help-26

[next aspen]

how we get the second line from the first one

[wild tapir]

yeah i agree notation here sucks

all they did was apply this with A = x + h, B = x

[next aspen]

ye it does it's an ss of quora my hw is on a paper

[wild tapir]

and write division by h as multiplication by 1/h

[next aspen]

oh so first step is removing the denom by putting the 1/h to the left

and we have the numerator remaining

so we just divide them

using the rule

[wild tapir]

yeah

[next aspen]

that was easy

aight thank u

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[neon iron]

help

[neon iron]

how do I make the 4 geometric things on there i just dont know how to do it

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[sonic bone]

anyone know where i mightve went wrong

[glossy jay]

how did you start, what was your first step

[noble laurel]

!show

[topaz sinew]

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

[sonic bone]

Idk how readable this is but

@glossy jay @noble laurel

[glossy jay]

you forgot a power when you used chain rule at the start

@sonic bone

quotient rule seems correct

[sonic bone]

power of 4?

[glossy jay]

yes

[sonic bone]

so do i start all over with it to the power of 4

[glossy jay]

it should be an easy fix you have done most of the work

but yeah you should put the power 4 back

quotient rule is correct so that doesnt change

[sonic bone]

ok

how do I get b

[glossy jay]

the tangent has to touch the graph, that means -7x+b=y

you plug x and y inside the formula

you will get b

[sonic bone]

can u refresh my memory on how I get y

I know x is -2

[glossy jay]

you have the formula at the top

the parabola

[sonic bone]

so just plug in -2 to get y there

[glossy jay]

yup you got it

[sonic bone]

no differentiating ?

[glossy jay]

no need

[sonic bone]

ok ty

What is going on

B not 21?

[glossy jay]

nope

you made a mistake at the end.

last line

-7(-2)+b=7

is correct

so b=?

[sonic bone]

ohh -14

-7

completely stuck on this

[glossy jay]

Use chain rule to find the derivative of h, and H

[sonic bone]

ik that but how do ik which numbrs to use cuz originally i did 1 * 3 * 3 and that was wrong

i just took th numbrs where it said x 2

[glossy jay]

the second row ?

[sonic bone]

yeah

[glossy jay]

see how h is the composition of two functions

when you plug 2 in g

you will get a new value

you plug that back in f'

so you will have to switch rows depending on the value you get

[sonic bone]

im noot getting it

[glossy jay]

What is the formula you got for h'

[sonic bone]

idk 3(3)

[glossy jay]

I mean the general formula

Before you plug your numbers in

[sonic bone]

f(g(x))

[glossy jay]

That s the formula of h

We need the formula of h' the drrivative

[sonic bone]

idk how to do that

im not getting it

[glossy jay]

$(f(g(x)))'=f'(g(x))g'(x)$

[thorny flame]

ThibaultF02

[glossy jay]

Have you seen this formula ?

that s the chain rule

[sonic bone]

yeah

[glossy jay]

Since $h(x)=f(g(x))$

[thorny flame]

ThibaultF02

[glossy jay]

we can find $h'(x)$

[thorny flame]

ThibaultF02

[sonic bone]

is h'x is f' * g * g'

[glossy jay]

yeah

and we want to find h'(2)

[glossy jay]

actually it's f'(g)*g'

g is inside f'

[sonic bone]

and how do I get the numbers for that

[glossy jay]

$h'(2)=f'(g(2))\times g'(2)$

you can find g(2) and g'(2) in the table

[thorny flame]

ThibaultF02

[glossy jay]

and when you get the value of g(2) you will have to plug it inside f'

to get the full value

[sonic bone]

is g'(2)=1?

[glossy jay]

no

[sonic bone]

damn

[sonic bone]

3?

[glossy jay]

yes

[sonic bone]

oh

so g(2) is 3

[glossy jay]

yeah

[sonic bone]

f'3 is 1

so 1 * 3?

[glossy jay]

yeah

[sonic bone]

ohh okay

[glossy jay]

yeah you got it

[sonic bone]

i think i understand now

thanks smm

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[plain sand]

need help with this question

[plain sand]

here’s how I got it

nvm got it

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[fallow chasm]

Can someone please help me find the stabiliser G7 in this question. I know it has order 6 but don't know how to find it 🙏🙏

[topaz sinew]

@fallow chasm Has your question been resolved?

[strong beacon]

I had a problem previous about finding the inverse of N(x). I believe I was told it is erf(x), but it doesnt look like an inverse, whats wrong?

[limber ridge]

@strong beacon

[topaz sinew]

@strong beacon Has your question been resolved?

[strong beacon]

This is confusing

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[worldly peak]

I need help with question 5, 6, and 9. Regarding the answers. I used b^2-4ac for 5 and 6 at first, the value was correct but the inequality sign was the opposite of the answer. This was fixed when our teacher used completing the square but I still dont get how CTS works here fully. And regarding 9. My work shows it is inbound -1.5≤m≤1.5 but the book answer says its out bound m≤-1.5 and m≥1.5

[worldly peak]

The working using completing the square

Working using b^2-4ac

[topaz sinew]

@worldly peak Has your question been resolved?

