help-26

[neon iron]

You are not opening the brackets properly

And your method is too long and hard on the eye

[onyx lance]

but I seem to have fixed them already

[neon iron]

I doubt it

Try doing it from the very beginning

Avoid fractions

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[icy rapids]

$$
\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt[3]n + (-1)^{n(n+1)/2}}
$$

[thorny flame]

Fluffy McGiggles

[icy rapids]

I have to see if the series converges or not

it seems impossible to do

pls help!

[split osprey]

Can you take cases on n = 4k, n=4k+1, n=4k+2, n=4k+3

As a starting point

[icy rapids]

you mean this?

[thorny flame]

Fluffy McGiggles

[split osprey]

Yup,

[icy rapids]

now?

[split osprey]

Now try to bound these

[icy rapids]

$$
s_1 \left( k \right) = \frac{1}
{{\sqrt[3]{{4k}} + 1}} - \frac{1}
{{\sqrt[3]{{4k + 1}} - 1}}
$$

[thorny flame]

Fluffy McGiggles

[patent light]

Yes

[split osprey]

Can you apply alternating series test on these?

[icy rapids]

no

[split osprey]

Why?

Okay, wait, you have 4 terms

Apply the test on

[icy rapids]

it doesn't end well

[split osprey]

1st and 4th term

And 2nd and 3rd term

Club 1st and 4th term

[icy rapids]

?

[split osprey]

this

You have four terms, right?

[split osprey]

In this try proving the convergence of
$a_{4k} + a_{4k+3}$
and
$a_{4k+1} + a_{4k+2}$
separately

[icy rapids]

???

[thorny flame]

fukwerint

[icy rapids]

ok

[split osprey]

Each of these is an alternating series

With subsequent terms decreasing

[icy rapids]

converges

[split osprey]

Yup

[icy rapids]

👍

thank you

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[karmic kindle]

would i be able to find the right answer if i change T(x) to y and solve from there?

using the reciprocal of 2.1 and so on?

[upbeat eagle]

Yeah if you just rearrange for x

[karmic kindle]

thx

$10/21y+20=x$

[thorny flame]

chilly

[karmic kindle]

this is the equation I got

its asking for the coordinates?

i don't know how to do that

[upbeat eagle]

I got a different equation

[upbeat eagle]

can i see your workings?

[karmic kindle]

y=2.1(x-20)

[upbeat eagle]

OHH

[karmic kindle]

2.1 convertes to a fraction 21/10

[upbeat eagle]

You can't add 2 to 0.1(x-20)

[karmic kindle]

oh

[upbeat eagle]

you could multiply that bracket out and continue from there

[karmic kindle]

T(x)=2+0.1x-2?

yes?

oh

[upbeat eagle]

yep :)

[karmic kindle]

wait then im left with 0.1x=T(x)

[upbeat eagle]

Yep, do you know how to plot that?

[karmic kindle]

sadly no

i have (0,0) as my first point

oh wait

nvm

OH

i forgot to switch the x and y

x=0.1y?

ill just close this

thanks so much

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[trim dove]

I couldn't understand how the part that i marked got simplified that way

[wooden moon]

tan70*tan20

= tan 70 * tan(90-70)

=tan70 * cot70

=1

[trim dove]

Ohhhhhh

Thank you

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[sage frost]

Uhm hello, how do they arrive to this conclusion from this

[sage frost]

to this

here's a bit of my process

I've just been... going in circles, I feel crazy

[wooden moon]

I haven't solved it, but check if any of this works

No I solved it, and nothing worked

[sage frost]

I think I got it... Ill write it down in a sec and put it here if anyone ever needs help with something similar

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[neon iron]

is this actually neither

[neon iron]

i feel like my math is just

wrong

neither does this

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[steep gull]

in the u*v shape is the u(x) derived or not bcs in the formula it says U'V

here for example is the x^n the u(x) or the u'(x)

[steep gull]

w this formula

[woeful drift]

Generally polynomials tend to be u

[split osprey]

You do know that you can assign any of those as u or v

[steep gull]

yeah but like by logic it s first polynomials

but like is the x^n the u(x) or u'(x)

[restive inlet]

choice usually depends on what's easier to integrate/differentiate
and stuff like whether one derivative is periodic or eventually gives 0

[split osprey]

I remembered it like this

Easiest way

[steep gull]

yeah but like the beginning is it not u'v or like uv'

[split osprey]

Forget your formula

[split osprey]

Just use this

And tell me if you have any confusion

[steep gull]

let me try

[steep gull]

but wait u didn t write primitive here

[split osprey]

Where

This formula is 100% correct

And same as yours

Just a better way to write

[steep gull]

how there isn t the crochets that indicates primitives

[steep gull]

ı don t understand this one

[split osprey]

crochets indicate derivative

[steep gull]

like where are the primitives or derivatives

[split osprey]

Not primitives, which are antiderivatives

[steep gull]

us it s primitives that s why i m confuse

d

[split osprey]

No sir

crochets mean derivatives

[steep gull]

not in french system lmao

for us this sign ' is derivative

[split osprey]

Are you calling this crochet: $\int$

[thorny flame]

fukwerint

[steep gull]

no

[split osprey]

Then?

