help-49

[safe onyx]

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[cloud garnet]

can someone pls help me with this problem im not sure where to go with it

[cloud garnet]

this is equation one

[waxen willow]

What did you try

[rustic garden]

nvm i see why it has to be odd, the number of numbers has to be odd

[cloud garnet]

i havent tried anything yet i dont really know what to do, i can prove the first equation but im not sure where to go with on this one

[radiant roost]

i think you can double equation (1) to increase the gaps

[waxen willow]

Formulate the problem using math

Write your statement mathematically

What you have , what you wish to prove

[radiant roost]

an odd number can be expressed as 2m+1 for some integer m

[cloud garnet]

1 + 3 + 5 + . . . + (2k-1) is divible by k ?

is this on the right path

[radiant roost]

yeah but i think it needs to start at an arbitrary odd natural number

[waxen willow]

Sum of any consecutive odd numbers

[radiant roost]

not just starting at 1

[cloud garnet]

how would i go adressing this problem then

[waxen willow]

Well notice what starting from an arbitary odd number means

[cloud garnet]

so just 2k + 1

where k is any intergar

[waxen willow]

You start from there sure

And you end on

[cloud garnet]

yeah im not sure

[waxen willow]

Write it out

[dusty portal]

I would recommend to prove it for n=2k+1.

Or, test a few.

[cloud garnet]

2k+1 + 2k+ 3 + . . . (2k + 2n - 1) where n is the amount of terms and k is any natural odd number

[radiant roost]

yeah that makes sense

[waxen willow]

Notice something?

[cloud garnet]

lemme write it down rq

in the end the sum of the terms comes to n(k+2) so therefor the sum of any consecutive odd numbers is divible by n

[waxen willow]

I get something else

How did you get that the sum is that

[cloud garnet]

whatd you end up with

cause yours is prpblly right tbh

[dusty portal]

This works.

[waxen willow]

n(2k + n)

[waxen willow]

The important thing. If you noticed is that first and last term dont matter

As that is divisible by n

Bc n is odd

[midnight plank]

@cloud garnet Has your question been resolved?

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[maiden forge]

im so lost

[maiden forge]

i did the SSS rule

but it took me nowhere

its soo annoying

[paper inlet]

SSS?

[maiden forge]

same sign subtract

[tacit rose]

Try multiplying each equation by some constant that helps you eliminate a variable

[knotty forge]

You should have scaled them so either the x or y coefficients are equal and opposite

[maiden forge]

and then use the sss or dsa rule?

[maiden forge]

do you do the sss rule now?

[tacit rose]

You want the coefficients to be equal and opposite

Sure having a coefficient of 6 for y is fine, but it's not opposite

Meaning +6 and -6

[maiden forge]

ah i see what ur trying to do

so i should multiply one equation by -2

and other by normal 3

[tacit rose]

Yes

[maiden forge]

ok 1 sec

like this

and then cancel out 6y

then do 8-30?

[tacit rose]

You add the equations

[maiden forge]

oh yea

dsa

[slender walrus]

last line is wrong

[maiden forge]

38x = 19 ?

[slender walrus]

the thing you boxed

[maiden forge]

oh

its 0.5

yeah, accidentally did 38/19

there

this should be right now

[tacit rose]

Not quite

Because you used x = 2 when you solved for y

[maiden forge]

oh yeah..

i think this is right now

[tacit rose]

You can verify by plugging in your solution

[maiden forge]

yep its right

is this one right?

[tacit rose]

Equations look good, math looks fine

[tacit rose]

You can verify still by plugging in your solution

[maiden forge]

[tacit rose]

Doing that, you can determine if your answer is right or wrong rather than asking. It helps especially during a test, since you technially aren't allowed to use discord

[maiden forge]

yea, during a test i would do it but its currently 3:25am and im trying to cover all the topics from grade 5 to grade 8-9

getting tired

so might go bed now

thanks for the help

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[maiden forge]

.reopne

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[midnight plank]

✅

[maiden forge]

this would be x= 2 and y = 4

right?

[last slate]

u mean the intersection pt?

[last slate]

yah

[maiden forge]

ok

aaa im so tired

ima j ust go bed tb

h

[last slate]

its irritating ik

but straight lines are one of the toughest topics xD

[maiden forge]

what is this 1 talkinga bout

the quseiton

[last slate]

intersection.

coordinates of intersection

[maiden forge]

so the y intercepts?

[last slate]

no no

you have two lines

leave out the x and y intercepts

[maiden forge]

oh that bit where they both meet

how do u even find that @last slate

[last slate]

hmm there must be a pt x', y' common between both of them

right

[maiden forge]

whats pt

[last slate]

point

[maiden forge]

oh

yeah

[maiden forge]

yea'

[last slate]

now

x' and y' lie on both lines

[maiden forge]

yea

[last slate]

@maiden forge u there

[maiden forge]

yeah

[last slate]

hmm then x' and y' will satisfy both eqns

like when u sub it in those

eqns given

[maiden forge]

yeah but how do u find the x and y

oh

[last slate]

got it?

[maiden forge]

i think so

u line them up like simultaneous equations?

y = 2x + 3 and and y = - 2 /3 x + 1

[last slate]

||since both are common and x'+y' have 2 equations, u can equate them, y = 2x + 3 and and y = - 2 /3 x + 1 ||

[last slate]

equate na

cuz

then it will give u the common pt

[maiden forge]

whats na

[last slate]

just a slang mb

[maiden forge]

for what

[last slate]

nothng leave that

equate both by subracting

[maiden forge]

ok

so like y-y and 2x - -2/3x

?

