help-23

[south harbor]

Brehhh

[thin bridge]

didn't catch the same mistake you made earlier
k - 27 isn't -27k
so you'll need to fix that first

[south harbor]

K - 27

Right?

[thin bridge]

yes,
k-27 is k-27

if doing this division with k present is confusing you,
i'd recommend using factor theorem first to determine k
before doing division

[south harbor]

Dont worry

I got the hang of it

Can you present me the factor theorem?

[thin bridge]

is this the first time you're hearing about factor theorem?

[south harbor]

Not really

It just got out of my mind

Or slipped

Cuz i was thinking or delusioning at class that time

[thin bridge]

it more or less applies the same principle behind synthetic division,
but less labour intensive at the cost of less info at the end.

[thin bridge]

if (x-a) is a factor of polynomial P(x),
then P(a) = 0

[south harbor]

Is it like the same as reminder theorem?

[thin bridge]

yeh

factor theorem is a special case of remainder theorem with remainder = 0

[south harbor]

Oh, so i can only do reminder theorem if the question got a reminder?

[thin bridge]

yeh

but its very similar

[south harbor]

Okie so what next?

[thin bridge]

instead of setting = 0,
you set = the remainder

[south harbor]

Yep ik that one

[thin bridge]

since you are told (x+3) is a factor,
the remainder is 0

(by definition of factor)

the remainder of the division here is
9k - 198
set that = 0
and solving will tell you the value of k

[south harbor]

Ight ight

Hol up

Just the 9k - 198 = 0?

[thin bridge]

yes

[south harbor]

Ight ight

-22

K = -22

[thin bridge]

no

[south harbor]

Wha?

[thin bridge]

how are you getting -22

[south harbor]

Oh ueah

I forgot to change the sign

When i swapped -198

Lol

K=22

[thin bridge]

yes

now that you know k, you can use that in the quotient

[south harbor]

Oh ok

In this fellas who has a k?

So like 2x⁴ -2x³ +9x² then the ones with k's?

[thin bridge]

yeh,

what is k-27 when k=22?

[south harbor]

How do i write it?

[thin bridge]

what is k-27 when k=22?

[south harbor]

Like in parentheses?

(k-27)?

[thin bridge]

if expressing in terms of k yes

[south harbor]

And like i add x there so like
(K-27)x

[thin bridge]

but that step can be skipped over

[south harbor]

Ok ok

[thin bridge]

if you evaluate directly from the result of your division

[south harbor]

What about 9k-198?

Do i still put the 22?

Oh wait

That'll be zero

Lol

Do i still write +0?

[thin bridge]

no need

[south harbor]

Ight then i guess im done right?

[thin bridge]

well what's your end result

[south harbor]

@thin bridge can you check if this is right cuz if yes, then i can conquer the other 4

[thin bridge]

yep

[south harbor]

Uessssss

Tyyy

😘

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[south harbor]

Guys, the question here was
(X+3) is a factor of 2x⁵ +4x⁴ +3x³ +kx² -14x +3

[south harbor]

The final answer is in the bottom

Do i still have to write
(X+3)(2x⁴ -2x³ +9x² -5x +1)

[thin bridge]

depends on the exact wording of the question

[south harbor]

So how does it go?

[thin bridge]

the question is on your end

take a pic of exactly what's being asked

[south harbor]

No. 6

[thin bridge]

oh.

you weren't even asked to factorise

just find k

[south harbor]

D:

[thin bridge]

the overarching question was

in which case, factor theorem, direct evaluation would've been sufficient

,w 2x⁵ +4x⁴ +3x³ +kx² -14x +3 when x=-3

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[thin bridge]

gives the same result of the synth division you did

and that's equal to 0 when k=22

[south harbor]

Damnn

So like i just put -3 on each of the x?

[thin bridge]

yeh

[south harbor]

Welp

[thin bridge]

sub directly

[south harbor]

That's okayy

[thin bridge]

pretty much what they do in the examples

[south harbor]

Damn i really thought i can solve them after no. 6

No. 7 gave me 2 k

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[zinc wigeon]

hey guys i need help with math and im just wondering in the equation

1/2f(x+2)+1 would i do the stretch first or the left 2 or up 1

[uncut cloak]

Well you’d have to stretch before shifting up by 1

Otherwise U’d multiply the whole thing by 1/2

But you can stretch or do horizontal translate in whichever order

[zinc wigeon]

OKAY THANK U

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[kindred hedge]

Hi, may I know how to solve 3b?

[umbral horizon]

its a separable equation. move all the terms with x to one side and y to the other, then integrate

[umbral horizon]

ye like that

[kindred hedge]

,rotate

[kindred hedge]

Yea, but I don’t know how to get C

[umbral horizon]

like solve for the constant?

