help-42

[potent lotus]

What should I do

[hushed sequoia]

what about n=1?

[remote mural]

$$n^5 - 4n^2 < n^5 -3n$$

[potent lotus]

What should I do

[hushed sequoia]

so, you just proved that it converges

[remote mural]

Yay

Was just

checking

if i did everything right

TY

[remote mural]

idk what this means

[hushed sequoia]

ok

nvm

i could've calculated wrong

[remote mural]

My book did it this way

I do not understand the middle bit

so confusing

where did that come from

oh

its the orginal

uuuuh

the / is annoying

okay

i see now

ty

.close

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[calm coral]

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[dusky jolt]

the HCF of two numbers is 12.
the LCM of the two numbers is 144
both numbers are greater than 20.
what are the two numbers

[prime skiff]

to start

lets denote these x, y

[dusky jolt]

yeah

[prime skiff]

do you know the relationship between hcf and lcm

xy = HCF(x,y) * LCM(x,y)

heres an example

[dusky jolt]

uh huh

[prime skiff]

makes sense?

[dusky jolt]

yeah

[prime skiff]

great

so we're given the HCF and the LCM

our hcf is 12, and our lcm is 144

since the HCF * LCM = xy

we can say

that xy = 12 * 144

so the product of our two numbers is 1728

[dusky jolt]

yeah

[prime skiff]

so now

xy = 1728

x > 20, y > 20

[dusky jolt]

yea

[prime skiff]

there are multiple ways u can do this part

trial and error, finding factors etc

but i think finding factors would prolly be the best

so through finding factors

you can list the factors of 1728 and find the ones that meet the criteria (both greater than 20)

[dusky jolt]

3^3 x 2^6

[prime skiff]

hm

27 * 64

= 1728

problem is, it doesnt satify the other rules

hcf = 12, lcm = 144

[dusky jolt]

yeah

[prime skiff]

oh wait no i see what your doing

so you prime factorised 1728

[dusky jolt]

ye

but how to go on

[prime skiff]

okay that makes sense

well you have

3 * 3 * 3 * 2 * 2 * 2 * 2 * 2 * 2

so you could try to like

mix and match

to satisfy the rules

[dusky jolt]

alright

thanks for the help 😄

[prime skiff]

all good

[dusky jolt]

i should be able to do it now

[prime skiff]

good luck 👍🏻

[dusky jolt]

ty

.close

[calm coral]

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[calm coral]

Please don't occupy multiple help channels.

[unkempt sphinx]

Are you given $x+\frac{1}{x}>0$?

[potent lotus]

lpieleanu

[cursive linden]

x is already > 0

(given in the question)

so when is logx > 0?

[odd egret]

when x is not 1 ?

[odd egret]

no , it is only given that x>0

[cursive linden]

are you sure?

log is both negative as well as positive, it is zero at x = 1

when is it positive?

[cursive linden]

also if x > 0 then (x + 1/x) is also > 0

[odd egret]

greater than 1 ?

[odd egret]

ohk

so i just have to solve for a , then take the least +ve value of a ?

[cursive linden]

yeah

you will get an interval

[odd egret]

mhm

https://tenor.com/view/thank-you-sticker-thanks-sticker-line-sticker-cat-sticker-orange-cat-gif-26476683

[cursive linden]

yw

[odd egret]

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[remote mural]

Q2b need helo

[remote mural]

Help

.close

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[dusty crater]

how to find d/dx of this?

[dusty crater]

i think i do quotient rule

for context, this is a part of a proof by induction

[glass heart]

quotient rule works

[dusty crater]

$\frac{-[-(1-x)^k k!]}{(1-x)^{2k+2}}$

[glass heart]

but you could instead also write it as k! (1-x)^(-(k+1))

[dusty crater]

does this look right

[glass heart]

maybe that's easier

no

[dusty crater]

is it right now?

that simplifies to $(1-x)^{-k-2}k!$

[dusty crater]

but the markscheme is

how come I have k! andthe answer key has (k+1)!

[glass heart]

remember that from the power rule the exponent comes down

[dusty crater]

yep ofc

but why did I get k! and not (k+1)!

[glass heart]

well cause you didn't move the exponent down

[dusty crater]

which exponent

[glass heart]

if you did you would get a factor of (k+1)

[dusty crater]

$\frac{k!}{(1-x)^{k+2}}$

fuck latex

[dusty crater]

ok finally

so how does k! become (k+1)!

[glass heart]

the exponent was k+1

that comes down

gives you a factor of k+1

[dusty crater]

what

[glass heart]

which together with the k! gives you (k+1)!

[dusty crater]

ohhh

[dusty crater]

$k!(k+1)(1-x)^{-k-2}$

[dusty crater]

where do I go from here?

how does k!(k+1) = (k+1)!

[glass heart]

basic fact about factorials

what is (k+1)!

[dusty crater]

pls explain

k!=k(k+1)(k+2)...

