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[cobalt bay]

something like this?

[light sonnet]

As long as you're not breaking the wires ,that's a legal thing I did

[cobalt bay]

how would u add the 3,4

[light sonnet]

In the green, R1 and R234 are in parallel

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[jade surge]

Hi, could someone help me factor this step by step? (MHF4U — Advanced Functions)

[jade surge]

I've tried using the rational root theorem but I think I'm doing something wrong when factoring the quadratic function

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[stark maple]

im confused if i did this correctly

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[dreamy eagle]

im stuck, could someone help me finish this? (gaussian elimination)

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[somber wharf]

A bowl contains four salted caramel Hershey bars, five cookie and cream Hershey bars, and six milk chocolate Hershey bars. If candy bars are selected one by one in random order, what is the probability that at least two candy bars must be selected to obtain one that is milk chocolate?

So is it asking for the first two draws are not milk chococlate right

[somber wharf]

so my first is 4/15 and since 11 candy left so 5/11

and I multiply the two

is that right?

,calc (4/15)*(5/11)

[glossy valve]

Result:

0.12121212121212

[somber wharf]

is it right?

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[somber wharf]

<@&286206848099549185>

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[inland dirge]

hello, im stuck on this question

[inland dirge]

i tried doing f(x+h) - f(x)/h but in the end I couldnt progress because I couldnt factor the h out to cancel the numerator h

nevermind I found it 👍

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[nova wasp]

Would iii be all the positive integers excluding 3 and including 0?

[proper hawk]

you mean ii?

[nova wasp]

yeah sorry

I meant ii

[proper hawk]

then yes

[nova wasp]

And iii is the all the natural numbers excluding 2 and 3?

[proper hawk]

umm

I don't think so

notice that $B-A^C$ is disjoint from $A^C$

[glossy valve]

WhereWolf(ping if needed)

[nova wasp]

and why is that?

[proper hawk]

every element of A^C is removed from B

so B-A^C and A^C have no common element

[nova wasp]

so A^c is all the natural numbers without 2, B is the set that contains 2 and 3

[proper hawk]

yeah

[nova wasp]

you remove all the natural numbers without 2 from the set that contains 2 and 3, you just have a set that contains 2?

[proper hawk]

yep

and they are disjoint

[nova wasp]

and you are trying to find the intersection of the set that contains 2 and all the natural numbers without 2

the two sets have nothing in common

so you have an empty set

[proper hawk]

exactly

[nova wasp]

oh okay thank you 🙂

[proper hawk]

no problem

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[spring marsh]

Is this correct?

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[pseudo cape]

@short bridge Are you here?
We had that for "Determine the probability that when throwing two dice repeatedly, a 5 occurs before a 7.", it's P = 1/9 + (1 - 1/9 - 1/6) P

[pseudo cape]

Why is that?

Why is the P at the end

There are 4 partitions of 5, 6 of 7.

So P(5) = 1/9 and P(7) = 1/6, yes.

But why is P = P(5) + (1 - P(5) - P(7)) * P?

[short bridge]

Yes

[short bridge]

Okay. Try to think of it in this way.

We are gonna try to divide our possibilities for occurrence of 5 before 7 in two broad categories. Which are -
It happens on the first roll,
OR
It happens on some later roll(not the first one).

I'm gonna call "occurrence of 5 before 7" the event X.
Now, P(X) = P(It happens on the first roll) + P(It happens on some later roll)

P(It happens on the first roll) = P(5) = 1/9

P(It happens on some later roll) = P(It doesn't happen on the first roll) * P( It happens on the any other roll except the first)

Now, think about what this "P( It happens on the any other roll except the first)" represent. Basically, we have rolled the dice once and guaranteed that it didn't roll 5 or 7.
Now, we have to make the second roll.
Does a previous role of a dice affect the next role in any way?

After this, ping me so i can continue.

[pseudo cape]

No, it doesn't

All probabilities remain the same

[short bridge]

Yes. So, for a second forget the first roll, since it doesn't affect our next roll at all.

What are we trying to find?

[pseudo cape]

(Assuming roll means throwing both dice)

[short bridge]

We are still finding the exact same thing - Probability that a 5 rolls before a 7.

[pseudo cape]

That we get 5 before 7

[short bridge]

Yes

[pseudo cape]

yes

[short bridge]

Which is just P.

So, we put it as P.

[pseudo cape]

oh

[pseudo cape]

We are allowed to multiply because they are independant, right?

[short bridge]

Yes.

[pseudo cape]

So to denote "Neither P nor 5 and then 5 before 7", we can multiply accordingly

Thanks!

[pseudo cape]

The answer in the MSE post just right away said it's 1/9/(1/9 + 1/6) or 4/(4 + 6) (see https://math.stackexchange.com/a/1828299), so I don't think they used a recursive way like this one though. Do you know how they got that right away?

[short bridge]

They used that only. They just generalised the formula.

Let's try to do that.

[pseudo cape]

Oh, wait

Yeah, get P to the LHS, then factor out P, there will be a +1 and -1, those cancel out, we can divide and we'll have

P(5)/(P(5) + P(7))

[short bridge]

P = P(5) + ( 1- p(5) - P(7))*P
P(1 - (1- p(5) - P(7))) = P(5)
P = P(5)/(p(5) + p(7))

[pseudo cape]

Yeah, thanks!

