help-0

[small trellis]

Looks like 77 to me

I can't tell exactly

[full dome]

77 and?

I can't read graphs very well.

[small trellis]

2018

That's your two axis

Your years, and your exchange

[full dome]

So that's a valid coordinate set?

[small trellis]

Depends what's considered valid

I don't know

[fiery berry]

Anderson is bored in physics class. His favorite numbers are $1, 7$, and $33$. He writes $0.$, and randomly appends one of his favorite numbers to the end of the decimal he has already written. Since physics class is infinitely long, Anderson writes an infinitely long decimal number. (An example of such a number is $0.1337173377133733 \ldots$) If the expected value of the number Anderson wrote down is of the form $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers, find $a + b$.

[full dome]

Can you possibly help for a few more minutes?

I have some more questions, I really want to sleep y'see.

[small trellis]

To find the slope you do change in y over change in x

Find two coordinates and subtract the y coordinate and divide that number by the change in the X coordinates

[full dome]

Can we move to dms rq?

[small trellis]

Ok

[ocean seal]

Initializationvector

[gritty cave]

Is this just using trigonometric ratios, is there another way to do this?

[jolly stone]

write equation of expected value x as the other way by dividing into 3 equally likely cases

x = 1/3 * (case1) + 1/3 * (case2) + 1/3 * (case3)

case 1 is writing 0.1 followed by the same procedure

the rest is trivial and is left as an exercise for readers

[alpine sable]

you find inverse function first

[elfin talon]

Is the ratio gonna be the average of these 3 numbers over 100 + the average over 1000… which is approximately 0.15(185)recursing

@fiery berry

[small cypress]

How do u do this

I don’t understand the second part

[thorny patio]

There's an answer for both:
Log x = .2

And log x = -.2

If you do absolute value both of those numbers satisfy the equation

[small cypress]

huh how do u get that?

[thorny patio]

| log x | = .2

You have calculator on the exam?

[small cypress]

Yep

This is int maths

Not normal

[thorny patio]

Ok so do 10 ^ .2

This is the x so log x = .2

That's one of the answers

[small cypress]

i got atound 1.59

[thorny patio]

Sounds right maybe xD

As long as you entered it correctly

There's another value for -.2

Do 10 ^ -.2

[small cypress]

wait so .2 bcs u wanna get 1.5 on the y axis?

[thorny patio]

You said pt 2 right

[small cypress]

ye

ii)

[thorny patio]

So no need even to look at the graph actually

Just the equation

| log x | = 0.2

[small cypress]

oh so it can be 0.2 or -.0.2

[thorny patio]

This equation works when either

A. log x = 0.2

or

B. log x = -0.2

[small cypress]

right?

[thorny patio]

Yea

Exactly

[small cypress]

oh crap im dum

so for b would it b the same

the bar indicates that it can be both - and + right?

[thorny patio]

Yes but that one is a little trickier i think for b

The | |

Is called absolute value usually

There are other names for this idea

It means give me the "size" or magnitude of a number

Size is always positive

So it gives you a non negative answer always

| - 2 | = 2

2 units away from zero, the size is 2 etc etc

[small cypress]

yep

[thorny patio]

Different ways you can think of it

[small cypress]

it doesnt mean sqaure root right just absolute o?

[thorny patio]

Yes correct

Square root has another symbol

The radical symbol

(You can use exponents to express roots too though)

[small cypress]

so for b would it be 1-x/4 or 1+x/4?

the x values

[thorny patio]

You have a calculator?

[small cypress]

ye

[thorny patio]

They are ok with decimals

?

I'd use that lmao

That's a hard one to solve algebraicaly

[small cypress]

how do u change it into decimal?

[thorny patio]

The calculator most likely would give you a decimal

If you have a graphing calculator

I might just be tired but I'm not sure how to solve that one easily

[devout sigil]

For b, log x=1-(x/4) or (x/4)-1

[small cypress]

ahhh

[small cypress]

?

i thought there r 2 diff signs

• and -

[devout sigil]

-(1-(x/4))≠ 1+(x/4)

[thorny patio]

Solving it is the hard part though

The negative sign distributes through the whole expression

When you get the 2 branches

From the absolute value

[devout sigil]

The negative expression would be (x/4)-1 not (x/4)+1

[small cypress]

what abt 1-x/4 n 1+x/4

[devout sigil]

1+(x/4) is wrong

[solid sage]

Pls anyone?

[devout sigil]

Busy channel

[solid sage]

@devout sigil Ok

[restive prism]

Given that point E lies on the line BC such that BE=4CE , find the coordinate of E.

Anyone know how to solve this ?

[cerulean vine]

Any diagram

[restive prism]

the concept i got is BE/CE=4/1

but i dont know how to slove it

nope

no diagram

and umm

B is (1, 14.4)

C is (7,7)

and e is (x , y)

[restive prism]

i think

my teacher

cropped it ot

out

[cerulean vine]

Is it midpoint or distance

[restive prism]

ummm

BCE is a line

its

BC=4CE

its ratio theorem

[wary valley]

How would u do 1 c ?

[vapid oak]

An average rate of change means the gradient of the line that joins the two points on the graph of the function

A bit confusing when worded like that but you take the two points (x1, y1) and (x2, y2) and work out the gradient of the line which joins them

so at t=0 the first point is (0, 300)

Try and work out the second point

[solid sage]

#help-1 if free

[thorny patio]

@wary valley

Just do

(Final value - start value)/(distance)

Plug in (Q(2) - Q(0))/2

[rich basin]

Can someone go through the steps in how to solve this?

[vale wigeon]

have you solved absolute value equations before?

[rich basin]

Yeah

This is just revision but i forgotten

[vale wigeon]

right, so you're familiar with things like breaking down into cases?

