help-0

[alpine sable]

ohh okay i kinda get it know but the qwuestion shows 2 types of lines, one w a singular line but the other with dash like lines

[median oar]

So dashed lines is like open circles

Have you seen open and closed circles on intervals?

[alpine sable]

no

[median oar]

https://qph.cf2.quoracdn.net/main-qimg-ac87a9130be0ab3d01cae07eb09048fb

These?

[alpine sable]

ohh yes i have

[gritty wave]

u know the > and >= signs

[alpine sable]

yess

[median oar]

Ok dashed vs solid lines is a 2 dimensional version of the open and closed circle

Cos how can we “open” the boundary of a line

It’s hard to even think about how to draw that

So we denote it with a dashed line

[alpine sable]

ohh okay yk how u said the line w y=x thingy, how do we know which side of it would be shaded?

[median oar]

Well

Let’s pick a point

You choose

[alpine sable]

uhhh

6,6

idk

[median oar]

Ok

Is 6 < 6

[alpine sable]

no

[median oar]

So is (6,6) coloured in

I had the inequality sign the wrong way but it’s still correct

[alpine sable]

oh

how do i get the answer to my question tho

[median oar]

Well do you colour in (6, 6)?

[alpine sable]

is it d and b??

[median oar]

Yes

But I hope you’re not guessing

[alpine sable]

im nottt

i used ur method

tyyyy

[median oar]

👍

[alpine sable]

i was stuck on that questuion fir so longg

[median oar]

Sometimes it does get confusing with all the letters and pictures

[alpine sable]

ahhh

ok ty

.close

[lone heart]

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[zealous lichen]

you can't just add them

when sinx is negative, |7sinx| becomes positive and the result is positive

it's easy after removing the absolute value btw

[gritty wave]

(the function will always be positive)

[weak field]

*non negative

[zealous lichen]

yep

[lone heart]

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[formal wraith]

can someone help me with this?:

[upbeat gorge]

Think of properties of squares

[formal wraith]

@upbeat gorge

I've done this, I gathered the distance of one side of the square, and drawed a diagonal line from point B, with the value that the distance gave me

I don't know if this good, or I just misunderstood everything

[upbeat gorge]

Yeah this doesn’t look right

Not sure why your D is on the x-axis

And yeah AC isn’t right

[formal wraith]

I draw it there because, i just drawed a diagonal from b

[upbeat gorge]

The length of BC and the observation that AB is a diagonal are the only correct things there

[formal wraith]

Okey, but i'm going on the correct way or this is just wrong?

[upbeat gorge]

Here’s the thing: there’s a locus (set) of points that are distance 5 from B

Like, it’s a circle

[formal wraith]

mmm

So i have to do the distance formula but with x and Y as incognita?

I mean, something like:

[upbeat gorge]

Yeah you can do that

Do that with both BD and AD and you have a system of equations you can solve

[formal wraith]

I haven't done that yet, im still in the term 1 of the book szprecalculus, and I just have seen the cartesian plane, and distance, midpoint formulas

[lone heart]

@formal wraith Has your question been resolved?

[lone heart]

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[fast crypt]

Is this wrong? According to a calculator it’s wrong but I’m not sure where it went awry

[alpine sable]

shouldn't this be [
\f{2c^{-1}}{ab} \divisionsymbol \f{3ab^{-2}}{c^{-2}} = \f{2c^{-1}}{ab} \by \f{c^{-2}}{3ab^{-2}}
]

[worn fox]

f

[fast crypt]

i dont follow

oh sorry

i reread it a couple times

[alpine sable]

[
\f ab \divisionsymbol \f cd = \f ab \by \f dc
]
is true. What you did meanwhile is [
\f ab \divisionsymbol \f cd = \f ab \by \f{d^{-1}}{c^{-1}}
]

[fast crypt]

im quite stumped about it

because i was taught that when a neg indice gets put on the bottom it becomes a positive

[alpine sable]

yes this is true. [
a^{-1} = \f 1 a \tss{and} \f1{a^{-1}} = a
]
but what you did is that you double "switched" the signs

[fast crypt]

so wait

i get what youre trying to get across

i j dont understand why it doesnt apply here

[alpine sable]

okay let me notate it

[fast crypt]

like when do i know if it applies and when it doesnt

okay thanks

[alpine sable]

,tex \env{alignat*}{{1}
&\f{2c^{-1}}{ab} \divisionsymbol \f{3ab^{-2}}{c^{-2}} \
={}& \f{2c^{-1}}{ab} \divisionsymbol \f{3ac^2}{b^2} \tqs{(from the fact that} a^{-1} = \f 1 a \tss{and} \f1{a^{-1}} = a) \
={}& \f{2c^{-1}}{ab} \by \f{b^2}{3ac^2} \tqs{(from the fact that} \f ab \divisionsymbol \f cd = \f ab \by \f dc)
}

@fast crypt

wait I messed up sorry

[fast crypt]

all goo

d

take ur time

[alpine sable]

okay yes this is correct

@fast crypt

take Ur time to digest this

I gtg but someone else can continue helping if u may

[fast crypt]

ok

thanks again

[lone heart]

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[low frigate]

I've been able to understand every question up until (bii).
I figured out that the inverse function = -f(-x) of the original function.
However, when I do the translation on paper, the signs of the numerator don't match to the inverse function.
Could someone please help me with this?

