help-33

[elfin ivy]

Something like that yeah

[loud eagle]

because you do
138.142/360 = .384
.384 * 785 = 301.22

[elfin ivy]

I get 301.38 but that's because I don't round part way through

Your calculator might not be able to do that

So that looks good

[loud eagle]

alright thanks i actually understand how to do it now

[elfin ivy]

No problem!

Do you want me to summarize what we did today?

[loud eagle]

sure

[elfin ivy]

So for the ferris wheel problem, the idea is this: Construct a right triangle connecting the point where the person stopped to the vertical axis through the ferris wheel. Find the segment along that axis that gives the height of the person. Then find either the height or the angle of the right triangle where appropriate, using your trig rules.

[loud eagle]

thanks man ill put that down in my notes

[elfin ivy]

Any time

[marsh citrus]

@loud eagle Has your question been resolved?

[marsh citrus]

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[eager robin]

How would the new function look like to connect g(x) and h(x)?
The function needs to be connecting the points without a kink and smooth
I am not sure if its f(x)=ax^5+bx^4+cx^3+dx^2+ex+f

[flat raft]

there are a lot of ways to do that

[eager robin]

could you show me a easy way and a one that is a bit more complicated?

[pine orchid]

g'(x)=0, g(-1)=0 & h'(x)=1 , h(2)=2

If y=f(x) connects the two points smoothly and without empty domain, the function is continuous in the given range.
f'(-1)=0, f(-1)=0 & f'(2)=1, f(2)=2

[bright jay]

f(x) could be a piecewise function

[eager robin]

i didnt had this yet, is it more complicated than the other ways?

[eager robin]

i figured this out too with
g'(x)=0, g(-1)=0 & h'(x)=1 , h(2)=2
f'(-1)=0, f(-1)=0 & f'(2)=1, f(2)=2

my question is more about if f(x)=ax^5+bx^4+cx^3+dx^2+ex+f would be a right way doing it or if i should use a different option

[bright jay]

No, a piecewise function is literally a function of restricted functions

[eager robin]

but how would i do this

[bright jay]

Find a function from x = -1 to x = 0 then x = 0 to x = 2

[pine orchid]

It's unique in cubic curve.
https://www.desmos.com/calculator/5rlo8ospve
This might help you.
https://en.wikipedia.org/wiki/Interpolation

[marsh citrus]

@eager robin Has your question been resolved?

[pine orchid]

Interpolation method finds smooth curves which connect the given points. It's used in approximation.

[eager robin]

i have a question about the cubic curve, i see the solution but i want to know how do you calculate it?
Atm I am at this point and i dont really know what to do next

[pine orchid]

You can substitute x=-1 and x=2 in the cubic function and its derivative.

[eager robin]

how do you get 12a and "= 1"?

should it be -1 or is it supposed to be +1

i understand these 3 but the last one is a bit confusing for me

[pine orchid]

We suppose that f(x) is a cubic polynomial.
The first two equations are f(-1)=0 and f'(-1)=0.
The last two equations are f(2)=2 and f'(2)=1.

[eager robin]

i forgot about f(2)=2 and f'(2)=1., Thank you!

Now i need to solve a b c and d right?

[pine orchid]

Yes.

[eager robin]

one more question before i do it, i need to go from right to left or left to right

like first remove a or remove d

[pine orchid]

Eliminating d would be simple.

(as subtraction of two equation eliminates d)

[marsh citrus]

@eager robin Has your question been resolved?

[eager robin]

yea i eliminated all of them now i got a 1/27

but i dont get 2/9 for b

[sudden portal]

Sup

[marsh citrus]

@eager robin Has your question been resolved?

[marsh citrus]

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[zealous canopy]

Can someone pls help me

[marsh citrus]

Please don't occupy multiple help channels.

[upper briar]

i think

i close this or ur prev

[zealous canopy]

.close

[marsh citrus]

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[sweet raptor]

How did he get this? I dont understand the formula

[next ravine]

He took the derivative

[crystal wraith]

s'(t) is v(t)

By definition

[sweet raptor]

alright

but how did he get these numbers

[next ravine]

Taking the derivative (power rule)

Derivative of t^3, t^2, and t

[sweet raptor]

alright ill look into the power rule

oh

okay! I understand the power rule

how did he get these numbers now?

[crystal wraith]

6 times 3 is 18
-45 times 2 is -90 and so on.

[sweet raptor]

no no

sorry

the underline red

[crystal wraith]

Those came from the t^3 and t^2 and so on

[next ravine]

Just the power rule

[crystal wraith]

Power rule, like Shen has been saying.

[sweet raptor]

oh

okay

so like this?

[next ravine]

Exactly

Yes

[sweet raptor]

okay!

[next ravine]

The new power is (old power - 1)

[sweet raptor]

yes

[next ravine]

What have you found for the power rule and what it is online?

[sweet raptor]

oh

this is what I see

but I dont really understand it

[next ravine]

Send a screenshot and we can talk about fit

*about it

[sweet raptor]

sure

[next ravine]

There you go!

Bring the power down front, and subtract 1

[sweet raptor]

wow

that makes so much sense

thanks

.close

[marsh citrus]

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[wary bluff]

What did I do wrong

[main idol]

the problem doesn't tell you which of the 7 is wrong?

[wary bluff]

nope

[main idol]

can you just plot it?

[wary bluff]

I graphed it

and checked

I cant find my mistake

[main idol]

x = 1

[wary bluff]

omg

-1, 1

fuck

still wrong

[main idol]

same for 2

[crystal wraith]

Because the second statment

[wary bluff]

ohh

tysm

I see

.close

[marsh citrus]

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[marsh citrus]

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[still temple]

only in english ?

