help-33

[cunning fiber]

jack, queen, king

[velvet surge]

(1 point) (Note that an Ace is considered a face card for this problem)

3/52 didnt work neither did 4/52

[cunning fiber]

Are you sure you don’t need to simplify

[velvet surge]

p sure

[cunning fiber]

Just try 1/13

[velvet surge]

its 8/52

[cunning fiber]

oh wait

there's two black suits lol

clubs and spades

💀

god I haven't played cards in so long

[velvet surge]

P( black or a face card )

[sleek lake]

yes, sometimes they do 4 colors, but 2 is way more common

[velvet surge]

what are u on

anyways

[sleek lake]

8 face cards in each color

26/52 + 8/52

wait that's d

b is 8/52

[velvet surge]

i know

LAST ONE P( face card or a number card )

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[marsh citrus]

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[elfin berry]

woods

[marsh citrus]

@fresh fiber Has your question been resolved?

[elfin berry]

woods

[marsh citrus]

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[cosmic raptor]

how would you guys find the integral for e^-0.5t?

[cosmic raptor]

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[cosmic raptor]

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✅

[atomic ruin]

i gotchu

oops

forgot the dt

at the top

[cosmic raptor]

sorry, i'm not that good at this one yet as i'm brushing up on integrals, is this change of var method?

[atomic ruin]

u-substitution

[cosmic raptor]

got it now, thanks!

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[cosmic raptor]

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[marsh citrus]

✅

[cosmic raptor]

actually, @atomic ruin how would you go about finding integral of c(t) = 16te^-0.5t ?

[atomic ruin]

Wait lemme do it

@cosmic raptor here by parts is used, as well as the integral earlier

i suppose this is a part a,b question?

[cosmic raptor]

exploring ways to solve this lol

i think on the solution provided, they did e^0.5t^2

[atomic ruin]

Ohh

did my ans help

[cosmic raptor]

it did help, learnt something new which kind of help me understand the question a bit more

[atomic ruin]

I dont know how you would do the question though lol

[cosmic raptor]

lemme do it, and send it here

[atomic ruin]

are you supposed to integrate it from 0 to 3 then multiply by 6

[cosmic raptor]

i somehow got the calculation right on the integral, but I'm not sure why it's 6mg/my answer

[atomic ruin]

Are you sure thats the problem

the question

[cosmic raptor]

yup, that's what is on the papers

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[sullen plank]

why is this true

[high ledge]

(2/3)^∞ = 0

[sullen plank]

yeah why is that true

is that because the fraction keeps becoming smaller as they multiply?

[shell ether]

2^∞ is way too small than 3^∞

[high ledge]

(2/3)^∞ = 2^∞/3^∞

[torpid orchid]

Approaches zero

[high ledge]

3 > 2 so the denominator is bigger

[sullen plank]

ah makes sense and 3^ inf grows faster so term goes to 0

[high ledge]

Yes

[sullen plank]

thanks

[high ledge]

Basically because the denominator will be a lot bigger

[sullen plank]

yes, thank you

[shell ether]

Actually you could say that 2/3 is 0,(6)

So number that is less than 1 bigger 0 in n power will approach zero with every step

Kinda

[sullen plank]

yeah

[torpid orchid]

Only happens when 0 < a < 1

[sullen plank]

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[jaunty dirge]

$$ 5\fraq{-1^n}{2^3n} $$ The series from n = 0 to inf (dont know how to do this in LaTeX) Am I correct when I say that the sum is 40/9?

[jaunty dirge]

$$ 5 * \fraq{-1^n}{2^3n} $$

[waxen remnant]

$5\cdot \frac {-1^n} {2^3n}$

[elfin berry]

znsn

[waxen remnant]

$5\cdot \frac {(-1)^n} {2^{3n}}$

this?

[jaunty dirge]

Yep :)

[elfin berry]

znsn

[jaunty dirge]

The series from n = 0 to inf (dont know how to do this in LaTeX) Am I correct when I say that the sum is 40/9?

[waxen remnant]

ill check give me a moment

[jaunty dirge]

Ty

[waxen remnant]

40/9

yeah

i got that

too

[jaunty dirge]

Nice

[waxen remnant]

because it's essentially $5\cdot \left(\frac {-1} {8}\right)^n$

[elfin berry]

znsn

[waxen remnant]

and then you use your geometric series formula

[jaunty dirge]

Yeah exactly

[waxen remnant]

👍

[jaunty dirge]

And this one?