[kindred hedge]

Yea

[umbral horizon]

put pi/4 in for x and 1 in for y, then solve for it

[kindred hedge]

But isn’t tan pi/2 undefined?

[umbral horizon]

yes, but you have tan(pi/2 + C)

not tan(pi/2)

[kindred hedge]

Actually I can sub in inverse tan right.

?

[umbral horizon]

sure

[kindred hedge]

Something like this?

[umbral horizon]

you have $1 = \frac12 \tan \left( \frac{\pi}{2} + C \right) \to \tan^{-1} 2 = \frac{\pi}{2} + C \to C = \tan^{-1} 2 - \frac{\pi}{2}$

[flat frigate]

mono

[umbral horizon]

oh thats what you wrote

sorry i misread

[kindred hedge]

Yea, I wonder do we need to write in a nicer form?

Or this will do?

[umbral horizon]

its probably fine

[kindred hedge]

Okay, thanks

[umbral horizon]

i dont think arctan 2 has a particularly nice form here

np

[kindred hedge]

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[winged warren]

y=sqrt(x-a) find coordinates of the closest point to the origin, solve in terms of a, find assuming a is a negative value

[peak estuary]

what is the distance from the point (x,y) to (0,0) ?

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[grave tangle]

The only way I can see that the answer would be pi/10 is if the limits of integration for theta was from 0 to pi/2. However, this doesn't make sense to me because solving for theta with x = rcos(theta), I get x = sqrt(2-x^2)cos(theta).

Substituting in 1 for x makes it 1 = cos(theta), so theta = arccos(1), which is 2pi.

Symbolab agrees with the textbook but won't tell me how they got pi/2 as a limit of integration for theta.

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<@&286206848099549185>

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[trail otter]

let W be a vector space

[trail otter]

how do i find all diagonal matirices which belong to W?

[quartz prawn]

could somebody tutor me??

[trail otter]

.occupied @quartz prawn

[quartz prawn]

ok

[raven heart]

well take a matrix in W, what would it mean for it to be diagonal ?

[raven heart]

well the middle terms could be 0 technically

there's no constraint on them

but yeah the off-diagonal terms are 0

[trail otter]

oh

thats the catch

soo how should i find it then?

[raven heart]

the 0 matrix is diagonal

kinda boring but it is

[raven heart]

well you have a system of 2 equations haven't you?

[trail otter]

a+2b+c=0 and b=0?

[raven heart]

yup

[trail otter]

so then a=-c but that yields 0 on position 2.2

[raven heart]

so be it

[trail otter]

can i have a matrix with some non zero number at position 1x1 and 0 at 2x2 and still call it diagonal matrix?

[raven heart]

yeah

I literally told you 3mins ago

[trail otter]

xdd sorry my bad

ok thanks then that cleared it up

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[shrewd topaz]

how is the slope not 3x+3

touches y axis at 3 and x at -3 but because its positive the x intercept is as well

[pure oxide]

You have y=3x+3.
Okay, fill in y = 0, and solve for x.
This would then be a coordinate in your graph, is that the case?

[shrewd topaz]

wdym

-1 would be a coordinate?

[pure oxide]

(-1,0) would be a coordinate if your formula was correct yes

Is that the case?

[shrewd topaz]

no

[pure oxide]

The the formula of the asymtope is not y = 3x + 3.

[shrewd topaz]

but the oblique asymptote is just the slope

right

[potent bay]

slope is the ( difference in y / diff in x ) between two points

[shrewd topaz]

so then how do i find the slope from just looking at the graph

[pure oxide]

Take two points on the slope.
(X1,Y1) and (X2,Y2)
Slope = (Y2 - Y1)/(X2-X1)

[shrewd topaz]

alr i found it

i thought there was a shortcut

whats my domain for the graph

domain is everything except the x intercept right?

[pure oxide]

From what you are given in the picture or the domain of the function?

[shrewd topaz]

in the picture

[potent bay]

domain is the interval where the function can be evaluated

[pure oxide]

I think that would be from what to what the x-axis goes in the picture.

If they ask the domain of the graph in the picture.

[shrewd topaz]

Yes they're asking for the domain solely from the graph

[potent bay]

knowing this, what would you say the domain is?

[shrewd topaz]

well i dont know what the function is

i could plug x into the denom and figure it out

but i dont have that

[potent bay]

we neither

[shrewd topaz]

it doesnt look like something's excluded

[pure oxide]

You don't need the function to determine the domain in the picture.

[shrewd topaz]

well how do i find it

[pure oxide]

.

[shrewd topaz]

can you explain that

it goes between -1 and 1

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[potent bay]

"The domain of a function is the set of its possible inputs, i.e., the set of input values where for which the function is defined. In the function machine metaphor, the domain is the set of objects that the machine will accept as inputs."

