes

[tall moon]

so if we factor i out
b+sqrt(144-b^2)=9

[verbal finch]

um it is +b^2

not -b^2

[tall moon]

oh ok

i thought it would become sqrt(a^2+(bi)^2)

alr then

[verbal finch]

nop

|z| is basically the distance of point (x,y) from origin if z=x+iy

[tall moon]

oo

[verbal finch]

yep

[tall moon]

so is b 3,5?

that doesent feel right

[verbal finch]

idk i didnt solve

just a sec

[tall moon]

$$144+b^2=(b+9)^2$$
$$144+b^2=b^2+18b+81$$
$$63=18b$$

[potent lotus]

Skill_Issue

[verbal finch]

9-b

u send b to right

[tall moon]

oh right

so 12-3.5i?

[verbal finch]

prolly

ull get b

wait

[tall moon]

still no?

[verbal finch]

see

ull get

81 + b^2 - 18b = 144 + b^2

mhm

yes thts right

-3.5i

[tall moon]

how bout this

dont know much about arcsin

[verbal finch]

send pi to tht side

u writing some exam or smthg?

[tall moon]

alright to be honest, im playing a roblox game

[verbal finch]

bruh then y doing math and playing roblox

[tall moon]

the game is about math

(x-2)(x-1)

[verbal finch]

the wht

[tall moon]

the game

[verbal finch]

roblox is a math game?

from when

[verbal finch]

oh mah gawd

fine

so ok whts the q

[tall moon]

this

[verbal finch]

tht is pretty simple

convert tht arccos into arctan

[tall moon]

how?

[verbal finch]

consider arccos(5/13) = x

then cosx = 5/13

draw right angle triangle

from there get tanx

[tall moon]

is it 5/12?

but how about the arcsin one

[verbal finch]

yes

[verbal finch]

i think they are missing something in arcsin one

[tall moon]

wdym

[verbal finch]

the question

u just cant add a number and an angle

[tall moon]

i difress

[verbal finch]

oh nvm sorry i did not see tht its whole tan

ok same thing

convert arcsin into arctan

[tall moon]

is it the same process?

[tall moon]

also isnt this supposed to be 12/5

[verbal finch]

yes

[verbal finch]

oh yea

[verbal finch]

yes

do it

[tall moon]

i got a 345 triangle

can you explain to me what we have?

im not quite getting it at how this works yet unfortunately

[verbal finch]

yep

get tanx from there

[tall moon]

3/4?

[verbal finch]

yes

now what did u get?

[tall moon]

wdym

[verbal finch]

tan(arctan(12/5)+arctan(3/4))

[tall moon]

no clue

also isnt the outer supposed to be tan?

[verbal finch]

u just converted them

[verbal finch]

oops mb

[tall moon]

i have no idea 😭

i dont really know trig that well

[verbal finch]

then how did u convert them 😭

[verbal finch]

hmm

i think u need to learn inverse trigonometry

but u converted them and they are indeed correct

[tall moon]

what do i do now

[verbal finch]

arctana + arctanb formula

it is arctan(a+b/1-ab)

[tall moon]

ok

[verbal finch]

mhm

then its done

cuz u have tan(arctan(a+b/1-ab))

remove tan and arctan

ull be left with

a+b
---
1-ab

[tall moon]

-63/16?

oh yeah its correct

[verbal finch]

ig

hm

[tall moon]

wtf the next question is about matrices 😭 ima close this for now, tysm

[verbal finch]

np

[tall moon]

.close

[calm coral]

Channel closed

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[calm coral]

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[honest swallow]

Idk what to do after this tho

<@&286206848099549185>

[remote mural]

2/35?

[honest swallow]

how tho

[remote mural]

break log(a)/log(2-root 3), now ffactorise 2-root 3 in such way you get 2+root 3

similarly for logb/log(whaterver term)

[honest swallow]

thats the main problem how do i get 2 + root 3

[remote mural]

okay wait

[honest swallow]

alr

[remote mural]

log(a)/log(2- root 3)=log(a)/log[(2-root 3)*2+root 3/(2+root 3)]=log(a)/log[(4-3)/(2+root 3)]=log(a)/log(1/ 2+root 3)

@honest swallow

similarly try for logb

[honest swallow]

ohhhh

[remote mural]

yes do it first and then after tht take 2+root 3 =t or any variable and then take log on both side

2+root 3 will cNCEL YOU WILL LEFT WITH LOG(AB)=LOG(1/35)

and since a and b is prime

[honest swallow]

7 and 5

[remote mural]

yes

[honest swallow]

got it

tysm

[remote mural]

no prooblem

[honest swallow]

.close

[calm coral]

Channel closed

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[calm coral]

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[remote mural]

i just need the asnwer

[austere hazel]

is this roblox wtf

[remote mural]

im tryna get the badge

help me out

[austere hazel]

do they mean find x

?

