help-28

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[zinc dome]

Where are they getting 71 from? T is in minutes and it said 3 hours, 3x60 is 180 so t must = 180 minutes right?

When I calculated 199.457(0.994)^180 I got 67.51- rounded to 68. But it said 71 was the answer? I am confused

[light sonnet]

,calc 199.457(0.994)^120

[glossy valve]

Result:

96.875824168526

[light sonnet]

Using that answer from part a, isn't accurate enough

[eager violet]

You did 120

[shrewd hamlet]

Webassign has a margin of error that is permissible

[shrewd hamlet]

Yea check the table

[light sonnet]

I know

[eager violet]

Oh

[light sonnet]

I was checking how close the mg value is with 120

And it's 3 mg off

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@zinc dome Has your question been resolved?

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[severe basin]

The question mentions the random variables to be iid.

$Var(A)=\frac{\sigma^2}{3}$

[glossy valve]

QuantumBee

[severe basin]

$Var(B)=(0.1^2+0.3^2+0.6^2) \times \frac{\sigma^2}{3}$

[glossy valve]

QuantumBee

[onyx glen]

times sigma^2/9

[severe basin]

How?

[onyx glen]

wait, hold on. no.

that's wrong too

it should be (0.1^2 + 0.3^2 + 0.6^2) sigma^2

you were off by a factor of 3 but i was wrong about the direction in which you were off

[severe basin]

Oh, is it because there are three RVs identically distributed, so the variance gets multiplied by 3?

[onyx glen]

no

what

no

[severe basin]

I'm not understanding this part

How is Var(B) = (0.1^2 + 0.3^2 + 0.6^2) sigma^2 ?

[onyx glen]

Var(A) = (1/3^2 + 1/3^2 + 1/3^2) sigma^2

[severe basin]

=(3/3^2) sigma^2

=(1/3) sigma^2

Isn't that right?

[onyx glen]

yes

and Var(B) accordingly is 0.1^2 sigma^2 + 0.3^2 sigma^2 + 0.6^2 sigma^2

[severe basin]

Oof, sorry

I get my mistake

I have one more doubt

[onyx glen]

yes?

[severe basin]

It's a different question

I will post.

Let the random variables X and Y be uniformly distributed in the triangular region D, where D is bounded by x=0, x=2y, x=2.
Find $f _{Y∣X=1}(y)$

[glossy valve]

QuantumBee

[severe basin]

$f {Y∣X=1}(y)=\frac{f{XY}(1,y)}{f_{X}(1)}$

[glossy valve]

QuantumBee

[severe basin]

This will be the graph of the triangular region

[onyx glen]

oh this one

eugh

i don't want to do thsi

this*

[severe basin]

Umm, please say why

[onyx glen]

i don't want to do this, that's why.

[severe basin]

I doubt the question to be wrong

[severe basin]

Probably this question could be made by the instructors.

I can report it to the professor

[onyx glen]

I DIDNT SAY THE QUESTION WAS WRONG I SAID I DIDNT WANT TO DO IT

[severe basin]

Ok

I can't force anyone to help me

I will wait till someone arrives to help me with this

[full forum]

@severe basin Has your question been resolved?

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@severe basin Has your question been resolved?

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@severe basin Has your question been resolved?

[severe basin]

<@&286206848099549185>

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@severe basin Has your question been resolved?

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@severe basin Has your question been resolved?

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@severe basin Has your question been resolved?

[torn jolt]

people can't answer for you

breh

this is help

[severe basin]

[short crest]

no, i'm help! this is help-28

[torn jolt]

people help you to solve the robnlem

they dont solve it

[severe basin]

It's alright

It's been almost two days

Nobody came for help.

[severe basin]

I can wait

[torn jolt]

isn't this a graph of y=0, x=2y , x=2

[severe basin]

no

it is graph of x=2y for x lying between 0 and 2.

[torn jolt]

,calc 0.000001^0.000001

[glossy valve]

Result:

0.99998618458488

[torn jolt]

[lavish fractal]

,calc 0.000001*23456

[glossy valve]

Result:

0.023456

[severe basin]

THIS IS NOT THE CHANNEL FOR CALCULATIONS

[lavish fractal]

sorry

was helping a feind and this was the one opened

[severe basin]

Go to #latex-testing

[lavish fractal]

ik

sorry

[full forum]

@severe basin Has your question been resolved?

[wraith barn]

still unsolved lmao

close the channel @severe basin

[severe basin]

no

[primal owl]

I'll try to solve it

@severe basin Are you there?

What does it mean by $X$ and $Y$ being uniformly distributed over the triangular area?

[glossy valve]

384920

[primal owl]

I assume that $f_{X,Y}$ must be constant over that triangular area

[glossy valve]

384920

[primal owl]

and the integral over that region is equal to one

So then $f_{X,Y}(x,y)=\begin{cases} 1 & 0\leq y\leq\frac{1}{2}x\leq 1 \ 0 & \text{otherwise} \end{cases}$

[glossy valve]

384920

[primal owl]

Then $f_{X,Y}(1,y)=\begin{cases} 1 & 0\leq y\leq\frac{1}{2} \ 0 & \text{otherwise} \end{cases}$

[glossy valve]

384920

[severe basin]

Yes

My specific doubt is, can conditional density be more than 1?

