help-26

[long lake]

what

4C4 is 1

combination of 1?

[stuck hearth]

And the only combination

Is abcd

Correct

[long lake]

what do I understand from thsi

[stuck hearth]

That you’re over counting

Try to list all combos

When you do 4c2 2c2

It’s different from 4c4

[long lake]

can you tell me the wrong thing in this

i don't get anything from abcd example

<@&286206848099549185>

[pallid sigil]

so the result is positive if and only if:
• 0 of your choices are negative
• 2 of your choices are negative
• all 4 of your choices are negative

So in others words, we need to find
P(0 negatives and 4 positives) + P(2 negatives and 2 positives) + P(4 negatives and 0 positives)

which i think equals:
(8C0)(6C4)/(14C4) + (8C2)(6C2)/(14C4) + (8C4)(6C0)/(14C4)

hopefully that makes sense

[topaz sinew]

@long lake Has your question been resolved?

[cunning snow]

She isn't asking the answer she is asking her method is wrong

[pallid sigil]

o ok

mb

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[dull escarp]

I Have the function f(x)=x+1/x and I have to find f^-1([-2;4)) , I want to know if what I did is correct

[topaz sinew]

@dull escarp Has your question been resolved?

[dull escarp]

<@&286206848099549185>

<@&286206848099549185>

[unreal quest]

Your function is (1+x)/x, right?

[dull escarp]

no

x+1/x

x+(1/x)

[unreal quest]

Alright

[unreal quest]

Aren't they asking you for an analytic expression of the inverse function for that interval?

[dull escarp]

no

just the interval

[unreal quest]

Then pretty sure you phrased this wrong

Well, regardless, I'll leave this one to someone else

[topaz sinew]

@dull escarp Has your question been resolved?

[dull escarp]

<@&286206848099549185>

[thorny flame]

Pupu&Bubu

Pupu&Bubu

Pupu&Bubu

[cunning snow]

Wait arnt they asking the range of inverse function of x+1/x over the interval -2 to 4

Yea so it is asking a expression for inverse function

[dull escarp]

si pupu

è corretto quello che ho fatto?

[thorny flame]

Pupu&Bubu

[dull escarp]

bene perchè non ho la soluzione di questo esercizio e quindi mi serviva che qualcuno confermasse che il risultato fosse corretto

thanks

[thorny flame]

Pupu&Bubu

[dull escarp]

prendi (-infinite; 2-3^1/2) e (2+3^1/2,+infinite)

io ho x^2-4x+1 >0

vabbè in ogni caso il mio ragionamento è corretto quindi va bene ; grazie per la conferma

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[pulsar dust]

I need help understanding this

I get that all the sides are the same, but I am confused on how to get the height

[restive inlet]

there's a formula for area of equilateral triangle if you're allowed to apply that
if not, bisect the triangle and apply pythag

[pulsar dust]

can you show me the work to bisect the triangle?

hello

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[fair copper]

area of equilateral triangle = (√3)a^2/4

where a is side length of eq triangle

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[deft holly]

help with 19

what should i sub in to show its divisible

c and -c and replace a with b+c

and make the eq P(b), sub c or smth in?

[snow nimbus]

Sub a = b+c in and use binomial formula to make simplifications

[deft holly]

how would i use binomial formula to simplify?

[snow nimbus]

Do you know what it is?

[deft holly]

break them up into their thingggsss

yeaa

[snow nimbus]

Then use it for the (b+c)^n

[snow nimbus]

Great

Now do you see how to continue?

[deft holly]

sub k=2 and simplify?

nah

[snow nimbus]

You can immediately simplify. Some terms appear in the sum and also on the right side

[deft holly]

sorry, how so?

is it just like

in the sigma notation theres the

uh

terms

[snow nimbus]

Yes

[deft holly]

and every pther term when crossed out is divisible by c^2

cause itll be c^2 term till c^n?

ah

[snow nimbus]

I mean what we want to show here is that every term where k =< 1 here cancels as every term with k > 1 already has a c^2

[deft holly]

can i have help with a closing statement

c^k, k>=2 therefore divisible by c^2?

[snow nimbus]

Can you further simplify T0 and T1?

[deft holly]

ah yea i will

is this alright?

[snow nimbus]

Yeah

[deft holly]

tysm

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[snow nimbus]

Don't forget for a complete proof you need to consider n = 1, too

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[quaint bone]

I don't know how to start this question

[quaint bone]

This is an activity we're doing for our class and we already had 2 quizzes on this unit, but for this question

I'm at a complete loss

How would I even start?

[snow nimbus]

The idea here is to use a decreasing "basis function", which gives us the general behavior, that goes from the top left to the bottom right and add sin x to it

So f(x) = b(x)+sin x

What would you recommend for b(x)?

[quaint bone]

hm

maybe -1?

[snow nimbus]

That's just a horizontal line

[quaint bone]

I was playing around with this

and this is pretty similar to sinx-99/100x

[snow nimbus]

So you meant -x

[quaint bone]

ye

WAIT

I said -1

mb

I meant to say -x

[snow nimbus]

Yeah, I think -0.5x looks even better

But this is already not bad

[quaint bone]

ye

the goal is for it to look like this

[snow nimbus]

Yes, I think -0.5x+sin(x) does the job well enough?

