help-26

[gilded fractal]

If we multiply by 2

Its 2

[winged moat]

$2 \cdot \frac{x}{2}$

[gilded fractal]

2/2 is 2

[thorny flame]

USS-Enterprise

[winged moat]

We multiply the whole number with the top (the numerator)

$= \frac{2x}{2}$

[thorny flame]

USS-Enterprise

[winged moat]

And the 2's cancel

Because 2/2 = 1

[gilded fractal]

2/2

Oh

[winged moat]

and we're left with x

so, to get rid of fractions

Multiply by the denominator

If we had $\frac{3x}{5} = 3$

[thorny flame]

USS-Enterprise

[winged moat]

The denominator is 5

We multiply both sides by 5

$5 \cdot \frac{3x}{5} = 3 \cdot 5$

[thorny flame]

USS-Enterprise

[winged moat]

$\frac{5 \cdot 3x}{5} = 15$

[thorny flame]

USS-Enterprise

[winged moat]

Notice how 5's again cancel, because 5/5 = 1

$3x = 15$

[thorny flame]

USS-Enterprise

[winged moat]

divide by 3, $x = 5$

[thorny flame]

USS-Enterprise

[winged moat]

get it?

[gilded fractal]

No

💀

[winged moat]

hm

[gilded fractal]

What do you mean with cancel out

What

[winged moat]

you gotta simplify the fraction

$\frac{5 \cdot 3x}{5}$

[thorny flame]

USS-Enterprise

[winged moat]

You divide both top and bottom by 5

Or 5's cancel

we get $\frac{1 \cdot 3x}{1}$

[gilded fractal]

Then were left with 3x

[thorny flame]

USS-Enterprise

[winged moat]

$3x$

[thorny flame]

USS-Enterprise

[gilded fractal]

So if we try a now

We got 3x/8 first

We multiply by 8?

[winged moat]

Wait a minute

I wanted to introduce what happens if we have multiple fractions with different denominators

[gilded fractal]

Get 8times 3x/8

[winged moat]

Say we've got $\frac{2x}{3} + \frac{x}{5} = 9$

[thorny flame]

USS-Enterprise

[winged moat]

We first gotta get rid of fractions

But one denominator is 3, the other is 5

[gilded fractal]

Times 3

[winged moat]

That will only get rid of the first fraction

Is there such a number we can use to get rid of both fractions?

[gilded fractal]

M

Idk

Dont think so

[winged moat]

We can.

We need to find the lowest common denominator

This means finding the multiple which both numbers share

So write multiples of 3 = 3, 6, 9, 12, 15, 18, 21

And 5 = 5, 10, 15, 20, 15

[gilded fractal]

15

[winged moat]

Exactly

[gilded fractal]

Ohhhh

I get it

And then division 15

[winged moat]

So if we multiply the sides by 15 we'll get rid of both fractions at the same time!

Let's do it the long way

[gilded fractal]

Times 3

[winged moat]

We get $15 \cdot \left(\frac{2x}{3} + \frac{x}{5}\right) = 15 \cdot 9$

[thorny flame]

USS-Enterprise

[winged moat]

We just multiplied both sides by 15 for now, you get this right?

[gilded fractal]

Yes

[winged moat]

Now let's distribute 15, slowly

$15 \cdot \frac{2x}{3} + 15 \cdot \frac{x}{5} = 135$

[thorny flame]

USS-Enterprise

[winged moat]

Get this as well?

We just distributed 15 through the parantheses

[gilded fractal]

Yes i get it

[winged moat]

Okay now we do what we did before

[gilded fractal]

Division

[winged moat]

$\frac{15 \cdot 2x}{3} + \frac{15 \cdot x}{5} = 135$

[thorny flame]

USS-Enterprise

[winged moat]

Now we simplify the fractions

Which means dividing both the top and the bottom by the same number

[gilded fractal]

But why does the 15 only go to the upper part and not the thing down there

[winged moat]

This is the definition of multiplying a whole number with a fraction

You multiply it only with the top part

If we say 3 times 1/2, we get 3/2 right

Not 3/6

[gilded fractal]

What

[winged moat]

Say $3 \cdot \frac{1}{2}$

[thorny flame]

USS-Enterprise

[winged moat]

3 times 'one half'

This is the same as $\frac{1}{2} + \frac{1}{2} + \frac{1}{2}$

[thorny flame]

USS-Enterprise

[winged moat]

And you know how to add fractions right?

[gilded fractal]

Whats a fraction

[winged moat]

This

$\frac{a}{b}$

[thorny flame]

USS-Enterprise

[winged moat]

Is a fraction

Means a divided by b

[gilded fractal]

Mhm

[winged moat]

So yeah it's the definition

Multiplying a fraction (a/b) with a whole number (c), you multiply the whole number with the top part (the numerator)

ac/b

Get it?

