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[lone heart]

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[autumn steppe]

Hi

[autumn steppe]

@rigid smelt

Hey

Sorry for ping you but that channel got occupied

Anyways how can I know that function is actually a diameter

I have never dealt with a situation like this before

[rigid smelt]

Im sorry but could you post your problem again, "that channel" can be any channel and i dont remember what you asked

[autumn steppe]

This

[rigid smelt]

Right

So basically the idea is that you want to integrate the function that represents the area of the cross-section to give you the volume, was this your first intention? Or did i assume too quickly

[autumn steppe]

Yeah

[rigid smelt]

Ok good, so 1/2pi*(-4x/5+4)^2 would technically have represent the area of the cross-section, if it had the radius -4x/5+4.
Here it didnt, try to think of the picture like this, the is a slice of a semicircle perpendicular to the x-axis specifically, and xy-plane generally, the semicircle is restricted to the region bounded by y=-4x/5+4, x and y axis, so the line at the bottom of the slice is simply the diameter of the semicirc, not the radius, and the length of the line (aka the diameter) is represented by
y=-4x/5+4.
Wait for me to draw a rough diagram

Here's a rough diagram

The integral you had would have given the volume of this solid

[lone heart]

@autumn steppe Has your question been resolved?

[autumn steppe]

Can you elaborate more

[rigid smelt]

Like i said, 1/2pi(-4x/5+4)^2 would have represented the area of a semicircle with a radius of length
-4x/5 + 4

In the picture, it is the red firgure

And the length of the radius is the length from the blue line to the x axis (which is the y value of the line y=-4x/5 + 4 at an arbitrary x coordinate in [0, 5]

Which is half of the red line perpendicular to the x axis on the xy plane

[rigid smelt]

And this is what the exercise wanted

Where the y value of the line represents the diameter of the semicirc, not the radius

[lone heart]

@autumn steppe Has your question been resolved?

[lone heart]

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[crisp kindle]

Can someone please explain how to do question 7

[crisp kindle]

.cancel

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[alpine sable]

does a relation have to be an equivalent relation for it to have equivalent classes?

[worn fox]

yes

thats why they're called equivalence classes

[alpine sable]

thanks

.close

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[indigo crown]

so for I have A = xy

x = (x - 20)

A = (x - 20)y

[gray isle]

where's x = x-20 coming from

[indigo crown]

because on the other side, it's 20 ft

[gray isle]

and how do you have both

A = xy
and
A = (x - 20)y

x = (x - 20) → 0 = -20
something that's clearly false

[indigo crown]

^substituting (x - 20)

[indigo crown]

well... any help is appreciated lol

[gray isle]

it feels like you're randomly changing how x and y is defined during your work

so start by clearly stating/defining what your variables are supposed to represent and clearly mark it on your diagram

[indigo crown]

okay

done that

[gray isle]

show me

[indigo crown]

^

[gray isle]

ok

what's the equation for the area of the field from that diagram

[indigo crown]

A = xy

[gray isle]

yes

now use the given restriction that there is only 1000 feet of fencing to set up another equation

[indigo crown]

so far, I have 2x + 2y = 1000

[gray isle]

no

[indigo crown]

otherwise i'm stuck

[gray isle]

fencing isn't needed along the building nor the part that's in common with the building

the only places that need fencing are the parts that you marked in red

[indigo crown]

yeah I understood that

but I don't get how to approach it from here

[gray isle]

don't overthink it

[indigo crown]

do I take the whole rectangle and subtract a 20 somewhere ?

[gray isle]

no

don't overthink this

the only places that need fencing are the parts that you marked in red

what is the sum of the parts you marked in red (in terms of x and y)

[indigo crown]

x + y + (x-20)

[gray isle]

yes

and what is the numerical value of that

[indigo crown]

1000

[gray isle]

yes

and combine those two simple facts and you'll have your second equation

[indigo crown]

2x + y = 980

[gray isle]

no

[indigo crown]

1020

[gray isle]

trying to jump ahead / skip steps / not doing the proper manipulation

[indigo crown]

smh

yeah 😄

[gray isle]

1020
is more appropriate (but still skipping ahead)

anyway isolate one of your variables and substitute into the other equation

[indigo crown]

i did it in my book 😅

[indigo crown]

I've done some working out but I'm a bit confused

[gray isle]

what exactly are you confused about

oh, your made some errors near the end

you subbed the value of A'' at x=255 instead of the value of x=255 itself

[indigo crown]

oops okay I got it now

this was the answer I was given. But I'm not sure if my answer matches this

after substituting the correct x value, I got y = 769 for x = 255

[gray isle]

1 sec, let me double check everything

you still made errors substitutitng

how are you getting y = 769

[indigo crown]

nvm i found my error

y = 510

[gray isle]

that would be coorect now

[indigo crown]

thanks for your help

.close

[lone heart]

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[trim horizon]

It looks to me like the minus sign disappears in Be^(2ix)

does someone know if that's a mistake?