If so, why?

[primal owl]

And is $f_{X}(x)=\int_{0}^{\frac{1}{2}x},dy=\frac{1}{2}x$ for $0\leq x\leq 2$ and $f_{X}(x)=0$ otherwise?

[glossy valve]

384920

[primal owl]

So $f_{X}(1)=\frac{1}{2}$

[glossy valve]

384920

[severe basin]

Correct

[primal owl]

$\frac{f_{X,Y}(1,y)}{f_{X}(1)}=\begin{cases} 2 & 0\leq y\leq\frac{1}{2} \ 0 & \text{otherwise} \end{cases}$

And are you asking whether the integral over those values of $y$ are supposed to be $1$ or not?

[glossy valve]

384920

[severe basin]

How is this right?

The answer is 2.

[primal owl]

Oh

Ohhh

[glossy valve]

384920

[primal owl]

It was an arithmetic mistake sorry

$1 / (1/2)=2$ not $1/2$ ...

[glossy valve]

384920

[primal owl]

Anyway integrating that over the region gives an integral of $1$

[glossy valve]

384920

[primal owl]

Over those possible values of $y$

[glossy valve]

384920

[primal owl]

How does this look @severe basin ?

[severe basin]

Fine

[severe basin]

.

[primal owl]

The value of the integral over the specified region?

In general?

I think not

[severe basin]

No

[primal owl]

Oh

The conditional density refers to the probability of some event if you integrate over some region right?

And the probability of integrating over the whole region

That should correspond to any outcome

Which should have an integral of 1 right?

Even in the case where it is conditional?

I don't know if this is your question or not

Because it is still a density function from what I understand

Integrating over the whole space of outcomes should given an integral of 1

[severe basin]

That is understood

How is the answer "2" feasible?

Probability should lie between 0 and 1 only right?

@primal owl

[primal owl]

Oh

You shouldn't get 2

From integrating over any valid subset

Only values in [0,1]

[severe basin]

I'm talking about the FINAL ANSWER

[primal owl]

Are you referring to the function value being 2?

Like at a point, it can be 0 or 2?

Is that what you are asking about?

[severe basin]

$f_{Y∣X=1}(y)=2$

[glossy valve]

QuantumBee

[severe basin]

We just got this ^

[primal owl]

Yes for a certain range of y

It's ok for that value to be greater than 1

[severe basin]

Why?

[primal owl]

All we need is that

This is a probability density function

[severe basin]

Because, it's a PDF?

[primal owl]

Yes

[severe basin]

Oh yes

[primal owl]

We just need the integral of any reasonable subset to be in [0,1]

[severe basin]

THANK YOU FOR HELPING

[primal owl]

You're welcome!

[severe basin]

Nobody cared to help me

[primal owl]

Oh

[severe basin]

And, you came

[primal owl]

I did this two years ago and forgot it all

[severe basin]

.close

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[primal owl]

Hopefully that helped!

[severe basin]

Have a nice daay sir

[primal owl]

Thanks you too!

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[stiff oak]

Don’t know anything about sas and sss properties

[hallow cedar]

sas just means side-angle-side, it means that if 2 triangles share 2 sides, and have the same angle between them, they are congruent

sss means side-side-side and it means if 2 triangles have all the same sides they are congruent

[stiff oak]

That makes sense

For question 2, how come it isn’t similar by sas

[hallow cedar]

,rotate

[hallow cedar]

well see, in question 2 we see that 2 sides are equal

[stiff oak]

We know 2 sides are equal, and the angle between those 2 angle sides should be equal too… or am I wrong?

[hallow cedar]

no it doesn't say that

the other length determines how big the angle si

is*

[stiff oak]

but doesn’t it have to be, like mathematically

[hallow cedar]

no let me illustrate

[stiff oak]

Okay

[hallow cedar]

suppose all the sides I marked with a line are equal

[stiff oak]

Okay

[hallow cedar]

the triangles are clearly different

since the other side can still change

[stiff oak]

That actually makes a lot of sense okay

[hallow cedar]

👍 that's why we need an extra angle / length to determine if they're congruent or not

[stiff oak]

thank you for that explanation

[stiff oak]

I was able to solve all the other questions because of that

thank you againnn

[hallow cedar]

no problem

[stiff oak]

.close

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[raven spruce]

Can you please help me with the following homework question?
Posting question and work below......
Solve the following differential equation by using integrating factors. y ' = 2 y + x^2
dy(x)/dx - 2y(x) = x^2
e^-2x dy(x)/dx -(2e^-2x)y(x) = e^-2x x^2
e^-2x dy(x)/dx = d/dx (e^-2x) y(x0 = e^-2x x^2
e^-2x dy(x)/dx + d/dx (e^-2x) y(x) = e^-2x x^2
integral e^-2x x^2 dx
-1/4e^-2x(2x^2+2x+1)+C
1/4 (-2x^2-2x+4Ce^2x - 1)
Final answer - > [(-x^2-x)/2] + [Ce^(2x)] - [(1/4)]
FYI here is what I put

[nocturne creek]

try y= your answer

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[full forum]

Please don't occupy multiple help channels.

