#help-13

428200 messages · Page 525 of 429

plucky pawn
#

this is calc 2????

unkempt aspen
#

I only ruled it out because it doesn't really make sense

unkempt aspen
plucky pawn
#

OOOOOOOOOOOOOOO

unkempt aspen
#

relative rates of growth is in our cal 2 class for some reason

plucky pawn
#

I thought calc 2 was integration

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not this but idk

unkempt aspen
#

it is mostly

plucky pawn
#

get ready integration is a whole new ball game compared to diff

unkempt aspen
#

i have to finish out this section then do integration by parts by tomorrow

#

oh we're already well into integration

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they just have an ungodly scheme for this class

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satan made out syllabus

plucky pawn
#

sounds like my calc 1

unkempt aspen
#

you de facto have to have a 100% on every homework to do the quiz, and without doing the previous quiz you can't look or see what the next topic is

plucky pawn
#

I swear proff covering calc 2 in calc 1

unkempt aspen
#

i wouldn't have minded 1 semester for the entirety of single variable calc

plucky pawn
#

I couldn't care for it much... just gotta take em as a preque

unkempt aspen
#

alright so according to symbolab which I'm now learning has a limit function (a funky tool which will help with tomorrow's exam) this evaluates to infinity

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which i knew had to be the answer due to nlnx being slow as hell from comp sci

#

mystery solved boys

#

.close

cedar kilnBOT
#
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unkempt aspen
#

.reopen

cedar kilnBOT
#

unkempt aspen
#

actually let me explain WHY, work-wise this is the answer

#

constant multiple rule makes it 4/3 * the limit

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x^2/xlnx

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x^2/xlnx = x/lnx

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1/1/x

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4/3*inf = inf

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hence the function is slower

#

.close

cedar kilnBOT
#
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cedar kilnBOT
#
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mental summit
#

Hello so i'm trying to reduce a fraction (i'm not sure that's the correct term)
but the fraction is:
x^2-2xy

2y^2-xy
The answer i got is:
x

y
The correct answer is:
-x

-y
The way i solved:
x(x-2y) x
--------- = ---
y(2y-x) y

-Probably missed an important step or rule but i'm lost to why it's minus and not plus

cerulean star
mental summit
#

Sorry my bad

cerulean star
#

$\frac{-x}{-y}$

wraith daggerBOT
#

Disorganized

mental summit
#

it's just - x/y

#

That's the correct answer (according to the answers in the book i'm learning)

cerulean star
#

Ok whats the issue

mental summit
#

i got x/y without the minus

cerulean star
#

You just didn't show one step

#

Under "way I solved"

#

2y-x is the negative of what you need

#

So

#

Mutliply the denominator by two factors of -1

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And distribute one of them INTO the parentheses

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This will let you write the parenthesized part backwards

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And it will match the factor in the numerator

mental summit
#

oh i can't substract what's in the parentheses before i match them to look alike? (bad english sorry)

cerulean star
mental summit
#

i didn't know it's not possible if it's not the same

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i thought i could just substract what's in the parentheses

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i'm sorry if i sound stupid, but what is the way to make (x-2y) into (2y-x)

cerulean star
#

First multiply through by -1

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(Do it, write it)

mental summit
#

-1(-2y+x)

cerulean star
#

Uh uh and that becomes?

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Wait no

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You are doing it all at once

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This is correct

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So it's fine

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Moving fast then, lol

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What you have done

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Is multiplied by (-1)(-1)

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...which is just 1

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But one of those -1's

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...

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Wait

#

No.

#

Let's go slower

#

Start here

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(-1)(x-2y)

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Distribute the -1

mental summit
#

-x+2y

cerulean star
#

Now commute the terms

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(Commutative property)

#

@mental summit

mental summit
#

commute as to explain what i did ?

cerulean star
#

No, "commutative property", Google it

mental summit
#

-1x+(-1)-2y

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ughhhh i hope that's what u meant

cerulean star
#

No, Google "commutative property" so you remember what it is

#

You need to know it

#

$a+b = b+ a$

wraith daggerBOT
#

Disorganized

cerulean star
#

Commutative property (of addition)

cerulean star
mental summit
#

so -x+2y=2y-x

cerulean star
#

Yes

#

And this is the factor you needed

#

BUT

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We have changed what we started with

mental summit
#

no no you right that was important for me in general

cerulean star
#

Because we multiplied by -1 in the denominator only

#

We can fix this

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Just by multiplying by another factor of -1

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It won't matter if we do it in the numerator or denominator

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That will make the net result the same as if we had multiplied by 1 in the first place (since (-1)(-1)=1)

#

So your denominator could look like

-y(x-2y)

#

Now you can cancel stuff

mental summit
#

i kinda understood what you said, but by what you've said to me about commutative property

#

isn't y(x-2y) would be the same ?

shut reef
#

man I was confused

mental summit
#

i just having hard time to realise why it's not x/y and it's - x/y

past orbit
#

u had
x(x-2y)

y(2y-x)

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x-2y <> 2y-x

mental summit
#

Yes and just substracted

past orbit
#

x-2y = -1* (2y-x)

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when u cancel those 2 out, ur left with a negative sign

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so it becomes -x/y

mental summit
past orbit
#

nope

#

a-b is not the same as b-a unless both are 0 or the same value

#

so the reason it doesnt work

#

is because the commutive property states that a+b=b+a

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but a stands for ANY number

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that means u have to bring the negative sign into it

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so if a is negative thats fine

#

u would just have (-3)+(4)=4+(-3)

#

not 3-4 which is what u did

mental summit
#

Thank you very much for the explanation

cedar kilnBOT
#

@mental summit Has your question been resolved?

mental summit
cedar kilnBOT
#
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past orbit
#

yes exactly

cedar kilnBOT
#
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mental summit
#

that's why @cerulean star brought up commutive property, sorry for not understanding what u meant, thanks both !

worthy path
#

u have no question @mental summit

cedar kilnBOT
#

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cedar kilnBOT
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cedar kilnBOT
tropic oxide
#

the top one is in REF and the bottom one is in RREF, yes.

