help-28

[wide star]

Me and you both see tv+w is not an equation.

It's an expression.

So clearly the person who wrote this question is a moron who does not know what the word equation means like you and I.

But what's missing for tv+w to be an equation?

An equals sign and a vector on the lhs.

So, pick z for ex and let z=tv+w

You know v and w. If you pick values of t you will be able to compute z.

Idk how well you understand what I'm getting at with this?

Basically plugging in values for t gives you z as x,y coords on your line.

[analog kayak]

Not very much

From what I know the top number is x and the bottom number is y

[wide star]

Yeah pretty much

[analog kayak]

So how do I put those 2 values into 1 variable

[wide star]

You don't have to do that or really need to do that but if you wanna you can notice z=tv+w gives you a system of equations x=(some stuff involving t) and y=(some stuff involving t)

And then eliminate t using some algebra.

[analog kayak]

The correct answer for this is 1 and 4 (no solutions and they r parallel) I think

[wide star]

Yeah I agree

But it's important to understand why that's correct

[analog kayak]

Yea I was gonna say

[wide star]

Here's an example

[analog kayak]

I’m gonna come back to this tmr

Bc I’m so insanely lost

[analog kayak]

But I get the gist

[wide star]

The bottom three lines all say the exact same thing

If you look at the y=tm+b line

That is the general equation of a line in 2d

[analog kayak]

Yes

[wide star]

The m vector tells you the direction of the line (like the slope)

[torn jolt]

Easiest way to do this is to just turn change the line equations from vector form to Cartesian form

[wide star]

It points in the direction the line itself points

The b is a point on the line

[analog kayak]

This is exactly what I want to do, but don’t know how to

[wide star]

That's not the easiest way to do this problem

[torn jolt]

What would be easier?

[analog kayak]

I’m not familiar with the format or the notation

This is my first time learning vectors

[wide star]

The easiest way to do this problem is to notice BOTH lines have the same direction vector

Which is why I'm explaining the general equation of a line here lmao

[torn jolt]

Ah

I didn't see that

Yeah they're parallel @analog kayak

[wide star]

Granted one thing to note is that two lines might have different direction vectors and still be parallel

[torn jolt]

Ik

But when it's the same

Obv

[wide star]

I'm not saying that to you

[torn jolt]

Alr well

[analog kayak]

Is there any YouTube vid u can recommend to teach me ab the basics of this?

[torn jolt]

Bai hf

[analog kayak]

Thx for chiming in

[wide star]

So, the thing I was saying about direction vectors

[torn jolt]

I don't think there's a video for this tbh

Maybe searching for line equations in vector form would help?

[analog kayak]

Hmm kk

[torn jolt]

I think that'll just show lines in 3d space tho

[wide star]

There almost def are vids about this on youtube.

[torn jolt]

Not 2d

[torn jolt]

I've mostly seen them on 3d

[wide star]

Like, any multivar calc class usually reviews this stuff

This problem is in 2d..

[torn jolt]

Ik

[analog kayak]

If u could suggest any that would be great

[wide star]

Even then nothing changes in n-dim

[torn jolt]

Ik nothing changes

[wide star]

Other than the number of components

[torn jolt]

But @analog kayak would probably prefer to understand in 2d first

[analog kayak]

Does this ticket automatically delete?

[torn jolt]

Not as long as we're talking

[wide star]

I still need to point out something

[analog kayak]

It’s late and I’d prefer to come back to this tmr

[torn jolt]

Alr gn

[analog kayak]

So would it delete overnight

[torn jolt]

Not rlly

[wide star]

Do .close

[torn jolt]

It'll just move forward

Other people use help channels as well

[analog kayak]

Oh right

Thanks guys

[wide star]

If you type .close the channel will close

@analog kayak

[analog kayak]

I don’t want it to close

[torn jolt]

Type this

.close

[analog kayak]

So I can read again tmr

[torn jolt]

It will do that automatically then

[wide star]

You can still search it even if you close iirc

[torn jolt]

^^

[wide star]

It's just that somebody else will most likely use the channel between now and then is all

[torn jolt]

^ that is, irrespective of whether u close it

[wide star]

If you close it it's not just sitting up here in the way when helpers are looking for ppl to help

And other people can use the channel

[torn jolt]

Oh also @wide star

How's my new drip

Lol

(I mean my pfp)

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[bronze hazel]

Could I please have help with this quesition

so

how to I convert the first part of the equation

into a a whole number

[modest oar]

So the y value must be the same and true for both equations?

[bronze hazel]

yes

[modest oar]

One second let me see how can I explain it

What the

I’m sorry I tried solving it on my own

but like lowkey I’m dumb asf

I’m sorry ;-;

<@&286206848099549185> :) ^

[bronze hazel]

same lol

[modest oar]

I’m saying compared to everyone else

You not dumb :)

i hope the people here can help you

And enlighten you

[vital shadow]

add the two fractions?

