help-23

[lean otter]

u substitution

u/u+1

[dire crypt]

try doing it, you'll get an extra e^x term in numerator left

[lean otter]

yes

which is again

so overall u will (u)/(u+1)

after that rewrie u on the numerator of u+1-1

and distribute

[dire crypt]

oh ryt then +1 and -1 on numerator to get [1-(1/u+1)]

[lean otter]

ya prefer putting the bracket on the denominator

[normal hedge]

so u / u + 1 right

[lean otter]

yes

[lean otter]

u/(u+1) yes

$\int \frac{u}{u+1} du$

[flat frigate]

does calc even end

[lean otter]

😵

u can either do the longer way thats partial value decomposition

or be quick and rewrite u on the numerator and u+1-1 which is easier

[normal hedge]

wait im a little confused

[lean otter]

$\int \frac{u+1}{u+1} - \frac{1}{u+1}du$

[lean otter]

about the sub?

[flat frigate]

Alham

[lean otter]

split up fraction

Notice that if you combine fractions the 1's on top cancel

leaving you to what you had before

prefer not approach it that way

that makes it seems like a magical step

its better to show it as

$\int \frac{u}{u+1} du=\int \frac{(u+1)-1}{u+1} du = \int \frac{u+1}{u+1}-\frac{1}{u+1} du$

[flat frigate]

does calc even end

[normal hedge]

ok im not just a little confused how does u / u + 1 turn to the second?

[lean otter]

ok

did u take sub u=e^x?

and see what happens?

[normal hedge]

you mean did i substitute?

[lean otter]

yes

[normal hedge]

e^2x / e^2x + 1 du right

[lean otter]

no

$\int \frac{e^{2x}}{e^x+1} dx$

[flat frigate]

does calc even end

[lean otter]

taking sub $u=e^x$

[flat frigate]

does calc even end

[lean otter]

$dx=\frac{du}{u}$

[flat frigate]

does calc even end

[lean otter]

so $= \int \frac{u^2}{u+1} \frac{du}{u}=\int \frac{u}{u+1} du$

[normal hedge]

u^2?

[flat frigate]

does calc even end

[lean otter]

coz $e^{2x}=(e^x)^2=u^2$

[flat frigate]

does calc even end

[normal hedge]

ohhh ok i forgot about that

[lean otter]

do you understand whats going on now?

[normal hedge]

yeah it took me some time but yess

[lean otter]

nice

[normal hedge]

im kinda slow butyeah gotta try

[lean otter]

can u see what to do from here?

[normal hedge]

ummm not really

do i like integrate it by here

[lean otter]

yes

its easy integral

from here

[normal hedge]

Is it ln | u + 1 | + c?

[lean otter]

no

make sure u didnt missed the 1 in the integrand

[normal hedge]

u-ln | u + 1 | + c?

[lean otter]

correct

now just resubsitute

u=e^x

[normal hedge]

so id have e^x - ln | e^x + 1 | + c correct?

[lean otter]

yes

[normal hedge]

ohhh

so thats final

[lean otter]

u can remove the | bars

if x is an real number

[normal hedge]

absolute value?

[lean otter]

coz e^x is always positive then

[lean otter]

yes ln | |

[normal hedge]

ohhh

i was never really taught what that meant just directlyh used it lmao

[lean otter]

oh

[normal hedge]

yeah it was just like we saw bars

and it was said ti be absolute value

so we just did what we did

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[steel osprey]

There are 25 classmates in a class. The next year there are 28 classmates (2 of them new girls and one boy) with proposition 4:3. How much boys and girls will there be if they were 25.

[median comet]

x+y=25

4x+3y=28

now solve for x and y

ig

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[hollow siren]

I tried to input these values and they are not right. I also tried to plug in .1, .01, and .001 and that was also incorrect

When it says at t=1 and lasting that makes me think it’s 1.1, 1.01, 1.001

Oh wait do I need to take derivative of the height in the air

Ok I did that and the last 2 answers are correct but the first 2 are wrong lol

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[safe radish]

Please don't occupy multiple help channels.