[astral cloud]

How to get (a+b/2 ,c/2) and (b-a/2,c/2)

[zealous lichen]

show your work please

[zealous lichen]

find anther channel

[low frigate]

Alright, might take a minute.

[low frigate]

Is this alright?

[zealous lichen]

the last step wron

g

you distributed the minus sign to both numerator and denominator

[low frigate]

Was I meant to do it to just the denominator?
I don't see how else it would match the inverse function.

[lone heart]

@low frigate Has your question been resolved?

[lone heart]

@low frigate Has your question been resolved?

[lone heart]

@low frigate Has your question been resolved?

[lone heart]

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[left scarab]

can anyone explain this?

[left scarab]

idk how to really read it

I understand the symbols

but like is this how it's read?
"q is not q, and p will be true if q is true, therefore p is not p" ?

i do not even understand it even as i typed it

[lone heart]

@left scarab Has your question been resolved?

[lone heart]

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[quick egret]

Solving for y

[quick egret]

I don’t know exactly how to set it up

[fleet spire]

my best guess would be cross-multiplying

[quick egret]

Yeah, but when I do that the answer is wrong

[fleet spire]

that means 3*(y+13) = 1*(16y)

[quick egret]

Alright

[fleet spire]

so 3y +39 = 16y

39 = 16y -3y

[quick egret]

So the answer is 1 = y

[fleet spire]

39 = 13y

y = 39/13

=3

[quick egret]

Oh

[fleet spire]

sorry i messed up

did this answer your question?

[quick egret]

Yes thank you

.close

[lone heart]

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[thick lynx]

Find the equations of the lines, calculate their intersection and determine the areas a, b, c and d.

How would we determine the areas though?

<@&286206848099549185>

[slate bane]

Carefully defined integrals could do the job

[thick lynx]

Can we do it without integrals though?

[slate bane]

cant see how so i let that to other people

[thick lynx]

.close

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[serene pilot]

3:6=12:24

[serene pilot]

true or false

[vale wigeon]

what do you thin

think*

[serene pilot]

im not sure

[vale wigeon]

can you put both ratios in lowest terms?

[serene pilot]

i dont know at all

[vale wigeon]

https://www.youtube.com/watch?v=SJyNCvU6EUA

[serene pilot]

so would both ratios equal 2?

[vale wigeon]

1 : 2

[serene pilot]

the question is asking of they are equal

right

.close

[lone heart]

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[warm forge]

How can I simplify this further?

[vale wigeon]

can factor out (1-3x)(x^2-2)^2 by the looks of it

[warm forge]

How do I do that T_T

im not used to factoring big polynomials like this

[lone heart]

@warm forge Has your question been resolved?

[unkempt robin]

Try replacing terms with letters for a nicer visual look

Let's say

$a = (1-3x)$

[ocean seal]

USS-Enterprise

[unkempt robin]

And

$b = x^2 - 2$

[ocean seal]

USS-Enterprise

[unkempt robin]

Your equation now becomes

$-6 * a * b^3 + 6x * b^2 * a^2$

[ocean seal]

USS-Enterprise

[unkempt robin]

Do you see how you can factor out 3 things here

[warm forge]

yes but idk how

wait

am i on the right track?

[unkempt robin]

You started well

You factored out $(-6 * (1-3x) * (x^2 - 2)^2)$

[ocean seal]

USS-Enterprise

[unkempt robin]

What is being multiplied by this

[warm forge]

I dont understand

[unkempt robin]

Can we return to the simple letters

[warm forge]

Sure

[unkempt robin]

What can you factor out here

[warm forge]

$b^2$

and

[ocean seal]

OT

[warm forge]

$a$

[ocean seal]

OT

[unkempt robin]

Do you see that both terms also share a 6

[warm forge]

ohhhh

I didnt see that

[unkempt robin]

Yeah, so we can factor out $6ab^2$

[ocean seal]

USS-Enterprise

[unkempt robin]

Now this needs to multiply something

[warm forge]

Let me try again

[unkempt robin]

$6ab * (something ± something else)$

[ocean seal]

USS-Enterprise

[unkempt robin]

We use distributivity; when we multiply 6ab by something we have to get the first term

So what is something that when multiplied by 6ab gives us $-6ab^3$

[ocean seal]

USS-Enterprise

[warm forge]

$-1(x^2-2)$?

[ocean seal]

OT

[unkempt robin]

Correct!