[worn fox]

typo seems like yeah

[trim horizon]

okay, so would the solution be e^(3x) (C cos(2x) - D sin(2x))?

[worn fox]

the final line is still the same

[spark creek]

I think it could still be the same, since D is a constant

[trim horizon]

oh, that makes sense

okay thanks

.close

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[supple tundra]

yo

[supple tundra]

actually nvm its fine, figured it out

.close

[lone heart]

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[lone heart]

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[alpine sable]

dy/dx: y=x^8*cos(x)

Should it not be the following:

[alpine sable]

.close

[lone heart]

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[lone heart]

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[alpine sable]

(a)
(i) Trapezium

(ii) x ≤ 5
And y ≤ -1/1x + 6

(iii) y = 1x + 0

(iv) m² = -1/1
Hence,
y = -1/1x + 2

Can anyone just confirm? If their answers are matching mine

[lone heart]

@alpine sable Has your question been resolved?

[last ether]

i) Wrong shape
II) correct
iii) Wrong equation, use point-slope form
iv) correct. Simplify to -1

[alpine sable]

Let's talk about the wrong ones

@last ether

[last ether]

Yeah sure

Gimmie a sec

@alpine sable aight so how many sides does R have?

[alpine sable]

5

@last ether

It's a pentagon?

Ic

Part iii

[last ether]

It's a pentagon yes

For part. iii, use point slope form

[alpine sable]

Huh?

[last ether]

$y=m(x-a)+b$

Where $(a, b)$ is a point on the linear function and $m$ is the slope

[ocean seal]

Umbraleviathan

[alpine sable]

You mean y= mx + c

[last ether]

That's slope intercept form

[last ether]

This is point-slope form

Did you learn about the different ways to write lines functions?

[last ether]

And this is derived from point-slope form (if the y intercept is (0, c))

[alpine sable]

Hm

What will the answer be?

[last ether]

Let's work through that

[last ether]

What would m be

For iii

[alpine sable]

m= 1
c = 0

Is it?

[last ether]

Well

Slow down a bit lol

m = 1

And we're given a point that it passes through, (5, 0)

[alpine sable]

Yeah

[last ether]

Given this information, what would a be?

[alpine sable]

What's a

[last ether]

Read it

[alpine sable]

I ain't familiar with that formula

[alpine sable]

y= mx + c
0= 1(5) + c

5 = c

[last ether]

Don't use y=mx + c

That only works if a = 0

But a = 5

Because you have the point (5, 0)

(a, b) (5, 0)

[alpine sable]

That's x, y

[last ether]

Yeah I'm just replacing that with a and b, respectively

[alpine sable]

Ic now

[last ether]

Yeah

For trivia, here's the actual point-slope formula:

$y - y_1 = m(x-x_1)$

[ocean seal]

Umbraleviathan

[last ether]

But you don't need to worry about that

I simplified it for you

[last ether]

So if a = 5, what would b be?

[alpine sable]

0

[last ether]

Good

So now plug it into the formula:
$y = 1(x-5) + 0$

[ocean seal]

Umbraleviathan

[last ether]

Simplify that

[alpine sable]

I literally gave wild guess

[last ether]

Lmao well you guessed wel

[alpine sable]

But that formula

Is

Abit yk

[last ether]

Yeah

You'll get used to it

And it's more useful than slope-intercept

[alpine sable]

Okay lemme absorb the info

What is a slope point

[last ether]

It's the name of the form

[alpine sable]

Form of formula

When is it used

[last ether]

Because you have a slope (m) and a point ( (a, b) )

[last ether]

When you have a slope and a point that the function passes through

[alpine sable]

So u told me two formulas

One of slope

[alpine sable]

And other this

[last ether]

No they're the same

Wdym two formulas

[alpine sable]

Are they?

[last ether]

It's still y = m(x-a)+b

[alpine sable]

M= Y²-Y¹/ X²-X¹

This?

[last ether]

Oh yeah that is the linear slope formula

But that is what m equals

But if you already have what m is, you don't need to do that calculation

[alpine sable]

M = 1

We know it

[last ether]

Yeah

[alpine sable]

So what to do now

y = m(x-a)+b

Let's apply

Ane find b

[last ether]

You're given a point, what point is that?

[alpine sable]

5, 0

[last ether]

Yeah

So if we read this

(a, b) = (5, 0)

By logic, a = 5
b = 0

[alpine sable]

Y = 1(x-0) + 5

What's b

[last ether]

No no, keep x

x ≠ a

You don't replace x

You only replace m, a, and b

[alpine sable]

What are we trying to dig out

From that formula

y = 1(x-0) + 5

[last ether]

You switched the a and b values

[alpine sable]

Yeah I did

[last ether]

The formula essentially states "a line with this passes through this particular point"

Do you understand this at least?

[alpine sable]

Well

Hm

They asked to draw a line

I just noticed

[last ether]

Yeah, but you can use the point (5, 0) as a baseline

And then draw the line from there

[alpine sable]

Is it?