[lean otter]

pls

help

[worthy hemlock]

As the bot stated, don't open multiple channels

[lean otter]

ik im sorry

i didnt know before

could u pls help though

srry

[worthy hemlock]

So then close the other channels you have open

[lean otter]

how

[worthy hemlock]

.close

[lean otter]

okay done

could you help me now

pls

[worthy hemlock]

Do you know what a linear pair means?

[lean otter]

yes

they form a line

so 180 degrees right

its only been 2 weeks of school since we learned geometry adn hes alr quizing us on it

[worthy hemlock]

The second sentence, can you create an equation from that?

[lean otter]

uh yes

<1 + <2 = 180

@worthy hemlock

[worthy hemlock]

That's the first sentence

[lean otter]

oh

yes

<2= <1 x 2 + 6

[worthy hemlock]

Good

You have 2 equations now

[worthy hemlock]

This one

[lean otter]

now what should i do

[worthy hemlock]

And this

[lean otter]

oh wait im sped

i know

<1=180-<2

then i ca plug it in

[worthy hemlock]

Yes

[lean otter]

i got it

122

thank you

.close

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[violet brook]

what would this be

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[vivid ginkgo]

im not sure if im right on the first one

and on the second one how would i calculate that?

like i think the answer to the second one is 5

also

but not sure

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[vivid ginkgo]

.close

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[eternal mica]

I just need a bit of help understanding this formula. I have the other slides its from MIT.

starting from slide 13

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@eternal mica Has your question been resolved?

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@eternal mica Has your question been resolved?

[eternal mica]

<@&286206848099549185>

Plz

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@eternal mica Has your question been resolved?

[rigid inlet]

what about it do you not understand?

[eternal mica]

The numernator and denominator

[rigid inlet]

What about them

[eternal mica]

I dont unerstand them

What do they repersent?

[rigid inlet]

CF_n represents the cash flow at time n

r is the interest rate

[eternal mica]

what is v_0, what is (1 + r)

[rigid inlet]

V_0 is the present value of all cash flows at time 0

1+r is 1+r

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[terse ether]

How many even numbers greater than 300 can be formed from 1,2,3,4,5

[upper badge]

attempt?

[bright wharf]

5 is ه?

[terse ether]

Yes my bad I did an Arabic 5

[peak estuary]

what are you allowed to do with the numbers?

[terse ether]

What do you mean

Form numbers

Put them together

No operations

[peak estuary]

so just put them next to each other? or are you allowed to add them or something?

ok

well lets start with three digits numbers

what can the last digit be

[terse ether]

I dunno, you can only use 3 numbers in the hundredth, 4 numbers in the tenths, and 2 in the units, so 3*4*2 was my attempt

[terse ether]

554

[terse ether]

3x4x5* discord

[peak estuary]

\* for *

[terse ether]

Ah got it

[peak estuary]

can you repeat the numbers?

[terse ether]

Its asking for both cases where you can repeat and not repeat the numbers

[peak estuary]

why do you say there are only 4 options for the tenths?

[upper badge]

well, for 3 digit numbers, your first digit should be ≥ 3 and your subsequent digits can be anything

[peak estuary]

ok lets do repeating first then

[upper badge]

solving for repetition case first

[terse ether]

Ok so if each set is like the numbers we can use in the digit's respective places then is it like {3,4,5}{1,2,3,4,5}{2,4}

3,4,5 in the hundredths, 1,2,3,4,5 in the tenths, 2,4 in the units

[upper badge]

why only 2,4 in the units?

[terse ether]

So that it's even

Or is that not how it works

[peak estuary]

yes

so how many options?

[terse ether]

3x5x2?