[last slate]

yea

[last slate]

u forgot the one

1*

[maiden forge]

where

the 3 -1?

[last slate]

see here my friend

yea

[maiden forge]

2x- -2/3x is 4/3x right?

right?

please tell me this is right

@last slate

so i can go sleep in peace

[last slate]

u hv just told what i have done... u havent understood

leave it for now then, gn

[maiden forge]

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[maiden forge]

thanks 4 the help tho

[last slate]

np

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[grim bison]

prove that [√(4-√7)-√(4+√7)]/√2 is a negative number

[grim bison]

i was thinking about [√(4-√7)-√(4+√7)]^2/√2^2

to raise at the power of 2

but i have no idea if it works

[bronze wigeon]

What do you get?

[grim bison]

not the right answer

[√(4-√7)-√(4+√7)]^2 this i believe it's 2

[bronze wigeon]

I think the issue with this approach is that when you square your number, you can’t determine if it’s positive or negative anymore

[grim bison]

yes

[bronze wigeon]

Actually

[lyric charm]

this will be useless

[bronze wigeon]

there’s a caveman approach to this

[lyric charm]

a negative number is by definition a number that's less than zero

try showing $\sqrt{4 - \sqrt{7}} - \sqrt{4 + \sqrt{7}} < 0$

[grand pond]

ann.in.a.teacup

[grim bison]

ok, so √(4-√7)= x and √(4+√7)=y and i have to prove x<y , now i believe i could square them and i will have 4-√7<4+√7 so x<y

[stiff pollen]

do you understand why that expression is less than zero?

[grim bison]

because y is bigger than x

[stiff pollen]

yeah but why

[grim bison]

because when you substract a bigger value from a smaller value it will automatically result in a value with - meaning negative value meaning less than 0

[stiff pollen]

could you theoretically define a function f(x)=sqrt(x)

then say f(x) is 1-1

and just have to prove the inner part

[grim bison]

f(4-√7)-f(4+√7)<0, f(x)=sqrt(x) so the inequality is true, only if 4-√7<4+√7

[stiff pollen]

pull it all to one side

[grim bison]

(4-√7)-(4+√7)<0

[stiff pollen]

go on

[grim bison]

umm

so if (4-√7)-(4+√7) is a negative number, then (4-√7)-(4+√7)/√2 will be a negative number as well

[stiff pollen]

yeah but like

4-sqrt7-4-sqrt7<0

-2sqrt7<0

[grim bison]

oh

[stiff pollen]

you can even divide by 2sqrt7 both sides to get -1>0 if anything else confuses u

[grim bison]

no, no i got it, i just didn't believe there was such an explicit demonstration to elaborate

[stiff pollen]

squaring both sides would also work though, no?

[grim bison]

that's what i tried the first time, and it seemed right

[grim bison]

like this

[stiff pollen]

you can tell both sides are positive

[slender walrus]

consider using conjugates

$\frac{\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}}{\sqrt{2}}\cdot\frac{\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}}{\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}}$

[grand pond]

ℝαμOmeganato5

[stiff pollen]

we know sqrt7 is smaller than 4 because sqrt of 7 is smaller than sqrt of 9 which is equal to 3

[grim bison]

yes, because if it were otherwise, when squared, we would have to change 4-sqrt7 to sqrt7-4

but it remains the same

[grim bison]

it seems a good approach as well

ok, i believe i can handle it now, i'm glad i know how to prove it rigorously

thank you

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[fleet sonnet]

im doing completing the square but i dont think ive ever added like the coefficient squared or whatever it called to both side

[fleet sonnet]

and i still get it right?

[flat spire]

can you show us an example

[fleet sonnet]

ill give random

so

3x^2=6x+4

u move everything to one side so its 3x^2 -6x -4 = 0

then x^2 - 2x - 4/3 =0

then its (x-1)^2 + 1 -4/3 =0

but it should be (x-1)^2 + 1 -4/3 -1 =0

or smth

wait if we do like a easy one x^2-2x-6=0

this is my working
x^2-2x-6=0
(x-1)^2 -7 = 0
(x-1)^2 = 7
x = 1+- square root of 7

the book says to add like the 2^2 in this case for both sides

but for the questions ive been doing i havent added it at all but the answer is the same?

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[feral sedge]

maybe i am missing something but plugging in m = 1 does not even seem to yield a solution

[atomic grail]

How do you know?

The relation for n=1 is this

[atomic grail]

@feral sedge

[feral sedge]

ah so what i missed was my inability to read. nice

[atomic grail]

How do I do this

[feral sedge]

is there any restriction on what the a_i are allowed to be

[atomic grail]

This is all the problem

They are natural numbers

Without 0

[feral sedge]

that's good

[feral sedge]

so you can take this and rewrite it

[atomic grail]

Yes and it gave me a second grade equation

With x=a1

And delta is with a2

Can I say a2>2 right?

If they are natural and not equal

[feral sedge]

so you are saying a_2 is strictly greater than 2?

[atomic grail]

Well yeah

[feral sedge]

can you explain how that conclusion arises?

