help-39

[sharp smelt]

what?

the identity

only for even ns

[rancid rune]

do u know the expansion for 1/1-x

[sharp smelt]

the approximation, yes

[pearl pond]

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[sharp smelt]

I suppose I could use the taylor expansion about 0

[sharp smelt]

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[pearl pond]

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[rancid rune]

this thing

[sharp smelt]

no

[rancid rune]

u havent seen this?

[sharp smelt]

I don't think so

[rancid rune]

ah ok

[sharp smelt]

unless I'm tripping

oh

that

yes

I know that

[rancid rune]

lol

it's all related

[sharp smelt]

I see

thanks a lot!

[glacial sequoia]

it's just a geometric progression btw

|u| < 1

[pearl pond]

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[worthy grail]

did i do this right?

it’s the last bit i’m unsure about

[worthy grail]

but usually you add the two previous terms to get the next so subtracting would be recursive…?

[autumn fossil]

nope, it should be adding

[worthy grail]

oh dang

i was totally wrong

[autumn fossil]

Fn = F(n-2) - F(n-1) means that Nth term is found by subtracting N-1th term from N-2th term

[worthy grail]

oh

so it’s b

but

f_0

[autumn fossil]

ye, it looks like b

[worthy grail]

is 0

f1 is 1

[autumn fossil]

sequences sometimes start at 0, i dont think its a big problem

[worthy grail]

so in the fibonacci sequence we’d have 0,1

[autumn fossil]

yes, thats how it starts

[worthy grail]

fn is fn-2 plus fn-1

that bit i don’t get

[autumn fossil]

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89

[autumn fossil]

this means that to calculate nth term, you add n-2th term and n-1th term

so e.g. to calculate 3rd term, you add 2nd term and 1st term

meaning you add the 2 previous terms

[worthy grail]

nth term is just whatever term we are trying to find right

[autumn fossil]

yep

it can be replaced with any number

[worthy grail]

ohhhhh

this makes sense

alright yeah it’s b

[autumn fossil]

e.g. if we take n = 41, then we get that the 41st term = 40th term + 39th term

excuse the grammar there

[worthy grail]

nooo problem

thanks a bunch

.close

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[pearl pond]

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[sterile tusk]

have you learned derivatives or no?

alright so do you remember what the turning point of a quadratic is also known as?

alright it’s basically just the vertex of the parabola

do you remember how to find that?

yep

that’d be the x coordinate

so we would find that first and then plug it back into the quadratic to get the y as well

yep exactly

yw!

hmm not really in this case

it just gives you -b/2a as well

it’ll make more sense where that comes from with derivatives

well yeah that too

[dreamy ibex]

the good thing about derivatives is they'll help you find the turning point for pretty much any function

in the case of parabola, you knew the theory that vertex is the turning point so yea x coord is -b/2a

but if i were to give you some random y=x^4-3x^2+14, then using derivatives would give you the answer without having to visualize the graph, more or less

[pearl pond]

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[green berry]

hi

[green berry]

how can i remove the yellow part

pls 😦

[tepid mauve]

Only define it above y=0

Put in the domain y<=0

[green berry]

where

im not good in this

@tepid mauve

[tepid mauve]

Just put in ${0 \le y }$

[jolly parrot]

Root

[tepid mauve]

With the brackets

[green berry]

$\left{0\le x\le2:\ \ x^{2}-2\right}\le y\le\left{0\le x\le2:x\right}\left{0\le y\right}$

[jolly parrot]

alee

[green berry]

where 😦

it doesnt work where i write it now

@tepid mauve

[brave schooner]

im not sure if its possible in desmos

[pearl pond]

@green berry Has your question been resolved?

[pearl pond]

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[prime viper]

hello

[prime viper]

How do I proceed from here

I tried 7 but it’s too low for 9

[old marsh]

Why don’t u draw a triangle diagram

@prime viper

[prime viper]

triangle diagram?

[old marsh]

Yea in quadrant 4

Draw a right triangle

With theta being the angle to the terminal side

(Below the x-axis)

[prime viper]

Like this?

Brah I did it wrong

Like this?

[old marsh]

Yes, but theta is the other angle

Not the bottom one

The one at the origin

(That’s how I like to do these)

[old marsh]

Good

So sin(theta) = -3/7

[prime viper]

yes

[old marsh]

What is sin theta a ratio of?

[prime viper]

csc

[old marsh]

No, I mean

[prime viper]

1/csc?