[tidal reef]

,rccw

[tidal reef]

have you made any progress?

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@hybrid marsh Has your question been resolved?

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@hybrid marsh Has your question been resolved?

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[torn jolt]

Line ST is shared between the triangles

You know two of their angles are congruent

That means the third is too and you can invoke the angle side angle postulate

[lament reef]

how do i write this out

@torn jolt

[torn jolt]

I just told you bruh

[lament reef]

hb this

[torn jolt]

You know both are right triangles

FG is the same between th

Them

They share hypotenuses too

[lament reef]

mkkm

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[outer pasture]

.close

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[cursive kestrel]

.reopen

[cursive kestrel]

How do i find the inverse

for this function

i tried but i couldn't

[vast fossil]

y = 3^x/(3^x + 1) means that y 3^x = 3^x + 1, first isolate 3^x

[cursive kestrel]

can u show on paper or paint i don't really understand ur meaning

[vast fossil]

$y = \frac{3^x}{3^x + 1} \ 3^xy = 3^x + 1 \$ isolate $3^x$

[glossy valve]

A Lonely Bean

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[median cedar]

$\lim ____ (x \to 0) \frac{sin({x\over2})}{2x(1-x\over2)}$

[glossy valve]

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

[sharp vine]

$$\lim_{x \to 0} \frac{\sin \frac{x}{2}}{2x\Big(1-\ \frac{x}{2} \Big)}$$

this?

[median cedar]

denominator is 2x(1 - {x/2})

[glossy valve]

Modus

[median cedar]

yep

[little forge]

lim = 1/4?

[median cedar]

I made this question about

if I can take sinx/2 / x/2

i'm bad at math

just consider the x in the denominator is an unknown value

(i really don't know)

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@median cedar Has your question been resolved?

[median cedar]

<@&286206848099549185>

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@median cedar Has your question been resolved?

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@median cedar Has your question been resolved?

[median cedar]

<@&286206848099549185>

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@median cedar Has your question been resolved?

[primal owl]

You can't do this

If you know that $\lim_{x\to 0}\frac{\sin (x)}{x}=0$,

[glossy valve]

384920

[primal owl]

$\lim_{x\to 0}\frac{\sin (x/2)}{x/2}=0$

[glossy valve]

384920

[primal owl]

And $2x=4\cdot (x/2)$

[glossy valve]

384920

[primal owl]

Try using those to split the limit into the product of two terms whose limit you know

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[grave marten]

Help

[grave marten]

I was given some knowledge relating to using the formulas of a rectangle, 2 semi circles of a small and big circle, the Area = the rectangle + small circle + big circle

and is suppose to be quadratic also apparently

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@grave marten Has your question been resolved?

[grave marten]

<@&286206848099549185>

Pls

🙏

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[wanton valley]

Uh if i factor x² + 9x + 3 the answer would be (x + 3)² right

[wanton valley]

<@&286206848099549185>

[vast fossil]

No, (x + 3)^2 = x^2 + 6x + 9

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[wanton valley]

Ok ty

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[buoyant epoch]

Can someone explain how my equation is wrong

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@buoyant epoch Has your question been resolved?

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@buoyant epoch Has your question been resolved?

[silver drift]

Looks good to me

as in, you got the right answer

[buoyant epoch]

Is there another way of writing it? It’s making it wrong

[silver drift]

only thing I can think is 5 sqrt(4-x).

[buoyant epoch]

Marking*

Lemme try that

Didn’t work

[buoyant epoch]

Is this correct too?

If it is it’s probably just Desmos fault

[silver drift]

You haven't shared the whole problem, I think. Did either of these include a line like "y = sqrt(x)" or "y=3root(x)"?

I'm more than happy to blame the errors on Desmos, but I can't be sure of that without the full problem statement.

[buoyant epoch]

Sure give me a moment

First problem

[buoyant epoch]

Second problem

[silver drift]

Huh. It's just visual. They didn't tell you the function was sqrt.

[buoyant epoch]

Is that important?

[silver drift]

Not really, it just makes it hard to know you're giving the right answer.

It looks like all your answers are correct.

And more importantly, you understand how transformations of the graph and transformations of the equation are related.

Even if you don't get points in class for this exercise, you'll have that skill.