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[solar garden]

what is the integral of cosec2x

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@full forum resolved

close

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[crimson terrace]

Would this be correct?

[tame bobcat]

to me it looks like you need to read the last sentence again

the one before "do not simplify ..."

think about that

[crimson terrace]

I didn’t simplify

@tame bobcat

[tame bobcat]

ok lol read what I wrote again

I meant the sentence before that

[crimson terrace]

Oh

Would this be right

[cyan kelp]

you made the same mistake as in your last channel

[crimson terrace]

Joe

How

[cyan kelp]

last time, it said to do some operations, then "divide <some number> by what you have" and you divided what you have by <some number>. you did the same thing again here, before you fixed it

it says "divide j by what you have" and so you should divide j by 3 * 3 + k

[crimson terrace]

So I’m right now?

[cyan kelp]

so you'd write j / (3 * 3 + k)

yeah

just gotta read more thorougly

[crimson terrace]

K

Am I right though?

[cyan kelp]

so i'm right now?
yeah
K
am i right though
just gotta read more thoroughly

[crimson terrace]

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[fair walrus]

Don't understand the equation I need to do

I understand I need to find KLM but don't know how to

I so far did 66 - 54 = 12 but this was incorrect

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[wild vapor]

assume f(x1) = f(x2) => prove that necessarily x1 = x2

this makes it injective

lemme think about it

alright

my original hint works

try to work from f(x1) = f(x2) with algebra

[spice orchid]

The function is $\frac{x}{1+|x|}$?

[glossy valve]

ΣΑCu

[spice orchid]

And what is it mapping between

Okay that's important to include bc obv it's not a bijection from R to R

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[grizzled lion]

I am a bit confused

[grizzled lion]

I factored the denominator and got x^2(x - 5)

So i thought i would need two fractions in the decomposed form

one for x^2 and the other for x-5

oh wait i see now

We have an x^2 and because of that we need another fraction for x^1

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[graceful merlin]

Is
X^2-x even function ?

[graceful merlin]

I wrote
F(-x)=(-x^2)-(x)
= even function

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[wild vapor]

you sure that's correct?

[wild vapor]

you cannot use this to prove f is injective

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[wheat garden]

$\lim_{x\to 0} \frac{\tan^3(2x)}{x^3}$

[glossy valve]

Jash

[narrow pawn]

Apply l’hopital rule

[wheat garden]

i would have to do it 3 times

[hollow sable]

You can cube root it and then use l’hoptal

[wheat garden]

with multiple product rules

[hollow sable]

And then cube it again

[wheat garden]

oooh so $\left(\lim_{x\to 0} \frac{\tan{2x}}{x}\right)^3$

[glossy valve]

Jash

[hollow sable]

Yes

[wheat garden]

does tan(ax)/(bx) have the same property as sin(ax)/(bx)=a/b?

[hollow sable]

Yeah you can prove it with l’hoptials

When x—>0

[wheat garden]

ty

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[pine pond]

when i do the arc tan of 30/360 i get 5 not 4.76

[pine pond]

also i thought j was cos and i was sin shouldnt it be arc tan 360/30 ?

nvm i fixed my calculator

but the other point stands in regards to i and j

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[gaunt flume]

I have a doubt, if there are 3 green balls and 2 red balls. And we are picking 2 of them with replacement.

then
Probability(green ball and red ball picked) = 2 * (3/5) * (2/5) = 12/25
Probability( 2 green ball ) = (3/5) * (3/5) = 9/25

In the second case we didn't have to multiply by 2. Can anyone explain why?

[torn jolt]

im not sure why youd multiply by 2

essentially in a probability equation, picking a green ball and a red ball:

3/5 [because there are 3 green balls] * 2/5 [because there are 2 red balls]

[gaunt flume]

doesn't the order matter? we are picking green first and red second here.

[torn jolt]

same thing with 2 green balls
3/5 [because there are 3 green balls] *
3/5 [because there are 3 green balls]

[torn jolt]

it only matters if its with replacement

because then the previous event determines the next event

example:

pick a green ball and a red ball with replacement

3/5 [because there are 3 green balls in a set of 5 balls]** * 2/4 **[because there are 2 red balls in a set of 4 balls. you've taken a ball out]

[gaunt flume]

Lets take a smaller sample space 2G, 1R

then Sample space = {GG,GG,GR,GG,GG,GR,RG,RG,RR}

here if we pick 2 of them:
P(one red and one green) = 2*(2/3)(1/3)
P(both green) = (2/3)(2/3)

[torn jolt]

2 of them with or without replacement?

[gaunt flume]

with

[torn jolt]

cool

[gaunt flume]

sorry with

[torn jolt]

wait

ok

[torn jolt]

im not sure why youre multiplying by 2 here

[gaunt flume]

It's with replacement. I have edited the question.

[torn jolt]

ohhhhhhh my apologies

[torn jolt]

but the correct answers would be:
1/3 x 2/3

and:
2/3 x 2/3 (this is right)

as far as i know, multiplying by 2 will result in an incorrect answer

[gaunt flume]

but if we check the sample space, {GG,GG,GR,GG,GG,GR,RG,RG,RR}

there are 4 occurrence out of 9 (gr,gr,rg,rg)
but probability gives only 2 of them

[torn jolt]

hmm

[gaunt flume]

I have a doubt, if there are 3 green balls and 2 red balls. And we are picking 2 of them with replacement.

then
Probability(green ball and red ball picked) = 2 * (3/5) * (2/5) = 12/25
Probability( 2 green ball ) = (3/5) * (3/5) = 9/25

In the second case we didn't have to multiply by 2. Can anyone explain why?