[quaint bone]

fair enough

Thanks

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[quaint bone]

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[gleaming bane]

I need help D: I don't know how to continue

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@gleaming bane Has your question been resolved?

[gleaming bane]

<@&286206848099549185> 🥺

[lone stone]

well,

(m+1)^n > mn. => (m+1)^n / m > n

(m+1)(m+1)^n > m(n+1) we can rewrite as (m+1)(m+1)^n / m > n + 1;
(m+1)^n / m is larger than n. m+1 is larger than m => (m+1)(m+1)^n / m is larger than nm which is larger than n+1 if n and m > 1. Lets look if they both = 1, 2 > 1 (true).

So it is correct for n and m >= 1

is it clear?

[neon iron]

would suggest inducting on n

[gleaming bane]

yees i think i got it

[gleaming bane]

what do you mean by that?

[neon iron]

treat m as a constant number and induct on n

[gleaming bane]

ooh yeah, that's what I was doing but I didn't know how to continue :c

[neon iron]

(m+1)(m+1)^n > (m+1)mn = mn+m^2*n >= mn+m = m(n+1)

[gleaming bane]

OMGG Ty very muuuuch!!

can i take that mn+m^2*n > mn+m? or must it be >=?

[lone stone]

i think u cant

[gleaming bane]

ty!!

I love you guys ❤️

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[rare marten]

i got up to 1/2(r^2+7r+10)sin(30)=(r+1)^2

[rare marten]

idk how to continue

[flint stump]

so can you sub in the value for sin 30?

[rare marten]

0.5

[flint stump]

yeah

so put it in the equation you have

[flint stump]

now rearrange all terms so you can see the equation in a for similar to what you need to show

[rare marten]

like this?

[flint stump]

yes

and simplify so you group together the coefficients for r^2 and r

[rare marten]

or is this form better

[flint stump]

But check for the signs. There is some mistakke there

[flint stump]

either one works

whichever form makes it easy for you to work with, take that

[rare marten]

whats the mistake abt

[flint stump]

when you shift the term on other side, its sign changes

[rare marten]

o yeah

hold up

this better?

[flint stump]

yep

now group them together

[rare marten]

waht should i do abt the 0.25

this?

15

my brain just doesnt work half of the time

ofc lmao

look at my pfp

[flint stump]

everyone works at their own pace. No need to call them out like that

[rare marten]

when my brain revives i can do it ez

[arctic briar]

没事练练就会了

u guys are lucky to have brains

superb chinglish right here

卧龙凤雏棒棒哒

ok time to botherhelp others

[rare marten]

alr thx for help guys

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[lucid junco]

how do i do part a) using the simultaneous eq?

[vernal matrix]

[lucid junco]

chaaaaaaaaaaaaaaaaaaaaaaaaaaaaartbit

help me

theyre making me manually find it without a calculator

[vernal matrix]

(put in the appropriate sums, and with n as how many points you have, no?)

[lucid junco]

[vernal matrix]

What that's painful asf

[lucid junco]

i don get it whats my a and b here

[vernal matrix]

You want to find the a's and b's, no?

[lucid junco]

yes

[loud oasis]

you plug in the known x and y and solve for the unknown a and b

[lucid junco]

so n =7

oops

wrong q im looking at

n=4

im confused

im looking for a1, b1

and then a2, b2?

[vernal matrix]

(hang on, I was gonna say, am I tripping or did they put the a2 and b2 as exactly the same )

[arctic briar]

lol it didn't ping cuz it was edit ping, also that man got blocked for u know what

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[nimble crown]

this is physics help again soz but how do I figure this out?

[haughty mural]

one ampere is defined as one coulomb per second

so what we need to know is how many coulombs each electron carries

coulomb being the charge

[nimble crown]

oh so we multiply each one by 1.602*10^-19

[haughty mural]

yeah

[nimble crown]

but from my notes, the formula to find current is Q/t

but t is 1 second, so is Q just electrons x charge?

[haughty mural]

so we get 1.59 * e^16 * 1.602 * 10^-19 coulombs per second which is amps

yes

[nimble crown]

huh

[haughty mural]

we have X electron per second

each electron carries a charge q which gives us a total charge Q that is delivered each second
so we get Q/t, t being 1s and Q being number of electrons * electron charge

[nimble crown]

but since it's in milliamps I assume we muliply by 10^3?

[haughty mural]

yes

[nimble crown]

perfect

perfect got my answer,

sorry if this is a bit much but could you help me on another? it's a circuit

also ty for the help

[haughty mural]

sure thing

[nimble crown]

I do not know how to read circuits well

but since they are parrallel I assume ampeter 2 and 3 read the same thing??

[haughty mural]

they might look parallel, but they are not the same

R2 and R3 are different

so lets assume R3>R2, meaning there is less resistance going the upper path

so there will be more current in the top

[nimble crown]

yes makes sense, current takes the path of less resistance

martin?

[haughty mural]

yep

[nimble crown]

oh thought you left

[haughty mural]

for number 2

U2=U3

does that make sense?

[nimble crown]

U2?

what is U referring to?