[gilded fractal]

No i dont get why only the upper

[winged moat]

Okay

Let's say

$4 \cdot \frac{1}{3}$

[thorny flame]

USS-Enterprise

[winged moat]

We can rewrite 4 as a fraction

[gilded fractal]

1/4

[winged moat]

$\frac{4}{1} \cdot \frac{1}{3}$

[thorny flame]

USS-Enterprise

[winged moat]

No, 4/1

1/4 is 0.25

[gilded fractal]

Ohhhh

i get ittt

[winged moat]

And this is 4 * 1 / 1 * 3

See why we don't multiply the bottom

Because it's always getting multiplied by 1

So it remains the same

[gilded fractal]

Just takes you to pull out the baby explanations thats what i need

[winged moat]

😂

[gilded fractal]

[winged moat]

Anyway

We've got this

$\frac{15 \cdot 2x}{3} + \frac{15 \cdot x}{5} = 135$

[thorny flame]

USS-Enterprise

[winged moat]

We gotta simplify each fraction

[gilded fractal]

Divided by 15

[winged moat]

Because in the start we multiplied both sides by the lowest common denominator

THis guarantees us that we will get rid of fractions

So if we divide both the top and bottom in the first fraction by 3

We get $\frac{15 \cdot 2x \div 3}{3 \div 3}$

[thorny flame]

USS-Enterprise

[winged moat]

Which is $\frac{5 \cdot 2x}{1}$

[thorny flame]

USS-Enterprise

[winged moat]

Get it?

[gilded fractal]

Whwt

Slower

Baby explanations is all i need

[winged moat]

Right, we've got $\frac{15 \cdot 2x}{3}$

[gilded fractal]

[thorny flame]

USS-Enterprise

[winged moat]

We gotta simplify this.

Because in the start

[gilded fractal]

3

[winged moat]

Wait.

Because we in the start multiplied everything by the lowest common denominator (15)

This guarantees us that we can get rid of fractions if we divide by the denominator on both the top and bottom

The denominator here is 3

[gilded fractal]

[winged moat]

So we divide the top and bottom by 3

[gilded fractal]

There stays 5times 2x

[winged moat]

Exactly

And we've gotten rid of the first fraction

$\frac{15 \cdot x}{5}$

[thorny flame]

USS-Enterprise

[winged moat]

What do we divide the top and bottom here with?

[gilded fractal]

5

[winged moat]

And we get

[gilded fractal]

3times x

3x

[winged moat]

Exactly

So in the first we have 5 times 2x, or 10x

In the second 3x

[gilded fractal]

[winged moat]

$10x + 3x = 135$

[thorny flame]

USS-Enterprise

[gilded fractal]

But dont we have to do the same in each stop on every number why it wouldnt work

[winged moat]

What do you mean?

[gilded fractal]

We have to do the same on each step

10+20x=500|-10

There we have to do the same

Thing on every number

[winged moat]

Ah I think you are mistaking two things

In the start

We multiplied both sides of the equation by 15

[gilded fractal]

[winged moat]

This means multiplying everything by 15

But then we simplified each fraction individually

Simplifying a fraction still has the same value overall

[gilded fractal]

OHHHH

[winged moat]

For example, $\frac{2}{4} = \frac{1}{2}$

[gilded fractal]

I get it

[thorny flame]

USS-Enterprise

[gilded fractal]

[winged moat]

So we can do whatever we want to each individual fraction

[gilded fractal]

yeaaa

[winged moat]

As long as its overall value remains the same

Okay, so we've got $10x + 3x = 135$

[thorny flame]

USS-Enterprise

[winged moat]

Now what?

[gilded fractal]

Divided by a number but idk wich one

[winged moat]

First, we collect like terms

[gilded fractal]

Because 10 is straight and 3 isnt

[winged moat]

WHich means adding 10x + 3x

This the same as 13x

Do you get this?

[gilded fractal]

Oh yeah we could do that

[winged moat]

Okay, we've got $13x = 135$

[thorny flame]

USS-Enterprise

[winged moat]

What's the final step

[gilded fractal]

E

[winged moat]

Think of how would you solve $2x = 4$

[thorny flame]

USS-Enterprise

[gilded fractal]

With a 2

[winged moat]

you've got 2 times x, but you need 1 times x

so we divide both sides by 2 right?

x = 2

[gilded fractal]

Ye

[winged moat]

So what do we do in $13x = 135$

[thorny flame]

USS-Enterprise

[gilded fractal]

Theres no number to devide with

[winged moat]

we've got 13 times x, we need 1 times x

How?

We don't care what happens on the right

It can be as ugly as we want it to be

We just need 1x = something

[gilded fractal]

OHHH

13

[winged moat]

what 13?