2ait no

I did it upside down

[last ether]

It passes through (5, 0)

That doesn't pass through (5, 0)

[alpine sable]

So tell me where will line go

[last ether]

(5, 0)

[alpine sable]

The points of that line

[last ether]

Goes through that

[alpine sable]

The new line

[last ether]

Quite literally from (5,0) parallel to y = x + 2

[alpine sable]

Dude it's literally 1 mark question

They asked to draw line

[last ether]

From (5, 0), it goes up 1, right 1

And so on

[alpine sable]

I would have given up 1 mark for this new.formula

I-

[last ether]

I'm surprised you haven't been taught the forms

[alpine sable]

Let's just add 1 point and done

[last ether]

_ _

[alpine sable]

I just know y= mx + c

And they won't give smth out of syllabus

[last ether]

That doesn't seem very useful lmao

If you're just given one form

[alpine sable]

DONE?

[last ether]

that's not parallel to y = x + 2

It's from 5,0

Up by 1

[alpine sable]

Then

[last ether]

Right by 1

So the new point will be (6, 1)

And so on

[alpine sable]

I did that?

[last ether]

Okay I think you're getting confused

That is another point it crosses through

And then you connect (5, 0) and (6, 1)

[alpine sable]

Cuz of it's + 2

Ic

[last ether]

No the + 2 has nothing to do with the new line

Parallel means "same slope"

[alpine sable]

Yea

[last ether]

So if you have the same slope

[alpine sable]

7.2

[last ether]

Same gradient

[alpine sable]

Is the answer

Ik

7,2

[last ether]

Well yeah that would be another point, yes

And 8, 3

[alpine sable]

Don't confuse me

[last ether]

And 9, 4

So

[alpine sable]

Its infinite

[last ether]

Just draw the line

[alpine sable]

What's the answer

[last ether]

That has the same slope, but it starts from 5,0

[alpine sable]

I mean I will keep on guessing?

[alpine sable]

Yea

[alpine sable]

6,5

I know it

That's it

Is it?

[last ether]

That's it

Because

Like I said

[alpine sable]

wot

[last ether]

Same slope

Starts at 5, 0

So you literally apply the same characteristics of the slope

Starting from 5, 0

[alpine sable]

What should I have done

To know that' I shouldn't have done that

[last ether]

Translate the line to fit the criteria is what you would've done

Basically

Start at 5,0

Draw a line with the same slope that passes through that

[alpine sable]

Yeah

U mean x= 5

I GET IT

[last ether]

(5,0) is x = 5, y = 0

Yeah

[alpine sable]

Y was 0
And I kept on guessing wrong

[last ether]

You just start at that point instead of the axis

[alpine sable]

Next time will try guessing better

Thanku

Thanku for bearing lol

[last ether]

Lol yeah np

[alpine sable]

.close

[lone heart]

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[rigid thistle]

What do I do for b?

[wanton turtle]

Since you got the values of a and b and have the entire equation, just follow the usual steps of finding stationary points

[rigid thistle]

So I just sub the a and b and differentiate?

[wanton turtle]

Yup

[rigid thistle]

Oh alright

[wanton turtle]

And solve the resulting equation

[rigid thistle]

Something's wrong

[wanton turtle]

Yeah your initial steps are wrong

b is a constant

What happens when you differentiate a constant?

@rigid thistle

[rigid thistle]

OOH it's like am number right

So there's nothing

[wanton turtle]

Yes

[rigid thistle]

Ooh

I got it, thank you

.close

[lone heart]

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[gray isle]

log(ab) = log(a) + log(b)

it isn't explicitly a product of two things
but it can be expressed as such

specifically 2+x^2 = 2 * (1 + x^2/2)

[lone heart]

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[gray isle]

just like how log(6) = log(2) + log(3)

[lone heart]

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[fast oak]

Help

[fast oak]

X- bounds

[shell narwhal]

What is your question?

[last ether]

Post a specific question

What about x bounds?

[shell narwhal]

Thats not a question

[fast oak]

The upper 0 and lower 4

[last ether]

Yeah

That's the same as $-\int_{0}^{4}(2x+2){dx}$

[ocean seal]

Umbraleviathan

[fast oak]

Idk how to do the limiy change

[last ether]

Wdym limit change

Okay, first, don't do u-sub

[fast oak]

Ok

[last ether]

What's the antiderivative 2x + 2

[fast oak]

3

[last ether]

No

[fast oak]

X^2+2x+c?

[last ether]

Yeah.

So now when you do the integral

[fast oak]

Ohh ok

[last ether]

$[x^2 + 2x + C]|_{4}^{0}$

[ocean seal]

Umbraleviathan

[last ether]

Evaluate this

[fast oak]

Wait

1/3^3+2×+C?

[last ether]

No

You know how to do integrals right

[last ether]

Just evaluate this

Do you know what this means?

[fast oak]

Yes

Ok wait

[last ether]

So evaluate that

[fast oak]

1^3+2x+C

[last ether]

No

That's not what it means

[fast oak]

The lower 4 and upper 0?