[remote mural]

maybe idk

i dont know anything about matrix

[austere hazel]

ok

[austere hazel]

what happens if u get it wrong

[remote mural]

nothing

[austere hazel]

then j trial and error it lmao

[remote mural]

i did

[austere hazel]

ok fine gimme 2 mins

[remote mural]

aight

[austere hazel]

does the answerr

have to be

an integer

[remote mural]

no

[austere hazel]

try

51.33

[remote mural]

can i get the fraction

[austere hazel]

154/3

[remote mural]

ty

[austere hazel]

did it work

[remote mural]

ye

[austere hazel]

less go

[remote mural]

wait i got another one

just 2 stages more

ppls

[austere hazel]

ggs

ok

3 mins

-14

try

[remote mural]

u the goat bro

[austere hazel]

lesss gooo

[remote mural]

bestiesss

i beat it

thanks

.close

[calm coral]

Channel closed

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[calm coral]

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[tough knot]

How do i find the sum of first nth term of the series of the term of sequence. 8+80+782+7784+77786

[agile yacht]

Have you managed to figure out the pattern yet?

[tough knot]

yes

its 7/9(10^n - 1) + (2n-1)

[agile yacht]

That's not how I was think about it, but if there numbers line up thats great!

Let me check real quick just be sure.

Yeah thats it! So are you struggling to sum the first 10 numbers of the sequence?

Or first n terms excuse me

[tough knot]

I am struggling to find the formula of summation of nth terms

[tough knot]

this formula is for general nth term not for summation

[tough knot]

mb i gave comma in the question

it was +

[agile yacht]

Yes okay makes sense, so lets break it down into smaller chunks, we can seperately determine a formula for summing 2n - 1 and 7/9(10^n - 1), and then sum those formula's right?

[agile yacht]

np

[tough knot]

yes

[agile yacht]

So summing the first n terms of 2m-1, how would we calculate that?

I am sure you can figure that out

[tough knot]

first term would be 1 and 2nd term would be 3

So its an arithmetic series
We know,
Summation of arithmetic series is n/2{2a+(n-1)d}
So it would be n/2{2+3n-3}

[agile yacht]

Let try it, the first 5 odd number sum to 1 + 3 + 5 + 7 + 9 = 25, and 2.5(2 + 15 - 3) = 2.5 * 14 = 35. So I think something is going wrong.

I am not yet sure what though.

[tough knot]

umm so what is wrong actually

oh d is 2

Im sorry

[agile yacht]

No worries!

[tough knot]

common difference is 2

[agile yacht]

I dont know the formula you used exactly, what would the new answer be? I know it looks nice.

[tough knot]

25 answer

[agile yacht]

I mean what the new formula would be

[tough knot]

n/2{2+2n-2}

[agile yacht]

Wonderful, that simplifies to n^2, which is indeed the sum of the first n odd integers.

Now we need to determine the sum of 7/9(10^n - 1), and lets again split thing up a little, and seperately the determine a formula for the first n terms of 10^n, and -1. Then we multiply with the constant later.

Do you get what I mean with this?

[tough knot]

yes im thinking abt the fomrula

1 min please

[agile yacht]

No hurry!

[tough knot]

10(10^n-1)/9

and for -1 it would be -n

so 10(10^n-1)/9 - n

[agile yacht]

That looks great. Can you find the final answer now (i.e. put it all together)?

[tough knot]

will it be 7/9{10(10^n-1)/9}-1 + n^2

[agile yacht]

I think the -1 is supposed to be -7/9 * n, right?

[tough knot]

mmm how?

[agile yacht]

because we were summing the -1, and and it has to be multiplied with the constant 7/9

[tough knot]

but it can be written seperated as -1 + -1 + -1 = -3 which is -n if three terms

oh that would be -n

-n*

but why would we multiply that though

[agile yacht]

Remember we had 7/9(10^n - 1)

So the 7/9 was multiplied with 10^n and -1!

[tough knot]

so what will be the final answer?

[agile yacht]

$\frac{7}{9} \left( \frac{10(10^n - 1)}{9} - n \right) + n^2$

[potent lotus]

Casper

[tough knot]

thats the final answer?