[rotund valley]

ax^2 + bx + c ≥ 0 for all x when a > 0

then b^2 - 4ac ≤ 0

not ≥

[still citrus]

you know wavy curve?

[worldly peak]

Wait bro please elaborate

@rotund valley do pardon me for the ping but i would appreciate it if you could elaborate on this

[rotund valley]

umm could you wait for a minute I'm helping someone else right now

[worldly peak]

Ah ye sure

[rotund valley]

In quadratic equation formula, b^2 - 4ac is in the square root

[worldly peak]

Oh yeah

[rotund valley]

so the b^2 - 4ac determines whether the root exists or not

if b^2 - 4ac > 0, there are 2 roots

if b^2 -4ac = 0, there are only one root

if b^2 - 4ac < 0, there are not any roots

right?

[worldly peak]

Yep

[rotund valley]

uh do you know the graph of quadratic function

[worldly peak]

Yeah

A parabola

[rotund valley]

yes

like U shape

if a > 0

n shape if a < 0

[worldly peak]

Yep

[rotund valley]

but we have a > 0, the function is U shape

[worldly peak]

Yea

[rotund valley]

the function has positive value for all real values of x

in problem 5

that means the graph of the function never pass below the x-axis

yes?

because if the graph passes below the x-axis, then the function has negative y-value

and even, the graph never intersect with x-axis

if the graph intersects with x-axis, then the function will take 0 as a value, but 0 is not positive

I'm talking about the function y = 2x^2 + 3x + k

in problem 5

So, the equation doesn't have any roots

otherwise the graph of the equation will pass below the x-axis or intersect with x-axis

that's why b^2 - 4ac should be negative

do you understand?

[worldly peak]

Gimme One sec to read

[rotund valley]

ok take your time

[worldly peak]

OHH

okay so

f(x) = y

And when graohinf we write y= (eqtion goes here)

Holy shit this helps

Thanks alot man

[rotund valley]

no problem

[worldly peak]

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[vernal vale]

still a little confused on this one

so say we start with E_1

[thorny flame]

jan Niku

jan Niku

jan Niku

jan Niku

[vernal vale]

so then E_1 is empty?

[sweet shard]

not empty

something less strict

[vernal vale]

sure measure 0

idk im trying to figure out how to connect it back to the larger proof

so you have ${x \in E_1 : f(x) > 0 }$ and likewise for the negative are both measure 0

[thorny flame]

jan Niku

[vernal vale]

whats the mechanism of tracking this back out to the larger thing were trying to show

do you element chase in E

[sweet shard]

find a relationship between E, E1, and E2

[vernal vale]

you can invent a new set

E = E0 U E1 U E2

where E0 is all the places its 0

but im not certain how to pull it back out

[sweet shard]

what do you mean "pull it back out"

[vernal vale]

well we want to argue that ${ x \in E : f(x) > 0 }$ is measure 0

[thorny flame]

jan Niku

[vernal vale]

but all we have is that ${ x \in E_1 : f_1 (x) > 0}$ is measure 0

[thorny flame]

jan Niku

[sweet shard]

what's f_1?

[vernal vale]

f restricted to E_1

[sweet shard]

f_1 and f are the same when f(x) > 0

[vernal vale]

and we can just replace E1 with E?

[sweet shard]

when f(x) > 0 those two sets are the same yes

[vernal vale]

that doesnt seem right but ill take it

[sweet shard]

you just need to connect all the scattered facts you've stated to prove

[vernal vale]

i mean, it seems true, but it doesnt seem like this is sufficient

[sweet shard]

sufficient for what

[vernal vale]

the argument is missing i mean not that what were asserting is false

thhat you can just say {x in E1 : f1(x) > 0} and {x in E : f(x) > 0} are the same

[sweet shard]

yea you use f1(x) = f(x) on E1

then E1 is a subset of E such that f(x) > 0

[vernal vale]

hrm

[sweet shard]

idk what you think is missing

[vernal vale]

idk either

ill just write it down

its maybe a waste of time to spend more time on this if i have it and just dont see it

[sweet shard]

indeed

[vernal vale]

thanks

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[river root]

I dont have a clue on what to do, wasn't here during this lesson

[sweet shard]

,rotate

[pseudo bear]

What is the area of the wall before the window is removed?

[river root]

how to get area

L + W

[eternal wagon]

no

[river root]

times?

[eternal wagon]

yeah

so what would it be

[river root]

3x^2 + 15x

[eternal wagon]

and the window?

[river root]

let me write it 1 sec

x^2 + 3x + 2x + 6

should I simplify?

[pseudo bear]

OK, combine like terms.