3x+5y = 3*15

6x+3y = 55

[bronze hazel]

where did you get 55?

oh

right

[vital shadow]

it's given in the ques?

[modest oar]

wait but how would they find the exact value of y without putting x in the equation

[vital shadow]

just solve the equations via elimination or substitution

6x+10y = 90

eliminate 6x and u'll be left with 7y = 35

y = 5

[modest oar]

o-

[bronze hazel]

wouldnt i have to make one negative

[vital shadow]

wdym

[bronze hazel]

for elimination

6+6=12

[vital shadow]

u minus it?

[modest oar]

I recommend multiplying -2 with (3x+5y=45)

To get rid of x

because you primarily want y

[vital shadow]

that's valid, but u can just minus the equations later

6x+10y-6x-3y = 90-55

[modest oar]

o wow you really simplified it

👍

Nice one cHad

[bronze hazel]

thought I would multiple 3x+5y=45 by -2

[bronze hazel]

7y=-55?

[vital shadow]

how did u get -55

[bronze hazel]

oops

-90+35

[vital shadow]

55*

[bronze hazel]

-90+35 us -55

[vital shadow]

it's -90+55

not 35

[bronze hazel]

oh yeah

so 7y= -35

y=

-5

[vital shadow]

-7y = -35

y = 5

[bronze hazel]

-10+3=-7

oh

yeah

right

thanks sorry for wasting your time

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[scarlet socket]

Im confused on how to solve for the angle

[sharp vine]

how many radians is a full rotation?

[scarlet socket]

2pi

[sharp vine]

okay, so here you have full rotation and a little more

can you tell me measure of that angle? maybe firstly in degrees if it's simpler

[scarlet socket]

135

and rad is 3pi/4

[hot knoll]

If you had a full rotation already, you had some rotation before reaching the "135" position

You had 360 degrees and then an additional 135 degrees

[scarlet socket]

so 495

[hot knoll]

Yea

And that 495 degree angle ends up being in the same place as 135

You can describe the same angle position infinitely many ways

You just need to keep adding full rotations

[scarlet socket]

so the answeer would be 495 or do i have to convert it to rad

[hot knoll]

You have to check for what they ask

In this case it's radian

[scarlet socket]

so it would be 11pi/4

[hot knoll]

Yes that's right

2pi + 3pi/4

[scarlet socket]

Thank you

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[hollow edge]

Have I done this correct?

I feel like

idk what I am doing lol

When I did it I didn't get anything listed

[sharp vine]

yes

[wild sparrow]

what did you do?

[hollow edge]

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[old arch]

Hi I want to ask you for help
This is the statement of the exercise: "Find three different quadratic equations whose solutions are -2 and 3."

[old arch]

I alredy find out one formula where a = 1, b = -1, c = -6

[deft tulip]

hi, yours would be correct yes, and it would factor out to y = (x - 3)(x + 2)

in general you could have any equation of the form y = A(x - 3)(x + 2) satisfy those solutions, where A is a non-zero number, the one you are giving is the one where A = 1 and probably the one that mostly jumps out. You can just pick any other nonzero value of A to get another equation. If you graph the equations to help you visualize what changing A does, it might be a good idea: flipping the sign of A flips the graph upside down, but it still goes through those root points, scaling the value stretches the graph vertically while still having it go through those roots

[old arch]

So that is all, just changing the value of a?

[deft tulip]

yes, basically multiplying the whole thing by a number

[old arch]

Ok thanks

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[bronze trench]

How do I write the expansion of 1/3^4?

[bronze trench]

I assume it’s same as 3^-4 right?

Wait

[atomic blade]

Yeah

[bronze trench]

is 1/3 * 1/3 * 1/3?

[atomic blade]

No

Because that's 1/3^3

[bronze trench]

Ik

I got lazy

So I type 3 only

How to turn (5/17)^8 into square root equivalent

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[strong crow]

Is there any way to recognize if this is a sine or cosine function? I can't tell if they're providing me with the fact that it's a sine function or not. https://i.imgur.com/5IsRWfi.png

[silk flame]

It tells you it’s a sin function

[strong crow]

yeah I figured because there's no other way to tell right?

[silk flame]

You need to work out it’s transformations

[strong crow]

because cosine and sine look way too similar.

I couldn't tell if it was trying to get me to understand by the graph or not

you know what I mean?

they say "form of asin"

same 'form' as acos

[onyx valley]

How many people here do math for fun

[strong crow]

well, it's all I do, when I don't work, so I guess?