[warm forge]

sorry

oh

nvm

[unkempt robin]

Now do the same for somethingelse

[warm forge]

$x(1-3x)$ for the other term?

[ocean seal]

OT

[unkempt robin]

Correct

[warm forge]

thats it?

[unkempt robin]

So we end up with

$6(x^2-2)^2(1-3x) * (-1(x^2-2) + x(1-3x))$

[ocean seal]

USS-Enterprise

[warm forge]

looks good to me

Thank you stranger!! Next time, I will replace stuff with variables. It really does help.

Have a good day!

[unkempt robin]

We aren't over yet!

[warm forge]

Oh

[unkempt robin]

You see how terrible everything looks in the second parenthesis

We must simplify that

[warm forge]

$-1x^2+2+x-3x^2

$-1x^2+2+x-3x^2$

[ocean seal]

OT

[unkempt robin]

Correct

Now collect like terms

And use commutativity to rearrange them

[warm forge]

$-4x^2+x+2$

[ocean seal]

OT

[warm forge]

Dont tell me we have to multiply with the other side

[unkempt robin]

No, we would be reversing the work

[warm forge]

ok good

[unkempt robin]

Now sometimes, though this is isn't as necessary

That 6 at the start is kind of ugly so we might want to distribute it to one of the paranthesis

Maybe distribute 6*(1-3x)

So we und up with this:

$(x^2-2)^2(-18x+6) \cdot (-4x^2 + x + 2)$

And there's your result

[warm forge]

Thank you so much

I appreciate you taking the time

[ocean seal]

USS-Enterprise

[unkempt robin]

No problem 🙂

[warm forge]

have a good day now

[unkempt robin]

You too

[warm forge]

.close

[lone heart]

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[quick moth]

Is this correct

[fiery forge]

Should be

[quick moth]

Are u sure

I don’t want to get it wrong

[mellow knoll]

its correct

[quick moth]

Thanks

[lone heart]

@quick moth Has your question been resolved?

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[lone heart]

Please don't occupy multiple help channels.

[rustic notch]

aleese

do u solve for x

than put x back in

in one of the equations

[nocturne lily]

first find the value of x

[rustic notch]

then set it up

[nocturne lily]

yuh

[rustic notch]

9y-5+wtv the number was =180

oh alrr

.close

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[proven robin]

how to prove that the product always converges to 0?

[proven robin]

<@&286206848099549185>

[lone heart]

@proven robin Has your question been resolved?

[muted hornet]

first, write out a few terms. say, $$\prod_{n = 1}^3 \frac{x + 2n - 1}{x + 2n}$$

[ocean seal]

ves

[proven robin]

alr

$\frac{x+1}{x+2} \cdot \frac{x+3}{x+4} \cdot \frac{x+5}{x+6}$

[ocean seal]

skittle

[muted hornet]

wait i should have written out a few terms before i helped this isn't what i thought it was

i will think, or maybe someone else will help before i'm done thinking

[proven robin]

oh

[muted hornet]

do you have any theorems that might help prove this? is it for a class :o

[lone heart]

@proven robin Has your question been resolved?

[muted hornet]

okay i THINK i have something

but i might have made some mistakes, and i haven't checked it over too much yet

You can write each term of your product as $$\left(1 - \frac{1}{x + 2n}\right).$$ So the question is really does $$\prod_{n = 1}^\infty \left(1 - \frac{1}{x + 2n}\right)$$ converge.

If you write it out, you see that \begin{aligned*} \prod_{n = 1}^N \left(1 - \frac{1}{x + 2k}\right) &= 1 - \sum_{n = 1}^N \frac{1}{x + 2k} + \dotsb &\quad+ (-1)^N \sum_{1 \leq n_1 < \dotsb < n_N \leq N} \frac{1}{(x + 2n_1)\dotsb(x + 2n_N)}\end{aligned*}

[ocean seal]

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

[muted hornet]

good enough

you can write this last sum as $$1 + \sum_{k = 1}^N (-1)^k \sum_{1 \leq n_1 < \dotsb < n_k \leq N} \frac{1}{(x + 2n_1) \dotsb (x +2n_k)}$$

[ocean seal]

ves

[muted hornet]

and i think that if you set $$ b_k = \sum_{1 \leq n_1 < \dotsb < n_k \leq N} \frac{1}{(x + 2n_1)\dotsb(x + 2n_k)},$$ then $\sum_{k = 1}^N (-1)^k b_k$ satisfies the alternating series test

[ocean seal]

ves

[muted hornet]

hmmmmmm this doesn't give that it's limit zero though just that it's convergent

maybe someone else will come in and tell me i'm bad and give the gigachad way to solve it

[muted hornet]

this is sort of weird to define because b_k depends on N

yeah its probably super bad actually. nevermind

good luck

[lone heart]

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[tight blaze]

Not sure where to start

[tight blaze]

i got the answer but honestly I forgot how to do it

[jagged cobalt]

x^2-4x-7=(x-2)^2-(-2)^2-7

split the x term coefficient

[tight blaze]

ok

[jagged cobalt]

when the coefficient of x^2 isnt 1 theres an extra step but not too different overall

[tight blaze]

ok how about to get -11

was it -4 -7

ohhhhhhhhhhhhhhhh ok i see it cancelled

alright cool thanks

.close

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[vernal hinge]

Needing help with class work

[median oar]

,rotate

[median oar]

Is -4 <=2 or > 2

[vernal hinge]

huh?