[still temple]

A fruit and vegetable retailer studies the evolution of his melon sales in order to anticipate his orders.
The retailer carries out a study on his customers. He finds that :

• among the customers who buy a melon in a given week, 90% of them buy a melon the following week
  the following week;
• among the customers who do not buy a melon in a given week, 60% of them do not buy a melon
  the following week.
  We randomly choose a customer who bought a melon during week 1 and, for n > 1, we note
  An is the event: "the customer buys a melon in week n" and pn = P(An). Thus p1 = 1.

1. Prove that for any integer n > 1, pn+1 = 0.5pn + 0.4.
   2.a. Show by recurrence that, for any integer n > 1, pn > 0.8.
   b. Show that the sequence (pn) is decreasing.
   c. Is the sequence (pn) convergent?
2. For any integer n > 1, vn = pn - 0.8.
   a. Prove that (vn) is geometric and give the reason and the first term v1.
   b. Express vn as a function of n.
   Deduce that, for all n > 1, pn = 0.8 + 0.2 × 0.5
   n-1
   .
   c. Determine the limit of the sequence (pn).

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

damn no one want to help me

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[forest kayak]

Je peux aider en français

@still temple

On note $p_n = P(A_n)$, donc $p_n$ est bien la probabilité de l'événement $A_n$

[elfin berry]

Epsilia aka Mellow

[forest kayak]

On veut démontrer que $p_{n+1} = 0,5 p_n + 0,4$.
Donc on veut une relation de récurrence qui lie la probabilité à la semaine n+1 avec la probabilité à la semaine n

[elfin berry]

Epsilia aka Mellow

[forest kayak]

Et sur ceux qui payent un melon à la semaine n+1, on a deux cas :

• Ceux qui ont payé un melon à la semaine n ;
• Ceux qui n'en ont pas payé à la semaine n.

Sachant que $p_n$ c'est la proba d'un événement, et que tu as un découpage des cas en deux, $A_n$ et $\bar{A_n}$ sont alors des événements complémentaires

[elfin berry]

Epsilia aka Mellow

[still temple]

merci bcp

[forest kayak]

Et la probabilité de deux événements complémentaires, il y a un lien que tu connais bien entre eux

$P(\bar{A_n})=1-P(A_n)$

[elfin berry]

Epsilia aka Mellow

[forest kayak]

Tu peux même représenter un arbre de probabilités

Qui aura juste un seul niveau, de la semaine n à la semaine n+1, certes, mais au moins tu le visualiserais sur quelque chose qui te serait familier ?

Toi tu veux avoir que ceux qui en achètent

Du coup tu veux avoir les 90% de ceux qui ont acheté

• les (100-60)% de ceux qui n'ont pas acheté mais qui vont acheter

J'ai eu ce 100-60 par le fait que tu as 60% des gens qui n'ont pas acheté et qui ne vont toujours pas acheter

Comme tu regardes seulement ceux qui achètent, tu regardes les 40% restants du coup

Donc ça te fait le $p_{n+1}=0,9p_n+0,4(1-p_n)$

[elfin berry]

Epsilia aka Mellow

[forest kayak]

Tu développes et retrouve le résultat

Ah et le fait que personne ne t'aide est sûrement parce que tu as posté tout un énoncé et pas juste une question sur laquelle tu butais

Ca montre d'un manque de volonté et y'a pas grand monde qui aime ça par ici

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[viral halo]

i got 2,700 would that be correct?

[tight furnace]

think so

[viral halo]

.close

[marsh citrus]

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[marsh citrus]

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[thick oasis]

alright so ik this is rly dumb but for some reason i literally forgot how to do this stuff and i cant find anything online on specifically how to do this

[thick oasis]

how did it go from the first equation to the second one

like from the one where -30 is isolated to the one where t is isolated

.close

[marsh citrus]

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[floral copper]

Help

[floral copper]

this is just an example
If im given say x^4-3x^3-3x^2-ax+b and given zeros at -2,1 (order 2) how do i find a and b?

[fallen galleon]

aye aye

wassup

so

hm

have u

tried anything

or nah

[hazy shard]

So you would plug in one of the values of x and equal the result to 0, do the same with the other, and then pass the constants you are left with to the other side

So now you're gonna be left with a system of equations

In this case:
{(-2)^(4)-3(-2)^(3)-3(-2)^(2)+2a+b=0
{(1)^(4)-3(1)^(3)-3(1)^(2)-a+b=0

Reducing:
{16+24-12+2a+b=0
{1-3-3-a+b=0

{2a+b+28=0
{-a+b-5=0

{2a+b=-28
{-a+b=5

From there, we can multiply the bottom equation by -1, such that:
{2a+b=-28
{a-b=-5

And now, we can add both equations up, which cancels the bs and leaves us with 3a=-33

Isolate a to have a=-33/3=-11

Then plug that into one of the equations in place of a. I'll use the bottom one for simplicity.
11+b=5
b=5-11
b=-6

Therefore, a=-11 and b=-6

[marsh citrus]

@floral copper Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[fierce belfry]

@hushed egret hi

[hushed egret]

hi

[fierce belfry]

U missed me

I know

[hushed egret]

i cant stay rn

[fierce belfry]

I took a break from srudying for like 3 days

Oh

1 min

Quick one

Pls

[hushed egret]

uh

[fierce belfry]

Pls

[hushed egret]

if its really really quick

because

[fierce belfry]

Yesmam

[hushed egret]

i have a class

in very soon

[fierce belfry]

I dont get why to get w we need to multiply by the base

[hushed egret]

okay so

whenever you write like

(x, y, z)

you're implicitly using the standard basis

[fierce belfry]

Okay

[hushed egret]

{(1, 0, 0),
(0, 1, 0),
(1, 0, 0)}

and x, y, z just tell you how much of each basis vector you have

and by no surprise

[hushed egret]

but if you're using a different basis

you gotta multiply by the basis vectors of that basis

[fierce belfry]

So like

[hushed egret]

and your (a, b, c)_S means a*v1 + b*v2 + c*v3

[fierce belfry]

Wait so to get the standard basis

I multiply by the

New basis

To get the new basis i

Uh

[hushed egret]

you invert

so basically you solve the system of equations

this one

its the same as multiplying (5, -1, 9) by the inverse of the matrix in the left

inverse of this

okay i gotta go

[fierce belfry]

Yesm

Mam

I yhink ibget

[marsh citrus]

@fierce belfry Has your question been resolved?