[haughty mural]

voltage

ah you use V maybe?

but there is no reason why U1 should be U2+U3
after all, since U=R*I, consider the following:

increase the resistances R2 and R3
this will result in less current overall in the whole circuit
thus I1 will decrease as well
but since we didnt change R1, U1 will go down because of lower overall current

[nimble crown]

I see

oh yea U2 and U3 are the ameter

but your point still stands

[haughty mural]

U2 and U3 are not ammeters

do you use V for Voltage? we can use V then

[nimble crown]

I mean that when you say U2 and U3 the question is asking that V1 is A2+A3

so mix up I assume

[haughty mural]

V1 can never be A2 + A3

V1 is a voltmeter that measures a voltage U in V (Volt)
A1 and A2 are ammeters that measure current I in A (Ampere)
and Ampere plus Ampere is not Volt

[nimble crown]

I see

[haughty mural]

as for number 3:
U=R*I, thus U/I = R
so what we are looking at are the resistances
and it claims that: R1 = R2 + R3
is that true?

remember, that the resistors can have any values

so we can just choose some and see that it doesnt work out

[nimble crown]

could be...

[haughty mural]

for example 3 is not 5 + 6

so we found a counter exampple

[nimble crown]

oh

[haughty mural]

number 4, we already did

what do you think of number 5?
it claims that: I1 = I2 + I3

[nimble crown]

no we didn't do 4, we did 1, this is for voltage now

oh is it the same principle?

[haughty mural]

ah wait

suppose this is our setup

and we apply a voltage from left to right

now we want to measure the voltage over the upper lamp

so we might do this

which is the same as this

[nimble crown]

wait since it's parallel, wouldn't they have the same volt reading?

[haughty mural]

which is the same as this

so measuring the voltage over the upper or lower part is the same thing

[haughty mural]

yes

[nimble crown]

so thats right ok I see

[haughty mural]

for number 5:
suppose 10 electrons go through the circuit per second

[nimble crown]

ok

[haughty mural]

meaning 10 electrons pass the first ammeter

now at the junction, they go up or down

they have to decide

meaning the split up

for example if 7 go up, only 3 go down

meaning that I1 = I2 + I3

note that this is more of a intuition thing and not a proof, but good enough here

[nimble crown]

wait that's how it works

huh

I see

ok perfect, thank you so much for the help martin!

also soz for asking in a math discord of all places lol

the physics discord isn't the quickest when it comes to phys, really good to discuss conceptual ideas though

[haughty mural]

nah its alright

this server has a better layout

physics discord takes longer, i agree

and there is alot of overlap

[nimble crown]

ye

ty ty for the help

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[shell orbit]

I thought I knew how to do this but the solution they gave is completely different to what I did 👀

First I wanted to find where ln(8-x) crosses the x axis for my limits of integration

So ln(8-x) = 0 gave me x = 7

Then I integrate this between 0 and 7:

(ln(8-x))^2 - (ln(2))^2

Right?

Nevermind I see my mistake

But im not actually sure what to do now

First I integrate ln(8-x)^2 between 0 and 7

Then I take something away from that, but idk what

<@&286206848099549185>

[wintry urchin]

Notice that you have two different functions to integrate here

[wintry urchin]

This would work too

[shell orbit]

Hmm ok the blue one is just ln(2) * the intercept of ln(2) and g(x)

Im not sure what the green one is tho...

[shell orbit]

Whats the next step for that method tho?

[wintry urchin]

I take back what I said. It is being rotated around the y axis

[shell orbit]

Oh what 💀

[wintry urchin]

yeah lmao

[shell orbit]

I didnt even notice that lmao

Thank you 😅

Hmm this makes it easier tho no?

[wintry urchin]

Yeah

[shell orbit]

I think...

Ok good

The y intercept of g(x) is ln(8)

So between limits 0 and ln(8) we integrate

[wintry urchin]

Not quite

[shell orbit]

Oh?

[wintry urchin]

We start by writing x in terms of y

[shell orbit]

But its g(x)^2 - ln(2)^2, right?

Converted to y

[shell orbit]

I can do that bit no problem im just not sure about what im actually converting to y first if you get me

[shell orbit]

Is it this?

OH WAIT

Is it just ln(8-x), write in terms of y and square to make it (-(e^y) + 8)^2

[shell orbit]

Then integrate that between ln(8) and ln(2), right?

[wintry urchin]

Yes

[wintry urchin]

integrate between 0 and ln(2)

[shell orbit]

Oh yeah lol

Ok I got it now

Thank you so much!!

❤️

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[eager root]

Help me please, is this correct?

[topaz sinew]

@eager root Has your question been resolved?

[eager root]

<@&286206848099549185>

<@&286206848099549185> like this I mean

[vestal sigil]

first line is correct, but second line is incorrect

you cant do it like that

to show you that it is incorrect, just notice the following..

the first line results in a function of x only

(a polynomial in x)

but the second line results in a function of x and n

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[trail timber]

how do i do this

[loud oasis]

write 5pi/12 as the sum or difference of two angles whose trig fucnctions are known

[cunning snow]

Sin(5pi/6) = sin(2×5pi/12)

[cunning snow]

he understands that much

[trail timber]

so pi/6 and pi/4

[loud oasis]

yes. then you can evaluate sin(pi/6 + pi/4) using the angle sum identity

[trail timber]

thats what i get

[loud oasis]

you need to multiply the denominators

[trail timber]

wym

where

[loud oasis]

,, \frac 12 \cdot \frac{\sqrt 2}2 \ne \frac{\sqrt 2}2

[thorny flame]

cloud

[trail timber]

oh

thanks bro

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[deep vector]

how would I go about solving this

[celest bough]

try to divide it in 2 branches

i think thats how you say it in English

then evaluate the lateral limits

[deep vector]

huh

[weak cliff]

Take the limit from both the left and the right is what izzys trying to say i think

[deep vector]

lim x->3 (|x-3|) and lim x->3 (x-3)?