I made a fun equation the other day

[silk flame]

It tells you f(x) = a sin (bx + c)+d

[strong crow]

489sin(pix/106)+69 I think that's the one I made

can you solve it? 😄

[hybrid geyser]

😭🤩🥳

[strong crow]

yeah my question is can I tell the function by the graph? no way right?

cosine and sine look the same

[deft crane]

😭🤩🥳

[hybrid geyser]

what does it =?

[deft crane]

I do!

[onyx valley]

Not gonna try and what grade are you in?

Or what do you do for a living

[strong crow]

I'm in community college

[hybrid geyser]

depends

[strong crow]

I'm a security guard

[hybrid geyser]

what part of the graph are they showing

[strong crow]

I do homework at my post

[onyx valley]

What degree are you trying to get

[hybrid geyser]

okg

omg

[strong crow]

software engineering + probably a minor in psychology

[onyx valley]

Okay

[strong crow]

https://i.imgur.com/5IsRWfi.png

so

I know how the graphs of sine/cosine are

sin(0) = 0

sin(pi/2) = 1

sine has a positive increase at the midline intersection of pi/2

or am I wrong fuck

my bad

midline intersection at (0,0)

has a maximum point following the midline intersection at pi/2

(pi/2,1)

cosine has a maximum point on the midline intersection following a minimum point

but the thing is, there can be so many transformations to the functions.

they can be reflected vertically...

they can be shifted there an there

I'm struggling to understand this 😦

if it was a sine function, the midline intersection would be shifted down 4 units and then pushed pi units to the right.

it has a period of 3pi apparently also

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[plain crypt]

I’ve found all the zeroes and the degree but I have no idea how to proceed with this problem

[light sonnet]

Recall that the function can be written as k(x - a)(x - b)(x - c) ... = 0

Where a, b, c... are the roots and k is the scaling factor

[plain crypt]

Oh I see now

I completely forgot about the k

Let me check real quick

Would the answer be -9(x+2)(x+1)(x-1)(x-3)?

[light sonnet]

Plot that and see if it matches the problem

[plain crypt]

no im missing an exponent

[buoyant tinsel]

double check your zero at x=3

[topaz valley]

bouncy bounce

[buoyant tinsel]

iirc the fun terms are pass, bounce, and slide

[plain crypt]

im sorry i still dont understand what im missing

[light sonnet]

FYI, a good tip to know, if it bounces on the axis, it has an even multiplicity, aka even amount of the same root. If it crosses, it has an odd multiplicity

[plain crypt]

ok

hold on

ok so i know that all the x intercepts have an odd multiplicity except at (x-3) right?

[light sonnet]

Yes

[plain crypt]

ok i think i got it finally

am i allowed to check my answer with you guys?

anyways i really appreciate the help have a good day everyone

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[wary nest]

I have five coins, one is double-sided. I pick one throw it five times and get heads each time. What is the probability that it’s the double headed coin?

[wary nest]

i feel like theres no way to solve this since they didnt give us any information, about the other coins

[eager violet]

I imagine it's implied the other coins are normal

[wary nest]

but we dont know that though

we cant just assume

[eager violet]

Well, either you put "can't solve" and risk getting a 0 for being pedantic, or you assume the other coins are normal and risk getting a 0 because your teacher is pedantic

[wary nest]

bro this isnt even a hw q chill out

im just trying to understand the problem

ik how to solve it

you use conditional probability

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[silk burrow]

I don't know what to do where to start

[torn jolt]

apply the definition of T

[silk burrow]

is it 1 0 0 -1 matrix?

T is?

[torn jolt]

T(x)=Ax

so A determines T

[silk burrow]

like this?

[torn jolt]

no

[torn jolt]

write out T(u) using the definition of T here

[silk burrow]

okay got it

in other words I can just multiply Ax and u

[torn jolt]

yes, T(u)=Au, so multiply A with u

[silk burrow]

Perfect!

every pivot on A was 0.5

can I change them into 1?

I believe I can but why would it imply that 0.5

[torn jolt]

what do you mean by "change"

if you are asking if A is the identity matrix, then no

[silk burrow]

okay got it

[silk burrow]

on 8 I have to find Ax=b what is x did I get it correctly?

[torn jolt]

yes, thats correct

[silk burrow]

Perfect! Thank you so much

[torn jolt]

yw :)

[silk burrow]

Have a wonderful day!

should I close it

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[swift bone]

am confused on how to find acc radial

i tried finding the radius

not quite sure if i found it right

?

but attempted do divide .8/2 = .4

then plugged into my A radial eq

which is 4pi^2(R)/T^2

where t=1.55s

and R=.4

then got 6.57m/s^2

which is obviously wrong

i also tried finding the radious by .8cos41.8

which got me .5963

i thought that could also be the radious, but apperently not, im confused

nvm i figured it out lmaoo

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[compact wasp]

How do I prove that

[compact wasp]

$||\dot||_w$

[glossy valve]

Maxerature

[compact wasp]

is a vector norm?

ummm hmm

[light sonnet]

What even is this?