[rich basin]

Yeah

[vale wigeon]

cause this ineq can be done with that pretty easily

just consider x > 0 vs. x < 0

[rich basin]

wouldn't it be x>= 0?

why is 0 undefined in sbsolute values?

[vale wigeon]

oh i mean you can do x ≥ 0 too

it's just that 0 itself is very clearly a non-solution

[rich basin]

so when i do x < 0

only the absolute values would be negative?

[vale wigeon]

if you're talking about the fact that the equation |x| + 3x > 1 would become -x + 3x > 1 and not -x - 3x > 1, yes

[rich basin]

how about solving a problem like (x+1)/||x-3| > 3

what would we do>

$\frac{x+1}{\left|x-3\right|}>3$ @vale wigeon

[ocean seal]

BorutoEyePower

[rich basin]

I managed to get 2 < x < 3 or 3 < x < 5

so it would be 2 < x < 5 but cannot equal 3

but why on desmos it does not get rid of the 3

[vale wigeon]

kinda busy sorry

[vapid oak]

well if you just put in 2<x<5 you havent specified anything to do with 3

if you want it to exclude 3 you should put together, 2<x<3, 3<x<5

[rich basin]

@vapid oak But then why does desmos get rid of it?

when you put in that equation up there

[vapid oak]

because why would desmos know what you want?

if you write 2<x<5 they will depict that region because that inequality doesnt mention 3 at all

if you want desmos to exclude 3 you have to specify it

[rich basin]

Okay, manage to get it out on wolfram

[placid zinc]

They've given you that:
a + b = -m
ab = n

What's the roots of nx^2 + (2n - m)x + n?

In terms of n and m

[rich basin]

@placid zinc no idea

[placid zinc]

I think you'll know that one, haha

[rich basin]

I don't know what to do next after finding the roots for the first equation

what should i do next find the roots for the other one?

@placid zinc So i made it alpha = 1/ beta

and then did (-2n-m^2)/n = 1/B + B

but what would i do next?

[alpine sable]

Sorry for bad handwriting
Does it make sense?

Nvm I am typing it

[rich basin]

what handwriting?

[alpine sable]

Here typing takes too much time

[alpine sable]

Does it make sense??

[rich basin]

no

[alpine sable]

Ah well

Wait

[rich basin]

i have osmething in my eye be back

back

[fierce frigate]

yo i found this questionaire on th internet,
x2 + 14x + 24 = 0
this is
x= 3 and x= -7 right ?

bvs they answer key is
x=-3 and x = -7

both negative

but using the quadratic formula

that fals

[sacred vortex]

shouldn't the roots be -12 and -2

[fierce frigate]

oop i missscalc

[loud grove]

how do i do this?

[fierce frigate]

i havent vn caalculated it ahahah

i forgot

idk why i said that number

dont even know how i got there

[rich basin]

<@&286206848099549185>

[dim acorn]

hlo i need help

[alpine sable]

can anyone help with this problem

-5/4 = t+3/-4

I got that from the numbers and putting it into slope equation

then I multiply both sides by -4 and get 5=t+3 -> t=2

where did I go wrong?

[spare fern]

$m_1 × m_2=-1$

[ocean seal]

The Godfather

[spare fern]

$\frac{t+3}{-2-2} × \frac{5}{4}=-1$

[ocean seal]

The Godfather

[alpine sable]

ok let me try that

why can we solve for t using -5/4 = t+3/-4

@spare fern

[spare fern]

Wdym

You have made an error

It's not -5/4 on the LHS

It's -4/5

[alpine sable]

ohhh

I'm such a retard

ok ty

[spare fern]

[dull onyx]

The sum of the length and width of a rectangle is 200 meters. If its length is x meters:

1. Prove that the area of ​​the rectangle as a function of x is E (x) = -x ^ 2 + 200x

i dont rly understand this :c

[glass lichen]

so you know that length+width=200

[dull onyx]

yes

[glass lichen]

so given length is denoted as x, what's width=?

[dull onyx]

200/x no?

[glass lichen]

no

[dull onyx]

ok wIt

wait

[glass lichen]

length and width are being added, not multiplied

[dull onyx]

oh

200-x

[glass lichen]

yes

so length=x and width=200-x, what's the area of a rectangle?

[dull onyx]

now we multiply

[glass lichen]

yes

[dull onyx]

200-x * x

ok

[glass lichen]

x(200-x)

[dull onyx]

yes

sorry

im just not sure how to write it down

[fresh bolt]

thats a scary profile pic

[dull onyx]

whose

[fresh bolt]

yours

[dull onyx]

man

[simple hamlet]

i will solve it for u guys

[simple hamlet]

i solve it

[alpine sable]

<@&268886789983436800> we got a troll

sorry for the ping

[sly mantle]

i don't really see it as trolling, but @simple hamlet pls don't answer leading questions meant for others

[sacred rock]

not exactly a math related question, but i was wondering if anyone had an idea of how much i should charge if i tutor 7th graders as a highschooler without any diplomas, etc. i really want to earn some money on the side by tutoring since english is my strongest subject.

any other advice from experienced tutors would also be appreciated : )
<@&286206848099549185>

[night geyser]

prices vary regionally, maybe search some listings on facebook and compare

i charged $25 cad/hr (roughly $20 usd) as a high schooler in a relatively affluent part of a medium-sized canadian city

that was like a decade ago though

also, at least from my experience, english tutors are less in demand than math/physics/chem ones

[vale wigeon]

perhaps i should reconsider my pricing

[sacred rock]

@night geyser yea, my personal tutors charge around 40/hr but they're pretty specialised plus they teach math, piano, etc which i feel is in higher demand, so i was thinking maybe 20/hr or a lil less would be good to begin with?

[topaz scaffold]

How much were you charging?

