help-0

[slender girder]

right?

[oak chasm]

Yes, and then you show that if it works, the next number has to work as well.

So, it's if a number works (we're not sure, but if), the next number works as well.

[slender girder]

right

[oak chasm]

And then it's like a machine.

53 works.

Since 53 works, 54 works.

Since 54 works, 55 works.

And so on.

That's why the inductive step doesn't know what the number is.

[slender girder]

Yeah

[oak chasm]

It has to work for any number that worked.

So, it's like a variable we don't know the value of.

Or the input to a function.

[slender girder]

Right

[oak chasm]

So, for your problem, you proved it for m = 1.

Then you prove that if it works for m, it works for m + 1, regardless of the m.

Then that machine idea works for it.

[slender girder]

Ahh

[oak chasm]

It works for 1, therefore it works for 2.
It works for 2, therefore it works for 3.
And so on.

[slender girder]

Right

[oak chasm]

And on that, you have:

Inductive hypothesis:

ruₘ₋₁ + s = rᵐ⁻¹ u₁ + s((rᵐ⁻¹ - 1)/(r - 1))

Thing to prove:

ruₘ + s = rᵐ u₁ + s((rᵐ - 1)/(r - 1))

[slender girder]

yeah

[oak chasm]

The inductive hypothesis is the "if it works" for one number. We just assume it works and leave that for the base case and machine to prove that part.

So, you can replace uₘ₋₁ with the recursive formula version.

[slender girder]

right

[oak chasm]

Solve this for uₘ₋₁:
uₘ = ruₘ₋₁ + s

Replace uₘ₋₁ in the inductive hypothesis:

ruₘ₋₁ + s = rᵐ⁻¹ u₁ + s((rᵐ⁻¹ - 1)/(r - 1))

Solve for ruₘ + s.

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[queen sierra]

Helo with question 4please

Help*

[harsh swallow]

what's the problem

[queen sierra]

I can do part a just cant do part b

[harsh swallow]

you need to find the point where the distance of the moth to the origin is the same as the radius of the lamp

the diameter of the lamp is 2 pi

and so the radius is pi

you need to find where $|\vec{r}| = \pi$

[ocean seal]

Katharine

[harsh swallow]

that will be at some value for t

[queen sierra]

i dont get the notation you jsut wrotw

[harsh swallow]

and you can then fill it into $\dot{\vec{r}}$

[ocean seal]

Katharine

[harsh swallow]

the size of r

the distance of the moth to the origin

that's the vertical lines

it means the size of the vector

[queen sierra]

so do i make rdot = to pi

[harsh swallow]

You start with $|\vec{r}| = |(4 \pi - t) \sin(t) \hat{i} + (4 \pi - t) \cos(t) \hat{j}| = \pi$

[ocean seal]

Katharine

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[supple tundra]

This seems reasonable, but I am struggling to draw a diagram that shows why this is

[ancient saddle]

Maybe you can draw it like a sum of "vectors", the resultant vector of a+b is always between a and b so the argument is between the arg of a and arg of b

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[supple tundra]

ok thanks leonardo

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[wispy salmon]

Is this an actual identity or the app tripping?

[wispy salmon]

And if it is can anyone explain it to me?

[buoyant kayak]

yes, it's the triple angle identity

[supple tundra]

You learned $\sin{2t} = 2\sin{t}\cos{t}$?

[ocean seal]

azeem321

[supple tundra]

$\sin{3t} = \sin({2t+t}) = \sin{2t}\cos{t} + \cos{2t}\sin{t} = (2\sin{t}\cos{t})\cos{t}+(\cos^2{t}-\sin^2{t})\sin{t}$

[ocean seal]

azeem321

[supple tundra]

and then u can finish it off 😄

[wispy salmon]

im still trying to understand how this works

ah i just cant get it, will go watch some videos i guess, thanks for the help guys

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[wispy salmon]

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✅

[wispy salmon]

just came here to drop this xD

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[fossil valley]

Not sure if the answer I have selected is correct

Is divergent the same thing as indeterminant?

Because the second integral would equal infinity/infinity right

which is indeterminant

[quasi scarab]

why?

[fossil valley]

Because when you sub in infinity for x in the original result it gives you that?

I'm not sure if that is correct but that is my reasoning

[quasi scarab]

how did you calculate?

[fossil valley]

mentally

[quasi scarab]

oh

then think it through again

(your answer is wrong)

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@fossil valley Has your question been resolved?

[small bear]

When you have an improper integral, make sure to solve the necessary limit.

Don't just plug infinity. You need to check the arithmetic again using the limit.

[fossil valley]

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[brittle path]

There were no solutions provided for this question. So I have no way of understanding how to do it. Without a relationship between the 2 squares, besides one is longer than the other.

[hushed loom]

Let's say the first piece is xcm and the second is ycm.

The first statement says x + y = sqrt(300)

The second statement is a bit tricky. When we bend the x piece into a square, this square will have sides of length x/4, right? Then its area will be (x/4)^2.

Then you do the same for the other piece. Do you know what is the area of the first plus the second square?

With this you can build a system of equations to solve

[brittle path]

So what I have and I'm stuck on is (a 1/4)^2 + (b 1/4)^2=13cm^2

[hushed loom]

Use the first statement too!
a + b = sqrt(300), then
b = sqrt(300) - a

Have you tried that, then substituting in the one you have?