[steep gull]

this is crochet = primitives

for me

[split osprey]

Ohhh lol

Okay

This is nothing

See carefully

$[f(x)]^b_a = f(b) - f(a)$

[thorny flame]

fukwerint

[split osprey]

It doesn't mean derivative, nor primitive

[steep gull]

But like here for different parts

How does it work

[split osprey]

Yup, so it doesn't contain that because my formula is for indefinite integrals

[steep gull]

I m trying to figure out how to know if it s u'(x) or u(x)

[split osprey]

Yours is for definite integral

There's not much difference

Wait

I'll send you one for definite too

[steep gull]

Thank youuu🙏🏻

[split osprey]

Can't find it for definite integrals

But its the same

Just evaluate it at b and a and subtract

[steep gull]

But like what is u'(x) and u(x) here pls

Like is the n^x the u'(x) or u

[split osprey]

Suppose you assume that x^n is the first function, i.e u(x)

[steep gull]

And so do i have to derive it to use this formula

[split osprey]

No

You first need to integrate e^x to use the formula

Since e^x = v'(x), right?

[steep gull]

Ok but what about the u'(x) in the beginning

[split osprey]

In the second term?

[split osprey]

Are you using this

[split osprey]

Or this

[pallid copper]

This is the correct one. I think in the last image you mixed concepts. Here, you take :

---> u(x)=x^n as it is the one which helps you to simplify the expression, as you are gonna derivate it and decrease its degree by one. Note that if you take as u(x) = e^x you dont obtain any benefit as the derivate of e^x is e^x and you dont advance anything

---> v'(x)=e^x dx. That's the function you will integrate in the second part of the expression

[steep gull]

I m trying to use this

[pallid copper]

then you have to call v(x) = e^x

as its the one u gonna derivate

[steep gull]

But it says u' in the beginning of the formula

So do i have to derive x^n or ?

I m so lost

[pallid copper]

yeah baby

[split osprey]

Drink milk

[steep gull]

thanks

[steep gull]

U have a thing for milk

[pallid copper]

https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/326abc19e334ea879a2c03b2171648ad_MIT18_01SCF10_Ses76e.pdf

you can take a look at this maybe it helps, idk

[steep gull]

Have u done this ?

[steep gull]

Thanks

[pallid copper]

no no , i found it

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[ivory island]

What is the general way to construct the integers which uses a axiomatic approach like the peano axioms for natural numbers?

[ivory island]

or like a way which utilizes the peano axioms definition of natural numbers

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[ivory island]

@helper

<@&286206848099549185>

also feel free to close the channel once you've answered, I don't have battery left and I don't wanna leave it occupied for too long

[sonic escarp]

peano defines a smallest number and a successor function.

[modest burrow]

https://youtu.be/IzUw53h12wU
this is a great video on the subject
obv you dont need to watch the whole thing, defining then integers will be the first major step
although the whole thing is definitely worth the watch

[whole sparrow]

I don't understand the question.

[modest burrow]

if you want a brief answer its something like this:
the positive integer c is defined as the collection of pairs (a, b) where a = b + c
the negative integer -c is defined as the collection of pairs (a, b) where b = a + c
and 0 is (a, a)

[whole sparrow]

what how?

[modest burrow]

its a definition, a construction

starting with only the naturals

[whole sparrow]

Is it like constutive intergers or smth?

[modest burrow]

what ?

[whole sparrow]

nvm

[modest burrow]

then we can define addition of such pairs (a, b) + (a', b') as (a + a', b + b')
and multiplication (a, b)(a', b') = (aa' + bb', ab' + a'b)

for some intuition you think of (a, b) as a - b

and from there you can go on to prove and confirm all of the nice properties you know about integers

[whole sparrow]

i meant like if there are 3 constutive intergers = to 10 you to
x+ x+1 +x+2 = 10

[modest burrow]

oh consequtive

ah i dont see what that has to do with this?

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[whole sparrow]

oops i read it wrong

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[flat tapir]

how is the bottom left element not k-4

oh nvm

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[strong furnace]

Can someone resolve this equation please

2^(x-1) + 2^(1-x) = 5/2

The answer is x=0 but im strugling with the resolution

[torpid sparrow]

Where are you stucked?