[eternal wagon]

yeah

[river root]

x^2 + 5x + 6

[eternal wagon]

so what is the area of the grey shaded area

first

what would u do to solve for that

[river root]

area right

[eternal wagon]

yeah but what would you do to find that

[river root]

wdym

[eternal wagon]

how would you get that area

[river root]

L x W no?

[eternal wagon]

so we found the area of the wall

and we found the area of the window

[river root]

yes

[eternal wagon]

we are supposed to find the area after the window is removed

how would we find tha

[river root]

yes

[eternal wagon]

t

[river root]

you'd combine like terms with both?

[eternal wagon]

wouldnt you subtract the area of the window from the area of the wall>

?

[river root]

I suppose you would

[eternal wagon]

so...

what would you get

?

[river root]

well how would I do that?

[pseudo bear]

If you remove some area away from a large area, the area you removed is subtracted from the large area.

You have the area of the large rectangle.

You have the area of the small rectangle.

You're taking away the area of the small rectangle from the large rectangle.

So, you do large area - small area getting removed.

[river root]

2 + 9 + 5x

thats the answer

[pseudo bear]

That's your answer?

Or the answer from a book or something?

[river root]

my answer atleast yeah

[pseudo bear]

Oh, OK. Can you show your work?

[river root]

its wrong right

[pseudo bear]

OK, so you want to write it like this:

Wall: 3x(x + 5) = 3x^2 + 15x
Window: (x + 2)(x + 3) = x^2 + 2x + 3x + 6 = x^2 + 5x + 6

Wall without window: (3x^2 + 15x) - (x^2 + 5x + 6)

So you got the first two lines.

The label I put there like, "Window" is there so I don't forget what x^2 + 5x + 6 is supposed to be for.

Then I write out the subtraction I talked about before.

You're starting with the wall's area and then taking out the window's area, so you subtract.

Does that make sense so far?

[river root]

yes

[pseudo bear]

OK, now putting it like that makes it easier.

What is (3x^2 + 15x) - (x^2 + 5x + 6)?

[river root]

2x^2 + 10x + 6

[pseudo bear]

Almost.

[river root]

brahh

[pseudo bear]

With a minus sign outside the parentheses, you change the sign of everything inside.

You did that for x^2 and x.

But you forgot to do it for the 6.

[river root]

wdym

[pseudo bear]

Like you did 3x^2 - x^2 and then you did 15x - 5x.

Then you have 0 - 6.

[river root]

yes but there no commons for 6

where'd 0 come from

[pseudo bear]

Well, when you have a polynomial, the parts you leave out are 0s.

Like if I have 3x^3 + 8x - 7, that's the same thing as 3x^3 + 0x^2 + 8x - 7.

Do you see why?

[river root]

no I dont

why would a 0x^2 come from nowhere

[pseudo bear]

OK, so when you add 0 to something, it stays the same, right?

[river root]

yes

[pseudo bear]

And 0x^2 is 0, right?

[river root]

yes

[pseudo bear]

OK, so you can put that in without changing anything.

Does that make sense so far?

[river root]

yes but what difference would it have made to the 6?

[pseudo bear]

OK, so let's try it with the 0.

(3x^2 + 15x + 0) - (x^2 + 5x + 6)

You got 2x^2 and you got 10x. What do you get for the constant term?

[river root]

what do you get?

oh

-6

[pseudo bear]

Right!

[river root]

so I have to turn the number negative in every situation like this?

[pseudo bear]

Not always.

You do when you're subtracting it.

[river root]

what if it was a lonely x

like just 5x

would I do the same to get negative

0x

but that wouldn't make sense because x has a invisible q

[pseudo bear]

Like if you had (3x^2) - (x^2 + 5x + 6)?

[river root]

1*

[river root]

yes

[pseudo bear]

OK, then that's the same as (3x^2 + 0x + 0) - (x^2 + 5x + 6).

So, for the x part, you'd do 0x - 5x.

[river root]

you have to write 0?

[pseudo bear]

You don't have to.

That's kind of like training wheels.

It helps you to understand what's going on, and then you can later do it without needing that.

[river root]

x itself is considered to have a 1 how does that work out thought

though

[pseudo bear]

OK, if you have x, it's 1x. If you have no x, it's 0x.

So, like 3x^2 has no x, so it's 3x^2 + 0x.

3x^2 + x has an x, so it's 3x^2 + 1x.

[river root]

but id have to write 0x so the teacher understands right

[pseudo bear]

Well, you should at first.

Then, once you get the hang of it, you won't need to.

[river root]

than he'd realize yeah okok

so final answer to A is 2x^2 + 10x -6

[pseudo bear]

Right.

Actually, it's the answer to part B.

[river root]

what

[pseudo bear]

Well, you've simplified it and gotten it in the form of ax^2 + bx + c.

Like you have 2x^2 + 10x - 6.

So, that's what part B asks for.

[river root]

so the answer to A is

(3x^2 + 15x + 0) - (x^2 + 5x +6)

[pseudo bear]

Almost, it's like 3x(x + 5) - (x + 2)(x + 3).