[upper badge]

sorry I didnt read the question properly yea mb

[peak estuary]

yes

[terse ether]

30

[peak estuary]

ok. 4 digit numbers next

[terse ether]

...oh shit I forgot it doesn't have to be 3 digits only

Wait what prevents me from going up to like 555554 then

Like I could just go up to infinity

[peak estuary]

yes

and also 5555555555555555555555555555554 or something

so we can form infinitely many numbers. but it's still interesting to know how many numbers with a certain number of digits we can form

[terse ether]

But what number of digits do we stop at

Huh

Maybe it's implied that since it gave 5 digits it only wants up to 5 digit long numbers

[peak estuary]

well I don't know the original phrasing

later when we don't allow repeating digits we won't have this problem

[terse ether]

If we have 1,2,3,4,5 find how many even number greater than 300 can be formed from these digits, if the digits:
Can be repeated
Can not be repeated

Yeah ig it just goes up to 5 digits only

Ok so 3 digits is 30 possible numbers

4 digits would be 3x5x5x2

5 would be 3x5x5x5x2

Combine these sum up to 930

But the answer is 1530

[peak estuary]

no

we now have more options for the first digit

[terse ether]

Ahh

[peak estuary]

1112 is already bigger than 300

[terse ether]

Yeah got it thanks, I will try quickly doing the non repeating one

3 digits would 3x3x2
4 digits 4x3x2x2
5 digits 4x3x2x1x2
?

[peak estuary]

why?

remember that if you for example started the number with a 4, you can't also end it with a 4

[terse ether]

Idk how to calculate with this

[terse ether]

Doesn't that justify using 4 possible numbers instead of 5 at the first digit

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@terse ether Has your question been resolved?

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[earnest trout]

I need help

[weary goblet]

do the : mean division in this context?

[worthy hemlock]

$x:4 \equiv \frac{x}{4}$?

[flat frigate]

dldh06

[thick sparrow]

Ratio can become fraction, so your problem easily becomes solving for

$\frac{2 + x}{4 + x} = \frac{6 + x}{9 + x}$

[flat frigate]

jimmy1234

[earnest trout]

Idk

It gave that question

[worthy hemlock]

This could be wrong

[earnest trout]

But I don’t think so

[worthy hemlock]

Could be $2 + \frac{x}{4} + x$

[flat frigate]

dldh06

[earnest trout]

Ye

[worthy hemlock]

Because no parentheses were given

I suggest asking your teacher for confirmation

[earnest trout]

Ok

Thx

.closed

[thick sparrow]

In math context, one seldom mix ratio colon in the middle of arithmetic operation I guess…

[earnest trout]

Ye

.close

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[dire crypt]

f(c+h)-f(c) was already tending to 0

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[proven falcon]

hey guys could someone help

im pretty sure my amplitude is right

i did 6 -(-2) to get 8

then divided by 2

the max / min values

and got 4

but i dont know how to get equation of central axis

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[proven falcon]

<@&286206848099549185>

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[fringe oar]

I need help with tips on what to think about with this one

[analog flare]

About what?

[fringe oar]

i needed to trim the pic

[velvet monolith]

okay

[analog flare]

What does the question ask for?

[fringe oar]

"calculate"

[analog flare]

Solve for x ig

[velvet monolith]

its would be the best if you start with $\sqrt{\frac{a}{b}}$

[flat frigate]

kjmkty

[velvet monolith]

do you know how you can write that using exponents

[fringe oar]

like 1/2 as exponent?

[velvet monolith]

yess

[fringe oar]

but what with the multiplication?

in the left side

[analog flare]

$\frac{b}{a}$

[flat frigate]

Pluton

[analog flare]

Write this in (a/b)^n where n is your exponent

[fringe oar]

aahh

[analog flare]

Is that aahh i get it or aahh i dont get it

[fringe oar]

but after (a/b)^(1/2)

?