[normal tapir]

potato time

[strong furnace]

At the begining lol, I dont even know how to isolate x

[torpid sparrow]

Do you know power properties?

[strong furnace]

My teacher told us to try to find

2^x = 1

To affirm that x=0

But idk how to arrives at this step

[torpid sparrow]

Forget about that

Answer me

[ember jasper]

can somn help with my question in help-34😅

[strong furnace]

No sir

[torpid sparrow]

If I tell you that x^2 = x * x

You think this is true or false?

[strong furnace]

Yes

[torpid sparrow]

The question is true/false, not yes/no

[strong furnace]

True

[torpid sparrow]

Ok

What about x^3 = x^2 * x

[strong furnace]

True

[torpid sparrow]

Ok try the same with 2^(x-1)

What would you do here

To make something similar

[strong furnace]

2^x * 2^-1

[torpid sparrow]

Perfect

Do you know what 2^(-1) means?

[strong furnace]

1/2

[torpid sparrow]

Good

So we can rewrite that like

2^x/2

Correct?

[strong furnace]

YeS sir

[torpid sparrow]

Now do the same with the other one

2^(1-x)

[strong furnace]

2 * 1/(2^x)

[torpid sparrow]

Nice so we have

2^x/2 + 2/2^x = 5/2

Bow we have clearly 2^x as a thing

We can use a subtitution to solve

Lets call t=2^x

Write the equation in terms of t

What do you have?

[strong furnace]

t/2 + 2/t = 5/2

[torpid sparrow]

Solve that equation for me please

[strong furnace]

I should * by 2t everything ?

[torpid sparrow]

Have you ever solved an equation with fractions?

[strong furnace]

Probably but ngl idk how to actually do it
I'm ashamed lol

[torpid sparrow]

Start adding the two fractions

t/2 + 2/t

In a single fraction

What do you get?

[strong furnace]

(t^2)/2t + 4/2t = (t^2 + 4)/2t

Right ?

[torpid sparrow]

Perfect

Can you solve that quadratic?

[strong furnace]

What is quadratic pls ?

[torpid sparrow]

that equation is called quadratic because it has 2 solutions

Now you have (t^2+4)/(2t)=5/2

[strong furnace]

Ohh it’s with delta ? B^2 - 4ac ?

[torpid sparrow]

Start by multiplying both sides with 2t

To simplify the fraction

[neon iron]

Cmon imran

[torpid sparrow]

Do you have it?

[strong furnace]

We obtain

t^2 - 5t + 4 = 0

Then we do delta to obtain x1 x2 ?

[torpid sparrow]

Are you sure?

Lets be sure first

(t^2+4) = 5t

Perfect

Now solve for t

[strong furnace]

Yes imo

[torpid sparrow]

Obtain t1 and t2

[strong furnace]

Okay give me 1min

T1 = 4 and t2 = 1

[torpid sparrow]

Before continuing you have to check thise answers are correct by plugging them into the equation t/2 + 2/t = 5/2

Lets verify

4/2 + 2/4 = 2 + 1/2

Which is 5/2

Second one

1/2 + 2

Also 5/2

Now that you found both solutions for t

You have to find the solutions for x

[torpid sparrow]

Do you relember what was t equal to in terms of x?

[strong furnace]

2^x if my brain is still good

[torpid sparrow]

Ok so

One of the t solutions was 4

So 2^x = 4

[strong furnace]

So x = 2

x1*

[torpid sparrow]

And the other

2^x = 1

[strong furnace]

0

[torpid sparrow]

So now you have the two solutions for x

X1 = 2

X2 = 0

Last step

Plug your answers to check if you are correct in the original equation

2^(x-1) + 2^(1-x) = 5/2

2^(2-1) + 2^(1-2) = 5/2

2^1 + 2^(-1) = 2+1/2=5/2

2^(0-1) + 2^(1-0) = 5/2

1/2+2=5/2

Verified

Did you understand the whole process?

[strong furnace]

Yes sir tysmmmm !

[torpid sparrow]

You are welcome

[strong furnace]

Can I give you something in return ? Maybee a tip ?

[torpid sparrow]

No need, I do this with pleasure.