[brittle path]

wow, never thought of that. But yeah I see it now.

so I can insert b = sqrt(300) - a to the second statement right? And do the same for a?

[hushed loom]

you don't have to, if you just do it for b then you have a quadratic equation, right?

[brittle path]

So (sqrt(300) - a) + a = sqrt(300), but now how would I solve for a?

wait nevermind

I have aha moment, but thank you for helping me

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[ashen igloo]

This is infinite number of solutions right

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[steep rover]

what part of it do u need help on

[torn girder]

Just those questions

I suck at these problems

[steep rover]

i can walk u thru them but i wont give u the answer

since that kinda destroys the learning process

so for the first one it gives u what units represent what variables

and it says that she can buy no more than 6

[torn girder]

Wouldn't it be c then?

[steep rover]

so what do u think that means? like which sign would that be

yeah

its c gj

[torn girder]

nice

[steep rover]

for the shading part for number 2

just image what the lines would look like

and the shaded region is all of the points that meet both conditions

[torn girder]

I picked a on that one but Idk if it's b I'm stuck between those two

The way they setup the shaded areas just confuses me on that one

[steep rover]

well if its less than -2, is that the right side of the graph or the left side

[torn girder]

Right side?

[steep rover]

not so quite

if its shaded to the left, it means the values are smaller

because the farther you move left, the smaller the integer/value of the number becomes

think of a number line

[torn girder]

So it would be a because x is smaller than 2?

[steep rover]

yup

[torn girder]

Ah alr

[steep rover]

do u still need help with the other 3 is it kinda making sense

[torn girder]

Yea kinda

The pair ones

like 4 and 5

I have no clue how to do

same for number 1

[steep rover]

you know that in point notation or whatever its called the order is (x,y) right?

[torn girder]

I know 2 is 13 and 6 is D

[torn girder]

Not really

[steep rover]

oh well

when its a point

and its given in the parenthesis

its always (x,y)

where the first number represents the x value and the y obv represents the y value

so for number 4 just try plugging in the points it gives u

and find what point makes the equation true

[torn girder]

Alr gimme one sec

B?

[steep rover]

yeah

u understand 5 right?

[torn girder]

5 tho how would I detirmen the correct one

[steep rover]

you just plug in

-5 times -2 is what?

[torn girder]

10

[steep rover]

so y has to be greater than 10

since the question is asking for y

[torn girder]

So would it be D then?

Since it can't be b or a

[steep rover]

the sign is greater than or equal to

u need ur y value to be larger or equal to 10

8 isnt larger than 10

[torn girder]

So then wouldn't it be c since 2 is larger than -3

[steep rover]

are u still on 5?

[torn girder]

Yes

[steep rover]

for the right side of the equation u got 10 right??

since u plugged in -5 for x

[torn girder]

Oh so 5 is D?

[steep rover]

so whatever y is, it has to be larger than 10

8 isnt larger than 10

so it cant be D

[torn girder]

Yea so it can't be D

[steep rover]

only one of the number is larger than 10

btw what grade are u in?

[torn girder]

10th

I'm usually good at math but when it comes to inequalitys and graphs I just can't understand them

[steep rover]

i see

[torn girder]

It's like the one thing I suck at

[steep rover]

i thought u were in like elementary or something lol

[torn girder]

Yea they keep on making me do em even know I'm in honors classes

It's just I can't figure them out

[steep rover]

anyways what did u get for 5

[torn girder]

I think it's C

[steep rover]

why?

[torn girder]

Since I don't think it can be A because it cancels out cause 5x0

and 1/2 was is a false in the true or not

If it is C can you explain how it is?

[forest bear]

1 is B and 2 is A that’s all Ik. 3rd I forget

[steep rover]

ok i think i have to re explain it

[torn girder]

So I had those 2 right thanks

[torn girder]

Ok

[steep rover]

so the question is asking for which value of y, does it make the ineqaulity true

if u plug in -5 for x, it becomes -2(-5) = 10

[forest bear]

I think 3 is also A as it’s the only one that falls within the shaded area

[steep rover]

^

for 2, only 13 is greater than 10, so it makes the statement true

so A

[torn girder]

Wait so 5 is A?

[torn girder]

Were talking about 5 still right?

[steep rover]

yeah

for 6 that guys reasoning is correct

[torn girder]

Your going off the numbers on top of the problem right?

1 is B, 2 is A, 3 is A, 4 is B, 5 Idk, 6 is D, 7 is C, 8 is A

Thats right correct?

[steep rover]

i only see 6 problems

oh wait

theres a 3rd screenshot

[torn girder]

Which numbers did u not see?

[steep rover]

i was going off the order of the screen shots

[torn girder]

Based on the numbers on top of the problems is this right?

[steep rover]

1 is b, 2 a , 3 a, 4b, 5 is hella weird gimme a sec, 6D, 7 c, 8a

[steep rover]

i got the order of the problems wrong since i didnt see they were numbered for a sec

but the list above is the solutions i got

5 should give u a graph or something

[torn girder]

Yea so I got all them right except 5 atm

[steep rover]

not points

[torn girder]

Ik they question these weirldy sometimes

[steep rover]

i think the question is wrong

like its asking for graphs but gives points

its the same problem as 4

[torn girder]

Yea

Which one do you think I should pick for it?