[velvet monolith]

‚think of this first

[velvet monolith]

we will come back to this later

[fringe oar]

?

i only wrote (a/b)^(1/2)

for now

[velvet monolith]

$\frac{b}{a}= (\frac{a}{b})^n$

[flat frigate]

kjmkty

[velvet monolith]

what would be n

[fringe oar]

lemme think

[velvet monolith]

basically this is what pluton said

[velvet monolith]

take your time

[analog flare]

$$\sqrt{\frac{a}{b} * \sqrt{ \frac{b}{a} * \left(\frac{a}{b}\right)^{\frac{1}{2}}}}$$

[flat frigate]

Pluton

[analog flare]

This is what you got now on left side

[fringe oar]

yes thank you

[analog flare]

Ignore the (a/b)^1/2 for now

[fringe oar]

yep

[velvet monolith]

you should only work b/a out

for now

[fringe oar]

is n= -1?

[analog flare]

Yes

Now you can multiply them

[fringe oar]

is that a mathematic formula?

cause it is hard to translate from the swedish terms to english

[analog flare]

Pretend a/b is x

Whats

$$x^{-1} * x^{\frac{1}{2}}$$

[flat frigate]

Pluton

[fringe oar]

x^(-0,5)?

[analog flare]

Yes

$$\sqrt{\frac{a}{b} * \sqrt{\left(\frac{a}{b}\right)^{-0.5}}}$$

[flat frigate]

Pluton

[analog flare]

This is your left side now

[fringe oar]

yes

so a/b^(1/2) times bla the right

and then 1/2+-0.5?

exponents

or wait

[analog flare]

Do it step by step

[fringe oar]

but if i convert the root to an exponent do i add to the already existing exponent?

[analog flare]

No you multiply it

[fringe oar]

okay

would -0,5 be -1/2?

[analog flare]

Yes

[fringe oar]

so -1/2 x 1/2 which is -1/4?

= -0,25

and then we got (a/b)^(1/2) x (a/b)^(-1/4)

which gives us 1/4 as x?

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[fringe oar]

.close

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[limber birch]

Hi, quick math question. To settle an argument with coworkers.

We have a snail with a speed of 10cm on day one. On the second day the speed doubles, then the next day it doubles again.
How many days would we need (with decimals) to get to 700cm?

[cedar rover]

cm is not a unit of speed, do you mean cm/s? or is that length?
The equation is 10 · (2)^([Number of days - 1]), I'll call "x" the [Number of days - 1], and we get
10 · 2^x = 700 or 2^x = 70
And the result is about 7.13 days

[limber birch]

If it was a speed it would be 10cm/day

[cedar rover]

Oh ok, still the result shouldn't change

Oh my bad, I made a little mistake

[limber birch]

Ooh, thanks, I what's the mistake?

[cedar rover]

Day Speed
1 10
2 20
3 40
4 80
5 160
6 320
7 640

I fixed it

I was considering the first day as day 0, I forgot we were counting it as day 1

[limber birch]

OOh, but wouldn't the previous days add up to it being under 7 days total?

[cedar rover]

It depends on how you are counting days
If you don't consider the first day, it's 6.13, if you do, it's 7.13

[limber birch]

At day 6 it should have a speed of 320cm/day and before it reaches day 7 it should have walked 6 something, I bet 6.13 is the correct

[cedar rover]

If you count days like that, it gets to 700 a little after day 7, so it's 7.13

If you count days like in the photo, at day 6.13 it's going at ~350 cm/day

[limber birch]

Ooh,

I get it thanks!

[cedar rover]

If you have more questions feel free to ask, someone will answer, but now I have to go

Bye!

[limber birch]

Thank you!

.close

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[oblique wigeon]

Two circles S and T meet at two points A and B. Pick any point P on AB. Let CD be any chord of S passing through P and EF be any chord of T passing through P. Show that CEDF are concyclic

[oblique wigeon]

It seems to be the equivalent of proving one set of concyclic points from two others.

not sure how to proceed

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[oblique wigeon]

<@&286206848099549185>

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@oblique wigeon Has your question been resolved?