#

does your textbook forbid non-1 leading coefficients?

#

show me the definition of REF as given in your textbook.

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row echelon form, yes.

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orifyoulikeomittingspacesbetweenwordsit'srowechelonformyes

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you can and should check it yourself lol

#

okay then let's walk through the definition step by step

#

_Recall that a matrix is in row-echelon form if:

  • Rows of all zeroes are at the bottom.
  • The first entry of any nonzero row is a 1.
  • All leading 1's are to the right of the leading 1's in the rows above it._
#

is it true that in your matrix all rows consisting of nothing but zeros are at the bottom? Y/N

#

@plain bridge are you still here?

#

actually you're wrong

#

it is true that all rows of zeros are at the bottom

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precisely because there are no rows of zeros

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just to be clear, what you're wrong about is your assessment of the first criterion as false.

#

the first criterion is true; it's the second which fails, for a reason you have already mentioned.

#

no, wrong again

#

how is the top one in ref when there's a leading 5
in the third row

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we've been talking about the first one all along.

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yes, it is neither.

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yes, you were wrong about the reason.

#

yes

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what's wrong with them?

#

oh, they made a typo

#

you made it seem like there was something wrong with it fundamentally

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there's a typo in either the rref or the parameterized solution

#

ok then there's a typo in the solutions

cedar kilnBOT
#

@plain bridge Has your question been resolved?

#
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cedar kilnBOT
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crimson obsidian
cedar kilnBOT
crimson obsidian
#

Which one of the above 2 images is the correct version?

#

First one or second.

#

I just learned this determinant and properties so I'm not confident

random shale
crimson obsidian
#

It doesn't match

random shale
#

Yes

crimson obsidian
#

Is not the double

random shale
#

$det(2A) = 2^ndet(A)$

wraith daggerBOT
random shale
#

Where n is the dimension of A

crimson obsidian
#

What is dimension?

random shale
#

It is a 2 by 2 matrix

crimson obsidian
#

Ohh

random shale
#

So dimension is 2 in this case

crimson obsidian
#

I see

#

Which is 4 delta A

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Right?

random shale
#

Yea

#

but not delta

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Det(A)

crimson obsidian
#

That triangle is delta right?

random shale
#

You can prove this for any number lambda using the fact $det(A\times B)= detA \times detB$

wraith daggerBOT
crimson obsidian
#

I see

#

Its kind of not taught yet

random shale
#

$det(\lambda A) = det((\lambda I) A) = det(\lambda I) \times det(A)$

wraith daggerBOT
crimson obsidian
#

So I was having trouble but thanks

random shale
#

And the determinant of I is the product of the diagonal and the product of the diagonal is just $\lambda^n$ and the result follows

wraith daggerBOT
crimson obsidian
random shale
crimson obsidian
#

I didn't learn them but thanks

#

.close

cedar kilnBOT
#
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crimson sedge
#

do i just have to plug the numbers in or is a different?

elder pewter
#

wdym plug numbers in?

#

your mass defect is when the LHS is not equal to the RHS

#

you get a difference in mass

crimson sedge
#

oh

#

Then how would i do this

crimson sedge
#

im a bit lost

elder pewter
#

first calculate the total mass on the left

#

in terms of u

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then calculate the total mass on the right in terms of u

#

find difference

#

that is your mass defect

crimson sedge
#

Oh okay

crimson sedge
elder pewter
#

you have been given mass of everything

#

so just add them up

crimson sedge
#

oh okay

#

well i got 1.25 x 10^-1

elder pewter
#

u?

crimson sedge
crimson sedge
#

the mass defect

elder pewter
#

I got that too

#

as in the units is u

crimson sedge
#

Ohh okay

#

now for the yield am i suppose to (1.25 x 10^-1) (

#

what am i suppose to be multiplying it by?

elder pewter
#

do you know the energy mass equation

#

famous one by einstein?

crimson sedge
#

Yea

#

I got it

#

I got 1.17045 x 10^16

elder pewter
#

that doesn't look quite right

#

did you convert to kilograms first?

crimson sedge
elder pewter
#

need to do that first

#

mass must be in kilograms first

crimson sedge
elder pewter
#

yes

#

because that is currently in units of u

crimson sedge
#

ohh okay

cedar kilnBOT
#

@crimson sedge Has your question been resolved?