[last ether]

It's $(0^2+2(0) ) - (4^2 + 2(4))$

[last ether]

Remember your fundamental theorem of calculus

[fast oak]

Ohh

[last ether]

Where F(x) is the antiderivative of f(x)

[fast oak]

Ohh ok i get it now

[last ether]

Technically there shouldn't be a C

In here

[fast oak]

Cancel?

[ocean seal]

Umbraleviathan

[last ether]

There

[fast oak]

-24

[last ether]

Yeah

[fast oak]

Ok ty u

.close

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[lone heart]

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[devout latch]

Hi, I have a problem with an exercise with geometry in three dimensions (x, y, z)

1. Verify that the triangle A(4; 9; 0), B(0; -4; 0), C(0; 0; -4) is equilateral.
2. Determine the angle α that the height CH forms with the Oxy plane.

[devout latch]

I did the first question, but I can't solve the second.

[knotty hull]

yo

[devout latch]

hi

[lone heart]

@devout latch Has your question been resolved?

[nova quail]

1/2x|a x b|

= area

= a.b.sin(alpha)X1/2

[lone heart]

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[devout latch]

.reopen

[lone heart]

✅

[devout latch]

thank you very much

now the result is correct!

.close

[lone heart]

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[nova quail]

Okay bruhh

[lone heart]

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[snow coyote]

.open

[bronze ibex]

anyone knows how we can share mathtrainer.org score

[pliant current]

we don't have anywhere dedicated for sharing mathtrainer.org scores - or any scores of that matter

nor do i think we plan to do such a thing

if you have a question regarding a math problem, feel free to use these channels; these kinds of questions arent meant for the help channels :p

[lone heart]

@bronze ibex Has your question been resolved?

[lone heart]

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[magic kelp]

what if I have two limites... lets say lim x->a f(x) and lim x->a g(x)... doesnt matter where the limit is... what if I sum them so I have lim x->a f(x) + lim x-> g(x)... if lim x-> a f(x) doesnt exist but the g(x) one does... the sum exists?

[magic kelp]

ping me when I am answered pls

and thx for helping

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[magic kelp]

.reopen

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✅

[magic kelp]

<@&286206848099549185>

[lament glen]

I don't think so

let me see if I can find examples

[magic kelp]

for example...

the one from the left

I dont think it exists but wanted to make sure about it

[worn fox]

Sure it exists

[lament glen]

the limits exist

on their own too, I think

oh wait

[magic kelp]

In their own I was taught that they doesnt exists

[lament glen]

oh nvm

[magic kelp]

But together... why would they

[worn fox]

The first question is a one sided limit, so both exist on their own

[lament glen]

is it? it just says $-2$ not smth like $-2^{+}$

[ocean seal]

Doggo

[magic kelp]

I mean... I think yeah... they mean -2, not 2-

[worn fox]

Oh yeah dumb

[magic kelp]

Np

[worn fox]

Read -2 as 2^-

Oop

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@magic kelp Has your question been resolved?

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@magic kelp Has your question been resolved?

[proper tangle]

calculate left-hand limit $\to a^-$ and the right-hand limit $\to a^+$, and you'd find out the ans

[ocean seal]

vin100

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[dreamy kraken]

Can someone confirm that T = [-1, 2, 0] and [T]b = [-1 0 0; 0 2 0; 0 0 0]

<@&286206848099549185>

[placid zinc]

T = [-1, 2, 0]? That doesn't parse.

T(1) = 1
T(x) = 2x - 1
T(x²) = (2x - 1)² = 4x² - 4x + 1

I don't see any 4s in your matrix so this isn't looking good

[dreamy kraken]

So T is a 3x3 matrix ?

[placid zinc]

[T]b is, yeah.
T isn't a matrix at all, but a function.

T can be given a matrix representation, for that you need to choose a basis. Then, [T]b is born.

[dreamy kraken]

Right

Is there a way I can write what T is equal to

or is it just equal to a polynomial trasnformation

Can you also just explain the process of T(1) and so on, I'm not sure i have fully understood this concept

[placid zinc]

Since T is a linear function, T is completely determined by how it acts on a basis

So, we check what it does to the elements in b

[dreamy kraken]

So are basically saying (2x - 1) multiplyes by 1 = 1?

[placid zinc]

T takes a polynomial, and substitutes x out for 2x - 1.

T(x²) = (2x - 1)²
We're taking every x, replacing it with 2x - 1

[dreamy kraken]

Ahhhh right

[placid zinc]

T(1) = 1
Because that's just how 1 rolls

[dreamy kraken]

so for example T((x-1)^2) = x^2 -2x

[placid zinc]

T((x-1)²)
= ((2x - 1) - 1)²
= (2x - 2)²
= 4x² - 8x + 4

[dreamy kraken]

Rightttt that makes so much sense now thank you

so T (x-1) = 2x - 2

[placid zinc]

Note that we're abusing the fact that T is a linear function. That is, T distributes:
T(x - 1)
= T(x) - T(1)
= (2x - 1) - (1)

Because of that, a matrix form can be found. That's [T]b

[dreamy kraken]

Yeah I wasn't aware that T alone couldn't be written as a matrix

I am asked questions about the egienspace of T

do you think thank refers to [T]b

[placid zinc]

One way to get such a space is to find [T]b then do all the determinant shenanigans to it

Sometimes it's a little obvious though

For example, we already found that:
T(x - 1) = 2x - 2
So x-1 is an eigenvector with eigenvalue 2.