[old marsh]

In terms of hypotenuse, adjacent, and opposite

[prime viper]

or im confsing them

oh i dont know about that one chief

sin is y

[old marsh]

Sin theta is opposite/hypotenuse

[prime viper]

so im thinking maybe opposite?

oh

[old marsh]

I’m surprised ur doing these problems without covering this tbh

But anyways

sin(t) = -3/7

and

sin(t) = opp/hyp

U can fill in the side lengths of the triangle accordingly

[prime viper]

?

im not sure if im doing this right

[old marsh]

Close but not quite

Do u know what hypotenuse is

[prime viper]

its B?

like A^2 + B^2 = C^2?

[old marsh]

No, its C

So here, we have

sin(t) = -3/7

and that’s opposite/hypotenuse

u did the opposite part right

Now it’s just the hypotenuse

[prime viper]

yes

7 is C then?

wait yeah you just said that

[old marsh]

Yes

Show me what u have now

[prime viper]

okay

[old marsh]

Good

So now

Label theta

And the question is asking for

Cos(theta)

Cos theta is adjacent/hypotenuse

See if u can figure it out

[prime viper]

i labled theta near 7

on the top

[old marsh]

Lmk what u get for cos(theta)

[prime viper]

what if you dont have your hypotenus?

wait

we do have it, its 7

but what about adjacent?

we dont have that

opp is -3

[old marsh]

We can figure that out using this

[prime viper]

3364

thats what i got

[prime viper]

is this wrong

i squared it too

58

wrong too

oh wait

58/7?

[old marsh]

Show me how u got that

[prime viper]

omg

i almost doxxed myself

[old marsh]

Why are u solving for C

We have C

C is 7 remember

Also u know SOHCAHTOA

[prime viper]

nooo

i just googled it and wrote it down

like barely

[old marsh]

Ah ok xd

[prime viper]

sorry

[old marsh]

It’s fine

So we know C, and we know a side length

We are just trying to find the length of the last side

[prime viper]

wait im so confused

Okay we are trying to solve for COS ADJ/HYPO

right?

great we are given this function which is 7/-3

Great.

So looking at COS = ADJ/HYPO

How do we know 7/-3 goes where???

They don’t tell us 7 is hypo or opp or adj

Neither for -3

[old marsh]

,rotate

[old marsh]

sin = opp/hyp

we know sin is -3/7

therefore

opp is -3, hyp is 7

do you see that?

[prime viper]

yes thank you for clarifying

im so silly

[old marsh]

ok now

cos = adj/hyp

find adj, and u will get ur answer

[prime viper]

oh c^2-B^2=A^2?

[old marsh]

yes

[prime viper]

wait

sin = opp/hypo which mean -3/7

-3 is our opp and 7 is our hypo

but we need to find cos

[old marsh]

yes

[prime viper]

hello

sorry there was a spider

black widow on my keyboard

[old marsh]

[prime viper]

haha

shes okay, i released her outside

okay so we know that sin = opp/hypo and it is 7/-3

yes?

okay and they want cos =adj/hypo

we already have hypo

[old marsh]

-3/7

[old marsh]

yes

[prime viper]

im blind

[old marsh]

yes

[prime viper]

ok we want to find adjacent

adjacent is B

okay great, we use formula which can be uhhhhhhhhhhhhhh B=C^2-A^2

yes?

[old marsh]

B^2 = c^2 - a^2

[prime viper]

yes

[old marsh]

find b

that will be ur adj

[prime viper]

1600

[old marsh]

show me how u got that

ur correct until the last line

so we have

$7^2 - (-3)^2 = B^2 \newline 49 - 9 = B^2$

[prime viper]

do we always have to put parenthesis on a negative number

[old marsh]

yes

[prime viper]

ok

so we have 49-9=B^2

[old marsh]

so solve for B

[prime viper]

but what now?

40

i got 40

but its 40^2

[old marsh]

whats the value of B that u got

[prime viper]

1600

do i square it

i got 40 when its squared

[old marsh]

if B = 1600

then

$49 - 9 = 1600^2$

[old marsh]

is that what ur saying?

[prime viper]

no

i got 40

[old marsh]

so are u saying

$49-9 = 40^2$

[old marsh]

?

[prime viper]

yes

[old marsh]

does that not raise any alarm bells for u

what do u get when u simplify the left side and the right side

[prime viper]

well it looks weird because i already subtracted 49-9 equals 40

but why do i need to simplify? there is no x on 49 or 9

i just need to subtract

[old marsh]

yes, subtract, same thing in this context

what do u get upon subtracting

[prime viper]

40

[old marsh]

so

$40 = 40^2$

[old marsh]

is that what ur saying?

[prime viper]

yes

because on the formula it says B^2

so after i subtract 49-9 i get 40 and then i put on 40^2

[old marsh]

look at this

what is 40^2 equal to

[prime viper]

oh wait

1600

[old marsh]

is 40 = 1600 ?