It's where you realize that you're subtracting out the window's area.

So, you write the subtraction out.

[river root]

oh yeah 3 is out

[pseudo bear]

And it's where you don't multiply to get the area of each part quite yet.

So, in future problems, you'd presumably realize that it's an area subtraction, then you'd write out the multiplications for each area and subtract.

You don't do any multiplications or simplifying yet.

They want you to be able to see right away that it's an area subtraction.

And that's how you'd start it off.

Then, you'd do all the multiplying and subtracting and simplifying.

Does that make sense?

[river root]

but that aint what we did

[pseudo bear]

Well, you broke it into parts.

[river root]

we distributed

[pseudo bear]

You did the area multiplication for the wall.

[river root]

and the window

[pseudo bear]

You did the area multiplication for the window and simplified it.

And then you wrote out the subtraction.

[river root]

then we subtracted to the wall without the window

[pseudo bear]

But they want you to write out the subtraction first, before the multiplying.

[river root]

so they are trying to trick me?

[pseudo bear]

Not really.

It's just a different way to solve the problem.

[river root]

I prefer this way it was easier

[pseudo bear]

Oh, OK.

[river root]

ok let's do c?

[pseudo bear]

OK.

Do you know how to do it?

[river root]

no

[pseudo bear]

OK, so you have the area of the wall with xs: 2x^2 + 10x - 6.

They're saying that x = 5 ft.

So, what you do is to fill in x with 5 ft in parentheses.

So, it would be like 2(5)^2 + 10(5) - 6.

Does that make sense?

[river root]

what does parentheses mean

[pseudo bear]

It means multiplying it.

Like 5(10) = 50.

[river root]

oh okay

[pseudo bear]

But you use the parentheses to avoid mistakes.

[river root]

10^2 + 50 - 6

[pseudo bear]

Almost.

Do you know PEMDAS or BODMAS?

[river root]

bedmas

brackets exponets division multiplication addition subtraction

[pseudo bear]

OK, so you have 2(5)^2. The 2 in front is a multiplication. The ^2 is an exponent.

So, which goes first?

[river root]

exponent

[pseudo bear]

Right, so what would 2(5)^2 be after you do the exponent?

[river root]

2(5) 2(5)

[pseudo bear]

No, the exponent is written to the right of the only thing it's doing the exponent for.

[river root]

oh yeah sorry

[pseudo bear]

So, like 2(5)^2 has the ^2 only squaring the (5), since that's what's next to it.

[river root]

id be 2(10)

no 35

25*

[pseudo bear]

Right, it would be 2(25).

[river root]

yes then you would multiply

[pseudo bear]

Right.

[river root]

50

[pseudo bear]

Good.

What's 10(5)?

[river root]

50

[pseudo bear]

OK, and then you have -6.

So, you have 50 + 50 - 6.

What does that become?

[river root]

94

[pseudo bear]

Right.

,w 3x(x + 5) - (x + 2)(x + 3) where x = 5

[thorny flame]

Results provided by WolframAlpha

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Requested by chai.t.rex

[pseudo bear]

Now the units are important.

Do you know about dimensions?

[river root]

units?

[pseudo bear]

Like 3D or 2D?

[river root]

ah no sorry

[pseudo bear]

OK, 1D is like a line.

[river root]

yeah

[pseudo bear]

2D is like a surface, like a piece of paper or a movie you don't wear 3D glasses for.

[river root]

yes

[pseudo bear]

3D is like real space.

Like if you wear 3D glasses, the things you're watching can fill up space rather than being flat.

Now area is a flat kind of thing.

[river root]

yes I understand

[pseudo bear]

So, it's 2D.

[river root]

yes

[pseudo bear]

So, when you're doing feet, you do feet to the power of the dimension.

Like ft^2.

That's what the units are for area.

Volume, like how much space is inside a container, is ft^3.

And then feet are like ft^1, which is just a measurement of how long a line is.

Which is 1-dimensional.

Does that make sense?

[river root]

sure

[pseudo bear]

OK, so the answer is 94 ft^2 or 94 square feet.

Does that make sense how I got that?

[river root]

why the ^2

[pseudo bear]

Because it's two dimensional.

[river root]

ohhhhh

now it makes sense

[pseudo bear]

OK 🙂

Now, the way they wanted you to do part A is something you should try to learn.

The reason is that if you do it your way, you can do small problems like that.

But when you get into bigger problems, you'll have too many parts to do one at a time.

Plus, you have to remember what each part is for and all that.

[river root]

how me they subtraction method then

shoe*

[pseudo bear]

But with an expression like 3x(x + 5) - (x + 2)(x + 3), you only have to do simplifying.

[river root]

show***

[pseudo bear]

Well, you know that area is length times width.

[river root]

yes

[pseudo bear]

And you now know that when you take away some area from a larger area, the new area is the larger area minus the smaller area you removed.

