help-0

[hallow thicket]

Can’t seem to get the right answer. Don’t know where I am messing up

[lone heart]

@hallow thicket Has your question been resolved?

[hallow thicket]

<@&286206848099549185>

[tacit arch]

,calc 3*81/4

[ocean seal]

Result:

60.75

[hallow thicket]

which step were you talking about?

[hallow thicket]

.close

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[prime spindle]

how do this

[lone heart]

@prime spindle Has your question been resolved?

[gilded citrus]

going to use a, b, c instead of alpha beta gamma

\begin{align*}
& a^2b + a^2c + b^2c + b^2a + c^2a + c^2b \\
=& ab(a+b) + ac(a+c) + bc(b+c) \\
=& ab(a+b+c) + ac(a+b+c) + bc(a+b+c) - 3abc \\
=& (a+b+c)(ab+ac+bc) - 3abc
\end{align*}

[ocean seal]

hsp

[gilded citrus]

the second one just write it in terms of (ab+bc+ca)^2

[prime spindle]

@gilded citrus why does the 2nd line have a extra c

and -3abc

[gilded citrus]

i added abc to each term so i could get all of them to have (a+b+c)

[prime spindle]

but

why is there a -3abc

also

[gilded citrus]

i cant just add abc out of nowhere

i have to do + 3abc - 3abc

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[prime spindle]

tf

[gilded citrus]

wut

[prime spindle]

.reopen

[lone heart]

✅

[prime spindle]

but

you only added c

wouldnt its be -3c?

[gilded citrus]

no?

[prime spindle]

o

i see

you added 1

letter in the each one

[gilded citrus]

yes

[prime spindle]

you added those letters

so it can match the equations

right

[gilded citrus]

yes, because i know how to find a+b+c and abc from the cubic

and ab+bc+ca

[prime spindle]

for the second one

it would be (αβ+βγ+αγ)^2

and u would just sub in αβ+βγ+αγ

into the bracket right

so (-2)^2

right

?

[gilded citrus]

no

wait

$$a^2b^2 + b^2c^2 + c^2a^2 = (ab+bc+ca)^2 - 2(ab^2c + bc^2a + ca^2b)$$

hmm

the second term is weirder than i thought

hm

oh wait no its not

[ocean seal]

hsp

[gilded citrus]

which is $$(ab+bc+ca)^2 - 2abc(a+b+c)$$

[ocean seal]

hsp

[prime spindle]

how i would i easily simplify this

from the original

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[tidal pendant]

Pre-cal. Im so at a loss, i don't know what context to give

[gilded citrus]

$(\cos(\theta), \sin(\theta))$

[ocean seal]

hsp

[tidal pendant]

Thank you, I'm trying that

is cos and sin always the same like this?

[gilded citrus]

yes

it follows from the definition of cos and sin

[tidal pendant]

with that, i plugged it in and answered this. It said it correct but its reasoning makes no sence

this is the information given after answering it

[gilded citrus]

id say the trigonometric way is far simpler

[tidal pendant]

I found the page on my textbook connecting x^2 to cos and y^2 to sin

Yea, the way you showed makes more sense.

Thank you!

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[gloomy vigil]

trying to figure out how to do solve this

basically tryign to do (1/15) + 2

= 2.066

but getting an incorrect answer

please advise. thanks

[gray isle]

don't use a calculator

work with fractions

get a common denominator etc

supposedly they want you to arrive at an exact value

[gloomy vigil]

what do you mean exact value

so im assuming this answer will result in a decimal number if they're asking not to round the number

if anyone could help out just send me a DM. thanks

[gray isle]

no

don't use a calculator
work with fractions

the answer will be a fraction

[gloomy vigil]

i already have the answer but i dont know how it got to it

[gray isle]

do you know how to add fractions?

[gloomy vigil]

the answer is is 31/15

you have to find a common denominator

[gray isle]

yes

do that here

so already simplified the 1/(3*5) to 1/15
are you able to express 2 as a fraction so that your two terms have a common denominator?

[gloomy vigil]

1/15 + 2 translates to 1/15 + 30/15?

[gray isle]

yes

[gloomy vigil]

2 = 2/1

right?

[gray isle]

yes

[gloomy vigil]

ok so

[gray isle]

you already got to

1/15 + 2 translates to 1/15 + 30/15?

[gloomy vigil]

arent you supposed to multiply the top and bottom by the same number to get the common denominator?

[gray isle]

and you're just 1 step away from the answer

[gloomy vigil]

how did they get from 2/1 to 30/15

in order to get the common denominator?

[gray isle]

well like you said

arent you supposed to multiply the top and bottom by the same number to get the common denominator?

2 = 2/1 and multiplying numerator and denom of that by 15 gets you 30/15

which i thought you did already when you said

1/15 + 2 translates to 1/15 + 30/15?

[gloomy vigil]

oh i was multiplying by 2 for the other fraction

god im dumb

[lone heart]

@gloomy vigil Has your question been resolved?

[lone heart]

@gloomy vigil Has your question been resolved?

[restive hearth]

@gloomy vigil

You have (1)/(3 • 5) + (2)/(1)

Which becomes (1)/(15) + (2)/(1)

To get the common denominator, you have to multiply one of the fractions or both in some cases to find something that is in common.

You can multiply 2/1 by something to get 15.

What would you do?