[weak cliff]

No that wont work

[deep vector]

oh

so like 3^+ and 3^-

but how would i do that

make up values like 2.9 and 3.1?

[weak cliff]

Yes

[weak cliff]

You could do that to get a feeling yes but the idea is to notice the plus or negative sign when doing those

[weak cliff]

Because the upper part will always be positive but the lower part of the fraction will change sign if you approach it from the left or from the right

[barren zodiac]

That's one way to look at it, another way is what izzy was trying to say, write it as a piecewise function

[deep vector]

so I dont want a sign change when doing piecewise

[weak cliff]

First evaluate the limit for the first part of the piecewise function worry about the other part later

[weak cliff]

So for x < 3

So you would approach from the left

[deep vector]

is there a reason why it is -|x-3|

[weak cliff]

I was trying to figure that out as well haha

[celest bough]

think about a number less then 3

like 2.9

2.9-3 = -0.1

absolute value (-0.1) = 0.1

[weak cliff]

I would just write it as 3-x for the case of x between 0 and 3 which is thus valid for 3 from the left (2.999…)

[celest bough]

if you plug a number less then 3 you will always get a negative number inside the absolue value

and the absolute value turns it positive

[weak cliff]

This way youll skip the abs value signs

[celest bough]

the samething with the "-" sign i put in the expression always takes a negative number and turns it into positive

[deep vector]

the thing is i dont know if im supposed to do it that way

[celest bough]

my English is not that good i am sorry

[deep vector]

because I was never taught to convert it to a piecewise

[celest bough]

hmmm , so how would you typically aproach a problem like this ? with absolute value ?

[deep vector]

i dont know

[celest bough]

its the first time you are seeing something like this ?

[deep vector]

yes

i was only taught, root, polynomial, fraction

not absolute value or trig yet

[celest bough]

well from my experience when we are talking about limits and absolute value , you should always treat the original function as a picewise function

and then evaluate the lateral limits and see what happens

[deep vector]

no.16 is the one i sent

ok thank you for ur help

i will just email my teacher about this

.close

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[barren zodiac]

You can always write functions with absolute value as piecewise functions, in |x-3| case:
f = x-3 if x≥3
f = -(x-3) = 3-x if x < 3

[topaz sinew]

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[olive lichen]

I have a function such as $\lim_{R \to \infty} \int_{-R}^{R} f(Re^{i\theta}) e^{-iRe^{i\theta}t}i Re^{i\theta} d\theta$. In what situation is interchanging the integral and limits allowed?

[thorny flame]

KySquared

[olive lichen]

I’m aware for function that are absolutely integrable interchanging integrals are allowed. So does that logic extend to interchanging a limit with an integral?

[worthy storm]

the interval of integration depends on R

you can't just move the limit inside

but you could rewrite as

$$\int_{-R}^R = \int_{-\infty}^\infty \chi_{[-R,R]}$$

[thorny flame]

Bungo

[worthy storm]

where $\chi_{[-R,R]}$ is the characteristic function of $[-R,R]$

[thorny flame]

Bungo

[olive lichen]

[worthy storm]

then you have the usual tools available for sufficient (not necessary) conditions to move the limit inside: you could try the dominated convergence theorem for example

[olive lichen]

Gotcha
My initial idea was to just..
WAIT NO I WROTE THE INTEGRAL WRONG

[worthy storm]

oh, what should it have been?

[olive lichen]

Sorry I was thinking about the bounds prior to the parameterization

$\lim_{R \to \infty} \int_{0}^{\pi} f(Re^{i\theta}) e^{-iRe^{i\theta}t}i Re^{i\theta} d\theta$

[thorny flame]

KySquared

[olive lichen]

This is it

[worthy storm]

ohh

that changes things

you can still try the DCT

that R near the end of the integrand might give you trouble

so whether you can move the limit inside will depend on the specific f, i think

[olive lichen]

Thats what I was thinking

[worthy storm]

if |f| decays at least as fast as 1/R as R->infty, you should be all right

(assuming |f| is integrable)

[olive lichen]

Since $e^{-iRe^{i\theta}t} = e^{yt }e^{-itx}$

[thorny flame]

KySquared

[worthy storm]

oh right, crap, i overlooked the e^(itheta) in the exponent, that means its magnitude is not just 1

yea you'll have to take that into account as well

[olive lichen]

the magnitude of the weird exponential is e^(yt)
So as long as f scales slower than that I’m set right?

[worthy storm]

yea

where y depends on theta of course

[olive lichen]

Yeah I just rewrote it in cartesian

Didnt feel like writing Rsin(theta) lol

[worthy storm]

oh wait, theta is the integration variable

ah you can just bound e^(yt) on [0,pi] then

[olive lichen]

So the max value of e^(yt) on [0,pi] is e^(Rt)

Aaaa this is so much more work compared to just moving the limit into the integral

I have to do like 2 layers of integral inequalities

[olive lichen]

Wait how did you get this?