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[livid sparrow]

Quick question.

Is the sqrt(12)/4 the same as sqrt(3) ?

[livid sparrow]

Or, is sqrt(12)/4 == sqrt(3)*2 ?

[ornate acorn]

$\frac{\sqrt(12)}{4} = \frac{2\sqrt(3)}{4} = \frac{\sqrt(3)}{2}$

[flat frigate]

clip

[livid sparrow]

How come sqrt(3) multiplied by 2?

If there's an explanation

[ornate acorn]

$\sqrt(12) = \sqrt(4*3) = \sqrt(4)\sqrt(3) = 2\sqrt(3)$

[flat frigate]

clip

[livid sparrow]

Interesting. Thank you 🙂

Also what do you suggest I type into google to find more information on this?

[ornate acorn]

https://www.mathwords.com/s/square_root_rules.htm

[livid sparrow]

Ty! You are a very helpful person

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[open urchin]

how to solve

[open urchin]

?

[dense tiger]

Almost

Make a formulae for Celsius first with variables

F=2C+30

We subtract the 30

F-30=2C

Isolate the c

F/2-30/2

F/2-15=C

[open urchin]

tysm ur the best let me solve

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@open urchin Has your question been resolved?

[open urchin]

not yet

[open urchin]

@dense tiger

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[strong parcel]

Help with top one please

[pure agate]

SR + RI = SI

All you need to do is substitute the equations given to you and solve for x

[strong parcel]

Yeah I figured that out it’s just solving for x

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[trim birch]

why is this the case?

[trim birch]

from what i know, in order for two vectors to be perpendicular, the dot product has to be 0

in this case wouldnt the dot product be (1 x 1) + (-1 x -1) + (3 x 3) = 11?

this is the original question btw:

<@&286206848099549185>

pls halp :/

oh wait what does normal mean in this case

[pure agate]

It's looking for any vector that is normal to the plane, it just so happens that a plane given by $Ax + By + Cz = D$ will have have a normal vector that is a scalar of $<A,B,C>$. To find a parametric that goes through a given point, $P$, you just add a scalar of the normal vector to $P$.

[flat frigate]

Kookiemon

[trim birch]

what does normal mean in this scenario

does it j mean the vector w/o the = D basically

[pure agate]

A line can have a perpendicular line in 2D space, but in 3D space, normal specifically means perpendicular to every point on the plane.

[trim birch]

hm

sorry still a lil confused tho

so w the whole dot product = 0 means perpendicular thing

when does that apply

[pure agate]

When you are determining if two vectors are perpendicular.

But this question is about a plane and a vector.

[trim birch]

oh

a plane is like

a plane is like a window right

kind of

like it has area to it rather than it j being a line

[pure agate]

Yes.

[trim birch]

oh but in this case it would be volume not area correct, since we're in 3d

[pure agate]

https://www.geogebra.org/classic/vesgqnnw

Mess around with that to see what I mean.

[trim birch]

thanks

[trim birch]

but j to check this is correct correct

[pure agate]

No, the plane would be area.

[trim birch]

wahhhh

i thought that area of the plane would be the volume since its 3d

i.e. its not just x and y with a rectangle but instead x y z, making a rectangular prism

[pure agate]

Nope, here you are defining an area in a 3-dimensional space.

You can orient a piece of a paper any way you want and it will not have any volume per se.

[trim birch]

ohhh

so a plane is basically kind of like a rectangle in 3d terms

[pure agate]

Correct.

[trim birch]

hmm

so a rectangle with x, y, and z coordinates?

j tryna wrap my head around it sorry

[pure agate]

Yes.

[trim birch]

hmmm i should go back to the basics

[pure agate]

Any you can see that in the link I sent.

[trim birch]

so j to check one last time

planes have no volume correct

its j a flat thingie majiggie with x y and z coordinates?

[pure agate]

Yes.

[trim birch]

gotcha thanks for the help

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[stray urchin]

Is this correct? Are there any other methods I can do?