[restive hearth]

Hint ^

https://youtu.be/SAgVrzkSsU4

[gray isle]

there's no point with just responding with ❌,
you need to tell us what issue you still have with this @gloomy vigil

[gloomy vigil]

I understand the solution now. Thanks

You guys can close this.

[gray isle]

.close

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[desert tusk]

In an arithmetic progression, the twelfth term = sum of the first twelve terms in the arithmetic progression. Find the 6th term in this progression.

[desert tusk]

I try to equate the formula of both things, but not much is obtained.

UPDATE: The answer is 0.

I then form the formula of finding the nth term
Tn = a + (n-1)d
Tn = -5 + (n-1)d
T6 = -5 + (6-1)d
T6 = -5 + 5d
T6 = 0

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[ornate condor]

can a closed contour cross itself?

[naive valley]

yes

if it does not, it's called a simple closed contour

[ornate condor]

then how about this

the closed contour anticlockwise

can it also cross itself?

[keen plinth]

the orientation of the contour could possibly be ambiguous

depending on how it encircles poles

[ornate condor]

no

i do not get it

.close

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[keen plinth]

looks like you got it

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[bright tree]

Find the derivatives by 1st principle 1/3-2x

[median oar]

do you know what is first principles?

[bright tree]

Yes I did some calculations but I couldn't reach the answer

[median oar]

$m=\left(\frac{f(x_1)-f(x_0)}{x_1-x_0}\right)\$
We let $x_1=x_0+\Delta{x}\$
$m=\left(\frac{f(x_0+\Delta{x})-f(x_0)}{x_0+\Delta{x}-x_0}\right)\$
$m=\left(\frac{f(x_0+\Delta{x})-f(x_0)}{\Delta{x}}\right)$

[ocean seal]

Frosst

[median oar]

which step are you stuck on?

[median oar]

do you have more working out for 2nd last and last line

i cant quite follow you've done a lot of steps there

,rotate

[bright tree]

No but at last I forgot to write 1/delta x

[median oar]

there should be a bracket on the bottom

[bright tree]

I'm not getting you

[median oar]

$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{2x+2(x+\Delta{x})}{(3-2(x+\Delta{x}))\cdot (3-2x)\cdot \Delta{x}}$

[gray isle]

Delta

[bright tree]

1/Delta X is missing

[gray isle]

) in wrong place

before the \cdot, not near the end

[median oar]

wait what happened to lim

[gray isle]

\lim_{\Delta x \to 0}

[ocean seal]

Frosst

[median oar]

aha

i got there in the end

but there's supposed to be a bracket there on the 3-2(x+delta x)

that's my solution you should be getting the same i think

[gray isle]

didn't need the expand the denom like that
also don't be lazy by not writing Lim where appropriate

[bright tree]

Thanks

[median oar]

oh yeah that's true

sorry >.>

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[alpine sable]

How can I solve this?

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[oak gust]

hey

[oak gust]

Say R is to be the Leuteo Foundation reserve on day n;

R =100,000,000

Let E be the amount vested on day n :

R =R-1 (1-d) = E (1-d)n

what does this mean?

<@&286206848099549185>

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[past bloom]

what is the difference between stationary point of inflection and point of inflection?

[gray isle]

stationary points of inflection are also stationary points
points of inflection don't have to be stationary points

(0,0) would be a stationary point of inflection
(2,-3.2) would be an ordinary non-stationary point of inflection

[oak gust]

Can you please answer mine @gray isle

[gray isle]

your what

[oak gust]

Say R is to be the Leuteo Foundation reserve on day n;

R =100,000,000

Let E be the amount vested on day n :

R =R-1 (1-d) = E (1-d)n

[gray isle]

make a new channel, this is someone else's now
also your question is unclear and has very little context

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[lament basalt]

It's about the sum and difference identities in trigonometry

[lament basalt]

So I reverted it back to being tan (65 degrees - 110 degrees) which became tan (-45 degrees )

And I did this:

-[ tan(45 degrees) ]

And did this

• [tan (45 + 0)]

And then eventually did the equation in the pic again

I got - 2

Is that correct? Thanks

[hard mountain]

Are you sure you can do that with tan(x) ? Sine and cosine functions at least arent linear operators

[gray isle]

how did you end up with -2 from -tan(45+0)

[lament basalt]

• ( tan (45) + tan (0) / 1 - tan(45) tan(0) )

[gray isle]

isn't -2

[lament basalt]

What number it should end up?

[gray isle]

not -2

considering just

-[ tan(45 degrees) ]
you shouldn't most definitely know the value of tan(45°)

[lament basalt]

Oh I just realized yes bruh

-1

Sry

-1

[gray isle]

tan(45°) itself is just 1

the negative of that is -1

[lament basalt]

Yeah

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[wind kraken]

What is the formula of calculating Particular Integral, when F(a) = 0 ?

The question is, find the particular integral of (4D² + 4D + 1)y = 8e^(-x/2)

[lone heart]

@wind kraken Has your question been resolved?

[slender marten]

I can only assume you mean particular solution and that F(a) is meant to mean F(D) = 4D^2 + 4D + 1 when a is -1/2. Use the exponential shift theorem for differential operators to get your desired result.

y_p = e^(-x/2)1/4(D -1/2)^2 + 4(D - 1/2) +1 = e^(-x/2)D^(-2)(1). @wind kraken

D^(-2)(1) = x^2 so y_p = x^2e^(-x/2).