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[main mural]

because division by zero is not defined

that's enough of an argument

oh absolutely

nah

[warm stirrup]

any number multiplied by 0 is 0

so 0/0 can be anything

[main mural]

assume division by zero is defined. then 1/0 = a, where a is some number
multiply both sides by zero
1 = a*0
1 = 0
contradiction

therefore, division by zero is not defined

show

but he is definitely making an error

um

he saying it in a very weird way and trying to sound smart

but is failing miserably

if we have f(x) = 0x, then its integral is F(x) = 0 + C = C where C is a constant

trying to define the inverse of f is wrong because it involves division by zero that i just shown to be invalid

the first sentence is from calculus, that i assume you haven't started

oh

well

okay he is just contradicting himself there

if 0=1/u, and 0/0 is u^-2 = 1/u^2
then 0^2 = 0/0

which would make u = 0

and i assume he goes off the premise that u≠0

if this is a game of flexing who knows more math, there's a proof from abstract algebra that a field doesn't have multiplicative inverses for its zeros

np

lol

ask if anything

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[solid mortar]

help

[solid mortar]

what do i do for the +- sign

im trying to implement this in code

all the variables are defined im just confused with the +-

https://mathworld.wolfram.com/Circle-LineIntersection.html

[tulip saddle]

it means there are two values for x and y respectively

assuming this is the solution for a coordinate pair, then you have two ordered doubles as a solution

[solid mortar]

ohhh

thank you i forgot

because there are multiple intersections

[tulip saddle]

yup

sometimes they'll be the same though

bc r^2d_r^2-D^2=0

which means the line is tangent to the circle & passing cleanly through a point

in a program that'll basically never happen though

ok I gtg gl

[solid mortar]

thank you very much astral

also the tangent collision is going to happen a lot of times because im using it for physics

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[west hedge]

What is it called?

[viral venture]

a-th root

[west hedge]

So the symbol is called root

[viral venture]

e.g. ⁶√x = sixth root of x

Yep

[west hedge]

I see

What it is called root

[viral venture]

√

[obtuse plover]

Or radical

[west hedge]

Is it related to a root of sth metaphorically

Like a tree has a root

[lean otter]

No

More like root of polynomials

[west hedge]

But they don’t necessarily mean that

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[unreal river]

I'd like to confirm I did this properly

[obtuse plover]

My oh my

[unreal river]

Hello again stephen

Lol

[obtuse plover]

Yo

[ember sphinx]

what is that

[unreal river]

SORRY

I just wanted to know if I did this right lol

I just plug in 2pi in all of this right

If its all correct

[obtuse plover]

Yo Mortta I got this

[unreal river]

Lol

[ember sphinx]

i aint denying that u dont

in what scenario would u need to solve that

[obtuse plover]

I’m peeing on the tree

[unreal river]

Uhh

I mean its asking for the slope of the tangent line

So I have to differentiate the equation

Then plug in 2pi

For x

[obtuse plover]

Yea lemme take a look

[unreal river]

Unless theres an easier way lol

[unreal river]

It was in reponse to this though

[west hedge]

Yee

[unreal river]

Nw

[obtuse plover]

Ye looks good

Hold on lemme check

,w differentiate (2+cos(x))^(4+sin(x))

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[unreal river]

Hm

[obtuse plover]

Ye it’s the same

Ye now just plug in 2pi

[unreal river]

Thats

A ridiculously long equation to solve

[obtuse plover]

Nah

It should be a clean answer

Cuz 2pi gives clean numbers

[unreal river]

Thankfully

But like

On an exam,

It would be ridiculous if it didnt

And I get no calculator :D

[obtuse plover]

Just know unit circle well

[unreal river]

Hm

I get 81*ln(3)

[obtuse plover]

Ye shd be right

[unreal river]

Cool

Well thank you

Wasnt so bad

But def long for no reason

[obtuse plover]

Your skills have increased

Not for no reason

[unreal river]

Hm

Thats true

I appreciate the encouragement

And help

[obtuse plover]

Np

[unreal river]

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[lean otter]

not sure I understand this, how do they determine when f(x) and g(x) is >=0

[lean otter]

without subbing the asymptotes/points in

is there a way to determine this graphically?

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[rose oracle]

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[rose oracle]

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[lost patrol]

Okay so do you know how to convert

[rose oracle]

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[lost patrol]

Well thats basically what you have to do. Convert the units

Okay so I'll go by the method your images show.

So...

1cm = ? km

[rose oracle]

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[rose oracle]

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[lost patrol]

Okay just try

[rose oracle]

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[rose oracle]

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[lost patrol]

1km = ? metres

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[timber basin]

Log 0 base e is (-infinity) or not defined?