[compact wasp]

$\parallel \dot \parallel$

[glossy valve]

Maxerature

[compact wasp]

wtf latex

$\parallel . \parallel_w$

[glossy valve]

Maxerature

[light sonnet]

Could you just write in in paint and screenshot and send that?

[compact wasp]

the weighted p-norm of . (. being any vector, I guess)

The most recent one is actually not wrong

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@compact wasp Has your question been resolved?

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@compact wasp Has your question been resolved?

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@compact wasp Has your question been resolved?

[normal tree]

does checking that it satisfies the properties not suffice?

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[copper yew]

Hi, can someone help with this problem? It was solved on the whiteboard but I couldn't quite follow well. I think the answer was 17/6 or 17/8, not sure

[copper yew]

Give me a second

[fast peak]

split the fraction into s^4/s^2 and 1/s^2

then integrate both individually with the (reverse) power rule

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[copper yew]

Wait still trying to do it. Thanks for being here

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[copper yew]

I feel like I did it right this time, not sure though.

[fast peak]

yes

[copper yew]

Yay thanks!

In studying Calculus like this, what are some tips you'd like to give?

e.g. focusing on theorems, making lots of notes

[fast peak]

practice

an integral like this for example shouldn't take more than just a couple of minutes

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[half sparrow]

Quick question, does the distributive property only applies If the variable is monomial and being multiplied to Binomial, Trinomial or Polynomial?

[hot herald]

no

[half sparrow]

Please explain.

[hot herald]

its pretty much applicable any time something is being multiplied to a sum/difference of many things

[honest totem]

[half sparrow]

what

[hot herald]

$\br{x^{\frac12} +y^3 - z^{\frac13}}\br{x^\frac{7}{3} + 5 - p^{2} + q^{\frac52}}$

[glossy valve]

ℝamonov

[half sparrow]

WHAT

[hot herald]

distributive property would be applicable when expanding that

[half sparrow]

Yes, thank you for explanation. I need to sleep to review this message again before veins in my head would start popping off.

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[copper yew]

Thanks so much

[atomic blade]

Bruh

[copper yew]

.close

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[copper yew]

Sorry, just had to say thanks to previous response

[atomic blade]

This dumbass school wifi aint fucking sending shit

[fast peak]

you're welcome

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[torn jolt]

,rotate

[torn jolt]

how could we simplify

[onyx glen]

could probably do something like factor out b/(3a) but not much beyond that...

[torn jolt]

okay!! thanks

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[broken rain]

Let the matrix A, a_1 - a_2 = 3 and a_2 + a_3 / 2 = 2. Find the determinant.
I try this but I doesn't match with the answers options :´(

[broken rain]

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[glossy thistle]

Don’t get why there’s a minus?

[fleet cedar]

isn't that just the laplace expansion?

[topaz valley]

laplace has an expansion named after him?

LOL

[fleet cedar]

,w laplace expansion

[glossy valve]

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[wild sparrow]

shee

[topaz valley]

did not know it was called that

[fleet cedar]

sheeesh

do you know it as cofactor expansion?

[topaz valley]

yeah

[fleet cedar]

icic

[topaz valley]

or just

"that way to calculate determinants" lol

[fleet cedar]

LOL apt name

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[glossy thistle]

thanks

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[willow girder]

hey

[willow girder]

Identify like terms in the list.

2a to the 4th is a dead giveaway

hello?

[clever idol]

I think those are the only like terms here?

[willow girder]

thats what i thought but its wrong

[clever idol]

Oh wait I am dumb

All terms with the same variable

Are like

So 4/11 r 9r are like

[willow girder]

ohhh

[clever idol]

7.2q, 5q and q/6 are like

They just have to be the same variable

[willow girder]

Ok

[clever idol]

I am sorry I answered incorrectly initially

[willow girder]

but if one problem has the same variable but is to the power of something other than the original number's power than it would be wrong

right?

[clever idol]

Yep

For like terms the exponent has to be the same

[willow girder]

alright thank you

[clever idol]

so x and x^2 are not like

[willow girder]

yeah

[clever idol]

But 8x and 34x are

[willow girder]

mhm because same variable

[clever idol]

Yep

[willow girder]

but different numbers its fine

so like

8x to the 4th wouldnt be alike with 13x to the 2nd

[clever idol]

Yeah it won't be (not like terms)

[onyx glen]

heres a little tip

[willow girder]

?