[atomic grail]

No, I mean in my 2nd grade equation, a2 needs to be >=2 for the equation to have solutions

I think there are 2 cases

[feral sedge]

uh...

okay, so

you have something like

$a_1a_2(a_1 - 2) + 2 = 0$

[grand pond]

Mqnic_

[feral sedge]

yes?

[atomic grail]

Yes

[feral sedge]

right

so it is true you can use cases here

but i'd consider a_1

[atomic grail]

<@&286206848099549185>

[feral sedge]

bro

[atomic grail]

Bro it doesn t give nothing

[feral sedge]

what happens if a_1 > 1?

[atomic grail]

And this case is just for a1 a2

It gave me this

And it s with plus

Because if it s with minus it probably won t be natural anymorw

[feral sedge]

ok wait so

are the a_i natural numbers

[atomic grail]

Yes all of them

[feral sedge]

then you don't need to solve the quadratic

being integers is an extremely restrictive condition

[feral sedge]

observe the way this equation is presented

[feral sedge]

.

[atomic grail]

Well we have a1-2 there

So that could be negative

Ah because it is

It s -2

So a1 is between 1 and 2

Right? @feral sedge

[feral sedge]

if a_1 = 2

then you have 2 = 0

contradiction

[atomic grail]

Yes

So 1<a1<2

[feral sedge]

but a_1 = 1 is possible

[atomic grail]

Ah yes

Then a2=2

@feral sedge

So could an=n be the string?

Let me try

[feral sedge]

if you make the assumption that all of a_n is positive integers, then yes

as it turns out, that is the unique string under this assumption, and it is not hard to prove, assuming you are familiar with a simple trick

[atomic grail]

What simple trick?

[feral sedge]

well

try the problem

[atomic grail]

Is this a sum?

What simple trick @feral sedge

[feral sedge]

well i call it a trick but you still need to do the work

i was referring to something similar to a telescoping summation

[atomic grail]

Ah yes i thinked about that too

So the assumption should be correct?

Ah yeah the trick works

Wait a sec

Yes bro it works @feral sedge

For a_n=n

And because i demonstrated that 1=<a1<2

And a1 natural

Then this one was the only possibility

Right?

Is it enough to demonstrate @feral sedge

?

Yeah i think it s good

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[twilit field]

Trying to analyse the logical form here

Nobody in the calculus class is smarter than everybody in the discrete math class.

[twilit field]

I was thinking let C(x) mean x is the calculus class, D(y) mean y is in the discrete maths class. S(x,y) mean x is smarter then y

$(\forall x(C(x)) \exists y (D(y)))( S(y,x))$

[grand pond]

What a wonderful world it is !

[dreamy lichen]

Too many parens perhaps

[twilit field]

Too little I think , some are missing closing parens

[sinful topaz]

why not just say C is the class of calculus students and D the class of discrete students?

saves on brackets

[twilit field]

I can't use sets

otherwise I would have

[dreamy lichen]

or just leave out the unneeded brackets

[sinful topaz]

[twilit field]

This is not my decision, okay?

[dreamy lichen]

FOL technically doesnt have builtin set theory

though using it to translate english is stupid (i get its not your fault)

[twilit field]

except for a few set theory questions, these are the kind of questions I'll get on my exam

luckily for me , it's a half sem course

[sinful topaz]

does that emote have bad connotations or smth?

[twilit field]

Yea

I think?

[sinful topaz]

ok i'll replace it

😔

[dreamy lichen]

those are 2 statements, not joined by any connective

Are you using multiplication instead of $\land$? That's usually only done in propositional logic afaik

[grand pond]

MathIsAlwaysRight

[twilit field]

no

okay, so it's missing an and

[dreamy lichen]

more than just one actually

it seems like you are using parenthesis instead of and

try rewriting it while keeping in mind that you should use logical connectives

[twilit field]

Unless I'm remebering things wrong, that's allowed

no?

[dreamy lichen]

I wouldnt do that in first order logic

and even if your teachers do allow it, you still need to fix your nesting

[dreamy lichen]

the blue statement should be nested inside the orange one

the ∧ should be between C(x) and the blue statement

[twilit field]

I see

okay

$(\forall x (C(x))) \land (\exists y(D(y))$

[grand pond]

What a wonderful world it is !

[dreamy lichen]

(everyone takes calc) and (someone from disrete math is smarter than x)

wait no

its even worse than that

the parenthesis dont make sense at all

Let's just build the statement from bottom to top

how would you translate
"x is smarter than everybody in the discrete class"

[twilit field]

$[\forall y(D(y))] \land (S(x,y))$

[dreamy lichen]

It's either wrong or just a non-standard notation

[grand pond]

What a wonderful world it is !

[dreamy lichen]

why those [] brackets?

$\forall y(D(y) \implies S(x,y))$

this is how I would write it

[twilit field]

Hmm, okay

yeah, that makes a lot of sense

[dreamy lichen]

wait actually no

this

[grand pond]

MathIsAlwaysRight

[dreamy lichen]

is this how you were taught to do it?

[dreamy lichen]

For all y (if y is in discrete math class, then x is smarter than y)

[twilit field]

Yeah, this makes much more sense

thanks

[dreamy lichen]

Statements of form
"Everyone who does A does B" are translated as
For all x [A(x) -> B(x)] where A(x) and B(x) have interpretation "x does A" and "x does B" respectively

[dreamy lichen]

Anyway, this statement is now supposed to be taken as one unit

if you wish, you can put parenthesis around it

but you should not put parenthesis inside of it

[dreamy lichen]

now you can proceed to translate the nobody in the calc class... part

[twilit field]

I think I should do some easier questions first

[dreamy lichen]

Yeah, that might be a good idea actually

"Nobody in the calculus class got an A"

Is this easier?