Lets take a smaller sample space 2G, 1R

then Sample space = {GG,GG,GR,GG,GG,GR,RG,RG,RR}

here if we pick 2 of them:
P(one red and one green) = 2(2/3)(1/3)
P(both green) = (2/3)*(2/3)

<@&286206848099549185>

[torn jolt]

in a mutually exclusive event, the outcomes cannot occur simultaneously

therefore RG and RG, GR and GR are possible

which means its 2/9*2/9

so yea your correct

sorry i overlooked that

[split swift]

in the first question, we multiply by 2 to account for the possible permutations of choosing reds and greens (RG or GR), but in the second, the greens are indistinguishable, so we don't need to permute them

more specifically,
P(green ball and red ball) = P(green ball, followed by red ball) + P(red ball, followed by green ball)
= (3/5) * (2/5) + (2/5) * (3/5)
= 2 * (3/5) * (2/5)

but P(green ball) = P(green ball, followed by green ball)
= (3/5) * (3/5)

[torn jolt]

^^

[gaunt flume]

🫂 thanks guys

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[torn jolt]

YI

[torn jolt]

TOO SLOW

[cloud snow]

bruh

[calm tinsel]

DAMMIT

[torn jolt]

Can this situation be solved exclusively with vectors?

[calm tinsel]

oh... this is physics

[calm tinsel]

technically yes

[torn jolt]

how

[calm tinsel]

I sadly cannot answer that

because that type of physics is hard

because its a mathematical application

and I never done it with vectors

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@torn jolt Has your question been resolved?

[torn jolt]

I'm trying to fill out the vectors but i'm not sure what the components of V would be

lets just say

its

uuuuh

no

I quit

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[grand spire]

hey

[soft axle]

!15min

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[mental wolf]

need help solving this i can show an attempt if needed

[mental wolf]

<@&286206848099549185>

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[tawdry horizon]

id start with sketching the graph

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[proven chasm]

my question is #9

[proven chasm]

i’m just making sure my work is correct, idk how to “de assemble” composite functions

or would f(x) and g(x) be swapped?

<@&286206848099549185>

[torn jolt]

If f(g(x)) = x, and g(f(x)) = x, then the two functions are the inverse of one another

I think that's the rule you're looking for?

[proven chasm]

i’m not sure

i think im supposed to decompose the decomposition function

is that how u do it?

[torn jolt]

Question 10 right?

[proven chasm]

no it’s 9

[torn jolt]

Oh one sec my bad

[proven chasm]

phew i was scared for a second lol

i thought i was super lost

[torn jolt]

Could you type what H(x) is? I can't see it very well

[proven chasm]

H(x) = (x^2+1)^50

[torn jolt]

I think the best way to decompose this kind of stuff is to work from inner to outer

[proven chasm]

the inner function is G right?

[torn jolt]

We are given that H(x) = f ° g

Yes

[proven chasm]

and the outter is f

[torn jolt]

Yeah you got it then

g(x) = x^2 + 1, f(x) = x^50

[proven chasm]

i think i mixed up my answers

yeah i think that’s what i got when i redid it

let me check

yeah that’s what i got

that’s not too bad then

i guess i can check my answer by just plugging it in right?

[torn jolt]

Yep

[proven chasm]

that was actually relatively simple which probably made me confused, but ty for the help brother

[torn jolt]

Ofc, no problem

[proven chasm]

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[valid ginkgo]

The figure shows the graph of a function and its tangent at the point P.
What value does the derivative of the function have at the point P?
Do I take two points of the tangent and use delta x and y?

[proper hawk]

yea

[valid ginkgo]

Is that it? The answer is -1/4?

[proper hawk]

how

[valid ginkgo]

How do I get -1/4 or what do you mean?

[proper hawk]

hou'd you get -1/4

how

[valid ginkgo]

Woops

[proper hawk]

lol

[valid ginkgo]

-1/2

[proper hawk]

yep

correct

[valid ginkgo]

So that's my final answer?

[proper hawk]

yes

[valid ginkgo]

alr thank you

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[crude lake]

Find K so that the terms K-3, K+1, and 4k-2 form a geometric sequence? What formula should I use?

[torn jolt]

are those terms consecutive?

if yes then use the def of a geometric sequence where r will be the same for each of them

[crude lake]

yes they are consecutive

[torn jolt]

t2/t1=t3/t2

[crude lake]

thank you

[torn jolt]

np

[crude lake]

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[silver valve]

if two lines have the same y coordinate but different slopes that go parallel, whats the point of intercection?

[reef dust]

how does it have a different slope but goes parallel

and have the same y coordinate

[onyx glen]

damn you got the whole gang confused

[silver valve]

Gang ion know

Here lemme like put in a Cartesian plane

[solar flare]

[silver valve]

What’s the point of intersection

Is it 3

Or no poi

[neon basin]

Well that doesn't look much like two parallel lines

[silver valve]

Oh well yk what I mean gng

What’s the point of interrsdction 😭

No poi or is it 3

[neon basin]

I mean you literally have (0, 3) marked

[silver valve]

So is that the poi

[neon basin]

Do you know what an intersection of two lines is

[silver valve]

When it intersects the lines

Yuh

When it crossed

Crosse

Crosses

Goes over

Laps over

[neon basin]

Yes ...