[steep rover]

same answer for 4

ig

[torn girder]

Alr I'ma sumbit this

yea we got all em correct accept 5

It was a

Still 14/16

poinysd

Thank you very much mate

Saved my butt and helped me learn some of this better

[steep rover]

np

[torn girder]

It's a 87.5 I'll be good all my other ones i did bymyself and got 100 it's just those problems get me man

[torn girder]

Hey you have a great night gn man

[steep rover]

you too

[torn girder]

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[tropic veldt]

What is the answer?

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@tropic veldt Has your question been resolved?

[tropic veldt]

<@&286206848099549185>

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@tropic veldt Has your question been resolved?

[tacit arch]

.close

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[gray ingot]

nvm you missed 1

I think you forgot to count set A itself

set A is also a subset of itself

[alpine sable]

I would try finding all the subsets without 11 and 12

[gray ingot]

that'd take too long

2^10-9

💀

[alpine sable]

Thats the universe

[pale reef]

wouldnt that be more complicated?

[alpine sable]

Actually it might be

You could do it directly too

Anyways

whats your answer?

or work

[pale reef]

i tried using sum of binoms

so (8c0) + (8c1) + ... + (8c8)

so the sum would be 2^8

[alpine sable]

Uhhhhhh

Hold up lemme think

[elder grail]

isn't this the answer?

[pale reef]

no ans so was checking if itd be correct :/

[elder grail]

yeah ig that's correct

[alpine sable]

Yeah it would be

[pale reef]

okk thank you :)

if they changed the qn to A being a 3 element subset then itd be 8 right

[alpine sable]

It'd be 7 no?

[elder grail]

yeah 7

no wait

8 only

[alpine sable]

Cause you're just doing this

[pale reef]

but there are 8 rem numbers to choose from

[elder grail]

13->20 8 numbers

[pale reef]

so isnt it j 8c1?

[alpine sable]

wait

I think

there's something wrong here

Its 11 and 12

[pale reef]

yeah

[alpine sable]

so you can have a subset with every number except 12, which is choosing from 9 numbers

[pale reef]

but isnt this more straightforward

[alpine sable]

Yeah i think so

im just seeing if we're under-counting

Yeah i think its fine

2^8 is right

[elder grail]

yeah

if it were 'or'

then jswatj's method better ig

[alpine sable]

If you had a subset of 3 numbers then it would be 14->20

[elder grail]

why not 13?

[alpine sable]

11, 12, 13

oh assuming its those 3

[elder grail]

(11,12,(anything from 13->20))

is this right?

[alpine sable]

that would be if it was 11, 12, x, if x was some number from 13, 20 in the set A

[elder grail]

yeah

so for a subset of 3 that's right?

with 11,12 included

[alpine sable]

yes

you could just find how many there are for 3 and multiply it by 7

anyways

does this answer your question?

[elder grail]

@pale reef

[swift shore]

@alpine sable powersets are counted like
{true, false, false, true, false}
so there are two options (include/exclude) for each element of the set, for a total of 2^n possible subsets

[alpine sable]

I know

[swift shore]

oh lol

it's 2^8 right

[alpine sable]

yes

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[shell narwhal]

how do i find the solution for each function ?

[alpine sable]

wait are you solving both at the same time or each one separately?

[shell narwhal]

doesnt matter

well

both

[alpine sable]

if its both, i would add one to the other

you'll immediately cancel out the 1's and 2xy's

[shell narwhal]

i end up with y^2=-x^2

then what ?

[alpine sable]

you'd end up with y^2-x^2=0

[oak chasm]

When you square a real number, the result is nonnegative.

So, both sides have to be nonnegative.

[shell narwhal]

do you mean + ?

[oak chasm]

y² is already nonnegative, but -x² is nonpositive, so it can only be nonnegative when?

[alpine sable]

oh my bad

wait no

thats right

[oak chasm]

What numbers are both nonpositive and nonnegative?

[shell narwhal]

0 ?

[oak chasm]

Right.

[oak chasm]

So, the left side is zero and the right side is zero, assuming x and y are real numbers.

[shell narwhal]

btw the asnwer should be (1,1)

[oak chasm]

1² = -1² is incorrect according to y² = -x², so if (1, 1) is the solution, y² = -x² is invalid.

[shell narwhal]

so did i make a mistake

[alpine sable]

It should be

y^2-x^2=0

so y^2=x^2

[shell narwhal]

can you show me your work ?

[alpine sable]

I dont have any

Next square root both sides

notice that |y|=|x|

use the absolute function

[shell narwhal]

how did you find this ?

[alpine sable]

because

i solved the system

I added one equation to the other

[shell narwhal]

since 0=0

[alpine sable]

?????

No

[shell narwhal]

what did yo udo ?

[alpine sable]

You can add one equation to the other

[shell narwhal]

how

[alpine sable]

a "copy" of it

??

[shell narwhal]

a copy ?

o.O

[alpine sable]

its a rule in systems of equations

you can check out #help-7｜zen1thxyz

[shell narwhal]

ughhh

[alpine sable]

they're learning about it

[shell narwhal]

i dont get it

[alpine sable]

You could also do substitution I guess

solve for one variable

then substitute it into the other equation

[shell narwhal]

could you explain the copy method ?

[elder grail]

basically a=b and c=d then a+c=b+d

[alpine sable]

If you have a system of equations

and you add a copy of one equation to another

it doesn't change the solutions that the system produces

[shell narwhal]

can you write it down ?