[prime viper]

no

40^2=1600

[old marsh]

ok so that means u went wrong somewhere

[prime viper]

yeah

[old marsh]

what do u get on the left side of equation 2

[prime viper]

this one?

40

[old marsh]

yes

$40 = B^2$

[old marsh]

how would u solve for B

[prime viper]

i put 40 on B^2

which is 40b???

what

i dont know

[old marsh]

this is very problematic, u really need to go review ur algebra fundamentals

[prime viper]

no

i didnt realize that after getting -9, youre supposed to change that to a positive

49+9 equals 58

[old marsh]

what?

how would u get -9

[prime viper]

3x3=9

but since there is a negative, i let it slip over me

and thought it was -9

its not

[prime viper]

question

i did a different problem

[old marsh]

❓

[prime viper]

I don’t know what to do here

[old marsh]

u need to give more context on the problem

also how did u get 36 - 1 = 37

...

[prime viper]

i see

okay here is the problem

[old marsh]

u were on the right path until u did 36 - 1 = 37

i have no idea why u did that

[pearl pond]

@prime viper Has your question been resolved?

[prime viper]

whys that?

Thank you for your help Stephen, I appreciate it

.close

[pearl pond]

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[obsidian grail]

How’d calculate the area if you know that each has 2cm?

[versed mica]

trig

draw an altitude

what will each angle be

if all the sides are the same

[obsidian grail]

Huh? @versed mica

Isn’t it a^2sqrt2

[jagged burrow]

A = 1/2 ab sin(C)

[obsidian grail]

Huh. That’s not what I’ve taught in school.

[versed mica]

each angle is 60 degrees

if you drew an altitude you’d have a right triangle

with hypotenuse of length 2

and angle of 60

you could find out the height from sine ratio

[pearl pond]

@obsidian grail Has your question been resolved?

[obsidian grail]

I’m not on that level.

[versed mica]

🤔

[eternal tulip]

do you know pythagorean theorem?

[obsidian grail]

No

[eternal tulip]

what do you know about triangles

are you familar with heron's formula?

[pearl pond]

@obsidian grail Has your question been resolved?

[pearl pond]

@obsidian grail Has your question been resolved?

[pearl pond]

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[midnight coral]

where did i go wrong

i’m sure it’s very little

[eternal tulip]

,rccw

[midnight coral]

sorry i wrote semi annually wrong

[balmy helm]

wait why did you write 0.08333

[midnight coral]

cause that’s 8.3 percent

[balmy helm]

is it not just 0.083?

[midnight coral]

yeah

would it not be the same

[balmy helm]

0.083 is 0.08300000000......

[midnight coral]

oh.

[balmy helm]

that could be where u went wrong, i got my answer as 8.2

[midnight coral]

this is what i got on the calc

[balmy helm]

gimme a sec

[midnight coral]

okie

[ashen temple]

It's EAR = (1 + APR/n)^n - 1

[ashen temple]

.083 is supposed to be on the left side of the equal sign

[midnight coral]

why don’t u use this form

cause i did and i got 8.47

[midnight coral]

oh ywha

[ashen temple]

They did

[midnight coral]

oh

was the way i calculated it wrong

[balmy helm]

ah that was just how i remembered it

[midnight coral]

if it’s 8.2

[balmy helm]

wait let me recheck

actually it's 8.1%

[ashen temple]

,w 0.083 = (1 + (x/2))^(2) - 1

[balmy helm]

i rounded off in my workings too many times

[ashen temple]

Rip

🤦‍♂️

[midnight coral]

yeo bro

this mad complicatwd

[jolly parrot]

Results provided by WolframAlpha

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

Requested by akumaenjeru

[ashen temple]

Oh there we go

[midnight coral]

where did i write

how did i write it wrong

[ashen temple]

You plugged 0.083 into the spot for APR

Instead of EAR

[ashen temple]

Notice where x is

That's what you're solving for

[midnight coral]

oh

awnser is 8.1?

[ashen temple]

Depends on how your teacher wants it

Did they tell you to round to 1 decimal place?

[midnight coral]

no

it’s fine i got it tho

tysm

.closw

.close

[pearl pond]

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[sinful flare]

help me in my 8th grade geometry assesment 🙏

[covert pivot]

oo mcap

brings back memories

[sweet junco]

it can help writing in paper the things stated there

or visualizing

[old marsh]

Ur name is on the top corner, just to lyk

[sinful flare]

i know

dont do anything

its not the 1st one

[sweet junco]

now that makes me wanna do something

[sinful flare]

gulp

not the 2nd

[sweet junco]

PR = RS
PS = PR+PS = 2PR = 2PS
ST=3QR
PQ=5 units
PT =25
PT=PQ+QR*+RS+ST

pardon me

[sinful flare]

your pardoned

[sweet junco]

we know that PQ+QR=RS

what the fork these are so hard

[sinful flare]

i know

i think the last one is right

cause i measured it with my hands

[sweet junco]

substitute them in the equations I gave

and check if it works

[sinful flare]

nvm

last one is wrong

because qr isnt 3

and 3x3 is 9

so its not last one

boom

[pearl pond]

@sinful flare Has your question been resolved?