[buoyant epoch]

Ah ok, thanks for trying anyways

.close

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[smoky compass]

where did i mess up 2

[stable heart]

$-3(-7)=21$ not $-21$

[glossy valve]

Duh Hello

[smoky compass]

i didn't put -21

[stable heart]

$\frac{10-3(-7)+4}{-5}=\frac{10+21+4}{-5}$you have $-21$ here

[glossy valve]

Duh Hello

[smoky compass]

why would it be + not -

[stable heart]

because minus times minus is plus

[smoky compass]

it's one negative tho

10 - 3(-17) goes to 10 -21 ?

[stable heart]

10-3(-17) is 10+51

[smoky compass]

what??

where did you get +

[stable heart]

because there are 2 minuses

-3 and -17

[smoky compass]

but isnt it saying 10-3

[stable heart]

no, remember the order of operations

[smoky compass]

yes i know

[stable heart]

you gotta do multiplication first

[smoky compass]

i did

[stable heart]

$-3(-7)=-3\times(-7)$

[glossy valve]

Duh Hello

[stable heart]

thats what it means when there is a number directly in front of a parenthesis

so after doing the parenthesis you have to do the multiplication before doing the addition

[smoky compass]

yes i know

but i don't see any addition

[stable heart]

but $-3\times(-7)=21$ which is positive, so addition

[glossy valve]

Duh Hello

[stable heart]

if you dont understand this then i would recommend going through some books or watching some videos on it. this is very essential

[torpid stag]

-3(-7) = -(3(-7)) = -(-21) = 21

[smoky compass]

where can i find these videos

[stable heart]

just a quick google search gives me this. looks like it should be alright https://www.youtube.com/watch?v=O6bRgxVRoZ4

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[last locust]

How to do this energy conservation problem?

[last locust]

google tells me this but I dont know how to apply it

[sharp flame]

This is a physics problem, I could help you, but I'm not sure if it's permitted

[last locust]

I think its all related to a point

Ive asked statics problems before

[sharp flame]

Well...

I don't have an issue helping you

So at the point of release, what would the initial velocity be?

[last locust]

v 0

[sharp flame]

Excellent, that makes our kinetic energy 0

i.e. K1 = 0

What would the potential energy (U1) be?

[last locust]

no clue

the formula for pot energy i guess

[sharp flame]

$\text{Potential Energy } = \text{mass} \times g \times \text{height}$

[glossy valve]

NEONPerseus

[sharp flame]

What would the mass and height be in this case, looking at the question?

[last locust]

mass m and height h0

[sharp flame]

Yup so we have the LHS all sorted

Let's look at the RHS now

What would be the kinetic energy of the mass at the instant of hitting the ground?

[last locust]

ve

[sharp flame]

That's the velocity 😅

[last locust]

thats velocity

[sharp flame]

We're looking for the kinetic energy yes

$K = \frac{1}{2} \times \text{mass} \times \text{velocity}^2$

[last locust]

yess i was gonna write that

[glossy valve]

NEONPerseus

[sharp flame]

Alright cool

So what would the kinetic energy be at that instant?

[last locust]

$\frac{1}{2} \times \text{m} \times \text{ve}^2$

[glossy valve]

Stefan

[last locust]

havent tried this before lol

[sharp flame]

Yup cool

[sharp flame]

It's alright

And what would the potential energy be at the bottom?

[last locust]

0?

[sharp flame]

Yup

If we were to string the whole thing together

$mgh_0 = \frac{1}{2}mv_{e}^2$

[last locust]

1/2mv(0)^2 +mgh(0)=1/2mv(e)^2

[glossy valve]

NEONPerseus

[last locust]

?

[sharp flame]

There is no initial kinetic energy 😅

The object starts from rest

[last locust]

is this wrong then?

[sharp flame]

You got it right for the most part

It's just that the initial velocity is 0

Since it's been released at a height of h-nought

[last locust]

okay I get it now

thanks

a lot

[sharp flame]

np

[last locust]

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[sharp igloo]

ive worked until task 3 but im stuck

[eager violet]

What have you tried

[sharp igloo]

im not sure if i did it right but for task 2 i got square root of (16+6t+t^2) as the height

and then for the midpoint i just substituted that into the midpoint formula

i didnt understand how i can use substitution to eliminate t

[eager violet]

For the midpoint, you should have (x(t), y(t))

x = some function of t, y = some function of t

[sharp igloo]

i got this

so do i need to make this into functions of t

ngl im clueless

[eager violet]

So, assuming those formulas are correct, you have x = (3 + t)/2 and y = √(16 + 6t + t²)/2

Solve for t in one equation and plug it into the other

[sharp igloo]

oooooooooooh

is it okay if i send u my working out to get this answer

to check if its correct

[eager violet]

I'm boutta leave so I can't check it

[sharp igloo]

ah thats okay

thanks for ur help

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[cursive coral]

Hi, could someone help me understand how the last expression was derived from the one above it?

I don't understand how you just differentiate that double integral wrt z to get the last expression

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@cursive coral Has your question been resolved?