[alpine sable]

No you can google this

[shell narwhal]

:c

[alpine sable]

plus its too long to write

[shell narwhal]

thanks

[elder grail]

yeah now ur a and c in this case are 0 each

and u should be able to get the rest

[shell narwhal]

yhea i got x^2=y^2

so the answer is (1,1)

[elder grail]

done

?

[shell narwhal]

thanks

yes

thanks so much

btw

nvm

[elder grail]

?

[shell narwhal]

well

[elder grail]

go ahead

[shell narwhal]

is (-1,-1) good ?

[elder grail]

yeah

[shell narwhal]

so there are 2 answers ?

or 4 ?

[elder grail]

wait lemme do properly

[alpine sable]

there's more steps to this

[elder grail]

yeah

[alpine sable]

use absolute function and stuff

|y|=|x|

[elder grail]

u need to check conditions

[alpine sable]

^

[elder grail]

like domain n all

[shell narwhal]

wut

[shell narwhal]

what is this ?

[alpine sable]

absolute value of y equals absolute value of x

[shell narwhal]

i meant

what is it for ?

[elder grail]

do u know absolute value func?

[shell narwhal]

yea

[elder grail]

yeah so x^2=y^2 u need to use that

[shell narwhal]

but i dont know why we are using it in our context y^2=x^2

[alpine sable]

It's because

im breaking down the square

You could also look at it like this

let y^2-x^2=0

then we can factor as such

(y+x)(y-x)=0

[shell narwhal]

i see

[alpine sable]

so y=x or y=-x

look at both cases

[elder grail]

yeah n absolute value thing gives the same answer

kinda compressed

*representation

[shell narwhal]

so (1,1), (-1,-1), (-1,1), (1,-1) are good answers ?

[elder grail]

check if these values satisfy both eq u have

[shell narwhal]

which equations ?

the originals /

[elder grail]

the original ones

yeah

[shell narwhal]

oh kk

il lcheck

[elder grail]

that's also a reason why 0,0 isn't a solution

[alpine sable]

you should be solving for these

[elder grail]

tho 0^2=0^2

[alpine sable]

In each case

[shell narwhal]

hmm no

doesnt work

[elder grail]

which ones do?

[shell narwhal]

(-1,1) and (1,-1) gives 0=4

which isnt true

[elder grail]

that's it

so they not solutions

[shell narwhal]

i see

[elder grail]

makes sense now?

[shell narwhal]

yhea thank you so much

.close

[lone heart]

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[lone heart]

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[lone night]

is there any formula to multiply combinations

[grizzled gyro]

i don't think so, just decompose the combinations into the formula and go from there

very plug and chug

this is also a very googable question: "can you multiply combinations", "how to multiply combinations", "multiply combinations formula"

[lone heart]

@lone night Has your question been resolved?

[elder grail]

there's a faster way using coefficient of x^4 in the expansion of (1+x)^5 * (1+x)^7

tho idk if i could explain it well

if u've seen a similar approach before then u could use that

[lone night]

okay

[lone night]

didnt fight a direct answer

[lone heart]

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[lone night]

thanks

[lone heart]

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[echo briar]

help

[gray ingot]

what is your question

[echo briar]

Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99%

: (

i am confused a lot

[swift shore]

interesting

if a coin is tossed once, the probably of getting heads is 1/2

[echo briar]

yea

[swift shore]

if it's tossed twice, the probability of getting heads at least once is 3/4

TT, TH, HT, HH

[echo briar]

what is meaning of atleast once

is it more than 1

or less than 1

[gray ingot]

more than or equal to 1

so basically TH, HT, HH

the probability of getting at least 1 head is the compliment of getting all tails

[swift shore]

@echo briar do you understand this

[echo briar]

that's the problem my dude

i dont understand why we are taking complement

rather than finding the all heads

[gray ingot]

all heads is just HH

[swift shore]

it's not all heads

it's at least one heads

[gray ingot]

at least one head is TH, HT, HH

[echo briar]

a

[swift shore]

so you could roll HeadsTails, or TailsHeads, or HeadsHeads

[echo briar]

oh

[swift shore]

do you see now

the only way to not get at least one heads would be for all of them to be tails

[echo briar]

so if we consider all heads which means we are considering only one case ?

[swift shore]

yeah

and we don't care about all heads rn

[swift shore]

so we can deduce that
(probability of at least one heads) = 1 - (probability of all tails)

do you understand now

[echo briar]

https://tenor.com/view/crying-black-guy-meme-sad-gif-11746329

confusion cleared

[swift shore]

nice lol

so do you think you can do it from here

[echo briar]

the probability of getting all heads means that we only gett like TTT

no wiat

[gray ingot]

we don't get TTT*

[swift shore]

what lol

[echo briar]

all tails i mean

[swift shore]

yea yea

so basically

[gray ingot]

P(at least 1 head) = P(not all tails)

[swift shore]

we're asking how many times do you have to flip a coin, in order for P(all tails) to be less than 1%

[echo briar]

https://tenor.com/view/house-river-flooding-floating-gif-9580205

my house is flooded by knowledge

[swift shore]

yes bruh

yes it is

[echo briar]

NOW

Thanks

[swift shore]

can you figure this out now

@echo briar

[echo briar]

now the other question is

why is porbability of getting all tails (1/2)^n

[gray ingot]

if you flip 2 coins what is the probability of getting 2 tails?