[twilit field]

Let $A(x)$ mean x got in A in calculus

[grand pond]

What a wonderful world it is !

[twilit field]

$\not \exists x(A(x))$

[grand pond]

What a wonderful world it is !

[dreamy lichen]

\lnot makes the logical symbol if you want that

But yep

that does the job

[twilit field]

If Jones can do it, anyone can.

[dreamy lichen]

actually, can you do this again but this time use
A(x) - x got an A
C(x) - x is in the calc class

[twilit field]

Let $D(x)$ mean $x$ can do it.
\
We then have $D(J) \implies \forall x (D(x)$

[grand pond]

What a wonderful world it is !

[dreamy lichen]

one bracket is missing (at the end) and J means Jones i guess

but otherwise its good

[twilit field]

yeah

If nobody failed the test, then everybody who got an A will tutor someone who got a D.

Let $F(x)$ mean x failed the test. Let $A(y)$ mean y got an A, let $D(z)$ mean z got a D.
$\lnot \exists x(F(x)) \implies [(\forall y)(A(y)) \exists z D(z)) \implies T(y,z)]$

[grand pond]

What a wonderful world it is !

[dreamy lichen]

i dont like the bracketing

[twilit field]

yeah, lem'me fix that

[dreamy lichen]

there is at least one extra bracket

[twilit field]

$\lnot \exists x(F(x)) \implies [(\forall y)(A(y)) \exists z (D(z)) \implies T(y,z)]$

[grand pond]

What a wonderful world it is !

[dreamy lichen]

you should group these together

the parenthesis around forall y are probably redundant

[twilit field]

Why

[dreamy lichen]

same with the ones around A(y)

[dreamy lichen]

why you should group them together?

[twilit field]

Shouldn't this be one thing

[dreamy lichen]

nope

[twilit field]

why not

[dreamy lichen]

because if you group these together, z will have no meaning outside of that block

and thus you cant use it in T(y, z)

[twilit field]

I see

okay

[dreamy lichen]

1. If nobody failed the test, then everybody who got an A will tutor someone who gota D.
2. Nobody failed the test ⇒ everybody who got an A will tutor someone who got D
3. ¬(somebody failed the test) ⇒ ∀y(if y got an A, then y will tutor someone who got D)
4. ¬∃y(y failed the test) ⇒ ∀y(y got an A ⇒ y will tutor someone who got D)
5. ¬∃yF(y) ⇒ ∀y(A(y) ⇒ ∃z(z got D and y will tutor z))
6. ¬∃yF(y) ⇒ ∀y(A(y) ⇒ ∃z(D(z) ∧ T(y, z)))

this is how I would go about translating it part by part

[twilit field]

I see

okay

thanks

[dreamy lichen]

I'd suggest you to not try to translate it by reading it left to right and just adding symbols as you go with the reading

[dreamy lichen]

rather try doing it symbol by symbol like i did here

its long, but with practise you will be able to skip most of that and find the translation in only a few steps

[twilit field]

All solutions of the inequality $x^3 − 3x < 3$ are smaller than 10.

[grand pond]

What a wonderful world it is !

[twilit field]

$I can re-write this as if x is a solution of x^2-3x<3, then x<10

$\forall x( x^3-3x<3) \implies x<10$

[dreamy lichen]

the quantifier isnt right

otherwise its fine

[twilit field]

oops

right

[grand pond]

What a wonderful world it is !

[dreamy lichen]

now it just needs a few brackets

[twilit field]

Ah

[dreamy lichen]

specifically, it needs brackets right here

[twilit field]

right

[dreamy lichen]

in terms of where you need brackets and where you dont, the PEMDAS alternative for logic is:
Parenthesis - quantifiers and negation - AND and OR - IMPLIES and EQUIVALENCE

[dreamy lichen]

This without parenthesis would be interpreted like this:

[twilit field]

If there is a number x such that $x^2 + 5x = w$ and there is a number y such that $4 − y^2 = w$, then w is strictly between −10 and 10.

[grand pond]

What a wonderful world it is !

[twilit field]

$\exists x(x^2+5x=w) \land \exists y( 4-y^2 = w) \implies (-10<w<10)$

[grand pond]

What a wonderful world it is !

[dreamy lichen]

sorry its fine

[twilit field]

Cool, thanks!

I think i'mma eat dinner now

,ti

[grand pond]

The current time for math_rocks is 08:21 PM (IST) on Sat, 22/02/2025.

[twilit field]

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[white pendant]

Hello everyone :] I’m very confused on how to solve this. Wouldn’t the upper bound be the cube of 24? And why would we subtract from 2? What am I missing here?

[white pendant]

<@&286206848099549185>

[true horizon]

Is the ans 2pi/5?

@white pendant

[white pendant]

The answer is the second image. We don’t have to solve it exactly

[true horizon]

Oh alr

Yeah I can't solve the 1st one

Mb

[white pendant]

You’re fine :’)

[true horizon]

Alrrr

[midnight plank]

@white pendant Has your question been resolved?