[fierce belfry]

Thanks

.close

[marsh citrus]

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[tropic leaf]

Real talk I got no idea wtf they talking about

[tropic leaf]

Image sending

⁉️

how do you sketch some stuff like that 😭

never mind figured it out

!done

.close

[marsh citrus]

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[quiet tree]

124 people attended a concert. The ticket price was 40 USD for children, and 70 USD for adults. The concert got a total of 6310 USD. Call the amount of children for x, and set up an equation and solve it in order to find how many children and adults were in the concert

[quiet tree]

Doesn't this question have multiple answers?

What I've done so far is made the following equation:

40x + 70y = 6310
where x represents the amount of children and y represents the amount of adults

not sure how to continue, though, as we have two unknown values and no additional given information

[cloud iron]

Try constructing an equation to describe one variable in terms of the other

Then you'll have a system of equations to work with

[still temple]

remember x and y together must be 124

[cloud iron]

Guys let's not do it for them, just put them on the right track

[quiet tree]

My god, I've completely forgotten that the total amount of people was listed...

having
40x + 70y = 6310
by itself of course won't make any sense

but i've totally forgotten there were 124 total at the concert

that makes
40x + 70y = 6310
x + y = 124

then multiply the second equation by 40 and subtract it from the first equation to find the amount of adults

and use that to find the amount of children

[cloud iron]

👍

[quiet tree]

So that would be

40x + 70y = 6310
40x +40 y = 4 960

subtract second equation from first equation to get

30y = 1 350
y = 45

45 + x = 124
x = 79

there are a total of 79 children

Thanks a lot, guys

However if I was for instance given only this equation with no further instructions:

40x + 70y = 6310

How would I come up with an exampe of what y and x can be?

(there are multiple ways this can be solved, assuming you only have one equation), but how do you exactly solve for x and y in that case?

if x = 1

70y =6270
y = 89.571428571428571428571428571429

is there any specific method for doing this and getting whole numbers on both variables?

[cloud iron]

Evaluating that equation given any x value would yield an appropriate y value, and vice versa

[quiet tree]

my bad, i'm european

[quiet tree]

sure. However I tried plugging 1 into x and solving for y to find out that it's a decimal

although it's a correct solve,

i believed it's prefered to have a whole nunber for both variables, if possible

[cloud iron]

I guess you could assume the total people to be anything but 124 and it should generate integers if you solve the same way you did originally

Like if you say c + a = 240 it'll give you two whole numbers when you solve the system

Or in that case $c + a = n, \forall n \in \mathbb{Z}$

[elfin berry]

lexitorius

[cloud iron]

Would generate a pair of integers for c and a afaik

Actually I could be wrong there

[quiet tree]

Let's assume x + y = 120

40x + 70y = 6310
x + y = 120

which means that

40x + 70y = 6310
40x + 40y = 4800

30y = 1 510

which leads a conclusion
y = 50.3333

yep, doesn't seem like there's a way of solving these types of equations and getting an integr, other than throughout guessing

I'll give that a search

Thanks a lot for the assistance, folks

.close

[marsh citrus]

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[cloud plinth]

i tried solving it by finding the first term -> -1 and the last term -> 2498
then doing
Sn = (-1 + 2498 / 2 ) 50
then i got 62,425

[cloud plinth]

but the answer is A

[still temple]

that only works if Sn is a sum of an arithmetic progression

[cloud plinth]

here its a geometric seq ?

[glass perch]

You can't apply Gauß Formula here as it's not 1+2+3+...

[still temple]

no, its quadratic

[cloud plinth]

fair

how am i gonna solve it then

[still temple]

theres a formula for the sum of n^2

[glass perch]

Try splitting it into two sums first

[cloud plinth]

which is

[glass perch]

Sum 1 to n^2 = n(n+1)(2n+1)/6

[cloud plinth]

mind explaining the process ?

[glass perch]

If you want to prove it, you can use induction

[cloud plinth]

the question is just asking to evaluate it

[cloud plinth]

oh i see

[glass perch]

Yes

[cloud plinth]

how did we come up with that formula exactly ?

[upper briar]

its google

[glass perch]

https://youtu.be/aXbT37IlyZQ

[upper briar]

hm yea good q that isnt exactly intuitive

[glass perch]

Hi 🙂

[upper briar]

hi

[upper briar]

stil not green

[glass perch]

I'm still not considered helpful

[bright jay]

Because you're unhelpful \s

[glass perch]

What?

[upper briar]

thats whats sometimes call a joke

[glass perch]

Anyways, do you know how to use this formula here?

[cloud plinth]

yeah

thank you for the help i get it now

[cloud plinth]

weirdest flex i've ever seen

[glass perch]

Not sure. Some people here would already blame our talk in a help channel

[upper briar]

that was a comment

[cloud plinth]

solved

hm

close

[upper briar]

o rly

[cloud plinth]

helped

[upper briar]

ok lemme do the honors

.close

[cloud plinth]

uh sorry i'm new

[marsh citrus]

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[upper briar]

uve no green

[bright jay]

.reopen

[marsh citrus]

✅

[bright jay]

.close

[marsh citrus]

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[upper briar]

wait

[glass perch]

Can't helpers close anymore?