[wind kraken]

Thank you so much!

[slender marten]

🙂

There's a good chapter in Ordinary Differential Equations by Morris Tenenbaum and Harry Pollard.

[wind kraken]

whats the book called?

[slender marten]

It's called Ordinary Differential Equations. 😆

[wind kraken]

alright! I'll check it out

[slender marten]

I made this a while ago using the book so it might help you in your studies:

[wind kraken]

It for sure will help. Thankyou:)

.close

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[slow wasp]

https://cdn.discordapp.com/attachments/983587105349468160/1013073109701640223/756355377926504568.png

[slow wasp]

.close

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[rough egret]

how do i solve this

[alpine sable]

$u = \log_{\frac{1}{2}}\left(x\right)$

[ocean seal]

Aude

[alpine sable]

then vertex form

[rough egret]

uh

2u^2 + 7u + 3 <= 0?

[alpine sable]

yeah

you have a quadratic equation now

[rough egret]

(2u + 1)(u + 3) <= 0

[alpine sable]

nice

[rough egret]

-3 <= u <= -1/2

what do i do with that

[alpine sable]

what do you think

[rough egret]

i think its also -3 <= log_1/2(x) <= -1/2

but how do i get x

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@rough egret Has your question been resolved?

[past bloom]

so origin is the stationary point of inflection? and the other one is point of inflection?

but why (2,-3.2) is a point of inflection?

[ornate condor]

u shld start ur own help channel, just copy the Q

[past bloom]

ok

sorry to interupt jn

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[rocky crag]

(√6 -√2) (√3 + 1)

[rocky crag]

simplify

[median dirge]

Hi.

[rocky crag]

hi

hello??

[median dirge]

Hi...

You could simplify the whole thing by maybe taking out sqrt{2} common in the sqrt{6}-sqrt{2} term

Then solve for (a+b)(a-b)

[rocky crag]

wait lemme try

[sullen dome]

I would evaluate this to 2√2

[rocky crag]

umm

can i multiply them??

like (a-b) (a+b)
a(a+b) -b(a+b)

[sullen dome]

Expand the expression. To 6+√6*√3-√2-√2*√3

Yes ofc

[rocky crag]

ok

[sullen dome]

Then factor the roots

√2×√3+√2×√3×√3-1×√2-1×√2×√3

Then see that √3√3√2 = 3√2

Then see that √2√3 - √2√3 cancels

[rocky crag]

i got 2 √2

[sullen dome]

So you're left with 3√2 - √2

Yeah

[rocky crag]

3-1 root 2

[sullen dome]

Doesn't get more simple than that

[rocky crag]

2 root 2

gg

[sullen dome]

Hope you understood the process

[rocky crag]

yes

i did

thanks a lot for helping

umm

r u free?

√2 x 7 * cube root(7)

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[heady finch]

Can anyone help me out with this?

Adam was born on 22nd March 1982. On what day of the week was he 14 years 7 months and 8 days of age?

[heady finch]

There were 8 days, so the number of odd days would be 1. And 22 March 1982, as per my calculations, is Monday, so a day after Monday is Tuesday. But the actual answer says Sunday

[mortal trellis]

well what exactly counts as a month in this context?

22nd october?

or 7*30 days later?

either way, first I would work out on which date he is 14 years, 7 months and 8 days old and then work out which day of the week that is

given that either way you have to figure out the day of the week for some random date, better to first figure out the date and then do it

[ornate condor]

i would.. pull up a calender

[heady finch]

Yeah we don't have that luxury in the exams

[heady finch]

Yeah this

I've already figured out the day of 22 March 1982

[ornate condor]

but what other info do u have? do u work back from todays current day/date?

[heady finch]

Yeah all I have is when he was born, so now i have to calculate what the day of the week would be when he'll be 14 years 7 months and 8 days old

So 22 March 1982 is a Monday which is the first step

And then in 14 years there are 3 leap years

So the total number of days in 14 days is (11x365) + (3x366)

[mortal trellis]

84, 88, 92 and 96 are leap years

[heady finch]

Which is 5113 days

[heady finch]

Add that to 7 months and 8 days

[heady finch]

(30+31+30+31+31+30+31) + 8

Which is 222

So 5113+222=5335

5335/7 leaves a remainder of 1

And it was Monday on 22 March 1982

So bc of the 1, Monday+1 is Tuesday

[mortal trellis]

so, you forgot about 1 leapyear, so you are one day off. the 30th october 1996 was actually a wednesday

my best guess is that they treat "month" as 30 days

[heady finch]

Ah yeah you're right

[mortal trellis]

but even if they treat "month" as 30 days the answer would be saturday. hmm

[heady finch]

One sec

[heady finch]

No i think it'd be 30, 31, 30 and so on

The month after March has 30 days

After that 31

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@heady finch Has your question been resolved?

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@heady finch Has your question been resolved?

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[analog gulch]

I have this function here.

[analog gulch]

$f(x)=\max\left(-\left(x-0.5\right)^{2}\cdot 16+1,0\right)$

[ocean seal]

Game Dungeon

[analog gulch]

I want to find the average value of the function between 0, 1

I'm really not sure how to do it with the max there

[lone heart]

@analog gulch Has your question been resolved?