[silver valve]

SO IS IT 0,3 THE POI

[neon basin]

And where does one line intersect, or cross, the other

[silver valve]

idkkkk

😭😭😭

[reef dust]

also the lines arent parallel what r u on

[neon basin]

Though does aren't lines anyway*

[silver valve]

oh well same thing

[brittle steeple]

Really these aren't lines but rays

[silver valve]

anyways

[brittle steeple]

Ye

[silver valve]

What’s the point of intersection

wait

ohh I get it now

so the poi is 0,3

oo

[neon basin]

Yes ...

[silver valve]

TYSM

[neon basin]

I guess ...

[silver valve]

hope u have a god blessed rest of ur day

YURR

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[storm apex]

i cannot quite understand what is being done here

[storm apex]

we have a periodic function, sin, and we are interested in only when it is zero

could someone explain what 'k' really is?

[left bone]

k is any arbitrary integer

[storm apex]

ok, but how is it related here? because sin can only be zero at 0 and pi, right?

[glacial pasture]

if the argument of sin() is any integer multiple of pi then sin is 0

[glacial pasture]

sin is 0 at ...,-2pi,-pi,0,pi,2pi,3pi,....

[storm apex]

okay, i see. I didn't disclose that part of the question but the range used here is -pi to pi

the interval sorry

[glacial pasture]

then you take any solutions for t from 2t+pi/4=kpi s.t. t is in that interval

and k is just any integer

[storm apex]

to me then it seems like the first equation 2t + pi/4 = k*pi is more useful for solving the problem. What is the purpose of isolating t

[glacial pasture]

because t is what you want to find

[glacial pasture]

here

[storm apex]

yeah, i get that, but we want to find when 2t+pi/4 is a multiple of pi

i don't question you're right but

i can only imagine how to begin to solve that with the first equation not the last

you say k is any integer, should we ignore it?

or should we treat both t and k as input parameters and try make them match basically?

[glacial pasture]

theres two options really

you can find the range of values that 2t+pi/4 can be,
in this case this range is -7pi/4<=2t+pi/4<=9pi/4
in which case k can only be -1,0,1 or 2

or you can rearrange to get t=kpi/2 - pi/8
and you know -pi<=kpi/2 -pi/8 <=pi

then find k values that satisfy that

which will give k=-1,0,1 or 2

[storm apex]

what you're saying is that RHS is within interval of -pi to pi

[glacial pasture]

yes because t is

and the rhs = t

[storm apex]

so do we ever even apply, knowing when sin is zero here?

[glacial pasture]

the whole thing is based on that

[glacial pasture]

thats where this equation comes from

everything else comes after

sin(LHS)=0 if the LHS is an integer multiple of pi

[storm apex]

so how do we start figuring out the t values, i still can't see it

[glacial pasture]

once you figure these possible values of k

[glacial pasture]

you can directly plug them into this

[storm apex]

and why not for example k = 3?

[glacial pasture]

k=3 is outside the range

3pi=12pi/4

upper limit of 2t+pi/4 is 9pi/4

[glacial pasture]

if you plugged k=3 into this, that would be apparent that this interval doesnt hold

[storm apex]

i am now struggling to see how we found this interval

"in this case this range is -pi/4 <= 2t+pi/4 <= 9pi/4"

[glacial pasture]

-pi<=t<=pi, yeah?

[storm apex]

yes

[glacial pasture]

so -2pi<=2t<=2pi

then -2pi+pi/4<=2t+pi/4<=2p+pi/4

so -7pi/4<=2t+pi/4<=9pi/4

[storm apex]

so we plug lower and upper bound of interval into what's inside the sin function

into t of that, that is

[glacial pasture]

you dont plug the upper and lower bounds in
however note that 2t+pi/4=kpi
so then -7pi/4<=kpi<=9pi/4

which can be made into -7/4<=k<=9/4

since k is an integer just find those that lie in there

[glacial pasture]

then you can just plug the k values into this to find the t values

[storm apex]

i am still not seeing this jump from 2t+pi/4 to -7pi/4

[glacial pasture]

wdym?

[storm apex]

i understand 2t+pi/4 = kpi

and thus if we modify LHS we can modify RHS accordingly

and thus find k

[glacial pasture]

-7pi/4 is from this working

[glacial pasture]

then to this

[storm apex]

2(-2pi)+pi/4 ?

[glacial pasture]

just -2pi+pi/4

[glacial pasture]

here i multiply by 2, then the next line is +pi/4

then i just simplify

[storm apex]

so -2pi+pi/4

that's still not -7pi/4

[glacial pasture]

it is

-2pi+pi/4 = -8pi/4+pi/4=-7pi/4

[storm apex]

oh wait

obv OOO

yeah okay i agree

so t is -pi, 0 or pi, as defined by our interval. 2t is then -2pi, 0 or 2pi. this allows us to create the range of what 2t + pi/4 can be. once we know that range, we can find out what k can be, accordingly. Then on the second equation, we can simply plug those k values in. This is how i understood it so far

BUT

[glacial pasture]

no

t is not -pi, 0, or pi

[storm apex]

okay yeah

that confused me too

[glacial pasture]

t lies in the interval -pi<=t<=pi

[glacial pasture]

these results all fall in there

[storm apex]

okay, so it lies in the interval. It CAN be -pi, 0 and pi

[glacial pasture]

it can yes

they are possible values of t

[storm apex]

and k is integer values

[glacial pasture]

yes

[storm apex]

so it cannot be anything other than the integers of its range

[glacial pasture]

what do you mean by that?