[upper briar]

..

[upper briar]

no now its green ppl i think

[bright jay]

Try it

[upper briar]

helpers just get pinged

lol

[bright jay]

Reopen

[cloud plinth]

.close

[upper briar]

imagine

[cloud plinth]

hm

[upper briar]

they try it

n get banned

LOL

[glass perch]

I feel like being very active for a week is much more efficient in gaining ranks than what I do

[upper briar]

no just type faster

n more frequently

LOL

[marsh citrus]

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@rugged breach Has your question been resolved?

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[marsh citrus]

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[oblique acorn]

I'm stuck on those limits if anybody could help i'd be very grateful

<@&286206848099549185>

[marsh citrus]

@oblique acorn Has your question been resolved?

[oblique acorn]

.close

[marsh citrus]

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[marsh citrus]

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[bitter hamlet]

how i do this

[bitter hamlet]

i used to getting a number

[hidden plaza]

Write this as $\sum_{i=1}^n i+1 -\sum_{i=1}^4 i+1$

[elfin berry]

Pure

[hidden plaza]

Then $\sum_{i=1}^n i+ \sum_{i=1}^n 1 -\sum_{i=1}^4 i -\sum_{i=1}^4 1$

[bitter hamlet]

ok i see thanks

[elfin berry]

Pure

[bitter hamlet]

.close

[marsh citrus]

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[eager turtle]

consider the polynomial P(x) = 3x^3 - 2x^2 + ax + b

Knowing that the remainder of dividing P(x) by x + 1 is -10 and that P(x) is divisible by x - 1, determine the values ​​of a and b

[paper fiber]

what have you tried?

[eager turtle]

i've tried using ruffini's rule

i've gotta go sleep now so ill just close this.

.close

[marsh citrus]

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[paper fiber]

hav

o

[eager turtle]

?

[paper fiber]

i was gonna ask if you have learnt about remainder theorem

[eager turtle]

oh, not yet

i was preparing for a test tomorrow and that question popped up

but its fine

anyways, have a nice day

[paper fiber]

well remainder theorem will make this question simple so you should learn it when you have time

its a pretty simple theorem anyway

you too

[marsh citrus]

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[silver star]

hihi

[silver star]

stuck on this

[eager dome]

$\log_e(x)=y \iff e^y=x$

[elfin berry]

Rhaenys

[silver star]

ok so i take

still confused

where do i start

[eager dome]

Rewrite that formula using the values and variables you're given

ln is just log_e btw

[silver star]

oh

so it would be

so its just 2?

like a = 2?

ok i got it lol

how do i to this

[eager dome]

The laws of logarithms are dual to the laws of exponents

$\log(x) + \log(y) = \log (xy)$ also $\log(x)-\log(y)=\log(x/y)$

$p\log (x)=\log (x^p)$

[elfin berry]

Rhaenys

Rhaenys

[silver star]

ohh so it would be

ln (a+b times a-b)?

[late geode]

read the question carefully

[silver star]

it says in a

[late geode]

also you should probably leave stuff in factored form

it says in a
no it doesn't,
it says
first write as ln A, ln(A); the natural log of A
and the question wants the A in that

(my is that you're trying to give ln(A) instead of A)

you're also missing b in the second term of your expansion but

also you should probably leave stuff in factored form

[silver star]

still al ittle lost

oh wait

so i keep(a-b)

A = (a + b)(a - b)3/c4

ooo

got it

ok

<@&286206848099549185>

[marsh citrus]

@silver star Has your question been resolved?

[silver star]

no

[silver star]

<@&286206848099549185>

.close

[marsh citrus]

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[high gust]

Need help

[high gust]

Forget the numbers I already wrote

[nova totem]

Im@gonna need more information

You have a line and you have a line segment

[high gust]

Shi my bad

It’s asking to solve for x

[nova totem]

Okay yeah but you have a line and a line segment

Is the length of that partition of the line equal to the length of the line segment?

[high gust]

The line on top is suppose to be equal to the line on the bottom

[nova totem]

Lines go on to infinity, line segments don't

If it's a partition then $12x = 4 + 7x + 1$

[elfin berry]

Umbraleviathan

[high gust]

Ohhh

X= 1 theb

Thanks man

[marsh citrus]

@high gust Has your question been resolved?

[marsh citrus]

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[low aurora]

Doing my trigonometry homework and I am stuck here in said picture:

There is a linked video attached, but in it the equation us something different and has + between both fractions which allows them to do something like 5(3-1) + (3+1)5 into 5(3-1+3+1) however my question has subtraction and I'm unsure if I can do the same thing in said video.

I'm also unsure if 1+cos * 1-cos is 1-cos^2 or cos^2-1. Again, basing it off the video...

Said video: https://www.youtube.com/watch?v=WJL7KjEz7a4

[worthy trellis]

Do you know (1+cos(x))(1-cos(x))=sin^2(x)

[worthy trellis]

…

[worthy trellis]

And this is not reallly correct

Try doing this on paper and then type it up

[low aurora]

What do you mean?

I got a common denominator and had to multiple the top and bottom

The website even recognizes it as a correct (first?) step

[worthy trellis]

It is

What I am asking you to do is simplify the denominator

[low aurora]

I'm also unsure if 1+cos * 1-cos is 1-cos^2 or cos^2-1.

I don't know if the bottom turns into
1-cos^2
or
cos^2-1

I know 1-cos^2 = sin^2

[worthy trellis]

Are you aware of what is this is, (a-b)(a+b)

?