T(1) = 1
So 1 is an eigenvector with eigenvalue 1.

We're missing a dimension, so there's probably another one

[dreamy kraken]

Right that makes sense

I am also trying to calculate [T]c using the change of basis formula but I can't seem to get the values to match my previous calc of [T]c using the other method

[placid zinc]

We need a matrix that takes members of B to members of C then

Or do you already have the change of basis matrix?

[dreamy kraken]

I have it but I am not sure its correct

Pc,b = [1 1 3; 0 1 2; 0 0 1]

[placid zinc]

I'll call that matrix A

Just because lazy

A•(1,0,0) = (1,0,0)
A•(0,1,0) = (-1,1,0)
A•(0,0,1) = (1,-2,1)

That's how A should multiply members of B

[dreamy kraken]

I made a sign error

So am i right in saying B to C is Pc,b

[placid zinc]

I may not use the same notation, haha

We're interested in a matrix that takes members of be B to members of C

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[dreamy kraken]

Thanks so much for your help @placid zinc !

[cold zenith]

lol

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[plain cipher]

Would the bounds of integration be from -2 to 2 and it would be the integral of 2nd function minus 1st function?

[undone hinge]

yea

[plain cipher]

What about problems like this?

[undone hinge]

the second integral:

integrate from -1 to 0 and from 0 to 1 and add them up

[plain cipher]

ah ok

[undone hinge]

so -1 to 0 second function minus first function, 0 to 1 first function minus second function

it actually doesnt really matter wether you subtract the first one from the second one or vice versa since you use the absolute value

[plain cipher]

o

1 min

@undone hinge what the heck would my integrals be

did I even graph it right lol

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[lean finch]

how do I convert this into cylindrical ?

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[elder lark]

https://cdn.discordapp.com/attachments/582707120789782551/975589576678711346/unknown.png
How would I find the average wavelength?
I was thinking maybe weight the wavelength values by the absorbance and then average that weighted value, but then how would I get wavelength back?

[elder lark]

I have a large table of the data

This feels suspiciously like an integral

ooo is this center of mass? How would I get the center of mass with data points instead of a function?

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[kind sierra]

integral over interval / length of interval?

this seems to be a smooth function

so try this

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[cunning trout]

I don’t understand the part where it says “passes through the same x-int”

[slate jolt]

it intercepts the x axis at the same point?

[vale wigeon]

the two lines are supposed to have the same x-intercept but it was phrased poorly

[cunning trout]

So our point is (0,1)?

[devout summit]

x-intercept is the point where the line intersects x-axis where y=0

[devout summit]

This is the y-intercept of given line

[cunning trout]

Then how would we find the x-int

[devout summit]

I gave the step here. find x where y=0

Given that y=mx+c is perpendicular to y=x/8 + 1 what would the slope "m" be

[cunning trout]

-8

[devout summit]

Yes

[devout summit]

Did you find the x-intercept?

[cunning trout]

Yeah

[devout summit]

That should be (-8,0)

[cunning trout]

Yes

[devout summit]

Now, we know our line takes form y=-8x+c

What should c be so that it passes through (-8,0)?

[cunning trout]

Ok I got the equation

[devout summit]

Good!

[cunning trout]

Ok thanks

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[stiff anvil]

isnt the 13 the area

it says its wrong

[real solar]

Pythagorean theorem seems to be the way

[gray ingot]

no

find the area of one triangle then x 4

[stiff anvil]

what about this tho

why is 13 in the exact place of where area is

[gray ingot]

no? a doesn't mean area

it means side length

[real solar]

[stiff anvil]

its means area

look

[gray ingot]

how is the diagonal 2.6?

it's clearly 24

[stiff anvil]

i did that wrong

[gray ingot]

🗿

[stiff anvil]

but

still a is area

[gray ingot]

no

[stiff anvil]

how

[gray ingot]

capital A is the area

a is the side length

a = p^2/4 + q^2/4

[stiff anvil]

got it

so 13x4

?

[gray ingot]

no

[stiff anvil]

then what do i do

[stiff anvil]

huh

uh helllo

.closer

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[karmic quail]

what is b if we have vertices (-6,2) (14,2) and foci (11,2) (4,2)

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[stone ruin]

Is there a simple formula that tells the y coordinate where a line segment that crosses x=0 will cross over x=0

[stone ruin]

given only the endpoints of the line segment

[slate jolt]

well from the endpoints you can derive the line equation

then find for which y cordinate x=0

and then find if that point is in your initial boundaries

[slender gull]

Section formula.