[glass heart]

no

k! is the product of all integers up to k

[dusty crater]

oh shit

i messed up the properties

k(k-1)(k-2)...

oh it makes sense now

[glass heart]

(k+1)! is the product of all integers up to k+1. so first you multiply all the numbers up to k and then you multiply k+1

so k!*(k+1)

[dusty crater]

now it's just (k+1)(k)(k-1)...

thanks

factorials suck

can't wait to get to tetration

[glass heart]

that's not a thing that's commonly used

not in any way comparable to factorials

[dusty crater]

ok

ty for helping

.close

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[knotty pagoda]

I have done till here till now

unsure where to go from here?

[celest igloo]

have you tried synthetic division

i got this but my coefficient on the b^2 * a term was -9

#❓how-to-get-help

[tropic sky]

oh my fault

[celest igloo]

all g

[calm coral]

@knotty pagoda Has your question been resolved?

[knotty pagoda]

Idk how to do that-

[tight dock]

it's supposed to be 6ab² here

[calm coral]

@knotty pagoda Has your question been resolved?

[knotty pagoda]

No?

Also nvm I'm taking too long to reply like this 😭

I'll ask maths teacher 👍

[calm coral]

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[remote mural]

Are all power series in the form $$\sum_{n=1}^\infty A_n (x-c)^n$$

[potent lotus]

What should I do

[calm coral]

@remote mural Has your question been resolved?

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[remote mural]

Increasing/decreasing qt/

[remote mural]

My answer was increasing: (1.2,inf)U(-1.2,0) Decreasing: (-inf,-1.2)U(0,1.2)

also just wanted to clarify, are we finding an increasing y-value by how it correlates with x or vice versa?

[polar jetty]

It must be increasing y-values but i can't be certain unless you show me the full question.

[polar jetty]

About x, we just give intervals in terms of x.

[polar jetty]

Yeah. y-values.

[remote mural]

so you look at y, and write what x is doing during y's movements

in simple words

[polar jetty]

Not exactly. You see the region where y-values are increasing and then you write the corresponding x-interval for that. Then you can say that y-values are increasing in that interval.

[remote mural]

alright, so would this be correct in this instance?

[polar jetty]

If the roots are indeed 1.2 and -1.2, then yes.

[remote mural]

Thx!

@polar jetty this one was tricky, will be surprised if I got it right lol

Increasing: (-2,6)U(-1,1.5)
Decreasing: (-inf, -2)U(1.5,inf)

[polar jetty]

Increasing: (-2,6) ?

Check again.

[remote mural]

oh right, (-2,-1)

[polar jetty]

Also, are you sure your maxima and minima points are correct?

(-1, 1.5) also seems fishy from the figure.

If there's more information in the question, share.

[calm coral]

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[calm coral]

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[cyan stratus]

i need help

[leaden thunder]

...

[pure kayak]

doing what

[leaden thunder]

Just work on one problem at a time

They're all pretty similar

[cyan stratus]

I can't do this and it has to be submitted by the 19th, if I don't submit it I could fail the year

[pure kayak]

cant do what? youve just presented equations

[leaden thunder]

Yea ok but that's still not informative about what you want help with

Show the instructions or original question

[cyan stratus]

ok

I need to make a graph with this

similar to this

but I can't do the math

[pure kayak]

are you checking which of those equations corresponds to this graph?

or are you graphing each equation

[cyan stratus]

I am graphing each equation

[pure kayak]

theres 3 things youll be looking for:
the y intercept: x=0
any roots: where y=0
and the vertex - you can put it into vertex form or you could use derivatives to find the critical point which will be the vertex

[cyan stratus]

So, I just don't know how to do it

[pure kayak]

how to do what

[cyan stratus]

the accounts to find

[pure kayak]

do you understand any of this?

[cyan stratus]

no much

[pure kayak]

the y intercept (where the function crosses the y-axis) is found when x=0, the graph you sent is of y=x^2+x-12 so ill use it as an example,
when x=0 y=0^2+0-12=-12 so y=-12 the y intercept it (0,-12)

the roots are where y=0 so 0=x^2+x-12,
you will have to solve this equation, one way is through factoring:0=(x+4)(x-3)
which is satisfied when x=-4 or x=3, as you can see on the graph this is where it crosses the x axis

for the vertex - tip of the graph:
either put it in vertex form y=a(x+b)^2+c

or use derivatives: dy/dx=2x+1=0 x=-1/2 then find y

[calm coral]

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[late torrent]

having trouble with this

[hearty musk]

factor out e^(2x) from top and bottom

[late torrent]

lim x-> e^2x(3e-1

how do i factor out the 2

does it become - sqrt1

[hearty musk]

what

[late torrent]

-e^x

[hearty musk]

no

wait

[hearty musk]

factor out the numerator like this: e^(2x)(3-1/e^x+5/(e^2x))

and then do the same kind of thing for the bottom

[hearty musk]

it's like what you did for the earlier problem

[late torrent]

can u fix ur parenthesis

im just trying to understand what u wrote exactly

[hearty musk]

oh sorry

uh

(e^(2x))(3-(1/(e^x))+(5/(e^(2x))))

is the best i can do

[late torrent]

lol ok let me try to read that

[hearty musk]

like this

[late torrent]

[(e^(2x))(3-(1/e^x))+(5/e^2x)]

thats the numerator

[hearty musk]

the five should be divided by e^(2x)

[hearty musk]

what you're doing is dividing the terms in the numerator by e^(2x) and then putting it at the front

it's the distributive rule

[late torrent]

ik but its a weird thing to factor out of a whole number. im just trying to wrap my head around it

[hearty musk]

ah gotcha

[late torrent]

lemme try to do the denominator

[(e^(2x))(3-(1/e^x))+(5/e^2x)]/[(2/e^(2x)-3

wait im not done yet but

just to clarify

[hearty musk]

?