[onyx glen]

you can use ^ for exponents in plaintext

[clever idol]

Yep

[onyx glen]

like 8x^4 instead of "8x to the 4th"

much more convenient

[willow girder]

oh

8x^4

thanks

[clever idol]

6x² i fortunately have exponents built into my keyboard

[willow girder]

wow lucky

numpad should be exponents

[clever idol]

Yep

[willow girder]

i also need help with Simplify 5x 3 + 5 (6x 3 − 5b 4 ).

wait

it didnt copy as exponent

lemme rewrite that

[clever idol]

Is it -5x4- oh

Oh yeah

[willow girder]

Simplify 5x ^3 + 5 (6x ^3 − 5b ^4).

[clever idol]

Yeah so open ur brackets first and see the 5 outside it?

[willow girder]

yeah

[clever idol]

Multiply it to e a c h individual term

To open the bracket

[willow girder]

Ok

so 5x^3 is 125x multiplied

[clever idol]

So it's 5 * 6x³ - 5*5b⁴

[willow girder]

whattt

i do not understand

[clever idol]

Oh here

See the exponent is over x

[willow girder]

yes

[clever idol]

It's not (5x)³

But 5x³

So the 5 stays alone

[willow girder]

Ohh

[clever idol]

We do not touch it unless something else is multiplied

[willow girder]

so 15x

not 125x

[clever idol]

It's 5*6

That is 30

[willow girder]

O

[clever idol]

Here a little help

4 into 6x² is 24x²
5 into 7x⁴ is 35x⁴

[willow girder]

5x^3 so x . x .x

[clever idol]

You multiply the coefficients (the number stuck to the side of the variable)

[clever idol]

Yeah but you don't need to expand it like that

Maths usually likes its variables smol

[willow girder]

yes

[clever idol]

I meant * not italics in this msg

[willow girder]

Ok so according to that wouldnt it be 35x intead of 30

[clever idol]

So in 5(6x³ - 5b⁴) you multiply 5 to the numbers to the side of each term that is,

5 * 6x³ - 5 * 5b⁴

[clever idol]

Well in ur question it's gonna be 30x³ - 25b⁴ (don't forget the exponent on variables)

[willow girder]

Ok

[clever idol]

Now simplifying it all

[willow girder]

35 x^3 - 25 b^4

[clever idol]

5*6 is 30

But if it was 5(7x³) you would be correct

[willow girder]

Oh

hmm

[clever idol]

Yeah

[willow girder]

alright thank you

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[noble grail]

how would i go abt this problem?

<@&286206848099549185>

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[noble grail]

im not sure how avg and instantaneous relate to each other

[deft zodiac]

eh

they don really

[noble grail]

word

[deft zodiac]

but they want order

[noble grail]

yes

how do i

find the order

<@&286206848099549185>

how do i calculate the two types of values?

[topaz valley]

woah you're green

[steel cove]

green is pain

[deft zodiac]

yea goodbye bloo

[steel cove]

do not become green

green is ping simulator

[deft zodiac]

i just wanted

[topaz valley]

lol

[deft zodiac]

close perms

[topaz valley]

i don't want green

[deft zodiac]

omg i see it now

[noble grail]

😦

[noble grail]

even though they are not related how do i find the order?

[topaz valley]

they are each a numerical value

calculate/approximate their numerical value

then you can order

[noble grail]

how do i calculate their value?

[topaz valley]

im on a phone D:

gradient

and tangent

[noble grail]

huh?

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[agile thunder]

Can anyone show me how to do all of these problems?

#4 I just don't get what it means
#5 I followed some tutorial but the tutorial doesn't explain how to do it if two points don't have the same x value and I'm honestly confused af still
#6 no idea how to do this

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@agile thunder Has your question been resolved?

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@agile thunder Has your question been resolved?

[deft flower]

Let's start with #4. Looking at what we're given, could you answer the following:

• What can you tell me about the equation supplied by the problem?
• Do you recall what it means for two figures to be congruent?
• Do you recall what it means for a figure to be translated?

[agile thunder]

1. It's the same x^2 & y^2 but different integer values that shift the graph
2. Yeah, same length and angles
3. Yes I think when you shift a figure a certain amount of points

Sorry for late reply I was walking

[deft flower]

it's all good

With regards to 1: What does that integer tell you about the circle?