!done

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[strong furnace]

❤️ have a good day/night

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[flat tapir]

how would i get the answer to b

i know it's the dimension of the corresponding eigen space but

how am i meant to reach that

this is the answer key but i dont really get it

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[tame cypress]

(b) + (Some knowledge of linear algebra and abstract algebra required.) The set A consists of n + 1 positive integers, none of which have a prime divisor that is larger than the nth smallest prime number. Prove that there exists a non-empty subset B C A so that the product of the elements of B is a perfect square. can someone

[tame cypress]

tell me which topics in abstract algebra and linear algebra this question is referring to so i can come back to this question after reading up on those

please dont tell me answer or how to do problem

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[fallow igloo]

Factors and multiplies

[tame cypress]

ty i got it

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[fossil osprey]

How would u do this

[coarse tusk]

Consider three cases

1. Neither the man nor the woman are chosen
2. Only the man and not the woman has been chosen
3. Only the woman and not the man has been chosen

We specifically omit the 4th case as it’s given to us that that case isn’t possible

[fossil osprey]

what is the posiibility

of 2 and 3

[coarse tusk]

Just omit that one man/woman from your choice

[fossil osprey]

for case 2 i did 2(1)(11C5)

[coarse tusk]

Why 2(1)?

[fossil osprey]

I mean 2x4x11C5

[coarse tusk]

Not quite

[fossil osprey]

what would it be

[coarse tusk]

You need to choose 2 men out of 4 men and 5 women out of 11 women

[fossil osprey]

yes

[coarse tusk]

So what would it be?

[fossil osprey]

so would it be 1x4x11C5?

for only the man

and not that one female

[coarse tusk]

Ahh wait you were doing the 2nd case

My bad

[coarse tusk]

Yeah

What about case 1?

[fossil osprey]

5C2x12C5

if this was the case

Do I do each case individually?

[pearl fog]

i think we can just disregard the men for now and just focus on the women, i think its the same as before

so 5c3 (neither) + 5c2×2c1 and thats the possibility from women

thencmultiply for mens

[fossil osprey]

.close

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[grave hedge]

Can someone help me with part B, why is the correct value of b = -4 and not a positive 4?

[tame void]

When you look at a n-degree polynomial with constant term c, the product of roots is ((-1)^n)×c.

So here you'll have (-b)=product of roots

[opal vault]

$\alpha^2 \beta^2 = \text{product of roots} = -b$

[thorny flame]

rafilou2003

[grave hedge]

Oh okay thanks

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[neon iron]

I'm quite confused on what im supposed to be doing here

[neon iron]

so this is what i have so far

but im not sure if thats correct

[woeful drift]

One way to check is to see what the height is at time 0

[neon iron]

ok

[neon iron]

sorry for the wait

[woeful drift]

This doesn't look at all like the equation you have for y

[neon iron]

whoops yes

[woeful drift]

Oh wait that's a 100, nvm

[woeful drift]

Yeah I thought the red curve covered up a 400 not a 100

Ok anyway, according to the question, what should the height at time 0 be?

[neon iron]

3 💀

[woeful drift]

Alright then

[neon iron]

my equation is clearly wrong

[woeful drift]

Unfortunately

[neon iron]

how woudl i be able to fix it

[woeful drift]

It's going to take a little more than just plugging values into a premade template

According to the dimensions of the wheel, what is the range of heights?

[neon iron]

yes ik

[neon iron]

if the wheel starts at 3 meters, then [3, 103]

[woeful drift]

Ok, so we have an amplitude of 50

The amplitude of cos is?

[neon iron]

what do you mean by that

[woeful drift]

By what

[neon iron]

amplitude of cos

[woeful drift]

What is the definition of amplitude?

[neon iron]

heigth

[woeful drift]

Be more specific please

[neon iron]

are oyu asking me to find the amplitude of y=cos(x)

[woeful drift]

Sure

[neon iron]

which is 2

[woeful drift]

Is it?

[neon iron]

uh

oh

1

[woeful drift]

So with this in mind, your vertical scaling factor is

[neon iron]

50

did you just divide by 2 to get 50

as a cos function's amplitude is 1, so 1*2 = 2

[woeful drift]

That follows from how amplitude is defined

Now is it going to be 50 or -50?

[neon iron]

why would it be -50

oh to start a 0

[woeful drift]

Well, consider cos

At x=0, are you at the top or bottom of the range?

[neon iron]

top

[woeful drift]

And with -cos, are you at the top or bottom at x=0?

[neon iron]

bottom

[woeful drift]

Does that answer this?

[neon iron]

yes

[woeful drift]

Alright

The vertical offset is probably the easiest thing to find next

[neon iron]

3

or

hold on i need to check

[woeful drift]

That would be wise

[neon iron]

so right now we have y=-50cos(x)

[neon iron]

as the amplitude is 50

[woeful drift]

So far so good

[neon iron]

it needs to be shifted to ground level, so +50

and then shifted up 3 more

so +53

[woeful drift]

Much better

[neon iron]

yay

[woeful drift]

Now all that's left is to consider the period

[neon iron]

sorry 1 sec lemme write all thsi down

y=-50 cos⁡(t)+53

[neon iron]

if 1 revolution is 2pi

[woeful drift]

They ask you to use x not t but sure

[neon iron]

did they ask to use x

oh they did

nvm

[woeful drift]

Anyway continue

[neon iron]

then every 30 minutes it travels 2pi

so every 60 its 4pi

uhhh

so the wavelength is.. 4pi

?