[late torrent]

if i try to factor e^2x out of 7x it becomes 1/7e^2?

[hearty musk]

where is 7?

[late torrent]

no like if i have 7x

and i try to factor out e^2x

[hearty musk]

oh

uh

this might not be the best method to use if you have e^x and x

it works best if you have a polynomial of some sort (it can be a polynomial in a function of x, like e^x)

[late torrent]

just not sure how to factor the e^2x out of -3e^x

bc of the ^2

[hearty musk]

e^(2x)=(e^x)^2

[late torrent]

-3/e^x?

[hearty musk]

why is the -3 in the denominator there?

[late torrent]

o.o

[hearty musk]

yeah that seems right now

[late torrent]

[(e^(2x))(3-(1/e^x))+(5/e^2x)]/[(e^(2x))(2/e^(2x))-((3/e^x)+1)]

i hate parenthesis

[hearty musk]

uh

the e^2x is getting factored out of the entire numerator and denominator

[late torrent]

yes so i took it out of both

[hearty musk]

so it should be multiplied by the whole thing instead of just part

[late torrent]

i took it out of numerator and denominator

[hearty musk]

your current numerator looks like this

which might just be because typesetting sucks

[late torrent]

how is it supposed to look

y r u the one crying u didnt have to type that

[hearty musk]

like this

[hearty musk]

lol

[late torrent]

hold onnn

[hearty musk]

?

[late torrent]

[((e^(2x))(3-(1/e^x))+(5/e^2x))]/[((e^(2x))(2/e^(2x))-((3/e^x)+1))]

how does that look?

[hearty musk]

looks right

ok great

now cancel the e^2x in the numerator and denominator

[late torrent]

how does the denominator look

[hearty musk]

oh I think you have the same error there as you did in the numerator

[late torrent]

i tried to fix it

look at it again

[((e^(2x))(3-(1/e^x))+(5/e^2x))]/[((e^(2x))(2/e^(2x))-((3/e^x)+1))]
[(3-(1/e^x))+(5/e^2x))]/[(2/e^(2x))-((3/e^x)+1))]

now what-

it looks so weird ._.

[hearty musk]

well

now we can evaluate the limit

because at -infinity

the e^2x and e^x go to -infinity

so the terms divided by those go to 0

[late torrent]

wait so now i just plug in

[hearty musk]

basically yeah

[late torrent]

(3-0+0)/(0-0+1)

3/1
3

OML I DID IT

[hearty musk]

GJ

YOU DID IT

[late torrent]

lemme write that down now : D

[hearty musk]

*btw this should be positive infinity not negative infinity im dumb

[late torrent]

same thing

[hearty musk]

well for this problem it is yeah

nah he actually did good tho 😭

[late torrent]

wait

how am i supposed to solve that (looking for horizontal asymptotes) without a limit

ur listening to travis scott u cant talk

[hearty musk]

well what are the asymptotes of arctan(x)

[hearty musk]

lmao fr

[late torrent]

arctan 1 is pi/2

um horizontal?

it goes from pi/2 to 3pi/2 i think

[hearty musk]

so then those should be the asymptotes

[late torrent]

oh-

wait

actually arctan 1 is pi/4

bc tan = y / x

ru sure its pi/2 and 3pi/2

[hearty musk]

should be pi/2 and -pi/2

[hearty musk]

seems right

[late torrent]

OH

by f(x) it means find the limits of both positive and negative infinity

ok ok

[hearty musk]

f(x) just means that it's a function

[late torrent]

shush

[hearty musk]

wait no the asymptotes of this function should be different

it should be pi/2 and pi/4

at infinity and -infinity respectively

[late torrent]

so its pi/2 and pi/4

but how do i write that

does arctan(2e^x) = pi/2

if so then tan(2e^x) = 1/0

2e^infinity is infinitely large

so how does that = 1/0

wait nvm i got it 1 sec

.close

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[naive temple]

how did x+3/x-1 become x+3

[scenic nova]

cuz x-1*0 is 0

hence x+3 = 0

[naive temple]

so i simply multiply the denominator with y

?

[scenic nova]

multiply both sides with x-1

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[glad delta]

what have i done wrong here?

[calm coral]

@glad delta Has your question been resolved?

[glad delta]

if yk anyone who knows applied maths please lmk. IDK what to do

ping them plz

[calm coral]

@glad delta Has your question been resolved?

[calm coral]

@glad delta Has your question been resolved?

[glad delta]

Nope

[calm coral]

@glad delta Has your question been resolved?