[grand yacht]

they just solved the inner integral

[cursive coral]

still don't get how :/ sorry

how did d/dz of integral (fxy (x,y) dy) turn into fxy(x,z-x)

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@cursive coral Has your question been resolved?

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[smoky compass]

how do you find the rest of the variables

[woven sand]

Take a look at the 70 degree angle and angle b

How do they relate to each other?

[smoky compass]

they make a supplementary angle?

[woven sand]

Correct

Which always add to...

[smoky compass]

180?

[woven sand]

Correct

[smoky compass]

110

[woven sand]

You can use that information to find b, and then go from there

[smoky compass]

tysm you’re so helpful

[woven sand]

👍

[smoky compass]

how do i solve 8

[woven sand]

Let's start with a triangle and figure out what the formula is

Unless you already know it

[smoky compass]

no idea

[woven sand]

The formula for the total sum of interior angles will be $(n-2)(180)$ where n is the number of sides

[glossy valve]

lexitorius

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[undone yew]

How do i rewrite the second half

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[unreal temple]

function increasing when x > 4 decreasing when x < 4

im pretty sure that’s correct

[torpid stag]

That is correct, yes

[unreal temple]

.solve

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[unreal temple]

right thank you

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[scarlet galleon]

was wondering how they got A^-1

and isnt AB mulltiply BA just BA^-2?

[light sonnet]

No

You can't swap matrices around like that

Because $AB \neq BA$

[glossy valve]

dldh06

[light sonnet]

In terms of matrices

[scarlet galleon]

oh right

because matricies are non-cumative right?

so how did they get this

how is AB multiply BA^-1 AB^2A^-1?

[light sonnet]

When you distribute AB(BA^-1 + B^-1)
You get ABBA^-1 + ABB^-1

[scarlet galleon]

right

[light sonnet]

The BB simplifies to B^2

BB^-1 = I

[scarlet galleon]

oh

right

so I have

X + A = AB^2 A^-1 + A

then I multiply the whole thing

by the inverse of A

giving me

AB^2 A^-2?

[light sonnet]

Why?

[scarlet galleon]

because its X + A

oh shit

yea

its X + A

so I multiply A by its inverse to get the idenity matrix right

i do that on the left i do that on the right

multiply the last A gives me I

but what about AB^2 and A^-1?

[light sonnet]

You don't need to do this

It's X + A

Just subtract A on both sides

[scarlet galleon]

I can do -A on the right side?

ah okay

oh I see

because B^-1 A^-1

in the original equation is multiplyig one another

but here its simply adding

so this would be the final answer

[light sonnet]

Yes

[scarlet galleon]

ah okay

what if I had

an inverse multiply its inverse

that would be A^-2?

[light sonnet]

If it was XA then that's when you can multiply by A^-1

[scarlet galleon]

yea

but say I have XA = A^-1

what would

A^-1 multiply A^-1 be?

[light sonnet]

A^-2

[scarlet galleon]

ah okay

and how would I go about getting A^-2 to its identiy matrix

i would have to multiplyA twice right?

or A^2?

[fickle path]

how do u solve this (2x)/(x-2)=(1)/(x+2)-2?

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@scarlet galleon Has your question been resolved?

[scarlet galleon]

@fickle path u can first cross multiply the values

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@scarlet galleon Has your question been resolved?

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@scarlet galleon Has your question been resolved?

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[quick heath]

What is the definition of rg(A) your working with?

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[severe basin]

$F_{z}(0)=0.5, F_{z}(−0.06)=0.47608$

[glossy valve]

QuantumBee

[severe basin]

How do I find the value of F_{z}(1.96) using the above information?

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@severe basin Has your question been resolved?

[severe basin]

<@&286206848099549185>

Z is the standard normal distribution

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@severe basin Has your question been resolved?

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@severe basin Has your question been resolved?

[full forum]

@severe basin Has your question been resolved?

[primal owl]

You don't need that information to calculate it

It will be hard to do it by hand though

[severe basin]

Ok

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[waxen warren]

Hello, can anyone help me factor this -3x^2 -2x +1 ?

[shrewd hamlet]

First I’d factor out a -1 so u have -1(3x^2 + 2x - 1)

[coral crypt]

doesn't that change the sign of all the terms in the parentheses?

[shrewd hamlet]

Yea mb typo

Ty

[coral crypt]

happens

[coral crypt]

look for 2 integers that multiply to become 1

positive 1*

[waxen warren]

-1(3x^2 + 2x - 1)

to become -1?

[shrewd hamlet]

Yea use mystics method, mine takes a bit of explaining

[waxen warren]

Okok

-1 and -1 = 1

Sum is -2

[coral crypt]

yours is basically just the same thing but easier to follow

[shrewd hamlet]

o

[coral crypt]

yes, try those first

also be cognisant of the coefficient of the 2nd-degree term altering ONLY the 2nd term in the quadratic

[waxen warren]

what do I do with those two in -3x^2 -2x +1

[coral crypt]

(-3x -1)(x-1)

try it

[waxen warren]

-3(x-1)(x-1)

[waxen warren]

How did you get that?