[karmic quail]

no

[stone ruin]

i figured it out ty

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[livid bison]

find the length of x

im trying to solve ot using the formula

m<A = 1/2 ( arc mCE - arc mBD)

but idk what arc mCE is?

like theres no stated angle just
those numbers

[gray isle]

finding measures of arcs is unnecessary for finding the length x

[livid bison]

oh then

wait heres a clearer picture

how am i supposed to solve it?

[uncut plank]

https://en.m.wikipedia.org/wiki/Intersecting_secants_theorem

[gray isle]

consider that you have two secants,
the secant-secant theorem will be applicable

[livid bison]

i see! ok

thank u

ive been lost as to what to search

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[livid bison]

what theorem should i use for this?

[livid bison]

i thought it would be tangent-secant segment but its not?

nvm i found a tutorial for it :))

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[sterile vale]

could somebody please explain to me the logic behind this please?

[vale wigeon]

when is the product of two real numbers positive?

[sterile vale]

Perfect, got it

thanks a lot

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[dawn quail]

I don’t understand this answer

The formula is as follows

So a few things… p squared is 2p?

[harsh girder]

$(a-b)^2=a^2-2ab+b^2$

[dawn quail]

I thought it is p * p?

[ocean seal]

秋水

[dawn quail]

My formula is wrong? I need one for a + b and one for a - b?

Also, when it says a and b are you not supposed to include the negative symbol for each term?

[harsh girder]

in this, a is p^n, b is 1

[dawn quail]

Why is it not -1 for b?

For Quadratic Formula it includes + or - for each term for a b and c

[harsh girder]

if you use (a+b)^2 , then b is -1, is ok

then is 2*(p^n)*(-1)

[dawn quail]

Can’t I just use this formula if I include + or negative for each term? Or would output be different

[harsh girder]

The left side has ±, which represents two equations, and your right side should also have ±

[dawn quail]

Ah OK

[vale wigeon]

yes these are the same

[dawn quail]

OK

I think I prefer 1 formula instead of memorizing two

If it’s a + b or a - b I mean

And I always include the symbols + or - with terms

That was the way I was taught the symbol in front of the term actually belongs to that term to show if it’s positive or negative

So to remove the symbol for this formula is confusing for me

By removing the symbol I mean for 2ab the second line has positive 1 for b, which doesn’t make sense to me…

But the third line has -1 which makes sense to me

Can someone explain to me why (p^n)² is p^2n?

This is my last question for this solution I don’t understand why it’s 2n. 2n is not the same thing as n*n

[harsh girder]

do you know $(a^b)^c=a^{bc}$

[ocean seal]

秋水

[dawn quail]

Right that makes sense to me, using the distribution method

[harsh girder]

p^n is the product of n p and (p^n)^2=(n^p)*(n^p) is the product of 2n p

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@dawn quail Has your question been resolved?

[dawn quail]

Top or bottom is correct? Or both wrong?

[harsh girder]

i mean p*p*p*p*...*p have n times

[naive valley]

top is right, not sure what you are doing with the bottom calculation

[harsh girder]

my English is not good

p^(2n) = (p^n)*(p^n) = (p*p*p...p)*(p*p*p...p) there are 2n times

[dawn quail]

Ohhh

One sec

Do I have this correct?

[harsh girder]

correct

[dawn quail]

The number in front of the exponent n just means how many times it is repeated?

[harsh girder]

a^n means a*a*a*a...*a

[dawn quail]

Thanks

It is just confusing to see a number in front of an exponent like that, I feel like it’s saying multiply 2 with n for ^2n

But it’s actually saying ^n ^n

Two times

Thank you!

[harsh girder]

^n means n times

[dawn quail]

Right but n is unknown so I feel like the two in front is multiplying with that

And it is, but just not in the way I imagined it at first

It’s saying repeat this exponent for the term itself two times rather than multiply the exponent itself by 2

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[uncut jackal]

Is there an horizontal asymptote?

[naive valley]

how do you usually find horizontal asymptotes?

[uncut jackal]

X's possession

But here there’s no x at the top

[naive valley]

not sure what you mean by possession

what is the limit of this function as x -> infinity?

[uncut jackal]

I meant x’s strong

[uncut jackal]

Legit I’m only grade 11 we haven’t learned infinity yet

[naive valley]

oh

i'm not sure how you find asymptotes without limits...

[uncut jackal]

This is the method we use

The top one is “no asymptote”

[naive valley]

well you can always put x^0 in the numerator

that just equals 1 for any nonzero x

[uncut jackal]

Oh yeah thanks

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[rapid bloom]

hi, reviewing a past paper but i don't understand this question, help please?

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[rapid bloom]

<@&286206848099549185>

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[valid moth]

Help?

[kind sierra]

...

This is a math server

[valid moth]

Oh I forgot

[kind sierra]

and there is only one formula

@valid moth close this thread if you are done.

[valid moth]

.close

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[paper owl]

Hello

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[kind sierra]

@paper owl please close this thread!

Do not open help channels for no reason

use .close

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[cosmic zenith]

trigonometry

[cosmic zenith]

need to get C

what i know is that
a=61 m
b=30 m
c=80 m
A=43 degrees

[languid river]

on what?