[calm coral]

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[calm coral]

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[subtle nexus]

bisects means cuts in half

so DE=EF

[wheat ginkgo]

actually nevermind

.close

how

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[graceful kelp]

what have i done wrong?

Apprently im supposed to type in ordered pair

How do i do that?

.close

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[vast timber]

I’m not sure what to do from this point

[pallid halo]

do you know the intermediate value theorem?

[vast timber]

yes

[pallid halo]

ok

so you found a point where f(x) = 0

namely when x=0

now find a point where f(x) >= 1900 (or explain why there must be such a point)

then use the IVT

[vast timber]

that is the problem i am having, i dont know how what to use to get a number > =1900

[pallid halo]

well you have a bound for sin

namely -1 <= sin(x) <= 1

whereas x^2 has no upper bound

use that to argue why f has no upper bound

[vast timber]

so could i make x something like 1000 and it will be greater than 1900?

[pallid halo]

yea, for example if x = 1000 then 2x^2 is 2 million

and whatever sin(x) is, it's betwen -1 and 1

so at overall f(x) is at least 2 million minus 8

[vast timber]

how would i find what sin(1000) with out a calculator?

[pallid halo]

you don't need to, that's the point

you know f(x) is at least 2 million minus 8

so it's certainly bigger than 1900

knowing that sin(1000) is at least -1 is all you really need

[vast timber]

Would this be right?

[pallid halo]

yes

[vast timber]

ok thank you!

.close

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[steel jolt]

checking my answer for 244 b and c

[calm coral]

@steel jolt Has your question been resolved?

[steel jolt]

<@&286206848099549185>

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@steel jolt Has your question been resolved?

[whole hinge]

lemme take a look!

[whole hinge]

looks good! I think you forgot to write down the equals signs for c though

[steel jolt]

like equals t?

[whole hinge]

no I mean

[\frac{x-4}{-2} = \frac y3 = \frac{z-5}{-4}]

[potent lotus]

propERICly_embedded

[steel jolt]

oh i see ty

Is this right?

[whole hinge]

I'll check!

what is PM? do you mean PA?

[tame tusk]

1 = (x-1) / (1-x)
pleae solve and explain to me

[steel jolt]

oh yea oops my b

[whole hinge]

!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).

[whole hinge]

I didn't look through all the arithmetic, but the method looks right!

[steel jolt]

ok tyvm

.close

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[rigid kettle]

Im tryna (for no particular reason) turn this contrapositive proof into a direct one but struggling. Or an alternate direct. Maybe impossible...?

===

Statement to prove:
Take N, naturals.
Suppose I have a collection of finite subsets of N where any finite intersection within it is non-empty.
Then the intersection of the entire collection is non-empty.

===

Contrapositive. Suppose intersection of collection empty. Let A = {x1, ..., xn} in collection.

Exists Ai in collection with xi not in Ai for each i from 1 to n.

Then intersection {A, A1, ..., An} empty. Hence there exists some finite empty intersection.

[rigid kettle]

idk if mind block or it just cant be reversed

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[remote mural]

Could someone help me go abt solving this question

[fallen oriole]

Which part are you having trouble with? It looks quite trivial

It’s asking for the set of x’s for which f(x) is negative

[remote mural]

I dont even know where to start LOL

oh wait

its legit just asking for every single x that is negative?

but wouldnt it just be any negative number?

[fallen oriole]

Every single x for which the value of the function is negative

Not every single x that is negative

Value of the function is on the vertical axis, X is on the horizontal axis

Notice there are regions of x for which the y values (vertical axis) dips below 0

[remote mural]

wait im confused now

the final eqution would just be x is any number below 0?

[fallen oriole]

The answer is the set of x values, for which the blue line goes below y = 0, the vertical axis

For example, -1 would be in the answer set. Because for x = -1 you can see y is about -5

[remote mural]

so how do i write that in the answer my treacher wants

[fallen oriole]

Hm, do you not know the set builder notation?

[remote mural]

No

[fallen oriole]

It might be best if you maybe google it. But the gist of it is that you can describe the x values

The example they gave would be read as “The set of x’s which are an element of the real numbers such that x is bigger than -8 smaller than 2”

[remote mural]

ok ty could u help me w this question too

[coral osprey]

You can use desmos

Might as well use desmos

[remote mural]

How does desmos help LOL

could u check my math for the question?

First, we need to rewrite the inequality so that one side is equal to zero. To do this, we can subtract 3 from both sides of the inequality: x^4 - 5x^3 + 2x^2 + 8x + 3 - 3 < 3 - 3

This simplifies to: x^4 - 5x^3 + 2x^2 + 8x < 0 
Now, we need to factor the polynomial on the left side of the inequality. We can do this using a graphing calculator.

Doing this we find the polynomial has the following approximate roots: x ≈ 0.317, x ≈ 1.000, x ≈ 2.683 
Now that we have the roots, we can determine the intervals where the polynomial is less than zero. We will test the intervals between the roots and the intervals to the left of the smallest root and to the right of the largest root. 