$ \fraq{3 \cdot 2^n}{n!} $

AHHH WORK

[waxen remnant]

$ \frac{3 \cdot 2^n} {n!} $

$\frac{3 \cdot 2^n} {n!}$

[elfin berry]

znsn

[jaunty dirge]

I got 3e^2

[waxen remnant]

ngl i havent learned this yet 😭

[jaunty dirge]

Ohh, no worrys

[waxen remnant]

how did u do it

just curious

[jaunty dirge]

So the series for $ e^x = x^n \cdot 1/n! $

$ e^x = x^n \cdot 1/n! $

$e^x = x^n \cdot 1/n!$

[elfin berry]

Kiip256

[jaunty dirge]

So it thought that its would be 3 * e^2

[waxen remnant]

that does make sense using that formula i think

[jaunty dirge]

Okay nice

Thanks for the help!

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[winged pier]

I need some help on where to go next, solving for theta for 0<theya<360

Trig identities but idk what to do

[still temple]

factorise

i think

[winged pier]

I can factor everything

Can’t

[still temple]

i think you can

[winged pier]

Or maybe I can I guess

I will factor sin out of cos so it’s cos/sin right

[still temple]

with sinx(2cosx - 1) + sqrt(3)/2(2cosx - 1) = 0

[winged pier]

what

How’d you get the 2nd part

[still temple]

i subtracted sqrt3/2

on both sides

and factored out sqrt3/2

[winged pier]

ok

OHHHH

Wow how did you know though

[serene basalt]

Awesome bro i am also trying to figure out

How you do that

[winged pier]

I’m wondering how you knew that you could get similar terms

[still temple]

its like a pattern

[serene basalt]

Are you alien

[still temple]

you can get there with practice

no

[serene basalt]

Ya

[winged pier]

You have good foresight I suppose

Ok I factored

I got pi/6

Pi/3*

[native sage]

Nope

[still temple]

make sure you get all of the roots

[native sage]

The angle lies in 4th quadrant

[native sage]

As cos is postive and sin is negative

[winged pier]

Theta needs to be from 0 to 360 degrees

[native sage]

Yes so the angle should be 300 degrees

[winged pier]

I got 330 degrees 210 degrees

These are all of my solutions does this seem correct

[native sage]

How did you get 330?

For sin?

[winged pier]

-pi/6 + 2pi

[native sage]

Why pi/6?

pi/3 right?

[winged pier]

because sin(-pi/6) = -sqrt(3/2)

[native sage]

pi/3 gives √3/2 right?

[winged pier]

Pi/6 does

For sin

[native sage]

Pi/6 is what degrees?

[winged pier]

wait

[native sage]

Yes

[winged pier]

AAA

I mixed it up I think so

just switch the cos and sin

[native sage]

Take your time

So 300 degrees

And for cos we can't write 120 degrees

Because then it will give -1/2

[native sage]

Cos is postive in 1st and 4th quadrant only that is between 0 and 90 and 270 and 360 degrees

[winged pier]

Is this amazing

,rotate

[native sage]

Yep

[winged pier]

Wow

Great solution

[native sage]

So the common solution is 300 degrees

[winged pier]

Does that mean anything

That’s just (0)(0)

[native sage]

Yes

[winged pier]

is that the final answer? Why can’t we have 3 answers

[native sage]

It means that at 300 degrees, both the conditions are satisfied

[winged pier]

But that doesn’t mean anything right

[native sage]

Because we need to have sin negative and cos positive simultaneously

[winged pier]

So the other answers are wrong?

[native sage]

It means that at 300 degrees, sin is -√3/2 and cos is 1/2

[winged pier]

If I put in 60 degrees in the original equation would it be incorrect

[native sage]

Try doing it

[winged pier]

Please no

[native sage]

Wait I will do it for you

Sin2theta will give sin120 which is √3/2

[winged pier]

wait it’s correct

I did it with calculator and it’s correct so why isn’t it a solution

Btw the question defines theta as 0<theta<360

[native sage]

Oh yes, then include the others too

[winged pier]

👍

Nicee

ok tytytytyty

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[south lance]

I need to solve via PMI without using the binomial thereom and I’m struggling for the first couple steps

[marsh citrus]

@south lance Has your question been resolved?

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[stuck solar]

so this is a explanation for this question

[stuck solar]

i dont get how in the 3rd line 2 is there before sin

but in the next line its gone

[novel fjord]

zero product property

oh it says that

lmfao that probably doesn't help

[stuck solar]

right but how does the 2 just disappear

[novel fjord]

if you have 2ab=0, you set 2a=0 and b=0 to solve

2a=0 simplifies to a=0

[stuck solar]

oh that makes sense

they just assumed i already knew that

lol

ty tho

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[kind goblet]

i have a graph function, but it does not output the precise output i want. I need the values to get rounded the nearest multiple of 0.2 precisely
desmos:https://www.desmos.com/calculator/jgdzhlq3je
feels like a simple thing my brain is choking at

[marsh citrus]

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@kind goblet Has your question been resolved?