[coral crypt]

may not be correct though but a good start

I'm back, sorry

[coral crypt]

from the coefficient of the term to the power of 2

you just grab it

[waxen warren]

Of the term

Which term

[coral crypt]

x^2

its coefficient for this problem is a -3

[waxen warren]

Yes

[coral crypt]

therefore factoring must result in (-3x (blank))(x - (blank))

have you tested the two -1s yet?

[waxen warren]

(-3x -1)(x-1)

(3-1)(-2)

[coral crypt]

that's not the right step

heard of FOIL?

[waxen warren]

Nop

[coral crypt]

yeah, I can't remember what it stands for either

but you multiply the first term by both of the terms in the next factored thingy

so, (-3x)(x) and (-3x)(-1)

-3x^2, seems right so far

and 3x

so far:
-3x^2 + 3x

now for the last multiplying we need to do

(-1)(x) + (-1)(-1)

-1x + 1

sum all the terms extracted from the multiplication

-3x^2 + 2x + 1 = -3x^2 -2x + 1?

[coral crypt]

True or false?

[waxen warren]

That’s some abracadabra for me, never seen anyone do that

Let me check

[waxen warren]

False

[coral crypt]

Coreect

correct*

now let's check our other possible test case

(-3x+1)(x+1)

because (1)(1) = 1, resulting in our c

of ax^2 + bx + c

[coral crypt]

verify this through the method I showed

[waxen warren]

Solve that?

[coral crypt]

yes

want me to walk you through it again?

[waxen warren]

-3x * x = -3x^2

-3x

x

1

[coral crypt]

that part is correct

[waxen warren]

1 *x = x ?
1 *1 = 1 ?

[coral crypt]

1 * x does equal x

and 1 * 1 does equal 1

[waxen warren]

-3x *1 = -3x

[coral crypt]

now lay out all the terms you extracted and add like terms amongst one another

yep

[waxen warren]

Didn’t I just do that :p

I wrote it in vertical

[coral crypt]

yes

[waxen warren]

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

[coral crypt]

now just add them all together

no

remember that 1*x = x

x with a coefficient of 1 is x itself

[waxen warren]

-3^2 -2x + 1

Wait we had the factored form and we brought it back?

[coral crypt]

yes

to verify if it's correct

which you just did

-3x^2 - 2x + 1 = -3x^2 - 2x + 1 is a true statement

do you need the roots (answers)?

[waxen warren]

Alright but I’m still lost on how we factored it

[coral crypt]

alright

let's say we have

[waxen warren]

No all good

[coral crypt]

x^2 + 3x - 18

factor it

[waxen warren]

One sec

x * x + 3 * x - 18

[coral crypt]

not the best way to approach it

[waxen warren]

Okok

x^2 + 3x - 18

[coral crypt]

you see the constant?

Negative 18

-18

[waxen warren]

Yes

Oh wait I think I remember this

[coral crypt]

what two integers multiply to 18?

[waxen warren]

One negative and a positive

[coral crypt]

yep

what two numbers multiply to a product of 18

[waxen warren]

1 18
2 9
3 6

3 and 6 it is and they differ by 3

[coral crypt]

correct

and one must be negative

because we have a -18

[waxen warren]

-3 and 6

[coral crypt]

sure

so now we have

(x-3)(x+6)

[waxen warren]

Yeah and that’s it right

[coral crypt]

yes

just to factor

[waxen warren]

But we have a -3x^2 :d

If it was -3(x^2 something something) it would be easier

[coral crypt]

then re-expand it and combine alike terms to revert back to the original quadratic

[coral crypt]

I said this is an example problem

you said you had a problem with foiling

only when the coefficient of the x^2 term is greater than 1?

[waxen warren]

Yeah because my math problems before this one would let you factor out k(x^2 bx + c)

But now x^2 has -3 in front of it

[coral crypt]

mm alright

now try this:

2x^2 + 5x - 18

[waxen warren]

divide everything by 2?

[coral crypt]

I wouldn't do that

[waxen warren]

Because of the 5x term?

[coral crypt]

leaves us with fractions in this case

I suppose

alright

so first

look at the last term

negative 18

what two numbers multiply to it?

[waxen warren]

1 18
2 9
3 6

[coral crypt]

let's try combinations of 9 and 2 this time

one has to be negative now

because of our negative "c" value

[waxen warren]

Yes but I can’t see them to differ by 5?

9-2 = 7

2-9 = -7

[coral crypt]

they don't, yeah

because the coefficient of the first term alters the 2nd

so you can't use that method for this one

[waxen warren]

now try this:

2x^2 + 5x - 18

Okok

[coral crypt]

you see the 2x^2?

[waxen warren]

Yeah

[coral crypt]

that means one of our first factor terms can be (2x )

[waxen warren]

You just picked it right?

[coral crypt]

so: (2x + _)(x - _)

[waxen warren]

Why is the second one x?