[agile thunder]

I believe the size? I can go on desmos and double check

And I think to move it left/right, you do something like (x-h)

And up/down is (y-k)

I think

OH RIGHT

It tells us the squared radius

Ok its coming back

(x - h)^2 + (y - k)^2 = r^2

Damm if that's correct I'm proud of myself

Yay

Ok ok I think I got it

(x + 3)^2 + (y + 5)^2 = 36

@deft flower

[deft flower]

yeah that looks good

Alright, so let's swing down to #6, since it'll end up being easier

• What do we know about the situation?
• What does it mean for something to be tangent to a line?
• What can you tell me about the circle, given this information? (I strongly recommend drawing a picture, it'll become really apparent)

[deft flower]

ope sorry forgot to let you know, @agile thunder that looks correct for #4 (good job!), now i've got a couple questions for #6 if you're still needing/wanting help with that one

[agile thunder]

@deft flower omg I'm so sorry

😭

I was helping my mom cook

Mb

[deft flower]

no you're completely fine, no worries

if you need/want a break to work on other things, that's completely fine.

[agile thunder]

Well the circle is an unknown radius but we're given the center (5, 1). The line 7x + 2y = 17 (Ax + By = C) is tangent or touches the circle at one point. What we know about the circle is that we have to find the radius from it to the line so we have to use the distance formula so it would be good to graph this

[agile thunder]

Kk

Lemme get my notes out

I think I get t

I'll try this on my own

[deft flower]

Alright, sounds good!

Huh, I actually didn't know that formula

[agile thunder]

Uh dumb question lol but do I need to convert 7x + 2y = 17 to slope intercept form to graph it lol

[agile thunder]

Never heard of it either

[deft flower]

It's usually the easiest way, though not the only

[agile thunder]

Alr

[deft flower]

If you'd like, I can walk you through an alternative route through the problem in a little bit, though it'll take me a few minutes to walk home

[agile thunder]

Yeah sure

No problem

[deft flower]

Unfortunately I'm going to need to do some cooking shortly, but I can still provide a more step-by-step solution to #6 if you'd like

If you're fine with just the formula, that's alright too

[agile thunder]

No problem

Yeah sure if you can

[deft flower]

Alright

So we know that this circle has a center at (5, 1), and that it's tangent to a line. We're only missing one piece of information to make our circle; what do we need to solve for?

Ah nvm you already mentioned the unknown radius

Alright; we're given that it's tangent to a line. We know the line must intersect the circle to be tangent. It can only do that if at some point the radius lies on the line.

If you're up for a drawing exercise, plot a center point and draw a line. Draw a radial line from the center to that tangent line, and draw the circle about it. Repeat this a few times until you see a trend

• What can we determine about what the radius must be (in relation to our point and tangent line?)
• What can we determine about the line(s) that connects the circle to the tangent line?

[deft flower]

@agile thunder sorry here's the walkthrough above

at least the first part of it

you may be able to see where this is going as well so if you'd like to finish on your own just lmk

[agile thunder]

Yeah btw wdym repeat?

I just drew a line from the radius to the line perpendicular

[deft flower]

Ah well that's the only one that matters

[agile thunder]

Oh

[agile thunder]

Alr

Would we make a triangle

[deft flower]

Technically there's infinite radiuses but only one is tangent

[agile thunder]

Use Pythagoraen Theorem

[deft flower]

You actually don't need to in this situation

If we can find two points, we can find a distance

[deft flower]

you've already mentioned a very important fact of this line here

If we have a line perpendicular to a line with the equation 7x + 2y = 17, can we determine anything about our line?

[agile thunder]

Oh yeah the real distance formula

But idk how to do that

[agile thunder]

Yeah

Uh

We can find the slope of that and take the because reciprocal?

[deft flower]

Yeah, exactly

Now we have two lines, but we need a point still to use the distance formula

Oh sorry I skipped a step

We have a line, do we have any way to tell where it might be connected (perhaps a y intercept or a point we know it passes through?)

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@agile thunder Has your question been resolved?

[agile thunder]

Omg I'm so sorry

I keep not responding

[agile thunder]

Well we know the perpendicular line passes through the center

Alr so

We can find out what the line is, easy

Then use point slope form

Then plug it in

And for problem #5 I think i use a system of equations

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@agile thunder Has your question been resolved?

[deft flower]

Yeah exactly

[deft flower]

that's probably the best way to move forward with that one

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[agile sand]

im confused with this, if its skewed right wouldnt that be variable 4 and skewed left is variable 3? and the 2 on the top are symmetric?

<@&286206848099549185>

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[hot knoll]

Skew right and left are the opposite of the way my mind imagines them

Skew right means the "tail" or trailing portion of the data is toward the right.

Skew left means the "tail" is to the left

[hot knoll]

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[tawny reef]

hey

[tawny reef]

i need all trigonometry formulas/identities

A and B ones

ok just give me all the identities that are helpful in integration

[sweet ingot]

Did you try google?