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gritty mulch
#

how many sides has a regular polygon whose number of sides increases by 2 when its angle increases by 9 degrees

gritty mulch
#

I'll be back in a few minutes

crimson sedge
#

write it out as an equation

#

will help

gritty mulch
#

sorry but how?

livid hound
#

no

cedar kilnBOT
#

@gritty mulch Has your question been resolved?

livid hound
#

its a little more convenient to work with exterior angles rather than interior as the angle sums of those is constant

#

also don't just hang in the background otherwise i'm gonna assume you're not there

gritty mulch
#

the answer is 8

#

if anyone can find that solution

livid hound
#

start by introducing a variable to denote the number of sides in your original polygon

#

or don't...

gritty mulch
livid hound
#

sarcasm

gritty mulch
#

ok

#

heh

livid hound
#

i'm attempting to guide you through the problem

#

and that takes effort and responses from you

#

otherwise i'm potentially gonna bail and do other things

gritty mulch
livid hound
#

pick a letter/symbol or whatever

#

and use that to represent the number of sides of the original polygon

#

let "insert choice here" be the number of sides of the original regular polygon

#

at this point all i really want from you is to pick an appropriate letter/symbol

#

which is pretty much standard for all questions where you're trying to find an unknown and no variable has been given to you (so you introduce one yourself for ease of calculations)

gritty mulch
#

.close

cedar kilnBOT
#
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bright sorrel
#

very quick q: is (5-x)^2=(x-5)^2

cedar kilnBOT
bright sorrel
#

it is right?

sacred stone
#

yes

bright sorrel
#

ok ty

#

.close

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cedar kilnBOT
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knotty tartan
#

.close

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regal fog
#

I know (a^b)^x is (a)^b×c but what about

cedar kilnBOT
regal fog
#

(a^b×k^r)^x

crimson sedge
#

use the product of power rule to move the power of x in, then use power of power rule

cedar kilnBOT
#

@regal fog Has your question been resolved?

regal fog
#

Thanks.

cedar kilnBOT
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fossil thorn
cedar kilnBOT
fossil thorn
#

Can someone help me understand why this is incorrect

still hare
#

You multiplied with positive number 3 ... No need to change the inequality sign.

#

In 2nd last step

fossil thorn
#

But aren't we multiplying (-3*3)?

#

So there is a negative number

still hare
#

Basically you are multiying +3 on both sides. So the inequality doesn't change

fossil thorn
#

Oh okay so if it was X/-3 then we would change signs

cold briar
fossil thorn
#

Kk tysm

#

.close

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late halo
#

Can anyone tell how it was solved that x,y,z was obtained?

dire geode
late halo
dire geode
#

.close

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urban plover
#

hello can anyone check my answer

cedar kilnBOT
urban plover
balmy apex
#

$\bar{z}$ denotes the complex conjugate of z?

wraith daggerBOT
#

illuminator3

urban plover
#

you can see under the second picture

crystal raptor
urban plover
#

so nothing?

#

<@&286206848099549185> ?

vernal shell
# urban plover

It looks correct to me 🤔, last line is 4abi + 2a-b = 8i, right?

urban plover
#

yeah

#

but I have no clue what should I do next because the instructions say define every complex number of z for which is true:

vernal shell
#

The real part of 8i is 0

#

What's the real part of the left-hand side of the equation?

urban plover
#

Umm

#

I don't know honestly

cedar kilnBOT
#

@urban plover Has your question been resolved?

cedar kilnBOT
#

@urban plover Has your question been resolved?

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round magnet
#

i need to get A valor & B valor

cedar kilnBOT
shut reef
#

no way ..

hazy minnow
#

Multiply both side by x^2 - 1.

#

Recall x^2 - 1 = (x+1)(x-1).

#

The rest follows easily since your dealing with two polynomials that are equal.

round magnet
#

I need to solve smth like this?

hazy minnow
#

Yep, try expanding stuff out.

#

In standard form.

round magnet
#

Oh

hazy minnow
#

We have two polynomials over R that are equal.

#

What should that tell you?

round magnet
#

um?

urban plover
#

yeah exactly why are people here answering in fingquestions

cedar kilnBOT
#

@round magnet Has your question been resolved?

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cedar kilnBOT
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torpid fulcrum
#

can you get x+2z=0 and y+z=0?

hazy minnow
#

This ^ is it.

#

z is the free variable.

#

So write x = -2z and y = -z.

#

The solution space is (-2c, -c, c) for any scalar c.

#

That is one solution.

#

Did you mean c(-2,-1,1)?

#

@plain bridge

#

Yes.

#

“free non leading variable” trying to parse this.

#

If the system you have has no solution then it’s not true. But if a solution exist your statement is false.

#

False I meant.

#

Yep.

#

Sure.

#

rank means number of linear independent columns, which is the columns that have pivot entries, so yes we have one free variable.

#

Remembering my past LA course, I am kinda of rusty. Currently relearning.

#

(c,c,c,c,c) yes.

#

Not sure what you mean by “making our answer just 1”.

#

Ok then yes.

#

Are we talking about the same matrix but doing two different elementary operation on them?

#

If not then yes the system of equation are not the same.

#

If so then it will have the same solution space not matter what elementary operation you do.

#

But it possible the two matrix might be row equivalent.

hazy minnow
#

Yep.

#

Unless we know the two different system are row equivalent.

#

Which we can check by row reducing.

hazy minnow
#

A solution exist.

#

@plain bridge

#

Yep.

#

x = 1, y = 0, z = 0.

#

Yep.

cosmic steppe
#

x = 1, y = 0, z = 0 if you're using x y and z

hazy minnow
#

Why’s did you think a solution did not exist?

#

Yes.

cosmic steppe
#

I guess it works

hazy minnow
#

Even if we had 1 on the last rows of the last column we still have a solution.

