help-10

[bleak storm]

and ive crossed out the x+9 from the denominator of 4/x+9 and crossed out the x from 7/x

all thats left is to multiply 1 by x+9, leave 4 as is, and multiply 7 by 9

I feel like I sorta got the steps down but im still lost

[lost tree]

what do umean

[bleak storm]

oh im so dumb

i forgot to say what the question wanted

[warm shale]

𝕾ilver𝕾oldier

[bleak storm]

Its solving a rational equation

[bleak storm]

tbh im not sure

im supposed to find the LCD

[marble shuttle]

I would get the denominator the same for all fractions

[lost tree]

lcm im assuming

[marble shuttle]

Including 1

[lost tree]

least common multiple

[bleak storm]

yea

which is x+9

[lost tree]

yes

so make the denominator

of both fractions

x+9

[bleak storm]

because 1 goes into anything

and x is in x+9

[marble shuttle]

1 / 1

[bleak storm]

My teacher was telling us to multiply the lcm after

multiply it by all of the fractions

[lost tree]

$1-\frac{4}{x+9}=\frac{1}{1}-\frac{4}{x+9}$

[warm shale]

𝕾ilver𝕾oldier

[marble shuttle]

^^^

[bleak storm]

and it also wants 2 answers

[marble shuttle]

yes there are 2 correct answers

[lost tree]

okay with this?

[bleak storm]

yes

[warm shale]

𝕾ilver𝕾oldier

𝕾ilver𝕾oldier

[bleak storm]

yep all good with that

[lost tree]

coz the denominator of $\frac{4}{x+9}$ is already $x+9$

[bleak storm]

im at the part where i multiply the fractions with the lcm

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

u dont need to do anything to that fraction

[bleak storm]

wouldnt i cross out the x+9 from 4/x+9?

[lost tree]

the denominator of $\frac{1}{1}$ however is not $x+9$

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

uhm im not sure what u mean, cross out with what?

but are u okay with what i said so far?

[bleak storm]

im so sorry i am not good at explaining things so bare with me but, from what I know im supposed to find the LCM then multiply all the fractions by it and cross out like terms

so for 4/x+9 cross out x+9

for 7/x cross out x

[lost tree]

so u must try to make the denominator of $\frac{1}{1}$ equal to $x+9$. You can do that by multiplying both the numerator and denominator of $\frac{1}{1}$ with $x+9$

[bleak storm]

then multiply by what is left of the lcm thats being multiplied

[warm shale]

𝕾ilver𝕾oldier

[bleak storm]

yeah

[bleak storm]

yea im good with that too

makes sense

[lost tree]

ie. $\frac{1\times(x+9)}{1\times(x+9)}$

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

then this becomes equal to $\frac{x+9}{x+9}-\frac{4}{x+9}$

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

are u okay with that?

[bleak storm]

yes

[lost tree]

okay

[bleak storm]

wouldnt x+9/x+9 just be x+9

[lost tree]

now that u have two fractions with the same denominator on ur left hand side, u can just add/subtract the numerators

so $\frac{x+9}{x+9}-\frac{4}{x+9}=\frac{x+9-4}{x+9}$

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

okay?

[bleak storm]

yup

[lost tree]

right so then ur equation is now

$\frac{x+9-4}{x+9}=\frac{7}{x}$

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

so $\frac{x+5}{x+9}=\frac{7}{x}$

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

oky with that?

[bleak storm]

Yea that part makes sense

[lost tree]

now u want to try to get rid of the denominators

[bleak storm]

But I was taught to do these kinda problems like this

sorry if its hard to see

like see what i mean by cross out

and it worked for the problems with one solution

but its not working out now for some reason

[lost tree]

are u multiplying both sides by $x+9$ here?

[warm shale]

𝕾ilver𝕾oldier

[bleak storm]

no

im taking the LCM

and multiplying it by all of them

and crossing out like terms

[lost tree]

right so u r multiplying $\frac{7}{x}$ by $x+9$

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

on the right hand side

$\frac{7}{x}\times(x+9)$ right?

[warm shale]

𝕾ilver𝕾oldier

[bleak storm]

yep

then i crossed out the x

and the 9 was left

so i did 9 x 7

[lost tree]

u cant cancel out the x now coz the x in the brackets is not a factor of the expression in the brackets

$\frac{7}{\cancel{x}}\times(\cancel{x}+9)$ is not valid, coz theres a plus sign there and not a multiplication sign

[warm shale]

𝕾ilver𝕾oldier

[bleak storm]

oh my lord

i completely forgot about that

the crossing out was multiplication

rightttt

[lost tree]

if u had $\frac{7}{x}\times x\times9$, then u cud cross out the $x$'s like that

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

but thats not what u have

[bleak storm]

i had another problem where the LCM was 5x

and thats multiplication

so thats why it worked

[lost tree]

right

[bleak storm]

and now its not working since its not multiplication

yeah so far this is making sense thanks

[lost tree]

so u cud find the lcm of the denominators of all the fractions

so the lcm of $1$, $x+9$ and $x$

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

rather than just the lcm of $1$ and $x+9$

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

and then multiply all the fractions by that

[bleak storm]

im still kinda confused

after I get the LCM

which is x+9

and multiply 1 by it

what next?