[coral crypt]

yes, you can just take it

[coral crypt]

because (2x)(x) = 2x^2

giving us our first term

[waxen warren]

so: (2x + _)(x - _)

Okok

[coral crypt]

now add in the numbers 9 or 2 into our current blanks

[waxen warren]

Does it matter which of them is negative and positive?

(2x + 9)(x - 2)

[coral crypt]

i think so

yes, it does

because of the 2nd term

[coral crypt]

this is correct

[waxen warren]

The second term is 5x?

In the problem

[coral crypt]

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

(9)(x) = 9x
(9)(-2) = -18

2x^2 - 4x + 9x - 18 = 2x^2 + 5x - 18

[waxen warren]

You have to try if the numbers you put in the brackets should be negative or positive?

Like you did here

[coral crypt]

yep

[waxen warren]

Alright, then I think I’ve got it

Damn this took 40 minutes

[coral crypt]

or you can just find the roots by setting both factored thingies equal to zero

then plug the x-components into the equation

[waxen warren]

That’s if I’ve gotten the roots?

(2x + 9)(x - 2) = 0

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[torn jolt]

how can i represent a summation but difference

[torn jolt]

like 5^2 - 4^2 - 3^2 ... 1

so we start at n^2 and keep subtracting until we get to 1

so it would look like n^2 - (n-1)^2 - (n-2)^2 ... 1

but im not sure how to put that in a formula like one used for summation

[pallid gorge]

I believe that would be the summation with a -1 involved in the process

^ that too

[torn jolt]

sorry can u give an example cuz im having trouble visualizing it

oh i see what u mean the only thing is im trying to put a recurrence relation in closed form

so i found the pattern which is what i said before but like i need to put it in a closed form formula

<@&286206848099549185>

im still stuck 😭

this is what i am trying to put in a closed form formula

.close

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[torn jolt]

hey, Im doing differentiation and Ive found the derivative of the formula for the area of a sphere and set it to 972pi but im not getting the right answer... ( this is the right answer: 324 cm^3/s)

[torn jolt]

dy/dx of the area of a sphere i found to be 4(pi)(r^2)

[haughty nebula]

the radius has a constant rate of change

[torn jolt]

should I be finding the differentiation of the inverse of the volume of a sphere?

[torn jolt]

yes

[haughty nebula]

wait this question is weirdly worded

[torn jolt]

lmao

[stable heart]

note that this isnt asking for area of a sphere, its asking for the volume

[haughty nebula]

oh I see

[torn jolt]

sorry i meant voluome

[stable heart]

so $V=\frac{4}{3}\pi r^3$

[glossy valve]

Duh Hello

[torn jolt]

yeah

[stable heart]

not $4\pi r^2$

[haughty nebula]

so the rate of change of volume would be?

[glossy valve]

Duh Hello

[torn jolt]

I differentiated that to get this

[stable heart]

ah, ok i see lol. nvm

[haughty nebula]

note that r is a function though

albeit changing at a constant rate

[torn jolt]

lol this question should be easy for me but im just being dumb

[torn jolt]

should I be doing this?

[stable heart]

so you need to solve for when $V(r)=972\pi$ and then plug that value for $r$ into the derivative

[glossy valve]

Duh Hello

[torn jolt]

yeah but it seems wrong...

[stable heart]

and yeah you need implicit derivative, since $\dv{r}{t}=\frac{1}{\pi}$

[glossy valve]

Duh Hello

[torn jolt]

wh-

OHHHH

I gotta turn it upside down?

[torn jolt]

after doing this?

sorry idk why im so confused

[stable heart]

am brain lagging a bit here myself lol. give me a moment

[torn jolt]

haha its okay

r=9√3,−9√3 is what my calc gives me when I set it to 972pi

when I multiply that by the rate of change its not right

[torn jolt]

using this to go back to right answer

[stable heart]

ok yeah, since r is also changing then $$\dv{V(r)}{t}=4\pi r^2 \dv{r}{t}=4\pi r^2 \frac{1}{\pi}$$

[glossy valve]

Duh Hello

[torn jolt]

yup

ima try differentiating the inverse so that r is on the right side

[stable heart]

i get r=27, but for some reason my brain decided to die rn so dont know if im doing anything wrong

[torn jolt]

still wouldnt lead to 324 unfortunately

[stable heart]

oh i just made a typo, wrote in $r^2$ instead of $r^3$. so if you have $$V(r)=972\pi=\frac{4}{3}\pi r^3$$ then you should get the solution $r=9$ from here you plug $r=9$ into the derivative of the volume to get 324

[glossy valve]

Duh Hello

[stable heart]

not sure why that was so hard for me, sorry about that lol

[torn jolt]

oh

its okay

thank you

how come you need to use the original equation?