#

Can you explain why you still think for this case that a solution does not exist?

#

If we had that.

#

If the last row was (0,0,0,1) then we have no solution.

#

Is that what you meant?

#

np.

cedar kilnBOT
#

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restive tapir
#

quick question, how do i transform a multiplication to a fraction form?

restive tapir
#

id like to transform sin(x)*ln(x) to a fraction so i can apply the hospital rule

dire geode
#

$y = 1/(1/y))$ where $y=sin(x)$ or $\log(x)$

wraith daggerBOT
#

riemann

dire geode
#

Probably the former

#

Since $1/\sin(x)=\csc(x)$

wraith daggerBOT
#

riemann

restive tapir
#

just to confirm that i got it right because i felt lost for a good minute, transform sinx to 1/cscx, multiply by lnx = lnx/csx then put back the sin or just keep as is correct?

#

basically ln(x)/csc(x) or ln(x)/1/sin(x)

cedar kilnBOT
#

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vast wolf
#

The letters of promise are arranged randomly in a row. Find how many solutions there are if the letter p and r are seperated by at least three letters

vast wolf
#

Anyone able to help

#

It's a permutation since it has order to it

#

Should I somehow find out when p and r are in 2 letters or less in proximity then subtract that from 7!

#

🤷‍♂️

cedar kilnBOT
#

@vast wolf Has your question been resolved?

cedar kilnBOT
#

@vast wolf Has your question been resolved?

sick ruin
#

@vast wolf I think you should go for the complement here

#

Take the number of rearrangements that have p and r separated by exactly 0, 1, or 2 letters, and subtract that from all possible rearrangements

#

Hmm

#

Maybe there’s something better here

vast wolf
#

Yeah

#

It's okay haha

#

I've given up on it and just asked my teacher

#

.close

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balmy beacon
#

Hi. I want to prove that a + b = a' + b' mod d.

Please tell me if this is correct:

So let be:
a, b, a', b' in Z and d in Z+
with (a = a mod d) and (b = b' mod d).

Then let be k, k', k'' in Z
(a = a' mod d) <=> (a = a' + kd)
(b = b' mod d) <=> (b = b' + k'd)

So
a+b = a' mod d + b' mod d
= a' + kd + b' + k'd
= a' + b' + kd + k'd
= (a' + b' + k''d)
= a' + b' mod d

Is thiss proof correct?

cedar kilnBOT
#

@balmy beacon Has your question been resolved?

torpid fulcrum
#

assuming that k''=k+k', the proof looks good

#

just as a side note, the "a' mod d + b' mod d" isn't strictly necessary

#

a+b = a' + kd + b' + k'd is sufficient

balmy beacon
#

thank you @torpid fulcrum !

#

I'll close this then

#

.close

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craggy anchor
cedar kilnBOT
crimson sedge
#

Ooo fancy question

#

I like it

#

Build systems of 180-a

#

180-b

#

Bla bla

#

And you will get it

craggy anchor
#

cant i just use exterior angles of polygon add to 360? i know that if i can get to x-180 + a + b + c = 360 then x= 540-(a+b+c) but i dont know what that last angle is

#

your method does work tho so ill just use that ig

#

.close

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crimson sedge
#

How to increase your chances of winning at games of chance?

crimson sedge
crimson sedge
fair geyser
#

not every game is gameable, and it wouldn't be the same strategy anyway

crimson sedge
#

I know I'm going to have to calculate the probability of winning the games and see if it's over 50% etc but I don't know where to start

crimson sedge
balmy apex
#

choose coin flip

#

ez

crimson sedge
#

Coin flip game I see

fair geyser
#

it doesn't make sense to do that no

crimson sedge
#

Do what?

fair geyser
#

oh you mean you are going to calculate it before choosing the game and then just play it?

crimson sedge
#

Yes, it can be a benefit over a certain number of rounds or in general

crimson sedge
fair geyser
#

i meant like, it doesn't matter if your chance to win is 58% if you pay x to win 1.75x

#

something like that

crimson sedge
#

But, if the winrate is 58% if I play a lot of times I'm sure to make a profit

fair geyser
#

no, that depends on what the prize is and what you pay to play

crimson sedge
#

Ye but in most of game u pay less than the profit right

fair geyser
#

of course you pay less than the prize, it's not enough

crimson sedge
#

The probability of winning is 53% and we pay 2€ we can win 3€ if we play 1000 times we are sure to make a profit I mean

fair geyser
#

yeah that's what i think you mean

#

that loses you money big time

crimson sedge
#

So, u make profit right

fair geyser
#

you lose

#

what do you mean

#

why would i use words "that loses you money big time"

crimson sedge
#

How u lose money

fair geyser
#

are you used to people saying the opposite of what they want to say lmao

crimson sedge
#

english not my native language haha

balmy apex
fair geyser
#

you pay 2€ 1000 times

#

you win 530 times, -2000 paid, 1590 won

crimson sedge
#

wtf am i that dumb

#

Ye i mean ur right

#

forgot that the gain must be above x2 at least

fair geyser
#

slightly below x2

#

50% and x2 means you can play for fun, it's free

#

53% and x2 means you make profit, and it can be a bit below x2, still profitable

crimson sedge
#

Oh ye

#

depend of the winrate

#

so how can u know the profit u need to make with the winrate

fair geyser
#

when you win you gain (prize − cost), this happens 53% of the time, rest of the time you lose the money, you can write an equation
0.53((multiplier * x) - x) = 0.47x