And sorry I am on my phone so slower to type

[low aurora]

I haven't touched math in a few years. That looks familiar, looks like FOIL right?

a*a a*b
-b*a -b*b
?

[worthy trellis]

Okay

Go on

a^2 + ab - ab - b^2

Yes

Can you clean it up further?

[worthy trellis]

No

That is incorrect

[low aurora]

at the end that should be -cos^2?

[worthy trellis]

You should write the t too

This kinda makes me uncomfortable seeing XD

[worthy trellis]

But yea

[low aurora]

I'm the opposite and I hate writing all the unnecessary t's until the end since I'm trying to narrow it down to specifically the info I need atm

Okay so the denominator then is

1-cos^2

which means sin^2

[worthy trellis]

Yes

Can you simplify further?

[low aurora]

I don't think so?

[worthy trellis]

Okay, split it into two fractions

[low aurora]

I know I need to do the top now

[worthy trellis]

Like sinx(1-cosx)/sin^2(x)

[low aurora]

Yes

[worthy trellis]

Yes

Can you simplify this?

[low aurora]

I think the sin on the top becomes 1 and the sin^2 becomes just sin?

[worthy trellis]

Yes

So what do you get?

[low aurora]

1-cos / sin - 1+cos / sin ???

Darn website isn't taking this though for a step...

Took that though

Wait, would I split them up even more...?

1 / sin - cos / sin - 1 / sin + cos / sin?

[worthy trellis]

Can you simplify this further?

[worthy trellis]

And no

[low aurora]

Hrm

[worthy trellis]

Check the signs

[low aurora]

I mean I could combine the denominators

[worthy trellis]

Since that is not correct

Wait

[worthy trellis]

…

Check the signs

[low aurora]

I am very lost

[worthy trellis]

Read what I am saying

[worthy trellis]

…

You need to correct the expression

[low aurora]

Idk what you are saying by check the signs. You mean - and +?

[worthy trellis]

Yes

When you have -(a+b) you have -a-b, yes?

[low aurora]

Wait, the - inbetween the fractions is multiplication to the top as a whole?

so it's -(1+cos) ?

[worthy trellis]

Yes

[low aurora]

so then am I looking at...

1 - cos 1 - cos

---

 sin              sin

[worthy trellis]

No

How’d the second frac become 1-cosx/sinx

?

It’s a plus ain’t it?

[low aurora]

-(a+b) you have -a-b

so it's -(1+cos) ?

[worthy trellis]

Yes and that would be?

[low aurora]

If you distribute -(1+cos) you get -1-cos

no?

[worthy trellis]

Yes

Good, what next?

[low aurora]

1 - cos 1 - cos

---

 sin              sin

So then?

[worthy trellis]

Again the negative?

[low aurora]

Are you trying to say it should be

1 - cos (- 1 - cos)
-------- + --------
sin sin

[worthy trellis]

Lol yes

[low aurora]

I guess that makes sense. The negative meaning has to be somewhere still

[worthy trellis]

So now if you combine the denominators…

[low aurora]

1 - cos + (- 1 - cos)

           sin

[worthy trellis]

Keep going

[low aurora]

I'm not sure what the next step would be? I don't imagine I could distribute the first 1 - cos into the (-1-cos)

[worthy trellis]

Okay

So

It’s

(1-cosx-1-cosx)/sinx

[worthy trellis]

And what do you mean by that?

[low aurora]

I am so confused as to how you are achieving

1 - cos - 1 - cos

      sin

I'm trying to think of any algebraic stuff I've learned years ago that allowed me to do that but I can't see how

Where did the plus go?

[worthy trellis]

If you have +(-a-b) = -a-b

[low aurora]

You're distributing the invisible +1 outside?

[worthy trellis]

Yes

So as to open the parentheses

[low aurora]

Okay, that makes sense. Then, don't we end up with the 1's canceling out

[still temple]

deol can you help me after

[low aurora]

and -2cos?

[worthy trellis]

Yes

[still temple]

can i ask my question now

[worthy trellis]

I need to go in like 5 minutes max

[still temple]

ok

its super quick

and easy

[worthy trellis]

No get a channel

[still temple]

bro no one is answering

i swear its super quick

[low aurora]

Okay so

-2 cos

sin

[worthy trellis]

They did answer

[low aurora]

-2 cos
--- x ----
sin sin

wait no

[worthy trellis]

The denom are wrong

[low aurora]

but it's still

-2 cos

sin

[worthy trellis]

Yes

You know that tanx = sinx/cosx

[low aurora]

I have no idea how to get the cos down to the sin to get the tangent

yeah

but it looks like then

-2 tan

but I need

[worthy trellis]

Okay wait if you don’t see it yet

[low aurora]

divide by cos?

[worthy trellis]

Cosx/sinx= cot x

[low aurora]

Sorry, yeah

[worthy trellis]

Cotx = 1/tanx

[low aurora]

that's cot

So it's currently
-2cot(t)

[worthy trellis]

Yes

[worthy trellis]

…

[low aurora]

Well, I see it now but

The stupid website won't take my work for the last point

tried

-2cot(t)

-2

tan(t)

Unfortunate

[worthy trellis]

That’s not a step worth considering a step

That’s why ig

[worthy trellis]

So put cotx in that

See if it takes that?

I gtg

[low aurora]

-2cot(x)

I'll try, but i think it didn't

thank you

.close

[marsh citrus]

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[clear blade]

why do these have different domains

the first one would just be (x+5)(x-5) >= 0, and the second one would be (5-x)(5+x) >=0 right?

but wouldnt both just be x=+-5 in the end?