[echo briar]

(HH) , (HT) , (TH) and (TT) so it's 1/4

[gray ingot]

we can think of it like
getting a tail on the first coin and getting a tail on the second

[echo briar]

yeah

[gray ingot]

or in other words 1/2 x 1/2

[echo briar]

O.O

hmm

[slender gull]

Well the number of possibilities when n coins are tossed would be 2^n.
And out of all those only one will have all tails.
Hence 1/2^n

[echo briar]

dang

i was thinking the same

[slender gull]

Good for you.

[swift shore]

nice

so

whaddya think

@echo briar

[echo briar]

it was interesting problem

you guys are smart

[vale sapphire]

Another way to see it is that 1/2 means that you divide the amount of possibilities by 2. When you get heads on one throw, that divides the amount of possibilities by 2, since every possibility where you land tails is now excluded.

[echo briar]

oo

oh

my teacher is very excellent in maths but he cant explain problems nicely

[vale sapphire]

In my experience the lack of either is direly felt

"probability 1/2" means "half of the cases are out"

[echo briar]

i have another question

[vale sapphire]

Although, I want to draw your attention to something

Here it's easy to conceptualize "1/2" because the amount of cases is neatly divided in two

But suppose we had a rigged coin

That landed a heads with a probability 1/3

[echo briar]

what ?

[vale sapphire]

So like it'd be weighed in such a way that when you throw it a lot, it only comes up heads a third of the time

[echo briar]

oh

[vale sapphire]

Then the probability of getting two heads in a row would be (1/3)². 1/3 for the first throw, and then 1/3 for the second throw.

[echo briar]

yea

it's same like above he explained

[vale sapphire]

Yeah, except you just change the probability

[echo briar]

yes

[vale sapphire]

I just wanted to make sure that was okay, because here we're not just splitting the amount of cases in two and calling it a day

We're also saying that one half of the cases has less chance of occuring overall

[echo briar]

yea no problem , i have another question if you can help

[vale sapphire]

go ahead

[echo briar]

The probability hitting a traget is 1/3 . Then minium number of times the traget should de hit so that probability of at least one success is greater than 5/6 is ?

[vale sapphire]

here, there is a fruitful observation to be made, and i want you to keep this in mind

"at least one" sucks to calculate

[echo briar]

ture

[vale sapphire]

It's "1 or 2 or 3 or ...."

Once you know how to manipulate series you can make more sense of it, but still, it's not very wieldy

[swift shore]

trager ☹️

[echo briar]

our exams wont ask excatly

[vale sapphire]

Instead, the very common trick used here is to consider what the opposite of that is

What's the opposite of "at least 1"

[echo briar]

🗿

atmost

?

[vale sapphire]

at most what

[echo briar]

atmost 1

[vale sapphire]

1 is at least 1

But also at most 1

So that doesn't work

[swift shore]

lmao

[vale sapphire]

I want the logical opposite

[echo briar]

huh

[vale sapphire]

What does it mean to NOT be "at least 1"

Just think about which numbers don't satisfy this, and find a way to describe them

[echo briar]

it's 0 or 1

cause atleast means more than 1

[slender gull]

Atleast also includes 1.

[vale sapphire]

oh, my bad

yes

[echo briar]

yea

[vale sapphire]

should have made that clear

so you're telling me that 1 is not at least 1

Something doesn't check out

[echo briar]

no

[swift shore]

if I DON'T have "at least one apple", how many apples do I have?

[echo briar]

it's 0

[swift shore]

yeah

exactly

[echo briar]

so what is your point

[vale sapphire]

the point is that the opposite of "at least 1" is "strictly less than 1"

"at most 1" includes 1, it isn't strict

[echo briar]

yeah

[vale sapphire]

Using symbols, the opposite of "x ≥ 1" is "x < 1"

[echo briar]

yes

[vale sapphire]

Since we're working with integers, "x < 1" means "x ≤ 0"

And since we're working with natural numbers, it just means "x = 0"

So the opposite of "getting at least one success" is "getting no successes"

Put like that it doesn't look like much

[echo briar]

oh

now you are coming to point

[vale sapphire]

The thing is, you can easily calculate "getting no successes", and you can calculate the probability of the opposite of an event

[echo briar]

yes

[vale sapphire]

What's the probability of failing n times in a row?

[echo briar]

are you asking about the question i asked

[vale sapphire]

Yes, because "failing n times in a row" is the opposite of "getting at least 1 success over n attempts"

So we can go from one to the other easily

[echo briar]

The probability hitting a traget is 1/3 . Then minium number of times the traget should de hit so that probability of at least one success is greater than 5/6 is ?

now according to you we need to calculate the probability of not hitting the traget then subtract 1 from it ?

[swift shore]

P(at least one success) > 5/6

since P(no successes) = 1 - P(at least one success), then we can substitute:

1 - P(no successes) > 5/6
P(no successes) < 1/6

do you understand

[echo briar]

let me think

[echo briar]

yes

[vale sapphire]

This is very common, if you see something like "at least X" inside of a question, consider the opposite

Because that will turn an infinite "at least X" (hard to work with) into a finite "strictly less than X" (easier to work with) when dealing with natural numbers, which is common as well

[echo briar]

yes

i finnaly understood the point

[echo briar]

https://tenor.com/view/luffy-one-piece-anime-cry-crying-gif-5378935

I finally sloved the question

answer came out right

[swift shore]

niceee

[echo briar]

thanks for help

.close

[lone heart]

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[lime bronze]

Question

[lime bronze]

Imagine a circle then using 2 chords make a central angle now if the intercepted arc of that central angle is straightened will the angle increase or remain the same?