Interval 1: x < 0.317 Test value: x = 0 Plug this into the polynomial: (0)^4 - 5(0)^3 + 2(0)^2 + 8(0) = 0 Since 0 is not less than 0, the polynomial is not less than zero in this interval.
Interval 2: 0.317 < x < 1.000 Test value: x = 0.5 Plug this into the polynomial: (0.5)^4 - 5(0.5)^3 + 2(0.5)^2 + 8(0.5) ≈ -0.6875 Since -0.6875 is less than 0, the polynomial is less than zero in this interval.
Interval 3: 1.000 < x < 2.683 Test value: x = 1.5 Plug this into the polynomial: (1.5)^4 - 5(1.5)^3 + 2(1.5)^2 + 8(1.5) ≈ 2.34375 Since 2.34375 is not less than 0, the polynomial is not less than zero in this interval.
Interval 4: x > 2.683 Test value: x = 3 Plug this into the polynomial: (3)^4 - 5(3)^3 + 2(3)^2 + 8(3) ≈ 9 Since 9 is not less than 0, the polynomial is not less than zero in this interval.

Therefore, the polynomial is less than zero only in the interval 0.317 < x < 1.000. 

[craggy zodiac]

Get rid of the 3 and see when $x^4-5x^3+2x^2+8x$ the polynomial is inferior to 0

[potent lotus]

Joseph.P

[remote mural]

For this

[fallen oriole]

You write the inequality, desmos literally shows you the solution 😄

[craggy zodiac]

IA is proscribed

[remote mural]

I have it in desmos but it isnt showing me

[fallen oriole]

It’s literally showing you the regions of x that satisfy the inequality

[remote mural]

Wait so its just the shaded regions?

So it would be -1 to 0 2 to 4?

But thats not what i got

[remote mural]

For this question i got this {x ∈ R | x < 1} ∪ {x ∈ R | 3 < x < 4}

is that right

proper formatting

.close

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[median schooner]

Prove that for all $a > 0$ and for all $x \in \mathbb{R}$ there is exactly one value of $y \in \mathbb{R}$ such that $x^2y + ay + 2x^3 = 0$

[potent lotus]

o.O

[median schooner]

How should I begin to approach this?

So far, I have written down:
$$ \forall a, x \in \mathbb{R} , : , a > 0,, \exists ,\text{exactly one}, y \in \mathbb{R}, : , x^2y + ay + 2x^3 = 0$$

[potent lotus]

o.O

[median schooner]

Im curious whether I should attempt to do this directly, e.g. suppose a and x are real numbers s.t. a > 0, then proceed with some algebraic reasoning, or if I should attempt the contrapositive and then try a large series of if and only if statements. or maybe there is another way instead of a direct proof or by contrapositive

I feel like if I do the contrapositive it is possible to factor it some how into an obviously true statement, such as (a + b)^2 >= 0 is obviously always true for all real numbers a and b

I also considered trying various options for what y could be. e.g. "Suppose y = a" or "Suppose y = x" but I dont think I got into any forms that would've helped me

e.g. suppose $y = x$, then we have $$x^3 + ax + 2x^3 = 0 \iff 3x^3 + ax = 0 \iff x(3x^2 + a) = 0$$

[potent lotus]

o.O

[median schooner]

and maybe we proceed by cases to apply zero product identity?

$3x^2 + a$ doesn't necessarily have any real solutions, but we know $x = 0$ will satisfy

[potent lotus]

o.O

[median schooner]

though I don't think this works because we have only shown that if x = 0 and a > 0 then there is a y

[calm coral]

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[remote mural]

Hi! I need help really badly I been stuck with 8 problems for past 2 days 😢

[remote mural]

it has 3 steps to it

[remote mural]

this is an example

need help so bad

<@&286206848099549185>

🙏

[fallen oriole]

It even tells you what to do, plug the numbers into a calculator as it tells you

Or you can do it the long way by hand, with these or with the formula they gave, whatever floats your boat

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[wicked tapir]

just trying to get a refresher on imaginary numbers. How do I simplify sqrt(-26)

[wicked tapir]

i got the answer -2i(sqrt(13))

not negative sorry

[visual nimbus]

sqrt(-26) = i sqrt(26)

[wicked tapir]

how do you get that?

[visual nimbus]

sqrt(-1) = i

[wicked tapir]

oh got ya

thanks

.close

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[shell token]

is this question bugged? they all seem wrong to me

the key says a but the median is 3....

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[eager acorn]

Here can we not do this?

[civic dirge]

That's the probability that exactly 5 of the 100 are defective, given a 10% chance that any is defective

Independent of the 5 you've grabbed

Consider:

• How many ways are there to grab 5 working bulbs?
• How many ways are there to grab 5 bulbs in general?

However, looking at the answers, I think they used an approximation

Probability of grabbing a working bulb is 9/10, so the probability of grabbing 5 is (9/10)⁵ (but not really)

It's the only answer that's even close though, so that's what they did

[calm coral]

@eager acorn Has your question been resolved?