So, you do this:

wall - window
3x(x + 5) - (x + 2)(x + 3)

The first thing you realize is that it's a subtraction since you're taking away area.

So, you write that out.

Then you fill in the expressions for the area.

Which is length times width.

[pseudo bear]

And then you get what I have here ^

Do those steps make sense?

[river root]

what are expressions

[pseudo bear]

Expressions are anything that's a number. Like 4x^2 + 5x - 3 is an expression because you can fill in x and get a number.

Also, 5 is an expression.

5x = 3 isn't an expression, though.

Because it's not a number.

You can get a number on one side and a number on the other side, so each side is an expression.

But the = part just tells you that the numbers on both sides are the same.

So, the expressions (which are numbers) on both sides are the same.

[river root]

so this isn't a expression

[pseudo bear]

Right.

[river root]

but it also is?

[pseudo bear]

Because you have expression = expression.

The things to the left and right of = are expressions.

But the whole thing isn't.

[river root]

oh yeah yes

[pseudo bear]

5 = 5 isn't a number.

[river root]

yes

[pseudo bear]

It's just a question or a statement.

Is 5 = 5? Yes.

I'm telling you that 5 = 5, which I'm right about.

Or something like that.

[river root]

I get now gea

yea

[pseudo bear]

So, it's more a true or false kind of thing rather than a number.

Like 3 = 5 is false.

But 3 = 5 isn't a number, so it can't be an expression.

So, back to part A.

You get:

wall area - window area
(wall width)(wall height) - (window width)(window height)
3x(x + 5) - (x + 2)(x + 3)

And you notice you haven't done any of the work of multiplying or subtracting.

You just got an expression.

So, now, instead of solving each part separately, you can just simplify that.

It's only one part rather than many.

[river root]

after simplified?

wouldn't you getting answer?

[pseudo bear]

Right, you write that for part A, then you simplify to get part B.

So:

A: 3x(x + 5) - (x + 2)(x + 3)
B: 2x^2 + 10x - 6

A is before you do any of the work.

B is after you simplify.

Then, C is just after you substitute (5) for x and simplify.

[river root]

ok

I have GCF and LCM word problems if you could also help me out with

[pseudo bear]

OK.

[pseudo bear]

,rccw

[pseudo bear]

OK, do you know what a divisor or factor is?

[river root]

what

[pseudo bear]

OK, so let's look at the number 39.

One way you can multiply two numbers together to get 39 is 1 times 39, right?

So, factors of 39 are any numbers you can multiply to get 39.

Like 3(13) = 39.

So, 3 and 13 are factors of 39.

1(39) = 39, so 1 and 39 are factors of 39.

So, the factors of 39 are 1, 3, 13, and 39.

Does that make sense?

[river root]

hold on

im going to try and solve it

[pseudo bear]

OK.

[river root]

ok

the unknown number is 6

and the LCM for 45 81 and 63 is 2835

[pseudo bear]

OK, so why is the unknown number 6?

[river root]

uhh

I dont know how to answer that

gcf and lcm both match

[pseudo bear]

OK, can you show your work to get 6?

OK, what is the LCM of 12 and 6?

[river root]

12

[pseudo bear]

Right. What does the question say the LCM should be?

[river root]

36

my bad

it is 36

36 is the unknown number

[pseudo bear]

OK, what's the GCD of 12 and 36?

Sorry, 12.

[river root]

gcd?

[pseudo bear]

Sorry, GCF.

[river root]

36

and 12

have a gcf of 6

[pseudo bear]

No, the GCF can't be larger than either of the numbers.

The GCF of 12 and 36 is 12.

[river root]

oh that works aswell

[pseudo bear]

What does the question say the GCF is?

[river root]

6

[pseudo bear]

Right, so 36 can't be the unknown number.

[river root]

its under 36

because it can't be bigger then the lcm

[pseudo bear]

Right.

[river root]

nothing under 6

in between

[pseudo bear]

Do you know how to factor numbers?

[river root]

what way

[pseudo bear]

Like 25 = 5^2?

[river root]

oh yeah no I dont

[pseudo bear]

OK, do you know about prime numbers?

[river root]

yeah

[pseudo bear]

OK, what are the first few prime numbers?

[river root]

2 3 5 7 9 13

[pseudo bear]

Almost.

[river root]

what

oh not 9

[pseudo bear]

It should be 11 instead.

[river root]

2 and 11 make 6

oh that's 12

[pseudo bear]

OK, so what we want to do to factor a number is to see if they're divisible by 2 and then 3, and then 5, and so on through the primes.

Like let's take 39.

Is it divisible by 2?

[river root]

no

[pseudo bear]

OK, so we go to the next prime number.

3 is the next one.

To know whether to stop, we square it.

3^2 = 9.

Is 9 bigger than the number we're factoring?

No.

9 isn't bigger than 39.

So, we try 3.

[river root]

even numbers are always perfect squares

[pseudo bear]

No, 14 isn't.

