help-17

[honest pollen]

The is another rule

The exponents can multiply to get x^2/3

knowing this, try rewriting this equation in a way that'd make it easier to plug in numbers

[dawn latch]

(x^1/3)^2 + 1/ sq 2x

[honest pollen]

correct

you can replace the 1/3 with a cube root if you know how cubic root works

But the sq 2x is wrong

[dawn latch]

i meant

[honest pollen]

It's also cube root x like this

[dawn latch]

square root x

but theres liek a 2

in the little gap

[honest pollen]

oh

that would've been true if it was x^1/2

but with x^1/3 it's a 3 in that gap

[dawn latch]

oh

so i did it wrong

[honest pollen]

You can try writing it again

[dawn latch]

(x^1/3)^2 + 1/ cube root x

there we go

all equal 0

[honest pollen]

yep

you can replace the other x^1/3 with another cube root

and try plugging in an easy number for x like 1 or -1

[dawn latch]

so it would be (cube root x)^2

[honest pollen]

correct

[dawn latch]

i still dont get how it will equal -1

LOL

[honest pollen]

lol

Did you try plugging in -1 for x

[honest pollen]

yep

[dawn latch]

wait y were we doing this

[honest pollen]

-1 ^ 2/3 isn't easy to find without a calculator

[dawn latch]

oh

[honest pollen]

but cube root of -1 is -1, -1 ^2 equals 1

so if you do the math, you'll get 1-1=0

which means that x=-1 is a critical point

sorry bro, but I'll hafta leave soon (it's been a while)

[dawn latch]

after that im done

or how will i do c

o

okay goodbye thanks for helping

[honest pollen]

yeah, technically you got x=-1 as an answer

you need to find the coordinates for the critical point, based on the question

so plug in x=-1 and x=0 to find the two critical points in the original equation

[dawn latch]

how do i use first derivative test to classify them

[honest pollen]

you just used the derivative to proof it

[dawn latch]

idk how to do that

will u be back

[honest pollen]

this is the derivative equation, show that plugging in -1 will make it equal to 0 to prove

yeah, after my classes are done I'll be back

good luck, and if you still need help tag helpers

[dawn latch]

okayy ty

[vocal sleet]

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[tough monolith]

Hi, I have a question from inverse of a function

[tough monolith]

If

$f(x) = x + \lfloor 3x \rfloor$

[twin meteor]

Arian

[tough monolith]

Then what is $f^{-1}(x)$

[twin meteor]

Arian

[vocal sleet]

@tough monolith Has your question been resolved?

[vocal sleet]

@tough monolith Has your question been resolved?

[tough monolith]

<@&286206848099549185>

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[tough monolith]

If

[tough monolith]

$f(x) = x + \lfloor 3x \rfloor$

[twin meteor]

Arian

[tough monolith]

Then what is $f^{-1}(x)$

[twin meteor]

Arian

[vocal sleet]

@tough monolith Has your question been resolved?

[tough monolith]

<@&286206848099549185>

[vocal sleet]

@tough monolith Has your question been resolved?

[loud pivot]

Do you know how to find an inverse function?

[vocal sleet]

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[tough monolith]

$f(x) = x + \lfloor 3x \rfloor$

[twin meteor]

Arian

[tough monolith]

Then what is $f^{-1}(x)$

[twin meteor]

Arian

[tough monolith]

Yes

But I can't use that here

[flat whale]

Why

What method do you know?

[vocal sleet]

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[tough monolith]

$f(x) = x + \lfloor 3x \rfloor$

[twin meteor]

Arian

[tough monolith]

Then what is $f^{-1}(x)$

[twin meteor]

Arian

[tough monolith]

Changing the place of y and x

[real shore]

then solve for y=x

[vocal sleet]

@tough monolith Has your question been resolved?

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[full arch]

hello

[full arch]

was taking a look at this , and I'm not sure how the negative or the placement of numbers shuffle around through multiplication? I have a strong sus is the identity matrix but I only made somin that doesnt really hold for all powers

[rugged orchid]

That looks like a rotation matrix

Look at what entries are in a Rotation matrix of θ degrees

That looks like θ = pi/4 or something

[full arch]

omg

TRUE

[rugged orchid]

The first thing I thought was diagonalise it

[full arch]

you get an imaginary binomial

;-;

[rugged orchid]

Doesn’t matter

Just diagonalise over C

(It’s a rotation matrix of course it has no real eigenvectors)

[full arch]

lemme post this real quick

[rugged orchid]

Well, very unlikely at least

[worthy citrus]

You don't even need to diagonalise, find the angle of rotation and then thanks the gods it divides 360 so you know how often powers of B repeat

[rugged orchid]

Yeah the first thing I saw was probably diagonalise

But upon further inspection it can be observed to be a rotation matrix

[full arch]

yeeee

thank you guys

im blind

also the thing I was gonna post seems kinda irrelevant so

get blue balled

[rugged orchid]

See it more and you’ll see it yourself!