[median oar]

Your function’s definition depends on the value of x

When x <= 2 it’s the top one

When x > 2 it’s the bottom one

We want to find f(-4) so we want to first check if -4 belongs to the first part or 2nd part

[vernal hinge]

Oh

[vernal hinge]

My teacher tried helping me but I still don’t get it

[lone heart]

@vernal hinge Has your question been resolved?

[lone heart]

@vernal hinge Has your question been resolved?

[lone heart]

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[alpine sable]

why does the expansion go like that

its similar to this

i just wanna know how can u expand these

<@&286206848099549185>

i get how the quadratic work

just higher degree expansion, i cant wrap my head

[wanton umbra]

to find all coefficients of x^(n-1) take r1 and at the rest factors take x, take r2 and at the rest x etc

[alpine sable]

wdym? cam u rephrase

[wanton umbra]

so u get (r1+r2+...+rn)x^(n-1)

[alpine sable]

lemme try to understand

srry

"take r1 and at the rest factors take x, take r2 and at the rest x etc"

i dont get this part

oh

so i first take r1

and multiply r1 by the rest of x

[wanton umbra]

yup

[alpine sable]

and take r2, multiple the rest of x?

OMG

ty

that makes sense

damn nice

[wanton umbra]

then the same with r1r2 and rest x

and so on

for x^(n-2)

[alpine sable]

yes i understand it

this is better than all trash explanations online

ty

.close

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[fresh rune]

need help

[fresh rune]

i know its chain rule but the question is confusing me

is f' = cos(cosx) since y=sin u and u = cos x ?

[dim oasis]

Mhm

[fresh rune]

im right??

[dim oasis]

then add g'(x) to the end of that

[fresh rune]

is g'(x)= -sin x?

im having trouble putting it all together

[dim oasis]

yup

carry the negative symbol to the front

[fresh rune]

of everthing?

would it be dy/dx=-cos(cosx)sinx?

<@&286206848099549185>

[wary stream]

Yes

[fresh rune]

thank you

how would I do 16?

do I just use y=u^2 and find f' and do the same thing as the last question?

[wary stream]

Yes

[fresh rune]

ok ill try

i got

dy/dx= 12u (36x-24) ?

is that right?

[wary stream]

Why did there a u still?

[fresh rune]

im not sure

[wary stream]

Take it one step at a time, you have f(u) and g(x), find f'(u) and g'(x)

[fresh rune]

what is f'(u)?

[wary stream]

The derivative of f(u)

[fresh rune]

so

is it

the derivative of 6x-4?

[wary stream]

No

That's not f(u)

[fresh rune]

then what is?

[wary stream]

y = f(u)

y = u^2

[fresh rune]

how do I differentiate that then

im lost

[wary stream]

Do you know power rule?

[fresh rune]

wait so its 2u^1?

[wary stream]

Yes

[fresh rune]

thats why I got 12u

or was I supposed to do 2(6x-4)?

[wary stream]

.

[fresh rune]

ok

[wary stream]

You found f'(u)

Now find g'(x)

[fresh rune]

g'(x) is 6

[wary stream]

Now plug that into the f'(g(x)) * g'(x)

[fresh rune]

so is it

2u(6x-4)*6?

[wary stream]

No

Why is there a u still

[fresh rune]

no?

because 2u is f'(x)

[wary stream]

No

2u is f'(u)

[fresh rune]

oh

no

not getting it

[wary stream]

f'(u) = 2u

g'(x) = 6

g(x) = 6x - 4

[fresh rune]

yea

[wary stream]

u = g(x)

f'(g(x)) * g'(x)
If u = g(x) and you have f'(u) which was 2u

Plug in u

[fresh rune]

so

oh

is it

12x-8?

[wary stream]

That's the f'(g(x)) part

Now you need the * g'(x)

[fresh rune]

so its dy/dx=72x-48?

[wary stream]

Yes

[fresh rune]

yess

thx

can you help me with a worded problem

[wary stream]

s stands for position, what is the derivative of position?

[fresh rune]

uhh

[wary stream]

Perhaps you should look it up

[fresh rune]

look up what?

[wary stream]

what is the derivative of position

[fresh rune]

oohhh

is it velocity?

[wary stream]

Yes

The question is asking for the velocity at t = 1

If you are given position, what should you do to get velocity?

[fresh rune]

idk..