[olive lichen]

My initial logic was “oh after parameterization the bounds aren’t dependent on R moving the limit into the integral is fair game” . I got the right answer in the end but my professor told me that it’s not the “correct way to do it

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@olive lichen Has your question been resolved?

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@olive lichen Has your question been resolved?

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[clear pike]

Hi for the first part I've done the following:

[clear pike]

n-1 and n+1 are both prime

So n must both be even and a multiple of 3

So n must be a multiple of 6

let n=6k

(6k)^2( (6k)^2 + 16 )

144k^2(9k^2+4)

So I can see that this expression is a multiple of 144

However I can't figure out how I would also show it to be a multiple of 5 (144*5=720)

[mint crescent]

Note that n can't be 0, 1 (since n-1 is prime), or 4 (since n+1 is prime) mod 5

So n is 2 or 3 mod 5

I'll let you fill in the rest

[clear pike]

what is mod 5?

[mint crescent]

remainder when is divided by 5

basically n can't be 5k, 5k+1, or 5k+4 for some integer k

so it's of the form 5k+2 or 5k+3

Anyway, I gtg, but you should get that the converse ||is false|| because ||n=78 is a counterexample since n-1=77 isn't prime||

[clear pike]

wait why cant n be 5k?

[mint crescent]

oh wait

I read the question wrong

but it's still the same logic from then on

n=5k -> n^2 div by 5

[clear pike]

29, 30, 31

oh

[mint crescent]

n=5k+2, 5k+3 -> n^2+16 div by 5

[clear pike]

ah right

is there an easier way to show that the expression with 5k+2 or 5k+3 ends up being a multiple of 5, without expanding the whole polynomial out by hand then factorising a 5 out of it

<@&286206848099549185>

[undone summit]

this is only barely easier but you can deduce without expanding that the first 2 terms in the expansion of (5k+2)^2 both have factors of 5 and the last term is 2^2=4 which becomes a multiple of 5 after adding 16 and the same goes for 3

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@clear pike Has your question been resolved?

[clear pike]

cheers

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[sharp dew]

Fk

[sharp dew]

Am I wrong

adj

Help me please

Please

I got 6/5

Why

Why

Please

[restive inlet]

why divide by 5

[sharp dew]

I assume 5 centimeter as x for calculation sake

[restive inlet]

unclear definition

can you try again in clearly describing what your x represents

and you shouldn't have used g and s in your equations like that

[sharp dew]

x = 5t, where t is the length of the belt

[restive inlet]

is wrong

[sharp dew]

Why

[restive inlet]

why are you multiplying to the total length

and you shouldn't have used g and s in your equations like that
306 - 40x = 210 - 24x
the way you're using x in your equation
x would represent the number of 5cm pieces of belt being sewn

[sharp dew]

Exactly

[sharp dew]

Then you got x=6

Am I wrong

[restive inlet]

yes, but then you did the wrong thing with that value

[fallow igloo]

[restive inlet]

number of belts / (length/belt) doesn't give you total length

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[hollow granite]

hi i need help please

i dont understand anything at all

wait i got it

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[queen dagger]

How do I do the 3rd question?

[queen dagger]

And 4

[neon iron]

How far are you so far?

[queen dagger]

Idk

Where I am

[neon iron]

Your issue is with the discount

You didn't calculate a 20% discount, you calculated 20% of the price

[queen dagger]

How do u calculate discount

[neon iron]

So its a 20% discount right, in more common terms 20% off

So take 20% off of 100% and you get 80%

So you just gotta change your fraction

[queen dagger]

Oh so its 0.8 x 185

[neon iron]

Yep

Then x35 again and you should have a profit

[queen dagger]

So thats it? How do I find the profit for all 200 seats?

[neon iron]

So that's asking for a ratio, profit per 200 seats can be simply written as Profit/Seats, or Profit/200

So what's your profit?

[queen dagger]

Oh ok profit by 200

[queen dagger]

Its 26100

[queen dagger]

And then how do I do 4?

@gray sand

[neon iron]

A bit different than I got, how did you get 26,100?

[queen dagger]

Idk

[neon iron]

Yeah you just evaluated that part wrong,
27,750 + 5,180 = 32,930
32,930 - 22,000 = 10,930

[queen dagger]

Divided by 200?

[neon iron]

Yep

[queen dagger]

How do I do 4th? Like idk what deposit or anything is

[neon iron]

Deposit basically just means putting money somewhere, like depositing money at a bank (putting money in your bank account), or in this case putting a deposit on the car, which is putting money towards the total cost you have to pay. Cars are expensive at dealerships so you gotta do stuff like this to afford them

A one third deposit means up front you are paying 1/3 of the cars price
After this we pay $420 a month 12 times
Once you put all that together, you have to compare the prices and that's your answer

[queen dagger]

Wait lemme do tha

[neon iron]

Question is a good example of how car dealerships always want to scam you lol

[queen dagger]

Is the answer 1440?

[queen dagger]

That is why I dint understand

[neon iron]

Close you just messed up a digit

2440

Are you using a calculator or working by hand?