[full arch]

https://tenor.com/view/cat-glasses-on-cant-see-gif-15684379

.close

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[kind loom]

lets say we have an argand diagram like this, the point -24 -7i is z

[kind loom]

where is the argument actually measured from?

[rugged orchid]

Positive real axis

And it goes counterclockwise

[kind loom]

so what, like this?

[rugged orchid]

Yes that

[kind loom]

right, so that confuses me. third quadrant means that arg(z) is theta - pi, and given the point here, it means the angle would be -2.858 radians, how does this work? going from the positive real to the negative real (i'm presuming you'd call it) is 180 degrees, or pi radians, so why would it immediately become so negative?

afaik it goes from pi -> -pi radians

but if it was in the fourth quadrant, wouldn't it be a greater negative angle? or is that not how it works

is it measured in a way that i'm interpreting incorrectly?

[solar saffron]

Hmm I'm not so sure if I understood your interpretation. But here's an insight:

Like what Frosst said. It starts from positive real axis, at 0 degrees, then when you are measuring a positive angle or argument, it will go and measure counter-clockwise around about the origin.
If you are measuring a negative angle or argument, you will realize that it will go and measure clockwise around about the origin.

[vocal sleet]

@kind loom Has your question been resolved?

[kind loom]

so...eh? since this value is negative, the angle is basically what the red is? it would make more sense to me for it to be like that, but it conflicts with my interpretation of frosst's message

[patent fractal]

no the angle should be blue

but yo ucan could the red

if the angle is negative

[kind loom]

what

[patent fractal]

the red

the blue = theta

the red = -omega

[solar saffron]

omega? xD

[kind loom]

eh

[patent fractal]

hahah

say a and b then

what are you guys discussing

[kind loom]

i understand blue is theta, that was my original interpretation of it

[patent fractal]

angles are counted coutner clockwise

[solar saffron]

look, it always starts at real positive side. That is 0.

[kind loom]

but i guess i have misinterpreted hqppy's explanation

[solar saffron]

0 degrees, 0 rads

[kind loom]

yes.

[solar saffron]

When you go anti-clockwise... Your angle is positive

When you go clockwise... your angle is negative.

Here's a small proof

making sure that ur calculator is in degrees, find:

• sin(0)
• then sin(90),
• sin(180)
• sin(270)

or u can do

• sin(0),
• sin(pi/2),
• sin(pi),
• sin(3pi/2)
  for radians

[patent fractal]

here

a picture

[solar saffron]

After you do this, try finding the sin of -90, -180, -270. You notice that you go the other way around the unit circle.

[vocal sleet]

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[stark otter]

Not sure if i'm off or not

[loud pivot]

⚠️ This person is trolling ⚠️

[stark otter]

Dude i be here like every day 💀

your the most unhelpful person i've seen here in like 3 years

[loud pivot]

I don't know, you were trolling me in another channel and created a new one now

[stark otter]

cus u stopped helping me told me i was trolling and i said "i guess im trolling"

your corny

[vast shale]

!show

[vocal sleet]

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

[stark otter]

gotta redo it rq

[vast shale]

12sqrt(2) = 6
what

Not only is it wrong, where is it even coming from??

[stark otter]

square root is wrong symbol but i mean divide

[stark otter]

where is what coming from

[vast shale]

Where'd you get 12/2 from?

Oh wait were you just dividing the top and bottom by 2?

[stark otter]

ye

[vast shale]

Right, you were just simplifying it

[vast shale]

18 +- 1 = 17/19
That's uh... not how +- works

[stark otter]

i didn't want to draw it out twice jus for the sake of our time

18-1=17
18+1=19

17/19

[loud pivot]

I am telling you they are trolling

Told me you get 9x^2 by factoring 3 from 3x+21x

[stark otter]

ight whatever i'm jus gonna redo it

[vast shale]

okay

[stark otter]

.close

[vocal sleet]

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[vast shale]

Right... I don't know what was going on with the previous one but glad you solved your own problem

[stark otter]

🫡 thanks for helping

[vocal sleet]

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[broken geyser]

If sqrt(x) - 1 = x/4 proof that x = 4

[broken geyser]

Someone

[sly sierra]

try converting the problem into a quadratic

[broken geyser]

It will be x - 2√x +1 = x²/16

[sly sierra]

no that's not gonna help

[broken geyser]

Okay

Which formula

[sly sierra]

try to get an actual 2nd order polynomial (one without any square roots in it)

try a variable substitution

let y = ...

[broken geyser]

Lemme translate

I Can't understand

Can u explain more @sly sierra

[sly sierra]

what if you tried y = sqrt(x)

[broken geyser]

I will got

y-1 = y/4

And

4y -1 = y

[sly sierra]

wait

how can both sides have a y

[broken geyser]

What

Oh yeah

[sly sierra]

there was no sqrt(x) on the right side

[broken geyser]

Oh yeah

Y -1 = x/4

Then i need to solve two equation?

[sly sierra]

you can express x in terms of y

if y = sqrt(x), then what is x

[broken geyser]

x = y²

[sly sierra]

yea

try that

[broken geyser]

y-1 = y²/4

y -1 = y/2

Right?