[lost tree]

x+9 is the lcm of the numbers 1 and x+9

but im asking u to find the lcm of all the denominators, there are 3 different denominators

1, x, and x+9

[bleak storm]

yea and doesnt x go into x+9?

[lost tree]

no coz u cant divide x+9 by x

again coz theres a plus

its not 9x

[bleak storm]

that keeps confusing me

so what can i do?

[lost tree]

do u know how to find the lcm of 3 numbers? say 10, 15, 5

[bleak storm]

i know the long way

not really the short way

[lost tree]

what is the long way u know

[bleak storm]

isnt there a method where you branch each number out

[lost tree]

10 = 5 * 2

15 = 5 * 3

5 = 5 * 1

this u mean?

[bleak storm]

yea

[lost tree]

well yeah okay this wud do

[bleak storm]

is there a better way?

[lost tree]

so x = x * 1

[bleak storm]

i think my teacher taught a better way but i forgot it

[lost tree]

x+9 = (x+9) * 1

1 = 1 * 1

so the lcm is ?

[bleak storm]

1?

[lost tree]

no u find the distinct factors

and u multiply their highest powers together

okay i can tell u this though, if $a$, $b$, $c$ are any three numbers, then $abc$ is some multiple of all $a$, $b$ and $c$, if not the least one

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

and u can use that to simplify fractions too

so when it comes to the 3 numbers $10, 15, 5$ i gave u earlier, $10\times15\times5=750$ is a multiple of all $10$, $15$ and $5$, but not their least common multiple, but it wouldnt quite matter, if u were simplifying fractions whose denominators were $10, 15$ and $5$. you could just multiply all the fractions with $750$ and u shud be able to proceed

[warm shale]

𝕾ilver𝕾oldier

[bleak storm]

ah

[lost tree]

so when u have 1, x and x+9

u can multiply all the fractions by 1 * x * (x+9)

[bleak storm]

so (x+9) and (x)?

[lost tree]

in this case, it just happens that this quantity is also the least common multiple

[bleak storm]

(x+9)(x)

[lost tree]

yeah

multiply all the fractions by x(x+9)

[bleak storm]

to be real with you I had this same LCM 20 minutes ago but i over guessed myself

it just didnt feel right

so just multiply for now?

no crossing out?

cant I cross out the (x) since its not adding anything to it

like for 7/x i can cross it out

[lost tree]

in $\frac{7}{x}\times x(x+9)$

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

yes u can cross out the x coz that x is a factor of x(x+9)

so $\frac{7}{\cancel{x}}\times\cancel{x}(x+9)$ is valid

[bleak storm]

but then I would multiply 7 by x+9 which is 63 + 7x?

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

yes

7(x+9) = 7x + 63

[bleak storm]

also if the LCM is (x+9)(x), cant I cross out the x+9 from 4/x+9 since the lcm has multiplication going on?

[lost tree]

yes u can

$\frac{4}{\cancel{(x+9)}}\times x\cancel{(x+9)}$ is valid

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

again coz the entire (x+9) is a factor of x(x+9)

[bleak storm]

alright so so far I have (x+9)(x) - 4x = 7x+63

[bleak storm]

yeahhh

[lost tree]

yes

[bleak storm]

by itself it wouldnt work since its addition

[lost tree]

?

[bleak storm]

I mean like if the lcm was only x+9

[lost tree]

ah right

[bleak storm]

i wouldnt be able to cross out

[lost tree]

yes

[bleak storm]

so now Im back to where i died pretty much, (x+9)(x) = 11x + 63

[lost tree]

no that wud still be okay

$\frac{4}{\cancel{(x+9)}}\times\cancel{(x+9)}$ is okay

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

coz x+9 up there is still a factor of x+9; every number is a factor of itself

x+9 = (x+9) * 1

$\frac{4}{\cancel{(x+9)}}\times\cancel{(x+9)}\times1$

[warm shale]

𝕾ilver𝕾oldier

[lost tree]

$\frac{4}{\cancel{(x+9)}}\times[x+\cancel{(x+9)}]$ is not valid

[warm shale]

𝕾ilver𝕾oldier

[bleak storm]

ah

but x+9 would not work for 7/x

that is where you said it wouldnt work if im correct

[lost tree]

yes

[lost tree]

yes

[bleak storm]

yea makes sense

so far so good

[lost tree]

this is wrong

[bleak storm]

however if it was 9x it could work

right?

[lost tree]

yes

[bleak storm]

ok ok cool

[lost tree]

with 9x, it would not work with 4/(x+9) though

it wud only work with 7/x

[bleak storm]

makes sense

[lost tree]

this is now just a quadratic

[bleak storm]

so whats the best way to solving it?

would I factor?

[lost tree]

expand the brackets

and write it in the standard form first

[bleak storm]

alright

so FOIL?

x^2 + 9x

= 11x + 63

[lost tree]

yes

now simplify

write it in the standard form

ax^2 + bx + c = 0

[bleak storm]

x^2 = 3x + 63

oh

what

im lost again lol

oh right

[lost tree]

11-9 is not 3

[bleak storm]

-3

wait what

im an idiot

it is too late

2

222222

2

[lost tree]

yes

[bleak storm]

its 12 am for me I apologise

[lost tree]

so x^2 = 2x + 63

[bleak storm]

yes 2

yep

[lost tree]

so now move everything to one side and have 0 on the other side

[bleak storm]

oh this looks factorable

x^2 + 2x + 63 = 0

[lost tree]

x^2 - 2x - 63 = 0

minus

not plus

[bleak storm]

why?