OK, is 39 divisible by 3?

[river root]

my teacher really be lying to us

[pseudo bear]

Sorry, 3.

[river root]

yes

[pseudo bear]

OK, so we write 3 down.

What's 39/3?

[river root]

13

[pseudo bear]

OK, so now we see if 13 is divisible by 3.

Is it?

[river root]

yes

wait

ah no

[pseudo bear]

Right, so we go to the next prime number.

5^2 = 25.

Is 25 higher than 13?

[river root]

ye

[pseudo bear]

OK, so we stop.

39 = 3^1 * 13^1.

Let's try another.

We start with the first prime, 2.

What's 2^2?

[river root]

4

[pseudo bear]

Is that higher than 60?

[river root]

no

[pseudo bear]

OK, so we do 2.

Does 2 divide 60 evenly?

[river root]

yes

[pseudo bear]

OK, what do you get after the division?

[river root]

30

[pseudo bear]

OK, so we write down the 2.

Then, we try 2 again with the remaining part (30).

Does 2 divide 30?

[river root]

15

[pseudo bear]

Right, so we write down another 2.

We have 2 * 2 written down so far, and we have 15 left.

[river root]

your just doing gcf

[pseudo bear]

Not exactly.

[river root]

I know this method i use

[pseudo bear]

We only have one number we're dealing with here: 60.

GCF needs two numbers.

[river root]

I use this method to find gcf

[pseudo bear]

Oh, OK.

[river root]

its called birthday cake

[pseudo bear]

So, you go through the primes and such?

60 -> 30 -> 15

2 * 2

[river root]

yes

if it doesnt take w

fo 3

until u hit a prime number

[pseudo bear]

What do you mean by "fo 3"?

[river root]

if it doesn't take 2 for 15

do 3

[pseudo bear]

Oh, OK.

[river root]

which is 5

and 5 is prime

[pseudo bear]

Right, but a way to save work is that squared trick.

Like, let's say you have 39 like before.

And let's say you don't know that 13 is a prime.

So, you have to keep going until you get to 13.

Like "Does 5 divide 13? No. Does 7 divide 13? No. Does 11 divide 13? No."

That's a lot of work.

But with that trick, you have "5^2 is 25. Is that larger than 13? Yes, so we're done because 13 must be prime."

Now, even if you didn't know that 13 was prime, you can stop earlier.

But anyway.

We have 2 * 2 * 3 * 5 = 60.

What we do next is we put exponents in.

2^2 * 3^1 * 5^1.

[river root]

2^2

[pseudo bear]

Right.

Now, let's say we want to know the GCF of 39 and 60.

We have 39 = 3^1 * 13^1. We have 60 = 2^2 * 3^1 * 5^1.

Now, for each prime, what's the lowest exponent?

With 39, what's the exponent on 2?

[river root]

wdym

idk from the top of my head

[pseudo bear]

Well, what we want to do is this:```
39 = 3^1 * 13^1
60 = 2^2 * 3^1 * 5^1

Does that make sense how I got that?

[river root]

oh yes

[pseudo bear]

OK, now if there's no 2 for 39, it's counts as 2^0.

Does that make sense?

[river root]

yes

[pseudo bear]

39 = 2^0 * 3^1 * 5^0 * 13^1
60 = 2^2 * 3^1 * 5^1 * 13^0

So, we have this.

What's the lowest exponent for each prime number?

[river root]

0

[pseudo bear]

OK, so for 2, you have 2^0 as the lowest exponent.

What about 3?

[river root]

q

1

[pseudo bear]

OK, what about 5?

[river root]

0

[pseudo bear]

OK, what about 13?

[river root]

0

[pseudo bear]

2^0 * 3^1 * 5^0 * 13^0

OK, now let's do the same thing for the highest exponents.

What are the highest exponents?

[river root]

q

1

oh wait

2

[pseudo bear]

OK, what about the other primes?

[river root]

same for each

2 3 5 15

[pseudo bear]

No, look here ^

What's the highest exponent on 2?

[river root]

highest prime

[pseudo bear]

No, not prime, exponent.

[river root]

2

[pseudo bear]

OK, what about the other primes?

3^? * 5^? * 13^?

[river root]

1 1 1

[pseudo bear]

39 = 2^0 * 3^1 * 5^0 * 13^1
60 = 2^2 * 3^1 * 5^1 * 13^0

GCF(39, 60) = 2^0 * 3^1 * 5^0 * 13^0
LCM(39, 60) = 2^2 * 3^1 * 5^1 * 13^1

So, you factor the numbers like that.

Then, the GCF takes all the lowest exponents.

The LCM takes all the highest exponents.

Does that make sense?

[river root]

yes

[pseudo bear]

OK, let's look at your problem.

What does the GCF of 6 factor into?

[river root]

36

[pseudo bear]

No, the GCF is 6. Factor 6.

[river root]

6 factor into?

[pseudo bear]

Right, what does it factor into?