[sly sierra]

what happened to y^2/4

it's not equal to y/2

[broken geyser]

Sqrt

Sqrt y²/4 = y/2

Oh

Yeah

y-1 = y²/4

[sly sierra]

ok

do you know how to solve that

[broken geyser]

yeah probably

Lemme try

@sly sierra
y-1 = y²/4
4y/4 - 4/4 = y²/4
4y - 4 = y²
y²-4y+4 = 0
(y- 2)² = 0
y-2 = 0
y = 2

y = 2 then sqrt(x) = 2 then x = 4

[sly sierra]

nice

all correct

[broken geyser]

tysm bro

.close

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[sharp gust]

Helloo

[sharp gust]

in this case, can i apply this rule?

(a+b)^2

a^2 + 2ab + b^2

[light aurora]

Yes

Do that and it's the answer

[sharp gust]

so its x^2 + 2(x*3y) + 3y^2

right?

[light aurora]

I think so I can't be sure because I'm on my phone rn and don't have paper

Yes it's right

[fair plover]

It isn't

[light aurora]

Oh yeah mb

[light aurora]

x^2 +3xy +3xy +9y^2

[halcyon wing]

!nosols

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

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[vast shale]

Hello could you please help me with this maths question !

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

<@&286206848099549185>

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

@vast shale Has your question been resolved?

[exotic mason]

It's going to shrink, and it will be relative to the center. So, since K is 3 away from (7,3) in the x direction, it will become 1 away. Since K is 6 away from (7,3) in the y direction, it will become 2 away.

So what's 2 up and 1 right from (7,3) ?

[vast shale]

Wait no 8,5

[exotic mason]

I think that's correct.

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

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[vast shale]

I was wondering how you would be able to find the standard deviation of a histogram and what would the answer even be please help

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

<@&286206848099549185>

.close

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[vocal dagger]

Need help with last part

[vocal sleet]

@vocal dagger Has your question been resolved?

[hard atlas]

write out the sum explicitly (not in sigma notation)

use that for example P(X>0) is the same as P(X=1)+P(X=2)+...+P(X=n)

[vocal dagger]

Is this right ?

Sorry for rotated photo

[hard atlas]

exactly

[vocal sleet]

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[real jasper]

hello, im hella confused with this

[real jasper]

a. Write an expression that calculates the perimeter of the rectangle.

b. Write an expression that calculates the area of ​​the rectangle.'

c. Calculate the perimeter and area of ​​the rectangle,
when a = 9 cm and b = 6 cm.

[real jasper]

this looks wrong, i'm not sure what to do

[outer warren]

a + a isn't just 2

if one person gave you one apple, and another person gave you another apple
you don't have the "number 2"

[real jasper]

ohhhhhh, then its aa

[outer warren]

no

you wouldn't have the product of two apples either

also another issue, is that (a+b) and (a-b) are only two of the sides of the rectangle

the formula for perimeter isn't just the sum of two sides

if one person gave you one apple, and another person gave you another apple
how many apples would you have?

you'd have 2 apples

[real jasper]

yes

[outer warren]

in this case, a + a = 2a

now to address that other issue

the formula for perimeter isn't just the sum of two sides

[real jasper]

(a+b)+(a-b)+(a+b)+(a-b)

[outer warren]

yes

[real jasper]

so do i need the other sides too

[real jasper]

should i use (a+b)+(a-b)+(a+b)+(a-b) for b part

[outer warren]

formula for area is different

[real jasper]

then how do i do it?

[outer warren]

do you know the formula for area of a rectangle

[real jasper]

yes times both sides

[outer warren]

but first
simplify

(a+b)+(a-b)+(a+b)+(a-b)

[real jasper]

a+b+a-b+a+b+a-b

or

(a+b)^2+(a-b)^2

[outer warren]

no

[real jasper]

oh i can't do that?

[outer warren]

yes to

a+b+a-b+a+b+a-b
those () weren't needed

simplify that

combine like terms etc,
don't overthink

[real jasper]

a2+b2+a2-b2+a2

[outer warren]

no

[real jasper]

i'm sorry

[real jasper]

like aa+bb+aa+bb

[outer warren]

no

[real jasper]

then i don't know

i'm sorry

[outer warren]

think of these as apples and bananas

take this one step at a time if needed

starting with
apple and a banana
what do would you have after someone gives you an apple and takes away a banana

[real jasper]

then i have the apples

[outer warren]

how many apples

[real jasper]

2

or aa

[outer warren]

no

aa is NOT the same as 2a

[real jasper]

ohhhhhhh

so the answer is 2a

[outer warren]

aa algebraically represents the product of a and a, or a^2

which is NOT what you want

well the first half would be 2a

$\underbrace{a+b+a-b}_{2a} + a+b+a-b$

[twin meteor]

ℝam()n()v

[outer warren]

simplify the whole thing

[real jasper]

like this

4a?