[pearl pond]

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[hardy fable]

Can someone check that this is correct? I am trying to prove the series is convergent. To do this I am using the ratio test, but don't think that my dividing by 6^(k+1) is entirely correct

[plush oriole]

Split 6^(k+1) into 6^k*6

[hardy fable]

how does that help me

ok so I have done that for each of the components

and divided by 6^k, top and bottom

or is there another convergence test I can do?

[pearl pond]

@hardy fable Has your question been resolved?

[pearl pond]

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[pure creek]

1.there are 900 000 thousand 6 digit numbers
180 000 of them are dividable by 5 (900 000/5) could someone tell me what this method is called?
and what are the number of digits dividable by 5 greater than 500 000? I tried doing this with 400,000/5 but I'm not confident
2.How many 6 digit number without repeated digits exist?
i just used product rule 9x8x7x6x5x4
how many are dividable by 5? I sort of fucked up here but I did 9x8x7x6x5x2
and then they ask how many such numbers greater than 500 000 are dividable by 5
i did 4x8x7x6x5x2

[pearl pond]

@pure creek Has your question been resolved?

[pearl pond]

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[pure creek]

.reopen

[pearl pond]

✅

[pure creek]

1.there are 900 000 thousand 6 digit numbers
180 000 of them are dividable by 5 (900 000/5) could someone tell me what this method is called?
and what are the number of digits dividable by 5 greater than 500 000? I tried doing this with 400,000/5 but I'm not confident
2.How many 6 digit number without repeated digits exist?
i just used product rule 9x8x7x6x5x4
how many are dividable by 5? I sort of fucked up here but I did 9x8x7x6x5x2
and then they ask how many such numbers greater than 500 000 are dividable by 5
i did 4x8x7x6x5x2

<@&286206848099549185>

It's been 40 mins someone help😭

[spice lotus]

The number of 6 digit numbers that exist without repeating digits would be 9 × 9 × 8 × 7 × 6 × 5

[pure creek]

why 9 twice?

[spice lotus]

Coz u can't take zero in the 6th place , so there are nine options available

[pure creek]

ahh fuck truee

[spice lotus]

After using one of the nine options

There are 8 options other than zero

[pure creek]

so it should be 9x8 tho

[spice lotus]

Including zero , there are 9 again

[pure creek]

no wait it should be

8x8

cuz 8 options at first. I used a number then it's back to 8

[spice lotus]

0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9

Except zero , u pick one number

There will be 9 digits left

[pure creek]

oh they are 10 digits not 9

alright

[spice lotus]

Yes

[pure creek]

so are the rest correct?

well I should use 9x9 in the others so its

2.How many 6 digit number without repeated digits exist?
i just used product rule 9x9x7x6x5x4
how many are dividable by 5? I sort of fucked up here but I did 9x9x8x7x6x2
and then they ask how many such numbers greater than 500 000 are dividable by 5
i did 4x9x8x7x6x2

[spice lotus]

Out of these numbers, u want the number of them divisible by 5 right?

[pure creek]

yes

therefore the ones who end with 0 or 5

[spice lotus]

It should be 8 × 8 × 7 × 6 × 5 × 2

[pure creek]

cuz I can't use 0 or 5 till the end?

I thought it's okay because 5 is still a possibility

0 too like they can come before the last number

but 1 of them

[spice lotus]

Yea

[pure creek]

has to be in the end uk?

like I can have 156380

or 104835

which is why I took them both with 9x9

[spice lotus]

Wait a min

[spice lotus]

It should be 17 × 8 × 7 × 6 × 5

Use cases

In case 1 : where the last digit is 5
The number of approaches is 8 × 8 × 7 × 6 × 5
In case 2 : where the last number is zero ,
The number of approaches is 9 × 8 × 7 × 6 × 5

adding the two results up

[spice lotus]

U get this

[pure creek]

damnn

alright

and the nums greater than 500k are the same but I just change the first num

okay what abt number 1?

i know the first 2 answers are true

but what about the rest?