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[still temple]

i found a solution to my problem but idk how they get after they rationalize the limit can someone help me i will guess i am stupid and its very easy

[paper hornet]

Find the limit

[glacial hedge]

!occupied

[marsh citrus]

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).

[still temple]

i dont know with what they rationalize

i would really appreciate if someone will help me

nvm found

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[exotic matrix]

How do i direvate this equation?

[quaint elm]

,tex .diff rules

[elfin berry]

ヘイリー

[quaint elm]

quotient rule

[exotic matrix]

we got tought that we should move the denominator up and then direvate ti

but i dont know how to do that

do you know how to do that? @quaint elm

[quaint elm]

i mean you could write it as like

$f(x) = x(x^2+1)\inv$ sure

[elfin berry]

ヘイリー

[quaint elm]

and then use the product rule

[marsh citrus]

@exotic matrix Has your question been resolved?

[exotic matrix]

ok im gonna try to solve then ill get back to u

i cant figure it out

ffuck it

.close

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[fathom scaffold]

Hey! I am looking at a couple of series that "diverge in slightly different ways." I am curious as to what is hapenning to the sequence of partial sums in each respective case.

1. n = 1; E; (infinity) (2n)
2. n = 0; E; (infinity) (-1)^n
3. n=1; E; (infinity) (((-1)^n)(2n)
   sorry i am not very sure how to do sigma notation on a computer haha

[marsh citrus]

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[fathom scaffold]

sure man whateber

[marsh citrus]

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[valid cape]

$\int (sin^8x-cos^8x-4sin^6x)dx$

[elfin berry]

FungusDesu

[valid cape]

i would like to check my answer

i got $-\frac1{48}\sin{6x}-\frac7{16}\sin{2x}+\frac16\sin^5x\cos{x}+\frac5{24}\sin^3x\cos{x}-\frac5{16}x+\frac5{32}\sin{2x}+C$

[elfin berry]

FungusDesu

[still temple]

,w integral sin^8(x) - cos^8(x) - 4sin^6(x)

[elfin berry]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Jan 24, 2024, 1:10 AM

[still temple]

hmmmmm

[valid cape]

interesting, how is it so short

[valid cape]

im confident with the first 2 terms, what im concerned is thr sin^6x

[still temple]

they use identities to simplify the integral

[valid cape]

ok while typing i suddenly found the issue

i was too focused on the reduction that i forgot to multiply the whole thing by 4

[marsh citrus]

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[marsh citrus]

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[still temple]

sin(x)/1-sin^2(x) + tan(x) = tan(x)(1+cos(x))/cos(x) I need to prove that its equal

[still temple]

Can someone help me prove that this is equal :

sin(x)/1-sin^2(x) + tan(x) = tan(x)(1+cos(x))/cos(x)

I dont need to find x

just have to make either sides equal

for example sin(x) = sin(x)

after that if yall got time

can someone do this one too

(1+tan^2) / (1-tan^2) = 1/cos^2

,w (1+tan^2) / (1-tan^2) = 1/cos^2

[elfin berry]

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Requested by adqm0001

[dusk coral]

Sure!

what have you done so far?

[marsh citrus]

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[static quarry]

help

[static quarry]

for u sub which one do I use

[upbeat pasture]

Lol why do you want to sub

Just do 7/e^3x - e^-2x

And integrate

[static quarry]

wait then when do you want to use u-sub for these type of problems?

[upbeat pasture]

Well we use u sub when we don't know the results

But we can simplify it easily in this ques

[static quarry]

wait how did you get the -e^-2x

[upbeat pasture]

e^x / e^3x ?

[static quarry]

oh im so dumb ty

[upbeat pasture]

np

[marsh citrus]

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[wet peak]

How did theyu get this value of 6sqrt2

[delicate saffron]

well

[wet peak]

well?

[delicate saffron]

a 45-45-90 triangle always has the same ratio

and that is 1-1-sqrt(2)

if im not mistaken

now because 12 is the hypotenuse

you divide by sqrt(2) to get the legs

[wet peak]

ok

so 12/sqrt2

?

[delicate saffron]

yeah

[wet peak]

ok

[delicate saffron]

cause the hypotenuse is always sqrt(2) times bigger than the leg

[wet peak]

ok thanks

can you help wioth another?

[delicate saffron]

sure which one

[wet peak]

how does that turn into that

[delicate saffron]

power rule

when you take the derivative

3 comes down

[wet peak]

can you show me

[quaint elm]

,tex .diff rules

[elfin berry]

ヘイリー

[static quarry]

are you asking where dh/dt comes from?