[outer warren]

yes

[real jasper]

okay thanks for the help

but

[outer warren]

do you know the formula for area of a rectangle
yes times both sides
so do that

[real jasper]

a. Write an expression that calculates the perimeter of the rectangle.
b. Write an expression that calculates the area of ​​the rectangle.' =4a
c. Calculate the perimeter and area of ​​the rectangle,
when a = 9 cm and b = 6 cm.

[real jasper]

so times both sides

a and b

[outer warren]

no

the sides aren't a and b

the side lengths are
(a+b) and (a-b)

[real jasper]

and time it

right?

[outer warren]

multiply those like the area formula says

[real jasper]

wait

i forgot something

[outer warren]

no

you only multiply each side once

area = l * w
NOT l * w * l * w

[real jasper]

ohhhhhh

i think i did

[outer warren]

missing ()

[real jasper]

(a+b) *(a-b)

[outer warren]

yes

[real jasper]

it says

0

wait

[outer warren]

what's saying 0

[real jasper]

(a+b) *(a-b)

=0

[outer warren]

where's 0 coming from

[real jasper]

okay my machine is broken

[outer warren]

don't use whatever machine you're using that's outputting those results

nor should you be using a machine for this

[real jasper]

okay what machine do you use

[real jasper]

please show me

[outer warren]

these questions don't require any calculator whatsoever

[real jasper]

wait is it a2-b2

[outer warren]

use ^ to denote exponentiation

if you mean
$$a^2 - b^2$$
yes

[twin meteor]

ℝam()n()v

[real jasper]

c. Calculate the perimeter and area of ​​the rectangle,
when a = 9 cm and b = 6 cm.

so a2*9cm^b26

wait

i need some help

[maiden spindle]

Is Perimeter 30cm and area is 54cm^2 ?

[real jasper]

?

.close

[vocal sleet]

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[blissful needle]

wait for a hyperbole to not intersect a curve its asymptote has to be more or less than the oblique asymptote of the curve?

[blissful needle]

@spice sune

<@&286206848099549185>

[vast shale]

hello

[blissful needle]

Hello

[blissful needle]

Isit more or less

[vast shale]

hm, well what has you confused to far?

what is not clicking?

so*

[blissful needle]

For 6iii am I supposed to make it more than 2 or less

I kinda forgot

2x

[vast shale]

are calls allowed on here? i’m new to discord

[blissful needle]

I'm not available tk call tho

[vast shale]

ohh i see

okay

i’ll work the problem out

[blissful needle]

Wait actl dunnid

I'm jst asking if j shld make it more or less than the oblique asymptote

[vast shale]

more

[blissful needle]

Okok thanks

[vocal sleet]

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[vocal sleet]

@elfin moon Has your question been resolved?

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[elfin moon]

.reopen

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✅

[sand condor]

What is your idea of proceeding it

[vocal sleet]

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[south ingot]

how would i do these

[south ingot]

.close

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[undone patio]

Help me

[undone patio]

I wanna learn like so serious right now tired of using math way

[elder forge]

what have you tried? Where are you stuck?

[undone patio]

I don’t know what to do

Same with this

I’m stuck everywhere and I got test next class on basically all this

[karmic imp]

Have you tried asking your teacher for help?

Or classmates?

[undone patio]

I’m in health rn

And it’s way to late for that ngl

[karmic imp]

Afterschool help

[karmic imp]

https://www.youtube.com/watch?v=lzqTD0JWwhY&ab_channel=TheOrganicChemistryTutor

[undone patio]

Appricate it

Found the answer to this but don’t know how to solve for it

[elder forge]

Cramming this an hour before a test isn't very tactical no.
The two point formula and some basic algebra is probably all you have time for at this stage.

In general it's all about finding two numbers to make a function.
The standard form (for writing functions) is;
f(x) = a*x + b

Where a is the slope of the line, meaning for each step along the x-axis(the horizontal one), how much does the line move along the y-axis. (the vertical one)
And b is the point where the line crosses the y-axis.

Calculating this for two points P(j, k) and Q(m, n)
Can be done like this;

A = (n-k)/(m-n)

To find B, you need A first.
For the points in the thumbnail,
P(-2, 1) and Q(2, 9)

A = (9-1)/(2-(-2)) = 8/4 = 2

From here, you can find B by working backwards from one of the points.
If we choose P = (-2, 1), we take two steps along the X-axis from -2 to 0, for each step, add A, which is 2, to the Y value.
So from -2 to 0, we take two steps, so we add 4 to the Y value. This gives us b = 1 + 4 = 5.

So our function is f(x) = 2x+5

[undone patio]

U a W

We got a lot to talk about

With you I can pass the test

[undone patio]

This for number 2 right

[elder forge]

It's the method for solving number 2 ye

[undone patio]

So b=1 + 4=5 is my answer

And as u said a is the slope for this formulae and h is the y axis correct

[elder forge]

For your last picture.
With equations, we have a few legal "moves".

You can multiply both sides of the equation by pretty much anything.
I.e. for 2 = x, if we multiply both sides by 5, so we have 2*5 = 5x, the equation is still true.

You can move parts from one side to the other by addition and subtraction
If we have
2x + 3 = y
If we want to move this to standard form, we can subtract the number 3 from both sides.