[lost tree]

coz u r moving it to the other side of the equals sign

[bleak storm]

do i move 2x+63 to the left instead of x^2 to the right?

[lost tree]

when u have x^2 = 2x + 63

to get rid of the 2x on the right hand side

u wud subtract 2x from both sides

so x^2 - 2x = 2x - 2x + 63

x^2 - 2x = 63

[bleak storm]

makes sense

[lost tree]

now to get rid of the 63, u subtract 63 also from both sides

so x^2 - 2x - 63 = 63 - 63 = 0

[bleak storm]

i made a mistake by subtracting x^2

[lost tree]

u cud also have done that, then u wud have had 2x + 63 - x^2 = 0

from x^2 = 2x + 63, if u subtract x^2 from both sides, u get
x^2 - x^2 = 2x + 63 - x^2
0 = 2x + 63 - x^2

[bleak storm]

if i did it the first way you told me, is x^2 - 2x - 63 = 0 correct?

[lost tree]

yes

[bleak storm]

then factored it would be (x+9)(x-7)?

[lost tree]

(x-9)(x+7)

[bleak storm]

is there a difference?

[lost tree]

if u expand this u get a +2x

not a -2x

like there is in ur original expression

[bleak storm]

thats so weird

my answer sheet says x+9 is correct

i just peeked at it after solving and it said answer is D

[lost tree]

yes d is the answer, and u get this answer with (x-9)(x+7)

[bleak storm]

thats true

because +9

-7

[lost tree]

u shud have (x-9)(x+7) = 0

[bleak storm]

yea

[lost tree]

so either x-9 = 0

and x = 9

or x+7 = 0 and x = -7

[bleak storm]

its just late now that I look at it makes sense

have to add to get -2

multiply to -63

-9 + 7 = -2

so (x-9)(x+7)

which solved becomes x = 9 and x = -7

[bleak storm]

and - 9 x 7 = -63

[lost tree]

yes

[bleak storm]

honestly thank you so much

you helped me understand this finally

after struggling on it for a while

is there ANY chance you have 5-10 minutes for one more problem?

[lost tree]

yeah what is it

[bleak storm]

i think ill get it fast thanks to this recent knowledge

alright

thanks

[lost tree]

okay what comes to mind

[bleak storm]

well

x goes into x^2

[lost tree]

again the denominators have pluses and minuses

[bleak storm]

yes

alright

so I need an LCM

between x-3, x^2 -9, and x+3

[lost tree]

yes the lcm of x-3, x^2-9, and x+3

notice that x^2 - 9 is a difference of two squares

[bleak storm]

yeah I do

(x-3)(x+3)

OH

and thats perfect for the other denoms

because they are x-3 and x+3

[lost tree]

yes

[bleak storm]

and x^2 - 9 is the same as (x-3)(x+3)

alright so this is the LCM

[lost tree]

yeah

multiply thru by the lcm

[bleak storm]

this took so long lol

alright

24x+72 = 32x^2 - 16x^2 - 48x

[lost tree]

Well it's -16(x-3)

U must be careful of the signs

[bleak storm]

ah right

so -16x^2 + 48

[lost tree]

Yes

[bleak storm]

in your opinion what would be the wiser move? move 24x+72 to the right OR, move 32x^2 - 16x^2 +48x to the left

i kinda have trouble deciding with these

in these situations

[lost tree]

Well first simplify 32x² - 16x² to 16x²

[bleak storm]

yea

alright

[lost tree]

It really doesn't make a difference after thay

[bleak storm]

done

[lost tree]

But to me it's more convenient when the coefficient of x² is positive

[bleak storm]

alright looks like I did that

16x^2 + 72x - 72

but this looks so hard to factor

lol

[lost tree]

How did u get a 72x

Ah

[bleak storm]

24x+48x

[lost tree]

Wait ur equation is 24x + 72 = 32x² - 16x² + 48x right

[bleak storm]

yea

then I reached 24x+72 = 16x^2 + 48x

and then moved left to right

[lost tree]

Okay so 0 = 16x² + 48x - 24x - 72

[bleak storm]

holy crap

I am actually dumb

the simplest thing I did wrong

yes

[lost tree]

😅

[bleak storm]

midnight studying lol

somehow added 24 but subtracted 72

[bleak storm]

alrighty reached that now

looks MUCH better to factor

except for the fat 16

[lost tree]

So 16x² + 24x - 72 = 0

[bleak storm]

yea

[lost tree]

U can cancel a 2

Every number is divisible by 2

8x² + 12x - 36 = 0

Actually another 2

4x² + 6x - 18 = 0

And another one lol

2x² + 3x - 9 = 0

Get it?

[bleak storm]

yea lol

thanks

but I forgot what to do when theres a variable other than 1

for x^2

[wooden marten]

you could divide everything by that but it makes it look ugly

[lost tree]

U can multiply the constant term and the coefficient of x²

And find two numbers that multiply to that quantity, and add up to make the coefficient of the x term

So multiply 2 and -9 to get -18

Find two numbers thay multiply to that

And add up to +3

[bleak storm]

so should I basically change the -9 to a -18

like forget about the -9?

and do everything else normally

[lost tree]

Uhm well

U don't forget abt the -9

If u had something like x² + 3x - 9 = 0, u wud have tried to find two numbers that multiplied to -9 and added up to +3

When the coefficient of x² is not 1, so maybe 2x² + 3x - 9 = 0, u do the same thing but with 2 × (-9)

U don't forget abt the -9

Can u find two such numbers and tell me

[bleak storm]

add to 3 multiply to -18?