[tacit arch]

split the domain where the function changes definition

should help by plotting

instead of writing max, you want to write f(x) peicewise

[analog gulch]

Is there a way to get the pieces without plotting?

[tacit arch]

yes

just solve for when the first term is negative

[analog gulch]

So in this case it's $1/2\int_{0.25}^{0.75}(\max\left(-\left(x-0.5\right)^{2}\cdot 16+1,0\right))dx$ ?

[ocean seal]

Game Dungeon

[tacit arch]

show your work?

[analog gulch]

To start I don't really know much calculus. I'm doing this here for practical reasons.
Here's the function image.
it's a curve that spends between 0.25x and 0.75x above 0. It spends half the interval above 0.

[tacit arch]

i guess you're supposed to use algebra to show

$-16\left(x-\tfrac{1}{2}\right)^2 + 1 = 0$ implies $x=\tfrac{1}{4}$ or $\tfrac{3}{4}$.

[ocean seal]

riemann (eric tao for honorable)

[tacit arch]

but if it's just for non-hw, you can just plug those x values to the quadratic to confirm

i.e. check $-16\left(\tfrac{1}{4}-\tfrac{1}{2}\right)^2 + 1 = 0$

[ocean seal]

riemann (eric tao for honorable)

[analog gulch]

That's correct

[tacit arch]

once you do that, you can simplify this integral and drop the max and only integrate the quadratic

[hidden fable]

I tried this, fun exercise btw!

[tacit arch]

yea pretty fun

[hidden fable]

involves a lot of different types of math

[analog gulch]

You see the value 16. I want to be able to get the average over that interval of the function for any value there. (I'll call it b). I immediately got stuck as I couldn't even find the average for any value b.

16 was the simplest for me to go for 🙂

[analog gulch]

So knowing this how could I get the average of the function?

[tacit arch]

ultimately it comes down to solving for $x$ in $-b(x-\tfrac{1}{2})^2 + 1 = 0$

[ocean seal]

riemann (eric tao for honorable)

[tacit arch]

you can either use the quadratic formula, or solve for $x$ directly just using algebra

[ocean seal]

riemann (eric tao for honorable)

[tacit arch]

subtract 1 from both sides, divide by -b, take square root, then add 1/2

[analog gulch]

I know how to do that. I mean finding the average value of the function of the interval.

[hidden fable]

to get the avg value you need to integrate the function from 0 to 1 (and divide by 1, but it's not like that does anything haha)

so you need to define the function by parts first, so that's why finding where the function > 0 is useful

[tacit arch]

do you have an equivalent integral?

[tacit arch]

like this

[analog gulch]

Basically on my own I've learned how to get the definite integral of a simple function. Past that I'm lost.

That there was a random guess based on what I saw

[tacit arch]

can you separate what you know for sure and what is a "guess" to you?

[analog gulch]

I can do an integral. I don't know how to break them up into parts, hence why I got lost with the max.

[tacit arch]

can you show your integral once you found the roots?

your integral limits should have b in them

[analog gulch]

Not really. First I wanted to figure out how to deal with the max by just replacing the b with 16.

[tacit arch]

did you do this?

[analog gulch]

Yes.

[tacit arch]

what'd you get

[analog gulch]

oh wait that's the b one. One sec

[lone heart]

@analog gulch Has your question been resolved?

[analog gulch]

I had to leave for a bit

Though It has been a bit since I've done algebra and that wasn't fun.

$x=\frac{1}{2}(1\pm\frac{2}{\sqrt{b}})$

[ocean seal]

Game Dungeon

[analog gulch]

so what next?

[hidden fable]

you need to add 1/2

not multiply

[analog gulch]

Correct me if I'm wrong here.

$-b(x-\tfrac{1}{2})^2 + 1 = 0$

[ocean seal]

Game Dungeon

[analog gulch]

Becomes
$-b(x^2-x+1/4)+1=0$

[ocean seal]

Game Dungeon

[hidden fable]

you don't need to expand though, it's much easier if you do this

[analog gulch]

I feel some dumb lol. Expanding caused me so many issues.

[hidden fable]

dw, just remember it's easier to work with factorised expressions

[analog gulch]

Honestly though I don't even know how this helps the original problem lol

[hidden fable]

no idea, either

[analog gulch]

lol

[hidden fable]

I did something very different to solve this haha

so what I did was find the definition by parts of the function and integrate from 0 to 1
then you'd divide by 1 (the length of the interval), which does nothing, to get the avg value

[analog gulch]

!close

[lone heart]

@analog gulch Has your question been resolved?

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[weary void]

is this correct?

[chrome kite]

yup

looks good to me

[weary void]

i got 1 more question

[chrome kite]

shoot.

[weary void]

sry for messy writing

Var(x) = 1-(0.3+0.8+0.9)

nvm

[chrome kite]

i'm ngl, i'm not very good at stats

[weary void]

its -1

ok

all good

[chrome kite]

i just know the basic, E[x]

[weary void]

i think i got it

alright

[chrome kite]

nice!

[weary void]

thank you

.close

[lone heart]

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[weary void]

.reopen

[lone heart]

✅

[kindred sparrow]

.close

[weary void]

is this correct

testing

.close

[lone heart]

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[potent inlet]

Should be easy enough. I am getting 2. Idk how

[wary stream]

Post work

[potent inlet]

I take the derivative of the top, then of the bottom and then stack them.

so 2x/1

Then i plug in x=1

I get 2

[slender marten]

L'hopital's rule is not applicable here.