[wet peak]

so I just took the deriv of it

[wet peak]

yea

and how it turned to 9

[delicate saffron]

oh your deriving with respect to time

any variable thats not t has to be written as dh/dt

or like whatever the variable is over t

[wet peak]

but how com eit just shows up

[delicate saffron]

let me try explaining it this way

[wet peak]

ok

[delicate saffron]

any time you derive, your technically doing it with respect to a certain term

but you see how when you do this, dx/dx is just 1

[wet peak]

oh

yea

[delicate saffron]

so thats why you dont typically write it

[wet peak]

ok yea I remember this

[delicate saffron]

so same concept

but both variables are with respect to time

so since time isnt any of the variables you have to put them over dt

[quaint elm]

it took me so long to realize this

[wet peak]

ok thanks

I think I am good for now

I might have another give me like 5 min

[delicate saffron]

well i gotta go so too bad lol XD

someone else will probably help

[wet peak]

ok

[marsh citrus]

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[marsh citrus]

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[dark cairn]

Is the problem asking for X?

[sinful thistle]

yes

use the pythagorean theorem

[dark cairn]

so

like

[sinful thistle]

?

[dark cairn]

83 sqr + b squared

= 158 sqr?

[sinful thistle]

yes

[dark cairn]

That seems like too big of an answer

are you sure?

[sinful thistle]

yes

[dark cairn]

Can you tell me the answer and we compare ours?

[sinful thistle]

sqrt (158^2 - 83^2)

[dark cairn]

so I don't get the problem wrong?

[sinful thistle]

is x

[sinful thistle]

okay, what's yours

[dark cairn]

18075

?

[sinful thistle]

what-

okay look

[dark cairn]

@sinful thistle

[sinful thistle]

x SQUARED

plus 83 squared is 158 squared

so you need to take the root as well

[dark cairn]

wait

so whats ur answer

18075 sqr root?

@sinful thistle

[sinful thistle]

yes

[dark cairn]

so

is it 355??

is that your answer as well??

[sinful thistle]

no

[dark cairn]

what is ti?

[sinful thistle]

how is that the sqrt of 18085

[dark cairn]

*it?

[sinful thistle]

*18075

use a calculator

dude

[dark cairn]

ok

[sinful thistle]

what?

no-

that'd be around 120,000

;-;

[dark cairn]

@sinful thistle can you just tell me the answer you got??

Because I understand now

idk abt the calculators and stuff

@sinful thistle

[sinful thistle]

my god

yes lol

dont @sinful thistle me so much XD

but yea ur correct

gj

[dark cairn]

sorry

[sinful thistle]

do .close if ur done

[sinful thistle]

nah lol

[sinful thistle]

np

[dark cairn]

wait

so wahts the answer?

[sinful thistle]

...
355 squared is approx 120,000

[sinful thistle]

134.443

cm

[dark cairn]

rounded

to nearest tenth

so

134.4 right??

[sinful thistle]

yes

[dark cairn]

thanks

.close

[marsh citrus]

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[dark cairn]

bye @sinful thistle 😉

[sinful thistle]

bye

[marsh citrus]

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[lime night]

just wondering if i did these 2 limit questions correctly thanks!

[quiet anvil]

@lime night I'm reviewing the first one, and I'm not sure it's correct.

Specifically, if you know the chain rule, I would expect the limit to need to change as well

[still temple]

in part(a) when you differentiate f(2+5x), you won't be getting just that there would be an extra term think with chain rule

similar mistake has been done in part(b) as well

also in part(b) you wrote the derivative of f(2) to be f'(2) which is not correct

[lime night]

hmmm okay i'll try and fix it thanks!

here is my work with my attempt for accounting the chain rule!

i still feel like i’m forgetting something though 😿😿😿

[marsh citrus]

@lime night Has your question been resolved?

[still temple]

...

[lime night]

wait how come the derivative of f(2) isnt f'(2) 😅

[marsh citrus]

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[cosmic raptor]

(\int_{0}^{8} 700+4/5*x^3,dx) - lol i tried

[cosmic raptor]

what's the integral of 700+4/5(x^3)dx

0 --> 8

[umbral scroll]

integrate 700
integrate (x^3), times it by (4/5)
add them together
use the limits on that

[bleak ibex]

$\int_{0}^{8} 700+4/5*x^3,dx$

[elfin berry]

Jelle

[iron meadow]

you need to surround tex code between $ to get the bot to render it

[bleak ibex]

$\int_{0}^{8} 700+\frac45\cdot x^3,dx$

[cosmic raptor]

Ohhh gotcha

[elfin berry]

Jelle

[cosmic raptor]

This is what I gor

but apparently it's wrong

[bleak ibex]

Your calculation in the second integral is wrong

It isn't 655.36

[cosmic raptor]

ahh so it is right, it's just the calculator

i'm not sure why but my calculator is acting up

I did first integral + second integral altogether and it gave me a whole different answer

[bleak ibex]

What did you put in?