[wary stream]

You should reread the last 10 minutes of this conversation then

It was mentioned

[fresh rune]

i reread

but I still dont know

worded questions are literally the worst 😭

[wary stream]

Why do you think I made you look up what the derivative of position was

[fresh rune]

i mean I didn't really look it up

it was in my notes from todays class

[wary stream]

Either way it's mentioned it that screenshot

[fresh rune]

do I differentiate?

[wary stream]

s stands for position, what is the derivative of position?

what is the derivative of position
is it velocity?
Yes
If you are given position, what should you do to get velocity?

[wary stream]

Yes

[fresh rune]

so I differentiate sqrt5+4t?

[wary stream]

Yes

[fresh rune]

not sure how to

differentiate sqrt

;-;

my teacher said it can be rewritten

as (5+4t)^1/2

but then its like

fractions ;-;

[wary stream]

Fractions are numbers too

[fresh rune]

i knnow

I’m not sure where to go from there

[wary stream]

Power/chain rule

[fresh rune]

alright

wait

why chain rule

[wary stream]

Because there's an inner function

[fresh rune]

there is?

[wary stream]

Yes

[fresh rune]

is it the 5+4t?

[wary stream]

Yes

[fresh rune]

so do i write 1/2(5+4t)^-1/2 * (4)

[wary stream]

Yes that's the derivative of position

[fresh rune]

so

do i times the 1/2 in front by the 4?

[wary stream]

You can but it's not necessary

Because the problem asks for the velocity at t = 1

You have the function for velocity, now you need to do t = 1

[fresh rune]

ok

would it be 1/2(5+4(1))^-1/2 * 4?

[wary stream]

And that simplifies to?

[wary stream]

Actually what happened to the ^-1/2?

[fresh rune]

1/2(9)^-1/2 * 4?

yea i just saw that too

[wary stream]

Yes

And that equals?

Simplify the entire thing

[fresh rune]

18^-1/2?

[wary stream]

No

[fresh rune]

sqrt 18?

[wary stream]

Where did 18 come from?

[fresh rune]

mmm

i did 1/2 * 4

and got 2

[wary stream]

What happened to the (9)^-1/2?

[fresh rune]

i did 2 * (9)^-1/2

[wary stream]

No

And what does (9)^-1/2 equal to?

[fresh rune]

idk..

[wary stream]

Do you know your exponent rules?

Do you know what 9^1/2 means?

[fresh rune]

yea but fractions ;-;

[wary stream]

.

[fresh rune]

no

[wary stream]

You made sqrt(5 + 4t) be (5 + 4t)^1/2

You can do the reverse process and make 9^1/2 have a sqrt

[fresh rune]

oh

i see

[wary stream]

What is 9^1/2?

[fresh rune]

but its ^-1/2

[fresh rune]

sqrt 9

[wary stream]

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

[ocean seal]

dldh06

[wary stream]

You can apply that

Where you flip it

[fresh rune]

so

[wary stream]

So 9^-1/2 what does that become if you apply that rule?

[fresh rune]

1/sqrt 9?

[wary stream]

Yes

And that equals?

[fresh rune]

1/3

[wary stream]

Now don't forget the 2

[fresh rune]

yea

[wary stream]

.

[real ice]

sorry to bother but do u guys know the formula for the pythagorean theorem?

[lone heart]

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).

[wary stream]

And google it

[wary stream]

So what does this equal to?

[fresh rune]

2/3

[wary stream]

Yes

[fresh rune]

yuss

pls help with this as well

[wary stream]

Find the derivative

[fresh rune]

can I do the same thing

with the 1/2

[wary stream]

Yes

[fresh rune]

so chain rule ?

[wary stream]

Yes

[fresh rune]

im stuck

This where I’m at

[wary stream]

Multiply that altogether, apply that exponent rule

[fresh rune]

so do I multiply

the (19-3r^3) by 1/2

?

[wary stream]

Yes

[fresh rune]

hm

how would I find

19 * 1/2?

is it 9 1/2?

[wary stream]

I'm not sure where you are confused, you have (1/2)(19r - r^3)^-1/2 * (19 - 3r^2). Apply that exponent rule to the (19r - r^3)^-1/2, mulitply everything together, that's your answer

[fresh rune]

its just the part where I multiply everyhthing by 1/2

[round geyser]

that is the answer. you just have to express it in the same way as the options given to you

[fresh rune]

im lost

[round geyser]

look you have found $\frac{1}{2} (19r - r^3)^{-\frac{1}{2}} (19 - 3r^2)$

[ocean seal]

nalin

[fresh rune]

yea

[round geyser]

how do you express (19r - r^3)^{-1/2} as a square root

cause the answers are all in sqrts

[wary stream]

Basically apply that exponent rule

[fresh rune]

1/sqrt(19r-r^3)^2?

[round geyser]

that's almost right

why are you squaring it tho??