[potent inlet]

why

[slender marten]

The numerator is not zero when x = 1.

[potent inlet]

ahh so I just plug 1 into the equation

[slender marten]

It's not defined for x = 1 so you can't do that either.

[potent inlet]

what would I do them

[slender marten]

I figure you can rewrite it as (x^2 - 1)/(x - 1) - 8/(x - 1) and take the limit of that as x goes to 1.

The limit of the left term is 2 but the other has different sided limits, - infinity and + infinity.

[alpine sable]

So both numerator and denominator have to be 0?!

[slender marten]

Yes.

[wary stream]

For L'hopital? Yes

It needs to be in an indeterminant form

[slender marten]

Let me give you a short proof from my book. Once you see it you'll never think it could be anything otherwise.

[wary stream]

IE 0/0, infty/infty, etc

[alpine sable]

Right.

[slender marten]

Actually, have you ever seen f(b) - f(a) = f'(ξ)(b-a) for some a < ξ < b?

[alpine sable]

I have not

This sounds very complicated for me haha

[slender marten]

Well basically in L'hopital's formula it is relying on that f(a) to be zero so that f(b) = f'(ξ)(b - a) and you just let ξ go to zero. Since you do the same for the numerator the b - a part cancels to get that result.

[potent inlet]

where is the 8 come from

[slender marten]

In the numerator x^2 - 9 so I rewrote it as (x^2 - 1) - 8 and split the fractions.

[potent inlet]

Honestly, I still dont get it lol

[slender marten]

Another troll has entered the chat. <@&268886789983436800>

[potent inlet]

lol

[junior blade]

Please don't join the help channels and troll... For fucks sake

[slender marten]

6ix9ine. You know how it is.

[potent inlet]

so any idea how this problem works lol

[potent inlet]

?

[slender marten]

Snitched him up.

[slender marten]

You sometimes need to manipulate what you have to make it applicable to the things you know.

[potent inlet]

so what makes a limit not exist

[tall wing]

banned

thx

[urban pine]

when the limits from the left and right don't agree

usually with jumps

vertical asymptotes are another one (some people call it infinite jumps)

[lone heart]

@potent inlet Has your question been resolved?

[lone heart]

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[rose path]

Complete the square and find the minimum or maximum value of the quadratic function 𝑦=8−(9𝑥^2+𝑥). I did -b/2a where I put -1/2(9) which gave me an answer of -1/18 but my homework program is saying its wrong. not sure what i'm doing

[last ether]

That's not the maximum value

Or a minimum value

That's where it is located

[rose path]

so then what do i do to get the maximum value

[last ether]

Right here

It says how

[rose path]

what's the difference between that and what i did

[last ether]

You only found -b/(2a)

You didn't plug it into f(x)

The maximum value refers to the maximum output; you only found the input that gives the maximum output

[rose path]

so would it be -1/18=8−(9𝑥^2+𝑥) ?

[lone heart]

@rose path Has your question been resolved?

[last ether]

Bruh no you plug -1/18 into the quadratic

You know what function notation is?

You would plug -1/18 into your quadratic, and the output is your extremum (in this case, it's a maximum because a < 0)

[rose path]

not really, i havent done this in years.

[lone heart]

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[half nexus]

Hi!

[half nexus]

I am trying to implement a low pass filter with a DFT

However, the DFT I am currently using seems to utilise a different normalisation (phi to be precise)

Why is this happening?

[lone heart]

@half nexus Has your question been resolved?

[tacit arch]

Are we supposed to know what your dft is

[half nexus]

Python scipy DFT

[tacit arch]

Well you have to look up that definition and understand that one using the docs

It's not guaranteed to be the same as wiki's

[lone heart]

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[modern topaz]

i think i need to make the entire inequality (smth)^2 (smth)^2 (smth)^2 <0 to solve it

[lone heart]

@modern topaz Has your question been resolved?

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[worthy magnet]

not exactly sure how to approach this problem since this was the only way i could logically think of doing it

[worthy magnet]

work

[lilac nest]

For the y component of F1+F2, you made a mistake in the sum. Check it again. For the x component, you sum the forces; think about it: are they acting in the same x direction?

[worthy magnet]

so i’d subtract the sums? how about finding the magnitude, would i still be adding the sums there?

[lilac nest]

You should choose a coordinate system before starting. That way you know which components are negative and which are positive

If you're finding the magnitude you have to use the Pythagorean theorem with the x and y components, so the sign goes away anyway.

As you can see the x component of F1 is negative

[worthy magnet]

did i go about finding the magnitude and direction correct?

[lilac nest]

Which component of the F1 vector is negative? X or Y?

[worthy magnet]

y?

[lilac nest]

Look at the drawing here

Are you sure it's the y component? Isn't it above the x axis?

[worthy magnet]

if it’s just the x component that’s negative, wouldnt the answer be the same as the original since 16cos(-45) is still positive?

[lilac nest]

The first half of your message is correct

But -16cos(45) is not the same as 16cos(-45)

You shouldn't use -45 as an angle

[worthy magnet]

ah okay I see. would the way i went about, not necessarily the answer, of the magnitude and direction of F still be correct?