[eager acorn]

I didn't understand your explanation

Properly to this question

[eager acorn]

<@&286206848099549185>

[exotic falcon]

@eager acorn

We will prevail

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[royal ridge]

is this correct

[royal ridge]

im not sure if i answered all the questions correctly

.close

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[twilit delta]

I need some help finishing this problem

[twilit delta]

https://cdn.discordapp.com/attachments/830088413419798538/1154564054648623215/20230921_204429.jpg

I have the first part of the piecewise function

I kind of have the answer here on the graphing calculator, but I got this just by luck

I don't know where that -12 comes from

[calm coral]

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@twilit delta Has your question been resolved?

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@twilit delta Has your question been resolved?

[twilit delta]

<@&286206848099549185>

[cyan flume]

hey u there?

so as we can see for x >=1200, the min cost would be always 10 + (0.06*1200)

we can write the extra cost over 1200

as 0.07(x-1200)

so adding min cost and extra cost together, we get

10 + (0.06 *1200) + 0.07 (x -1200)

we get 10 + 0.07x - 12

[twilit delta]

oh, so that's how it works

thanks friend

I can go to sleep now

.close

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[knotty flare]

9 people are sitting in a room. What is the probability that exactly 3 of them have bdays in the same month of the year and no other two persons have bdays in the same month of the year?

[calm coral]

Please don't occupy multiple help channels.

[knotty flare]

Can someone help

[calm coral]

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[floral tulip]

what does the sigma mean there?

I don't understand notation $\int_K f(dx) \quad \int_{dK} f(x)dx$

[potent lotus]

redve

[floral tulip]

second one is Stokes theorem, first one is something from my probability lectures, but it hasn't explained what does it mean

[lofty mantle]

$\frac{1}{b^n} = b^{-n}$

[potent lotus]

Gabe

[floral tulip]

man, this channel is actually occupied

[lofty mantle]

Ohh sorry I meant to open a new channel

I'll delete

[floral tulip]

np

[calm coral]

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[past hawk]

Given a square grid with two objects that have a uniform probability of moving in all four directions, is the average probability for the whole plane for the two objects meeting the same if with or without a divider down the middle?

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[raven mulch]

Help I still didn't get the right answer 😭

[raven mulch]

Question 2

[frank gull]

So last time we said that $716.6 = \frac{875(13) + 13706}{13 + n}$. What did you get for $n$?

[potent lotus]

MellowDramaLlama

[raven mulch]

Yuh

8.06

Which doesn't make sense

[frank gull]

correct.

it does not lol

Ok can you show me your work on how you got that answer?

[raven mulch]

Ok

Gimme a min

Wtf

I got the right answer

🤦

Calculation error

Mb

Thx @frank gull

😄

[frank gull]

np!

nice work

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[remote mural]

Hy

[remote mural]

Look at the SA

The oblique asymptot is what am tryna calculate

[hollow perch]

$y=\frac{3x^2+x+1}{x-5}$ right?

[remote mural]

I got 3x-14 and u can see my calculations i use euclidische deling in dutch

[potent lotus]

chlamydia

[remote mural]

Yes

[hollow perch]

$y=\frac{3x^2-15x+16x+1}{x-5}$

[potent lotus]

chlamydia

[hollow perch]

i've got 16 instead of 14?

[remote mural]

Tf how am I supposed to know i have to devide x terms like that

[remote mural]

The answer is with 16 i got -14

[hollow perch]

right

this is just a way of splitting up the terms to match the divisor

[remote mural]

But ur supposed to use Euclidean division

[hollow perch]

well i don't

[remote mural]

Ye but i have to on my test there is no point using ur method

[hollow perch]

then use euclidean

[remote mural]

I did didn’t get the right answer

[hollow perch]

then you've done something wrong in your division

[remote mural]

I ask for help cuz idk what i did wrong

[hollow perch]

can i see what you did then

[hollow perch]

nvm

[remote mural]

Do i need zooomed in?

[hollow perch]

it's fine

why'd you make it 3x^2+15x

when the divisor is x-5

[remote mural]

Just look at this

[hollow perch]

,rotate

[remote mural]

So 3x^2 devided by x is

3x

Then 3x multiplied by x and -5

Its 3x^2 and +15 x sorry

[hollow perch]

and -15x?

[remote mural]

Then minus

0-14x

And drop the 1 then -14x devided by x and so on

[remote mural]

What u think i did wrong?

[hollow perch]

you even say -5, but then why does that turn into a +15

[remote mural]

Cause

3 multiplied by 5 is 15

[remote mural]

.

[hollow perch]

fr bruh

[remote mural]

Ye so u think am wrong somewhere?

[remote mural]

.

<@&286206848099549185>

[gritty mantle]

What do you get when you multiply them?

[calm coral]

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[calm coral]

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[molten ruin]

how to solve this

[molten ruin]

<@&286206848099549185>

no?

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[opaque cape]

can someone help me solve these step-by-step?

[calm coral]

@opaque cape Has your question been resolved?

[formal chasm]

you're solving for z?

[calm coral]

@opaque cape Has your question been resolved?

[opaque cape]

yes

[formal chasm]

excellent

let's start with a

do you have any idea where to start or are just getting stuck on a step?

@opaque cape

[opaque cape]

expanding the left side?