[gilded fractal]

Divide

[winged moat]

exactly

$x = \frac{135}{13}$

[thorny flame]

USS-Enterprise

[winged moat]

We don't care what's on the right

We've got the value of x

Now we of course have to simplify the fraction generally

but 135/13 can not be simplified

So we leave it as is

Anyway

To your original example

[gilded fractal]

what

How

What

We divide be 13

[winged moat]

$\frac{3x}{8} - \frac{x}{6} = 5$

[thorny flame]

USS-Enterprise

[winged moat]

Yes

We divide both sides by 13

We get 13x divided by 13 = 135 divided by 13

[winged moat]

Yes, x is approx 10.39

We don't write approximations unless specifically asked to though

We leave the most simplified exact value

$x = \frac{135}{13}$

[thorny flame]

USS-Enterprise

[winged moat]

Is the most simplified exact value

[gilded fractal]

I dont get it

[winged moat]

If you insert 135/13 instead of x into our original equation you'll see that the left side equals the right side

This means solving an equation

$\frac{2x}{3} + \frac{x}{5} = 9$

[thorny flame]

USS-Enterprise

[winged moat]

This was our original equation, correct?

We solved it.

We got x = something

[gilded fractal]

OOHHHH

I get it

But why

[winged moat]

If we insert that something instead of x

We'll see that the left side equals the right side

[gilded fractal]

Why just not completely write it down

[winged moat]

What do you mean?

[gilded fractal]

Instead of 135/13

Just the x

And the 10,39 something

[winged moat]

Because that's an approximation

[gilded fractal]

And what does that mean

Approximation

[winged moat]

An exact value is something like $\frac{1}{3}$

[thorny flame]

USS-Enterprise

[winged moat]

An approximation is saying 0.333

Because it goes on forever

[gilded fractal]

Oh its a forever going number

[winged moat]

Yes, $\frac{135}{13}$

[gilded fractal]

Yeah get it now

[thorny flame]

USS-Enterprise

[winged moat]

goes on forever

[gilded fractal]

Lets do a

[winged moat]

$\frac{3x}{8} - \frac{x}{6} = 5$

[thorny flame]

USS-Enterprise

[winged moat]

So what did we say? First things first we get rid of fractions

We have two fractions with two different denominators, so what do we find?

[gilded fractal]

Times 24

[winged moat]

Exactly

$24 \cdot \frac{3x}{8} - 24 \cdot \frac{x}{6} = 24 \cdot 5$

[thorny flame]

USS-Enterprise

[winged moat]

$\frac{24 \cdot 3x}{8} - \frac{24x}{6} = 120$

[thorny flame]

USS-Enterprise

[winged moat]

Simplify both fractions

[gilded fractal]

Divide by 2

[winged moat]

What did we say earlier

We can and must divide the top and bottom by the denominator in each fraction

[gilded fractal]

I forgot

[winged moat]

Because we are guaranteed to get rid of fractions

[gilded fractal]

Oh

Wait

So first one is 8

Divide by 8

[winged moat]

Exactly

What do we get

[gilded fractal]

8times3x

[winged moat]

How'd you get 8?

You divide 24 by 8

[gilded fractal]

Oh

4

Mb my brain isnt braining

[winged moat]

Try again

[gilded fractal]

Getting pretty late

[winged moat]

😅

[gilded fractal]

3

😭😭😭

[winged moat]

there you go

[gilded fractal]

Suye

[winged moat]

3 times 3x

[gilded fractal]

Dude

[winged moat]

And in the second fraction?

[gilded fractal]

4times x

So 4x

[winged moat]

There you go

$3 \cdot 3x - 4x = 120$

[thorny flame]

USS-Enterprise

[winged moat]

Solve this

[gilded fractal]

Wait my friend wants to learn it to let me explain it to him too

After we solve this a

[winged moat]

Yeah, I'll have to go after this one unfortunately

[gilded fractal]

[winged moat]

Been here for over two hours now 😄

[gilded fractal]

[winged moat]

Anyway, solve the equation

$3 \cdot 3x - 4x = 120$

[thorny flame]

USS-Enterprise

[tidal glade]

hello

[winged moat]

Hello

[gilded fractal]

Thats my friend

[tidal glade]

just watching

[winged moat]

👍

[gilded fractal]

Wait what how do i do it

Divided by 3?

[winged moat]

The same way we've been doing it this whole time

No

Simplify the most you can first

Can you multiply 3 times 3x?

[gilded fractal]

OH FORGOT THAT EXOSTS

Exists

[winged moat]

😂

[gilded fractal]

9x

[winged moat]

Okay

Then can you substract 4x from 9x?

so 9x - 4x = 120?

[gilded fractal]

No

[winged moat]

Really?