[vale wigeon]

25 usd/hr is my current rate

i now wield a bachelor's degree though

[topaz scaffold]

Ah nice nice

[sacred rock]

if you don't mind me asking, do you teach groups or private lessons?

[vale wigeon]

private

[sacred rock]

ah, i should definitely change under 20$ then for groups haha

[vale wigeon]

well thats the thing maybe i am undercharging

i dont know tho

like, i'll keep my current rate for now unless something major enough happens that may warrant me adjusting it in either direction

[topaz scaffold]

Private lessons for instruments like piano are a lot more expensive...

How do you find people to tutor btw?

[vale wigeon]

i cast my net here occasionally

well

metaphorically

every now and then someone comes along whos in need of math tutoring

[topaz scaffold]

You do it online?

Tutoring I mean

[vale wigeon]

yes

[arctic steeple]

At what level of maths do you Tutor?

I’d also consider changing the rates based on how advanced the stuff is

[vale wigeon]

highschool / early uni, mostly

[arctic steeple]

Tutors for uni stuff can charge more than tutors for HS stuff

[topaz scaffold]

Which sucks cause uni students are more broke than hs students

[sacred rock]

im pretty inexperienced so im starting with ms, so probably around 15$/hr, or is that too expensive?

[arctic steeple]

Since the more advanced it becomes there are less people who can teach it reliably

Less supply you could say I guess

Hence higher rates

[topaz scaffold]

$17 I would say

[arctic steeple]

It all depends on what you’re teaching, who you’re teaching etc

Know your worth

[night geyser]

honestly dont listen to random discord speculation and compare your prices to local listings

[sacred rock]

hehe, i js wanna have clients to begin with first since this is my first tutoring gigs, but i don't wanna be broke T-T

[night geyser]

theyre not hard to find on facebook/whatever

[arctic steeple]

^^

This

[sacred rock]

okie okie

tysm everyone

[unique tiger]

can someone explain to me how to differentiate from this given

[oak chasm]

@unique tiger @unique tiger Differentiate cos(x) repeatedly until you get what you started with. Figure out how many times you differentiated to get back to where you started.

@unique tiger Did you find out how many times you need to differentiate cos(x) to get back to cos(x)?

[unique tiger]

im having trouble

[oak chasm]

OK, what's the derivative of cos(x)?

[unique tiger]

sin x?

[oak chasm]

Almost.

[unique tiger]

-sinx

[oak chasm]

Right.

What's the derivative of -sin(x)?

[unique tiger]

• cos x?

[oak chasm]

What's the derivative of -cos(x)?

[unique tiger]

sin x

[oak chasm]

What's the derivative of sin(x)?

[unique tiger]

cos x

[oak chasm]

So, we have:
cos(x)
-sin(x)
-cos(x)
sin(x)
cos(x)

Every 4, it repeats.

Now you have 86.

In your problem.

So, do you remember dividing and getting a quotient and remainder from elementary school?

[unique tiger]

yeah

[oak chasm]

OK, what's the quotient and remainder when you divide 86 by 4?

[unique tiger]

21.2

[oak chasm]

OK, so 86 = 4 · 21 + 2.

Now those 21 fours all go back to cosine.

It goes to 4, cosine. It goes to the next 4, cosine.

It does that 21 times.

Then it does 2 more derivatives.

Does that make sense so far?

[unique tiger]

so it repeats 21 times since it repeats every 4 times

[oak chasm]

Right.

[oak chasm]

@rich basin Sorry, channel is busy.

@unique tiger So, on the 84th time (21 · 4), it's back to cos(x).

Then it does two more derivatives of it.

Now we got all the derivatives.

0: cos(x)
1: -sin(x)
2: -cos(x)
3: sin(x)

From before.

So, what's two more derivatives after cos(x)?

[unique tiger]

• cos?

[oak chasm]

Right.

So, the remainder after you divide by the cycle length (how many times until it repeats) is how many times to differentiate.

So we differentiate twice since our remainder was 2.

And we get -cos(x).

Does that make sense?

[naive musk]

what will be the answer to this given that a is constant

[unique tiger]

so our final answer will be - cos x=2?

[oak chasm]

Nope, just -cos(x).

When you differentiate an expression, you don't get an equation.

[unique tiger]

ohh the two is just for indication of 2 more derivative

[oak chasm]

Right!

[unique tiger]

ohh okay .. I get now thank you

[oak chasm]

You're welcome.

[unique tiger]

can I ask 1 more question

[oak chasm]

Sure.

[unique tiger]

How

[oak chasm]

Do you use d(...) or (...)' for what's left to differentiate?

[unique tiger]

with prime

[oak chasm]

OK, so:

a⁴ - t⁴ = 3a²t
(a⁴ - t⁴)' = (3a²t)'

So we use the sum rule on the left.

And the constant multiple rule on the right.

(a⁴)' - (t⁴)' = 3(a²t)'

See how I simplified what's left to be differentiated?

[unique tiger]

yes you seperated the variables.. as I understand

[oak chasm]

OK, so what's (a⁴)'?

[unique tiger]

4a3

[oak chasm]

Right. You can use ^ for powers, like 4a^3.

What's (t⁴)'?

[unique tiger]

4a^3-4t^3? right?

[oak chasm]

Almost.

So, you want dt on the bottom, right?

The question asks for that.

So, with (t⁴)', you get 4t³.

With (a⁴)', you use the chain rule to get 4a³·a'.

You really do the same thing to both.

(t⁴)' → 4t³t'
(a⁴)' → 4a³a'

Do you understand why I got those?

[unique tiger]

how?

[oak chasm]

OK, do you know how the chain rule works?