[buoyant shadow]

either one

if you don't define it as −inf it's undefined

you don't usually define it as −inf because it's not a number

you like when you get numbers

[timber basin]

So both works?

[buoyant dragon]

not defined is better to say

[timber basin]

Okay thanks

.close

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[rain turtle]

hello plz anyone help me with this from the start what formula should i use and why from the start i am realy new to trigonometry

[stray socket]

Which part

Which question*

[rain turtle]

27 a

[stray socket]

Well did you graph the two functions

Or are you stuck with that from the get go

[rain turtle]

no i dont know how

i need help withg it too

plz

[stray socket]

Not even with cosine?

[lean otter]

the only "formula" you use is 1/cos(x)

for secant

[stray socket]

Start with cosine

Do you know how to graph cosine?

[rain turtle]

on the y axis right ?

[stray socket]

Well

It wants you to graph y = cos(x)

So it's a wave of amplitude 1 whose midline is 0

Starts at (0,1)

And you know it's period is 2π

[rain turtle]

okay so i start y = 1 and x = 0 right ?

[lean otter]

yes

are you aware of even/odd functions?

cos is symmetrical about the y axis, it'll be the same on the -x and +x axis

cos is also an oscillatory function its range is [max, min]

[safe radish]

@rain turtle Has your question been resolved?

[rain turtle]

.close

[safe radish]

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[obsidian acorn]

How do I do 5c and 5d

I have both answers for a and b a is 7 and b is 14

[safe bluff]

right

so

if we have the set {16,5,11,19,7,14,23,17,14,8,6,2,25}

what is the median

[obsidian acorn]

14,23

Wait no

23,17?

[safe bluff]

a set doesn't have 2 medians

if there are an even number of elements, you take the average of the two middle ones

[obsidian acorn]

23 is the medain?

[safe bluff]

but this should have an odd number of elements

i think that is too large

[obsidian acorn]

What should we do if it's too large

[safe bluff]

no i mean

i think that is larger than the correct answer

[torpid fable]

U need to sort before taking middle

[obsidian acorn]

Wait give me a min

Let me arrange

2,3,4,5,6,7,8,11,14,16,17,23,25

[hearty egret]

just use bubble sort

[obsidian acorn]

What's that

[safe bluff]

its only 13 elements

just

do it by hand

[hearty egret]

ah sorry

[torpid fable]

Just hope it was already sorted

[obsidian acorn]

11?

[torpid fable]

No

[safe bluff]

shouldn't be a 3

[torpid fable]

Whyd u give him so many data points lol

[safe bluff]

thats the number of data points in the problem

why are you asking me lmao

[torpid fable]

Arent there 6???

[safe bluff]

it said "pool this 6 element set and this 7 element set"

so you get a lot of elements

[torpid fable]

Nvm I did not see he said c and d

[obsidian acorn]

2,,4,5,6,7,8,11,14,16,17,19,23,25

Sorry I accidentally added 3 and forgot 19

[safe bluff]

yeah

so what's the median

[obsidian acorn]

11

[safe bluff]

seems right

gj

[obsidian acorn]

🥳🥳

Here's comes d

Do I combine all 3 answers a b and c and find the medain in all of them or

How do I explain my answer

[safe radish]

@obsidian acorn Has your question been resolved?

[safe radish]

@obsidian acorn Has your question been resolved?

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[pulsar pecan]

The original question: Let n be a natural number and A and B to n x n matrices. Assume both A and B are invertible matrices. In this case show that (A * B)^-1 = B^-1 * A^-1

above is my answer. I wanna know if its good or has some thinks that should definitely be improved.

[quasi bison]

is this in word

select this entire text and mark it as english, so that all these annoying red squiggles go away

[pulsar pecan]

sure

[quasi bison]

anyway this is pretty clunky to read

like your idea is good

[quasi bison]

also "associate ability" is not what people say generally. it's either "associative property" or "associative law"

but like otherwise...

[pulsar pecan]

its translated from danish.

x)

[quasi bison]

ok yeah

it is a translation artifact

still im trying to think of a better way to word this

cause what i would do is just

$(AB)(B^{-1}A^{-1}) = ABB^{-1}A^{-1} = AI_n A^{-1} = AA^{-1} = I_n$

[flat frigate]

Ann

[quasi bison]

like this is almost self-contained

you use the definition of inverse, the definition of identity matrix, and the associative law

[pulsar pecan]

yeah, but maybe too much at once?

im not sure

[quasi bison]

what do you mean "too much at once"

[pulsar pecan]

my main fears are that I wrote something thats just not correct or doesnt explain itself well enough

[quasi bison]

you wrote too much, imo.