[river root]

idkk

[pseudo bear]

OK, let's start with the first prime, 2.

What's 2^2?

[river root]

4

[pseudo bear]

Is that higher than 6?

[river root]

no

[pseudo bear]

OK, so we try 2. Does 2 divide 6?

[river root]

yes

3

[pseudo bear]

OK, so we have ```
Remaining number: 6 -> 3

Factors: 2

Does 2 divide 3?

[river root]

no

3 prime

[pseudo bear]

OK, so we have 2^1 * 3^1.

Does that make sense?

[river root]

yes

[pseudo bear]

OK, what does the LCM, 36, factor into?

[river root]

I've gotta go to work rn

[pseudo bear]

OK.

[river root]

but thank you for the help a bunch

[pseudo bear]

You're welcome.

[river root]

I really appreciated it

[pseudo bear]

Do you have a few minutes?

[river root]

sure

[pseudo bear]

I can show you where I was going.

12 = 2^2 * 3^1

GCF(12, x) = 2^1 * 3^1
LCM(12, x) = 2^2 * 3^2

Now the GCF has the small exponents and the LCM has the large exponents.

2^1 and 2^2 are the two exponents of 2.

2^2 is used up for 12, so the other must have the 2^1.

x = 2^1 * ?

Then, you do the same with the 3.

Min: 3^1, max: 3^2.

Well, the 12 has 3^1, so the other number must have the 3^2.

x = 2^1 * 3^2.

Does that make sense?

Edit: See also below after you say thanks for a quicker method that works only for this problem

[river root]

not at moment if I were to take time probably

for 5b the answer is 2835 rights

[pseudo bear]

No, you want the GCF.

[river root]

its lcm

[pseudo bear]

Each row has only one type of plant, and there's not 2835 tomatoes to put in one of the rows.

[river root]

hm

[pseudo bear]

Here's a link to the above part in case you want to look at it later when you have more time: #help-26 message

You can right click and copy it to somewhere.

[river root]

okok

thank you again

[pseudo bear]

You're welcome.

[river root]

take care

[pseudo bear]

You too.

[tawdry dagger]

!done

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If you are done with this channel, please mark your problem as solved by typing .close

[pseudo bear]

That's a method (factor into prime powers and the GCF is the smallest exponents and the LCM is the largest exponents) that works in a lot of problems, so it's good to learn it and practice it so that you can do it easily.

Another technique for just this problem is:

GCF(12, x) = 6
LCM(12, x) = 36

Divide everything by the GCF:
GCF(2, x/6) = 1
LCM(2, x/6) = 6

If the GCF is 1, the numbers multiply to give the LCM:
2*x/6 = 6
x/6 = 3
x = 18

.close

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[hot plank]

Forget where to go from here

[cursive patrol]

something rather important here is that 1/2 + 5/6 does not equal 11/12

[hot plank]

Oh

I'm so done

Sorry 1 sec then

[hot plank]

How would I solve for sin 11pi over 12?

In my book the answer shows root 3 over 2, but no answer

[sand echo]

its 2pi/3 + pi/4

[hot plank]

Will try that

[hot plank]

Can't solve

[sand echo]

why not

[hot plank]

Wait no I'm dumb

That worked

[hot plank]

My bad, thanks so much

[sand echo]

okk

[hot plank]

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[nimble quartz]

how can l hospitcals rules be applied here?

[nimble quartz]

i know for the 1/2 (t/e^-2t) that infinity over infinity

but for the -1/4 (e^2t) im not sure on what indetermint form can be used

[loud oasis]

that one isn't indeterminate so there's no need

[nimble quartz]

oh ok

so would that just be negative infinity then?

[loud oasis]

no

[nimble quartz]

oh wait i see the problem

cuz its apporaching -infinity but plugged into 2t that makes the entire e^-2t go into the denominator right?

making it 0?

[rustic finch]

it is 0 but… what?

[nimble quartz]

my fault lemme write it out 😭

Like this right?

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[dusty cargo]

Solve for ?

That is a 2 variable equation

Its a function

Well why did you send only this one ?

[worthy storm]

!original

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[remote flare]

I wish I listened in class tbh

looks easy ash

I just want an overall explanation for these questions and how to do them.

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[remote flare]

<@&286206848099549185>

ok

ty for the help

.close

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[fading urchin]

Thought I was done but this quiz got some tough questions

[craggy haven]

yeah unfortunately we won't be able to help with quizzes. good luck though!

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[neon iron]

2

you have to multiply and divide by (sqrt(3)+2)

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[neon iron]

my first toughts were to use the definition

(In + AB) * C = C * (In + AB) = I

In + BA is invertible <=> (In + BA) * D = D * (In + BA) = I

but probably wont work

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[atomic wadi]

Integrate 1/y(y+2)

[atomic wadi]

Can I do partial fraction for this?

[flat kindle]

partial fractions

[atomic wadi]

So A(y+2) +By=1?