2x + 3 -3 = y -3
Now the +3 and -3 on the left cancel out and we're left with
2x = y -3

Then we can subtract y on both sides to move that
2x - y = y - y - 3
The y and -y on the right cancel out, so we're left with
2x-y = -3

[undone patio]

Bro I appricate it a lot I’m understand a little but don’t understand a lot and trying my best I’m most likely going tell the test

[elder forge]

I'm educated in Norway so we might use different symbols, idk about H but i'll explain again to be safe:
A function is denoted by f(x) = bla bla something
A linear function, meaning a function that is just a straight line, no curving, look like this with the symbols i'm used to
f(x) = a*x +b

A is indeed the slope of the function, B is where the function crosses the Y axis

[undone patio]

I sent the whole paper in DMs or do u want me to send it here

[undone patio]

X1 x2 y1 y2

[elder forge]

I can't really solve all the tasks for you, just wanted to try and explain some of the most fundamental things you can use to solve them

[undone patio]

Bet got it

I can just use math way for the rest I appricate you

[elder forge]

np, good luck

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@undone patio Has your question been resolved?

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[vernal agate]

<@&286206848099549185> please help

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[vast shale]

what have you tried

[twilit hound]

nothing idk what to do

[deft mortar]

You have to rewrite first

Cot = cos/sen

[twilit hound]

does the 3x go on both top and bottom

for both

[deft mortar]

Sorry I have to go, I'm at uni

[twilit hound]

so it would be (cos3x/sin3x)(2x/sin2x)

[twilit hound]

youre good

[deft mortar]

But try to do: senx/x

So you LL know that's 1

Start cleaning stuff

Like that

[twilit hound]

kay

i ended up getting cos0/sin0

which would beeeee

1/0 right?

so dne?

@vast shale ^^

[deft mortar]

Don't replace with 0 yet

Bc you would get an indetermination

That's why you have to rewrite and play with that

Trying to get at anywhere a form like senX/X so you will cancel bc that's 1

Next if you can't do anything else so then replace with 0

If you have a 0 at denom you still missing a step

Or it is a trap and doesn't have a lim

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[vocal sleet]

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[glad obsidian]

problem: What is the probility the 4 random drawn cards form a deck are different values given they are different suits?

[glad obsidian]

My first thought was that it would be (1/13) (1/12) (1/11) (1/10) but I know this is wrong because we can still pick the value after it has been picked in a different suite

[dull maple]

That is indeed wrong. Although your logic is a little flawed.

Foremost, you are forgetting that you have multiple ways in which 4 different values can be there on 4 cards.

[glad obsidian]

to get the number of ways 4 different calues can be chosen, should we do (13 choose 4) ?

[dull maple]

Yes. You choose 4 values that can be there on the 4 cards all of which are from the different suits.

Then remember that these chosen values(let's say 1, 2, 3, and 4) can be on four cards in multiple ways as well.

[glad obsidian]

to choose the value of the 1st card

will it be (13 4) (4 1)

and 2nd card (13 4) (3 1)

3rd (13 4) (2 1)

and 4 (13 4) (1 1)

[dull maple]

(13 4) - means that you have already decided which 4 values will be there. So, why are you doing (13 4) for all card?

[glad obsidian]

oh i see

in the way i multiplied for all card,

it would be like i was selecting a new set of 4 cards each time right?

[dull maple]

Yes. Select the card. Distribute them among four suits. Then multiply with the probability of picking those 4 values in a particular distribution among 4 suits.

[glad obsidian]

ok i think i understand

.close

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[half yew]

Okay, need help with this

I understand the general premise, but I really am lost when it comes to thinking about where to start

nvm figured it out

.close

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[mortal flint]

hello
can someone please helo me check my proof for first one
i reached result but i feel that i was lost while writing it
and if someone help me on how to approach 2

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@mortal flint Has your question been resolved?

[mortal flint]

<@&286206848099549185>

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@mortal flint Has your question been resolved?

[mortal flint]

anyone lol

[mortal flint]

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[severe wharf]

hey I don't understand how my teacher went from this stage to the next.
I don't understand why it's not d^2/cosa^2

[vocal sleet]

Please don't occupy multiple help channels.

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[vast shale]

In part 4b at the very end, would it be the column or row space of A?

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@vast shale Has your question been resolved?

[vast shale]

Hello, may I get some help please?

I suppose not 😔 . Have a nice day.

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[autumn bear]

Help

[sly sierra]

what do you have available to work with?

do you know the rank-nullity theorem?

[vocal sleet]

@autumn bear Has your question been resolved?

[autumn bear]

No. It wasn't introduce to us yet. Huhu

[pale perch]

Suppose v, w, p belong to R^3
such that the linear transformation maps both v and w to p, so
Av=p and Aw=p

show then that v=w

[vocal sleet]

@autumn bear Has your question been resolved?

[autumn bear]

How do i start it?

[pale perch]

the above, then Av=Aw, go from there

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[fathom glen]

Need help with question 4
Idk what triple product is,not even sure if I have to take dot Product or cross Product
I'm currently doing vectors

[lethal bough]

triple product is typically, given vecs a,b,c

a • (b x c)

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@fathom glen Has your question been resolved?