[-5,5]

[eager dome]

You want the argument of the square root to be positive, so for the first you want x>=5 while for the second you want x<=5

[clear blade]

ohh

thank you

.close

[marsh citrus]

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[edgy wasp]

Why can't you square root the equation earlier?

[cobalt sentinel]

Like in which step

They need to have the term with the “x” isolated

[edgy wasp]

Oh, so you need to isolate x before you square root the equation?

[cobalt sentinel]

The term with x, yes

Otherwise u can’t solve for x only

[edgy wasp]

Ok

Thanks

[cobalt sentinel]

Sure np

[edgy wasp]

.close

[marsh citrus]

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[still temple]

Can someone help me with this? I don't understand the khan academy explanation.

[marsh citrus]

@still temple Has your question been resolved?

[oblique horizon]

The equation they show can be a little confusing. The question asks to find what temperature in Celsius is equivalent for 18° increase of Fahrenheit. You you use the original formula (F=1.8C+32), We can solve for Celsius by plugging in 18° Fahrenheit. We don’t use the “+32” because that is the rate we start with that temperature (The trick is the question is only asking for the INCREASE, not the entire temperature). Since we are no longer using the “+32” we are left with 1.8C= 18. Using some algebra we have a INCREASE of 18 degrees Fahrenheit is equivalent to to increase of 10 degrees Celsius

[still temple]

AHHHH i see

now i understand

that clears it up a lot

[oblique horizon]

No problem

[still temple]

.close

[marsh citrus]

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[still temple]

Can someone simply the fraction with the y

[brisk scarab]

Put the values

$\frac{3}{2} \cdot 6$

[elfin berry]

n11

[brisk scarab]

What do you not understand?

[still temple]

what do I do with them that’s my main problem

[brisk scarab]

You just have to put the values of given variables and simplify it

For example, to find $3x+4$ at $x=2$ would mean substituting $x=2$ in $3x+4$. Which means $3(2)+4=6+4=10$.

[elfin berry]

n11

[brisk scarab]

That's it

[still temple]

so do I do 3(6) or 2(6)

I’m not the best with fractions that’s my problem with it

[brisk scarab]

It'll go like this: $\frac{a}{b} \cdot c=\frac{a}{b} \cdot \frac{c}{1} = \frac{a \cdot c}{b \cdot 1}$

[elfin berry]

n11

[still temple]

So for the one I’m doing it would be: $\frac{3}{2} \cdot y=\frac{3}{2} \cdot \frac{y}{1}$

[elfin berry]

-TT-

[brisk scarab]

Yes

[still temple]

then how dose That = 9

[brisk scarab]

$\frac{3}{2} \cdot \frac{y}{1} = \frac{18}{2} = 9$

[elfin berry]

n11

[still temple]

Oh so you times them and then divide them

[brisk scarab]

Yep

or you can cancel the 6 out by 2 and it becomes 3×3 = 9

[still temple]

thank you so much honesty I’ve ask like 3 other people but then never explain it in a way I can understand so thank you

Can someone explain this

[dry nest]

what step are you struggling with

[brisk scarab]

no worries! I'm glad to have been of help :)
And sorry for a late reply... I was offline

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

0.25 and how does it = 4

[dry nest]

which step specifically

[marsh citrus]

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[marsh citrus]

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[jolly hare]

Please Help

[worthy trellis]

did you try anything?

[jolly hare]

i did

they said its wrong

[worthy trellis]

can you share your working?

and maybe the solution?

[jolly hare]

ok

[marsh citrus]

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[frosty bronze]

Can someone help me with this?

[frosty bronze]

Is the number 0 fixed?

[twin girder]

Wdym fixed?

[marsh citrus]

@frosty bronze Has your question been resolved?

[frosty bronze]

I dont understand where that zero came from

[marsh citrus]

@frosty bronze Has your question been resolved?

[twin girder]

Testing mu_1=mu_2 <=> mu_1-mu_2=0

[frosty bronze]

if i want to test if the expenditure increase for this, do i get 2019 - 2020 or 2020 - 2019

[twin girder]

Doesn’t matter just have your alternative be correct

[frosty bronze]

hmm ok cuz i read the textbook and it was pretty confusing

[twin girder]

mu_2020-mu_2019>0 <=> mu_2019-mu_2020<0

[frosty bronze]

oh ok so if i want to test if there is an increase then I would select the second option?

wait no the first option *

[twin girder]

?

I just told you they are the same

[marsh citrus]

@frosty bronze Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[plucky bone]

is there difference in finding the Cartesian Products when there are brackets between A and B? or shouldn't it not matter and give the same answer?

[marsh citrus]

@plucky bone Has your question been resolved?

[marsh citrus]

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[plain current]

A hexadecimal string consists of hex digits:
0,1,2,3,4,5,6,7,8,9,𝐴,𝐵,𝐶,𝐷,𝐸,𝐹
Questions and what I have tried so far:
a) How many hexadecimal strings of length 𝑛 are there? 16^n
b) How many hexadecimal strings of length 𝑛 do not have letters? 10^n
c) How many hexadecimal strings of length 𝑛 have letters? 6^n
d) How many hexadecimal strings of length 𝑛 have exactly 𝑘 letters? k^n
e) How many hexadecimal strings of length 𝑛 read the same forward and
backward? Not sure what to do.
Is what I have done so far correct? Also, for part E I am unsure what to do but I assume I have to consider 2 cases.

[mystic minnow]

C and D are not correct

C is asking for the string to "have letters," not "only letters." So A013C is a valid string

[plain current]

for d, would it be (15 6)^k * 10^(n-k)

(15 6) is a combination

then to the power of k cause amount of k letters then the rest of the string would be n-k numbers

[mystic minnow]

you have the right idea, but things are muddled.