[gilded fractal]

I dont know

[winged moat]

hm

Say we have 9 apples minus 4 apples

how many apples do we have

[gilded fractal]

25

[winged moat]

What

[gilded fractal]

5

[winged moat]

Yes

So then why can't we do 9x - 4x?

think of x as apples

9 apples minus 4 apples

[gilded fractal]

Oh i thought u said divide

[winged moat]

"substract"

[gilded fractal]

Yea just me being smart

So 5x

5x=120

And that divided by 5

x=24

[winged moat]

there you go 🙂

[gilded fractal]

[winged moat]

I have time for one more I'd say

Can you repost please

Don't want to scroll 200 messages back

[winged moat]

$8,4z - 4,3 - 0,9z + 6,2 - 7,5z = 1,9$

[thorny flame]

USS-Enterprise

[winged moat]

What would you do here?

[gilded fractal]

Divide by 0,3

No

Nvm

[winged moat]

Why do you always want to go straight to dividing 😅

Try to simplify as much as you can first!

[gilded fractal]

I dont know

[winged moat]

Again, think of apples

[gilded fractal]

Apple

The z turns into 0

[winged moat]

You've got 8.4 apples minus something minus 0.9 apples + something - 7.5 apples = 1.9

Collect apples and collect something

[winged moat]

Correct

We end up with zero apples

or 0z

or just nothing

What about normal numbers

-4.3 + 6.2

[gilded fractal]

1,9

[winged moat]

So we end up with $1,9 = 1,9$

[thorny flame]

USS-Enterprise

[winged moat]

Now this is very important

What does this tell us

[gilded fractal]

That we shortened it

[winged moat]

Sure

Let me ask you this

Is the equation true or not?

Is 1.9 equal to 1.9

[gilded fractal]

No

[winged moat]

Do we have any z in there?

[gilded fractal]

No

[winged moat]

So

If I pick z = 3 or z = -9345

Does it matter?

Or will the equation always be true

[gilded fractal]

Huh

[winged moat]

This can be a little tricky to understand

[gilded fractal]

[winged moat]

You see how all the z's added up to 0, right?

[gilded fractal]

Yes that means that we got no z's

[winged moat]

So no matter what z we pick we always end up with 0z, which is 0 right?

[winged moat]

And we end up with an equation 1.9 = 1.9

Which is always true

Okay let's try this

Let's say z = 2

let's solve the equation

$8.4 \cdot 2 - 4.3 - 0.9 \cdot 2 + 6.2 - 7.5 \cdot 2 = 1.9$

[thorny flame]

USS-Enterprise

[winged moat]

$16.8 - 4.3 - 1.8 + 6.2 - 15 = 1.9$

[thorny flame]

USS-Enterprise

[winged moat]

$12.5 - 1.8 + 6.2 - 15 = 1.9$

[thorny flame]

USS-Enterprise

[winged moat]

$10.7 + 6.2 - 15 = 1.9$

[thorny flame]

USS-Enterprise

[winged moat]

$16.9 - 15 = 1.9$

[thorny flame]

USS-Enterprise

[winged moat]

$1.9 = 1.9$

[thorny flame]

USS-Enterprise

[winged moat]

The equation is true!

So

If we, after solving

Get rid of the variable

[gilded fractal]

I get it

[winged moat]

And get an equation that's true (1.9 = 1.9)

This means that no matter what z we pick

[gilded fractal]

But this one isnt the same why is it true

[winged moat]

The equation is always true

[gilded fractal]

We just tested it by replacing the z's

[winged moat]

16.9 - 15 = 1.9

[gilded fractal]

Oh

[winged moat]

Yeah.

[gilded fractal]

Yeah

[winged moat]

how about this

[gilded fractal]

But why did u do times 2

[winged moat]

Because we picked z = 2

[gilded fractal]

And not times 1,9

[winged moat]

So we replace z with 2

if we picked z = 3432154 we would have gotten the same result

we just replace z with 3432154

[gilded fractal]

What

[winged moat]

Say this equation

$2x = 4$

[thorny flame]

USS-Enterprise

[winged moat]

We divide by two

$x = 2$

[gilded fractal]

So its not important what to replace jt with?

[thorny flame]

USS-Enterprise

[winged moat]

If we replace x in the original equation we get 2 * 2 = 4, 4 = 4 which is true

[winged moat]

In this case, because there's no z in our solution. I'm trying to show you that no matter what z we pick the equation is always true

Our solution to the equation was $1.9 = 1.9$

[thorny flame]

USS-Enterprise

[winged moat]

Rember that?

There's no z there

[gilded fractal]

But with the z it would be important

[winged moat]

And 1.9 equals 1.9

This means that no matter which z we pick the equation is always true

If we had something like z = 1.9 the equation would only be true if z = 1.9,. If we picked z = 2 the equation would not be true

But here's another extreme:

If we also somehow got rid of z's

And ended up with 1 = 1.9

This means that no matter which z we pick the equation is never true

Because we would always end up at 1 = 1.9, which isn't true

[gilded fractal]

Alr

[winged moat]

So

What have we learned

A linear equation

Which means the highest degree of the variable is 1 (z^1)

[gilded fractal]

solution quantity its called

Right?