[pulsar pecan]

yeah we also get penalized for that

but is it wrong? Thats a good start

[quasi bison]

again your logic's fine

[pulsar pecan]

alright

[quasi bison]

it is style that i am criticizing

[pulsar pecan]

and the conclusion isnt bad?

[quasi bison]

yeah it's fine

[pulsar pecan]

i will re-write probably then

I was also suggested by someone else to shorten it

[quasi bison]

i'd grade this somewhere between 60% and 75% of full credit.

[pulsar pecan]

Right, okay

tryna think how I would write it then

[pulsar pecan]

so you would write this entire thing out and then explain ?

or explain before

@quasi bison

[quasi bison]

probably easiest if i show you how i would write it

[pulsar pecan]

Yes, thanks

[quasi bison]

with... probably moderate fluff

which im fine doing, since this is just style guidance

ok so

Let $n \in \bN$ and let $A$ and $B$ be invertible $n \times n$ matrices. We wish to show that $(AB)^{-1} = B^{-1}A^{-1}$. To do this, we need to verify that $(AB)(B^{-1}A^{-1}) = (B^{-1}A^{-1})(AB) = I_n$, where $I_n$ denotes the $n \times n$ identity matrix.

We use the associative property of matrix multiplication, as well as the definition of the inverse (for $A$ and $B$) and the identity, in the following way:

$$(AB)(B^{-1}A^{-1}) = ABB^{-1}A^{-1} = AI_nA^{-1} = AA^{-1} = I_n.$$

Analogously,

$$(B^{-1}A^{-1})(AB) = B^{-1}A^{-1}AB = B^{-1}I_nB = B^{-1}B = I_n.$$

We have thus shown that multiplying $AB$ with $B^{-1}A^{-1}$ in either order yields the identity matrix, as desired.

[flat frigate]

Ann

[pulsar pecan]

Right, okay

What does "analogously" mean

[quasi bison]

in an analogous manner

i.e. "in a similar, or even almost the same way"

[pulsar pecan]

hmm

Not sure how to translate that back

I wrote it in a simlar way as you now with definition 7.18 and theorem 7.15 also referenced

"We use the associate property of matrix multiplication from theorem 7.15 as well as the definition of the inverse (for A and B) and the identity from definition 7.18 in the following way:"

@quasi bison would you say thats a good explanation ?

[quasi bison]

yeah sure if you want to reference specific thms in the book that's fine

[pulsar pecan]

yea

[quasi bison]

can't help you with the en>dk translation, i'm afraid.

[pulsar pecan]

Great

The question here being:

A matrix B is called symmetric if B = B^T
Let A be a n x n matrix. Show that the matrix A * A^T is symmetric

I think I shortened it enough? @quasi bison

[quasi bison]

yeah it's fine i guess

[pulsar pecan]

not too clunky and stuff?

@quasi bison

D.2:
Assume that B is symmetric and invertible. Is B^-1 also symmetric? Explain

This one is super long and clunky, but I am not too sure on how to reduce it

If you could write me your explanation that would be very nice

[quasi bison]

there is a nonzero amount of clunk but not enough for me to care about rewriting it...

[safe radish]

@pulsar pecan Has your question been resolved?

[pulsar pecan]

How would you grade it

[quasi bison]

¯_(ツ)_/¯

[pulsar pecan]

Could I bribe you with a few discord nitro to oversee the rest of my assignment? I’m the paranoid type when it comes to making sure that I didn’t make mistakes

[quasi bison]

uh

no thanks

plus that would probably run afoul of the rules lol

[pulsar pecan]

Oh okay

Cuz I don’t wanna waste your time if you are busy with something important

[safe radish]

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[humble shore]

Hello, I have a quick question: can I convert cartesian coordinates into cylindrical coordinates if the Z depends on a variable?

[humble shore]

thanks in advance

For the cartesian part:

x = a·cosθ
y = a·sinθ
z = b·t^2

(side note just in case: θ=ωt)

would that be possible to convert to cylindrical coordinate system or does z compound strictly need to be a constant?

ok nvm i just noticed that my question doesn't make any sense

.close

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[silent scaffold]

can any function f(x, y, z) be rewritten as f(x(t), y(t), z(t)) ?

[silent scaffold]

and if yes, how would you prove it

[tame raft]

heh?

badly worded question

you can do that only if you can write x, y, z as functions of t

in which case it's easy

[silent scaffold]

oh ok

ohhh i misunderstood this

[main mural]

can you parametrise any function? i doubt

[silent scaffold]

no

yea ty

!close

/close

.close

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[uncut frigate]

For any $n\geq 1$, let $Y_n$ be a geometric random variable on a probability space $(\Omega,\mathcal{F},P)$ with parameter $1/n^2$. Set $X_n=n^{-2}Y_n$. Show that for any $x>0$, $$\lim_{n\to\infty} P(X_n\leq x)=1-e^{-x}.$$

[flat frigate]

jsidind810

[safe radish]

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[neat vector]

Hello, I need help on my homework, my homework is about logs and i have a couple of problems that i don't know how to start the first one says: Express the equation in exponential form. ln(x + 4) = 3
so how would i start that?