[plain current]

what would be changed

would it be (6 1)

[mystic minnow]

there's 6 letters, 10 digits. You want k letters of the string of length n to be letters
So you'd have (6^k)(10^(n-k)), but you also have to choose where the letters go

so you'd multiply by a combination

[plain current]

so it wouldn't be (6 1)^k * 10^(n-k)

[mystic minnow]

if the string is length n, how would you choose k places to put the letters?

[plain current]

oh i apologize i forgot that this question swaps the k and n where I am used to n representing the elements to choose from and k being the elements chosen

what do you mean multiply by a combination

would you multiply by a combination since order matters?

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[slow dock]

rewrite in rectangular form

[slow dock]

what i have done so far

making that to a cosine and i sine value seems a bit overkill, escpecially since i need to do it by hand

unless it is just simple and i am over reacting

[smoky siren]

Do you know Demoivre's theorem? Doesn't seem like overkill to me.

[slow dock]

cosine(-4995*pi/12) ??? how much is that

[smoky siren]

Oh I see what you mean.

I would not evaluate it

You can use periodicity to simplify it though

[slow dock]

-1665/4

yea?

[smoky siren]

I meant that cosine is 2pi periodic. Also it's even so the negative sign can be removed.

[slow dock]

ah

[smoky siren]

1665/4 = 415.25 so cos(-4995pi/12) = cos(pi/4)

I guess I would evaluate it after all

[slow dock]

ok yea i think i see what you mean

but how much is sqrt(2)^-999

[smoky siren]

I would just leave that as is

[slow dock]

ah ok

that makes sense

[smoky siren]

You could in principle simplify it a little bit more because cos(pi/4) is 1/sqrt(2) to get 1/sqrt(2)^1000, but whatever

[slow dock]

thx for the help

[smoky siren]

which in turn is 1/2^500

[slow dock]

👍

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[burnt abyss]

when adding 2 vectors that are at at specific angle
do we use a^2+b^2-2abcos alpha
OR
a^2+b^2+2abcos alpha
?

[burnt abyss]

can someone help?

im not sure if i should use + or -

[sick trail]

You should use +, if the angle is greater than 90°, then the cosine will give you a negative sign

[burnt abyss]

thats the thing

i want to find the angle

idk if it's >90

[sick trail]

You still use plus

If the result in the end is positive, then the angle is acute

[burnt abyss]

the angle gives me 120

and if i use - it gives me 60

[sick trail]

That makes sense if you think about it

[burnt abyss]

yeah but which angle is correct?

[sick trail]

120

The formula is with +

[burnt abyss]

do i always use + when adding?

and - when subtracting?

[sick trail]

You always use +

By using +, you let the cosine take care of the sign

[burnt abyss]

ok

thankj

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[lean heron]

Hai, I'm wondering if there's any way to do this question quickly? I.e. without multiplying everything out and then factorising again

[lean heron]

I've factored it to ((x+1)(x+3))² = 1

[sharp lichen]

i think it would be better if you take 1 to the left side

and then factorize

like f(x)^2-1^2

where f(x) is the quadratic

[sharp lichen]

does this remind you of something?

[lean heron]

Not quite

Completeing the square?

[sharp lichen]

you can apply the formula a^2-b^2

nah

[lean heron]

I can substitute ((x+1)(x+3))² with a

or something

[sharp lichen]

yes

like

[lean heron]

a² - 1² = 0

[sharp lichen]

and also factorize it after you do the a^2-b^2 step

like it would be better if you take a as the quadratic

that is x^2+4x+3=a

[lean heron]

Right so

a² - 1² = 0
(a+1)(a-1) = 0

a = -1, +1

[sharp lichen]

yes

or just keep it that way

but yes that is correct

[lean heron]

(x+1)(x+3) = 1 or -1

[sharp lichen]

noo keep it in the original quadratic form

it would be better to see the roots and stuff

like dont write it as (x+1)(x+3)

jut yet

expand it

[lean heron]

mmm, at this point I'm wondering why the middle recurring root doesn't show up

[sharp lichen]

x^2+4x+3=1, solve this quadratic, also solve x^2+4x+3=-1

[lean heron]

Ok give me a few mins

Thanks!

[sharp lichen]

thats not the exact function tho

[sharp lichen]

the correct function to plot would be (x^2+4x+3)^2-1=y

yes

you can see it has 3 roots

[lean heron]

hence middle root

Yeah

[sharp lichen]

anyways, solve this

and lmk if u have any problems with this

[lean heron]

Alright, I got it

Thanks :)

[sharp lichen]

yw

[lean heron]

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[silk plover]

I'm working on understanding how to apply the laws of Boolean Algebra to simplify Boolean expressions. This is the way the law is expressed in my textbook, but it doesn't look like the same format when I look at my example problem.

I see how we can get from Step 1 to Step 2, but I don't understand the thought process between Start and Step 1.

I think I'm confused because it feels like distribution works way differently in boolean algebra than it does in regular algebra because of the logic behind these types of problems, even if it isn't really all that different. It just feels like it is, to me.

Please help me connect the dots. Surely I'm missing a few basic things that will give me the lightbulb-moment I need!

[opal turret]

Do you how to expand brackets?

[silk plover]

I don't know. What do you mean by that?

[opal turret]

To expand a bracket means to multiply each term in the bracket by the expression outside the bracket. For example, in the expression 3(m+7) , multiply both m and 7 by 3, so: 3 (m+7) = 3×m+3×7=3m + 21 .

[silk plover]

Oh. Yeah I understand that.

Like I mentioned, I understand how Step 1 -> Step 2 work

But I don't understand how the first term for example... (x+!y) became (!x+y)x

For Start -> Step 1, I mean.