@formal chasm

[formal chasm]

nope

start by getting rid of the fraction on the right

[opaque cape]

ok

[formal chasm]

what do you have have now?

[opaque cape]

(z+1)^2(1+i) = -7 + 1

[formal chasm]

nope

not what I meant

multiply the right side by some a/a to simplify the denominator into a real number

[opaque cape]

ohhhh

my bad

[formal chasm]

no worries

[opaque cape]

one sec

ok

so

(1-i)(-7+i)/2 = (z+1)^2 right

?

[formal chasm]

looks right, just simplify now, and we can move on to the next step

[opaque cape]

u mean expand the side with i?

[formal chasm]

the one that used to be fraction

[opaque cape]

the one wiht the square

mb no

the other one

[formal chasm]

yup

[opaque cape]

one sec

-6 + 7i?

8

-6 + 8i

right?

[formal chasm]

don't forget to divide by 2

[opaque cape]

sorry

-3 + 4i

[formal chasm]

yup!

now, we have (z+1)^2=-3+4i

[opaque cape]

cool

[formal chasm]

we now take a square root on both sides

do you know how to do that or do you want help? (with finding the square root)?

[opaque cape]

so we put both sides under square root?

[formal chasm]

yes, don't forget a ±

[opaque cape]

could you please help me with that too?

[formal chasm]

yup!

so square root of (z+1)^2 is z+1 fairly obvious

[opaque cape]

yeah..

[formal chasm]

square root of -3+4i is ±(a+bi) for some real a and b

how do we find a and b?

[opaque cape]

wait what

[formal chasm]

well, squaring a+bi, we get (a^2-b^2)+(2ab)i

so we know a^2-b^2=-3, and 2ab=4 (because our number is -3+4i, and the real and complex parts have to match)

make sense so far?

[opaque cape]

yes

[formal chasm]

so we have ab=2 and $a^2-b^2=-3$, which gives us $a^2-\frac{4}{a^2}=-3$, or $a^4-4=-3a^2$, $a^4+3a^2-4=0$, $(a^2+4)(a^2-1) = 0$, $a=±2i$ or $a=±1$ (first is invalid because we said a and b are real numbers earlier)
note this is just the quadratic formula using x=a^2

so a=±1, right?

if a=1, then since ab=2, b=2. Otherwise if a=-1, then since ab=2, b=-2, This means our answers are (1+2i) and (-1-2i), or ±(1+2i), right?

[opaque cape]

i see

[formal chasm]

you understand it all though? Please speak up if ANY part of it is unclear

[opaque cape]

trying to write this down one sec

[formal chasm]

no worries

[opaque cape]

^ sign makes it harder to understand

[formal chasm]

fair enough

[potent lotus]

Astral
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[opaque cape]

wait why does it give us a^2 -4/a^2 = -3?

[formal chasm]

we started with ab=2 and a^2-b^2=-3 yes?

[opaque cape]

yes

[formal chasm]

we divide the first equation by a to get b=2/a

then plug that into the second equation

[opaque cape]

ohhhh got it

[formal chasm]

so b^2 becomes (2/a)^2

[opaque cape]

understood

[formal chasm]

excellent!

so we have z+1=±(1+2i)

z=-1+1+2i or z=-1-1-2i,
z=2i and z=-2-2i

ta da! a complete

the whole process kinda makes sense to you?

[opaque cape]

wow yeah

thanks a lot

😄

[formal chasm]

excellent!

still need help with bcd?

they look....

[opaque cape]

if u have the time to help i would appreciate it

[formal chasm]

well this is a great way for me to do more complex numbers ig

[opaque cape]

its the same process though, right?

[formal chasm]

each time yeah

more or less

the rest look harder to me

[opaque cape]

they are

[formal chasm]

d is easy though

[opaque cape]

how comer

[formal chasm]

because it's one step

and it's fairly intuitive

what does z look like it could be?

[opaque cape]

-i

[formal chasm]

yup

that's all there is to it

[opaque cape]

i see

[formal chasm]

for the others....

lemme check

ig you could try a convoluted z=a+bi approach and use the system of equations to try solving for z? hmmm

I want to try to complete the square but can't figure out how

oh wait I'm stupid lol

for b we have [z^2(1+i)=z+i-3]

[potent lotus]

Astral

[formal chasm]

divide it by 1+i

simplify any fractions, then come back to me

[opaque cape]

z^2 = z - zi - 2 + 2i

right?

oh and the right side

[formal chasm]

not entirely sure trying to do it in the textbox

[opaque cape]

,help

[potent lotus]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[formal chasm]

$z^2=(z+i-3)(1-i)/2 = \frac{1}{2}(1-i)z+2i-1$

I got this

[potent lotus]

Astral

[formal chasm]

oh yeah I can work with this

not sure how you got -2+2i on the right

it should be (i-3)(1-i)/2

[opaque cape]

not sure i got the same answer

[formal chasm]

you did not

not sure how

[opaque cape]

uhhhh

[formal chasm]

uh wait

no wait I did it right

divided by 1+i, then multiplied the whole right side by (1-i)/(1-i), which is like multiplying the original right side by (1-i)/2 in the first place

[opaque cape]

one sec

ok goti t

[formal chasm]

hooray!