You might have put it in incorrectly

[cosmic raptor]

i did 700*8

then + second integral

[bleak ibex]

And you are sure Ans has value 700 * 8 = 5600?

[cosmic raptor]

i'm not sure if it has value of 5600

but my previous calculation resulted in 5600

[cosmic raptor]

so Ans = 5600

[bleak ibex]

I think you added the second integral twice

You might have pressed '=' twice

[cosmic raptor]

i haven't pressed it twice

kind of weird

lost a couple of marks before on a test because of it

i might just get a new calculator

[bleak ibex]

Yeah, maybe try it on someone else's calculator first

[cosmic raptor]

i still keep getting wrong answer

anyways, as long as I know the method

lol

.close

[marsh citrus]

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[bleak ibex]

Yes

[marsh citrus]

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[leaden jewel]

How do I solve this

[cosmic raptor]

what's the answer?

[grim hemlock]

https://www.integral-calculator.com/

[sand fable]

oh that is terrible

[marsh citrus]

@leaden jewel Has your question been resolved?

[sand fable]

are you familiar with trig substitutions that look similar to this?

are you familiar with hyperbolic trigonometric functions (like sinhx and coshx)

in any case, i'd start by expanding the denominator

[leaden jewel]

Well I don't recognise any trig subs in that integral

I have read on sinhx and coshx before though

[leaden jewel]

Do I take x as cos?

[sand fable]

to be honest no matter how you do this integral it's pretty bad

but using hyperbolic trig here is probably the "best" way to do it..?

[sand fable]

when you see indefinite integrals with square rooted polynomials (my bad i mean quadratics) in the denominator, you want to expand the polynomial and complete the square

[leaden jewel]

Ohhh right

I did try that but I got a pretty weird equation

[marsh citrus]

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[still temple]

If P and Q are equivalent statements, then does non(Q) imply non(P)

[quaint elm]

yep

[still temple]

is there a specific reason

idk

lol

P is true if and only if Q is true, then if Q is false, P has to be false, correct?

yes

[still temple]

Are you talking about $P \equiv Q$ to be clear?

[still temple]

because thats shorthand for $P \Iff Q$ which is true for all P's and Q's

[still temple]

so like, P or Q don't have to be anything here

P being true means Q can be true or false, it doesn't matter; the outcome is still true

and vice versa

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

i dont know this notation

[still temple]

im talkimg abt this

[still temple]

They are the same

[still temple]

anyways what i said here holds

[marsh citrus]

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[primal heath]

where did the 1/7 come from

[marsh citrus]

Please don't occupy multiple help channels.

@primal heath Has your question been resolved?

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[primal heath]

where did i get this wrong?

[quaint elm]

look up inverse of 2x2 matrix again

[primal heath]

its not [-4 -10],[-4,-9]?

[quaint elm]

did you do what i asked you to do?

[primal heath]

[-9, 4] [10, -9]

[quaint elm]

are you just guessing now?

[primal heath]

no XDD

its stil wrong?

XDD im trying out everything

[quaint elm]

well the 10 seems to have run away...

[quaint elm]

yes that's what I mean by "just guessing."

[primal heath]

LMFAOOOO

sryyy typo

[quaint elm]

go physically look at your book and find the formula for inverse of a 2x2 matrix

[primal heath]

srys ry didnt realize

[quaint elm]

and then come back and write it down here so I know youve done that

[primal heath]

ITS STILL NOT THIS? XDDD

[quaint elm]

why are there two 9s? do what i say, or I'm done helping you

[primal heath]

i redid eveyrthin

[ −9 4]
​[10 -4]

[quaint elm]

^^

[primal heath]

its wrong...

[marsh citrus]

@primal heath Has your question been resolved?

[iron meadow]

original advice still applies lol

[marsh citrus]

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[wise oxide]

What’s wrong with this

I only got 2/4 points for it

And also this

[marsh citrus]

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[burnt geyser]

Can anyone give me some tips on where to start with this first question?

[burnt geyser]

Do I somehow need to use a×b=|a||b|cos(y)

This is what the triangle looks like

[burnt geyser]

<@&286206848099549185>

[marsh citrus]

@burnt geyser Has your question been resolved?

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[brave marsh]

It goes to infinity, but the limits tell you it'll look like y=x on the long run.

To the right at least

I'm not sure what you mean by coordinates.