[stray urchin]

<@&286206848099549185>

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[still charm]

Not really

Looks right, not much else to it

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[hoary birch]

Hi is anyone able to help me with questions 2.2 and 2.3?

[safe radish]

@hoary birch Has your question been resolved?

[hoary birch]

<@&286206848099549185>

[safe radish]

@hoary birch Has your question been resolved?

[hoary birch]

is anyone able to help still? 😭

[cedar rover]

Have you tried sketching the region R, to see what shape it has?

[hoary birch]

hmm i did that its a semicircle right?

a quarter of one

wait im literally so dumb

im so confused i might need help with 2.1 as well

[cedar rover]

Nvm, the idea I had is bad
Yes it's a quarter of a circle

[hoary birch]

okok

i have a question @cedar rover. for 2.1 would rotating R about x =4 make a sort of dome ahh i cant explain it

[cedar rover]

Kinda, it's a weird shape

[hoary birch]

yeah

like it goes in (or down) towards the middle?

[cedar rover]

I might have a solution, but it's weird and there might be better ways

[hoary birch]

ok sure! anything sounds great at this point

[cedar rover]

The quarter-of-a-circle part is R
You could find the volume of the cylinder rotated around the given axis with r = 1 and h = 3 to get the volume of the shape R from
x = 3 to x = 4 with the additional little light blue part (You could find it's volume and then subtract it from the whole cylinder), then ignore the part from x = 4, to x = 5 (the red part) because that part is already inside the previous part when rotating the shape, and then you could find the volume of the yellow part rotated around the given axis (notice that there's a hole in the middle of that)
and then sum all the needed parts of R

This may be overkill

[hoary birch]

what do you mean by "(notice that there's a hole in the middle of that)"

[cedar rover]

I mean, like a donut

[hoary birch]

where is the hole?

[cedar rover]

seen from "above", when the yellow part (only) is rotated, it looks something like this

it doesn't touch the rotation axis

[hoary birch]

Yes

I get that

What about the part from x =2 to 3

[cedar rover]

That's not in R, x > 3

[hoary birch]

Why what

[cedar rover]

Look at 2.1

This is what the shape R looks like

[hoary birch]

Mhm

why is the red part inside the previous part?

[cedar rover]

Imagine rotating the part from x = 3 to x = 4, you'll notice the whole part from x = 4 to x = 5 is inside of it, so it's volume doesn't matter because that volume is already calculated in the part from x = 3 to x = 4

[hoary birch]

oh

i see

so its basically the volume of x=3 to x=4 + volume of x=5 to x=6?

[cedar rover]

Yes I think so

this is a better graph

Don't forget that little light blue part on top of the part from x = 3 to x = 4

[hoary birch]

yeah

[hoary birch]

what do i do about the little blue part

[cedar rover]

(insert bad idea here)

[hoary birch]

mhm

[cedar rover]

But if you don't like this you can still and you want to stick to the original plan, you can still find the volume of that little light blue

[hoary birch]

wait

are you sure thats true

idk maybe im dumb

[cedar rover]

oh actually not, that's sus you are right

[hoary birch]

okokok

so how do you actually do reimanns sum

like the method

ngl my professor posted solutions they just make no sense to me

[cedar rover]

Oh wait the previous idea worked, the one I deleted

the only thing is that the axis should be x = 1, not the y axis

[hoary birch]

yeah i thought so

[hoary birch]

btw i think its meant to be the other side of the semi circle?

idk maybe im confused

[cedar rover]

I'm not sure what they did in the very last image, it feels like they didn't consider the part from x = 3 to x = 4, but I'm probably wrong

Anyways I'm sorry but now I have to go, probably someone will help

[hoary birch]

thanks for your help anyways! it helped me get a better idea of how to use riemann sum.

here is the question if anyone is able to help!

[hoary birch]

<@&286206848099549185>

[safe radish]

@hoary birch Has your question been resolved?