[lone heart]

@lime bronze Has your question been resolved?

[lime bronze]

<@&286206848099549185>

[lime bronze]

Anyone?

[echo briar]

hmm

[echo briar]

it will increase

and can you provide a diagram as well ?

[lime bronze]

Just a thought

[echo briar]

hmm

wait let me think i read the question wrong

[lime bronze]

I couldn't wrap my head around it lol

[echo briar]

let me make a digram on note

[lime bronze]

Aight

[echo briar]

do arcs intersect on circumference of circle or outwards ?

[lime bronze]

On circle

[echo briar]

ok

[echo briar]

Do you mean the angle alpha

[lime bronze]

The blue angle

[echo briar]

do you mean angle alpha

the one made with red and blue lines

[lime bronze]

Yes

[echo briar]

the angle will decrease

[lime bronze]

Oh

[echo briar]

cause any arc on circle has always less distance than diameter of circle

if a arc has equal distance than diameter then it's diamter always

[echo briar]

and if you are increasing the intercept of arcs which means you are trying to make arc a diamter

now in this case there are many possibilites

[lime bronze]

Im following

[echo briar]

in the diagram above if you want to increase the intercept you just shift the line white downwards

and the down white line upwards

to increase distance

and since if you create motion imagination in your mind the angle decreases

hope your doubt is solved @lime bronze

[lime bronze]

Wait give me a minute

[remote kettle]

hello

[lime bronze]

Ye actually

I get ut

Thanks

[echo briar]

the other way around is that you are trying to meet the line to center of circle

and it's way more simple to imagine

[lime bronze]

I see

Aight thank you

.close

[lone heart]

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[sturdy rune]

Hello, im trying to solve the differential eqn using a power series. i got to the coefficient of (x-1)^N and confused why it is not (N+2)(N+1)_aN+2-x(N+1)_aN+1-aN ?

[sturdy rune]

would we not make it N=n+2 and N=n+1 ?

[lone heart]

@sturdy rune Has your question been resolved?

[lone heart]

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[sturdy rune]

.reopen

[lone heart]

✅

[lone heart]

@sturdy rune Has your question been resolved?

[lone heart]

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[alpine sable]

„Karim works at the fashion store "Zora".
His monthly salary consists of a basic salary of 1600 € and additional
2% commission. The commission is a share of his monthly sales.
a) Write a function equation that Karim can use to calculate his total
monthly salary.“

Is f(x) = 0.02x + €1600 the correct answer?

[marsh rapids]

yes

[lone heart]

@alpine sable Has your question been resolved?

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[light kraken]

I’m having trouble finding the whatever u call it for the diamond

[light kraken]

I always have trouble with this

The assignment is called solving a quadratic equation by factoring

[woeful pulsar]

what you can do is

1. use quadratic formula
2. just figure out from there what to write in the factors

[lone heart]

@light kraken Has your question been resolved?

[light kraken]

I just have trouble finding the 2 numbers

Idk tat formula I’m kinda dumb

My teacher has me find 2 numbers that add up to the smaller number

[woeful pulsar]

yeah but sometimes the numbers are pretty sizable that it's not that easy to factorise

[light kraken]

And then when they’re times’d I guess they equal the bigger number I think

[woeful pulsar]

have you tried completing the square too?

[light kraken]

No I didn’t learn that section yet

That’s next chapter I think

[woeful pulsar]

actually it's the coefficient of x, not always the smaller number

[light kraken]

Dang it

[woeful pulsar]

have you seen something like this $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

[ocean seal]

Element118

[light kraken]

No

Wait yes

[woeful pulsar]

what does it do

[light kraken]

Idk I think it means everything in top times the 2a

my teacher did that but without the letter in the bottom

[stiff geyser]

To solve by factoring

1. Multiply a with c
2. So find two numbers that when multiplied gives this product ac and when you add them it gives b

[light kraken]

So that part is new o me

Ty 💛

[stiff geyser]

In this example
a=1
b=60
c=900
Find 2 numbers such that you get 900 when you multiple them and you get 60 when you add them

||in this case it's 30 and 30||, seems like it's -30 and-30

[light kraken]

How is it 2 of the same number

Brb teacher

[woeful pulsar]

yeah in that case the a is one since x^2 coefficient is 1

[light kraken]

I might have to close this teacher wants my phone away

Idk how to close

[woeful pulsar]

.close

[light kraken]

.close

[lone heart]

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[reef trout]

Define the quasi-primes as follows.
• The first quasi-prime is q1 = 2
• For n ≥ 2, the n
th quasi-prime qn is the smallest integer greater than qn−1 and not of
the form qiqj for some 1 ≤ i ≤ j ≤ n − 1.
Determine, with proof, whether or not 1000 is a quasi-prime.

[woeful pulsar]

Have you tried some examples?

[reef trout]

no i am completely stuck

[woeful pulsar]

like have you calculated p2, p3, p4....? to try to notice some pattern?