The limits you computed compare the functions sqrt(x^2+4) and x for large x, and the limit being 1 means that they will have the same growth asymptotically, i.e. there is an asymptote for y=x as x gets large.

The other limit essentially tells you that the asymptote as -x gets large is y=-x, asymptotically.

Both are the case on your graph.

[brave marsh]

The S.A. are y=x and y=-x

That's what the limits you computed tell you

In general, if you have some function $f$ and you can find some function $g$ such that $\lim_{x\to\infty} \frac{f(x)}{g(x)} = 1$, then you have an asymptote like $y=g(x)$.

[elfin berry]

Azyrashacorki

[brave marsh]

It's not like that. The asymptote is y=x, that's it.

If you draw y=x on the graph, you'll see it gets close to it as you go to the right.

y=x is a line.

Slope is 1

[marsh citrus]

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[neat mortar]

needing help with this strong math induction proof

[neat mortar]

the image on the right is where im at right now

[marsh citrus]

@neat mortar Has your question been resolved?

[neat mortar]

<@&286206848099549185> hello would anyone be able to help with this?

[mint gust]

Have you proven n+1?

[neat mortar]

no i havent

[mint gust]

Ok let me try that

[neat mortar]

okay thank you!

these are from my lecture slides. im loosely basing it off this

[mint gust]

Is your question missing a bit of information?

[neat mortar]

thats all i was given on the assignment sheet

[mint gust]

I don’t see a sequence given…

And it looks like you do have a sequence on your slides

Gimme a moment.

[neat mortar]

no problem! yea sorry it looks like thats everything ive been given for this question

[burnt minnow]

Gimme like a minute and I cab help if u want

[mint gust]

I think I just solved it?

I don’t know if I should be writing my answers, maybe I should give you some ideas to start with?

sorry that took too long

[neat mortar]

yea i think i just need to be pointed in th e right direction for where i left off

no worries i appreciate your help!

[mint gust]

Alright, as we know An=(An-1)+12(An-2)

Uh it’s actually hard to phrase inductions

I will send you a part of my proof

[neat mortar]

lol sounds good, i feel dumb asking for help with these cause i never know how to word any of it either

[mint gust]

Still with me?

[neat mortar]

yup i got to that part

[mint gust]

Just subbing in what the question told us

[mint gust]

Good

[neat mortar]

just simplifying it i kept getting messed up

[mint gust]

Now we extract powers from them

[neat mortar]

as in use logs to bring the exponent down?

[mint gust]

Nope

It’s much more simpler than that

[neat mortar]

ohhhhhhhhhhhhhhh

[mint gust]

U see?

[neat mortar]

yea i 100% wouldve never thought of that

[mint gust]

If you have some new ideas, you can try going on yourself

Collect some terms

And u should be good

[neat mortar]

yea i think i can finish it up from there, thank you so much that was a huge help!

[mint gust]

👍

Good luck

[neat mortar]

appreciate you taking the time haha

[mint gust]

Glad to help!

[neat mortar]

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[quasi wedge]

Can someone help me with this problem

[marsh citrus]

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[vagrant gull]

!msgdel

[marsh citrus]

The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.

[marsh citrus]

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[winged pier]

I saw a question on here that i also had a question about but i lost it

[winged pier]

It was |5-a| = |-a+3| I think

hold on

Aaa idk but I’m wondering how to solve problems with double absolute values like that

[main idol]

Do all 4 cases

[winged pier]

Is it +|+,+|-,-|+,-|- ?

[sand fable]

yes but there’s technically only 2 since 1-4 and 2-3 are just additive inversing both sides

[winged pier]

👍 so only 2 solutions?

[sand fable]

2 cases to consider, not necessarily solutions

[winged pier]

Right

5-a = -a+3

That doesn’t work

5-a = a-3

A = 4

ok very nice I guess that’s the only solve?

@sand fable

[sand fable]

yes

[winged pier]

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[fossil badger]

Could someone help me w the piecewise function (or however you spell that)

If the Og graph line thing is going to the left does that not mean that x rn (-3) is the biggest it gets and everything after would be less than -3 making x the bigger one?

[fossil badger]

x > -3??

Nvm I got it

.close

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[fossil badger]

Why would a be 1? should it not be 5?

[fossil badger]

Uh this is what I did

Sorry for my handwriting

[marsh citrus]

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[fossil badger]

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[light totem]

Heyo! I just need help figuring out what method to use to solve this ODE. I want to say its a homogenous equation and I need to substitute v=y/x, but I keep messing up the intermediate steps where I cant get a nice function to intigrate.

[light totem]

^^ Also solving for C = F(x,y) but that shoudn't matter much for the process.