[midnight plank]

@white pendant Has your question been resolved?

[umbral timber]

okay so can you tell me the question in full once
i did not understand the wording from the image

[white pendant]

Sorry about that. This is a cross section problem where the cross section of the area bounded by y=x^3, y=0, and x=24 is in the shape of a semicircle perpendicular to the y-axis

Is that better?

[white pendant]

I tried setting the equations in terms of x because it’s perpendicular to the y-axis, and using the semicircle formula (1/2pir^2, where r = b/2) but I’m not getting anywhere near what the answer is.

x=24 and x= cube root of y intersect at the cube of 24. But the answer says the upper bound is 8??

And I don’t even know where the 2 is from

[umbral timber]

so you need to find the area OR the expression for the semicircle?

[white pendant]

The expression for the area. The second image I sent is the answer for reference

[umbral timber]

okayy

[umbral timber]

and just to clarify
the area is the area under the curve bound by curve, x=24 and the x axis?

[white pendant]

And y=x^3

All three

[umbral timber]

un i really dont understand
this is a very simple integral
how are they getting such a complex solution

[white pendant]

Right?? I have no idea bro.

[umbral timber]

and the area under the curve is not in the shape of a semi circle either

[white pendant]

Nono the cross section is

Not the area

[umbral timber]

what do you mean
whats the cross section here

our situation looks like this no?

[white pendant]

Yes

[white pendant]

The cross section of the area under the curve is a semicircle. I was just taught this but I’m pretty sure it’s this weird thing where if you look at the graph from the z axis that’s the shape it makes or something. I’m not entirely sure 😭 I just know it’s weird af

[umbral timber]

yeah you're definitely working with R³ here
in that case ive no idea
but idk your function is purely in terms of y and x so its a 2D thing. do you recon the question has a problem or the solution?

[white pendant]

Could be yeah. I’ll just ask my teacher about it probably. Even my college friends were confused out of their minds

Thanks for trying tho

[umbral timber]

yeahh no problem
good luck. i hope you find the solution

[dusk pier]

This is what I got

I see what they did

[white pendant]

I got the same value in the parentheses but how did you get the cube root of 24?

[dusk pier]

They must've swapped the 24 for an 8

[dusk pier]

We are integrating with respect to y

Wait oopsies

It should be 24^3

Brain fart

[white pendant]

Yeah ok same

[white pendant]

Wait how tho

[dusk pier]

No clue

[white pendant]

https://tenor.com/view/crying-emoji-gif-21922016

[midnight plank]

@white pendant Has your question been resolved?

[midnight plank]

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[safe onyx]

velocity is the integral of acceleration

[safe onyx]

why is it At and Bt^2

(solving for A)

[dusk pier]

No initial velocity

[safe onyx]

like shouldn't B be a 0?

[dusk pier]

No?

A+2Bt

[safe onyx]

A just is acceleration

but acceleration is constant

[dusk pier]

Acceleration can change

[dusk pier]

No, we cannot assume such unless the problem says so

[safe onyx]

In this case it does though?

oh wait

[west iron]

A is velocity here

2B is acceleration

[safe onyx]

yea

[dusk pier]

It can

[safe onyx]

v(t) = v_o + int acceleration dt

[west iron]

The problem doesn't give enough info to solve if acceleration isn't constant (probably anyways, there's no actual problem here)

[safe onyx]

I need to find a and b

[west iron]

Actually

The form of the velocity function implies constant acceleration anyways

[safe onyx]

so B is 0?

[west iron]

Oh wait

That's velocity

[safe onyx]

yes we have velocity function not position

[west iron]

ok acceleration isn't constant then

[safe onyx]

so B is the changing of accleration

a = 1.7 + ? t

[west iron]

Calculate the velocity at t=1 in terms of A and B

[safe onyx]

v = 0 + 1.7 t + ?/2 t^2

v(1) = 1.7 + B

2 = 1.7 + B

.3 = B

v(t) = 1.7t + .3t^2
a(t) = 1.7 + .6 t

[west iron]

hmm

[safe onyx]

,calc 1.7 + .6(5)

[grand pond]

Result:

4.7

[west iron]

where are you getting this?

[west iron]

oh right

[safe onyx]

B = ?/2

so 2B = ?

or just derivative power rule

[dusk pier]

Whatever feels more comfortable

[safe onyx]

Thank you both

[west iron]

I would just derive v(t)

np

a(t) = A + 2Bt = 1.7 + 0.6t

[safe onyx]

agreed it's easier that way

(my course requires calc 1 so it is assumed that students are familiar with it)

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[vestal juniper]

hi!!

[vestal juniper]

i’m currently learning number theory and i’m stuck on a question

im doing a

i’ve done the regular algorithm
32x + 78y = gcd(32,78)
78 = (2 × 32) + 14
32 = (14 × 2) + 4
14 = (4x3) + 2
4 = (2x2) + 0

but i’m a little confused now

2 = 14 - (4x3)
2 = 78-(32x2) - ((32)-(14x2) x 3)? is that correct

[midnight plank]

@vestal juniper Has your question been resolved?

[midnight plank]

@vestal juniper Has your question been resolved?