[lilac nest]

The procedure is correct, but the results you got aren't, since F1+F2 is wrong

If you think about it, the angle should also be less than 90°, since F is in the first quadrant, but you got 107°

Try again to sum F1+F2 with the correct numbers and you'll see that you get different values

[lilac nest]

So yeah the way you did it was correct, plug in the correct numbers and you're done👍

[worthy magnet]

perfect, thank you so much

.close

[lone heart]

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[alpine sable]

Help please

[alpine sable]

I keep getting this error.. please help

[remote heron]

are the brackets showing up on their own?

dont these usually just have parens

[alpine sable]

Im putting them there

[ornate condor]

i think u need an operation before 5.5

[alpine sable]

Alright I’ll try that

[remote heron]

dont these have mathprint

[alpine sable]

Still getting an error

I downgraded from a Ti Nspire CX to a Ti-84 plus CE

[gray isle]

that's a weird - before the 21

[ornate condor]

add an operation after the first 2?

[gray isle]

looks like its superscripted, not a normal -

[ornate condor]

i think thats the negative minus

[alpine sable]

Since its the calculator my school uses, and it was getting confusing using such a convoluted device

hello

help

[gray isle]

i think thats the negative minus
its shorter and higher

[alpine sable]

its a negative number

if I put the subtraction sign it wouldn’t work

[ornate condor]

did u

[alpine sable]

i will now

Still getting a syntax error

[ornate condor]

try just entering the whole expression n take a pic of whats the current equ

[gray isle]

put the 5.5 before the (30-1) instead of after

[alpine sable]

Ok

Still getting an error

[gray isle]

can you show your current input

[alpine sable]

-470.5 is incorrect, by the way

[gray isle]

you got rid of those ()

[alpine sable]

The formula im using is the formula for arithmetic series’

[gray isle]

missing ] in the second one

[alpine sable]

Im still getting syntax errors

[lilac nest]

There's a missing bracket at the end

[alpine sable]

I’m new to this calculator, sorry

[alpine sable]

For one of them only.

: (

[gray isle]

maybe chuck () around the 30/2

[alpine sable]

Still getting a syntax error

Let me try putting brackets around the 5.5

😦

😭

[lilac nest]

Try with all the * in the middle of each multiplication I guess

[alpine sable]

Yeah, already tried that and it still doesn’t work

Thank you all for your help

[lilac nest]

Try to ask how to use it to your professor then, Idk how to help you. Use some calculator online for your homework

[alpine sable]

its not homework

Im testing out this new calculator and I’m doing math for fun

I wont meet my teacher until the 6th 😦

It worked on my ti nspire

[lilac nest]

Oh ok, then see if there are explanations on how to use it, either in the box you got it from when you bought it or online. Maybe there is something in the settings which shouldn't be that way, idk

[alpine sable]

Alright, thank you so much

[lilac nest]

I don't know those calculator, here in Europe we don't have them and we mostly use Casio's calculators

[lilac nest]

Yw

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

got it to work

Used {} instead of [], which is pretty unusual, but it works lol

[lilac nest]

??? Super weird

[alpine sable]

Solved

.solved

!solved

?solved

[remote heron]

.close

[lilac nest]

.close

[lone heart]

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[cunning trout]

For part f) of this question,

[cunning trout]

Do i find the range and IQR

And the more lower one out of these two people would mean that they’re more consistent then the other

[void niche]

IQR probably more important than range

[cunning trout]

Mm i see

[void niche]

if i score 10 points 99 games in a row, and 0 points one game, I'd say im super consistent, despite my range being large

[cunning trout]

True

Thanks again!

.close

[lone heart]

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[cunning trout]

Quick question, is it a lower range that means more consistency or a higher one?

[cunning trout]

For example lower IQR and standard deviation means more consistency

Is it the same with range too?

Also does higher median tell us the better performance?

[lone heart]

@cunning trout Has your question been resolved?

[cunning trout]

Ok i just want to know this

[cunning trout]

<@&286206848099549185>

[kind eagle]

lower range = generally higher consistency
although range is heavily affected by outliers

[cunning trout]

Yeah so its not really a good thing to use

What about higher range ?

[kind eagle]

higher median when comparing two sets shows which data set did better generally

[cunning trout]

Oh i see

[kind eagle]

higher range = generally lower consistency, but again, outlier has a big impact

[cunning trout]

I see

I get it now

Thanks!

.close

[lone heart]

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[serene heart]

Hello! So far I reached this point, if I am correct I should substitute this in one of the 2 equations

[finite flax]

why are you using pen 😱

[north glacier]

,rotate

[serene heart]

Cuz pen!

I forgot to cancel 5 hahaha

[finite flax]

what problem was this for?

5?

[serene heart]

Number 5

[finite flax]

and what is your question

[serene heart]

Here

I should substitute to eitherequations right

[lone heart]

@serene heart Has your question been resolved?

[finite flax]

it's (3,6), or (6,5), by inspection

I actually messed it up a little

there are 2 points that will work

I tried it with distance formula squared (circle formula) and messed it up, somehow

[lone heart]

@serene heart Has your question been resolved?

[serene heart]

I see

Alrighty

[lone heart]

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[lone heart]

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[native nebula]

Help

[native nebula]

How's this happenin'?