[woeful drift]

Not quite

You want the argument of the cos, ax, to be such that when x = 30, ax = 2pi

[neon iron]

ohhhh

30a=2pi
x=2pi/30

pi/15

?

[woeful drift]

Better

[neon iron]

so then it sjust y=-50cos((pi/15)x)+53

[woeful drift]

Seems like it

[neon iron]

tysmmm

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[neon iron]

completely stuck doesn’t make any sense to me, tried chatgpt tried google tried youtube

[versed cairn]

Find point of intersection of first 2lines
Find slope of 3rd line
Find slope of new line
Find equation of new line

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[meager dawn]

<@&268886789983436800>

[neon iron]

<@&268886789983436800>

[meager dawn]

lol

[agile harness]

<@&268886789983436800>

interesting thing to spam

[edgy wyvern]

<@&268886789983436800> .. there's something happening here..

[little ore]

nah it's fine

[thorn briar]

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[little ore]

wow ok

[agile harness]

bruh

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[spiral robin]

Can't really understand the logic behind dot product projection, which results to a scalar value, and why its different from the magnitude provided by the cross product. and why should i take the resultant perpendicular to only a vector, and why even bother take perpendicular at all. And why the heck should i follow that stupid right hand rule.

[spiral robin]

please help

[near crypt]

[spiral robin]

Cant even make people understand what i am trying to ask even in maths stack exchange

I cant find any resources over the net to understand it. My teacher is also not qualified/ cant make me understand what i want

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@spiral robin Has your question been resolved?

[split osprey]

Right hand rule is a convention, just to maintain uniformity

We could have been using left hand rule

Then someone could have asked

Why we use left hand rule

The answer is, we just stick to one

And everyone follows that conventino

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[agile root]

Find the flux of the vector field F=<y,-z,x> across the part of the plane z=1+4x+3y above the rectangle [0,4] x [0,2] with upwards orientation

[agile root]

Can someone help me with this calc problem :3

[rigid ivy]

If you can wait 10 hours for me to go to bed and wake up tomorrow, yes

[agile root]

okay :3

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[spiral robin]

.reopen

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[rugged idol]

hey everyone!

I am confused with this exercise and would appreciate it a lot if someone could help me understand it.

Q: You have a regular deck of 52 cards and permute them. What is the probability that within the first two cards at least one is an ace.

There is this "standard" approach over the counterevent . But I tried calculating the probability that exactly one is an ace + probability that exactly two are aces.

Now P(X=1) = (4 C 1) * (2 C 1) * 1/52 = 8/52 because we can have the ace at position 1 or 2 and we can have one of 4 aces. The probability that this position is an ace is 1/52.

For P(X=2) I got confused.
I thought of something like (4 C 2) * 2! * (1/52)^2 = 4 * 3 * (1/52)^2 but that was off.
Does anyone know how I would determine P(X=2) ?

[sonic escarp]

for "at least one is ..."-probabilities you should use "no one is ..." as counterevent

[rugged idol]

thanks, but as I said, I am trying to solve it without this method

or more specifically , I am trying to solve the "exactly 2" case

[sonic escarp]

whats the probability for "the first is an ace"?

[rugged idol]

4 / 52

oh hang on

4/52 * 3/51 should be it

[hollow plover]

When you consider atleast one ace separately you need to consider three cases

First draw ace second not ace
first not ace second ace
both ace

[rugged idol]

I am sorry, I dont understand what you siad

oh

wait but thats not what I am trying to do

[hollow plover]

Your way is correct !

[rugged idol]

no its not

it differs from the solution

[hollow plover]

But why will you do

4c2

it will be 4/52 × 3/51

[rugged idol]

yeah the point is that

I missed a -1 in the denominator

4c2 * 2! = 4 * 3

thats the same thing

[hollow plover]

Yeah

Your issue solved ?

[rugged idol]

not quite

somehow its still wrong

but I need to check it for myself again

[hollow plover]

Whats answer ?

[rugged idol]

well the "easy" solution would be 1 - (48/52) * (47/51)

[hollow plover]

when you consider one ace

You must consider the other one isnt an ace

Multiply by 48/51

48/51×8/52

[rugged idol]

ohhh

genius

genius

man , thanks a lot

appreciate it

[hollow plover]

Welc bro ! Im glad to help

[rugged idol]

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[dusky gorge]

Hey community

i was reading this book called
introduction to theory of computation pdf michael sipser
second edition, availible as pdf here.