[halcyon matrix]

The answer is:
da/dt = (4a^3-6at)/(3a^2+4t^3)

[oak chasm]

@halcyon matrix Please see the rules at #❓how-to-get-help. One of them is to not just give the answer.

[unique tiger]

derivative with respect to x

[oak chasm]

The chain rule works like this:

[tawny lion]

you essentially take a' when you differentiate implicitly, if it were dt/da you'd do the same thing for t

[oak chasm]

(a²)²

You have squaring on the outside, right?

[unique tiger]

yeah

[oak chasm]

So, what you do is you pretend the inside (a²) is a variable.

And you get 2(a²).

Then you multiply by the derivative of the inside.

[unique tiger]

so 2(2a)?

[oak chasm]

Right.

Now we're going to do that here:

a⁴

You pretend a is a variable.

You get 3a⁴.

Then you multiply by the derivative of the inside.

3a⁴a'

See how I followed the chain rule?

[unique tiger]

ohh okay a^4 is pretend variable

[oak chasm]

No, you have like (a)⁴

You get the derivative, pretending that the parentheses are a variable.

4a³

Then you multiply by the derivative of the parentheses.

4a³a'

Like before.

(a²)²

2(a²)

2(a²)(2a)

[unique tiger]

2a^2a

[oak chasm]

You get the derivative as if the parentheses are a variable.

Then you multiply by the derivative of the parentheses.

So you get 4a³.

(a²)²
2(a²)(2a)
4a³

Does that make sense?

[unique tiger]

yeah

[oak chasm]

OK, now you want da/dt, right?

Usually, you're doing dy/dx.

And when you have x as your variable, you don't need to do the extra step I'll show.

So, x² → 2x. No extra steps.

What if you're doing y²?

Well, y isn't the variable in the bottom of dy/dx.

So you have to do an extra step.

y² → 2yy'

That's because of the chain rule.

[unique tiger]

okay, then

[oak chasm]

x² = (x)² → 2(x)(x') = 2x x'
y² = (y)² → 2(y)(y') = 2y y'

Now x' = 1, so it goes away.

2x

y' doesn't equal 1 necessarily., so it doesn't go away.

2y y'

Like if y = 25x², then y' will be 50x.

It's not always 1.

But x' is always 1.

Does that make sense?

[unique tiger]

yes

[oak chasm]

OK, so back to the problem.

(a⁴)' - (t⁴)' = 3(a²t)'

4a³a' - 4t³ = 3(a²t)'

Does it make sense why I got that?

[unique tiger]

yes

[oak chasm]

OK, now on the right, we have a product left to be differentiated.

Do you know the product rule?

[unique tiger]

yes

[oak chasm]

OK, so:
(a²t)'
(a²)'t + a²(t)'

[unique tiger]

right

[oak chasm]

So, what's (a²)'?

4a³a' - 4t³ = 3((a²)'t + a²(t)')

[unique tiger]

2a

[oak chasm]

Almost.

[unique tiger]

2

[oak chasm]

Don't forget the chain rule.

[unique tiger]

is t gonna be 0?

[oak chasm]

No, hold on.

We're not done with (a²)'

Remember to use the chain rule.

(a)² → what?

[unique tiger]

2a a'

[oak chasm]

Right.

Now (t)'.

What's that?

[unique tiger]

1?

[oak chasm]

Right.

4a³a' - 4t³ = 3((a²)'t + a²(t)')
4a³a' - 4t³ = 3(2aa't + a²)

[unique tiger]

okay

[oak chasm]

Now, a' is another way of writing da/dt.

And that's what they want us to find.

So solve that equation for a'.

[unique tiger]

is it 1?

[oak chasm]

Nope.

t' is because it's da/dt and t is on the bottom.

a' isn't.

Pretend a' is a separate variable.

So you have three variables, a, a', and t.

Solve for a' using algebra.

[unique tiger]

from 2aa't?

[oak chasm]

No, with solving, you solve an equation.

From 4a³a' - 4t³ = 3(2aa't + a²).

[unique tiger]

i dont get it....

[oak chasm]

How would you solve 4a³x - 4t³ = 3(2axt + a²) for x?

[unique tiger]

divie by x?

[oak chasm]

No.

[unique tiger]

3a(2at+a) ?

[oak chasm]

No.

First, you expand on the right.

What do you get?

[unique tiger]

6axt +3a^2

is it?

[oak chasm]

4a³a' - 4t³ = 3(2aa't + a²)
4a³a' - 4t³ = 6aa't + 3a²

Now get all of the terms with a' in it to one side and all of the other terms to the other side.

[unique tiger]

4a²a'- 6aa't =4t³+ 3a²

then?

[oak chasm]

Good.

Now factor out the a' on the left.

[unique tiger]

2a'(2a-3at)=4t³+ 3a²

[oak chasm]

Sorry, I made an error above copying a line.

We have 4a³a' - 6aa't = 4t³ + 3a² (I had 4a²a' by mistake).

Now we want to factor out just the a', not 2a'.

@rough acorn Sorry, channel is busy.

[rough acorn]

oops sorry posted before looking if it was active

[unique tiger]

so ...a'(4a³ - 6at) = 4t³ + 3a² ?

[oak chasm]

Right.

Now, divide both sides by the part in parentheses.

That will get us a' = something without a'.

What do you get?

[unique tiger]

a'=4t³ + 3a²/4a³ - 6at

[oak chasm]

Yes, except when you write it in chat, you should surround the top and bottom of a fraction with parentheses.

Otherwise it looks like 4t³ + 3(a²/4)a³ - 6at because of PEMDAS.

So, that's your answer.

[unique tiger]

a'=(4t³ + 3a²)/(4a³ - 6at)

[oak chasm]

Right, so:
da/dt = (4t³ + 3a²)/(4a³ - 6at)

Since a' is da/dt.