[gray isle]

$a^{pq} = {(a^p)}^q$

[ocean seal]

ℝamonov

[native nebula]

That i know

But it's kinda awkward

How can we write the whole squared thing?

[slender gull]

Let 2^{n-1} = p and 2 = q

[gray isle]

$a^{\blue{2^{n-1}}\cdot \red{2}} = \br{a^{\blue{2^{n-1}}}}^{\red{2}}$

[ocean seal]

ℝamonov

[wind cloak]

redpenblackpen

or is it blackpenredpen

I don't remember

[native nebula]

Is it equal to a^2n?

[alpine sable]

no

it is equal to $a^2^{n}$

I can't make the indices

a^2^n

[native nebula]

Is it correct?

[slender gull]

Yes.

[native nebula]

Am i doing correct?

@gray isle

Am i doing this correct?

[gray isle]

yeh

[native nebula]

What will the next step?

Am confused

[native nebula]

But a^2^n is not coming

[gray isle]

depends on the end goal

all the manipulation of that expression so far has been valid

when you had
a^(2^(n-1+1)) that simplifies to a^2^n

[native nebula]

I am asking about this one
How this expression boils down to a^2^n?

[gray isle]

well going this route sorta works against you in simplification

you could factor out 2^(n-1) in the power and/or note that you have 2 multiples of 2^(n-1)

which gets you back to what you had earlier

which simplifies to a^2^n from considering other exponent laws

[native nebula]

Can you walk me through it?

Like how to proceed?

[gray isle]

2^(n-1) and 2^(n-1) are like terms

2^(n-1) + 2^(n-1) = 2 * 2^(n-1)

which is pretty much what you had ealier

considering the law you applied earlier would be the most efficient way to simplify that

2*2^(n-1) = 2^(n-1+1) = 2^n

[native nebula]

Yeah that i got but trying it this way too

[gray isle]

there are things that you "can" do,
and there are things that are helpful

going this route isn't helpful

[native nebula]

How can you write this?

[gray isle]

k + k = 2k

[native nebula]

Ohh ohh

Didn't see that coming

Thanx

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[charred pelican]

how do i dk this

[charred pelican]

ive already opened the brackets

just dont know what to do next

[lusty whale]

have you tried substitution

[charred pelican]

no

but i can since theres no equals to

[lusty whale]

ok

well what can you do when you divide two fractions?

[charred pelican]

i think reciprocate and cross multiply?

[lusty whale]

well you can 'flip' the right fraction and the division becomes multiplication

hence you would be able to cancel the g and h

[charred pelican]

ok thx

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[signal sable]

How do I do this?

[lone heart]

@signal sable Has your question been resolved?

[vague seal]

a. or b.

for a. you should probably notice it's measuring energy for other things as well, like the energy of water, room temperature (informal language).

for b. you should set up an integral modelling the total energy emitted by the model described in the question and solve the initial value problem

the result will be an improper integral from time t=1 to t=inf

hint: the value of k is immediately determined by the boundary condition; the value of the constant +C is determined by the t=2 case. Then you can calculate the improper integral and find the total energy from t=1 to t=inf

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[fervent ferry]

i have 52 cards, i choose 4 cards, in how many ways i can pick exactly 2 diamonds?

[fervent ferry]

how should i think about this problem?
i would do 13C2 for the first 2 of the 4 spots, and 39! for the other 2 spots.
so the tota number of combinations would be 13C2 * 39!

[marsh rapids]

39C2. No reason to be different

[fervent ferry]

auch, yes you're right

and if it were:
"at least 2 diamonds" ?

it would be the complement of exactly 1

which would be... 13C4 - 13

right?

no. wait.

i don't see how should i pick my sample space.

i'm struggling with this

13C4 - 13 it makes sense to me, but if i pick 2 diamonds i would have to mult by 39C2, if i pick 3 diamonds i would have to mult by 39

[spring gust]

since order isn't important, you just need to choose which two diamonds you are taking and which two non-diamonds you're taking

so 39C2 for the nons and 13C2 for the diamonds

if it were 'at least 2 diamonds', then you can do this two ways: add the samples where you choose 3 diamonds, and where you choose 4 diamonds, or do (full sample space) - choose 1 diamond - choose 0 diamonds

[fervent ferry]

ok so for first method it would be 13C2 + 13C3 + 13C4 ?

[spring gust]

there are 13c2 ways to choose two diamonds cards, but you also need to account for the ways of choose the rest of the non diamonds

[fervent ferry]

ok so i think is what i was suggesting before let me try again

13C2 * 39C2 + 13C3 * 39 + 13C4

right?

for second method, full sample space would be ... 52C4 right?

and would be
52C4 - 13C1 - 39C4

[spring gust]

-13C1 * 39C3 ***

but yeah

it's a good trick to check you're not making mistakes

if the two methods aren't equal you missed something

[fervent ferry]

damn ok

thanks a lot

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[alpine sable]

What have you tried?

[normal bridge]

I think I have to find the area of the larger triangle first

And for that I need the x intercept of the other line

[alpine sable]

Correct and then subtract the area of triangle which is in first quadrant

[normal bridge]

But I'm not sure how to find the equation of that line with just one point

[alpine sable]

It has intercepted x axis at 6

[normal bridge]

But I need this intercept to find the base length right

[alpine sable]

Oh my bad

Okay

So you can find the eqn of other line right?