[cosmic zenith]

on what??

[languid river]

is there a picture to go with it?

[supple tundra]

its a triangle?

[languid river]

ah

[cosmic zenith]

yes

[languid river]

sin rule

[supple tundra]

you need to get better at asking qs cuz that alone makes no sense lol

[languid river]

a/sin a = c/sin c

[cosmic zenith]

aint it other way round?

sin a / a

[abstract fractal]

It doesn't matter, so long as you're consistent

[languid river]

ye either probably works

[cosmic zenith]

ok, ill try it

i've tried it but idk what ive done wrong, the result i get is 63 degrees but it should be around 117 degrees

[abstract fractal]

Show your work

[abstract fractal]

Arcsin is a multi valued function

[cosmic zenith]

so how am i supposed to use it?

nvm got it, used arccos instead

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[misty bobcat]

,rotate

Is this big O proof correct ?

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[turbid stag]

I have a question:

A mass oscillates on the end of a spring suspended from a ceiling. The depth of the mass
below the ceiling is given by 𝑑 = 5 sin(24𝑡) + 12 where 𝑑 is measured in centimeters
and 𝑡 is measured in seconds.

What is the maximum depth of the mass below the ceiling?
``` and i wrote:
the maximum depth of the mass below the ceiling is 5 centimetres as the number that multiples the trigonometric function determines the amplitude of the wave.

[turbid stag]

it doesnt seem like a correct explaination but i cant figure out how to express it

[primal berry]

From what I can get from this is

The maximum depth is 5cm, but only of the harmonic motion of the mass oscillation. The +12 must be additional length of the spring at the equilibrium point, or something of the sort

[turbid stag]

this is very elementary, its very literal

its 5cm but i dont know the reasoning, i just know the multiplying integer effects the amplitude

because the number that multiplies the trigonometric function determines the length of the line until the line is curved due to the trigonometric function

[primal berry]

IF we are to just look at 5sin(24t) then yes, the maximum depth is 5cm. As we know that sinx will have a range of y values between -1 and 1. We multiply that by 5 and it goes between -5 and 5. The 24t only alters the x values, or in this case, the t values

[turbid stag]

oh wow thats a really good explaination

[primal berry]

I hope it's all good at least, harmonic motion is one of the only parts of applied I can somewhat wrap my head around

[turbid stag]

could you give me peice of mind for the next question? its asking " whats the minimum depth of the mass below" the minimum would be the amplitude plus the other expression which isnt effecting x-axis

which would then mean, in the equation 5 x sin(24t) + 12, it would be 5 + 12

which would equal 17, and using desmos its correct

could you please tell me how to express it please?

is that explaination good? the integer that determines the amplitude and the integer that directly effects the y axis

wait so using the harmonic motion's representation of the wave equation, technically the y axis is being directly being affected by +12, but because the amplitude is determined by multiplying the trigonometric function

[primal berry]

That's essentially it really.
The + 12 on the end of the equation will shift the sin part "up" by 12 units. In this case if you're looking at it from the ceiling, it should really be 5sin(24t) - 12 if you take upwards as positive.

With that last statement, yes. The amplitude of the sin wave is being scaled by factor 5 which we've been given. The +12 is simply shifting or moving the whole equation up or down your graph

[turbid stag]

man you are articulated, thanks mate really helps

[primal berry]

And a really poor drawing to go along with it

And blame the fact that I had an exam which involved harmonic motion, it's why it's stuck in my head lol

[turbid stag]

harmonic motion is an axiom in so many domains

physics, engineering, mathematics, statistics, game dev even, its good knowledge

[primal berry]

Shame that we can't have perfect harmonic motion, only damped

[turbid stag]

is that due to the trignometric functions?

[primal berry]

If we look at it in real life there will always be a damping force on a harmonic system, take this mass spring system.
There will be energy lost to say - Air resistance, spring modulus of elasticity and such

[turbid stag]

wb in a vacuum or space

[primal berry]

Trigonometric functions will keep going on forever, look at sinx for a nice example. It is always 1 at pi/2 + 2npi (n belonging to the natural numbers)

Vacuum/space is a difficult one. You'd lose the air resistance but then it depends on the system. If it's mass spring then you'd lose energy to the spring. If it was a pendulum...I can't think off the top of my head. Most likely due to gravity. Depends if you're in a perfect vacuum in space or say, in the ISS

[turbid stag]

yeah fair, the laws of thermodynamics still apply in space

...i think?

[primal berry]

Heat can still be lost in the slightest of movements

[turbid stag]

they do right? or is it a frame of reference sort of deal

well, the 5rd law of thermodynamics dictates that that perpetual motion cant exist because energy is expended due to motion

[primal berry]

The thing is with space is that there's not make particles moving around compared to here on earth. So it's harder to transfer energy from one thing to another