[fresh rune]

idk

[round geyser]

just dont square the whole thing in the end for no reason and you'll have your denominator

[fresh rune]

so just 1/(19r-r^3)^2

[round geyser]

no

1/sqrt(19r - r^3)

think about the power of -1/2 in 2 steps, first a 1/2 then a -1

so the 1/2 means you square root it

and then the -1 mean flip it into the denominator

[fresh rune]

I see

so its

1/2 * 1/(19r-r^3) ?

<@&286206848099549185>

[round geyser]

1/sqrt(19r-r^3)

you forgot the sqrt

[fresh rune]

1/2 * 1/sqrt(19r-r^3)

[round geyser]

yeah that's the first part and then you also have to multiply by (19 - 3r^2) cause of chain rule

[fresh rune]

how do I do that

[weak seal]

You actually already have it. Your second term of (19-3r^2)^1/2 is correct you just need to rewrite the first term

[fresh rune]

???

[weak seal]

Sorry I meant 1/2(19-3r^2)^1/2

You can rewrite the term to mske it simpler

[round geyser]

$\frac{1}{2\sqrt{19r - r^3}} (19-3r^2) = \frac{19-3r^2}{2\sqrt{19r - r^3}}$

[weak seal]

Yes like that

[ocean seal]

nalin

[fresh rune]

how does it be like that?

[weak seal]

Its the sqrt x identity

[fresh rune]

i know but I mean

when you multiply by

(19-3r^2)

[weak seal]

That's just multiplying the numerator by 1 so you'll get (19-3r^2)

[fresh rune]

oh ok

so thats the final answer?

[weak seal]

Yeah

[fresh rune]

thx

i tried but failed at answering this one

the derivative of log b is 1/xln b right

<@&286206848099549185>

[round geyser]

chain rule

what did you get until?

[fresh rune]

I did

1/(4x-3)ln10 * 4

the I got 1/x-3ln10

[round geyser]

the answer is 4/(4x-3)ln10

you cant cancel the 4 in the numerator with the 4 in 4x - 3

[fresh rune]

oh ok

can I get help with this question

<@&286206848099549185>

<@&286206848099549185>

You know what I'll deal with these later

.close

[lone heart]

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[magic ore]

How would I find the range of this

[magic ore]

<@&286206848099549185>

[lone heart]

@magic ore Has your question been resolved?

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@magic ore Has your question been resolved?

[lone heart]

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[cinder sundial]

given a bag have 7 balls, each of them are being named as number 1,2,3,4,5,6,7 respectively

[cinder sundial]

now we pick 3 balls in sequence from the bag, and with each picks we dont put the ball chosen back into the bag

what's the probability for the second ball to be 5 in the order of quantities?

[cinder sundial]

the number of sample space would be C7_3 * 3!

[cinder sundial]

is this correct?

[lone heart]

@cinder sundial Has your question been resolved?

[lone heart]

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[timber bramble]

How can i get the adjescent value?

[timber bramble]

is it

opposite × tan50

?

No no

Opposite × (1/tan50)

So we get the cos

How tf

I forgot

[alpine sable]

adjacent to what

Law of sines maybe

You have all three angles and a side

[timber bramble]

Its 90 degrees

Trig

So

[alpine sable]

so?

[timber bramble]

I got the angle

And opposite

[alpine sable]

Yeah, use law of sines

[timber bramble]

Whats adj then

What is it

is it

[alpine sable]

adjacent to the bar idk

[timber bramble]

opp × (1/tan50)

What would you do here

x = ?

[alpine sable]

law of sines

Or sohcahtoa

just set up equality using cosine of the angle and solve for x

[unkempt vapor]

why cos?

tan 50 = 69/x

or ctg 50 = x/69

[alpine sable]

yeah why cos

because im braindead u see

[alpine sable]

yes tangent is best option

[median oar]

Sin cos and tan all work

[unkempt vapor]

-_-

[median oar]

But yes tan is the simplest one

[unkempt vapor]

then not using trig also works

[lone heart]

@timber bramble Has your question been resolved?

[lone heart]

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[lone heart]

@cloud fox Has your question been resolved?

[lone heart]

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✅

[lone heart]

@cloud fox Has your question been resolved?

[placid zinc]

@cloud fox
It's been a little while, where are you at with it?

So note that if you do "plug (α,β,γ) in", and search for zero solutions, you get:
α + 2β + 7γ = 0
4β - γ = 0

[lone heart]

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[placid zinc]

Not hard to reverse that, now that it's row reduced. Let γ be free. We get:
β = γ/4
α = 15γ/2

Infinite solutions, just plug in a γ

[lone heart]

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[thick lynx]

Determine the radius of the three circles

How would you do this?

[worn fox]

Are you forgetting to provide some extra information

[thick lynx]

The radius of the big circle is 1

1 - r = 2/3 * h, right?