[spice lotus]

Yea the 180 000 is true

Number of digits divisible by 5 greater than 500 000 is 10^5

See in the 6th place , digits greater than 4 can be used
Thus there are 5 approaches
Now , in the 5th , 4th , 3rd , 2nd place , u can use any digit , so 10 approaches for each place and for the 1st place , u can use only 2 digits 0 or 5 , so there are two approaches
That brings us to
5 × 10^4 × 2
= 10^5

The actual answer would be 10^5 - 1

Since it said greater than 500 000 and we can't count 500 000

As for the 2nd part of ur second question

That is number of six digit numbers divisible by 5 greater than 500 000 without repeating digits

It should be 9 × 8 × 7 × 6 × 5

[pearl pond]

@pure creek Has your question been resolved?

[pearl pond]

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[naive pelican]

the second image is the question and the second is the part of the solving process im confused on

[winter heron]

Anyone PLS help me

[ashen geyser]

go to a free channel

[winter heron]

Can U solve my problem

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[naive pelican]

is putting (1/8)^x/(1/8)^2 the same as (1/8)^x times (1/8)^-2

also uuhere did the 1 come from ?

[autumn topaz]

from factoring

[naive pelican]

i figured!!
but houu does it factor to 1?

and uuhat are uue factoring in this instance?

one of the 1/8s?

[autumn topaz]

you are factoring (1/8)^x

[naive pelican]

and that factors to 1?

im confused on uuhat uue're doing thats turning it into 1

[autumn topaz]

let y = 1/8. then your equation is y^x + y^{x - 2}

if you factor y^x from those terms, the first one you'll be left with 1

so it'll be y^x (1 + y^{-2})

[naive pelican]

im unfortunately still confused...
just to be clear can you
define factoring
uuhat's the process uue are going through
in this case the only factoring i could think to do is ((y^x)^-2)

[autumn topaz]

you aren't familiar with factoring?

[naive pelican]

i thought i uuas!!!
i definitely feel like im missing information uuhen it comes to factoring exponents though

im just not sure? uuhat?

[autumn topaz]

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:intro-factoring/v/factoring-linear-binomials

[naive pelican]

i sauu this video!!
though i'll uuatch it again and get back to you to be sure

okay i!!!
get uuhat you mean a little better
i think i just forgot the term factoring referring to that process

houu do i knouu uuhen to factor though?

[autumn topaz]

when there are common factors in both/all the terms

e.g., how would you simplify 3x + 18

[naive pelican]

i guess the common factor is 3 so
3(x)+3(6)?

does factoring still uuork if the only common factor is 1?

i guess thats

uuhy the +1 is there

[autumn topaz]

what happens when you factor (1/8)^x out of (1/8)^x

[naive pelican]

you get 1 i suppose

[autumn topaz]

yes

[naive pelican]

so in this case the factored form being
(1/8)^x times (1+64)

okay i!!!

think i get it nouu
thank you so much!!!

.close

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[slender blade]

Help

[slender blade]

I don't understand

What this means

[vivid grove]

Do you know how to perform multiplication?

[slender blade]

Huh?

Holon

Let me translate that to my leng

[vivid grove]

first line to second line is multiplication

[slender blade]

Ohhh perkalian

Yeah i know

Soo its like 9 x 16

[vivid grove]

So you understand how to get to the first line to second line?

[slender blade]

Nah

I dont

[vivid grove]

What is 9*3

[slender blade]

9^3?

9x9x9

[vivid grove]

* is times

[slender blade]

Soo

Its uh

Wait

729

[vivid grove]

no

[slender blade]

So its not 9 x 9 x 9?

[vivid grove]

9*3, and * is times

[slender blade]

Oh 9x3

[vivid grove]

you said earlier you know how to perform multiplication

[slender blade]

27

[vivid grove]

yep

So in place of * they use a dot, the teacher actually forget to put it there for 9*3. But anyhow

So we have gotten from the first line to the second line

[slender blade]

Yeah

[vivid grove]

To go from the second line to the third line

The following theorem is used:

sqrt(xy) = sqrt(x)*sqrt(y)

So for the first term you have x=9, y =3

For the second term you have x=16, y=3

[slender blade]

Yeah

Is that it

Orr

Hello?

@vivid grove r u there

[vivid grove]

yes

that is it

After that you just simplify

[slender blade]

so this is the answer for the square question okay

alrrr

just need to simplify it'

[vivid grove]

square question?

[slender blade]

Yeah

[vivid grove]

This is the steps for the question yes

[slender blade]

Ok

Lowkey im failing math

[pearl pond]

@slender blade Has your question been resolved?

[pearl pond]

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[fickle atlas]

this one is C right?
cuz by finding the limit we get 3/4 which is < infinity

[sharp quest]

the limit of the sequence is 3/4
not the series

[fickle atlas]

so do i do a limit comparison test and figure it out?