[unique tiger]

ohh

[oak chasm]

Does it make sense how we got this?

[unique tiger]

yeah

[oak chasm]

OK, you had a bit of trouble with the solving. If you'd like, here's a bit of review to help with it: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:solve-equations-inequalities.

It has some video lessons and practice problems on solving.

[unique tiger]

okay thanks.

it was very helpful

[oak chasm]

You're welcome.

[unique tiger]

is it same process if x and y are fractions

[oak chasm]

You mean the implicit differentiation or the algebra solving?

[unique tiger]

implicit differentiation

[oak chasm]

You might have to use the quotient rule, but it's the same sort of thing.

(a/b)' → (a'b - ab')/(b²)

That's the quotient rule.

[unique tiger]

quotient rule is first step then find x or y right? like a' in our problem

[oak chasm]

No, it's like the product rule.

(ab)' → ab' + a'b

You don't have to find anything.

(x⁸y⁷)' → x⁸(y⁷)' + (x⁸)'y⁷.

I didn't find anything, just wrote the parts where they go according to:
(ab)' = ab' + a'b

I copied a over to both places where it appears in the product rule.

I copied b over to both places where it appears in the product rule.

(ab)' → ab' + a'b
(x⁸y⁷)' → x⁸(y⁷)' + (x⁸)'y⁷

Does it make sense how I just copied the things that were multiplied together over to their places on the other side?

[unique tiger]

like in this example

[oak chasm]

Right, so let's do that.

5/y + 35/x = 1
(5/y + 35/x)' = 1'
(5/y)' + (35/x)' = 0

Does it make sense what I've done so far?

[alpine sable]

its an easy process, why not google it

[unique tiger]

okay I got the application.. then next step

[oak chasm]

Now we do (5/y)'

(a/b)' → (a'b - ab')/(b²)

(5/y)' → (5'y - 5y')/(y²)

Does it make sense how I filled in the top for a and the bottom for b in the quotient rule?

[unique tiger]

yes

[oak chasm]

OK, so, what's (35/x)'?

[unique tiger]

(35/x)'→ (35'x-35x')/(x^2)

[oak chasm]

5/y + 35/x = 1
(5/y + 35/x)' = 1'
(5/y)' + (35/x)' = 0
(5'y - 5y')/(y²) + (35'x - 35x')/(x²) = 0

So now we have this.

[unique tiger]

is that right?

[oak chasm]

Yes, that's right.

Now, what's 5'?

[unique tiger]

0?

[oak chasm]

Right. What's 35'?

[unique tiger]

0

[oak chasm]

Right. What's x'?

[unique tiger]

is it 1 ?

[oak chasm]

Right. It's dy/dx we're finding, so x is on the bottom, so x' = 1.

5/y + 35/x = 1
(5/y + 35/x)' = 1'
(5/y)' + (35/x)' = 0
(5'y - 5y')/(y²) + (35'x - 35x')/(x²) = 0
(-5y')/(y²) - 35/(x²) = 0

See how I got the last line after filling in 5', 35', and x'?

[unique tiger]

yeah

[oak chasm]

OK, now everything is differentiated.

Now we solve for y'.

[alpine sable]

how the fuck do you do 6th to 0 power

[oak chasm]

@alpine sable Sorry, channel is busy.

[unique tiger]

is the chain rule in application to y'?

[oak chasm]

Not this time.

We got y' just through the quotient rule.

No need for the chain rule.

[alpine sable]

the way you're tutoring him, channel's gonna be busy 24/7

[oak chasm]

@alpine sable There are other channels.

[alpine sable]

(I don't see a problem with it)

[oak chasm]

@unique tiger So, solve for y' in (-5y')/(y²) - 35/(x²) = 0.

[unique tiger]

y'(-5)/(y^2)- 35/(x²)= 0.

[oak chasm]

OK, now what?

[unique tiger]

can we cancel y' to y or just move all non y' to the other side?

[oak chasm]

First, you move all the y' terms to one side and the other terms to the other side.

[unique tiger]

y'(-5)/(y^2)=35/(x²)

[oak chasm]

Good.

Now you factor out y', which is already done.

Then you divide both sides by what's multiplied by y'.

What did you get?

[unique tiger]

(-5)/(y^2)=35y'/(x²)

[oak chasm]

No.

[unique tiger]

i dont know if it is rgiht

[oak chasm]

What's multiplied by y' in y'(-5)/(y²)?

[unique tiger]

-5

[oak chasm]

No, (-5)/(y²) is.

[unique tiger]

okay

[oak chasm]

So, you divide both sides by (-5)/(y²).

[unique tiger]

y'=((35/(x²))/(-5)/(y²)

[ocean seal]

Chai T. Rex

[oak chasm]

OK, so how do you simplify the right side?

[unique tiger]

rationalize?

[oak chasm]

How?

[unique tiger]

(35/x^2)x(y^2/-5)

[oak chasm]

Right.

So, what do you get for that?

[unique tiger]

(35y^2)/(-5x^2)

[copper kite]

i skipped a grade so I have no clue about the problems. Can anyone please explain how to do these?

[oak chasm]

Good, now how can you simplify that?

@copper kite Sorry, channel is busy.

[copper kite]

oh alright

thanks

[unique tiger]

-7x^2y^2

[oak chasm]

Not quite.

How did the x² get into the numerator?

[unique tiger]

so it stays in the deno?

[oak chasm]

Yes.

[unique tiger]

(-7y^2)/x^2

[oak chasm]

Yes, and to make it look a bit nicer, the - can come in front of the fraction.

[ocean seal]

Chai T. Rex

[oak chasm]

But with fractions, you can cancel common factors, but you can't really move something to the other side.

So,
y' = -(7y²)/(x²)
dy/dx = -(7y²)/(x²)

[unique tiger]

ohh okay nice

[oak chasm]

There's also a trick.