[lost tree]

Yes

[bleak storm]

alrighty

6 and -3

(x+6) (x-3)

[lost tree]

Yes

[lost tree]

Well no

But 6 and -3 are correct

So ur original quadratic is 2x² + 3x - 9 = 0
U rewrite the middle term I'm terms of the numbers u just found

[bleak storm]

but why?we need 2 numbers that add to positive 3 and multiply to -18

[lost tree]

So 2x² + (6 - 3)x - 9

Then expand the brackets

2x² + 6x - 3x - 9

Now look at the first two numbers, 2x² + 6x. U can factor out a 2x; 2x² + 6x = 2x(x+3)

[bleak storm]

im sorry i got so lost at the factoring part

i understand how we get -18

but then what after that?

[lost tree]

And then u find 6 and -3

[bleak storm]

I just assumed I need 2 numbers that multiply to -18 and add to positive 3 so i thought (x+6) and (x-3 was right)

[lost tree]

Well no its not just that, that only works when the coefficient of x² is 1

[bleak storm]

how did we get 6 ?

[lost tree]

I asked u to find two numbers that multiplied together to give -18, and added up to give +3 (coefficient of the middle term)

These two numbers are 6 and -3

[bleak storm]

yup

[lost tree]

Then I can do this, is that okay?

I rewrite the middle term using the numbers u just found

[bleak storm]

why the middle term?

[lost tree]

6 and -3 do add up to 3 so just replace the 3 with 6 + (-3) = 6 - 3

[lost tree]

Well that's actually the same thing u r doing even when the coefficient of x² is 1

So if u had x² + 5x + 6

Then 3 and 2 multiply to 6 and add to 5

[bleak storm]

yea

[lost tree]

So (x+2)(x+3)

But when u do FOIL, (x+2)(x+3) = x² + 3x + 2x + 6

Right?

[bleak storm]

yup

so its the same with this problem?

[lost tree]

Look at the middle

There's 3x + 2x

(3 + 2)x

So the numbers U find are actually numbers such that U can rewrite the middle term as their sum

[bleak storm]

so we get to 2x^2 + (6-3) x - 9?

but how come the 9 stays normal while the 3 changes

isnt the 9 the thing we are changing by multiplying it by 2

why does the 9 not change

[lost tree]

No we don't change 9 by multiplying it with 2

We just temporarily multiply 9 by 2, and find numbers that multiply together to that and add up to the middle term

[bleak storm]

ah alright

so 6 and -3

[lost tree]

Yes

[bleak storm]

but thats not the final answer

we just use it

to get to the answer

[lost tree]

Yes

2x² + (6 - 3)x - 9 is then
2x² + 6x - 3x - 9

[bleak storm]

but we get to the same thing

[lost tree]

Take 2x² + 6x together first and factor it, so u have 2x² + 6x = 2x(x+3)

Then take -3x - 9 together and factor it

So u get -3x - 9 = -3(x+3)

Put them together so u have

2x² + 6x - 3x - 9 = 2x(x+3) - 3(x+3)

Is that okay?

[bleak storm]

yup

[lost tree]

Right but u can once again factor out an (x+3) from 2x(x+3) - 3(x+3)

So 2x(x+3) - 3(x+3) = (2x - 3)(x+3)

Get it?

[bleak storm]

yup

wait

[lost tree]

So 2x² + 3x - 9 = (2x - 3)(x+3)

[bleak storm]

but how are you able to factor out the x+3?

[lost tree]

So we had 2x(x+3) - 3(x+3) right
Take t = x+3
Then that expression turns into 2x × t - 3t

Now u can factor out a t from 2x × t - 3t right?

[bleak storm]

yes

but why did we choose x+3

[lost tree]

Coz that's the quantity we see appearing in both terms

[bleak storm]

yeah it is

[lost tree]

In 2x(x+3) we have a (x+3)
In -3(x+3) also we have an (x+3)

So it's that sort of number that u can factor out

[bleak storm]

got it

[lost tree]

So that's it

U must have (2x - 3)(x+3) = 0

And now u can solve for x

[bleak storm]

sorry one last thing how was x+3 factored out

i dont get how it changed from 2x(x+3) - 3(x+3) to (2x - 3)(x+3)

i get it was factored

but how

[lost tree]

Well how wud u factor out t from 2xt - 3t

[zinc plover]

does anyone have a chegg acount

[bleak storm]

my brain is working on 1 cylinder right now

uhhhhh

t(2x)

?

lol idk

t(-3)

[lost tree]

Yes so 2xt - 3t = t×(2x) + t×(-3)

[bleak storm]

no way

okay this is good

[lost tree]

Now u can introduce a bracket right

[bleak storm]

where?

[lost tree]

t × [(2x) + (-3)]

This simplifies to t×(2x) + t×(-3)

[bleak storm]

yupp

[lost tree]

So t×(2x - 3)

That is how u factor out t from 2xt - 3t

Get it?