[tawny reef]

yes but it just gives me the integral formulas not the A and B identities

i need all formula written in a single slide so i can paste it somewhere in my house

to keep it in my head

[onyx glen]

what are "A and B identities"

[woven snow]

You mean something like

,tex \trigsum

[woven snow]

?

[tawny reef]

yus and i also need sin^2A + cos^2A ones too

[woven snow]

,tex \trigid

[onyx glen]

deep why does your tex have all of these eye-searing colors for its bg

[woven snow]

While we’re at it

[tawny reef]

and 1+tanA and 1+cotA......and all

[woven snow]

,tex \trigsumprod

[woven snow]

It chooses randomly

[onyx glen]

ok but why

[woven snow]

Google them

[woven snow]

You know for fun

[tawny reef]

thanks for helping

[woven snow]

Okay I barely did anything but i’ll take that

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[mossy ember]

small doubt

[mossy ember]

when im doing induction

and im in the proof

when i add k+1 to the thing should i add how its in the series or just k+!

k+1

[onyx glen]

"the thing"?

[mossy ember]

like gthe series

1cube + 2cube + 3 cube +.......+k cube + k+1 cube

when im doing the proof should i use k+1 cube or just k+2

k+1

[devout inlet]

Maybe show us what you're trying to prove.

[onyx glen]

no don't you get it, we can't expect OP to communicate their issue clearly! didn't they teach you that in mandatory telepathy training?

[devout inlet]

I couldn't imagine asking for help personally and not being as clear as humanly possible.

[onyx glen]

(i was being tongue-in-cheek here)

[devout inlet]

I can only assume you want to complete this problem.

Step 1. Prove it true for n = 1.
Step 2. Assume it true for n = k.
Step 3. Prove it true for n = k + 1.

For step 3, you need to prove that

$\sum_{k = 1}^{n + 1} k^3 = \frac{(n + 1)^2(n +2)^2}{4}$.

[glossy valve]

stabulo

[devout inlet]

Don't reuse k like I did though.

[mossy ember]

ok

ill send pic

.close

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[scenic reef]

X = number of coffees sold

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[scenic reef]

What do you mean

What’s that got to do with my question

I didn’t even finish showing my working out

[torn jolt]

i think it's a troll

[scenic reef]

I was going to post an error

[severe lantern]

Pretty sure sayn is trolling

Not very helpful

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[noble pelican]

[Linear Algebra]
Good afternoon, our professor told us that for a matrix A to be invertible the $det(A) \neq 0$.
Can somebody explain me why is that so?

[glossy valve]

HolyCross

[spice orchid]

One way to see it is that the formula for the inverse of a matrix involves dividing by the determinant, if the determinant is zero we are in trouble

[noble pelican]

ah alright, thanks!

I guess that if the $det(A) \neq 0$ though, that its still not enough to verify that the matrix is invertible right?

[glossy valve]

HolyCross

[spice orchid]

It is!

Because then the algorithm works, giving you an explicit inverse

[noble pelican]

Oh neat-o.
So if its associated to a linear map f (in regards to a couple of basis), and its $det(A) \neq 0$ we can also say that f is invertible as well (which would imply that f is bijective)?

[glossy valve]

HolyCross

[spice orchid]

Indeed, a linear map is invertible iff a matrix that represents it is invertible

This does not depend on your choice of basis

[noble pelican]

you are giving me gold right now

[noble pelican]

Is there anyway to understand if a linear map is just injective/surjective?

[spice orchid]

It would depend on the map, there are ways to use the matrix to tell

If your map is from a vector space to itself then infectivity is equivalent to surjectivity

Two for the price of one

[fast peak]

if the map is V->V and V is finite dimensional, then injective iff surjective iff bijective

V being finite dimensional is important here, not true for infinite dimensional spaces

this follows from rank-nullity

[noble pelican]

ok thanks

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.reopen

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✅

[noble pelican]

Does a matrix being invertible has any relation with being diagonizable?

[onyx glen]

no

[noble pelican]

okay makes sense, since the $det(A)$ is completely different from the $det(A - I_3)$ and one being 0 shouldn't really affect the result of the other one.

[glossy valve]

HolyCross

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[lethal pasture]

q3

[deft zodiac]

whatve u tried

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[vocal current]

@onyx glen its me again. the one you helped with scoring the distance

[onyx glen]

yes?

[vocal current]

the (1-dist/5.05)*100 worked but how to I rate 0 distance as 0?

[onyx glen]

well you could use IF()

since you're doing it in excel

make a special case for 0 distance

[vocal current]

are you familiar with excel.. 😅 i'm getting an error

=if(a2=0,”0”,(1-a2/5.05)*100)

a2 is the cell

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[spice kelp]

.

@sharp vine sorry tag, but can you help me on task c?