[fathom glen]

Oh thnx

.close

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[quartz tendon]

help

[quartz tendon]

why is the least common multiple between (x + 1)(x - 1) and (x + 1) = (x + 1)(x - 1)

at first, (x + 1)(x - 1) was x^2 - 1

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[honest heath]

i am so sorry, ive come back here trying to fully understand this proof like 10 times today already, but how would you go about proving the backwards direction of the if and only if statement? i did the forwards direction with lots of help by contrapositive, but now im stuck on the backwards direction

[paper depot]

as in, $\gamma_1^r = \dots = \gamma_k^r = e \implies \gamma_1^r \dots \gamma_k^r = e$?

[twin meteor]

Ann

[honest heath]

yes

[paper depot]

is that what you're asking how to prove?

[honest heath]

yes

[paper depot]

isnt it obvious

multiply several e's together what do you get

[honest heath]

omg

fuck

[brisk moss]

that looks a little different than what’s on the pic?

[honest heath]

thank you

[brisk moss]

oh

nvm

[honest heath]

its been a long day 🥲 thank you

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[stiff hull]

hey i need some hepp with task B

[ebon rapids]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[stiff hull]

so what ive done is 48 x 10^9 = 3 • 2^4 • 10^9

then on the bottom its 2^6 • 10^6

then ive switched position of 10^6

so its 10^9-6 on the top

and now im unsure what to do

[split heart]

simplify the 2 stuff

you're doing good so far

[stiff hull]

2^4-6?

[tawdry harness]

just 4-4 and keep 2^2 downstairs

🙂

[split heart]

that's one way to do it

[stiff hull]

how can i tell im only supposed to take 4 up?

and not all 6?

[split heart]

it doesn't matter

it will all turn out the same if you do it right

[stiff hull]

alright so doing it the way with 2^-2 how do i go on with this

[split heart]

do what you're comfortable with

don't forget the 3

[stiff hull]

yeah 3 • 2^-2 • 10^3

[split heart]

yep, now simplify that

[stiff hull]

thats where im stuck

[split heart]

Wait why

It involves the exact same tricks

[ebon rapids]

What you did for the 48 earlier, can you do that with some other numbers?

[stiff hull]

what does 2^-2 equal

[split heart]

if you're unsure, turn that to a positive 2

how would you do that

[stiff hull]

need some help here

cant think of it

[split heart]

alright

you had 10^6 in the denominator earlier

you flipped it to the numerator and you made it negative

why is that

[stiff hull]

well to simplify it

but now i dont have anything in the denominater do i?

[split heart]

well, sure but I guess the point I'm making is that you're aware to some extent that moving an exponent to the other side of the fraction bar flips the sign

the denominator is 1

[stiff hull]

ohh right

[split heart]

1 is always there, you just don't write it

[stiff hull]

so i can put the 2^2

[split heart]

exactly

[stiff hull]

right

forgot i can move both ways my bad

[split heart]

so now what do you get

[stiff hull]

3 • 10^3 / 2^2

[split heart]

which is

[stiff hull]

3000 / 4

= 750

thanks

[split heart]

ok cool

[stiff hull]

much appreciated

[split heart]

👍

[stiff hull]

.close

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[severe wharf]

hey I don't understand how my teacher went from this stage to the next.
I don't understand why it's not d^2/cosa^2

[split heart]

i think you used cos theta = d/r

[severe wharf]

yes

[split heart]

So what's the problem then

[severe wharf]

but I don't understand how I'm supposed to arrive at the result.

[split heart]

the one in the red box?

you just substitute what 1/r^2 is

[severe wharf]

dor me cos theta = d/r render d^2/cosa^2

not cosa^2/d^2

[split heart]

how'd you get that

1/r = cos theta / d

[severe wharf]

cross product

[split heart]

ok

so use that then

that will give you the same thing as what's in the picture

[severe wharf]

use what

[split heart]

r = d/ cos theta

[severe wharf]

but where is the logic

[split heart]

what do you mean

[severe wharf]

why my cross product isn't correct

[split heart]

it is, i'm not saying it's not

[severe wharf]

oooo I'm so stupid I had a mental block

ty

[split heart]

so are you okay now?

[severe wharf]

yep

[split heart]

cool good luck

[severe wharf]

ty have a nice day

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@severe wharf Has your question been resolved?

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[inner cove]

helloooo i hope i don t bother anyone but could anyone give me some details on how I should enter the sequence Un+1 = 1.2* un - 100 with u0= 1000 on my calculator Tİ nspire cx II-T ? I tried a lot of possibilities but I couldn't do it

[inner cove]

maybe a picture of the screen before entering it

[vast shale]

are u trying to solve the recurrence relation or something

also to be clear, you have [
a_{n+1} = 1.2 a_n - 100, \q a_0 = 1000
]
right?

[inner cove]

yup i m trying to put the graphics on my calculator

of it

[vast shale]

the graphics?