[bleak storm]

so x+3(2x-3)?

[lost tree]

Yes but x+3 in brackets

[bleak storm]

ok this makes so much more sense

i was overthinking it

[lost tree]

It's the entire x+3 that is being multiplied

[bleak storm]

yeah thanks it makes sense

[lost tree]

👍

[bleak storm]

alright so 2x(x+3) - 3(x+3) = (2x - 3)(x+3)

then

2x² + 6x - 3x - 9 = 2x(x+3) - 3(x+3)

this is what u sent earlier

and then factored

2x^2 + 6x - 3x - 9 = (2x - 3)(x+3)

?

[lost tree]

Yes

But 2x² + 6x - 3x - 9 is again 2x² + 3x - 9, ur original quadratic

So 2x² + 3x - 9 = (2x - 3)(x + 3)

[bleak storm]

whatttt

so we go back to the original

[lost tree]

No like we have got the original thing in the factored form

Our problem was to solve 2x² + 3x - 9 = 0

[bleak storm]

and now its (2x - 3)(x + 3)

[lost tree]

But now we know 2x² + 3x - 9 = (2x - 3)(x + 3), so we need only solve (2x - 3)(x + 3) = 0

[lost tree]

Yes so if we can find where this becomes equal to 0 we are done

[bleak storm]

is there a faster way to do this?

[bleak storm]

(x+3) becomes x = -3

not sure about (2x-3)

[lost tree]

2x-3 = 0 gives 2x = 3

So x = 3/2

[lost tree]

Well once u get to 2x² + 3x - 9 = 0 u can use the quadratic formula

[bleak storm]

what about the (x+3)?

why do we disregard it?

[obtuse pebble]

@bleak storm Has your question been resolved?

[lost tree]

We don't disregard it

[lost tree]

U already aaid this

When (2x - 3)(x + 3) = 0, either 2x-3 = 0 so x=3/2, or x+3=0 so x=-3

[bleak storm]

Alright it all makes sense

Thank you so so so much

I genuinely appreciate all the help

Really needed it

[lost tree]

Happy to have helped 😄👍

[bleak storm]

Take care 😄

[obtuse pebble]

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[hollow quiver]

I need homework help on this section

[royal basin]

,rcw

[royal basin]

which of these objects are you having trouble finding the names for?

[hollow quiver]

Like matching them

For 18 I putted A

[cyan kite]

what have u got so dlfar

[hollow quiver]

I answered some

Is just that I am confused

[cyan kite]

show what u have so far

[hollow quiver]

Kk waot

The instructions told me to match them with the picture. Just that the definition confuses me.

[hollow quiver]

For F, G, and H are confusing

[obtuse pebble]

@hollow quiver Has your question been resolved?

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[timid silo]

How do i do this one im having some trouble with it

i was able to get the first two right

[supple quest]

isn't the third term -4?

4th term -12?

[timid silo]

i figured it out

i wasnt replacing the right terms in

[supple quest]

lmfao

🤣

[timid silo]

Yea first time learning this

[supple quest]

nah np man

it happens

[timid silo]

ight thanks ttyl

a1=2
a2 = 2(a1-2) = 2(0)=0
a3 = 2(0-2) = -4
a4 = 2(-4-2) = -12

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[obtuse pebble]

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[timid silo]

Ohhh you got it mb

[obtuse pebble]

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[south plover]

i need to plot the magnitude of the frequency response of this system.
i know how to figure out the general shape of the plot (poles blow, zeroes suck) but im struggling with figuring out how to find more exact values for my plot.

[south plover]

i got the following system function from a

and i rewrote it to

from there i tried using eulers formula to rewrite, and then separate the real and imaginary parts, in order to figure out the absolute value at different points

however, i think im doing something wrong in the rewriting step, because i get a division by zero when i plug in pi

[gilded needle]

well the abs value of the numerator is easy, factor out e^(iw)

for the denom, yea i guess use euler's formula

[south plover]

ok i think i figured out what was wrong, i had forgotten de moivres....
but may i ask a stupid question: how does factoring the numerator make it easy? 🤔 and have i misunderstood it when im thinking that the abs value would be given by sqrt(re(H)^2+im(H)^2)? am i supposed to take the absolute value of the denominator and the numerator separately?

[gilded needle]

that's one way to do it, since in general the abs value of a fraction equals abs(num) / abs(denom)

and the numerator can be done like this:

$$\begin{aligned}
|e^{i2\omega}-1| &= |e^{i\omega}(e^{i\omega} - e^{-i\omega})| \
&= |e^{i\omega}||e^{i\omega} - e^{-i\omega}| \
&= |e^{i\omega} - e^{-i\omega}| \
&= |2i\sin(\omega)| \
&= 2|\sin(\omega)|\end{aligned}$$

[warm shale]

Bungo

[gilded needle]

sorry about typesetting issues, i think it's fixed now

[south plover]

oooo. thats really neat and something i really didnt think of. im not really fluent in complex numbers
so if i rewrite the numerator like that, and then use eulers formula on the denominator... did i understand it correctly that i can then take the absolute values of numerator and denominator independently, which will get me a fraction that describes the magnitude?