[storm apex]

which here is -1, 0, 1 and 2

[glacial pasture]

if you mean this by range, then yeah

[storm apex]

as in, for t, if we defined it as an integer, it would've been -3, -2, -1, 0, 1, 2, 3 as possible values

[glacial pasture]

under that condition, yes

[storm apex]

it's okay i'm just verbalizing my thinking it's not relevant

so this is the purpose of k basically

thus we can, for isolated t equation, simply plug in possible k values

and that will give us possible t

so i see k is truly nothing but an integer but it is still crucial to figure out the answer

[glacial pasture]

yes k is just an integer, and its range is restricted by the range of t

[storm apex]

okay, so i can understand now for sure how to find possible k values for minimum and maximum value of our interval (-pi and pi)

but what about everything in between?

$t + \frac{pi}{2} = pi * k$

[glossy valve]

marty

[storm apex]

this is another question i am currently solving

i put t to be -pi and pi

and get k values -1/2 and 3/2

[glacial pasture]

you need to stop focussing on the end bounds of -pi and pi

they arent helpful except for saying what results to ignore

[storm apex]

okay, but then i'm back to zero, how can i figure out k values

[glacial pasture]

theres no reason to randomly plug them in

[glacial pasture]

from this line of reasoning down

[glacial pasture]

you end up here

and can then pull out your valid k values

-1,0,1,2

[storm apex]

okay so we never leave the interval basically

[glacial pasture]

you cant, the interval is how t is defined

everything else works from that basis

[storm apex]

$-pi+\frac{pi}{2}<=t+\frac{pi}{2}<=pi+\frac{pi}{2}$

[glossy valve]

marty

[storm apex]

before simplification

[glacial pasture]

is that for this

[storm apex]

yep

[glacial pasture]

seems fine to me

$-\frac{\pi}{2}<=k\pi<=\frac{3\pi}{2}$

[glossy valve]

AℤØ

[storm apex]

aha

[glacial pasture]

then you could generally divide by pi

[storm apex]

so -1/2 <= k <= 3/2

[glacial pasture]

yeah

[storm apex]

and then

we take integers within this range

[glacial pasture]

yup

[storm apex]

0, 1, 2, 3, 4

[glacial pasture]

no

2,3 and 4 are not in that range

[storm apex]

oh

wait

[glacial pasture]

3/2=1.5

[storm apex]

i was multiplying with pi still

0, 1

okay so we work with intervals until we get the k interval and from there it's fairly easy to get t values

[glacial pasture]

indeed

[storm apex]

so this in general was the method of finding possible t values. Sometimes this is sufficient for only x, such as finding y intersections, but when we want to find origin intersections, y(t) also needs to be zero

we already have possible t values for x, what is the easiest way to find a match for y ?

[glacial pasture]

these t values are the ones such that y=0

[glacial pasture]

at least from this question anyway

[storm apex]

can we simply try to plug our t values into the sin function, or should we also isolate for t for y(t)

lemme show the whole one sec

i didn't post this because i wanted to isolate the problems

oh wait

wrong problem

let's not actually solve it

but just in general

[glacial pasture]

this one does require x(t)=y(t)=0

so youd find the solutions to them individually, then see which t values satisfy both

[storm apex]

ah okay

so we couldn't just simply plug t values found for x into y(t)

and interpret that somehow

[glacial pasture]

i suppose you could actually

see which x solutions make y(t)=0

that would work

or vice versa

[storm apex]

okay so whatever makes (pi/16 - t/8) = 0 in this example

one of the t values from x should match

[glacial pasture]

its not pi/16 -t/8=0, pi/16-t/8=kpi for some integer k

[glacial pasture]

one or more

or perhaps none

its possible it never passes through the origin

[storm apex]

yeah, i understand

[storm apex]

okay, so this is important distinction

so here we are interested in whenever the sine function is zero, which can be any multiple of pi

[glacial pasture]

yeah, although what you said would be a solution, it wouldnt be all possible solutions

[storm apex]

as opposed to the higher level of wanting y(t) to be zero, which only happens with 0, obviously

[storm apex]

yeah exactly

extremely helpful

i was very very stuck with this

thank you

[glacial pasture]

no worries

[storm apex]

i will just keep this open until i solve my problem

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[cursive herald]

I want to calculate the amount of an action if it starts at 50 and increases by 10% every day and that the sum of it is 12,000 times after five days

[cursive herald]

I don't know how though

like I haven't leaned exponentials so I don't know how to write an equasion

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[cursive herald]

<@&286206848099549185>

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@cursive herald Has your question been resolved?

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[blissful nova]

in your context probably but generally not at all

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@rigid bough Has your question been resolved?