#

when it's that, you don't lose

#

1.89x is below 2x

#

so you want the multiplier to be 1 / winrate

#

,w 1/0.53

wraith daggerBOT
cedar kilnBOT
#

@crimson sedge Has your question been resolved?

cedar kilnBOT
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craggy anchor
cedar kilnBOT
rotund vine
#

what have you done

#

any thoughts or work

craggy anchor
#

well i constructed line CA thats literally it but im pretty sure i cant say its parallel to qp cause it wants me to prove that so im not sure

cedar kilnBOT
#

@craggy anchor Has your question been resolved?

craggy anchor
#

<@&286206848099549185>

elder pewter
#

do you know the requirements for proving similar triangles

craggy anchor
#

well i usually just use equiangular

craggy anchor
#

<@&286206848099549185>

velvet mortar
#

BQ/BC = BP/BA = 1/2

craggy anchor
#

does that prove that the triangles are similar?

#

that doesnt prove that pq is similar to ca does it?

craggy anchor
#

can anyone confirm

#

<@&286206848099549185>

cedar kilnBOT
#

@craggy anchor Has your question been resolved?

craggy anchor
#

.close

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vagrant current
cedar kilnBOT
vagrant current
#

will i have to find the distance between every point

#

or the gradient

#

since 2 sides

#

need to be equal

tropic oxide
#

you can either find the gradients of your lines and show that they're equal or you can find the distances and show those are equal

#

it's entirey up to you

#

you can do both if you're really paranoid

vagrant current
#

Ohh right , since if the gradients are the same it gotta be parallel and parallelograms have parallel opposite sides

#

didnt think i could use the gradient for that tbh

#

ty for help

#

.close

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cedar kilnBOT
lusty tiger
#

Yes

balmy apex
#

yes

lusty tiger
cedar kilnBOT
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bold snow
#

how do I prove that OBB' is congruent to AXX'?

bold snow
#

OABX is the original parallelogram, with a shear applied to it by some multiple of OA

silent yew
#

@bold snow can u prove that the 2 triangles are similar?

#

U can easily prove congruency after proving the similarity

bold snow
silent yew
silent yew
#

So from AA property, the two triangles OBB' and AXX' are similar

#

Now since it's a parallelogram, OB = AX

#

So the respective sides are in a ratio 1:1

#

Therefore triangle AXX' is congruent to triangle OBB'

cedar kilnBOT
#

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bold snow
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low rose
#

Hi, I am not sure how to solve these problems, but i attempted and got 6591 for the second one. Can someone check it?

crimson sedge
#

Do you know

#

The properties of powers

#

@low rose

low rose
#

ye

crimson sedge
#

Try to write them in base forms first

#

Do you know how to write square root as a power

low rose
#

i got the answer now

#

.close

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crimson sedge
#

What exactly is he doing here, what does he mean by "taking x * h^(n-1) n times" and "selecting h from atleast two factors"

crimson sedge
#

This seems like another simple thing worded confusingly which is what always gives me the most headaches

sage forge
crimson sedge
#

Oh, it's that thing again

#

Ok i think i know what i need to reread

sage forge
#

Are you familiar with it?

crimson sedge
#

Well, i forgot what the two numbers ontop of eachother mean but i know what the sigma means

#

Maybe that's a sign i should reread the previous book

#

It'll be easier to retain everything

sage forge
#

Those are binomial coeffizients

crimson sedge
#

.close

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sand sedge
#

Can I get some help with this

cedar kilnBOT
sand sedge
fierce shard
#

Find the y intercept and the gradient

velvet mortar
#

Do you know y=mx+b?

#

What is m and b ?

sand sedge
#

I actually go through it thank you ☺️

#

Now I’m struggling with another one

velvet mortar
#

What have you tried

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#

@sand sedge Has your question been resolved?

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cinder sierra
#

I'm writing a short paper where I analytically prove a few theorems of elementary arithmetic about the divisibility relation n | m \↔️ ∃p(m = np), prior to introducing division with remainder and/or its related algorithms, as a proof of concept for a friend. right now I'm trying to prove a lemma of Euclid's lemma, to in turn prove the uniqueness part of the fundamental theorem of arithmetic

the lemma is that, given nonzero integers m,n,p,q where p,q are coprime and mq = np (i.e. m/n = p/q), there is an integer k such that m = pk and n = qp (i.e. m,n are multiples of p,q by the same integer). I know that, if q | n, then the theorem follows (because then n = qk for some k, in which case mq = pkq, and so m = pk), so in turn I just have to prove that q | n, but I don't know where do I even begin

so far I've proven that divisibility is a pre-order over the nonzero integers having 1 as a minimal element over the positive integers, that every composite number is a product of primes (though not the uniqueness of this product, of course), and that the integers resulting from the division of m,n by their gcd are coprime (i.e. that if m,n are nonzero, and if m = gcd(m, n)p and n = gcd(m, n)q, then gcd(p, q) = 1)

mighty drift
#

To prove q|n, use mq=np. This means q|np, but since q and p are coprime, it means (Gauss's lemma) that q | n

cinder sierra
mighty drift
#

Oh

cinder sierra
#

so that would be circular

cedar kilnBOT
#

@cinder sierra Has your question been resolved?