That's how you would approach it, right?

h being sqrt(3)/2 * (2r)

1 - r = sqrt(3)/3 * 2r

r = 2sqrt(3) - 3

Is that correct?

[worn fox]

Not exactly easy to follow a string of equations that are presumably coming from geometric arguments without seeing those geometric arguments

[thick lynx]

wait

[noble sail]

This might help you

[thick lynx]

this should be the correct approach

Thanks

.close

[lone heart]

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[fleet spire]

"For how many natural numbers n does the following identity hold?"
I know through my factorization that 7 , 9 , 1 + sqrt2 and 1 - sqrt2 are solutions but found 2 as a solution through substitution.
I don't know where i lost this solution in my calculations. Any ideas?
https://cdn.discordapp.com/attachments/1018703193473024061/1156287827345166409/453cbc8c-a7cc-4f81-9d76-5f1e13ab5940.png?ex=65151534&is=6513c3b4&hm=a09af57727edb807ddfb9ea596b9a78a79b7d76305f70fd3634663ac87ed6cf4& https://cdn.discordapp.com/attachments/1018703193473024061/1156287827772981339/IMG_0247.png?ex=65151534&is=6513c3b4&hm=30e8f37a6f2543976e310a9200c6f1e275e44f8f70ba4ff671003482d4c5c535&µ

[cinder sundial]

What’s the language

Using to describe the question

[fleet spire]

dutch

[languid bolt]

Why can't you just take log_(n^2 - 2n) ?

so you'' get
n^2 + 47 = 16n - 16

[finite flax]

(x^2 - 2x)^{x^2 + 47} = (x^2 - 2x)^{16x - 16}

(x^2 - 2x)^{x^2 - 16x + 63} = 1

(x^2 - 2x)^{(x - 9)(x - 7)} = 1

(x - 9)(x - 7)log(x^2 - 2x) = 0

log(x^2 - 2x) = 0
x^2 - 2x = 1
x^2 - 2x - 1 = 0
(x - 1 - sqrt(2))(x - 1 + sqrt(2)) = 0

x = 7, 9, 1-sqrt(2), 1+sqrt(2)

You are correct I think

@fleet spire

Yeah I messed it up one sec

[languid bolt]

1-sqrt(2), 1+sqrt(2)
wait why?

1 is correct no?

[finite flax]

No, b.c. negative times negative is positive

[languid bolt]

(1^2 - 2)^48 = (1 - 2)^0

1 = 1

1 should be correct

haven't did the working tho.

[finite flax]

Agreed, this is interesting

[languid bolt]

1-sqrt(2), 1+sqrt(2) is correct tho

it makes (x^2 - 2x) = 0

i was just questioning you removing 1

[finite flax]

I wonder what demos says is happening at x=1

[fleet spire]

im in this class rn

he said 7 , 9 , 1 and 2 are natural solutions

[languid bolt]

the line doesn't load.....

in between 0 and 2

[languid bolt]

ah yes 2

(x^2 - 2x) = 0
if x = 1 or 2

[finite flax]

I really just wonder how we properly factor the 1

[lone heart]

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[lone heart]

@fleet spire Has your question been resolved?

[lone heart]

@fleet spire Has your question been resolved?

[lone heart]

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[lone heart]

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[trim tendon]

Can someone please check if my answer to a multivariable limit question is ok?

[trim tendon]

using the squeeze law

I'm not actually sure if my case-work is okay or not

since I'm breaking it up depending on the magnitude of the denominator relative to 1

|denom| ≥ 1 and |denom| < 1

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[lusty nimbus]

Can anyone check if my solving was correct
This is pre cal ellipse lesson (Stand Form to Gen Form)

[lusty nimbus]

The no 2 problem only

<@&286206848099549185>

[wanton umbra]

!15min

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[lusty nimbus]

.close

[lone heart]

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[ocean depot]

How do i write a system of linear equations and solve a system of linear equations to find the values of x and y

[fickle sandal]

property of cointerior angles

[ocean depot]

what is that 😭

[ocean depot]

Okay so

what do i do now

[fickle sandal]

you have two pairs of cointerior angles in the question

[ocean depot]

how

there are no lines

okay nvm

so

umm

ykw

.close

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[woeful ridge]

How would you describe the shape of this distribution? Like a skew? How do I know which way it goes? This is for b

[lone heart]

@woeful ridge Has your question been resolved?

[woeful ridge]

No

<@&286206848099549185>

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[onyx hound]

FP = fixed payments. It's about a fixed payment loan. I can't seem to find the right formula 😭 LV = loan value

[onyx hound]

I end up finding that the annual FP is 598,95 euro which is not possible when the loan itself is 450 euro

I don't get it

And I find that the loan value is 150,26 euro when the year payment is 200 euro

i don't get it

<@&286206848099549185>

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[last umbra]

A stone is thrown vertically upwards with a speed of 20 m/s from the ground. and at the same time a second stone is thrown vertically downwards from a sight of 40 m along the same line of motion. Find the point of collision

can someone help?