[gilded needle]

yep, that is correct

might be easier than trying to find the real and imaginary parts of the entire fraction

[south plover]

alright! thank you a lot! i think you just saved me for my exam tomorrow! ❤️

[gilded needle]

sure, gl with the exam!

[south plover]

yes!!

thanks!

!close

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[daring salmon]

can someone please guide me to get the derivative of this

[carmine quest]

dervative of cos is -sin

[daring salmon]

yes

[carmine quest]

so cos changes to -sin and then you times by the dervative of sqrt(5t)

[daring salmon]

sqrt(5t)(-sin(sqrt(5t)) + e^(3t)

[carmine quest]

dervative of sqrt(5t) is not sqrt(5t)

[daring salmon]

5t^-1/2

[carmine quest]

no

it's using chain rule

[daring salmon]

idk

[daring salmon]

sqrt(5)/2sqrt(t)

[carmine quest]

no

[carmine quest]

it works like this

[daring salmon]

that is the derivative of sqrt(5t)

wait

no

you are right

but where did you get x5 from

[carmine quest]

the deravative of 5t is 5

then

[daring salmon]

can you explain how you put this back into a fraction

[carmine quest]

does that make sense?

[daring salmon]

so only the numerator was multiplied by 5

[carmine quest]

yes because i took (5t)^-1/2 down

this is how chain rule works

[daring salmon]

what am i supposed to have right now

[carmine quest]

this

then do you know the deravative of e^3t

[daring salmon]

would it be 3e^(3t)

[carmine quest]

yes

correct

[daring salmon]

is that it

[carmine quest]

yes

what is your final answer

[daring salmon]

(5/2sqrt(5t))(-sin(sqrt(5t))) + 3e^(3t)

should the -sin part be apart of my numerator

[carmine quest]

yes

so it would look like this

[daring salmon]

can we go over getting the derivative of (5t)^(1/2) again

[carmine quest]

sure

[daring salmon]

what would my first step be

im multiplying 5t by 1/2 right

(1/2)(5t)^(-1/2)

where does the times 5 come from

[carmine quest]

5 is the deravative of 5t from inside the brackets

[daring salmon]

oh

[carmine quest]

does this make more sense????

[daring salmon]

yes

[carmine quest]

a step by step working always works

[daring salmon]

did you get the derivative of 5t because of chain rule

[carmine quest]

yes

[daring salmon]

chain rule hurts my head

[carmine quest]

it gets much easier with practice

do more questions like this

[daring salmon]

i have my midterm in like 8 hours from now

[carmine quest]

i see

well good luck

[daring salmon]

can we go over a quotient rule problem

[carmine quest]

sure

[daring salmon]

i think i know how to do a

[carmine quest]

what did you get for a

[daring salmon]

40

[carmine quest]

yes that's right

lets check out b

[daring salmon]

let me try b

i got 0/16

would that just be 0

[carmine quest]

yes

i got that too

[daring salmon]

can we do this?

i have never done this type of problem

also thank you for helping me

[carmine quest]

np

[carmine quest]

i guess you'll have to make y the subject

[daring salmon]

so replace 0 with y

[carmine quest]

no

no

like isolate y

in this equation

[daring salmon]

oh okay

so do i just move all the y values to the other side

[carmine quest]

yes

but there's one value with both x and y

[daring salmon]

do i divide by y

((x^2)y) - x -11 = 5y

[carmine quest]

they wouldn't be useful tho

[daring salmon]

what do i do

[carmine quest]

tbh idk

im trying to isolate y but that isn't going very well

[daring salmon]

i have the answer if you want it

i just dont have the steps

[carmine quest]

what's the ans

[carmine quest]

wait

is this multivariable calculus?????

[daring salmon]

no idea

[carmine quest]

yeah it must be

i haven't learnt this stuff yet

you'll have to ask a more advanced person

[daring salmon]

do you have time to go over something else?

[carmine quest]

yes

im on holidays

what is the question

[daring salmon]

i need help with both problems

[carmine quest]

ok let's see

so for 1) you find the deravative of the function V

[daring salmon]

how

[daring salmon]

70((1.2)^t) ln(70(1.2))

[daring salmon]

ohh

so only 1.2 is a

[carmine quest]

yes

[daring salmon]

why not 70

because no exponent?

nvm it makes sense

[carmine quest]

yeah

so that should be the ans

[daring salmon]

yes

[carmine quest]

is that right?

[daring salmon]

yeah

lets do the second problem

[carmine quest]

have you tired it?

[daring salmon]

idk how to even start it

[carmine quest]

so part a of q2 is asking for the gradient of f(x) when x=4pi/3

so you need to find the deravative of f(x)

[daring salmon]

3 -sin(x) - 2 cos(x)?

[carmine quest]

correct

[daring salmon]

and replace x with 4pi/3?

[carmine quest]

yes

[daring salmon]

would i be able to plug all of that in a scientific calculator

[carmine quest]

well you can do it manually

[daring salmon]

oh oka

so 3 -sin(4pi/3) - 2 cos(4pi/3)

[carmine quest]

do know how to do that??

[daring salmon]

nope

[carmine quest]

ok so this might be confusing

have you learnt the unit circle

[daring salmon]

yes

[carmine quest]

ok

then you must know this right???

[daring salmon]

i wouldnt remember it

[carmine quest]

so it goes like this

[daring salmon]

cant a calculator just tell me or would it give me a decimal

[carmine quest]

are calcs allowed in your exam???