[inner cove]

on the ti nspire i m trying to have the graphical representation of it

so i can see each results on the table afterwards

[vast shale]

I mean I don't know about the technicalities of your calculator, but I can instead help you with solving the recurrence relation if that's what your goal is

[inner cove]

but since it s a reccurence relation idk how to do it on my calculator

[inner cove]

nope it isn t but thank you

i hoped someone w the same calculator could helppp

[vast shale]

you can use online software such as wolfram alpha if you are willing not to restrict yourself to a specific, physical calculator

[inner cove]

i unfortunately cannot cuz it s something i have to do in class

but still thank you for the advicee

found itttt

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[plush sage]

If $a<b<c$ and $a,b,c belongs to R, then comment on the nature of roots of f(x)= (x-a)(x-b)-1$

[twin meteor]

Muffins

[plush sage]

How to start this one?

[cobalt ocean]

are you missing anything

there's no point in having c in the problem

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@plush sage Has your question been resolved?

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[sinful helm]

I know this is stupid but I cant help but feel curious. Why doesn't foiling work here?

[paper depot]

what foiling

you've got (2 * 3) * (2 * 3) and not (2 + 3) * (2 + 3)

multiplication doesn't distribute over itself

[sinful helm]

Ohh okay thanks

Now i get it why foiling wouldnt work in a past equation

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[tulip sleet]

Im stuck on how to find theta. Im given a complex number to solve, so I first convert it to exponential form, but I need to find theta to do that

[tulip sleet]

Something to do with quadrants but idk exactly what

[maiden iron]

are you sure r = sqrt2 ?

🤔

(1-j)³ = 1³ - 3(1)²j + 3(1)j² - j³
= 1 - 3j -3 + j
= -2 - 2j

r = √(2² + 2²) = √8 = 2√2

now for the argument

tan x = -2/(-2) = 1

x = π/4 + kπ (k an integer)

[tulip sleet]

Pause

[tulip sleet]

This is the final answer. You just expanded the ^3

[maiden iron]

yea I did

[tulip sleet]

We have to first convert it to exponential form

[maiden iron]

why not but you will still get what I got

[tulip sleet]

Its a dumb rule of us i know

I thought a and b were just Re and Im parts?

To calculate r here

[maiden iron]

thats what I did

[tulip sleet]

Real is 1, im is also 1

[maiden iron]

no

[tulip sleet]

Huh

[maiden iron]

z = (1-j)³ = -2 -2j

real part -2
im part -2

[tulip sleet]

Yes okay but just given 1-j ^3

Not -2-2j

How do I calculate r and theta then

[maiden iron]

you are really assuming (1-j)³ is in the algebraic form

[maiden iron]

wdym

you either find the exponential form or algebraic form of (1-j)³ and voila

[tulip sleet]

Wait

Let me give another example

This is what my teacher showed me

Left is given

[maiden iron]

sqrt2 is definitely not the module of (1+j)³
you didnt take your notes correctly

but sqrt 2 is the module of 1+j yes

the argument is also π/4 for 1+j

but not (1+j)³

[tulip sleet]

Ah

I was assuming this

So this is wrong

Ok

[maiden iron]

(1+j) is raised to the power of 3
so its not in the algebraic form yet for you to do this

algebraic form : a+ib

here you have (a+ib)³

wrong

[tulip sleet]

Ohhhh

Okay

We call it the cartesian form

Mb

Okay now back to the theta part

Is b/a

So 1 ?

[maiden iron]

(1-j)³ = -2 - 2j
a = -2
b= -2

tan x = -2/(-2) = 1

its just trigonometry at this point to find x

[tulip sleet]

I figured out how my teacher does it

She ‘forgets’ the ^3 at first

So then you do have a cartesian form

Calc r and theta

And then just add the ^3 at the end

[maiden iron]

she can

[tulip sleet]

Kinda like this

So then, is my theta correct now?

[maiden iron]

I dont see your theta

i only see tan theta

[tulip sleet]

Arctan 1 then?

[maiden iron]

which is equal to ?

[tulip sleet]

Ok i dont know that function

Lol

[maiden iron]

you never had a course on trigonometry ?

[tulip sleet]

Nope 🙂

[maiden iron]

and you are doing complex number....

[tulip sleet]

Thats why im also confused with the quadrants

When calculating thetas

[maiden iron]

[tulip sleet]

Im struggling man

[maiden iron]

I find it weird that you do complex number before trigonometry

cuz trigonometry is a fcking requirement for complex number

[tulip sleet]

Its my second time doing this calc class, so i studied ..some.. unit circle trig things

Like a year ago

So arctan is apperantly 45 degrees

Which is

Pi /2

[maiden iron]

lol no

[tulip sleet]

what

[maiden iron]

pi/4

[tulip sleet]

Okay….

[maiden iron]

tan pi/2 doesnt exist

[tulip sleet]

And in some examples I see them just adding a random pi, whats up with that?

[maiden iron]

wdym

[tulip sleet]

Just a random exmaple

Where does the pi come from in theta = …

[fluid pumice]

pi rad is 180 no?