[wild vapor]

because you have to first prove that there exists an inverse f-1

and such an inverse exists only if f is injective, which is part of what you're trying to prove

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✅

[wild vapor]

it's x/(1+|x|)?

just examine what happens if f(x1) = f(x2)

first of all we notice that f(x) = 0 only for x=0.

then, for all other x's you can see (1 + |x1|)/x1 = (1+ |x2|)/x2

so now you can take three cases: either x1 and x2 are positive. in this case you have 1/x1 + 1 = 1/x2 +1 . so 1/x1 = 1/x2 and x1 = x2.

similarly for when x1 and x2 are both negative

and lastly, from the original equation you can see that x1 and x2 cannot possibly have a different sign

so that covers all cases. if f(x1) = f(x2), x1=x2

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@rigid bough Has your question been resolved?

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@balmy igloo Has your question been resolved?

[soft axle]

<@&268886789983436800>

[slender onyx]

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[hallow moth]

hi does anyone know how to solve this? it is polynomial function. thank you !!

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@hallow moth Has your question been resolved?

[hallow moth]

<@&286206848099549185>

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[hallow moth]

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[prisma remnant]

Im not sure where to start with these. In the case of a, I know that if it was just δ(t) then the definite integral would be 1 but I don't know if that helps me here

[brisk imp]

$$\sin(t) \times \delta(t) = ?$$

[glossy valve]

Luna

[torn jolt]

what

[brisk imp]

$$\sin(t) \times \delta(t) = \sin(0) \times \delta(t) = 0$$

[glossy valve]

Luna

[brisk imp]

So in (a), you are integrating 0

[prisma remnant]

How do you know it becomes sin(0)?

[brisk imp]

it is a property

😃

[torn jolt]

The product of any function with the Dirac delta function is

[
f(t) \cdot \delta(t) = f(0) \cdot \delta(t)
]

[glossy valve]

luke

[prisma remnant]

Oh alright. Is it something similar with b where I am supposed to use a specific property before integrating?

Or do I just change it to the integral from 0 to t of sin(T) since its a step function

[brisk imp]

Well here, you have two cases

When t<0 and when t>=0

if t<0, the integral is just equal to 0, because u(t)=0 for negative values of t

[prisma remnant]

What would the case be if t>=0?

[brisk imp]

if t>=0, yes, the integral simplifies to from 0 to t of sin(T)

Because u(t)=1

For positive values of t

[prisma remnant]

Thank you

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[light sonnet]

You should ask your teacher

Because that's not it

Yes and you should ask your teacher on how that augmented got there

Because that's not it

We aren't your teacher to know what their thought process was

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[elfin lynx]

Hey guys, could someone help me out with this?

[elfin lynx]

Suppose the foci of an ellipse are (-2x, 0 ) and (2,0) and the sum of the distances from point P(x,y) of the ellipse to the two foci, denoted by F1P + F2P = 8. Use the definition of an ellipse to find the equation of the ellipse.

Here is my working:

$$\sqrt{\left(x--2\right)\ +\ \left(y-0\right)^2}+\sqrt{\left(x-2\right)^2-\left(y-0\right)^2}=8$$

[glossy valve]

Lex1729

[elfin lynx]

$$\sqrt{\left(x--2\right)\ +\ \left(y-0\right)^2}=8-\sqrt{\left(x-2\right)^2-\left(y-0\right)^2}$$

[glossy valve]

Lex1729

[elfin lynx]

$$\left(\sqrt{\left(x--2\right)\ +\ \left(y-0\right)^2}\right)^2=\left(8-\sqrt{\left(x-2\right)^2-\left(y-0\right)^2}\right)^2$$

[glossy valve]

Lex1729

[elfin lynx]

$$\left(x+2\right)^2+y^2=64-16\sqrt{\left(x-2\right)^2+y^2}+\left(x-2\right)^2+y^2$$

[glossy valve]

Lex1729

[elfin lynx]

.close

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[torn jolt]

Inverting functions. I think I’m on the right path and I sort of get the idea but for some reason it’s not clicking. Can someone pls help me with this?

[teal nimbus]

Yes you've the idea

And the beginning is correct

What would you do after that?

[torn jolt]

My apologies for the late response

I’m not quite sure I’m thinking I need to divide to get rid of the two

But not sure

[teal nimbus]

You have to do something so that y is at the numerator

to isolate it

[torn jolt]

Right. So maybe I should multiply it by 3/2?

[teal nimbus]

Actually, let's do it in several steps: first eliminate the 3

[torn jolt]

How do I do that again with a problem like this?

[teal nimbus]

Just, when you want to get rid of 3, what would be the operation to do

Here $\dfrac3{y+2}=3\times\dfrac1{y+2}$.

[glossy valve]

Statufi

[torn jolt]

I’d multiply it by the denominator

I think. I haven’t done this stuff in a while

?

[teal nimbus]

I mean you can multiply by the denominator by it won't eliminate the 3

To get rid of a multiplication (here $\times 3$), you have to divide (here /3$=\dfrac13$).

[glossy valve]

Statufi

[torn jolt]

Hmm I need to divide it by 1/3? Isn’t that 9?

[teal nimbus]

No, 1/3=0.333333333333...