[sinful topaz]

no

2 = 78 - (32x2) - (32 - (14x2)) x 3

after that try substituting that last 14 in there

[runic hamlet]

dont try to skip steps

do it slowly

and simplify along the way

[midnight plank]

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[sly horizon]

what am i doing wrong here? apparently the answer is root 3 over 6

[dusty portal]

So the original limit is $\lim_{x\to\infty}\frac{\sqrt{3x^2+3x}-x\sqrt{3}}{3}$, correct?

[grand pond]

;(

[dusty portal]

There is no 3 term on that variable

[sly horizon]

yes

[floral junco]

that's your only mistake

[sly horizon]

oh wait

[dusk pier]

$\lim_{x\to\infty}\frac{3}{3(\sqrt{3}+\sqrt{3})}$

[sly horizon]

did i forget to distribute the 3 to that

[grand pond]

mathisfun

[dusty portal]

Yep

[sly horizon]

oh ight

thanks guys

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[royal dirge]

$2(x+y)=\sqrt{xy} \
x^2+y^2=12 \
(x+y)^2-2xy=12 \
x+y=\sqrt{12+2xy} \
2\sqrt{12+2xy}=\sqrt{xy}$
Am I doing this right? Where do I go from here?

[grand pond]

UCYT5040

[royal dirge]

ohh wait I can square both sides

$4(12+2xy)=xy \
48+8xy=xy \
-48=9xy \
xy=\frac{-36}{9}$

I feel like that didn't get me anywhere useful

oh wait i made a mistake its 9 not 7

[lucid dew]

yeah i think there's a mistake from -48 = 7xy to xy = -36/7

[grand pond]

UCYT5040

[lucid dew]

no 7 was correct

[royal dirge]

oh wait yeah

[lucid dew]

$4(12+2xy)=xy
48+8xy=xy
-48=7xy $

[grand pond]

➤ ⌞ clair ⌝
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lucid dew]

uhh wait

[royal dirge]

well i think its just $\frac{-48}{7}=xy$

[lucid dew]

$4(12+2xy)=xy
48+8xy=xy
-48=7xy $

[grand pond]

UCYT5040

[lucid dew]

that looks right to me

i think you just made some small calculation mistakes

[royal dirge]

but then what do i do from here?

[lucid dew]

is that not what the question was asking?

product of x and y

there you go

[royal dirge]

ohhh true

i dont even need to find x and y

[lucid dew]

yeah exactly

[royal dirge]

damn i could have gotten this lol

thank you

[lucid dew]

yeah you got that tho

:))

[royal dirge]

.close

[midnight plank]

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[royal dirge]

.reopen

[midnight plank]

✅

[royal dirge]

wait no no no

the answer was 6

i just checked the key

[lucid dew]

ohh im so sorry there's a mistake here

the first line

the ratio of (x + y) : sqrt(xy) is 2:1

[royal dirge]

so doesnt that mean two of (x+y) is = to one of sqrt(xy)?

[lucid dew]

$\frac{x+y}{\sqrt{xy}} = 2$

[grand pond]

➤ ⌞ clair ⌝

[lucid dew]

multiply each side by $\sqrt{xy}$

[grand pond]

➤ ⌞ clair ⌝

[royal dirge]

wow that makes things so much easier too

[lucid dew]

yeah so so sorry I didn't check the earlier steps

follow the same logic and you should get the right answer

[royal dirge]

$\sqrt{12+2xy}=2\sqrt{xy}$

[grand pond]

UCYT5040

[royal dirge]

its okay, no worries

$12+2xy=4xy \
12=2xy \
6=xy$

thanks alot!

[grand pond]

UCYT5040

[royal dirge]

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[dusty portal]

This implies that sqrt(x)=sqrt(y) and thus x=y

[midnight plank]

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[twilit field]

Show that the statements A ∩ B = ∅ and A \ B = A are equivalent.

[twilit field]

Using logic

By idea was the first statement means $\not \exists x( x \in A \land x \in B)$

[grand pond]

What a wonderful world it is !

[twilit field]

$A \setminus B = A$ is confusing me

[grand pond]

What a wonderful world it is !

[twilit field]

Like A\B is the set of elements in A that aren't in B

so $\forall x( x \notin B \land x \notin (A \cap B))$

[dawn dagger]

Shouldnt it be x not in B

[twilit field]

oh right

[grand pond]

What a wonderful world it is !

[dawn dagger]

What no

x in A and x not in B

[radiant roost]

it's a set equality

when we have A=B i think it's

$\forall x : x\in A \iff x\in B$

[grand pond]

Axe

[twilit field]

yeah, it is

[radiant roost]

so i think we can use that here

[twilit field]

oh, so I assume the set equality

[radiant roost]

huh?

the statement we are trying to express is a set equality

A \ B = A

[twilit field]

$\forall x: x \in A \iff (x \in A \land x \notin B)$

[grand pond]

What a wonderful world it is !

[radiant roost]

yes

[twilit field]

I now have to expand and simplify I pressume

[radiant roost]

whatever you need to show the statements are equivalent

probably get rid of the "iff"

[dawn dagger]

I think <= follows immediatly

[twilit field]

Now, I don't have to prove it

$(x \notin A \lor (x \in A \land x \notin B) \land \neg (( x \in A \land x \notin B) \lor x \in A)$

[uneven violet]

hi

[grand pond]

What a wonderful world it is !

[uneven violet]

...