[hoary birch]

Anyone 🥲

[lean otter]

This figure is wrong I think

Because the question says x<= 3

[cedar rover]

I'm dumb

Good catch

[lean otter]

Tess has drawn the correct one nevertheless

[lean otter]

oh lol it's your professor's

[hoary birch]

Wait i literally said that too

@lean otter please help im scared for my exam tmr im gonna cry

[lean otter]

uhh so do you understand your prof's solution of part 1 and 2 ?

[hoary birch]

Yes

Let me recheck to be surr

[lean otter]

alright

[hoary birch]

Part 2 doesnt make sensr

[lean otter]

can you visualise the shape when the figure is rotated about x=4 ?

[hoary birch]

Its a half donut?

[lean otter]

not quite donut

it has a variable radius

[hoary birch]

Like half a sphere but a cylinder is cut from centre?

[lean otter]

yes that

[hoary birch]

What does that mean

[lean otter]

wait not radius exactly but the distance of the boundary points from the axis changes as you go up and down the axis

[hoary birch]

yes i get that now

can you explain reimanns sum and part 3? @lean otter

[lean otter]

it's long, but we have a reference so I would say got through each step of your prof's solution and tell me which step you don't understand

[hoary birch]

ok

[hoary birch]

i dont understand the first step :/

[lean otter]

small radius =1 ?

see the diagram of the partitioned ring

[hoary birch]

yes

[lean otter]

the circle in between has radius 1

[hoary birch]

oh

oh

oh

what about the big radius

where does xi come from

[lean otter]

ok so the figure has been partitioned into horizontal rings, and i told you earlier that the distance of the boundary points from the axis is variable and thus

the rings are going to have variable radius

[hoary birch]

right

so thats why its xi

that makes sense

why does she write the Radius in terms of yi

[lean otter]

because we need one variable only for integration or Riemann sum

the thickness will be in terms of y

so the Area should too be

[hoary birch]

oh

so whats the point of reimanns sum why cant i just use the volumes of revolution formula instead

[lean otter]

I don't know what you exactly mean by volume of revolution formula

[hoary birch]

that one

[lean otter]

this is correct for an individual ring like structure

i think

it does not take account of the variable radius

in terms of y

[hoary birch]

ohhh

okok

i think im starting to understand all this

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[kind hearth]

if im asked to make an array of 8 elements, can i make a column or does it have to be a row?

[lean otter]

it doesn't matter

you can view it as both

[kind hearth]

ok cheers

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[misty belfry]

Does the statement i did in square acrually just turn to t?

[misty belfry]

I know its absorption law but can it work like this?

Because absorption law obly has 2 variables here

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[oblique mountain]

please help me i dont quite understand polynomials:<

1).64d³+f³=(4d+_)
(16d²-_f³)

2).54x³y+2y=2y
(3+)(9x²-3x+)

help me factor and supply the missing terms.ty

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[ionic bison]

Can someone please help me with my complex numbers homework
Find "z"
Re(z+1)=5, Im(z*) = -3
Thank you

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[stiff inlet]

Yo

[stiff inlet]

The value of B should be something like 5.91 right?

But doesn’t that seem strange? It should be a whole number?

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[mortal sandal]

Pythagorean theorem if you want 5.91 is right

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[lethal ravine]

I need help with this one

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[eternal gazelle]

isnt that just the definition of derivative

[lethal ravine]

I’m not sure

I believe it’s asking for the derivative

[lethal ravine]

Yea

Finding a derivative using limits

[eternal gazelle]

ya so isnt it just asking derivative of tan(3pi/4)

[lethal ravine]

Yea

Derivative of the whole thing

My main problem is getting the h out of the tangent arguement

[eternal gazelle]

wait but why would you find the derivative using limits

its derivative of tan(x), and then u evaluate it at 3pi/4 so sec^2(3pi/4)

right? tan(3pi/4) isnt a function bruh its just a value

@lethal ravine

[lethal ravine]