[agile yacht]

Yes.

See how we made it much easier by breaking the problem up into smaller pieces

[tough knot]

is 70/81(10^n-1)-7n/9 + n^2

can be answer?

[agile yacht]

You just have to keep books well, but thats much easier to keep track off when you're doing it by yourself

[tough knot]

You have good teaching skills.Appreciate for the help

[agile yacht]

Yes you can make that simplication if you wish.

I am glad I could help!

Good luck with the rest of you exercises.

[tough knot]

thanks have a nice day

?end

[agile yacht]

Its .close

You too!

[tough knot]

.close

[calm coral]

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[calm coral]

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[unique willow]

is there a quick way to find what number is the same multiple given two numbers?
say A = 13, B = 22 (+1)

How do i find common multiple?

[calm coral]

Please don't occupy multiple help channels.

[prime yacht]

What does 22+1 mean

[unique willow]

22 times table (+1)

its because i want to find 13 * ? = 1 mod 22

[prime yacht]

13x = 22?

[unique willow]

no because it wouldn't give u 1 remainder from mod

[prime yacht]

Or are you asking about the lowest common multiple

[unique willow]

yes

[prime yacht]

Prime factors of 13 are 13
Prime factors of 22 are 2,11
They have no common prime factors so their lcm is 13*22

[unique willow]

not multiple of 22

[prime yacht]

Say A = 14
B = 8

[unique willow]

i want multiple of 22 (and +1)

so 23, 45, 67, ...

[prime yacht]

Take the LCM of 13 and 22 and add 21?

,w ((13*22) +1)mod22

[potent lotus]

Results provided by WolframAlpha

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

Requested by cats1cle

[prime yacht]

Sorry add 1

[unique willow]

hows it 1

thats say 13 and 23 is the same

[prime yacht]

Could you rephrase

[unique willow]

i think ik how to do it, u have to use extend euclid algorithm to find multiplicative inverse of 13 mod 22, which should get ?

[prime yacht]

Why do you need the extended euclid, basic should be enough?

[calm coral]

@unique willow Has your question been resolved?

[calm coral]

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[royal raft]

Having trouble with this one

[royal raft]

I tried (d-c)(b-a) > 0 but that didn't work out

[glad grotto]

first do ca < cb and bc < bd and then the result just follows from transitivity

[royal raft]

but ca isnt always less than cb

when c is 0

oh wait

yeah nvm

when c is 0 we get 0 < 0

[glad grotto]

in the case c = 0, 0 < bd is trivially true because the relation is closed under multiplication anyway

same if a = 0

[royal raft]

woudln't it only be triviallty true if the hypothesis was false

[glad grotto]

by trivially true i mean it directly follows from the ordering axioms

because 0 < d and 0 < b => 0 < bd (smaller relation is closed under multiplication)

[royal raft]

oh yeah but how would you prove that starting with the hypothesis

how do we know ac < bd

[glad grotto]

so a < b => 0 < b - a

then in the case c isn't 0

0 < c and 0 < b - a => 0 < c(b - a)

=> ac < bc

do the same for bc < dc

and then the result just follows from transitivity ( x < y and y < z implies x < z)

transitivity is something you can also prove fairly easily from the ordering axioms

[calm coral]

@royal raft Has your question been resolved?

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[wraith leaf]

how do you simplify this

[wraith leaf]

i messed up in my simplification

.close

[calm coral]

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[heavy cargo]

hello

[heavy cargo]

hello lads

I would appreciate some assistance with these questions

<@&286206848099549185>

[gleaming vapor]

☑️

[gleaming vapor]

☑️

[heavy cargo]

?

[gleaming vapor]

i replied to the image

im saying those asnwers r correct

[heavy cargo]

they arent mine tho

[gleaming vapor]

oh

[heavy cargo]

I wanted to ask how do you get to those answers

[gleaming vapor]

ohhh

for q8

you solve for the radius of the semicircle

so multiply the area by 2 since semicircle is half of area circle

you get 8((a^2)/2)pi

or just 4a^2pi

when since we know circle area is pi*r^2

we know radius = sqrt(4a^2)

which is 2a

then area of the square is side^2

we know that 4 radii is the same as the length of that square

so we do 2a*4

which is 8a

then square that to get 64a^2

[heavy cargo]

ohhhh

tysm

do yk the solutions to the other questions

[gleaming vapor]

for q13

a whole circle has 360 degrees

360/15 = 24

so you know there are 24 of that same arc within the circle

so 35mm * 24 = 840 mm

which is 84 cm

[gleaming vapor]

for q32 i did angle theorems

ill draw it and send

we know the entire angle on that side is 121 due to alternate exterior angles

then we know interior is 180-72 since they are complementary angles

[heavy cargo]

ahh i see

[heavy cargo]

so do you reccomend I go over angle theorems to solve these type of questions?