[daring salmon]

ye

[carmine quest]

then you can just use calc i guess

yeah i checked, the ans is the same

[daring salmon]

it gives me a decimal

[carmine quest]

3.598????

[carmine quest]

can your calc not convert to exact form????

btw you need to put calc in radian mode

[daring salmon]

how do i do that

also the answer doesnt have radians in it i think

[carmine quest]

wdym

4pi/3 is in radians

[carmine quest]

put calc into radians and then enter this to get the ans

also it's -3sin(4pi/3) -2cos(4pi/3)

[carmine quest]

that would depend on the calc

your test is tmr and you've never used radian mode in your calc????

[carmine quest]

what's the ans???

[obtuse pebble]

@daring salmon Has your question been resolved?

[daring salmon]

sorry im back

these are the answers

[daring salmon]

nope

we didnt work with radians this entire class

[carmine quest]

well then you can convert 4pi/3 to degrees

by timsing by 180/pi

which would give 240

and you will get same ans

if you sub x=240 in degree mode

[daring salmon]

oh i figured out how to put it in radian mode

and it gave me exactly the answer

lets do b

[carmine quest]

so for be the gradient at the point is the answer to part a)

3.6 is the decimal version

[daring salmon]

4pi/3 = 3.6?

[carmine quest]

no no

3sqrt3 +2 / 2 = 3.6

[daring salmon]

is the answer from part a = 3.6

ah okay

[carmine quest]

yes

so tangent is y=mx+c

y=3.6x + c

to find c we need the y-ccordinate at 4pi/3

so sub x=4pi/3 into f(x) to find y-coordinate

[daring salmon]

i got 6.2321

[carmine quest]

that's right

[daring salmon]

its not the same thing as the answer given

wait

what

[carmine quest]

9 + 2sqrt3 /2 = 6.2321

so you use that to find c

and then you're done

do you understand?????

@daring salmon

[carmine quest]

This is the y coordinate at 4pi/3

So (4pi/3 , 6.23)

Using this point find the c value as we already have the gradient from part a

[daring salmon]

sorry im back

no more interruptions

yes that is my answer

[carmine quest]

So you got c as -8.848???

[daring salmon]

yes

[carmine quest]

Nice

[daring salmon]

do you know L'hopital rule

[carmine quest]

Id think so

I'm sorry but I have to go now

[daring salmon]

ahh no worries

thank you for helping me

[carmine quest]

My pleasure

Hope you do well in your exam tmr

[daring salmon]

thanks

enjoy your holidays

[carmine quest]

Thnx

[daring salmon]

.close

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[frosty brook]

https://media.discordapp.net/attachments/499124802154659841/1096053110184624189/image.png?width=1207&height=160

[frosty brook]

how do i even approach this

i'm confused with the fact that there's two variables

the answers still proved this by induction (base case, assume n = k, n = k + 1) but i don't get it

[jolly ginkgo]

It's a combinatorics problem

You have to show that \begin{align*}{n \choose 0} + {n+1 \choose 1} + {n+2 \choose 2}+ \dots + {n+r-1 \choose r} = {n+r \choose r}\end{align*}

[warm shale]

fäf

[frosty brook]

???

combinatorics

[jolly ginkgo]

What are you struggling with?

[frosty brook]

are you certain?

this topic isn't on combinatorics

it's proof by induction

[jolly ginkgo]

Yes it can be done by induction

[frosty brook]

yeah

so how would i do the base case?

like

for n = 1

because there are two variables

n and r

[jolly ginkgo]

r will be considered as a constant in this

[frosty brook]

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[wise lava]

I am not sure if I did the (e) part correctly
Can anyone check for me

[obtuse pebble]

@wise lava Has your question been resolved?

[obtuse pebble]

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[scenic fjord]

Can anyone tell me how find this integral?

Is it possible to work from 0 to positive infinity?

[warm canopy]

Where did you get the problem from?

[scenic fjord]

I’m searching how to derive that because in my textbook, it’s only given the answer

[somber sedge]

Short answer -> there is no such function

[scenic fjord]

https://jakubmarian.com/integral-of-exp-x2-from-minus-infinity-to-infinity/

[warm canopy]

The standard derivation requires you to square the integral and use polar coords

[somber sedge]

But there is a value

[scenic fjord]

And I refer to this

[scenic fjord]

How do you know it will involves polar coordinates?

[warm canopy]

Because I've seen it before

This is a pretty famous integral

[scenic fjord]

Like this, what’s the purpose of doing this? Is it just because we wanna solve by polar coordinates?

[warm canopy]

Pretty much

[somber sedge]

I believe 3blue1brown recently released a video deriving this exact integral, if you prefer that form

[scenic fjord]

And only can be solved by -ve infinity to +infinity?

[warm canopy]

You get -(x²+y²) which is begging to have polar coords used on it

[scenic fjord]

Cannot 0 to positive infinity?

[warm canopy]

The function is even so from 0 to inf is just half of -inf to inf

[scenic fjord]

Yea, so why can I directly times with the second I?

[warm canopy]

Why not? You're just squaring something

[somber sedge]

Only because it is symmetric at x = 0

[scenic fjord]

Oh
In YouTube?

[somber sedge]

Yes

[warm canopy]

? I said it's even

[scenic fjord]

Wait, I thought is times 2?