[queen dagger]

Calculator

[neon iron]

I'd recommend you type a bit slower just cause you seem to do small typos or erros in your work, so even though you do everything correctly you get t the wrong answer :p

Just take a moment to read the numbers again and it'll be easy for you

[queen dagger]

Shit I am getting 240 now

[neon iron]

Lol can you type here what ur putting in the calculator?

Best you can ofc

[queen dagger]

Deposit: 7200/3 = 2400
She pays: 420 x 12 = 5040
Profit: (2400+5040) - 7200

Is that ryt?

[neon iron]

Oh 240 is right LOL I need to go to.sleep

Yep its right

gw

[queen dagger]

Ohhh ok byeee tysmm

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[timber root]

In the group of division in z, is the negation of an element a just a?

[timber root]

I just thought of this and Im not too sure if its right

[surreal mural]

whats the group of division in z?

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[timber root]

.reopen

[topaz sinew]

✅

[timber root]

Im so stupid division isn't in z it's in q

but my observation still stands

[prisma mesa]

then the inverse element, if that's what you mean, is 1/r

for r!=0

you actually have to exclude 0 from the group

because in group, all elements must have an inverse

[timber root]

Im confused

if we're doing a/b

and the identity is 1 cuz

a/1=a

so negation is defined as

a/(neg(a))=1

wouldn't negation be a since a/a=1?

[prisma mesa]

oh wait, yes, you're right. But you still have to exclude 0

0/0 is undefined

[timber root]

tru

so q-{0}

[prisma mesa]

yes, that would work

[timber root]

or we could even do it in q+

or r

thx so much

[prisma mesa]

hmm

shouldnt groups also be associative though

[timber root]

uuhhh

[prisma mesa]

I still dont get the idea of division group

[timber root]

(a/b)/c

=

a/(b/c)

crap

that doesn't work

[prisma mesa]

yes, that doesnt work

[timber root]

oof almost worked

[prisma mesa]

i thought that division group is somewhat weird idea lol

[timber root]

uhhh how do I close the channel?

.close?

[prisma mesa]

.close

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[prisma mesa]

ye

[timber root]

k

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[neon iron]

How do I tackle something like this

The only thing I can comment on here is that I need to change the sin^2(2x) into cosine but nothing I write it as give me something useful

Tag me please

[torpid sparrow]

So did you change denominator into cos?

[neon iron]

Yes

2 - 2cos^(x)

[torpid sparrow]

Why 2?

and what is cos^(x)?

Use pythagorean identity

$\sin^2 a + \cos^2 a = 1$

[thorny flame]

Samuel

[torpid sparrow]

@neon iron

[neon iron]

Oh I messed up

1-cos^2 (2x)
1- (2cos^2(x) - 1)^2

[torpid sparrow]

Why are u squaring up that?

[torpid sparrow]

Use this

$\sin^2 a = 1 - \cos^2 a$

[thorny flame]

Samuel

[neon iron]

Okay

1 - cos^2 (2x)

How do I proceed

Nothing I can do to the numerator

[torpid sparrow]

Yes you can expand the numerator

Product into sums identities

Or taylor series

[neon iron]

I haven’t been taught either

[torpid sparrow]

Then you should study them before doing this exercise

[cursive patrol]

to do this problem without those methods, expand the entire numerator and denominator in terms of cos(x)

[topaz sinew]

@neon iron Has your question been resolved?

[neon iron]

oh i'll look through it

but i dont know what this question is expecting me to use

[neon iron]

i dont know what to do with the numerator

[cursive patrol]

cos(2x) you should know (2cos(x)^2 - 1)
cos(3x) you can derive using cos(x + 2x). but the identity is also known (4cos(x)^3 - 3cos(x))

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[main oxide]

hi

[main oxide]

does anyone know what x and θ refer to inside this f() ?

.close

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[mystic siren]

how is the bijection correct

<1, 1/2>, <1/2, 1/2> is there, so it isn't one-to-one.

[surreal mural]

are you sure f(1/2) = 1/2?

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[silent olive]

Hi everyone! I'm struggling a lot with number theory questions. Here is one that is supposed to be simple:
Let a and b two integers. Knowing that 7 divides a^2 + b^2, show that 7 divides a and b.
I tried the following steps:

• If k, a and b are integers, if k divides a and k divides b, k divides a^2 and k divides b^2, so k divides any linear combination of a^2 and b^2 (a^2 + b^2 in particular). The problem is that the property is not reciprocal.
• 7 divides a^2+b^2, so there exists an integer k such that a^2+b^2 = 7k. After that, I don't know what I can do with the expression. There seems no application to the Gauss theorem (can't factorise it in an interesting way), doesn't look like Bézout can help too with this form.
  And I'm basically stuck here. I'd be grateful for your help.

[prisma mesa]

Hint:
||a^2 = -b^2 mod 7||

[wary tulip]

i was going to ask if you were familiar with modular arithmetic

[silent olive]

I tried that, can't really see what property can suit here... Division is not an allowed operation in modulus, right?
Fermat's little theorem will bring be a mod 2, which won't really lead anywhere.

[prisma mesa]

What residues can a^2 have modulo 7?

the residues only depend on a mod 7 btw

so you can just plug in first 7 natural numbers

so 0^2 mod 7, 1^2 mod 7, 2^2 mod 7...