[winged moat]

Can either have 1 solution, z = something, no solutions (something like 2 = 3), or unlimited number of solutions (something like 1.9 = 1.9)

[winged moat]

Number of solutions, yes.

[gilded fractal]

Got it

Lep stalking us

[winged moat]

😂

[tidal glade]

ohoh

[winged moat]

Anyway, I gotta go now.

Hope I've been of help

[tidal glade]

bye

[gilded fractal]

Alr thanks for the help mage

Mate

[winged moat]

And wish you both all the best

See you around!

[gilded fractal]

Yea bye

[tidal glade]

thanks you to man

[gilded fractal]

[topaz sinew]

@gilded fractal Has your question been resolved?

[gilded fractal]

@winged moat ik ur busy but how did we come to 120 in the task 3 a

[topaz sinew]

Channel closed

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Available help channel!

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Remember:
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[wraith breach]

I have no clue what to do

.close

[topaz sinew]

Channel closed

Closed by @wraith breach

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[echo fulcrum]

Can't figure out why my solution is wrong

[topaz sinew]

Show your work, and if possible, explain where you are stuck.

[echo fulcrum]

alr asked for help in another channel and they gave up

This was the solution i ended up with by substituting

[quick gorge]

I assume it’s because you can simplify

2.5 / 2 = 1.25

[echo fulcrum]

this software usually allows unsimplified answers

Yeah still wrong

[quick gorge]

hmm

[echo fulcrum]

Error says "Can't take sqrt of -8"

[quick gorge]

I see that you substituted sqrt(l - d^2) for h

[echo fulcrum]

Which means whatever method it uses to verify answers is coming into trouble cause of the sqrt

[quick gorge]

may want to check that again

[echo fulcrum]

I cant find a flaw with that step

other than +- beingignored

[quick gorge]

you used pythagorean theorem right?

[echo fulcrum]

but i ignored +- cause its height

[echo fulcrum]

yeah

[quick gorge]

h^2 + d^2 = l^2

solve for h

[echo fulcrum]

h = sqrt(l^2-d^2)

[quick gorge]

yep

just missed an l^2

[broken niche]

the l is a 1

[quick gorge]

oh you’re rifht

wait what

then why do they say in terms of 1 and d

I feel like they meant the letter L

[echo fulcrum]

even if they meant the letter L

we just end up where we started

[quick gorge]

wdym

[echo fulcrum]

in the corner i evaluated it that way

we just end up with the same thing

[quick gorge]

right

dL/dt = 0 as the ladder doesn’t change length

so it’s the same thing

but h should be sqrt(L^2 - d^2)

[echo fulcrum]

ngl

this problem is diabolical

they DID mean lower case l

why on EARTH would a lower case L look EXACTLY like a 1

[quick gorge]

😭

[echo fulcrum]

this was in fact the solution

[broken niche]

lmao

[quick gorge]

yeah petition to stop using l in math

[echo fulcrum]

no thats straight up a 1 they cannot gaslight me

thank u guys

[quick gorge]

np

[echo fulcrum]

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[toxic stirrup]

why are the values for the maxima wrong

[toxic stirrup]

this is a graph of the function

[chilly walrus]

your task is to find it on the interval [0; pi]

there's definitely a maximum on that interval

[toxic stirrup]

it goes to inf though

[chilly walrus]

nope

it does not

zoom out

also, think logically, why would it go to infinity?

neither e^x nor cos(x) ever go to infinity for any finite x

[toxic stirrup]

oh

right

how could have i found that value without a graph?

[chilly walrus]

what's cos(pi)?

[toxic stirrup]

-1

[chilly walrus]

mhm

[toxic stirrup]

like this?

[chilly walrus]

yes

that's the correct value

[toxic stirrup]

yeah

[chilly walrus]

and cos is multiplied with -5e^x

which is negative

[toxic stirrup]

right

[chilly walrus]

so negative and negative cancel out

[toxic stirrup]

thank you

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[toxic stirrup]

i found the solution to this problem with a graph, but how could i have found it without a graph

[royal fiber]

wait you from the academica server 😅

[toxic stirrup]

I am lol

[royal fiber]

basically your critical points are unchanged by the absolute value and you can find them like you find the critical points for any other polynomial

solve
f(x)=0
f'(x)=0
f''(x)=0

[toxic stirrup]

oh

so i have to do it for first and second derivatives too

thanks!

[royal fiber]

yup welcome

yeah first derivative finds extremums
second derivative finds the inflection point

[tough cargo]

wait what

inflection points aren't necessarily critical points

[royal fiber]

depends on your text book and context 🤷‍♂️

in any case here the inflection point is also a root so all good

[tough cargo]

idk

Can't find any source that says inflections are critical points

[royal fiber]

im pretty sure i had a class where that was the case

🤔

[tough cargo]

Anyways f(x) = 0 is a critical point too buuuuuut, only for this problem

[royal fiber]

but there are 6 input boxes

so in other questions where there is no overlap they might want it?