[spice kelp]

The graph of a colinear function f passes through the points P(0;2) and Q(2;6)

a) Determine the pre-shift of f(x)
b) Calculate the function values f(-1), f(1) and f (10)
c) Calculate x when f(x) = 3 and f(x) = -13

A point A = ( 4; f(x))
d) Calculate the coordinates of A

Task a and b = done

I'm missing c and d

How do we do c and d?

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[zinc igloo]

can anyone help me solve this? i dont understand it

[median dawn]

What dont you understand about it?

[zinc igloo]

the 4 x 10-3 i dont know how to multiply it and then get the final answer

[median dawn]

Do you know what 10^-3 is?

[zinc igloo]

yes its 0,001

[median dawn]

Ok good

You can cancel the 4 with the 8 in the numerator

To get a 2

[zinc igloo]

oh ok tysm

but how do i multiply the 0,001?

[median dawn]

What would you have to multiply the bottom by to get 1

After getting rid of the 4

[zinc igloo]

i dont really know

[median dawn]

Oh come on

Any ideas

[zinc igloo]

like the 10-3 or the 0,001?

[median dawn]

Those are the same right?

[zinc igloo]

yea

[median dawn]

So whats their inverse

[zinc igloo]

i dont really know what that means

[median dawn]

Ok

[zinc igloo]

im not from america

so i dont understand english that well

[median dawn]

1/a = b

a and b are inverses

Atleast multiplicatively

Anyways do you know how exponents work

[zinc igloo]

yes

10^4 is 10x10x10x10 i think

[median dawn]

It is

So 10^-3 is 1/(101010)

[spice kelp]

I need help

[median dawn]

Barda diff channel

[zinc igloo]

ok

[median dawn]

Oops

[zinc igloo]

so no multiply?

[median dawn]

Yes multiply

Sry

**

That does a weird thing to whatever it encloses

[zinc igloo]

oh ok

[median dawn]

You need to multiply that by something to get 1

[zinc igloo]

by 10?

[median dawn]

Yeah but 3 times

[zinc igloo]

oh so instead of -3 its just a 3?

[median dawn]

Yes a^-ba^b = 1

Oh my

It did it again

[zinc igloo]

a

so how do i get to the final answer?

[median dawn]

Multiply the top and the bottom by 10^3

[zinc igloo]

i know that the top numbers are 8000

do i add them together after?

[median dawn]

Add what?

All you have is multiplication

[zinc igloo]

oh so when i multiply them both i get the answer?

[median dawn]

Wdym them both?

[zinc igloo]

the top and bottom numbers

[median dawn]

Oh yeah so first you can the 4 with the 8 then you multiply the top and bottom by 10^3

And then you will have some number over 1

And your answer will just be that number

[zinc igloo]

oh ok tysm for the help

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[azure blaze]

Howd they get 18k

And what do they mean with e0.384

[onyx glen]

e^0.384 means euler's constant e raised to the power of 0.384

your calculator probably has a button that says [e^x]

[torn jolt]

they multiply it e^0.384 with 12 600

[azure blaze]

Thank you guys

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[pulsar chasm]

Determine the number (a) so that the equality holds.

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[sly ore]

ye bro

@pulsar chasm u stuck?

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[patent spindle]

how come my Casio cg 50 calculator does not show any of the sin cos tan. values as a surd?

[patent spindle]

have a question to put in exact form, but comes out as a number, do I need to manually type in the form?

even so, it keeps giving me a number, not the surd

[misty topaz]

Are you in decimal or radians?

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[deft zodiac]

teapot

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[patent spindle]

.reopen

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✅

[patent spindle]

Is that an option?

I’ll look into it now

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[torn jolt]

help im so confused

[torn jolt]

how do i graph this

[normal tree]

just two dots

[torn jolt]

well obviously but like

on what number

nevermind i got it

ty lol

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[elfin plume]

https://cdn.discordapp.com/attachments/903481105888452609/1019714169555136622/unknown.png

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[woeful haven]

simple question, why isnt (x+2)² = (x²+4)? wouldnt the ² distribuate?

[runic bloom]

it's because $(a+b)^2 = a^2 + 2ab + b^2$

[glossy valve]

Kihei

[woeful haven]

what

[runic bloom]

the ^2 does distribute just not the way you may think it does

[woeful haven]

where did you get the 2ab?

[runic bloom]

well this is just the way you multiply this

basically

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

[glossy valve]

Kihei

[woeful haven]

yes

[runic bloom]

and we first multiply the first terms

a * a

then the outer terms

a * b

then the inner terms

b * a

finally the last terms

b * b

so basically that's how you'd multiply 2 binomials

[woeful haven]

so back to the x and 2

$(x + 2)(x + 2) = x(x+2) + 2(x+2)$