(5/y)'

You can do quotient rule.

[alpine sable]

is that your answer bruh

[oak chasm]

Or you can convert to powers.

(5/y)' = (5y⁻¹)'

Do you see how 5/y is also 5y⁻¹?

@unique tiger

[unique tiger]

yeah if you put power to numerator from denominator it changes sign

[oak chasm]

(5/y)' → (5'y - 5y')/(y²) = (-5y')/(y²)
(5y⁻¹)' → -5y⁻²y' = (-5y')/(y²)

One is done with the quotient rule.

One is done with the power rule and the chain rule.

They both get the same answer.

[unique tiger]

ohh nice

[oak chasm]

So that's how you can skip the quotient rule if the power rule applies.

But you should learn the quotient rule because the power rule doesn't always apply.

[unique tiger]

i got last 2 questions if you can help me 🥺

[oak chasm]

OK.

[alpine sable]

can i interrupt a bit

[oak chasm]

y = 5x³ + x - 6

So the inverse will be:

x = 5y³ + y - 6

Do you remember that method of getting the inverse from algebra?

[unique tiger]

ye s

[oak chasm]

Now it wants you to use the formula dy/dx = 1/(dx/dy).

So we get dx/dy.

So, implicitly differentiate x = 5y³ + y - 6.

What do you get after the first step?

[unique tiger]

5y^2y' right?

[oak chasm]

Not quite.

[safe flume]

Hey, can i ask a question here?

[oak chasm]

The first step is this:
(x)' = (5y³ + y - 6)'

@safe flume Sorry, this channel is busy, but it looks like #help-4 is open.

[runic flower]

hey please , answer me
does anyone know the symbol ll f ll oo in calculus ?

[oak chasm]

@runic flower Sorry, channel is busy.

[runic flower]

what

[oak chasm]

@unique tiger The next step is the sum rule:
x' = (5y³)' + (y)' + (6)'

[runic flower]

so you mean you work only in one question here ?

[oak chasm]

@runic flower Yes, that's right. See the rules and tips for getting help in #❓how-to-get-help.

[unique tiger]

one question at a time

[runic flower]

ohh ok thx

[oak chasm]

So, now do the terms one by one.

What's (5y³)'?

[unique tiger]

sooo. 15y^2+0+0 ?

[oak chasm]

Nope.

What's (y)'?

[unique tiger]

15y^2y

[oak chasm]

No, slow down.

What's (5y³)'?

Right.

Well, 15y²y'

What's y'?

[unique tiger]

0

[oak chasm]

No.

[unique tiger]

1

[oak chasm]

Right, because we're doing dx/dy and y is on the bottom.

So, (5y³)' + (y)' + (6)' = 15y² + 1 + 0.

[unique tiger]

okay///

[oak chasm]

So,
x' = (5y³)' + (y)' + (6)'
x' = 15y² + 1
dx/dy = 15y² + 1

[unique tiger]

ohh okay ....what is the difference of f prime of 1 and of f prime of -1

when seen in a question

[oak chasm]

It's not f'(-1).

It's (f⁻¹(x))'

f⁻¹(x) is the inverse function of f(x).

Now, they said to find dy/dx by using dy/dx = 1/(dx/dy).

We have dx/dy, so what's dy/dx?

Sorry?

[unique tiger]

inverse of dx/dy

[oak chasm]

No.

What did we get for dx/dy?

[unique tiger]

15y² + 1?

[oak chasm]

Now use dy/dx = 1/(dx/dy) to get dy/dx.

[unique tiger]

(dx/dy)x(dy/dx)=1

[oak chasm]

I didn't say to get what (dx/dy)(dy/dx) was.

I said to get dy/dx.

[dapper cedar]

help

[oak chasm]

@dapper cedar Sorry, channel is busy.

[dapper cedar]

ok

[unique tiger]

you said use dy/dx = 1/(dx/dy) to get dy/dx

[oak chasm]

Right.

You got (dy/dx)(dx/dy) = 1.

That's not what dy/dx is.

That's what (dy/dx)(dx/dy) is.

[unique tiger]

ohh i get it

dy/dx=1/( 15y² + 1)

[oak chasm]

Right.

[unique tiger]

that's it?

[oak chasm]

Yes.

[unique tiger]

thanks for teaching me it was my first time chatting in the server ... Ill answer the others by myself

[oak chasm]

You're welcome.

It wants you to check by differentiating implicitly.

[unique tiger]

it differs from what i am used to so I struggle

[oak chasm]

That can be done like this:
x = 5y³ + y + 6
1 = 15y²y' + y' + 0
1 = 15y²y' + y'
Fill in the y' you got (y' is dy/dx).

Make sure the sides are equal after you do the math.

Does the checking make sense?

[unique tiger]

yes just substitute

[oak chasm]

Yes, that's right.

[oak chasm]

OK, don't forget that link I put earlier if you want to refresh your solving knowledge.

[unique tiger]

okay thank you again

[oak chasm]

You're welcome.

[teal epoch]

Hello
I think the given that the two of its roots gives the product of 6 is wrong
If it is true, then -2 should be a root to the function, but it gives f(-2)=-72 instead which it just contradicts the statement
So is the question wrong or something?

[oak chasm]

@teal epoch What is the degree of the polynomial?

[teal epoch]

3

[oak chasm]

OK, so there are 3 roots (zeroes).

Do you know Vieta's formulas?

[teal epoch]

I do

[oak chasm]

OK, what is the product of the roots from Vieta's formulas?

[teal epoch]

-12

[oak chasm]

Right.

So, they say the two zeroes' product is 12.

Since there are three zeroes (it's degree 3), one of the zeroes must be repeated.

So, ab = 6.