[tough cargo]

critical points apperantely in places I googled have derivative that don't exist or derivative equal to 0. It just happens to be here that derivative doesn't exist if and only if f(x) = 0 I think

[royal fiber]

ah right

[tough cargo]

o well I guess if he winds up needing an extra root inflection is worth a shot? I haven't checked if it's already a value or not

[royal fiber]

he should probably ask his teacher/prof

[tough cargo]

oh yeh I guess they are counting inflection points cus f(x) = 0 gives x = -1

[royal fiber]

since in some places inflection is considered critical

[tough cargo]

(and 2 more roots that are not 1)

[royal fiber]

the inflection point here is at 1...

and at 1 there happens to also be the root

[tough cargo]

oh yeh wrote it wrong in my notepad

the poly I mean

That's why I got -1 as a root

OK so inflection should ask teacher but it doesn't matter for this problem is final decision

[toxic stirrup]

well the derivative gives me 0 and 2

i believe using the equation as it is gives me the -0.732 and 2.732

[royal fiber]

no this is supposed to be the outcome

dont forget solving f'(x)=0 gives you the extremums

[toxic stirrup]

right

[royal fiber]

you can visualise it by the tangent on the line

[toxic stirrup]

but those count as critical points

[royal fiber]

at those points the tangent is horizontal or of slope zero

[royal fiber]

yes

[toxic stirrup]

wait so what exactly do i do to find all the points on paper

first solve f(x)=0

then f'(x)=0

[royal fiber]

it should also give 1

[toxic stirrup]

do i need the second derivative

[toxic stirrup]

yeah i realized that

[royal fiber]

it is a cubic polynomial with 3 roots

you can always know the number of roots by the degree of the polynomial

[toxic stirrup]

and can i kinda ignore the ||

[royal fiber]

you might wanna ask your teacher if the infelction point is considered critical

[tough cargo]

yeah but it doesn't matter for this problem since inflection point is actually a root of f already

[toxic stirrup]

well just using f(x) and f'(x) gives all the points in the answer

[royal fiber]

yes, you can see in the image i sent how the polynomial without the absolute (green)
still has the same critical points as the polynomial with the absolute (blue)

the absolute only takes whatever is under the x axis and reflects it to the top without changing its shape

[royal fiber]

the inflection point happens to overlap here with the root at x=1

i'd still ask if you can

@toxic stirrup dont forget to close the channel if you are done with it by typing .close

[toxic stirrup]

oh yeah

thanks for the help!!

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[late crypt]

Can I solve this with ^nsqrt method

?

[ember lodge]

Can u pls take a better photo?

I cannot see the question clearly

or u can type in LaTex is fine tho

[late crypt]

Im not sure how

Its |sin n|

/ 2^n

[ember lodge]

$\sum_{n=0}^{\infty}\frac{|sin(n)|}{2^n}$

[thorny flame]

DaWhiteXD

[ember lodge]

can't help oof

pced out*

[late crypt]

What

[opal vault]

I know how to do it without the absolute value, maybe I can try with

[late crypt]

What differs

Oh and btw

I csnt use integers

[opal vault]

integrals you mean

[late crypt]

Yes

[opal vault]

ok so without the absolute value, the way you would do it is write sin(n) = Im(e^(in))

[late crypt]

Wtf

Its in R not C

Oh wait

[opal vault]

Yes but there is no other way to compute it

[late crypt]

Maybe it is in C

Oh

RllyM

[opal vault]

sin(x) is the imaginary part of e^(ix), and we use that fact proudly

and so, the whole sum becomes :

the imaginary part of $\sum_{n=0}^{\infty}\Big(\frac{e^i}{2}\Big)^n$

[thorny flame]

rafilou2003

[late crypt]

Can u not just see it as |sin n| is between 0 and 1 so since its divided by a much higher number it will be not infinite

[opal vault]

we find the sum of geometric sum

[opal vault]

oh you just want to show it converges right?

[late crypt]

Yea

[opal vault]

then $0\leq \frac{|\sin(n)|}{2^n}\leq \frac{1}{2^n}$

[thorny flame]

rafilou2003

[opal vault]

use comparison test

[late crypt]

This cant be solved can it

[opal vault]

no it can't

which is why we need comparison test

[late crypt]

Ok?

[opal vault]

,rcwr

[late crypt]

???

[late crypt]

Ooh

[opal vault]

neither

[late crypt]

Whay neither

[opal vault]

won't work either

sin(n) is unpredictable

[late crypt]

But u have

Sinn/sin n

Its 1

[opal vault]

and what about sin(n+1)/sin(n)?

[late crypt]

Im talking about that

Thats 1

Cuz n is inf

[opal vault]

no

[late crypt]

What

[opal vault]

don't work like that

sin(n) has NO limit in +infinity

[late crypt]

Then wtf to do lil

[opal vault]

comparison test

[opal vault]

this

[late crypt]

But

Wait

Even if its sin n / sin n?