[main idol]

Show your substitution with v

[light totem]

Ok one moment

[main idol]

How did y^3 become y

[light totem]

Thats so true. Let me fix that.

hows this looking

@main idol

I'm just not sure what to do with that extra x. It doesnt make it nice to work with

[main idol]

Yea that substitution doesn't help

Try v = 1/y or 1/y^2

[light totem]

Do you know which step I would substitute at?

[main idol]

Doesn't matter

[light totem]

So I just learned how to do this today, and I'm really struggling to grasp this portion. If I start from dy/dx=(17y^3-8xy)/x^2 , how do I sub this?

@main idol Hey thanks for your help. I think I found steps to a similar problem using Bernoulli's equation.

.close

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[shy spear]

I keep getting this wrong and dont know why,

In question 2 it says For which value(s) of a does g(x) increase or decrease without bound as x approaches a or for which value(s) of a is g(x) not defined? (Select all that apply.)

Now its decreasing without bound at x= -1
and its undefined (when lhs Doesnt rqual rhs) so a x= 0 and x = 1 so theres discontinuity

[shy spear]

but it says this is wrong

and i dont get why

[marsh citrus]

@shy spear Has your question been resolved?

[marsh citrus]

@shy spear Has your question been resolved?

[shy spear]

.close

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[vague elk]

Quick question:
For a × a = b, √(b) = ±a?

[whole sleet]

√b is a positive number always

[marsh peak]

It's rather sqrt(b) = |a|

[vague elk]

isn't that just principal root?

[sleek lake]

same thing

[vague elk]

because the definition of root I heard is that a^2 = b then √(b) = ±a
for example √(4) = ±2 because 2 * 2 = 4 and -2 * -2 = 4

[quaint elm]

well uh
√4 = 2

it has one value

[sleek lake]

I've only seen the other way

like haley said

[quaint elm]

now it is true that if a^2 = b then either √b = a or √b = -a

[sleek lake]

like it's not two competing conventions

[vague elk]

oooh

either

I think that's it

thanks

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[lofty sun]

hi! so i have a question, so basically given abcd is a square, e is the mdpt of ad, and be is perpendicular to ef,

[lofty sun]

how would you prove angle aeb = angle efd

[unborn condor]

AEB + 90º + DEF = 180º. Thus AEB + DEF = 90º;
DEF + 90º + EFD = 180º. Thus DEF + EFD = 90º;
Thus AEB = EFD

first line is because AD is a line.
second line is cuz you got a triangle.

[lofty sun]

would def be similar to abe?

for similar triangles

since AAA

or would it not work?

[marsh citrus]

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[fossil tide]

how to integrate
3 times the integral

[fossil tide]

is thiz corrct

then this

[marsh citrus]

@fossil tide Has your question been resolved?

[main idol]

What the hell is this handwriting

[fossil tide]

ms paint m8

[main idol]

Your t and + look the same

[fossil tide]

uhh

[main idol]

Just write it on paper and take a picture

[fossil tide]

wait i re do

oh yeah i have no charge on phonexd

[marsh citrus]

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[marsh citrus]

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[fossil tide]

.reopen

[marsh citrus]

✅

[fossil tide]

yeah i got till here

[marsh citrus]

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[marsh citrus]

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[vivid river]

Im confused how I got this wrong?

[late geode]

that's the total area

[vivid river]

Is that not what this is asking

[late geode]

they want the average

value

[vivid river]

Oh

[late geode]

in which case you'd need to divide by length of the overall interval

[vivid river]

Was I supposed to do

1/2-(-1)

[late geode]

no

[vivid river]

Confusion

[late geode]

wdym by do
1/2-(-1)

[vivid river]

One sec

Yk

This little formula

[late geode]

yes

[vivid river]

Right

So I'd have 1/3*-9

Or -3

Which would be B

[late geode]

you missed () and used the vague word of "do" so it wasn't entirely clear

[vivid river]

Unless Im wrong

[late geode]

yes

[vivid river]

Yes sorry

Idk how to do the desmos thing

That people do

🥲

[vivid river]

So I basically stopped midway

I see I see

I appreciate your help

🙏🏽

.close

[marsh citrus]

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[vagrant gull]

how do you find the exact value of cot 135?

[void elm]

i would first rewrite as cos(135)/sin(135)

and then use the unit circle

[glass silo]

alt: note that cot(135) = 1/tan(135) and notice that tan is periodic every 180

[smoky plover]

^it also becomes evident if you visualize tangent as the slope

[late geode]

why not just use the period of cot

which is the same

[novel juniper]

Hint:- write it as cot(180-45)

[vagrant gull]

I tried with this

[late geode]

bad

very bad

TrigFunction(angle) is not the product of
TrigFunction and (angle)

or whatever you're trying to do there

[rigid falcon]

oh dear

[late geode]

TrigFunction(angle) does not mean
TrigFunction= angle

[rigid falcon]

on the right track

but as people said before

it's 1/tan(135)

cotx = 1/tanx

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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[worn aspen]

So I used unit vectors on this problem and WebAssign tells me the problem cannot be understood

I feel like Im in the ballpark but of course im not there

[marsh citrus]

@worn aspen Has your question been resolved?