[gleaming vapor]

probably

[gleaming vapor]

i didnt necessarily use angle theorems, its just easier to explain it that way

[heavy cargo]

i see

well tysm for your help!

.close

[calm coral]

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[candid island]

help me

[remote mural]

wow

[potent igloo]

!da2a

[calm coral]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

@candid island Has your question been resolved?

[calm coral]

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[dusty nacelle]

Is the recursive function $y_{n+1}=y_n+2$ homogenous? I know it's not when a factor of n is added, but this can be conveyed in the form of $y_{n+1}=f(y_n)$, $f(x)$ here being $x+2$. The results I found online had me conflicted

[potent lotus]

Duirky

[tacit lark]

There is a "2" involved so it's non homogeneous but it is still linear

[calm coral]

@dusty nacelle Has your question been resolved?

[calm coral]

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[scarlet saddle]

help?

[vital raven]

we want to find t given g = 9.81 and L = 67m

they have given u the formula to calculate t

so substitute the numbers into the formula and calculate

[scarlet saddle]

omg im so dumb

my bad, thank you tho

.close

[calm coral]

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[vital raven]

you aren't dumb

[scarlet saddle]

lol

[vital raven]

we all have those moments

dont worry

[scarlet saddle]

thanks

[calm coral]

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[proud kite]

on complex equations if i find x=1 y=2 does it mean that z=1+2i?

[calm coral]

Please don't occupy multiple help channels.

[proud kite]

what

[proud kite]

im not sure if i understand correctly this or am mistaken

[calm coral]

@proud kite Has your question been resolved?

[mint scarab]

You can ping helpers @proud kite if no one answers

[proud kite]

<@&286206848099549185>

[fierce tartan]

Send the complete question pls

[proud kite]

Its just a general example because im not sure if I full understand the rules but if u want an exercise here

[fierce tartan]

Uh

U can go for geometrical interpretation or yes if the general form of the complex eqn is x+iy then u can replace them in z

[proud kite]

Ok but

Then what if I find smth like

x=i

Then what does it mean

[fierce tartan]

Not possible

[proud kite]

But u get this by doing z=x+iy

[fierce tartan]

X has to be a real number

[proud kite]

Say

[fierce tartan]

And y also

[proud kite]

z=2

x+iy=2

[fierce tartan]

Then x is 2-iy

Yes but as i=√-1 , x can never be imaginary

Suppose u plot a graph with x as the real axis and y as the imaginary axis
Then only y can have imaginary values(values in terms of i)
X cannot

[proud kite]

Ok but if z=2 what is x and y?

[fierce tartan]

X=2 and y=0

Hope I could help 🙏

[proud kite]

Do u consider |z|^2 as x²+y²?

.close

[calm coral]

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[fierce tartan]

x²-y²

[proud kite]

|z|=sqrt(x²+y²)

[calm coral]

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[deft berry]

how do I convert rectangular coordinates to polar?

[deft berry]

i forogt

[wet glen]

u need to use pythag thereom to find radius

then find the angle that the radius creates

[deft berry]

oh shit yeah

ur RIGHT

thanks mate

[wet glen]

ofc ofc

[deft berry]

.close

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[royal relic]

I'm trying to find X in

X = 4097(mod 7936)
4228(mod 8192)
2053(mod 3968)
3973(mod 7688)
1925(mod 3840)

Obviously these numbers are not relatively prime so I cannot easily use the Chinese Remainder Theorem. So what can I do?

[calm coral]

@royal relic Has your question been resolved?

[amber bolt]

even the first two can't happen

[royal relic]

How can I determine if its possible or not?