[opal vault]

it's sin(n***+1***)/sin(n)

[late crypt]

U take n

N(1+0)

So its n

Sin m

Sin n

[opal vault]

no

[late crypt]

Why

[opal vault]

N+1 is not N(1+0)

[late crypt]

With infinity it is

[opal vault]

except sin has NO limit in infinity

so no

[late crypt]

Yes but we have sin n / sin n

U arent calculating sin

U are doing n+1/n

Its n/n

Then sin n/ sin n

[opal vault]

it's like saying $\frac{(-1)^{n+1}}{(-1)^n} = 1$ because $n+1 = n$ at infinity

[late crypt]

Whatever the value it has to be 1

[thorny flame]

rafilou2003

[opal vault]

do you understand your mistake?

even if 1 is small compared to n, it's sufficient to destabilize the whole thing

[late crypt]

Like what?

[opal vault]

If $0\leq u_n \leq v_n$ and $\sum v_n$ converges, then $\sum u_n$ converges

[thorny flame]

rafilou2003

[opal vault]

then this

[late crypt]

Ok

End

And*

[late crypt]

We use this

So what

Why does it converge

[opal vault]

why does the sum of 1/2^n converge is what you're asking?

[late crypt]

Yes

[opal vault]

well now you can use your fancy ratio tests

[late crypt]

?

[opal vault]

this

on 1/2^n

[opal vault]

or this

[late crypt]

? What about it

Oh

1/2?

[opal vault]

yes

[late crypt]

Thanks

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[pallid tartan]

For 9 and 10, what is the force parallel to gravity? I know it must be subtracted from mg (w) to get 0, but idk what it is

[tough cargo]

It's practically the same calculation as for finding the horizontal force

[pallid tartan]

For 9 my first thought was t1 * cos theta1 - w = 0, but that doesnt factor in t2 or theta2

[tough cargo]

OK not exactly the same

Shouldn't be cos

[pallid tartan]

OOPS meant sin

but still

[tough cargo]

Hmmmmm

Just ignore t1sin(theta1) for now

What's the vertical component from t2 then

[pallid tartan]

Its t2 sin theta2

[tough cargo]

yeah

now it's a sum of forces

So what do u think u should do

with all the vertical components u found

[pallid tartan]

Thats what i dont know, it cant be adding the two vertical components

[tough cargo]

why not

[pallid tartan]

because that would be like double the weight force

[tough cargo]

no

just directs what theta1 and theta2 are

Well assuming t1 and t2 are fized

[vagrant umbra]

Can y’all help me with bio or nah

[pallid tartan]

bro go to another channel

[pallid tartan]

wait i dont understand

[tough cargo]

t1 and t2 can vary too ig

[pallid tartan]

t1 sin theta1 is like the same magnitude as t2 sin theta2

[tough cargo]

Why

[pallid tartan]

wouldnt adding them up be double of the weight force

[pallid tartan]

cuz like the line they go to are the same

[tough cargo]

Why are they the same magnitude

[tough cargo]

so

[pallid tartan]

is this line not both t1 sin theta1 and t2 sin theta2

[tough cargo]

maybe you're thinking of t2cos(theta2) = t1cos(theta2) which is true cus the forces add up to 0

And they're the only two forces

Idk what that Red line proves

Just means they share the same direction

Not magnitude

I mean the vertical components share the same vertical direction yes

[pallid tartan]

but if they were different magnitudes

then these wouldnt be right triangles

right?

this isnt a right triangle

[tough cargo]

idk what you're getting at with right triangles tbh

Yeah that probably isn't one

[pallid tartan]

do you not need a right triangle to use sine and cosine

[tough cargo]

yeah

[pallid tartan]

im so confused

[tough cargo]

Me too lol

idk what you're implying now

Lemme try to come up with some example values

Let's say g = 10 and m =1

[pallid tartan]

ok

[tough cargo]

,calc atan((10 - 3sqrt(3))/3)

[thorny flame]

Result:

1.0125570639419

[tough cargo]

,calc (10 - 3sqrt(3))/sin(1.01256)

[thorny flame]

Result:

5.6636414007259

[tough cargo]

OK so if I did the math right

T2 = 6

Theta2 = pi/3

Theta1 = 1.01256

And T1 = 5.66364

Theeeeeen the thing will be at equilibrium

And you can check with a calculator that T1sin(theta1) and T2sin(theta2) will not be equal (probably, I actually didn't check this part but it would be exceedingly unlikely lol)

[pallid tartan]

wait T1 and T2 are both 6?