[alpine sable]

All primes are definitely quasi-primes

8 would be a first quasi-prime which is not prime

[reef trout]

i am still confused on the idea of quasi prime can u explain it to me plz

[alpine sable]

Well, it's just how in definition

Can't be written as product of two quasi-primes smaller than it

So I think we can show 125 is quasi-prime too

[woeful pulsar]

so we notice that there are 2, 3, 5, 7, 8, 11, 13... and you may notice something in common with these numbers

[alpine sable]

p^3 for p prime is quasi-prime, yeah

And now 1000 = 2^3 • 5^3 is a product of two quasi-primes so not a quasi-prime

[reef trout]

they are the prime numbers

[alpine sable]

No because there's 8 there

[reef trout]

oh

[alpine sable]

8 = 2•4 is the only way to really write this as a product

And 4 = 2•2 is not a quasi-prime

So 8 is a quasi-prime

Right?

[reef trout]

yes

[alpine sable]

Yeah. But this really works because 8 is a third power of 2

[reef trout]

oh ok

[alpine sable]

So now if we take any third power of a prime number, it's a quasi-prime

[reef trout]

ohh ok i think i understand

[alpine sable]

Yeah. But 1000 = 2^3 • 5^3 so it's a product of two quasi-primes, can't be a quasi-prime

[reef trout]

so this will be how u prove it right , and a product of 2 quasi-primes in not a quasi -prime

[alpine sable]

Yes

[reef trout]

thank you!

[alpine sable]

You're welcome

[reef trout]

sorry but what do u put the number in to calculate the pattern

[woeful pulsar]

i meant just try out calculating these numbers

whoops

I mean q2, q3, q4

idk how I thought of them as p

[reef trout]

oh to check to see if they only can be written as one product of prime right

[woeful pulsar]

the main idea is don't be afraid to try out some small cases to get a feel of how the problem is

[alpine sable]

We can try checking whetever or not something like p_1^k_1 • ... • p_m^k_m is a quasi-prime. This depends only on k_1, ..., k_m so more of a combinatorial problem

[reef trout]

oh

[alpine sable]

I'm not thinking about the problem of checking if 1000 is quasi-prime, just like in general would be neat if we were able to say something about this

Maybe best cases to check would be when m = 1 or when k_1 = ... = k_m = 1

Now to think about it, it doesn't matter what power is p_1 to for example, just if that power is or isn't quasi-prime

So it'd solve the problem completely

[reef trout]

what does the underscore means in this case

[alpine sable]

Lower index

I'm lazy and don't feel like using LaTeX

[reef trout]

oh ok

[alpine sable]

I think p^(2n+1) are all quasi-primes

[reef trout]

i dont understand how do we check

[alpine sable]

By induction maybe

If the power is even then we can write it as 2n = 2n-1+1

[reef trout]

yess thats what i am on about what would the induction be for this case

[alpine sable]

So p^(2n) = p^(2n-1) • p would not be a quasi-prime

And if the power is odd then p^(2n+1) needs to be written as p^(2m) • p^(2(n-m)+1)

By induction p^(2m) is not a quasi-prime

[reef trout]

oh why do we have to check if the power is odd or even

[alpine sable]

Because the theorem is

p^n is quasi-prime iff n is odd

[reef trout]

oh ok i didnt know about this

[alpine sable]

We're assuming this holds up to some number

[alpine sable]

I just came up with it and proved it

[reef trout]

oh impressive ! this is all new to me, my brain hurts

[alpine sable]

Oh

The same proof works if we assume m is a product of n primes

Not necessarily distinct

So looks like a number is quasi-prime iff it's a product of odd number of primes

Makes sense

[reef trout]

how do we write the induction for this like 2,3,5,7,8,11,13...q,q+1=q^(2(n-m)+1)is this right

[alpine sable]

For n = 1 we observed already any prime number is quasi-prime

Suppose that n>1 is such that any product of k<m primes is a quasi-prime if k is odd and not a quasi-prime if k is even

We'll prove that a product of n primes is a quasi-prime iff n is odd

That will complete the induction

[reef trout]

oh ok

[alpine sable]

Now we have to cases

a) if n is odd we need to prove any product of n primes is a quasi-prime

But if you write it as any product, then we need to split it into even and into odd amount of primes

The even amount of primes is not a quasi-prime by induction

[reef trout]

ohh right

[alpine sable]

So this number is a quasi-prime since it's not a product of two quasi-primes

Sothis case is done

b) n is even

Then we can write n as a sum of two odd numbers

For example n = 1+(n-1)

So it's a product of a prime and a product of n-1 primes

By induction each of those is a quasi-prime

So the product of n primes is not a quasi-prime

[reef trout]

oh ok i'll take all of this down

[alpine sable]

This completes induction

So we've shown that a product of n primes is a quasi-prime iff n is odd

[reef trout]

oh so 1000 is even so is not quasi prime

[alpine sable]

It's a product of even amount of primes, yes

[reef trout]

For n ≥ 2, the n
th quasi-prime qn is the smallest integer greater than qn−1 and not of
the form qiqj for some 1 ≤ i ≤ j ≤ n − 1.

i dont understand this statement

[alpine sable]

1000 is a product of 6 primes so not a quasi-prime

[alpine sable]

This is definition by recursion

Imagine you have a list of quasi-primes you've already checked

[reef trout]

oh ok i would be confused if i got this for an exam

[alpine sable]

Then the next quasi-prime is the smallest number which is not a product of any two numbers you already have on the list

Greater than all of the mumbers already checked

[reef trout]

ohh ok how do u know so much about quasi prime haha

[alpine sable]

I don't know anything about them

I didn't know until today anyway

[reef trout]

have this all taken down

[reef trout]

wait really thats impressive , i thought inorder to solve the question you have to know what is a quasi prime

[alpine sable]

I mean you written the definition of what it is

That was all I used

[reef trout]

wait could u paste the part i wrote

[alpine sable]

For n ≥ 2, the n
th quasi-prime qn is the smallest integer greater than qn−1 and not of
the form qiqj for some 1 ≤ i ≤ j ≤ n − 1.