[amber bolt]

idk

[royal relic]

lol

[amber bolt]

(4097 + 7936k) mod 8192

it quickly loops and never hits 4228

[royal relic]

I see

[amber bolt]

from math import lcm
a = 0
step = 1
for r, m in ((4097,7936), (4228, 8192), (2053, 3968), (3973, 7688), (1925, 3840)):

   while a % m != r:  a += step
   step = lcm(step, m)
   print(a, step)

you can sieve it like this, it should take like 1 second

if it's possible

so after 5 seconds i would assume it fails

well you can detect the looping, that's always the reason it fails i suppose

[royal relic]

Yeah that looped for over 3 minutes. No dice

[amber bolt]

from math import lcm

(a, step), *cond = ((4097,7936), (4228, 8192), (2053, 3968), (3973, 7688), (1925, 3840))

for r, m in cond:
   k = set()   
   while (s := a % m) != r and s not in k:  
     a += step
     k.add(s)
   if s in k: 
     print("can't")
     exit()
   step = lcm(step, m)

print("success\n", a)

loop detection

[calm coral]

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[calm coral]

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[prime mauve]

Any tips to find this limit without l'hopital?

[prime mauve]

Right now I know the answer is -1, but I have no idea how to get to the answer.

[quaint sphinx]

compute arctanx series centered around infinity

notably, arctanx = pi/2 - arctan(1/x) for x > 0

[prime mauve]

Thank you very much for your help, now I was able to solve it

.close

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[quaint sphinx]

welcome 🙂

[calm coral]

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[dense falcon]

Can anyone explain the reaction?

[tall moon]

shouldnt this be posted on like a chemistry server instead of math server?

[dense falcon]

It doesn't have a help channel

I have seen a few physics and chem so Ig it's alright

.close

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[wheat pine]

How can I solve |x^2+x|=|5x-4|

[cerulean cradle]

Hey

Dont you think its useless to apply modulus both sides ?

@wheat pine ?

[remote mural]

x when less than 1 and negative

[thorn trellis]

Hey

[cerulean cradle]

Yeah ik so to avoid complicatons

[thorn trellis]

Just square both sides

[wheat pine]

No the excercise has them

[thorn trellis]

And solve it like an equation

[cerulean cradle]

No need to square

Just consider two cases

[tall moon]

you want them to solve a quartic equation?

[thorn trellis]

Yup

?

[cerulean cradle]

Where 5x - 4 > 0 and 5x -4 <0

[wheat pine]

yeah

[cerulean cradle]

and solve

Okk ?

[wheat pine]

ok

.close

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[cerulean cradle]

And if any solution

Doesnt satisfy original equation dont use it

[calm coral]

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[remote mural]

want to check my answer

[remote mural]

is $f(x) \in \R \ [4,8)$ the same as $(-\infty,4) \cup [8,\infty )$

[potent lotus]

Emploice Muswashans

is $f(x) \in \R \\ [4,8)$ the same as $(-\infty,4) \cup [8,\infty )$
```Compilation error:```! Undefined control sequence.
l.49 is $f(x) \in \R
                     \\ [4,8)$ the same as $(-\infty,4) \cup [8,\infty )$
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.

)
Runaway argument?```

[remote mural]

is $f(x) \in \R \ [4,8)$ the same as $(-\infty ,4) \cup [8,\infty )$

[potent lotus]

Emploice Muswashans
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[remote mural]

$$ f(x) \in \mathbb{R} \setminus [4, 8) $$

[potent lotus]

Emploice Muswashans

[remote mural]

$$ f(x) \in (-\infty, 4) \cup [8, \infty) $$

[potent lotus]

Emploice Muswashans

[remote mural]

are these two the same?

just ping me if you are replying to this question, I would be on another tab :)

Thanks

[quaint sphinx]

looks that way

[remote mural]

ahh awesome

thanks so much

.close

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[calm coral]

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[split bronze]

i need help findind a tangent line to a curve f(x)=(ln(x))/x that goes through (0;0)

[remote mural]

you need to differentiate this

[split bronze]

nvm i got it

i mistook something else as a slope

.close

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[sacred comet]

Hello, Is the eigenspace belonging to the eigenvalue 1 always the image of the matrix in question, or is this true only for projection matrices?

[sacred comet]

I know it's true for projection matrices but I'm not sure if it can be generalized

[pale matrix]

Hey fatties

SORRY

[calm coral]

@sacred comet Has your question been resolved?

[calm coral]

@sacred comet Has your question been resolved?

[calm coral]

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[vernal crest]

which part

a and b?

[feral bay]

b

waitwait

ill show you my working

this way, i get

Sin(x) = 2 and Sin(x) = 1/2

but sin = 2 is impossible

then do i just exclude that?

Or did I do something wrong?

[vernal crest]

yeah just exclude that value then

[vernal crest]

nope

[feral bay]

ahh okayokay

thanks thanks

.close

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[modern hemlock]

answer is 5 or 6

[dense fog]

Can you show what you did? Not clear to me

[vernal crest]

you're right it is an infinite geometric sum