[tough cargo]

No

T1 is at the end

[pallid tartan]

oh ok

[tough cargo]

,calc 5.66364*sin(1.01256)

[thorny flame]

Result:

4.8038463892108

[tough cargo]

,calc 6*sin(pi/3)

[thorny flame]

Result:

5.1961524227066

[tough cargo]

see not equal

So if you will not believe any sort of normal physical intuition at least believe this counterexample shows they don't have to be equal lol

,calc 5.66364*cos(1.01256)

[thorny flame]

Result:

2.9999796530088

[tough cargo]

,calc 6*cos(pi/3)

[thorny flame]

Result:

3

[tough cargo]

And you can check the other two values above these two add up to 10*1 = 10

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[marble comet]

How do I turn P = 1+2+3…+ $\frac{p-1}{2}$ into P = $\frac{p^2-1}{8}$? I’m talking the Law of Quadratic Reciprocity in class, trying to prove a specific Lemma given.

[thorny flame]

Tofu

[marble comet]

<@&286206848099549185>

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[marble comet]

.close

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[kindred patrol]

T

[kindred patrol]

,rotate

[kindred patrol]

I’m not very confident about problem 1

[stuck forum]

👍

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[undone dawn]

How do u use strictly interval notation?

[hoary vigil]

Solve for x, then based on the inequality you would have () open or [] closed intervals.
Then you can imagine a number line and write the interval using the appropriate brackets

[undone dawn]

Ohhh i see

I’ll try thanks

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[simple orchid]

Hello

[simple orchid]

Not sure how to start

[vernal matrix]

If you could prove that f(x) = x^3 + x - 1 is monotonic, then you know that any roots that you know exist would be unique

You can show that a root does exist with the IVT

[simple orchid]

What does monotonic mean?

[simple orchid]

What numbers should I know to put for IVT?

[vernal matrix]

either increasing/decreasing

[simple orchid]

So basically I have to use MVT and IVT

Right?

[vernal matrix]

Exercise left to reader try out a couple and you'll probably find one

[simple orchid]

Okay

[vernal matrix]

You don't really need the MVT for this one, just the IVT, though there is a way you can show the function's either increasing/decreasing...

[simple orchid]

I am going to pick 0 and 1 because I have desmos 🙂

[vernal matrix]

that's fair (also should be kinda clear if you look at the equation why those would work too!)

So at least one root exists there, now for why it's the only one...

[simple orchid]

How do I factor the equation itself though?

I have f(0) = -1

and f(1) = 1

Which means f(c) = 0

But I do I find c

[vernal matrix]

Note that they ask you to show that a [unique] root exists, they don't ask you for its value

[junior pumice]

this only means that if you prove its monotonic increasing.

[vernal matrix]

What it actually is doesn't matter too much

[simple orchid]

Oh

How do I prove it then?

[vernal matrix]

[to be fair, that does prove that there's some c between 0 and 1, just not its uniqueness]

[simple orchid]

How do I tell them it's unique?

[junior pumice]

monotonic means that the function only decreases/increases.

[simple orchid]

Um

[junior pumice]

y=0 will be proven unique by monotonicity. thats what we are talking about here isnt it

y=x is monotonic.

similar behaviors

see a similarity here? all of these are monotonic.

[vernal matrix]

[alternatively, if you know how to characterise increasing/decreasing functions by their derivative, you can also do it like that too]

[simple orchid]

So what do I do here?

[junior pumice]

prove the derivative is only positive or only negative

[simple orchid]

wdym?

3x^2 + 1 = 0?

Why not

mb

I meant 1

[junior pumice]

okay but not = 0

[simple orchid]

Okay

What next though

[junior pumice]

does the derivative have any special characteristics?

[simple orchid]

Nope

[junior pumice]

calculus is about analyzing functions, you have to be thorough.

does even or odd functions ring a bell?

[simple orchid]

It's an even function

[junior pumice]

provided it is an even function, what is the range of the derivative?

we can definitely say 3x^2 + 1 > x^2, and we know the range of x^2.

[simple orchid]

y ≥ 1

[junior pumice]

which means?

[simple orchid]

Uhhh

I don't know

[junior pumice]

what were we needing to prove earlier? monotonicity.

what does it mean when the derivative is only positive

the derivative is the rate of change of the function. which mean's when the derivative can only be positive, it is strictly increasing for all choices of x.

now going back to the IVT, we can then for sure say that there is an f(c) = 0 which exists and is unique.

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[simple orchid]

Ohhhh that makes sense

Thank you so much for the help

.close

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[tepid cliff]

How to ggett the range

[hazy pumice]

Hi Royyy

The range of a function is like the "shadow" of the graph onto the y-axis

So that entire interval on the y-axis is the Range of outputs for the function

Similarly, the Domain is the shadow of the graph projected onto the x-axis

x^2 + 3 is not the range

It is the inverse function

However

You need to restrict the domain to match the domain of f(x) iirc

(for function f and inverse f^-1, the shared domain should be whichever of the two is most restrictive. Or, to be more precise, the resulting domain should have all the restrictions of either function)

[tepid cliff]

Hii

Ohhh wait i think i get it now

The hanks alot!

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[static yoke]

.reopen