[light helm]

no need

[fickle atlas]

What do I do then

[sharp quest]

what does the divergence test say?

[fickle atlas]

if lim does not equal 0 or does not exist then its divergent

[sharp quest]

does the limit of your sequence =0?

[fickle atlas]

ohhhh

Nope its 3/4

so its divergent

wow man acc thanks thats some good knowledge revision

i completely forgot that bit

[sharp quest]

it doesn't come up a lot, but when it does it saves a lot of time haha

[fickle atlas]

Yes i swear

im kinda stuck here can't seem to be able to get the 2^n and 3^n into a geometric series form

should I move the sequence back 1 step and move on from there?

[sharp quest]

this is a good place for a comparison test

[fickle atlas]

a limit comparison test works?

lemme try wait

yeah its convergent cuz when i did the test the limit was 1 which is positive and finite
then testing bn i got 2^n / 3^n so in a geometric form its 2/3 < 1 making it convergent

yo math is kinda fun if u start understanding it

[sharp quest]

yeah! these problems are all just puzzles and riddles. You have some tools and rules to put it together, but you have to know how those rules work.

[pearl pond]

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[tender ingot]

What happens when square rooting both sides of an inequality such as x > y, x^2 > y and x^2 > y^2?

[rancid rune]

you have to consider cases

for example with x^2 > y^2

if both x and y are negative then x < y

if both x and y are positive then x > y

[tender ingot]

umm but isn't sqrt(x^2) = |x| and same for y

[rancid rune]

yes thats true

so how do you solve |x| > |y|

it's the same idea

either x is more negative or more positive

[tender ingot]

well, we might have to consider different possibilities, if x > 0 and y > 0, then x > y, if x < 0 and y < 0, then -x > -y, if x > 0 and y < 0, then x > -y, and finally if x < 0 and y > 0, then -x > y, right?

[rancid rune]

right yes

[tender ingot]

and for x^2 > y, we have |x| > sqrt(y), assuming y > or equal to zero, and if we again consider the two possibilities of x: if x > 0, x > sqrt(y), similarly, if x < 0, -x > sqrt (y), right?

I solved my confusion, thanks for your help, .close

.close

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[formal leaf]

I am so confused

[formal leaf]

someone help\

.cancel

.end

uh

.close

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[formal leaf]

.close

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[red compass]

for chopping and rounding how would d) (ii) result with 0.455?

[red compass]

d) gives the result (i) 301/660 = 0.456060606060... and so on, i think the chopping should give for (ii) is 0.456

is it possible this is just a misprint?

[fluid axle]

you have to perform the whole computation w/ 3-digit chopping

you can't just say, oh the exact result is that and I'm chopping the exact result

[red compass]

im a bit confused, sorry what do you mean by whole computation?

[fluid axle]

1/3 has a certain representation in 3-digit chopped arithmetic

as 3/11 does

add these two

then substract the 3-digit chopped equivalent of 3/20 from that

that's how you compute in the 3-digit chopped system

[red compass]

ohh so, i should do for 1/3 = 0.333 (same for round and chop)

3/11 = 0.273 (round)

3/11 = 0.272 (chop)

then add these two

i got it

thank you so much

[fluid axle]

yeah and 0.605 - 0.150 is 0.455 yes

[red compass]

thats great, i appreciate your help

.close

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[pine tangle]

Hey, I have a small question about Linear Algebra

In my exercise, I am supposed to find Im(f). I have already found Ker(f) and shown that f WAS NOT bijective (hope it's the good word in english?)
Moreover, I have the dimension of Im(f) (2) and Ker(f) (2)
But how am I supposed to find Im(f) now?

[pearl pond]

@pine tangle Has your question been resolved?

[pearl pond]

@pine tangle Has your question been resolved?

[river marten]

Im(f) = Vect ( f(e_i) ) , e_i base de l'espace de depart

ici Im(f)=vect( f(e1),f(e2),f(e3),f(e4)) ou (e1,e2,e3,e4) la base canonique de R^4

[pine tangle]

Okay, I see ! Thanks!

[pearl pond]

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[timber shoal]

does anyone know where the digits come from here?

[pearl pond]

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[proven bane]

Prove that $(\mathbb{Q},+)$ does not contain any proper maximal subgroups.

[jolly parrot]

Casiel368

[pearl pond]

@proven bane Has your question been resolved?

[proven bane]

<@&286206848099549185>

[sour burrow]

Can you give the definition of a proper maximal subgroup that you are working with

Usually maximal subgroups are themselves defined to be proper

[proven bane]

A group H is maximal in G if for every K that contains H, then K = H or K = G

[proven bane]

It's just there to avoid ambiguity

[sour burrow]

Okay, well you can show that no proper subgroup is maximal

I would recommend reading the stackexchange site for this https://math.stackexchange.com/questions/234995/mathbbq-has-no-maximal-subgroups, as it has some nice solutions

[proven bane]

I don't quite understand the first answer

Even less the second and the third

[sour burrow]

The one By N. S. seems reasonable enough

[proven bane]

Why is this?