Anyway, this can be only solved by polar coordinates and no other method can be use already?

[warm canopy]

You should just look into it

It's called the gaussian integral

[scenic fjord]

Oh, got the video

Thanks

.close

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[left needle]

How could you solve this system of bitwise equations by hand

X OR (Y XOR 111000) = 110111 ```
im not sure if there is a "trick" to solve these faster because what im doing takes very long

[viral blade]

A XOR B = C

there's actually an operation you can perform to both sides to isolate A

[left needle]

how?

but also i mean in general could someone show me, or send a video because i cant find it, how you would solve these equations in the most efficient way

what?

[trim portal]

isnt XOR it's own inverse?

[trim portal]

If you are doing it by hand, just write
abcdef XOR 100001 = 001110,
now a XOR 1 = 0, so a must be 1
b XOR 0 = 0, so b=0
c XOR 0 = 1, so c=1 etc...

[left needle]

i think so

[left needle]

so no faster way other than to write out bit by bit and solve the system for each bit

[trim portal]

you could also just do 100001 XOR 001110

since XOR is its own inverse

[left needle]

wait what

[trim portal]

(X OR Y) XOR 100001 = 001110, so (X OR Y) = 100001 XOR 001110

since inverse of XOR is XOR

[left needle]

ooh

i didnt know you could do that

is there a proof for this bc i dont really understand why it is true?

[trim portal]

You could just construct proof for single bit xor operation

[left needle]

oh i see cool

[left needle]

and once you have this you also have to go bit by bit and check each equation right?

there isnt a way to solve it like a linear equation, extract x and y

[trim portal]

I think there would be more than one solution

[left needle]

i mean yea makes sense bc function or doesnt have inverse

[left needle]

yup

in this specific question you gotta find pair (x,y) which has the most zeros in total

for that you dont really have to go bit by bit

but i wanted to know in general if there was a quick way to solve these equations/systems of equations

[obtuse pebble]

@left needle Has your question been resolved?

[obtuse pebble]

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[warped sinew]

Can anyone help me with this question?

So far, I've tried replacing f(x) and e^x with their power series representation. So x^n/n! for e^x and f(x). I'm not sure how to actually get a_0 or a_1 though. f(0) = a_n and f(1) = summation of a_n.

[obtuse pebble]

@warped sinew Has your question been resolved?

[obtuse pebble]

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[dapper venture]

What does it mean in the second picture when it says if a = 1 the complementary function is a constant so you need a solution without any constant terms?

If a = 1 it would just be un = un-1 + g(n) but un-1 is not a constant as it changes with every iteration?

[raven badger]

this is for in homogeneous linear recurrences right?

[dapper venture]

Yes

solving first order recurrence relations

this is specifically non-homogenous linear recurrences where a is not equal to 1

[raven badger]

offtopic: i dont remember if this covers it but https://www.youtube.com/watch?v=TWBB-JlmYUc this is where i learnt linear recurrences from. pretty great lecture. start from 57:00 for in homogeneous iirc

[unreal musk]

Well for the homogenous version, you'd have $u_{n} = u_{n-1}$ and that would be constant

[dapper venture]

Tysm I will look at it now

[warm shale]

@unreal musk

[dapper venture]

howcome that is regarded as a constant

$u_{n} = 2u_{n-1}$ is not regarded a constant?

[unreal musk]

for subscripts

u_{n-1}

[dapper venture]

Sorry?

wdym

[raven badger]

u_{n-1}, LaTeX for subscripts

[unreal musk]

To make them appear in your LaTeX here, type them as

u_{n} = 2 u_{n-1}

[dapper venture]

$u_{n} = 2 u_{n-1}$

[warm shale]

Aorliei

[dapper venture]

oh thank you

[warm shale]

Aorliei

[unreal musk]

That one isn't constant as your terms double each time, but $u_{n} = u_{n-1}$ is because from term to term, your sequence doesn't change

[warm shale]

@unreal musk

[unreal musk]

The next term is the same as the last and so on and so forth

[dapper venture]

Oh ok thank you that makes a lot of sense

so why does that mean our particular solution has to be one with no constant terms?

[dapper venture]

^ as said before

[obtuse pebble]

@dapper venture Has your question been resolved?

[obtuse pebble]

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

[tepid urchin]

what is up party people. what is this abomination

[timid silo]

oh my

[tepid urchin]

yes

[timid silo]

it's the fucking $\pi$ that is annoying

[warm shale]

sppetsnaz

[timid silo]

otherwise I would have ideas

[tepid urchin]

y

<@&286206848099549185>

[timid silo]

oh sorry this a lot easier than I thought

[tepid urchin]

lol

[timid silo]

wait I'll write a solution

[tepid urchin]

ty

[timid silo]

the taylor expansion for the sine function is :

plug in x=pi/4 and you're done

the sum is sin(pi/4)

[tepid urchin]

The sum is 1/sqrt2

[timid silo]

yep

[tepid urchin]

huh

ok

how

[timid silo]

did you study taylor expansions ?

[tepid urchin]

nop

[timid silo]

wow really

by applying this shit to the sine formula

[timid silo]

we get this

[tepid urchin]

ok

so how far do I go

infinitly?

[timid silo]

yes it's infinity

an infinite sum

[tepid urchin]

how do I find the sum?