[twilit field]

hi

[uneven violet]

hello

[radiant roost]

i don't get it

[dawn dagger]

one step at a time preferably

[uneven violet]

i am only a 5th grader

[twilit field]

Okay, so I first look at $x \in A \implies (x \in A \land x \notin B)$

[grand pond]

What a wonderful world it is !

[dawn dagger]

enjoy it

[radiant roost]

oh i get it

i think the "not" is misplaced

¬

this guy

[uneven violet]

ok even though i only know pemdas, how to add subtract add divide and multply decimals fractions, and know basically the math that a 5th grader should know

gtg bye

[dawn dagger]

this channel is occupied, for own math questions go to #❓how-to-get-help

[twilit field]

yeah, sorry

Okay, so what about we instead look at $x \in A \land x \notin (A \cap B)$

[grand pond]

What a wonderful world it is !

[twilit field]

so $\forall x( x \in A \land (x \notin A) \land (x \notin B))$

[grand pond]

What a wonderful world it is !

[radiant roost]

you're looking for a different way to express A \ B = A?

[twilit field]

A simpler way

I don't instead to use too much set theory

mostly logic , as this is for a logic exam

[radiant roost]

is this right?

[twilit field]

Isn't it?

[radiant roost]

$x\not\in (A\cap B)\equiv x\not\in A \lor x\not\in B$

[grand pond]

Axe

[twilit field]

oops

righty

So this is $\forall x:( x \in A \land x \notin A) \lor ( x \in A \lor x \notin B)$

[grand pond]

What a wonderful world it is !

[dawn dagger]

why did you use or

[radiant roost]

$x\not\in (A\cap B)\equiv \neg x\in(A\cap B)\equiv \neg(x\in A\land x\in B)\equiv (\neg x\in A)\lor (\neg x\in B)\equiv x\not\in A \lor x\not\in B$

[grand pond]

Axe

[radiant roost]

demorgan's laws

[twilit field]

Is this fine

[twilit field]

This is the same as $\forall x : x \in A \lor x \notin B$

[grand pond]

What a wonderful world it is !

[twilit field]

which feels wrong

[radiant roost]

idk if i agree with the premise that A \ B = A can be expressed like this

[twilit field]

Why not

[radiant roost]

cause i don't get it

[radiant roost]

yes that's right

[twilit field]

so should I work my way from there

$x \in A \land x \notin B \iff x \in A$

[grand pond]

What a wonderful world it is !

[radiant roost]

if you want
i still don't see why you can express A \ B = A that way

[twilit field]

Then how am I supposed to express it

[radiant roost]

also isn't it missing a quantifier?

[twilit field]

foral, yea

[radiant roost]

that implies A = U

[twilit field]

nvm, just looked at the soln

Thanks

and sorry

[radiant roost]

no worries

[dawn dagger]

forgiven

[twilit field]

.close

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[narrow dragon]

are my answers correct ?

[dusk pier]

Very correct

[narrow dragon]

Ok thank u

[dusk pier]

Perfect

[midnight plank]

@narrow dragon Has your question been resolved?

[lethal path]

wait

[slender walrus]

c is wrong

[lethal path]

you've done c incorrectly, cause that's $m \in (-\infty, -2)$

[grand pond]

south

[lethal path]

can you stop giving incorrect answers

if you don't know just stop helping

but yeah a, b, d are correct

[dusk pier]

My bad

Sometimes I can't see

Cause humans make errors you know

[lethal path]

you've made an error with jofuu as well, just saying

[dusk pier]

Sure

It is late

[midnight plank]

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[magic crescent]

How would I solve this problem?

[magic crescent]

I tried using the formula 2π ∫(x-a)(f(x) - g(x))

from -1 to 1 after setting x^2 = 1

[quick nexus]

yes

did you square the functions?

[magic crescent]

um no

[quick nexus]

wait nah i'm tripping

[magic crescent]

and made it 2π ∫(x^2 - 2)(1 - x^2)

from -1 to 1

but after i did that

and solved it out

it said the volume was wrong

[quick nexus]

you're finding the volume of the solid. In this case, you basically found part of the surface area

notice how you're calculating geometrically the sum of all 2*pi*r , which gives circumference

instead you need to calculate pi * r^2, the area, which when summed up in an integral gives volume

[magic crescent]

So should it be π ∫(x-a)^2(f(x) - g(x))

instead?

[quick nexus]

not really, you want to square the radius, which would effectively be the functions you are given

[dusk pier]

$x^2$ should be outer radius, 2-1 should be inner radius, thus $\pi\int_{-1}^1(x^2)^2-(2-1)^2dx$

[grand pond]

mathisfun

[magic crescent]

I tried using this formula and after inputting the answer

it still says its wrong

;-;

[dusk pier]

Show

[dusk pier]

oops

[dusk pier]

I meant to say (2-x^2) oops

[magic crescent]

I have a question though

What formula is this using

like i know 2π∫(x-a)(f(x)-g(x))

and then i think the other one is

[dusk pier]

Washer method

[slender walrus]

you should first draw a diagram of what's going on

to see whether shells or washer/disks is more appropriate

[magic crescent]

oh alright

ty

[midnight plank]

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[last slate]

how do i find out how long it'll take Ans*1.1 to be >= 1e+100

[last slate]

if ans starts at 1

likw...on a calculator start with 1 and going with Ans*1.1 and pressing equals .........is there a way to see how many presses it takes til it's >1e+100?