[opaque cape]

so now we do the same thing we did at a right

[formal chasm]

so we have
[z^2-\frac{2}{4}(1-i)z=2i-1]

[potent lotus]

Astral

[formal chasm]

so we now complete the square

try yourself and I'll show you my try

[opaque cape]

yeah i dont know what the other square is

actually

wait

[formal chasm]

(ignore that)

talk

and to be clear by complete the square

I mean like the pre-algebra, algebra version we learn

where we take x^2+2x=0 and get (x+1)^2-1=0

or more generally x^2+bx=c and get (x+b/2)-b^2/4=c

[opaque cape]

so its gonna be

z^2 - 1/2(1-i)z + 1/16(1-i)^2 - 1/16(1-i)^2

[formal chasm]

yes but you have to factor it or else it doesn't help us

(also easier to write it in factored form)

[opaque cape]

you mean (z+ 1/4(1-i))^2

[formal chasm]

yes!

which leaves us with

[(z-\frac{1}{4}(1-i))^2-\frac{1}{16}(1-i)^2=2i-1]

[potent lotus]

Astral

[opaque cape]

oh

[formal chasm]

that's exactly what you wrote fyi

[opaque cape]

and we write this down as a^2 - b^2

[formal chasm]

not yet

please no

[opaque cape]

right

[formal chasm]

;-; first we simplify

[opaque cape]

oh haha

sorry

[formal chasm]

[(z-\frac{1}{4}(1-i)^2=\frac{1}{4^2}(32i-16+1-i)^2]

[potent lotus]

Astral

[formal chasm]

you see how I got to this form yes?

[opaque cape]

not really

[formal chasm]

multiply right side by 16/16 then group constants (subtract 1/16(1-i) from both sides)

[opaque cape]

okay

[formal chasm]

then rewrite 1/16 as 1/4^2 bc we're about to take a square root

so we have (z-(1-i)/4)=sqrt(31i-15)/4 which we can simply write as 4z=1-i ± sqrt(31i-15) for simplicity's sake, defer the /4 till later

a^2-b^2=-15, 2ab=31

with me so far?

[opaque cape]

yes

[formal chasm]

this is going to be painful but I believe in us

[opaque cape]

hahahah

i can see that

[formal chasm]

oh I see why

I forgot to square the 1-i

[formal chasm]

this is wrong

[(z-\frac{1}{4}(1-i))^2=\frac{1}{4^2}(32i-16+(1-i)^2)=\frac{1}{4^2}(32i-16-2i)]

forgot to square the 1-i oof

so it's

[4z=1-i\pm\sqrt{30i-16}]

[potent lotus]

Astral

Astral

[formal chasm]

there.... mistakes fixed

so a^2-b^2=30

2ab=16

ab=8, so b=8/1, so a^2-64/a^2=30

a^4-64=30a^2,
a^4-30a^2-64=0, (a^2-32)(a^2+2)=0

a^2=32, a^2=-2

with me so far?

[opaque cape]

wait wait

so we took it under the square root right

[formal chasm]

I made a typo earlier and it screwed everything up

yes

[opaque cape]

and it gave us this

[formal chasm]

yes

[opaque cape]

alright

[formal chasm]

after multiplying everything by 4 and rearranging

[opaque cape]

makes sense but y is there a +-

[formal chasm]

remember earlier how we got a ± because our answers were + and -?

[opaque cape]

yes

[formal chasm]

it's the case for all square roots it's just more literal with complex numbers

you'll see here we get ± answers too

[opaque cape]

ohhh

[formal chasm]

I just put in the ± ahead of time since we knew it always happens

wait no

I'm dumb

.... am I?

no I'm right

[opaque cape]

im still at the a^2 - b^2 part

[formal chasm]

kk

no my math was right

a^2=32, a^2=-2

if a^2=-2, then a is imaginary which is not possible since we have a+bi and both a and b are real

then a^2=32, and b^2=1/4 since we said ab=8

so b=1/2 and a=4sqrt(2)

hmmm no I screwed up again just a sec

[opaque cape]

whyd u put a^4 here

[forest ocean]

Please can you help me with this

[formal chasm]

#❓how-to-get-help

[formal chasm]

I had a^2-64/a^2=30, multiplied it all by a^2 after using the b=8/a sub

[forest ocean]

I went there it said go to available channels

[formal chasm]

....available help channels in its category

you went to the MATH HELP (OCCUPIED) category

the server is a tad busy rn so there are less channels than usual but you can snag one when it clears

or go to #1021175428326633542 which is unlimited

[opaque cape]

ok

[formal chasm]

@opaque cape I screwed up because it was 30i-16, I'm used to -16+30i format sorry (I read it as 30-16i)

it's a^2-b^2=-16, 2ab=30 oops

[opaque cape]

oh

kik

[formal chasm]

so ab=15, b=15/a, a^4-225=-16a^2, (a^2-25)(a^2+9)=0

a=±5 and a=±3i, a is real so just a=±5

I knew it was harder than it should be lol