[silent olive]

We haven't learned about quadratic residues, I'll have to Google about these.

[prisma mesa]

you dont need any fancy theorem or stuff

the only fact you will need to know is this: if a = k mod 7, then a^2 = k^2 mod 7

this shows that quadratic residues are periodic with period of 7

[silent olive]

Is the fact reciprocal?

[prisma mesa]

not really, you cant reverse it

[prisma mesa]

Just try computing this, trust me

until 6^2 mod 7

[silent olive]

0^2 = 0 [7]
1^2 = 1 [7]
2^2 = 4 [7]
3^2 = 2 [7]
4^2 = 2 [7]
5^2 = 4 [7]
6^2 = 1 [7]

[prisma mesa]

right

so what can a^2 be mod 7

and what can b^2 be, in order for a^2 + b^2 = 0 mod 7

[topaz sinew]

@silent olive Has your question been resolved?

[silent olive]

I can't really see where it leads...
If a^2 = 0 mod 7, b^2 should be 0 mod 7.
If a^2 = 1 mod 7, b^2 should be 6 mod 7.
If a^2 = 2 mod 7, then b^2 should be 5 mod 7.
If a^2 = 4 mod 7, then b^2 should be 3 mod 7.
But like a^2, b^2 can only be equal to 0, 1, 2 or 4 mod 7.

Oh, right!!!
a and b can only be equal to 0 mod 7 so a^2 + b^2 = 0 mod 7, so 7 divides a and 7 divides b.
Thank you very much!

.close

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[white bluff]

hi im doing past paper questions and i got the determinant of this matrix to be 1/2 (not sure if this is correct) but im not sure how then this indicates how A changes volumes

[white bluff]

would it half the volume?

i know that the volume is given by a.(bxc)

which is a 3x3 determinant

wait nvm i calculated the det wrong i think its 1

so would it not change the volume??

[ocean terrace]

the det is 2

so A doubles volume

[white bluff]

D:

how did i calculate the det wrong twice lmfao

[white bluff]

so its just volume changes by magnitude of det then?

[white bluff]

wait im pretty sure its 1 i just checked again... could i see how you found it to be 2? :')

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[wintry urchin]

1 is correct

[white bluff]

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[clear pike]

how could i get started with this q? i note that the height of the trapezium would be 2r, but not sure where to go from there

[wispy pier]

Think about

How to get the height

You are given the area

[clear pike]

just with 1/2(a+b)h = 600 right?

the height is 2r

but we dont know what a and b are

[clear pike]

not sure what you mean by this then

[twilit hemlock]

you can find SP + RQ

[clear pike]

ah yes, this was the trick

thanks!

[twilit hemlock]

npp

[clear pike]

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[boreal quest]

im getting 9!*3! as the ans can anyone confirm or deny this for me please

The question
No.of Arrangements of the word PERMUTATIONS such that there are exactly 4 letters b/w PandT

(T and P also counted here) in case you are confused

[autumn willow]

so essentially 2 letters?

[boreal quest]

4 letters between the two

[autumn willow]

im getting 9x10!

idk if its right

[boreal quest]

like AIONPRMUSTET

[boreal quest]

how tho?

[autumn willow]

.

so lets split the tasks

first we find how many spots there can be for P___T

there are 9

now there are 10 spots left

so 9X10!

do u have the answer to check weather it is right?

[junior fable]

question... is "TermuPaionsT" valid?

[boreal quest]

wdym by 9 spots there?
also the 10! wont work as the second T can come behind P and thus the rule being broken

[autumn willow]

o

mb

did not notice the second t

[boreal quest]

yes that would be valid

[junior fable]

what about PeTrmTuaions?

[boreal quest]

nope

[junior fable]

but the letters e,t,r,m are between one P-T pair

[boreal quest]

i think the wording might be bit bad in the questions It should have been 4 letters between P and one T and more than that between P and second T

[junior fable]

strictly more?

e.g. sTermuPaionT

[boreal quest]

yes thats what the teach said when i asked similat examples

[boreal quest]

yes like this

[junior fable]

so that doesn't count because there are two P-T pairs with a gap of 4

[boreal quest]

wait 4 is also allowed

[junior fable]

ah ok

alright sorry for hijacking

[boreal quest]

its good you did

gave the question more clarity

[junior fable]

just a comment while I'm working it out, quite happy that the only duplicate letters in this word are T's lol

[boreal quest]

yeah it would be waay too complicated otherwise

[junior fable]

I'm getting a number much greater than 9!3! by considering the individual positions where P can go

[boreal quest]

could you tell me how you managed for T

[junior fable]

My thought is, when P is not in a middly position, one of the positions for a T is forced

then we only have so many choices for a second T, and the rest of the 9 characters may be freely permuted

when P is middley, it's actually easy, because there are only 3 valid positions a T could possibly be in, and two of them admits a gap of 4 characters from P

we don't have to consider such silly thoughts as, which T goes where, because they're the same. and we can easily permute the 9 remaining characters because they're all distinct

[boreal quest]

oh thats actually genius

[junior fable]

the special number I'm finding is 46

so see if that gets to 46 lol

that is, 46 * 9!