[lyric charm]

after pressing the [Ans] key $n$ times you will have the number $1.1^n$

[grand pond]

ann.in.a.teacup

[lyric charm]

so you need to solve the inequality $1.1^n > 10^{100}$

[grand pond]

ann.in.a.teacup

[midnight plank]

@last slate Has your question been resolved?

[last slate]

oh omg

i'm stupid

thank u

.close

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[spark badger]

No

[midnight plank]

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[midnight plank]

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[molten bay]

The set of all the points where the function is differentiable is

( F(x) = \frac{x}{a - |x|} )

So here we can see that ( a = |x| ) is not will be in domain so ( (-\infty, -a) \cup (a, \infty) ) will he differentible

[grand pond]

Andy

[midnight plank]

@molten bay Has your question been resolved?

[midnight plank]

@molten bay Has your question been resolved?

[smoky walrus]

what about when -a < x < a

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

Ohh i have to say R-{a,a}

[smoky walrus]

that'll work
also, you need to consider if a=0

[molten bay]

What does a=0 mean?

@smoky walrus

[smoky walrus]

a is a constant that can be any value right?

[molten bay]

Yeah

[smoky walrus]

so it can be 0

[molten bay]

so you meant when it is 0

[smoky walrus]

yeah

[molten bay]

It will affect x/|x|

So here we can not use |x|=0

Umm yeah

[smoky walrus]

yeah, I realized that it doesn't matter after graphing it

[smoky walrus]

that domain will still be correct

[molten bay]

How?

We still can not put x=0

R-{a,-a,0}?

[smoky walrus]

yeah but a=0 so it becomes R - {0, -0} = R - {0}

[uneven violet]

hello

[molten bay]

Hello

[midnight plank]

@molten bay Has your question been resolved?

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[wide sluice]

How would you rigorously and mathematically prove that the single source shortest path problem exhibits greedy choice property

[wide sluice]

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[obsidian hedge]

im wondering how is it advantageous to represent a pair of straight lines using a single second degree equation

[obsidian hedge]

what can we do using the second degree eqn which we cannot with the individual eqns

[dusk pier]

Discriminant bs

[obsidian hedge]

such as?

[dusk pier]

THat's all I know

[fallow scarab]

If you have a problem just show the original statement/question

[obsidian hedge]

i dont i just need to know the purpose

there must be a reason they were introduced after all!

[fallow scarab]

Probably true yea

But we have no idea what you're talking about unless you give a problem

[unique juniper]

depends

somethings are done for the sole purpose of vanity

making two lines into a second degree relation is one of them

[obsidian hedge]

google said something along the same lines

"more concise"

[unique juniper]

yeah don't buy it

[obsidian hedge]

ive only had to find the angle between two lines so far. can be done just fine with two seperate eqns!

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[timber tartan]

Hi is my answer correct?

[timber tartan]

,r

.r

uhh

[dusk pier]

,rccw

[dusk pier]

Oh, long time no see.

[timber tartan]

Do i know u?😭

[dusk pier]

Looks good.

[dusk pier]

You were the first guy I helped

[timber tartan]

I think i remember you, tryijg a new merhod to teach right

explain

[dusk pier]

Yep

[timber tartan]

Oh yeah

[dusk pier]

You kinda didn't explain why the triangles are similar

💀

[timber tartan]

I checked on gauth ai it was talking ab some weird stuff i had no idea what it was talkin ab and our results werent the same

[timber tartan]

Ueah i was ab to say that LOL

[dusk pier]

Use AAA similarity

[timber tartan]

uh to find if its similar are we gonna do it by angle angle?

Since the lines are parallel

[timber tartan]

Oh that works

But thats not included in our subject

So ill js use angle angle i think

[dusk pier]

Ok

Same thing

[dusk pier]

,rccw

[dusk pier]

Yep

Transversal laws

[timber tartan]

Alrrr ty

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[muted ether]

knowing that both A and B are invertible, can we go from (A.B)^(-1) to [B^(-1)].[A^(-1)]?

[olive matrix]

yes, this is fairly straightforward to prove

[muted ether]

i tried multiplying (A.B)^(-1) by I in different ways but didn't really go anywhere with that

[olive matrix]

oh, AB has to be invertible as well

[muted ether]

right

[olive matrix]

AB . X = 1

and solve for X

[muted ether]

can we not go from the expression (A.B)^(-1) to [B^(-1)].[A^(-1)] somehow though?

[olive matrix]

i mean yeah i guess? this is how you show that that's true

[dusk pier]

Maybe algebra

[dusk pier]

Very neat method though

I will try this

[muted ether]

that i get
pre-multiplying by A^(-1) first, then by B^(-1) we get [B^(-1)].[A^(-1)]

[olive matrix]

that's an expression

you started with an equation

[muted ether]

yes, i was explaining the process for the derivation from the equation you gave

[olive matrix]

we get **X = ** [B^(-1)].[A^(-1)]

[muted ether]

yes, from the equation

[olive matrix]

so the thing that takes AB to 1 is B^-1 A^-1

therefore (AB)^-1 = B^-1 A^-1

and now you can use that rule

[muted ether]

should i not be worrying about not being able to somehow go from (AB)^(-1) to [B^(-1)].[A^(-1)]?

[olive matrix]

i mean

you're going to need to use the definition of ^-1 at some point

[muted ether]

i suppose so

well, thanks for accompanying me 🫡

bbye

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