[main mural]

what's the definition of log, or ln for that matter?

[neat vector]

umm i remember In was the same thing as e

[main mural]

$\ln(x)$ is the same as $\log_e(x)$, yes

[flat frigate]

artemetra

[main mural]

$\log_b(a) = c \iff b^c = a$

[flat frigate]

artemetra

[main mural]

for example, log_2(8) = 3 because 2^3 = 8

[neat vector]

yes

[main mural]

great

same with ln: $\ln(a) = c \iff e^c = a$

[flat frigate]

artemetra

[main mural]

so try to match this with your homework

what are your a and your c?

[neat vector]

i think 4 and 3

[main mural]

3 is correct, 4 isn't

compare ln(a) and ln(x+4)

'a' isn't 4

'a' is everything inside ln

[neat vector]

oh

so then would i have to put In_4(x)= 3 and find what gives me?

[main mural]

wut

no

[neat vector]

nvm i got confuse

[main mural]

your a is x+4

the whole thing

[main mural]

you don't use bases with ln, ln is log base e so putting another base doesn't make any sense

[main mural]

ln(x+4) = 3
therefore,
x+4 = e^3

do you see it from what i wrote above?

[neat vector]

yes, sorry i for some reason thought that we change it to log idk why

[main mural]

nah

but yeah

another way to think about it is that e^something and ln(something) are inverse operations of each other

$e^{\ln{x}}=\ln{e^x}=x$

[flat frigate]

artemetra

[main mural]

they cancel each other out

[main mural]

so another way to get there is to exponentiate both sides

ln(x+4) = 3
e^ln(x+4) = e^3
x+4 = e^3

[neat vector]

ohh ok i think i like cancelling each other, i just forgot

i think i gonne try it on my own now but i'll come back later for other problems

thank you

[main mural]

np

[neat vector]

.close

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[pure grail]

if i have this rational set including sqrt(2) and sqrt(3), then does sqrt(6) also belongs to this set? cuz i can write it as sqrt(2) * sqrt(3)?

[main mural]

yes

it does belong

[safe radish]

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[sudden sinew]

can someone help me with both parts

[dapper venture]

calculate acceleration

[sudden sinew]

for which part?

[dapper venture]

for second one

[sudden sinew]

is the first one correct?

[dapper venture]

for first one impulse is just force integrated

[sudden sinew]

with the limits set to 7.7 and 0?

[dapper venture]

yes

[sudden sinew]

im still stuck on the second part

[dapper venture]

F=ma

[sudden sinew]

do i use f= 0.032t^1.3

[dapper venture]

yes

[sudden sinew]

i dont think thst correct

[spare pebble]

I forgot what a function in an ordered pair was. Can someone give me a breif definition?

[dapper venture]

#❓how-to-get-help

[sudden sinew]

@dapper venture i did that but now what

[dapper venture]

show your work

I can't read your mind

[sudden sinew]

ok

i got 0.032t^1.3/0.022 = a

thats my acceleration

[dapper venture]

how to find velocity given acceleration?

[sudden sinew]

v=u+at

[dapper venture]

that's when a is constant

[sudden sinew]

ok

im not sure

[dapper venture]

important fact: s'=v, v'=a

' stands for derivative

s is distance

[sudden sinew]

oh do i integrate the a equation

[dapper venture]

yep

[sudden sinew]

replace x with t

but yh thats wha

what i got

@dapper venture

[dapper venture]

final velocity is when t=7.7

[sudden sinew]

yup so plug in 7.7 for

x

correct?

[dapper venture]

yes

[dapper venture]

wth is this though

[sudden sinew]

thats the acceleration inegrated

replace the x with a t

that ^

[dapper venture]

how

also it's missing x

[sudden sinew]

ye kinda fucked it

one sec

better?

i forgot to add the t

[dapper venture]

nice

[sudden sinew]

i got 69.17

is this correct?

[dapper venture]

wait a sec

Do you remember something important about integration

the constant C

[sudden sinew]

yes

[dapper venture]

velocity = 160t^(23/10)/253+C

find C then plug in t=7.7

[sudden sinew]

i dont know how to find c

[dapper venture]

when t = 0 , the velocity is 2.3

[sudden sinew]

or do i plug in initial velocity and t=0

ye

thanks

final answer = 71.5

@dapper venture

[dapper venture]

I'm too lazy to verify

[sudden sinew]

oh bro please

i need this

you've come this far with me 🙂

@dapper venture

[safe radish]

@sudden sinew Has your question been resolved?