On the page number 7 he talks of

The set of possible inputs to the function is called its domain. The outputs
of a function come from a set called its range. The notation for saying that f is
a function with domain D and range R is
f : D−→R.
In the case of the function abs, if we are working with integers, the domain and
the range are Z, so we write abs : Z−→Z.```

How can the range of abs value function be integers?

[hollow plover]

Modulus function

Acts like |x| = x when x >0
And |x| = -x when x<0

So the domain of modulus function may range from (-∞,∞)

But range is [0,∞)

[dusky gorge]

yesh but the author says the range to be also Z/ integer which is confusing

[hollow plover]

so its range is not integers but natural numbers including 0 according to question

[hollow plover]

Author means in computational terms, but concept of natural ,whole numbers

is mathematical purely

so maybe he has used in a casual way rather

[dusky gorge]

fair when i looked some up
this https://www.cuemath.com/algebra/absolute-value-function/

Defines the abs function as
f(x) = a |x - h| + k

So for this particular defination the function can take negative range if a is negative.

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[rapid sky]

please walk me through the process of finding the general term for this (question C)

[icy sky]

did you add 1 to each term?

[rapid sky]

you mean (n+1)?

that does not work

what im given is
1 1
2 3
3 7

[icy sky]

add 1 to 1,3,7, etc

[rapid sky]

f(n) = (n+1)
only works for term 2 but not the others

f(n) = 1 + (n+1)2 does not work either

yes

[rapid sky]

.

is there even a general formula for this

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[burnt swift]

In $\triangle PQR,$ the sum of $\angle P$ and $\angle Q$ is $60^\circ.$ What is the measure of $\angle R?$

[thorny flame]

938c2cc0dcc05f2b68c4287040cfcf71

[radiant tapir]

What does a triangle's interior angles have to add up to?

[burnt swift]

180

[radiant tapir]

bingo. So we know that P + Q + R = 180 and that P + Q = 60. So we have 60 + R = 180. So just solve for R

[burnt swift]

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[burnt swift]

.reopen

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✅

[burnt swift]

can I ask about this

?

[neon iron]

what about it

[dim pulsar]

what can you say about CED and CFG?

[burnt swift]

they have the same angle 1.5x

I think

[dim pulsar]

yes

since AD and FG are parallel

[burnt swift]

due to what? theorem

huh?

[dim pulsar]

uh i believe its a postulate or something in euler's element book

i forgot the name of it

[prisma mesa]

Corresponding angles

[dim pulsar]

yes

wow meth is always right

[burnt swift]

?

[dim pulsar]

if you search "parallel angle theorem" u will find stuff about it

[prisma mesa]

The name is corresponding angles theorem i believe

or similar

[dim pulsar]

yep

[burnt swift]

sure

so what?

[dim pulsar]

anyways, we know x + 2x + 1.5x = 180

solve for x, then find 1.5x

[burnt swift]

huh?

[dim pulsar]

the angles

x, 2x, and 1.5x form a line

[burnt swift]

okay

[dim pulsar]

so they sum to 180 degrees

[burnt swift]

yeah I agree

[dim pulsar]

x + 2x + 1.5x = 4.5x

4.5x = 180

remember, we are looking for 1.5x

[burnt swift]

,calc 180/(4.5)

[thorny flame]

Result:

40

[dim pulsar]

so we divide by 3 on both sides

1.5x = 60

[burnt swift]

,calc 40*1.5

[thorny flame]

Result:

60

[burnt swift]

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[burnt swift]

.reopen

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✅

[burnt swift]

what about this one?

[dim pulsar]

okay

[burnt swift]

PT || QR

[dim pulsar]

extend TR a bit on both sides

makes it easier to visualize

[burnt swift]

can you do a drawing for this?

[dim pulsar]

i cant draw on my device rn

anyways, what im trying to show is the relationship between the angles

2x

and TRQ

and the 128 degrees

,calc 180-128

[thorny flame]

Result:

52

[dim pulsar]

TRQ = 52 degrees

52 + 2x = 180

find x and the rest is simple

[burnt swift]

52 + 2x = 180?

[dim pulsar]

yes

[burnt swift]

sum of interior angles is 180

I need more help

please help me

[sharp orchid]

with what

wait let me see

[gritty birch]

I'll be honest, he already did a lot of the work/gave you a lot of clues, but it doesn't seem like you care to spend another couple minutes figuring the rest out

[dim pulsar]

a simple linear equation to solve for x

[burnt swift]

2x = 52

[unborn monolith]

no

[dim pulsar]

52 + 2x = 180, or 2x = 128

[burnt swift]

2x = 128

2x = 128

alternate angles

sure

[dim pulsar]

yes... so solve for x

[burnt swift]

x = 64

[dim pulsar]

then replace the values in the diagram with their corresponding values

we have all the angles we need to solve for PQR

[burnt swift]

huh?