cinder sierra
#

<@&286206848099549185>

cedar kilnBOT
#

@cinder sierra Has your question been resolved?

cinder sierra
#

asking the question once more

#

.close

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bold shadow
#

Hi guys, what equation should I use to build the following graph? The second screenshot is the closest I found so far

balmy apex
#

nvm

#

you can use something like

#

$y = -\frac{2a}{\pi}\arctan(\cot\frac{\pi x}{p})$

wraith daggerBOT
#

illuminator3

balmy apex
#

with a being the amplitude and p being the period

#

such things are called sawtooth waves

crystal raptor
#

also messing with $a\cdot\text{arctan}(\tan(b(x-c)) + d$ will get you there

wraith daggerBOT
bold shadow
#

.close

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rugged palm
#

They gave you two great answers
you can also use the floor() function, but I'm pretty sure that's not what you are meant to use

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odd seal
#

Am I doing this right?

cedar kilnBOT
odd seal
#

whoops

#

.close

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crimson sedge
#

i have one that makes no sense to me, and yes its math not science

crimson sedge
#

a radioactive material has a half-life of 20 Years

#

how radioactive is the material after 80 years, if the value is set to 1 at the year 0?

#

<@&286206848099549185> idk if the question is broken or im dumb

gaunt hamlet
#

What have you tried?

crimson sedge
#

Just different ways of trying to understand it

gaunt hamlet
#

Well, how radioactive is the material after 20 years?

crimson sedge
#

Literely no idea

#

I dont know where to subtract from

#

Like halflife

#

It doesent tell me how much there is, it just says that the value is 1 at year 0

gaunt hamlet
#

What does halflife mean?

crimson sedge
#

That it looses its radioactive power

gaunt hamlet
#

By how much?

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#

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quiet shadow
cedar kilnBOT
quiet shadow
#

How could I continue this?

#

Don't know how to make it meet l'hopital conditions

dire geode
#

log(x^2) = 2log(x)

#

But also it does meet l'hopital conditions

#

Recheck what those are in your notes or textbook

quiet shadow
#

I thought log of infinity was undefined

#

I thought incorrectly I suppose lmao

#

its inf/inf

#

nevermind, thanks @dire geode

#

.close

cedar kilnBOT
#
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frozen tendon
#

guys can anyone telll me how is (b,c) = (b,-c)?

vernal palm
#

we need context

#

what's the theorem

frozen tendon
#

@vernal palm

vernal palm
#

oh damn

frozen tendon
#

the euclidean algorithm

#

its proof is like a nightmare

vernal palm
#

yeah

#

idk if I can help u here

#

sorry

frozen tendon
#

@vernal palm btw one remark is also given

#

idk if its related to it

#

@vernal palm^

vernal palm
#

idk lol sorry

frozen tendon
#

oh np

#

@tropic oxide sry for the ping

#

ig i need help again

frozen tendon
tropic oxide
#

any common divisor of b and c is also a common divisor of b and -c, and vice versa

frozen tendon
#

so it meant that the gcd is always the same

#

?

tropic oxide
#

"the gcd is always the same" is too vague

frozen tendon
dusty hazel
#

The absolute value of gcd is always the same?

tropic oxide
#

perhaps "multiplying either argument of gcd by -1 doesn't change the value"

#

or "gcd doesn't depend on the sign of its inputs"

frozen tendon
#

Guys is it the general form of gcd in which the largen number will always be in the first place and be considered as the value of b in b=aq+r ?

#

@tropic oxide

tropic oxide
#

???

#

"general form"?

frozen tendon
#

I mean why is it that 4840 = b and 1512 = to a like yea it's quite common to divide the bigger number with the smaller one but still I should have some statement or prove in order to make it true right?

tropic oxide
#

......

#

what do you mean?

#

are you looking for a theorem that says "The evaluation of a gcd MUST proceed with the bigger number first and the smaller number second, and in NO OTHER WAY"?

frozen tendon
#

Kinda

tropic oxide
#

well there is no such theorem because such imperatives are very very rare in math

#

i think you're confusing the gcd function itself with the Euclidean algorithm that evaluates it in a certain way.

frozen tendon
#

My question is that why in b=aq+r, b=4840 and a=1512

tropic oxide
#

what "lemme"?

frozen tendon
#

Division algorithm or like what bruuhhhhhh

#

Now it's freaking confusing

tropic oxide
#

there should be a clear difference between "what gcd is" and "how to find gcd"

frozen tendon
#

Ok

tropic oxide
#

one method of finding the gcd of two numbers is to run the euclidean algorithm on them, recursively doing division-with-remainder until the remainder turns to zero at some point

frozen tendon
#

To find gcd I can use the Euclidean algorithm, using b=aq+r where the gcd of a and b can be represented as ax+by also which is Stated in thee bézouts theorem, right?

tropic oxide
#

...no, i never mentioned bézout's theorem...