It's 3²=9

[torn jolt]

Also this problem is not for graphing

[teal nimbus]

And no, you have to multiply by 1/3 OR divide by 3

Which is the same operation

[torn jolt]

Sorry I meant 3/ 0.3333_ is 9

Gotcha

[teal nimbus]

No

[glossy valve]

Statufi

[teal nimbus]

And that's not giving what you wrote

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

Oh. I see. But how do I progress from there?

[gritty rose]

Multiply by something else to clear denominators

[gritty rose]

^

[torn jolt]

Should I multiply by y+2?

[full forum]

@torn jolt Has your question been resolved?

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[torn jolt]

.close

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[barren iris]

Why isn't the asnwer to b)
Ax + B / x^2 + x+ 4

[formal scarab]

work backwards from the answer you got

oh do you mean (Ax+B)/(x^2+x+4)

[barren iris]

yes

[formal scarab]

you can factorise x2+x+4

[barren iris]

using quadratic formula?

[formal scarab]

sure

or completing the square

[winged grotto]

Can someone please help??
Find the average (vertical) height of the shaded area in Figure 1 below.Ca

[formal scarab]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[winged grotto]

oh sorry

[barren iris]

Before I do that though, does this fraction not fail on the first assumption ?

[formal scarab]

true

[barren iris]

Why isn't doing long division enough then?

[formal scarab]

which means polynomial division first

[barren iris]

indeed

[formal scarab]

because you'll still be left with a linear/(x2+x+4)?

which needs further decomposition

[barren iris]

I got:
1 + (-x-4) / (x^2+x+4)

Ah

sigh

I suppose I need to read more about this method

(my lectures are useless lol)

[formal scarab]

practice is good

[barren iris]

Yes

but he didn't explain the rule very well

[formal scarab]

questions on youtube, etc. as well

[barren iris]

yes yes

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[tawdry lotus]

How can I find the domain of an inverse trigonometric function without graphing or inputting known values? I have arccos(2x/(1+x)) and it has been a weird time to get the domain algebrically.

[full forum]

@tawdry lotus Has your question been resolved?

[tawdry lotus]

<@&286206848099549185>

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@tawdry lotus Has your question been resolved?

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[restive roost]

trying to do linear congruences, would love a step by step workthrough with someone

[restive roost]

watched countless videos and examples

I just don't get it

for example

6x ≡ 15 (mod 21)

gcd(6,21) = 3

3 | 15

6x + 21y = 15

now i can divide through the entire congruence by the gcd

2x = 5 (mod 7)

now what though

[stark ruin]

you can straight up say x is 6 from here

for 1 solution

[restive roost]

i don't wanna just use inspection or tricks

like

[stark ruin]

6+7k

[restive roost]

I wanna find it step by step and understand how

[stark ruin]

once u simplify it to that point right

[restive roost]

once i've reached 2x = 5 (mod 7) whats my next step?

[stark ruin]

theres no 'method' its only u get a specific solution then add 7k

[restive roost]

what

[stark ruin]

because its a diophantine equation

the solution to that is 6+7k

[restive roost]

so from 2x = 5 (mod 7)

is it just trial and error and guessing values?

[stark ruin]

you have to find just 1 value

just like for diophantines

[restive roost]

how is that any different from 6x = 15 (mod 21) and just finding one value

idk what a diophantine is :/

[stark ruin]

aah

the equation you wrote

6x +15y=21

rhis is a diophantine

diophantine eq is a eq whose roots are integers

theres a general solution to this

or that you need any one solution

[restive roost]

oki

[stark ruin]

to form a general solution

[restive roost]

so wait whats the point of dividing by gcd

[stark ruin]

youre doing that to simplify

its harder to think

a solution to

[restive roost]

the only reason is to make it as easy as possible?

[stark ruin]

that

yes

[restive roost]

oki so

we have 2x + 5y = 7

or

2x ≡ 7 (mod 5)

I do understand how to get there

and then i just inspect / think of one solution?

[stark ruin]

yes

you have to only get one solution

for general solution to pop up

[restive roost]

so a soln i can think of here is

x = 6

[stark ruin]

yes

[restive roost]

how / why can i form a general solution from this?

[stark ruin]

wait 5 mod 7 or 7 mod 5?

[stark ruin]

this is sus

[restive roost]

7 mod 5 no?

[stark ruin]

7 mod 5 is just 2 mod 5

whats the original ques

[restive roost]

6x ≡ 15 (mod 21)

[stark ruin]

2x=5 mod 7

[restive roost]

yes you;re right oops sirrt

yes, so x = 6 is a soln

[stark ruin]

write this in a diophantine form

[restive roost]

2x + 7y = 5

[stark ruin]

meh any works

now for a diohpantine

once u get a solution

6,-1 here

the one you told

[restive roost]

sure ye

[stark ruin]

let 6 be a and -1 be b

to a you add the coefficient of y and to b you subtract coefficient of x

to get a general solutions

so your answer becomes

x = 6+7k y=-1-7k

[restive roost]

can you explain why?

[stark ruin]

i cant really explain it clearly

but it comes from modulo

[restive roost]

wouldn't that also mean there are infinite solutions then?

[stark ruin]

because

yes there are

hence the general solution

say

x= 2 mod 7

if instead of 2

i have any number

of the form

[restive roost]

but it says the total number of distinc solutions are = gcd(a,m)?