Sorry for sending screenshot. Is it possible to get the source from a post so that I can copy paste it here?

[sour burrow]

There probably is a typo there as it ought to be $r/n$

[jolly parrot]

F_iend

[sour burrow]

No worries about screenshots, I don't mind

[proven bane]

Still I don't get why is that true

where does the r come from

The intermediate step is confusing. I think the argument still works even if you remove that step

Since Q is H + <1/n>, every rational has such a writing

And 1 is in H so the restriction on k is fine

[sour burrow]

Yeah exactly

[proven bane]

I don't quite like the WLG step. If I were to remove it, then it would be something like h + h_0 * k/n, right? where h_0 is some nonzero element of H.

It would still remain to prove that <h_0/1, h_0/2 ...> generates all of Q

[sour burrow]

Yes indeed

Though it works better for this particular flavour of proof to have the wlg step

As it requires less hassle

[proven bane]

Especially considering that it uses a group morphism

But it still bugs me a little

Oh is it because morphisms preserve maximal subgroups?

[sour burrow]

Yes iirc they do that

[proven bane]

Okay then I'm convinced

I wasn't sure why would it preserve anything but I see it now

Thank you

[sour burrow]

No worries

I didn't do that much tbh

[proven bane]

Even if I have a fuzzy idea and it does work, it does help to have someone else check because I wasn't sure

🙂

.close

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[sour burrow]

That is definitely true, Happy to help 🙂

[pearl pond]

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[naive trellis]

would anyone be please willing to explain how this works? I dont understand it.

[sterile tusk]

basically if $f(x)$ is odd, then $\int_{-a}^{a}f(x)dx=0$

[jolly parrot]

y0shi

[sterile tusk]

general rule for odd functions

[naive trellis]

this is always true?

[sterile tusk]

yes as long as f(x) is continuous from -a to a

[naive trellis]

thank you, I have been wasting my time trying to integrate.

[pearl pond]

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[harsh fern]

!help

[pearl pond]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[pearl pond]

Please don't occupy multiple help channels.

[harsh fern]

hi

does anyone know how to do this

also can u check to see if -16 is right

so idk how to do it cause theres no 6.022 on the table

[tall flint]

The first part is done using the definition of the inverse, ie you should know what f^-1(f(x)) is by default

-16 looks fine for part 2

[harsh fern]

yes

but how would you know what 6.022 is

if there is no 6.022 anywhere on the scatterplot

[tall flint]

Again, it doesn't matter

[harsh fern]

i see

[tall flint]

By definition, you should know what f^-1(f(x)) is for any x

[harsh fern]

is it

is it just -?

-6.022

oh wait

[tall flint]

Not quite

[harsh fern]

hmm

can u give me a hint

i dont see it]

[tall flint]

My hint is go back to the definition of the inverse function

"f^-1 is the inverse of f" means f^-1(f(x)) = ...

[harsh fern]

well

because 6.022 is close to 6

wait

[tall flint]

I reckon they picked a number not listed within the function on purpose

[harsh fern]

oh

its 6.022

that is the answer

[tall flint]

That is

[harsh fern]

?

[lavish portal]

Courrr y

Correct

But do you understand why

[harsh fern]

let me see

trying to draw it

my explantion

btw

mooncaik

@lavish portal

the question is asking you given the x value 6.022 what is the y value

which would mean f^-1(y)

which asks what y value gave you the x value of 6.022

so its =

[lavish portal]

Yes

Correct

[pearl pond]

@harsh fern Has your question been resolved?

[harsh fern]

can anyone explain

how they got the second equation

-2x = y + 3

from the first one

i dont understand what step they equivalence perseving operation they did

oh

nvm they multiplied both sides by -2

to get rid of the fraction

[pearl pond]

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[merry stirrup]

So this is the question

This question complexly asks you

To find The minima and maxima

The*

Which I get it

But the problem comes when I try to find the Maxima and minima

If I use derivatives to find it

,w d/dx(√3sin(0.57x)-cos(0.57x))=0

[jolly parrot]

Results provided by WolframAlpha

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

Requested by arnabpal

[merry stirrup]

Okay so the value of constant omega is 0.57 btw

Approximately

But if I use derivatives I get this

There's another method

You multiply by 2 and divide it by 2 then you get this

[merry stirrup]

This is a solution I got on internet

[merry stirrup]

This is the solution by using derivatives

This is the graph of the equation

[merry stirrup]

Then this is true

[merry stirrup]

But why the derivative method is failing here

Can I use derivatives to find maxima and minima here?