[boreal quest]

hey one question tho suppose P is in the edge
then 1 T must be on the 5th block right?
and the other t could be anywhere between 6 and 12

[junior fable]

right

like if P is in position 1 you mean

wait wait, sorry

the nice T is actually in position 6

because it admits a gap of 4 (2,3,4,5)

hence the other T can be in positions 7-12

[boreal quest]

oh yeah right my mistake

ok that seems to work in the way i expected

Im in a hostel so we arent allowed to use our PC'S past a certain time
so ill have to go now ill try to let yk if i get the same (likely will)
Thanks a lot for the help i really appreciate it

Hope you have a great day/night

[junior fable]

thanky, you as well

[topaz sinew]

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[topaz sinew]

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[potent blade]

No idea on how to do this

[agile harness]

draw a free body diagram

[potent blade]

I did it

[agile harness]

what did you get

[agile harness]

mhm

so which part are you struggling with

[potent blade]

everything after this

[agile harness]

what’s the formula for gravitational potential energy

[potent blade]

mgh

[agile harness]

right so the change in gravitational potential energy is mgdeltah

so if it starts at a height of 0

[potent blade]

yes

[agile harness]

you’ll need trig to determine the height after moving 3 meters up the incline

the angle is 30

the hypotenuse is 3

what’s the opposite side

which corresponds to the height

[potent blade]

but the arrow isn't on the full side of the triangle

[agile harness]

doesn’t need to be the entire incline

it’s a similar triangle

[potent blade]

oh, i didn't know that

[agile harness]

what is h then

[potent blade]

opposite in trig

[agile harness]

which is

[potent blade]

so, basically we use sine

[agile harness]

mhm

sin(30)=…

[potent blade]

=O/3

[agile harness]

or h

but yea

then solve for h

[potent blade]

1.5

[agile harness]

mhm

so then use that to find Ug

[potent blade]

49.81.5

[agile harness]

hmm

[potent blade]

4 times 1.5 times 9.8

[agile harness]

you could just say 40(1.5)

the weight was given as 40 N

[potent blade]

oh ok

[agile harness]

which is mg

so 60

joules

[potent blade]

yep

[agile harness]

now what do you think you should do for the next part

[potent blade]

I know it's something related to x component and y component

[agile harness]

your fbd will be useful

[potent blade]

but that topic is confusing

[agile harness]

what does constant velocity imply

[potent blade]

0 net force

[agile harness]

mhm or zero acceleration

so

set up your two net force equations

for x and y components

then solve for F

what will your Fy equation be

[potent blade]

F does not have a y component right?

[agile harness]

no it does not

it’s along the incline

[potent blade]

oh

[agile harness]

what are the two equations then

first you need to know how to split mg into its components

[potent blade]

oh ok, yea i saw the process, but it doesn't make sense to me plus i don't know why we should do it

[agile harness]

what doesn’t make sense

[potent blade]

spliting mg

[agile harness]

hold on

in these problems this is how we define the coordinate system

notice that mg has components along each direction

mg is always straight down

but since our coordinate system is shifted mg has components

the normal force is perpendicular to the surface and thus points along the direction of the y axis

and has no components

the force F is along the x axis in this coordinate system so it has no y component

but mg isn’t entirely along either

[potent blade]

what do you mean

[agile harness]

well because generally you define the x and y axes to be straight up/down and straight left/right but it’s much more convenient to define our coordinate system in this way for these problems

[potent blade]

oh

okay

[agile harness]

do you see the new axes i drew?

[potent blade]

but how is spliting mg related to F

yeah

[agile harness]

because it’s needed in the net force equations

[potent blade]

oh

[agile harness]

if we define up the ramp to be positive and down the ramp to be negative

then we can say F-mgx=0

[potent blade]

ah ok

[agile harness]

thus F=mgx

what is the x component of mg though

[potent blade]

let me check

how do we draw the axes onto the mg arrow>

[potent blade]

Basically like this

[agile harness]

sort of yea

but the y component is more perpendicular to the incline

[agile harness]

can you see this?

[potent blade]

it's so difficult to picture it mentally haha

[agile harness]

did the image i sent not help?

[potent blade]

it did help

but the arrow is weird

[agile harness]

anyways you should get F=mgsintheta

[potent blade]

did you do it mentally?

[agile harness]

it’s standard

F-mgx=0

for the net force equation

[potent blade]

oh, so i don't require this drawing steps

[agile harness]

on an incline mgx is almost always mgsintheta

and mgy is mgcostheta

[potent blade]

ohh

how much is tetha

[agile harness]

it’s the same theta as the theta at the bottom of the incline

[potent blade]

oh, i see the usefulness of splitting

ok, so i indeed got mgcos30 and mgsin30

what do I do now

[agile harness]

F=mgsintheta

[potent blade]

how do know that F=mgsin30?

they are not close to each other

is it because F is an x component also

i mean y

[agile harness]

…

from the net force equation

[potent blade]

im sorry for stupid question

[agile harness]

they are the two forces along the defined x axis

if there is zero acceleration

the net force must be zero

[potent blade]

oh both mg sintetha and F is going along x axis

ok, what's the other equation

[agile harness]

doesn’t matter

you solved for F

you’re done

F=mgsintheta

=40(sin30)=20

[potent blade]

oh

how to do the last question