Where a and b are the two zeroes.

One of the zeroes is repeated.

aab = -12.

a²b = -12

What's a?

[teal epoch]

-2

[oak chasm]

Right, and you tried f(-2) and it didn't give 0.

So, they must be wrong.

[alpine sable]

i don’t get it

[oak chasm]

@alpine sable Sorry, channel is busy.

[alpine sable]

i do care

[oak chasm]

<@&268886789983436800>

[alpine sable]

okay i will go

[oak chasm]

,w roots of x³ - 9x² + 20x + 12

[ocean seal]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Chai T. Rex#7323

[sly mantle]

[alpine sable]

thanks

[oak chasm]

@teal epoch As you can see, the roots are very complicated.

[alpine sable]

oh

[oak chasm]

No pair of the roots will multiply to 6, as the other root would then be -2, but none of the roots are -2.

Does that make sense?

[teal epoch]

yes sir/ma'am

[alpine sable]

One endpoint of a line segment is (8, −1).
The point (5, −2) is one-third of the way
from that endpoint to the other endpoint.
Find the other endpoint.
I don't even know where to begin

Let's name them

how

well

A=(8,-1)

b= (5, −2)

c = (x, y)

[crystal ivy]

In a triangle ABC, on AB and BC the points P and Q are located respectively, so that m <BPQ = m <BCA. If AB = 15 cm, BQ = 5 cm and AC = 21 cm; determine the measure of PQ.

[alpine sable]

right

you can draw it too

do u want me to draw it

?

Sure in a graph so that (AB) is 1/3 of (AC)

I'm pretty sure you can then just read coordinates, you can also find it without drawing it

ok so how do I find it without drawing

The only thing you know is the distance between A and B is 1/3 of the distance between A and C

yes

and it's a line

ok

So find the distance (AB) and *3

Then you have the distance (AC) and you have the coordinates of A so you can find C

[hidden veldt]

two hundred five thousand thirty eight what six digit number does that translate to

im having a brain fart and need the numbers

[alpine sable]

Then you have the distance (AC) and you have the coordinates of A so you can find C

what do u mean

what is the distance (AC) in coordinates

[wary stream]

Pretty sure if you type that into Google, it should tell you

[alpine sable]

I don't know

I got 3.16 for the distance of ab

[hidden veldt]

i tried

[alpine sable]

@alpine sable

I got 3.16

now what

x3= 9.4

if we multiply by 3 that becomes 9.4

then what

No you need to think in coordinates

So find the distance (AB) and *3

yes how am I supposed to get the coordinates

To go from (8,-1) to (5,-2) how do you do

how do I do?

what

yeah

how do I do what?

I can do lots of things

[hidden veldt]

brain fart shit all the way through i remember numbers now thanks

[alpine sable]

You're in (8,-1) and you want to go to (5,-2)

What is the "movement" you need to do

drop the x by 3 and drop y by -1?

exactly

ok so what do I do with that

So now you're in B right?

yes

But what you want is actually going from A to C

ok

which is the same as A to B 3 times

Ok so what am I supposed to do

I get this

@alpine sable

yes

Well you multiply by 3 the movement you did

Then you go from the coordinates of A and you go from A to C so you end up in C

[alpine sable]

lol theres literally a formula

tell me

[alpine sable]

+9 -3

what is that ab

[alpine sable]

coordinate of point that devides line segment at ratio m:n

zamarus that doesn't get to the answer

in your case, 3:1

yes it does

(5,-2)

becomes 14, -5

right?

It's the movement A-)C so you go from A(8,-1)-(9,3)

oh

C is (-1,-4)

it does work

[sage swift]

Guys im doing exponential functions right now but i dont think im doing my work correctly can anyone help me please

[alpine sable]

lozano I still have questions to ask zamarus

@alpine sable why do you have to multiply by 3

[sage swift]

My bad sorry

[alpine sable]

no prob

Well you multiply by 3 the movement you did
Then you go from the coordinates of A and you go from A to C so you end up in C

I don't get this

because you said yourself that B is at 1/3 of the segment [AC]

So [AC]=3*[AB]

oh right

ok

you're in A you do the move to get from A to C so you're in C

I see

ok ty

np

[knotty hedge]

hi

hi

dying

[alpine sable]

Where are you stuck

[knotty hedge]

no idea what to do

[wary stream]

FOIL

[knotty hedge]

WHAT

[alpine sable]

It says expand the bracket

[knotty hedge]

IK

BUT I SUCK AT MATH LOL

[alpine sable]

So expand it

[wary stream]

First, outer, inner, last

[gritty surge]

hi

[wary stream]

WHY ARE YOU IN CAPS? IS YOUR CAPS LOCK BUTTON BROKEN?

[knotty hedge]

YE

[gritty surge]

does anyone know engineering mechanics

[knotty hedge]

no

[wary stream]

Look up foil method on Google

That will help

[knotty hedge]

ye maybe later

[wary stream]

You asked for help, there's the help

FOIL is an important concept

[knotty hedge]

bruuuuuuuuuuh

[alpine sable]

We can't help with you learning the method

[wary stream]

You're going to use FOIL a lot in the future, especially when it says expand

[knotty hedge]

what do they mean w the second one tho bruh

[alpine sable]

initial is for t=0

i think

[wary stream]

It means find the roots of the equation

[alpine sable]

initial velocity is the constant after first differentiation

Yeah it says the coefficient of t

Make sens

[wary stream]

Don't think they know what differentiation is

So use layman terms

[knotty hedge]

ye

[alpine sable]

never heard that my entire life
except for in chemistry

[wary stream]

It means use simpler words for people who don't understand that concept/is not familiar with that concept

[alpine sable]

just wondering, if x^2 is x squared, and x^3 is cubed, then would it make sense to say x^4 is x tesseracted? and x^5 is x penteracted?