[pearl pond]

@merry stirrup Has your question been resolved?

[merry stirrup]

<@&286206848099549185>

[pearl pond]

@merry stirrup Has your question been resolved?

[pearl pond]

@merry stirrup Has your question been resolved?

[pearl pond]

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[vernal hill]

how to simplife this ((3n)!(n-3)!)/((3n-1)!(n-2)!)

[tender ingot]

Do you know that n! = n (n-1)!

[vernal hill]

yes

[tender ingot]

Use that

[vernal hill]

how

[tender ingot]

(3n)! = 3n (3n-1)! and (n-2)! = (n-2)(n-3)!, then just simplify directly

[vernal hill]

ok

thanks

.close

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[zenith basin]

Can someone please help me with question 3b

[sterile turtle]

hello my dr du man

if there are more rainy days then fine days

we gotta find all of the combinations that state this

5 winter days are chosen, so we gotta make
3rainy 2fine
4 rainy 1 fine
5rainy

[zenith basin]

so we multiply them and add them?

[sterile turtle]

yep add all the cases

[zenith basin]

alr ill try that

[zenith basin]

i got 0.16416

but the answer is 0.6826

[sterile turtle]

lemme see what you did

I did it myself and got the answer

[zenith basin]

3/5^5 + 3/5^4 * 2/5 + 3/5^3 * 2/5^2

[sterile turtle]

aH yea

i never said that the 3 rainy and 2 fine have to be in that order

remember we’re CHOOSING 3rainy and 2fine

[zenith basin]

so we use the combinations thing then

and add 3 of those

[sterile turtle]

yep

[zenith basin]

alr

[sterile turtle]

remember to use it for all of em

[zenith basin]

oh yea i got it

that makes sense

[sterile turtle]

nice mate

[zenith basin]

thanks dr du man

[sterile turtle]

👍

[zenith basin]

.close

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[bright holly]

In this inequality

4y + 8 > -48

[bright holly]

Why can't we just subtract 3 from both sides

And then later subtract8 from both sides

Wouldn't we still get our ans?

Chat gpt isn't really of any help

He doesn't explain why it's wrong despite asking it to ;-;

4y + 8 > -48

Subtracting 3 from both sides

4y + 8 > -48
-3. -3

y + 8 > - 51

Subtracting 8 by both sides

y + 8 > - 51
-8. -8

Y > -59

Is there a order we have to follow?

The video on khan academy says we should first

Subtract 8 by both sides

Then divide by 4 both Sides

[fickle schooner]

8 - 3 is 5

[tender ingot]

You made a mistake, when subtracting 3, you didn't actually subtract from both sides

[untold lintel]

No

I will not enlighten you

[bright holly]

;-;

Bruhhh

[last wave]

Step 1: get rid pf your worldly possessions

[bright holly]

Cmon

Someone

[fickle schooner]

Am I missing something?

[bright holly]

We subtracted 8 by 8

Not 3 by 8

On khan academy

[fickle schooner]

When you subtracted 3 from both sides, you dropped the 4 as well(and also didn't even subtract 3 from 8)

4y + 8 > - 51
-3 -3
4y + 5 > -54

[bright holly]

4y
-3

Y

[fickle schooner]

4y =/= 4

Yeah that's not how it works

[bright holly]

Wha-

Why

In previous vid he did that!!!

That's how I tried on this one

[fickle schooner]

Are you sure he didn't have a number with a y next to it?

[bright holly]

Lemme send screen shot incase am missing something

Cuz I am generally very ignorant

We can't subtract 4y by 3 cuz it has no y?

So it isn't right?

[fickle schooner]

Yours isn't, yes

He's right

[bright holly]

: o

[bright holly]

This correct?

[fickle schooner]

Yeah

[bright holly]

Oki tyyyyy

Am still new to maths so thanks for bearing with me!

[fickle schooner]

Don't worry about it 👍

[bright holly]

.close

[pearl pond]

Channel closed

Closed by @bright holly

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[frosty heath]

hello can anyone help me with this limit problem

[steady crescent]

Sure

It diverges right

So find something that is less than it the goes to +ve infinity

It might help to note that x^2+4x+1 = (x+2)^2-3

[frosty heath]

my prof wanted us to use this factoring method so that we take the x out like this,,

supa lost

[rustic gate]

seems good

[steady crescent]

Sorry my bad

[rustic gate]

,, \lim_{x \to \infty} \f {x\parens{4 + \df1x}} {x \sqrt {1 + \df4x + \df1{x^2}} + x}