[safe radish]

@sudden sinew Has your question been resolved?

[safe radish]

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[strange orchid]

Can someone tell me why this step is wrong ?

[strange orchid]

Gaussian elimination

[safe radish]

@strange orchid Has your question been resolved?

[hasty wagon]

hmmm
2x +3y -2z = 5
-) 2x -4y -6z = 8
_ _ _ _ _ _ _ _ _ _
0x +7y +4z =-3

[strange orchid]

-) 2x -4y -6z = 8
where did you get this ?

I might write ugly sry

I checked and its correct

[hasty wagon]

from your R2-2R1

[strange orchid]

its
R1(-2)+(2)

[hasty wagon]

?

[strange orchid]

well you need to get rid of x in both

[hasty wagon]

what is (2)?

[strange orchid]

(-2)

[hasty wagon]

i see

[hasty wagon]

that means R1(-(-2))+(-2)?

[strange orchid]

I need to get rid of x
and that way I did this

1x -2y -3z = 4 //okay so I need to get rid of lets say x , I will do this by multiplying this row with -2
2x +3y -2z = 5

[hasty wagon]

yes

so, you like
(-2)R1+R2?

[strange orchid]

1x -2y -3z = 4 //(this stays the same )
2x +3y - 2z = 5 //now the result of multiplying the first row goes here (-2x +4y +6z = -8)

[hasty wagon]

i see

[strange orchid]

and last step is doing
1x -2y -3z = 4

---

2x +3y - 2z = 5
-2x +4y +6z = -8

[hasty wagon]

so we have
7y+4z=-3

[strange orchid]

yup , I hate this method but I got to learn it for some reason

well thank you I will try to read my problems better XD

.close

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[bleak trellis]

i dont know where to go next, do i use elimination?

here my step:
s + b = 9000
s(1/20) + b(2/25) = 525

[torpid fable]

You can use elimination. I think substitution would be faster since s+b=9000 is easy to solve. Btw first fraction should be 1/20

[bleak trellis]

oh my bad

ok so

[bleak trellis]

how would substition work

wait hold on

s + b = 9000
b = 9000 - s

s(1/20) + 9000 -s = 525

s(1/20) -s = -8475

[torpid fable]

Yeah like that

[bleak trellis]

s-19/20 = -8475

[torpid fable]

This part is slightly wrong

[bleak trellis]

oh dang

ok let me take a step back

s(1/20) - s = -8475

i got

-19s/20

[torpid fable]

Yeah

[bleak trellis]

alright then i multiply -19/20 to 8475

8051.25 = s

[torpid fable]

Divide both sudes by -19/20 aka multiply both sides by -20/19

[bleak trellis]

wait where you get -20/19 from

[torpid fable]

When you divide a number by a fraction, that is the same as multiplying the number by its reciprocal

[bleak trellis]

oh alright

so

-19s/20 (-20/19) = -8475 (-20/19) ?

[torpid fable]

Yeah

[bleak trellis]

s = 8921.05

so

if b = 9000 - s then

b = 9000 - 8921.05
b = 78.95

8921.05 + 78.95 = 9000

if this answer is true, this man aint investing anything in bonds bruh

@torpid fable

[safe radish]

@bleak trellis Has your question been resolved?

[torpid fable]

I think you forgot a 2/25 multiplying 9000-s

I gtg now though

[safe radish]

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[hoary nimbus]

Can someone explain how to do question 13 this is what i have so far

[plucky elk]

Your piecewise derivative is right

The function is increasing when the derivative is positive

[glacial cairn]

No it's not

[hoary nimbus]

but the solution says this

[glacial cairn]

Go step by step

When is x^2-5 positive, and when is it negative?

[plucky elk]

Pain I was wrong

[hoary nimbus]

negative when x < sqrt 5 and positive when x > sqrt5

[glacial cairn]

No

It's a quadratic

The graph is a parabola

[hoary nimbus]

im notsure how to figure out when its negative or positive

[glacial cairn]

Did you do all the previous 12 questions?

[hoary nimbus]

yeah i did

or all the odd numbered ones atleast

[glacial cairn]

Then I'm sure you can figure out when a parabola is above the x axis and when it's below

What are the roots of x^2-5?

[hoary nimbus]

• and - sqrt5?

[glacial cairn]

Yes

Is it going to be positive at -inf? At +inf?

[hoary nimbus]

at -inf itll be positive?

and +inf itll also be positive

[glacial cairn]

Yes of course

So when is x^2-5 positive and when is it negative?