This

[reef trout]

oh the part i didnt understand no wonder i didnt know what a quasi prime was

[reef trout]

but anyways thxs for the lesson, it was interesting

[alpine sable]

Np

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@reef trout Has your question been resolved?

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[reef trout]

.close

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[brave atlas]

Does anyone understand the basics of dividing GCF?

[brave atlas]

That’s what I got but I’m surely wrong

[hard patio]

what you gotta do here?

2b-8 = 2(b-4)

[lone heart]

@brave atlas Has your question been resolved?

[brave atlas]

How did you get that

[hard patio]

factor the common 2

then you can try to cancel things out

[hard patio]

Do you understand now?

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@brave atlas Has your question been resolved?

[brave atlas]

I guess I’ll read into it

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[chilly inlet]

yo, I need help with a concept

[chilly inlet]

can a complex root have only a real part?

like, if the roots of a polinomial are 3, 2+i, 2-i

does the polinomial have 3 complex roots, or just 2?

[wide raven]

see dude

[chilly inlet]

I mean, 3 is the same as 3 + 0i

so I would imagine that they are all complex roots

but I'm not sure

[gray ingot]

all real numbers are complex

[wide raven]

every real number is a complex number but the reverse is not true

but yeah we say 2 complex and one real

just to be more clear

[gray ingot]

BUT, people often use complex to mean purely complex

[chilly inlet]

that makes sense

[gray ingot]

yeah its weird

[chilly inlet]

So, even though 3 is complex, we don't call It a complex root

I see

[gray ingot]

mhm

[chilly inlet]

I mean, considering 3 is a root of a polinomial, of course

we just call It a real root, then?

[gray ingot]

yes

[chilly inlet]

Oh, I see

so, if a + bi is a complex root of a given polinomial

It is implied that b is different from 0?

[gray ingot]

mhm

[chilly inlet]

I mean, because if b is 0, then a + bi = a

so "a" is a real root

[gray ingot]

yes

[chilly inlet]

nice!

thank you

:)

question solved

[gray ingot]

.close

^^

[chilly inlet]

^^

.close

.close

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[alpine sable]

Solve x^2(x-1) >= 0

So far can see that x can be 1 and 0 that satisfies the 0 other than that i'm stuck

[buoyant kayak]

make a number line and use test points

[alpine sable]

I tried but book says x >= 1

if I plug in 0 its 0^2(0-1) >= 0

how is 0 not valid?

[buoyant kayak]

because 0 isn't a valid test point

0 is a point on your number line

[alpine sable]

how is it not valid?

[buoyant kayak]

So far can see that x can be 1 and 0 that satisfies the 0

if 0 satisfies the equation x^2(x-1)=0, why would you expect to get anything out of plugging in 0?

you just said that it satisfies the equation

plugging in zero will yield 0

you need to plug in points in between your x values that satisfy the equation

(and normally undefined x values as well, but your function is continuous everywhere)

[alpine sable]

But if 0 is valid, so is 0.0001 right? then x >= 0? no? @buoyant kayak

[buoyant kayak]

what

[alpine sable]

x^2(x-1) => 0 is all values x >= 0

[buoyant kayak]

no?

[alpine sable]

WHAT

but 1 is , how?

[buoyant kayak]

because it satisfies the inequality?

(greater than 1, not just 1)

[alpine sable]

greater than 0 also satisfies?

[buoyant kayak]

how

plug in 1/2

[alpine sable]

oh

BUT it works for 0 tho ????

Shouldnt it be x >= 1 U {0} ?

@buoyant kayak

[buoyant kayak]

ah yeah

it should be

[buoyant kayak]

just not this

[alpine sable]

Okay thank you I feel even more stupid now tbh math makes me less smart, insane @buoyant kayak

.close

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[pale reef]

helluu my ans is 501 (mod 728) could i check if this is correct?

[oak chasm]

@pale reef Sure, get the remainder when you divide 501 by 7. See if it's 4. And so on with the other congruences.

[pale reef]

ah ok great! thank u :)

[oak chasm]

No problem.

[pale reef]

.close

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[cyan basalt]

[1 - \frac{x}{1!} + \frac{x(x-1)}{2!} - . . . + (-1)^n \frac{x(x-1)* . . . *(x-n+1)}{n!} = 0]

[ocean seal]

Ausländer

[lone heart]

@cyan basalt Has your question been resolved?

[cyan basalt]

who has some ideas?

[keen pasture]

Seems like something being related to binomial coeffizients

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@cyan basalt Has your question been resolved?

[cyan basalt]

ok, im close

.close

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