[last umbra]

this is what i did
s = ut + ½at²

S1 = 20t - 5t²
S2 = ut + 5t²

S1 + S2 = 40

t (20 + u) = 40..?

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[minor steppe]

how do i do this? im confused as shouldnt the double end up as -25?

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[minor steppe]

<@&286206848099549185> pls help

[safe fog]

what's the question?

[minor steppe]

this

im kinds unsure how to approach the problem

i can do these type of problems with 3 sets but not 4

im confused as it says a pair of sets share 50

but wouldnt that result in 50 - (25+25+4)

[cosmic tartan]

What are the asking in the que??

[minor steppe]

how many total elements in the union of the 4 set

s

[cosmic tartan]

Is the answer 50

@minor steppe

[minor steppe]

idk

do you want me to check it?

[cosmic tartan]

Yes

[minor steppe]

it doesnt make much sense if its 50 imo but i could be wrong

ok

[cosmic tartan]

Ik but whats wrong in giving a try

[cosmic tartan]

Idk dude

[minor steppe]

lol ty anyways

[safe fog]

is it 59?

[safe fog]

damn

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[minor steppe]

No help?

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[minor steppe]

I give up

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[junior warren]

not sure if we can get help for matlab

but

i need to solve 3, ive made this logical array, and i need to somehow find the 75th even number

which would be the 75th logical 1 in a

nvm!

.close

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[wide marlin]

isnt the answer b and c?

[dim oasis]

What's your reasoning?

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[wide marlin]

both equal infinity

[dim oasis]

Now let's consider which would be bigger before x reaches infinity

A limit, of course, describes a variable approaching a number, not necessarily equaling that number

If x were 10 for example, what would they be?

[wide marlin]

ok

[dim oasis]

Likewise for 100, 10000, 1000000, so on

One of those two will almost always be bigger, even though they both eventually diverge to infinity

[wide marlin]

ok let me try

@dim oasis so b is the answer

@dim oasis ?

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[wide marlin]

.close

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[warm brook]

im lost.

[alpine sable]

What does the value of y approach as x gets closer and closer to 4

Did you read anything about limits?

[warm brook]

-3?

[warm brook]

not really, im watching some videos for it but, still feeling confused

[alpine sable]

yes

[alpine sable]

read book

https://batch.libretexts.org/print/Letter/Finished/math-89151/Full.pdf best book

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[livid grotto]

Hey - I'm trying to see if I'm missing something, here, but this summation seems way harder than I think my professor may have intended it to be. When I try to evaluate this (first picture), I always end up with something like the second picture, and I have no idea how to do it. Apparently the third image is correct according to WolframAlpha but I have no idea how it gets there.

[livid grotto]

(sorry for the extremely pixellated images, I am screenshotting this from MathJax in obsidian because the assignment was in that format for some reason)

I do remember going over summation, nested summation, etc. in Calculus 2, which was a prereq to this compsci course, but I don't remember anything about adding up polynomials like that.

[pallid horizon]

I guess you're right for the 2nd picture

And then you can split up this second sum into a sum of squares and a sum of just i. The second may be easier and for the first one (the sum of squares) there still is a formula

You've to use the linearity of the sum

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[livid grotto]

Ah, yeah, thanks

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[olive solar]

Im confused

[pallid horizon]

!status

[lone heart]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[olive solar]

I don't know where to begin

[ocean seal]

Statufi

[pallid horizon]

Or at least how to prove it has a solution? a unique solution?

[olive solar]

I do not

set it equal to zero?

[pallid horizon]

Yes, but actually when you have such a polynomial + ln, you can't really solve it explicitely generally... You have to use theorems to say if it has a solution or not

And if it is unique

Do you know the intermediate value theorem or Bolzano's theorem?

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[errant dagger]

looks better now 🙂

Anyways, why does gcd(a, m) gotta be 1?

[barren portal]

Yep

you have to do some work with me to see it.

take a=2, and modulo 4

can you evaluate the powers of 2 modulo 4?

@errant dagger

[errant dagger]

yeah

[barren portal]

Can you type out the pattern you see?

[errant dagger]

1, 2, 4, 8, 16, 32, 64, 128

[barren portal]

modulo 4

[errant dagger]

so itll just be 1, 2, 0, 0, 0, 0,0...

i think

[barren portal]

Not quite

by powers i mean 2^1, 2^2 2^3

reduce to mod 4

[errant dagger]

hm

2^1 is 2 mod 4 right

[barren portal]

yes

[errant dagger]

2^2 is 0 mod 4

[barren portal]

2^2?

yes 2^3?

[errant dagger]

2^3 is 0 mod 4

[barren portal]

I didn’t mean mutliply

[errant dagger]

im not really following haha

[barren portal]

I thought you were doing abstract algebra

no issue.

trying evaluating powers of 2 modulo 3?

same way