[stable heart]

because it is telling you what the volume is at that moment

so you are trying to find the rate of change when it is at a certain volume

[torn jolt]

ah

so the volume it gives is for the start

[stable heart]

well, its just at that instant

[torn jolt]

yes nvm

tysm anyways

i keep stumbling on questions that shouldnt be challenging

[stable heart]

no worries, this problem could be made easier by making V a function of time. so instead of $V(r)$ we can write $$V(t)=\frac{4}{3}\pi\left(\frac{t}{\pi}\right)^3$$ instead since $r(t)$ is a function of time

[glossy valve]

Duh Hello

[torn jolt]

oh yeah

sometimes i forget to do the small details because we breeze over them so fast in class

[stable heart]

yeah it happens to me as well been a few years since i was doing this type of maths but still manage to struggle this much with it somehow

[torn jolt]

oh lol

how old are you?

[stable heart]

25, started uni late tho so am at my 4th year

[torn jolt]

ah nice, im 15

[stable heart]

huh, dont remember implicit differentiation being a part of high school

[torn jolt]

lmao im in a very advanced maths class

we are basically 2-3 years ahead

but its really hard to keep up with so much content

and my memory is really bad

when I say 3 years ahead I mean uni

[stable heart]

understandable. well then you have nothing to feel bad about, when i was 15 i was struggling to understand wtf a derivative is in

[torn jolt]

lmfao ty

i wanna perhaps study cs in uni tho

so I have to stay in this course

anyways

sorry for telling my entire life story

thank you for helping

[stable heart]

no worries, have a good one

[torn jolt]

baii 🙂

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[keen spruce]

what are u thinking @gusty tinsel

[gusty tinsel]

What you mean?

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[keen spruce]

ok

[keen spruce]

i meant like, what have u thought so far lol

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[torn jolt]

im back 😭

[torn jolt]

idk how the answer to this is supposed to be 1/480

i decided that the equation i would use is $$V = 5x^3$$

[glossy valve]

🍌Hannah🍌

[torn jolt]

and $$dy/dx = 15x^2$$

[glossy valve]

🍌Hannah🍌

[glass crystal]

dont forget x is a function of time

in this case

[torn jolt]

$$V = 5(4)^3 makes 320$$

[glossy valve]

🍌Hannah🍌

[glass crystal]

and you dont know anything about dV/dx

[torn jolt]

and then plugging that value into dy/dx is just massive

[glass crystal]

you only know about dV/dt

[torn jolt]

oh-

im confused

[glass crystal]

you know the rate of change with respect to time

of the volume

so you know dV/dt = 0.5

[torn jolt]

yes

yes

[glass crystal]

so you should try to differentiate this with respect to t

to use this fact

[torn jolt]

but when you multiply the massive value from dy/dt you dont get the answer

[glass crystal]

wdym?

why would you multiply the volume?

forget about actual values for now

lets just do maths and we'll calculate things at the end

you are right with your volume formula

V=5x^3

you are also right about dV/dx

[torn jolt]

well after you differentiate and plug in the value you multiply it by the rate of change no?

[glass crystal]

wdym?

it sounds like a cookbook recipee

[torn jolt]

lmfao

[glass crystal]

what is equal to what?

[torn jolt]

i swear ive been doing this for the past 15 questions

[glass crystal]

what are the mathematical statements?

[torn jolt]

ok one sec ill explain a little more formally

[glass crystal]

nice

[torn jolt]

this would be the chain rule multiplication

that you do after

im my mind

I have to do

[glass crystal]

okay so you use chain rule

[torn jolt]

$$15(4)^2$$

[glossy valve]

🍌Hannah🍌

[torn jolt]

wait no

sorry my bad

[glass crystal]

$$\frac{dV}{dt}=15x^2 \frac{dx/dt}$$

$$\frac{dV}{dt}=15x^2 \frac{dx}{dt}$$

[glossy valve]

Benjamin

[torn jolt]

240

and then the rate of change is 0.5

so you get 120

but im doing something dumb somewhere

[glass crystal]

the thing is its not very clear what you are doing

you are litteraly throwing numbers at me

[torn jolt]

sorry

[glass crystal]

look at this

we know $\frac{dV}{dt}=0.5 cm^3.s^-1$

[glossy valve]

Benjamin

[torn jolt]

yes

[glass crystal]

we and we want to know $\frac{dx}{dt}$ for $x=4$

[glossy valve]

Benjamin

[torn jolt]

mhm

[glass crystal]

so just solve for $\frac{dx}{dt}$ in $\frac{dV}{dt}=15x^2 \frac{dx}{dt}$ and plug in $x=4$

[glossy valve]

Benjamin

[glass crystal]

and $\frac{dV}{dt}=0.5 $

like make sure to have the mathematical expression for what you want

and then you can do the calculations in the end

that is if you are unsure

it feels like something is bugging you

if you dont explain what it is i cant help

[torn jolt]

im sorry im just confused

[torn jolt]

i find it hard to comprehend what you are trying to say with this

maybe because i cant visualise dx/dt and dv/dt

[glass crystal]

you mean physically?

[torn jolt]

yeah

[glass crystal]

or as derivatives?