And both of them are perpendicular. product of their slopes will be -1

Using that you will get slope of desired line then use slope point formula to find eqn of line

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@normal bridge Has your question been resolved?

[normal bridge]

thanks :)

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[tough plume]

I don’t know how to answer this question if they don’t give me the SD

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@tough plume Has your question been resolved?

[tough plume]

<@&286206848099549185>

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@tough plume Has your question been resolved?

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[lean nacelle]

please lemme know if blurry

[oak perch]

Just count how many x+2y+3z=7 non-negative solutions, are there

[lean nacelle]

...

eh?-

[oak perch]

?

[lean nacelle]

uhm- a bit for info please-

*more

[oak perch]

z=0, 4 solutions
z=1, 3 solutions
z=2, 1 solutions, so 4+3+1=8

[lean nacelle]

hmm

[oak perch]

Oh

I forgot order matters

Nvm

Rethinking

[lean nacelle]

the answer is actually 44- if it helps

[oak perch]

Please withdraw it

I want to calculate it

[lean nacelle]

withdraw?-

[oak perch]

Nvm saw it already

[lean nacelle]

...

oki-

[oak perch]

Divide it into different cases by how many 3-steps he makes

Case 0: he doesn’t make 3-step

[lean nacelle]

ok?

[oak perch]

Then 1+6+5C2+4=21 solutions

Case 1: he makes exactly one 3-step:

(1+3+1)+(1+2)+(1+2+1)+(1+2)+(1+3+1)=20 solutions

Case 2: he makes exactly two 3-steps

3 solutions

44 in total

[prime badge]

||you just compute tribonacci, 3 cases for which step was last||

[oak perch]

4 terms represent 0 many/1 many/2 many/3many 2 steps he makes

[oak perch]

5 terms represent the location of that 3-steps is (123),(234),…,(567) respectively

Should be clear now

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[lean nacelle]

i just came back-

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[crisp zealot]

Could someone please explain why I got positive two on the one side and negative two on the other?

[icy token]

wait

what have you done

[alpine sable]

stands to reason

0=2 $\to$ 0=-2

[ocean seal]

Zermelon

[crisp zealot]

On the left side I started with the constants but in the right side I started with the variables

[alpine sable]

you just swapped the places

[abstract fractal]

It's effectively the same as "multiplying both sides by -1." The equation is inconsistent, don't expect consistency

[alpine sable]

add 2 to both sides

and voila

[icy token]

why would you solve the exact same thing twice, but swapped sides?

[crisp zealot]

So when I write an exam how would I know which one is right, for this one my teacher said -2 is the correct one

[alpine sable]

both are wrong

[crisp zealot]

My method is the one on the left and the other method is my teacher's

[alpine sable]

you can work backwards to the first equation from either

[icy token]

the equation's not right

[crisp zealot]

The equation isn't supposed to add up

[alpine sable]

oh

[icy token]

ah that makes more sense

[crisp zealot]

It's supposed to be not equal

[icy token]

argue with your teacher

[alpine sable]

you're trying to prove a contradiction?

both are correct then

[icy token]

say that by taking away 2 from both sides, you'd get 0 = -2 anyways

[crisp zealot]

Ohh

Thank you guys so much

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[icy token]

.close

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[wind cloak]

Stop trolling people here give their time out to help those genuinely in need

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[alpine sable]

how do I test this integral for convergence?

[keen plinth]

$e^x \geqslant 1 + x$

[ocean seal]

夢雪

[keen plinth]

use that to create a bound on the integrand

you should find that it converges

[alpine sable]

hmm ok but how did you come up with this one

[keen plinth]

the specific form of e^something - 1

you usually want to use some inequality involving the taylor series expansion of e^x

[alpine sable]

what does taylor expansion have to do with this

[keen plinth]

1 + x is the first 2 terms of the taylor expansion of e^x

[alpine sable]

alright thanks

$close

.close

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[digital moon]

A car with a mass of 2,000 kg is moving at a speed of 16 m/s. If in case
The truck brakes with a force of 8,000 N. Determine how far the truck has traveled
starting to brake until it stops?

[digital moon]

i cant find the answer since the time wasnt mentioned

[slow hound]

is the car the truck

[alpine sable]

here we had been given force

we can use force to calculate accelaration

[alpine sable]

does it matter dude.😂

[slow hound]

yeah it does

you dont know the mass of the truck otherwise 😂

how would you find the acceleration

[digital moon]

i didnt make the question

[alpine sable]

hahaa..that's why its a typo error

[alpine sable]

look at it again ...maybe its only truck or car

[digital moon]

yes it is

[alpine sable]

what? both

[digital moon]

just change the car to truck

[alpine sable]

ohh fine..then

you can find accelaration(a) with the help of given force

can you??

[digital moon]

let me try

[alpine sable]

sure..

[digital moon]

wait

i already done that

i am stuck with the s formula

[alpine sable]

you can use the 3rd one

[digital moon]

i cant find the time so cant finish the equation

[alpine sable]

v^2-u^2=2as

no need of t

[digital moon]

what is that

is it from vf'2 = vi2 + 2as

?

[alpine sable]

yeah..

i just omited i and f

[digital moon]

its the same meaning

[alpine sable]

cz v represents final and u initial..genrally

[digital moon]

u is vf

yeah i know

[alpine sable]

nah u is vi

[digital moon]

wait what