[alpine sable]

i see now, maybe cause im not a native English

[knotty hedge]

bruh

so smart

[signal rampart]

Could somebody explain this to me? Appreciate it

[alpine sable]

impressive

[knotty hedge]

ikr

[signal rampart]

Im quite young and I struggle with math, im in 8th grade

[knotty hedge]

omg so your 12 :000000000000000000000000

13

[signal rampart]

no?

[knotty hedge]

14

?

[alpine sable]

$4^{2}\times 2\times \dfrac{1}{3}+2,4$

[signal rampart]

14

[ocean seal]

Zamarus

[knotty hedge]

AYO GOOD MAN YEEEEEEEEEEE

[alpine sable]

$\begin{aligned}4^{2}\times 2\times \dfrac{1}{3}+2,4\ 4\times 4\times \dfrac{2}{3}+2,4\ 16\times \dfrac{2}{3}+2,4\end{aligned}
$

[alpine sable]

fr, is that 8th grade maths?😳

[signal rampart]

I guess?

[ocean seal]

Zamarus

[alpine sable]

the level is way difficult in our country

[signal rampart]

There are harder I guess 1 min

[alpine sable]

$\begin{aligned}\dfrac{32}{3}+2,4\ \dfrac{3\times 10+2}{3}+2,4\end{aligned}$

[ocean seal]

Zamarus

[alpine sable]

=12,4+2/3

[wary stream]

90% sure this is incorrect

[alpine sable]

Well i'm retarded then

[wary stream]

That's a mixed fraction

[carmine girder]

please

[wary stream]

Just find x

[carmine girder]

what is it help

[alpine sable]

for real

[wary stream]

Also, weird placement for the negative signs

Use area of triangle equation

[carmine girder]

could you help me its a test

[wary stream]

Literally move everything that is not x to the other side

[wary stream]

Can't help with exams

[wary stream]

<@&268886789983436800>

[alpine sable]

lmao

[carmine girder]

okay sorry

[alpine sable]

Hey, currently learning analytic geometry and was wondering how can you find the rectangle points when all you know is the x and y values of only 2 points?

[night geyser]

those 2 points are opposite corners of the rectangle

[wary stream]

Such a beautiful diagram

[alpine sable]

wow,

[alpine sable]

Yes let's take that as an example

[night geyser]

if you assume its grid-ailgned you can fill out the rest.

note that the "new" points take their x coordinate from 1 point, and their y coordinate from the other point

[fringe depot]

help

😥

[wary stream]

Don't spam

[night geyser]

the woes of #help-0 .

dont spam and dont interrupt.

[alpine sable]

Yeah but what if it's like that?

true

[night geyser]

indeed, thats why i said "assuming its grid aligned"

[alpine sable]

then its not possible to find the other points

[wary stream]

It's because you did not read #rules or #❓how-to-get-help

[night geyser]

techniaclly there are infinitely many rectangles for any pair of points

just 2 points isnt enough info

if youre ONLY given 2 points, you will not be able to uniquely determine a rectangle.

[alpine sable]

Yeah makes sense, thank you so much

[alpine sable]

Yep thanks 🙂

why do generating functions work?

From what I know, the x^n term is just a placeholder and we almost never compute it, the sequence goes like that but why does it work I still don't get it

I know why we use generating functions, basically because functions are easier to deal with than sequences, and also that power series is not the only way to do it. But I still think I am confused.

Probably something which is easy to grasp is what I need right now

u need to know what functions are

Task "come up with an inequality, that its solutions would be [2;+∞)"
so -2+x>=0 is a valid answer?
Task "come up with an inequality, that its solutions would be (-∞;-3)"
so -x-3>0 is a valid answer?

[novel siren]

yes and yes @alpine sable

[alpine sable]

[-5;5]

what would be the inequality

?

cant come up with it

<@&286206848099549185>

[ionic jewel]

$-5 \leq x \leq 5$

[ionic jewel]

or

$x^2 \leq 25$

[alpine sable]

thank you

[sleek quest]

It's probably very stupid but shouldn't
(x^2+2x+4)(x^2-2x+4) equal x^4 - (2x+4)^2?

It isn't and I don't understand

[alpine sable]

-(2x+4) = -2x-4

[sleek quest]

Does the sign affect the entire expression here

I need to expand the special product first no?

[alpine sable]

the sign affects the whole term, that's why
you can't apply it to formula (a+b)(a-b)=a^2-b^2

[sleek quest]

So I can't apply the formula in that case?

I have to multiply each term?

[alpine sable]

(x^2+2x+4)(x^2-2x+4) ≠ x^4 - (2x+4)^2
(x^2+2x+4)(x^2-2x-4) = x^4 - (2x+4)^2

[sleek quest]

Ohh

Got it thanks

[quiet folio]

how do i solve this

[tall kayak]

D?

[quiet folio]

it turned out to be a@tall kayak

but it took lots of time

and the 2nd one is ez B

[tall kayak]

oh

ye

sorry

A

silly me

[frank cairn]

Can anyone help me solve?

[sleek quest]

@quiet folio can you explain how you reached 8 I am curious

Curious but stooped

[frank cairn]

The power in a circuit with a current of 1.25x10-⁶ and a resistance of 4.8x10⁵ ohms is given (1.25x10-⁶)² x (4.8x10) w.

What is the power in the circuit?

[alpine sable]

ops misread

[quiet folio]

ok 1 sec

[sleek quest]

Got it

Thanks

[quiet folio]

I hope u can understand my hand writing kinda trash

[sleek quest]

It's ok, no worries

[quiet folio]

No problem m8 btw
Are these questions harder than the math sat

[sleek quest]

No idea, not from the US

LMAO