#

again you are confusing two things that are very different: finding the gcd itself vs. representing gcd(a,b) as a linear combination of a and b

frozen tendon
#

No I mean there is a part of the question that requires the bezouts theorem

#

It's just the half question

tropic oxide
#

im trying to take you through the underlying theory

frozen tendon
tropic oxide
#

bézout's theorem states that such a representation EXISTS

#

but it does not "help" us beyond that

#

the theorem itself gives us no method of computing it

frozen tendon
#

And you Euclidean algorithm helps us to find the gcd of two number say a and b

cedar kilnBOT
#

@frozen tendon Has your question been resolved?

cedar kilnBOT
#
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frozen tendon
#

Guys can anyone explain me what are the definitions for bézouts theorem, division algorithm and eucliding algorithm and what is the use of gcd in all of them

torpid fulcrum
#

do you mean bezouts identity?

frozen tendon
#

Like I am a bit confused as all of this uses the gcd

torpid fulcrum
#

if we're just considering integers: if gcd(a,b)=c then there exists numbers x and y such that ax+by=c

#

and assuming you're referring to the Euclidean algorithm

#

the euclidean algorithm is used for finding the GCD

frozen tendon
# frozen tendon

Well one more question why in this question the lhs is 4840 and not 1512?

torpid fulcrum
#

(just posting it so i dont have to scroll)

frozen tendon
#

Haha sure

torpid fulcrum
#

that is part of bezouts identity

frozen tendon
#

No I mean in the part where the Euclidean algorithm is used

torpid fulcrum
#

we are trying to find the remainder when 4840 is divided by 1512

#

i.e. 4840 mod 1512

#

since the remainder is 304

#

next, we do 1512 mod 304

#

and so on

frozen tendon
#

No like I mean that

#

here it's in the form of b=aq+r

torpid fulcrum
#

yes

frozen tendon
#

So why is 4840 = b and a= 1512

#

This is what I am a bit confused about

torpid fulcrum
#

that is the process of the euclidean algorithm

#

4840=3(1512)+304

#

so 4840 mod 1512 gives 304

cedar kilnBOT
#

@frozen tendon Has your question been resolved?

torpid fulcrum
#

did you have another question?

frozen tendon
#

But just one thing

#

The larger number is always supposed to be on the left side right?

#

@torpid fulcrum

cedar kilnBOT
#

@frozen tendon Has your question been resolved?

torpid fulcrum
#

sorry, was away for a bit

#

the order doesn't matter

#

but if you start with the smaller number on the left

#

the next step will swap the two numbers

#

for example if we had 425 and 623

#

425 mod 623 = 425

#

so in the next step we do 623 mod 425

#

@frozen tendon

cedar kilnBOT
#

@frozen tendon Has your question been resolved?

cedar kilnBOT
#
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cedar kilnBOT
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fast turret
cedar kilnBOT
fast turret
#

hello, for this problem i have all inequalities set up for exponents of base 3 and 4, but cannot think of a way to compare exponents of base 3 and 4.

#

i will show you my reasoning so far:

#

3^3^4 > 3^4^3, because 3^4^3 = 3^64 and 3^3^4 = 3^81

#

3^4^4 > 3^3^4 just by looking as it as both are the same, except that for 3^4^4 its exponents are greater

#

so we have the inequality 3^4^4 > 3^3^4 > 3^4^3

#

4^3^4 > 4^3^3 because 4^3^4 has greater exponents

#

4^3^4 > 3^3^4 because 4^3^4 is of a greater base

#

now i know i must change bases of these exponents because there appears to be no other way to figure this problem out

#

i have done a bit of research but cannot find anything conclusive about how to change the bases of these exponents

elder pewter
#

Of thats really what you want to use

#

But you could just intuitively look at the numbers

fast turret
cedar kilnBOT
#

@fast turret Has your question been resolved?

ember python
# fast turret

Oh this is a bmo1 problem I’ve literally done this before

fast turret
#

i understand that to prove 3^256 is greater than 4^81 (without using a calculator) i would have to either find a common base or use some exponential/logarithmic rule which i do not know of

#

imma try and figure it out using a logarithmic rule rq

#

one sec

ember python
#

We don’t need bases or exponents to be the same to make comparisons

#

We could have both the exponent and the base of a number be less than another

#

4^81 = 2^162 and clearly 2^162 > 3^256

ember python
#

You can do a similar thing to fit the remaining numbers together

fast turret
#

yeah like with 4^3^3

ember python
#

Yea

fast turret
#

do you come from the uk then?

ember python
#

Yeah I do

#

I did this problem last December

fast turret
#

ah cool me too

#

where in the uk?

ember python
#

Surrey

#

What about you?

fast turret
#

im from the north west

#

what year are you in?

ember python
#

So first year of college

fast turret
#

ahh right im in y11

#

i really should be studying for science and history but this is way more fun lol

#

besides after this week exams will be done

#

ill have 3 months to do just competitive maths

ember python
#

Lmao yea you’ve got gcses I guess

#

What got you starting to do bmo1 problems

fast turret
#

well i really like the subject so it’s fun but there’s also the advantage that it will look really good for maths related degree courses and jobs and such

#

but it’s mostly recreational for me

#

do you do the olympiads or do you just look at the questions?

ember python
#

That’s cool, I did the most recent bmo1 although I only started preparing like a week in advance

#

2016/17 had some pretty doable problems imo

#

Q1 - Q3

#

It’s decent to pick up skills tbf

#

I didn’t even know about this stuff in year 11 so I just did a level stuff instead, I kinda wish I did tbh

cedar kilnBOT
#

@fast turret Has your question been resolved?

fast turret
cedar kilnBOT
#

@fast turret Has your question been resolved?

elder pewter
#

@fast turret college is fun. I became heavily interested in maths whilst at college