PQR = 180 - 64

,calc 180 -64

[thorny flame]

Result:

116

[burnt swift]

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[burnt swift]

for you guys everything is easy

[gritty birch]

that has nothing to do with easy or not

you don't even try

[burnt swift]

I tried but could not find answer

[gritty birch]

oh really?

[gritty birch]

you asked that

not even a minute after he gave you clues

you didn't even spend more than 1 single minute to think about what to do next after he helped for a couple steps

[dim pulsar]

😨

[gritty birch]

and you're telling me you tried

[burnt swift]

he didnt explained some of the properties I was still lost

he assumed I already knew everything

[dim pulsar]

i literally gave you the equation... just a linear equation

[gritty birch]

in that case you should ask him for more specific clarification, instead of just "more help" which seems like you just wants more answer

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[vivid saffron]

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[quaint badge]

i need help on how to logically deduce the answer for question #8, through brute force i can figure out that the side lengths are 3, 5, and 7 cm respectively, however i dont know what the intended process is. I would guess to use ((x)^2 + (x+2)^2 + (x+4)^2) * 6 = 381, however this will be incorrect as it doesnt account for the overlapping area, how would i model an equation that does correctly account for this area? (i know my equation is false because x must equal 3 in it, which will evaluate to 498).

[vernal matrix]

,rccw

[thick kindle]

so

umm

i comed up with this

assuming that consecutive odd meant 3,5,7 and 9,11,13

basically the covered area was technically covered twice

as the cube on top's bottom have been covered and the cube on the bottom's top have been covered

so I use

$6x^2+6(x+2)^2+6(x+4)^2-2x^2-2(x+2)^2$

[thorny flame]

calinn

[thick kindle]

idk about the bottom of the largest cube

so I assume its included and if it wasn't we could always go back and change the answer

and simplifies down to $14x^2+64x+112$

[thorny flame]

calinn

[thick kindle]

then we should use the quadratic formula

and came up with some random number

[quaint badge]

isnt the discriminant imaginary with this?

[thick kindle]

oh shoot

forget about 381

alr

umm

apparently I forget about the 381 and got imaginary

i think the intended process is $6x^2+6(x+2)^2+6(x+4)^2-2x^2-2(x+2)^2-(x+4)^2=381$

[thorny flame]

calinn

[thick kindle]

@quaint badge

since the bottom count as covered area

and simplifies down to $13x^2+56x-285=0$

[thorny flame]

calinn

[thick kindle]

sry if there is mistake

[quaint badge]

which should give x = 3 and some negative thats just rejected

alr thanks, i just wanted to know what the intended way of doing it was

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[sweet fog]

hello, im struggling with this problem

[sweet fog]

find a quadratic polynomial whose solution set contains all of the solutions of the equation

one sec for screenshot

Where would i even start with that?

This has been killing me, ive been studying and doing this assignment for over 6 hours today and it's like im just sinking in quicksand getting nowhere

[jade thunder]

What if you move the y to the right

[fallow igloo]

Hint: move y from the left to the right

[jade thunder]

I somehow what to get rid of the square root thing

[sweet fog]

for a solution set theyre asking for coordinates?

[jade thunder]

It’s really the big obstacle

[jade thunder]

No we just want to first rewrite this condition a bit better

[sweet fog]

one sec let me try this and take a photo

[jade thunder]

Right now I can’t tell what that condition looks like

[fallow igloo]

For 6 hours…

[sweet fog]

yes

i'm not exaggerating. i'm struggling hard

it hasn't been this one problem specifically, it's the homework assignment. i'm only 13 out of 30 done after all this time, and as soon as i get to class monday there's a quiz where i'm sure i'll choke and forget how to do everything

[jade thunder]

So now

I’m thinking we could refactorise the y’s

If I expand them, collect like terms, then complete the square

I’ll get another (y-a)² term with some extra + b at the end

[sweet fog]

so just expand the y's?

[jade thunder]

Looks like a circle to me!

[sweet fog]

ok one se c

am i on the right track

ok cool

this feels absolutely awful and it seems impossible to get help in a way that's understandable

and then people just vanish in the help chats

[mint crescent]

,w expand (y+3)^2-(y-2)^2

[thorny flame]

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[mint crescent]

the process to get from the first pink to the second pink is fine

[fallow igloo]

nah

[topaz sinew]

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[neon iron]

can someone help

[topaz sinew]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[neon iron]

o

[gritty birch]

do you know what's the derivative of ln(x)

and product rule

[neon iron]

hm

[neon iron]

i think 1/x

[gritty birch]

that's right

what about product rule

[neon iron]

1

_

x

wait

then that means

[haughty wren]

There‘s a rule for differentiating a product of function, do you know it?