[stark ruin]

7k+2

they said its related

also notice how they are adding

repeatedly

means its infinite

[restive roost]

hm

im not really following

it definitely says theres only gcd(a,m) distinct solutions

[stark ruin]

no

thats a different thing

[restive roost]

AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

[stark ruin]

its saying solution to x mod m to the congruence ax = c mod m

is

dependent on d

illl let someone else explain my own theory part isnt that good to explain 💀

i can just do question

[restive roost]

tyy

i appreciate your help so far 🙂

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[wicked zephyr]

Can somebody teach me ?

[pale jacinth]

Calculate the probability of there being 0, 1, 2, 3 smartphones present in the selected group plot it

[wicked zephyr]

I did this so far

[pale jacinth]

So if you keep going with that tree, what do you think the probability of all 3 having a smartphone is

Although, I believe the smartphones aren't replaced so actually the probability differs each time

You are correct with the first branch

Just look at your 2nd one

[wicked zephyr]

20 people doesnt hv a phone ?

[pale jacinth]

No you're second branch, if 80 people have a phone the first time, then only 79 people will the 2nd time

And actually looking at you're probability you're chosing out of the 100 so the first probability would be 80/100

[wicked zephyr]

ohh i see

[pale jacinth]

Not 1/80

[wicked zephyr]

doesnt it mean in 100 people inside has 80 ppl hv phone 20 people does not hv phone ?

[pale jacinth]

yes

So the chances of picking someone with a phone is 80/100

[wicked zephyr]

the second branch is 80/99 people who with phone ?

[pale jacinth]

no

if you take someone with a phone out, then you have 79 with a phone left

So 79/99

[wicked zephyr]

when X = 3 the probability is 80/100* x 79/99 x 78/98 as well ?

[pale jacinth]

yes

[wicked zephyr]

thats incorrect the ans given is 64/125

[pale jacinth]

for what? Probability of 3 phones?

[wicked zephyr]

yea when P(X=3)

[pale jacinth]

That works assuming the phones get replaced (they put the phone back after they take it)

Is there more context to the question cause that doesn't seem intuitive

They've done (80/100)^3

[wicked zephyr]

why it doesnt decrease smtg ?

[pale jacinth]

Either the answer is wrong or there's more info

[wicked zephyr]

alright

[pale jacinth]

If this is a hw you can ask your teacher abt that

[wicked zephyr]

okay thank you

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[cedar mulch]

What to do next??

[full forum]

@cedar mulch Has your question been resolved?

[ornate bobcat]

Actually uniting the denominators can be more easier than that
@cedar mulch

[cedar mulch]

Wdym

I did, in the near end

[ornate bobcat]

I don't mean that

[cedar mulch]

Can u elaborate it then

[ornate bobcat]

You can simplfie it by applying the division

Either before uniting or after uniting the denominators

[cedar mulch]

Dividing what

[ornate bobcat]

Like
$$\frac{x}{x+1} = 1-\frac{1}{x+1}$$

[glossy valve]

Sherif Player

[cedar mulch]

Hm?

[ornate bobcat]

Also we don't want to unite the denominator of the constant numbers just take them to the other side

Do you know how to divide functions
Like long division or synthetic division?
@cedar mulch

?

@cedar mulch ?

[cedar mulch]

Mb mb

[cedar mulch]

Yeah

I'm bit confused on how to apply it there though

[cedar mulch]

Oh okayyy... i think i get it

[ornate bobcat]

So we get
$$6 \geq 1-\frac{1}{x+1} - 4 + \frac{8}{x+2}$$

[glossy valve]

Sherif Player

[ornate bobcat]

$$9\geq\frac{8}{x+2} - \frac1{x+1}$$

[cedar mulch]

What

[glossy valve]

Sherif Player

[cedar mulch]

How did u-

How did u get that

[ornate bobcat]

Just divided x by x+1
And 4x by x+2

To get it simpler

Now as we deal with fractions it will be easier if we took 9 to the RHS

Then uniting the denominator

[cedar mulch]

Why

[ornate bobcat]

$$0\geq\frac{8(x+1) - x+2 - 9(x²+3x+2)}{x^2+3x+2}$$
$$0\geq\frac{8x+8 - x - 2 - 9x^2 - 27x - 18}{x^2 + 3x + 2}$$
$$0\geq \frac{-9x^2 - 20x - 12}{x^2+3x+2}$$

[glossy valve]

Sherif Player

[ornate bobcat]

Sorry for confusing you here

Your way is correct

Now to solve it we need to see which x values would make the right side negative

To do that we need to draw a number line

And see when the numerator will output + and when it output -
And do the same for denominator and take the part that the sign they output is different than each other

[cedar mulch]

Yeah but how do i get the critical points

If the num and deno is a quadratic

I mean, i get that i need to factor the deno, but how about the num? It's not factorable by rational eq

[ornate bobcat]

Oh yes but we can use quadratic formula

[cedar mulch]

Wdym

How

[ornate bobcat]

I meant formula

[cedar mulch]

How??

[ornate bobcat]

First we check the determine

If you check it you will see that it is negative

Which states that it doesn't have real solutions

Which means that it is either always positive or either always negative

What do you think about here

@cedar mulch

[cedar mulch]

Determine where

[ornate bobcat]

What is the quadratic formula

Do you know it ?

[cedar mulch]

Yes

[ornate bobcat]

So you know what is under the sqrt in the quadratic formula

[cedar mulch]

Yes

b^2-4ac