[opal turret]

Okay so
If it was
(4+5)(x+y)
When you expand brackets you gotta multiply each number like this
4x+4y+5x+5y right?

[silk plover]

Right.

[opal turret]

So basically what we did is we multiplied each number by the (x+y)
So it was like 4(x+y)+5(x+y)

[silk plover]

Oh okay so you're saying what they're doing is just FOILing lol.

I always do that in my head and just write the result on the next line. I don't rewrite it like they did so I guess that's why it didn't make sense to me.

[opal turret]

Yes

[opal turret]

I’m sorry if I didn’t explain it very well

My math is another language so I’m not used to English math terms

[silk plover]

No it's okay. You did well.

Now I'm trying to decide if that's enough for me to continue hmm

I think it's enough for now. I'm gonna try the rest of my problem and I'll return again if I need more help. Thanks Spark! 🙂

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[tough abyss]

correct?

not sure about gamma relation

[still creek]

tbh this is probably too advanced for this

go to real-complex-analysis I think

[tough abyss]

alright

. close

[marsh citrus]

@tough abyss Has your question been resolved?

[tough abyss]

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[misty bramble]

How can I find the zeroes of a polynomial that has 4 terms?
Suppose I had: x^4 - 4x^3 + x^2 + 6x, I already know that x can be 0, or -1 by factoring out the x.

[cloud iron]

If you already know two solutions you could do polynomial long division to reduce the degree of the equation

[misty bramble]

Are you talking about the factor theorem?

[cloud iron]

Like if x=-1 is a zero, then divide x^4 - 4x^3 + x^2 +6x by x+1

[noble nebula]

x+1?

[cloud iron]

thanks lol

[misty bramble]

And what do I do with the result of this?

[cloud iron]

The result would be a polynomial with highest power ^3, effectively reducing it to something that may be a bit easier to work with

If you then divide the result by x, assuming x=0 is a zero, then you'll have a quadratic

Which is much easier to factor or use quad orm

That does rely on 0 and -1 being solutions

[misty bramble]

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[minor glade]

How would one go about solving this problem?

[cloud iron]

Do you know slope-intercept form of a line?

[minor glade]

Yes

[cloud iron]

So if we're able to interpolate the y-intercept and slope from the graph, we can use them to find the equation

Where's the y-intercept on that graph?

[minor glade]

5

[cloud iron]

Bingo

Now our slope will be the ratio rise over run

[minor glade]

How do I choose from the given options tho? I get my answer in the y=mx+b form

[cloud iron]

We can rearrange the answer to be different forms

What answer do you get?

[minor glade]

Got it

Just found a vid on this

[cloud iron]

👍

[minor glade]

Haven't tried yet

y=x+5?

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[silk plover]

My question is about using boolean laws to simplify boolean expressions.

I was looking at the steps to solve this problem and my first arrow points to how DeMorgan's Theorem is applied in that particular step of the solution.

Is that the only acceptable way to apply DeMorgan's Theorem here?

Would the 2nd arrow be acceptable too? If not, why?

[marsh citrus]

@silk plover Has your question been resolved?

[silk plover]

Still here, not afk or anything

[marsh citrus]

@silk plover Has your question been resolved?

[marsh citrus]

@silk plover Has your question been resolved?

[silk plover]

To add, I'm gonna put the AC and !A back in after this step. I just wanted to figure out how DeMorgan's was able to be applied first.

[subtle mango]

question : why is the symmetric difference quotient exact for quadratics

[cloud topaz]

@untold depot

[silk plover]

🥲

https://tenor.com/view/spongebob-wait-gif-5140632

[cloud topaz]

@silk plover I'm not a helper, but I can take a crack at it

I assume you're using engineering notation?

[silk plover]

I don't know if its called that. It's from my Discrete Math textbook though! 🙂

[cloud topaz]

i.e. "•" is AND and "+" is "OR"?

[silk plover]

yes

Even if you're not sure, if you have any thoughts on it I am interested.

Im doing something atm so it might take me a few mins to respond.

[cloud topaz]

Does "A(...)" = "A AND (...)" ?

[silk plover]

Yes.

[cloud topaz]

So is this a correct restatement of your problem?

[silk plover]

That looks right, yes.

[cloud topaz]

Oh, I see.

[silk plover]

Sorry if I made it confusing!

[cloud topaz]

Double bar cancels.

[silk plover]

Right. Yes, we have the double complement law for some of those terms.

Here is the full set of steps for the solution to this example problem. I'm just trying to understand if I can do it both ways or if the first arrow in my original photo of my work is the only way you can do it.

[cloud topaz]

No, you cannot split both bars.

One sec

[silk plover]

Here is the isolated steps for the particular piece we're looking at.

[silk plover]

But you do split the bars for this law I thought?

[cloud topaz]

Yes, but only one bar at a time

Every time you split a bar, you flip the sign

So you cannot break two bars but only flip the sign once

If you break two bars, you have to flip the sign twice

And when you flip the sign twice, you get the sign you started with.

[silk plover]

I guess I'm confused because the operator for A!B flips but so does the operator between A!B and !(DB).

I didn't expect the + to change to a *

I thought we worked on 1 term at a time.

referring to this

[cloud topaz]

You can write it where you flip the sign on both terms

[silk plover]

Okay. Yeah. I wondered why they skipped your 2nd line in the steps process.

That's exactly what I wrote on my paper though! 🙂

I thought you said it was wrong lol

I didn't write line 2 either, but I came up with your 3rd line there.

That's all I wanted to know. I just wanted to make sure that I could do it my way if I wanted to and still come up with the same answer.

[cloud topaz]

Are you sure?