[fresh oyster]

you shouldnt have the sqrt(51)

it doesnt want the normal vector

it looks like it just wants the same vector v in terms of ai+bj+ck

that is, rewrite (7,1,1) in terms of the unit basis vectors

[worn aspen]

Ohh gotcha I appreciate the help

so a,b,c is the same as v1,v2,v3

.close

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[marsh citrus]

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[spiral fable]

These are the last questions for my homework but I’m really struggling. Would appreciate if someone could lend a hand or show workings on how to complete these questions. Thanks

[vapid oak]

wait I will try

[marsh citrus]

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[vast star]

divide 60y with 50

you will get how much the x is

plug it in the original problem

and you will get the y value

and at the end multiply with the original x value

gimme a sec i will get it on paper

wait actually im stuck

i think its 11

but not 100% sur

e

[dense epoch]

One sec let me check

[marsh citrus]

@dense epoch Has your question been resolved?

[dense epoch]

I can't check

[dense epoch]

<@&286206848099549185> is this correct

[pliant siren]

lemme see

left side makes sense

on the right side though

if you solve for y, you get y = 5/6*x

[dense epoch]

Can u write it?

[pliant siren]

so you have
$\frac{\frac{6}{5}y+\frac{5}{6}x}{\frac{6}{5}y-\frac{5}{6}x}$

[elfin berry]

DerTheo

[pliant siren]

because
$50x=60y$

[elfin berry]

DerTheo

[pliant siren]

so

$x=\frac{60y}{50}=\frac{6}{5}y$

[elfin berry]

DerTheo

[pliant siren]

and

$y=\frac{50x}{60}=\frac{5}{6}x$

[elfin berry]

DerTheo

[pliant siren]

right?

@dense epoch

[dense epoch]

Yep

But

[pliant siren]

so now you need a value

hm

[dense epoch]

Why isn't there a y for the 60

[dense epoch]

And an x for the 50

[pliant siren]

because 50x = 60y

and then you divide both sides by 60

so on the right side, the 60 goes away

and on the left, you get $\frac{50x}{60}$

[dense epoch]

Why do you divide by 60?

[elfin berry]

DerTheo

[pliant siren]

to isolate y

[dense epoch]

So to isolate y you divide both sides by the coefficient of y?

[pliant siren]

yes

60, in this case

[dense epoch]

Alr

And then you isolate x too?

[pliant siren]

yes

by dividing by 50

[dense epoch]

Then you get 60y over 50 so 6y over 5?

[pliant siren]

yes

because the 10 cancels out

[dense epoch]

Alr

Then what

[pliant siren]

let me think

im just bringing this down so I dont have to scroll \
$\frac{\frac{6}{5}y+\frac{5}{6}x}{\frac{6}{5}y-\frac{5}{6}x}$

[elfin berry]

DerTheo

[pliant siren]

$\frac{\frac{6y}{5}+\frac{5x}{6}}{\frac{6y}{5}-\frac{5x}{6}}$

[elfin berry]

DerTheo

[pliant siren]

$\frac{\frac{36y}{30}+\frac{25x}{30}}{\frac{36y}{30}-\frac{25x}{30}}$

[elfin berry]

DerTheo

[pliant siren]

im just multiplying the fractions so that they have the same denominator

$\frac{\frac{36y+25x}{30}}{\frac{36y-25x}{30}}$

[elfin berry]

DerTheo

[pliant siren]

ill be back in a minute

[dense epoch]

Alr

[pliant siren]

try continuing on your own though

[dense epoch]

I have no idea what to do 😢

[pliant siren]

alr

$\frac{36y+25x}{30}\times\frac{30}{36y-25x}$

[elfin berry]

DerTheo

[pliant siren]

$\frac{36y+25x}{36y-25x}$

[elfin berry]

DerTheo

[dense epoch]

Why are u multiplying

[pliant siren]

to make it into one fraction

[dense epoch]

Oh u reciprocaled so the division symbol turned to multiplication?

[pliant siren]

yes

and then I put them together

that didn’t really bring me anywhere

im kinda clueless on how to continue

yep someone else is helping

[fleet urchin]

buddy
x=6y/5

[dense epoch]

Damn this question is stumping a lot of people

[fleet urchin]

do u agree?