[turbid stag]

but I don't know if that's a inertial frame of reference thing

[turbid stag]

i suppose its harder to lose energy in space, thats why things stay in "perpetual" motion in space

[primal berry]

Getting into thermodynamics is one thing I've not touched. I can only go off what I've learnt in my lectures about forced harmonic motion and such

[turbid stag]

lol all goodie, i just watched a tedex vid on it xD

but anyways appreciate the help on the questions

.close

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[median mauve]

this is my teacher solution

[median mauve]

I understand everything there

but i am not sure how that prove KL = KB.

[merry depot]

Why would angles FBC and BCF be the same?

[median mauve]

that is a typo

i think he means FBC and BCK

[solemn grove]

I think CK and DL trisect BA

which means 1/3 BA = BK = KL = LA

[median mauve]

how you know that

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[median mauve]

<@&286206848099549185>

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[median mauve]

[median mauve]

Any one have any idea

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[karmic quail]

.reopen

[uncut sleet]

angle FBC = angle BCK , so BF // CK // DL, because CF = CD, so BK=KL

[median mauve]

how because CF = CD, so BK=KL

[karmic quail]

how do you factor 0=-16t^2+32t+8
?

[oak perch]

Are using coordinates allowed?
Cause you can let B=(-1,0),C=(0,1),A=(1,0),E=(-t,1-t),D=(t,1-t). Then you can solve that K=((t-1)/(t+1),0), L=(t-((t-1)^2/(t+1)),0) then clearly t-((t-1)^2/(t+1))-1=2(t-1)/(t+1)

[uncut sleet]

@median mauve basically if multiple lines are parallel, the lines they intersect would have the same proportions.

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[median mauve]

What this theorem called

[oak perch]

KL:KB=FC:CD=EC:CD=1:1

the first equation is true because triangular ACE and AFB are similar, so AC/AK=AF/AB, or equivalently, AC/AF=AK/AB. Similarly ADL and AFB are similar so AD/AF=AL/AB. Therefore FC/CD=(AF-AC)/(AC-AD)=(AF/AC-1)/(1-AD/AC)=(AB/AK-1)/(1-AL/AK)=(AB-AK)/(AK-AL)=BK/KL

[median mauve]

Thanks so much!

[oak perch]

Sorry made little mistakes I am editing

Wait a second

[median mauve]

Ok

[oak perch]

Fixed

[median mauve]

Thanks so much!

.close

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[lilac copper]

i need help with finding a and b 😭

[steel pewter]

Can you make another photo?

[lilac copper]

no 😢

[steel pewter]

I can’t read it

[lilac copper]

The graph of y=f(x) has a rule of the form f(x) = a(x+b)^3+2
Find the values of a and b
they gave 2 coordinates (1,2) and (3,6)

[harsh girder]

then you know f(1)=2 and f(3)=6

[lilac copper]

dont i need to use simultaneous equation to do it

[harsh girder]

solve the two equations you can get a and b

[lilac copper]

how

is it anything like the example

it got a diff cubic function

[harsh girder]

$f(1) = a(1+b)^3+2=2$

[ocean seal]

秋水

[lilac copper]

i done that

but i didnt know what to do next

[harsh girder]

then you get $a(1+b)^3=0$, right?

[ocean seal]

秋水

[lilac copper]

yeahhh

[harsh girder]

a is not 0, so b = -1

[lilac copper]

OH

null factor law

[harsh girder]

then use f(3)=6 to get a

[lilac copper]

let me try that

cant i sub b = -1?

since i got the b now

[harsh girder]

of course

[lilac copper]

yeah yeah

ok

thank you

🙏

[lilac copper]

is this correct now

[harsh girder]

correct

[lilac copper]

thanks

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[misty bobcat]

graph theory task:

graph G is k-edge connected => graph G is union of k-edge connected disjoint spanning trees

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[indigo jasper]

hi

i need help

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[amber sphinx]

hi, so i have little problems with graphing functions

i cant really graph it and i would like to ask for help

i think i can do the rest

just im not sure how to graph it, ive always had problems with that

ive thought of calculating the x, but now im not sure completely how to graph it, because ive looked up in the answers and its different from what i thought it would be

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[amber sphinx]

<@&286206848099549185>

<@&286206848099549185>

[amber sphinx]

<@&286206848099549185>

[alpine sable]

Hey!

So to be clear here, you want to give a sketch of the graph of the function given in exercises a

Do you have (access to) one of those graphing calculators?

Otherwise, what i might suggest is just calculate the function value for x=-3, x=-8, x=-13 etc.

And plot those points in your sketch plane

To sketch it you just have to find enough points to really make a good sketch out of it. Maybe its good to have very big ‘jumps’ between the tick marks of the plane, so you get a more accurate depiction of what a lot of the graph looks like.

[alpine sable]

So maybe tick marks bigger than these though

Like calculate for x=-3, then -69, -420 etc whatever

Also, i realised a probably important detail: the x-coordinate of the top of the parabola of the function (with domain R) is x=-b/2a = -6/2 = -3. So -3 is actually the top of the parabola if you sketch the function over R.

What that says is, when you draw the function from x is -infinity to -3, youre actually drawing half of the parabola.