[maiden iron]

$\tan(x) = \tan(y) \implies x = y + k\pi$ with $k \in \mathbb{Z}$

[twin meteor]

Herels

[tulip sleet]

So

Tan function is the same as any other tan function + k pi

Pi being 180 degrees

Makes sense

[maiden iron]

nah you didnt understand

when two tan are equals, their angles are also equal but with with a kπ term (because tan is periodic of period π)

[tulip sleet]

Im going to explode

Yeah okay that makes sense

I get this exercise now

Thxx

[maiden iron]

go do some trig before complex number

[tulip sleet]

I will

Or ill just learn it while doing complex

Like I did just now

😎

[maiden iron]

yea good luck when your exams are going to be tough

[tulip sleet]

Still got couple months

Enough time

[maiden iron]

take one week for trigonometry

[tulip sleet]

(Not)

[maiden iron]

complex number isnt that hard anyway

[tulip sleet]

We dont have a course for trig actually

First chapter is complex

[maiden iron]

You can learn it yourself

on the internet

[tulip sleet]

Any good resources?

[maiden iron]

khan academy I guess, or youtube

any course should be good

trig isnt that deep

[tulip sleet]

Ok

Thx

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[soft steeple]

Oh is it

is the top 5a?

[hybrid flicker]

I see what you tried to do, however you're supposed to multiply by the conjugate

So not × (sqrt(3)-1)/(sqrt(3)-1), but rather...

[soft steeple]

just replace it by - with +?

smth like this

then the bottom is like 3-1 which is 2

then the top is smth like 2x3+2 root 3 +a root3+a x 1?

[hybrid flicker]

Exactly

And usually now that you wrote it like this you gather the terms in root3 together, and the others on the right

[soft steeple]

so 6+2 root 3+a +root 3 plus a

or can i write it like 8plus root3+a squared

[fluid pumice]

its (6+a) + (2+a)sqrt3

divided by 2

its called a conjugate surd btw

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@soft steeple Has your question been resolved?

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[pastel swan]

Help with no.4

[pastel swan]

We are given two set of equations of a pair of planes. A pair of planes have a unique line of intersection. We need the two line of intersections to be coplanar. How do I do that.

[merry python]

Understand the term co-planar

The two lines lie on the same plane

i.e. either both on x-y plane or on y-z plane or on z-x plane

[pastel swan]

But I do not know equation of lines.

[merry python]

Well, can you solve a system of linear equations?

[pastel swan]

Yes.

[merry python]

well then solve the system of equations
2x-y+3z+4=0
x-3y+z=0
x+2y+z+1=0

[pastel swan]

Ok.

[merry python]

you should get unique values of x,y and z which must also satisfy the equation ax+y-z+2=0

[pastel swan]

Ok. Done.

Ty.

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[ruby nest]

Can anybody help?

[mild flower]

diagonalize it, or multiply by squaring

[ruby nest]

ok,ty

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[spark basalt]

How do I find the eigenvalue of this equation? 3=Acos(3sqrt(Lambda))-Bsin(3sqrt(Lambda)). If I only had one sin or cos I could set the inside of it = to n*pi and solve to see how it satisfies the equation. However, with both cos and sin I am not sure how to continue. Any help is appreciated!

[bleak sapphire]

Are you trying to solve for lambda?

[spark basalt]

yes

[bleak sapphire]

You know how to do it with just sin or cos. Can you think of any ways to make the RHS into that form?

[spark basalt]

The only thing I can think of is to separate them like this?

not sure im doing trig identities right though, its not my strongsuit

[bleak sapphire]

Doesn't seem quite right. Suppose we can write the RHS as $R\sin(3\sqrt{\lambda}-\alpha)$. Can you expand and match coefficients to find $R$ and $\alpha$?

[twin meteor]

Wheeler

[spark basalt]

How do you get it in that form? And would we go about it like this?

[bleak sapphire]

What I mean is, if we want to get the RHS to only involve one of sine or cosine, we can assume that it takes the form above, with undetermined parameters R and alpha. Try using a trig identity to expand the expression, and compare with the original RHS.

[spark basalt]

I'm honestly not sure what you mean by that

I haven't used any trig identities since HS so this stuff kills me lol

[bleak sapphire]

You know the one with subtraction?

Hold on

[spark basalt]

that one has both sin and cos attached to each though? I don't see how that looks like the equation

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@spark basalt Has your question been resolved?

[vocal sleet]

@spark basalt Has your question been resolved?

[vocal sleet]

@spark basalt Has your question been resolved?

[vocal sleet]

@spark basalt Has your question been resolved?

[tawdry grove]

Try using the $\sin^2 + \cos^2 = 1$ identity instead.

[twin meteor]

chencking

[upper willow]

Why did blud mention me

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[fiery oriole]

how can I solve for A4 with the missing variables?

[livid tapir]

U don't I think, just assume it's arbitrary real x and then say what kind of matrix it is under that assumption

[crimson jetty]

You would case on x.

[fiery oriole]

hm.. I don't really think it falls under any of them for an arbitrary x