I don't know what's blocking you

[hoary nimbus]

im trying to go back to my other questions to understand what youre asking for because im kind of lacking in my funamentals idrk what ur asking for

[glacial cairn]

This is x^2-5

The roots are where it intersects the x axis

[hoary nimbus]

positive when x isless than neg sqrt 5 and positive sqrt 5

and then negative when x is between those 2

[glacial cairn]

Yes

[hoary nimbus]

is there a way to figure that out without needing the graph?

[glacial cairn]

Of course

A continuous function changes sign every time it crosses the x axis

So if you have the roots, you know when it touches the x axis

Note that it could just touch it and go back

But in the case of a quadratic, if you have two distinct roots, then it obviously crosses

So then if you know whether it's positive or negative at one point, you know the rest

(and to find one point you just plug in whatever value you want, except a root of course)

You can also just solve x^2-5 < 0; you can take the square root of both sides of an inequality (unless they're negative) because sqrt is a strictly increasing function

Actually scratch that it will be more confusing

[hoary nimbus]

Ah ok i see thank you

alright thanks for the help 🙏

.close

[safe radish]

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[indigo tundra]

Idek for step 4

Triangle FDG ~= triangle FEG

I think

[safe radish]

@indigo tundra Has your question been resolved?

[indigo tundra]

<@&286206848099549185>

And uhhh for 7 I think SSS

And for 11 I think steps 9 and 10

<@&286206848099549185>

<@&286206848099549185>

[indigo tundra]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[indigo tundra]

<@&286206848099549185>

[fringe silo]

Hey

What do you need

I will try my best to help you

[indigo tundra]

Are my answers right

I said wat i think they are under the images

[fringe silo]

I think yes

What do you need to know

[indigo tundra]

Idk if it's right

I dont think so😭

I don't think this would be there twice

So I think something is def wrong

Maybe 4. Is CDF~=CEF

<@&286206848099549185>

[fringe silo]

Hmm

What represents the 2, 5, 6 and 1,2,3

[indigo tundra]

Step numbers

The ones it like includes I guess

Idk how to say it

The things u need to know for the step

[fringe silo]

And what is the difference between =~ and =

It is practically the sams

WAit

No it is not

I saw it

[indigo tundra]

= is for like angle measures

~= is congruence

[fringe silo]

I know

But There is not =

Both have ~=

[indigo tundra]

Wat r u talking abt

[fringe silo]

About those triangles

[indigo tundra]

Yes

They are talking Abt congruence

[fringe silo]

Yes

Then what are the options

[indigo tundra]

For which

<@&286206848099549185>

[fringe silo]

There is another triangle that can be in there

[fringe silo]

Oh

Ok

Let me check

CDF And CEF can be the answear too

[indigo tundra]

That's what I put

Since I realized

The one I had put was already on the proof

[fringe silo]

What the side side side means?

[indigo tundra]

.

SSS to prove congruence

If a triangle has all same side lengths as another triangle, they are congruent

Sum like that

[fringe silo]

What About angles

[indigo tundra]

A side, an angle in-between, another side

[fringe silo]

The side is not on the picture

[indigo tundra]

Wat

[fringe silo]

You have triangle FEG

Side g is not on the picture

Then how Will you proof congurence by something that is not there

[indigo tundra]

G is a vertice point not a side I think

[fringe silo]

But small g

The side how do you call the side

Side g is not here

[indigo tundra]

I think u would call it a line segment

Between two points

To talk about a side

[fringe silo]

It is an axis

Bro

Idk

Angle side side

Or side side side

[indigo tundra]

<@&286206848099549185>

[indigo tundra]

ASS💀

[fringe silo]

💀💀

Do you know how we call it in czcech

SUS

Bruh

How can you do it then

[indigo tundra]

Bro is asking me now💀 😭

[fringe silo]

By ass or by sss

What do you think

[indigo tundra]

you can't use SSA as a criterion for triangle congruence unless the angle is obtuse or right angle

But I'm pretty sure there's a right angle

[fringe silo]

Yep There is

Angle 1 and angle 2

[indigo tundra]

Then it becomes the hypotenuse criterion thingy I think

Idk

[fringe silo]

Ayo

It is the third option?

[indigo tundra]

Hypotenuse leg

Idk

Maybe

[fringe silo]

What is it fkr

For*

[indigo tundra]

Search it up

😭 😭 😭

[fringe silo]

Tbe thing that are you doing

To school or where?

[indigo tundra]

Personal

Study

[fringe silo]

Ok

I study goniometrics

[indigo tundra]

Never heard of it

[fringe silo]

?

Sinus

Cosinus

Tangens

[indigo tundra]

Do u speak another language

[fringe silo]

Czech

A little bit german

And English

[indigo tundra]

Ok😭

[fringe silo]

?

[indigo tundra]

R u talking Abt sins cosines and tangents

[fringe silo]

?