[neon iron]

1 + In (x) is the answer then

[neon iron]

i forgor

[haughty wren]

$\dd{x}(f \cdot g) = f^\prime g + g^\prime f$

[thorny flame]

ℑμΤ𝛄𝛗θ

[haughty wren]

Where f and g are your functions

[neon iron]

.

wait

[haughty wren]

I meant d/dx not just dx

Just noticed that

[neon iron]

wait

is this the answer

[haughty wren]

Yes, but do you understand why?

[neon iron]

u have to use the product rule f (x) = x and g(x) = In (x)

then the rest of this formuLa

which is equal to 1 + In (x)

all togetHer

i think

[haughty wren]

Yup!

[neon iron]

thank you!!

[haughty wren]

No problem :)

[neon iron]

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[neon iron]

can some one help on number 3.

[icy sky]

have you drawn your figure

[neon iron]

no

[icy sky]

do you need help with that

[neon iron]

sure

yea'

[icy sky]

do you know how to graph g at least

[neon iron]

i guess

only a little

actually yeah

hello?

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[fallow igloo]

Have you solved it?

[neon iron]

yeH

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[manic nebula]

a+b+c=10
a²+b²+c²=20
a³+b³+c³=30

a⁵+b⁵+c⁵=?

[manic nebula]

I can normally solve if there are 2 unknown values

But that just doesn't really get anywhere

[terse abyss]

3 equations in 3 unknowns, this should be solvable

[manic nebula]

(it's supposed to be solvable tho)

It is, a number between 0 and 9999 btw

(as if it changed much, lol)

[terse abyss]

Is it given whether a b c are integers/etc.?

[manic nebula]

Nope

But the answer belongs to N

[terse abyss]

Is there a method you’re supposed to use?

[manic nebula]

Nope

[terse abyss]

Yeah I put it in a solver and it won’t solve…

[manic nebula]

It's from a math contest

It just happened today

I really thought it wasn't solvable when I first saw it's probably either unsolvable or typed wrongly

[terse abyss]

I suspect it’s a typo somewhere

[manic nebula]

Yea

So uh

How do I end my help request?

A

Nevermind

Thanks

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[neon iron]

This text books first chapter covers these topics and is called “fundamentals”

Is there a whole book covering these topics in depth and what is the subject called

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[maiden wave]

wtf happened here

[maiden wave]

is that legal ?

[loud oasis]

substituted (vx) for y

[maiden wave]

yhea ik

but why can they do that

and what happens to the d

does it beconme dvdx/dx

so dv only

[loud oasis]

it is simply applying the derivative to the function $y = v(x)$: $\odv{(vx)}{x} = \odv*{(vx)}{x}$

[thorny flame]

cloud

[maiden wave]

oh

ok i see

thanks

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[lucid oar]

a and b are integers,

P(x)=x³+ax²+bx-2
And this polynomail has only one real root.

P(1)=0, what is the smallest value for the a?

[lucid oar]

Idk where to start, i just know a+b=1

[fallow igloo]

Have you learnt complex number?

[lucid oar]

Yeah but i dont remember much

[cursive patrol]

yes

now replace b in terms of a and compute the quotient P(x) / (x - 1)

this removes the root 1 and gives you a polynomial with the other two roots of P

[lucid oar]

Oh X³+ax²+(1-a)x-2

[fallow igloo]

Have you done removing 1 from the roots?

[lucid oar]

Is it like that ( sorry im not good with english math terms)

Delta is a²+6a+9=0 (a+3)(a+3)
A= -3

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[fallow igloo]

Glad you figured it out on your own

[fallow igloo]

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[meager dawn]

think about what's going on away from 0. What is 1/cot(x) (just normally) (by that i mean that there is a nicer form for this that might be more familiar/nice)

[restive inlet]

lh not applicable here

[upbeat minnow]

but why

if you plug in x into 0 its 0

and if you plug 0 into cot (x) you get undefined

[restive inlet]

state when you're allowed to use lh

[shut belfry]

cotx = cosx / sinx

we could use this

[upbeat minnow]

when you get an indeterminate or something

[restive inlet]

and what are the indeterminate forms

[upbeat minnow]

0/0 and inf/inf?

i think

so it doesnt matter that its undefined

ok

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[shut belfry]

sinx.3x/cos x = 0/1

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[lethal star]

what am I doing wrong?
the answer should be – (1/xy) + log |(x/y)| = c