Ok I think I got it

tq

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[edgy merlin]

Hi, how can i do this : Proove that (n/k)^k <= (n k) where (n k) is the binomial coefficient ? Thank you in advance to answer

[edgy merlin]

I test a lot of possibility but i don't understand how can i do this

Yes

[fair thorn]

*write

*now

*deeper

[edgy merlin]

I did everything in this exercise except this first inequality in 3) I did the second inequality , maybe I can use the previous questions but I don’t know how

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[narrow steeple]

how do I prove 21b

[worthy storm]

can you find B^T

[cyan python]

First find B^T

[worthy storm]

in terms of A

[narrow steeple]

is it A (A^(-1) )^T

[cyan python]

Great

Now try to find BB^T

While solving you have to use that AA^T = A^TA

[narrow steeple]

ohh thank youu

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[manic ravine]

i dont get how to do b and c, can someone walk me through the process of getting the answer?

[abstract wadi]

The tension by a string is a force that can only pull the ball, it has to be positive always.

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[stark tiger]

Can someone help me with converse, inverse, and contrapositive statements?

[stark tiger]

I have this unusual question in my book:

All fish are not mammals

How do i write the converse, inverse , contrapositive of that

Like if it had the “If, then” i would have answered it easily

But i dont know how because i only know the if p, then q

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[narrow acorn]

What is this sign?

[vital moth]

this is called an integral

[narrow acorn]

?

[vital moth]

do you know derivatives

[narrow acorn]

no

[vital moth]

do you know limits

[narrow acorn]

like limit, for example like a bag can only hold 10 balls. that is its limit?

[vital moth]

so for example do you know what this means $\lim_{x\to 0}{\frac1x}$

[narrow acorn]

no

never saw something like that

i try to make something and i saw that sign in a formula

[vital moth]

in which grade are you

[narrow acorn]

8

[vital moth]

ohh ok

you see these concepts at 10-11

[narrow acorn]

...

oh

[vital moth]

but dont wait until then

[narrow acorn]

ill try to learn the concepts online for what i need to do rn, and at school at 10 or 11 ill complete them

[vital moth]

ok so do you know trigonometry

[narrow acorn]

like sin, cos, ctg, tg?

[vital moth]

wait are you interested in physics ?

[narrow acorn]

no but i try to make a game

[vital moth]

ohh ok

do you like reading books or watching vids to learn

[narrow acorn]

vids

books only if i have the real book in my hand

[vital moth]

ok go to #book-recommendations ask for vid resources to learn calculus

[narrow acorn]

alr. ty

[vital moth]

i can recommend books but idk about vids

[narrow acorn]

oh

[vital moth]

bc i prefer learning from books

thats why idk vids which are considered good

[narrow acorn]

alr ty

[vital moth]

yes these

do you know these well ?

[narrow acorn]

ik they were used to the triangle that had the angle of 90 degrees (idk how to say how it was named in english)

[vital moth]

ok you need to learn some stuff before calculus they are called precalculus

[narrow acorn]

i kind of know them but i must read them again. i never used them in 8th grade and in 7th grade just like 3 hours

[vital moth]

you should be familiar with these and i am pretty sure that you are already familiar with many concepts in precalc

[narrow acorn]

maybe i should ask a friend to help me with that part of the game...

[vital moth]

but check these before calc bc they are a must before you start

[vital moth]

as you wish if you enjoy math or want to learn some new entertaining stuff then go for these

you will see calculus concepts too often if you get into STEM majors at uni

also in high school

anyways explain what you already know briefly in #book-recommendations and ask for resources to study calc and whether you are ready or you still have some gaps in precalc

[narrow acorn]

allright. thank you

[vital moth]

they will be more helpful than me bc they answered such questions many times before

np i didnt do anything

when you are done please type .close

[narrow acorn]

wait

wydm explain everything i know? like do i just say i know trigonometry there?

.close

[topaz sinew]

Channel closed

Closed by @narrow acorn

Use .reopen if this was a mistake.

[vital moth]

.reopen

[topaz sinew]

✅

[vital moth]

ig its is better to ask for resources about precalc and calc

and check if you know all stuff in precalc yourself

before jumping into calc

[narrow acorn]

how would it may take to learn precalc and calc

[vital moth]

depending on the pacing that you follow

precalc wont take long

[narrow acorn]

so a few days?

[vital moth]

bc you are prob familiar with some

depending on how much you know

but yes it should be like this

for calc however you will need sometime

not too much if you work hard and grasp the ideas fast

[narrow acorn]

by myself i learn slow. usually all the math i know, is from class. i can understand it very well if someone explains it to me in real life. so it will take some time

so ig ill just find another way to fix my issue, and then learn calculus and precalculus for myself overtime. i must end the thing i need to do very fast.

.close

[topaz sinew]

Channel closed

Closed by @narrow acorn

Use .reopen if this was a mistake.

[narrow acorn]

thanks

[vital moth]

ok as you wish