#

I found further maths so much more interesting than normal A level maths

cedar kilnBOT
#
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cedar kilnBOT
#
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crimson sedge
#

f(x)=x²-4x+8
g(x)=-|-x-2t|+9
h(x)=f(x) (f(x)<g(x))
g(x) (f(x)>=g(x))
for some t y=2k and h(x) has 3 different intersection points
find every integer k

elder pewter
#

$f(x) = x^2-4x+8$

wraith daggerBOT
#

IntelligentCake

elder pewter
#

$g(x) = -|-x-2t|+9$

wraith daggerBOT
#

IntelligentCake

elder pewter
#

what is t

#

just a parameter?

crimson sedge
#

constant

elder pewter
#

value of constant??

crimson sedge
#

idk

#

i edited the question

#

idk what to call t

#

maybe its not called constant

#

<@&286206848099549185>

crimson sedge
cedar kilnBOT
#

@crimson sedge Has your question been resolved?

cedar kilnBOT
#
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crude scroll
cedar kilnBOT
shell lance
#

420 hah

wraith daggerBOT
#

Kurama

crude scroll
#

Im trying to generate some contradiction here

#

Not sure what the next step should be

cedar kilnBOT
#

@crude scroll Has your question been resolved?

crude scroll
#

<@&286206848099549185>

cedar kilnBOT
#

@crude scroll Has your question been resolved?

tough aurora
#

You will need to provide more information

crude scroll
#

Z(An) is the center of the alternating group

#

epsilon is the identity here

cedar kilnBOT
#

@crude scroll Has your question been resolved?

dire geode
#

.close

cedar kilnBOT
#
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cedar kilnBOT
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lethal moat
#

”Circle: (x+2)² + (y-3)² = 20 Line: y = -⅓x + 3. Give me the coordinates of the points were they intersect”

lethal moat
#

what do you get if you substitute y = -⅓x + 3 into (x+2)² + (y-3)² = 20

cedar kilnBOT
#

@lethal moat Has your question been resolved?

lethal moat
#

Help me w it

#

<@&286206848099549185>

dire geode
gleaming cloud
#

he just wants the answer

dire geode
#

Oh psh

cedar kilnBOT
#
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lethal moat
#

.reopen

cedar kilnBOT
#

lethal moat
dire geode
cedar kilnBOT
#

@lethal moat Has your question been resolved?

cedar kilnBOT
#
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delicate ember
#

this calculation seems wrong, shouldn't the power of 10 be -1

gaunt hamlet
#

It shouldn't be -5, but it shouldn't be -1 either

cedar kilnBOT
#

@delicate ember Has your question been resolved?

tropic oxide
#

,calc 1.7 * 5/14.9

wraith daggerBOT
#

Result:

0.57046979865772
tropic oxide
#

should be 5.7 * 10^-2 surely

delicate ember
#

Ah yes its 10^-2

#

alright

#

thanks

#

.close

cedar kilnBOT
#
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crimson sedge
cedar kilnBOT
crimson sedge
#

Would this be present value formula what formula would I use to solve this

cedar kilnBOT
#

@crimson sedge Has your question been resolved?

crimson sedge
#

<@&286206848099549185>

sick ruin
#

Let’s just figure it out @crimson sedge

cedar kilnBOT
#

@crimson sedge Has your question been resolved?

cedar kilnBOT
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cedar kilnBOT
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pastel sonnet
cedar kilnBOT
pastel sonnet
#

can I find the gradient for both first

#

then use m1 m2=-1

#

then I can use simultaneous eqn to check parallel?

oblique loom
#

ok so the gradient for this first like

#

is -3 as shown

pastel sonnet
#

yea

#

well no

oblique loom
#

and are you aware of the formula for finding gradient with two points

#

no

pastel sonnet
#

cuz u want it to become a mx+c

oblique loom
#

it must be in y = mx plus b form

#

mb

#

put it into y = mx + c form

#

and then

#

use this formula

elder pewter
#

first what is gradient for l1

#

tell use that first

oblique loom
#

y2 - y1/ x2-x1

pastel sonnet
#

3/4

elder pewter
#

ok good

oblique loom
#

if the gradients are the same than they are parralel

elder pewter
#

now for l2

pastel sonnet
#

and then its 9/6

elder pewter
#

what is gradient

#

no

pastel sonnet
#

-9/6

elder pewter
#

yes

#

or -3/2

#

is -3/2 the same as 3/4

pastel sonnet
#

No

#

so its neither?

elder pewter
#

so therefore they are not parallel

pastel sonnet
#

cuz using m1 m2=-1

oblique loom
#

no

elder pewter
#

now to check for perpendicular, m1m2 =-1

pastel sonnet
#

its not true as well

oblique loom
#

thats for perpendicular lines

elder pewter
#

ok good

#

so therefore it is neither

oblique loom
#

didnt read the question

elder pewter
#

so you are correct @pastel sonnet

pastel sonnet
#

okok

#

thx

#

I was over complicating stuff earlier

#

thats why I was a little confused

#

I thought about inplict differentiate the first one

#

implicit*

#

but I didn’t see I could just change sides lol

#

.close

cedar kilnBOT
#
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cedar kilnBOT
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wide light
cedar kilnBOT
wide light
#

i dont understand this at all

#

on its own the question is weak tbf, what im asking

#

but consider this following question:

#

so in this question

#

its first order because ratios are 0.1

#

but in that question 7 with sarah, its between 1 and 2 because its 0.35?? i dont understand