Yea

[eternal gazelle]

yeah its just that

[lethal ravine]

It’s just sec^2(3pi/4)

[eternal gazelle]

yeah

[lethal ravine]

Why though

[eternal gazelle]

cuz if you use the definition of derivative, the expression is literally just asking you what's the derivative of tan(x)

then it asks you to evaluate that when x = 3pi/4

[lethal ravine]

I think I am understanding it now

[eternal gazelle]

f(x) = tan(x)
f'(x) = sec^2(x)

x in this problem is 3pi/4

so just evaluate f'(3pi/4) = sec2(x)

[lethal ravine]

Yea

So is it just 4

That’s what I got

[eternal gazelle]

i think u did a calculation error

[lethal ravine]

Sec of 3pi/4 is -root2 right

[eternal gazelle]

no

its -root2/2

i mean

uh

cos

wait

what

im tripping

you're right

[lethal ravine]

Yea

[eternal gazelle]

but -root2 *- root2 doesnt give 4

[lethal ravine]

2root2 right

Not that it’s just 2

[eternal gazelle]

ya

cuz idk about u but when i calculate sec(x) i just do 1/cos(x) that's why i got confused

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[novel fox]

Pre-calc im stuck on, I just dont understand these concepts

[dense lynx]

All you need to do is to look at the table

[cedar rover]

f(g(4)) means to plug g(4) into f(x), so see what g(4) is and plug that value into f(x)

[dense lynx]

Its pretty basic you might want to re-learn that, or you'll struggle

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[iron sundial]

how do I do this

[dense lynx]

Find the value of x first

[iron sundial]

how?

I uh didn't Pay attention in class

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[daring kelp]

How do I factor this $1 + 2x^-^1 +x^-^2$

[flat frigate]

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

[cedar rover]

^{-1} is the way to not get any compilation errors from the bot

anyways, substitute something like t = x^-1 and then express that expression in terms of t instead of x

then just factor it normally

and then resubstitute t = x^-1 to get back to xs

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[eternal gazelle]

(1/x +1)^2

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[small mortar]

hi i just wanted to confirm these modular arithmetic properties. are they correct? i haven't taken a proper course, i just learned it gradually over time so i just wanted to make sure my knowledge is correct

in the expression a ≡ b (mod c), c can never be 0.
a ≡ a (mod b) for any a, b E N
a ≡ b (mod c) <=> a + nc ≡ b (mod c) <=> a ≡ b + nc (mod c) for any n E Z
a ≡ b (mod c) <=> a + n ≡ b + n (mod c) <=> an ≡ bn (mod c) for any n E Z
a ≡ b (mod c) <=> a^n ≡ b^n (mod c) for any n E N

[hardy lion]

Technically, (you have to be a little careful with this) you can also have n be any real number for the additive and multiplicative ones

Afaict the statements i know are all true except maybe the first one

[small mortar]

what are the technicalities though?

[small mortar]

but you can't have mod 0 though?

[hardy lion]

Im not sure

[small mortar]

it's like dividing by 0

[hardy lion]

Technicality: 6 cong 2 (mod4)
6/2 !cong 2/2 (mod4)

If the modulus and the denominator are coprime you can get away with it

[small mortar]

which number is the modulus?

[hardy lion]

(mod m) m is the modulus

[small mortar]

the mod 4?

ahh alright

so like

it doesn't really make sense in this case though
12 ≡ 5 (mod 7)
12/2 ≡ 5/2 (mod 7)
6 ≡ 2.5 (mod 7)

[solar hazel]

ummm what do you mean by the additive and multiplicative ones?

[hardy lion]

Statement 3

[hardy lion]

Oh right i believe this only works for integer cases

[solar hazel]

a ≡ b (mod c) => an ≡ bn (mod c) ?

[hardy lion]

Yes

[solar hazel]

so...
3 \equiv 8 mod 5 implies 1.5 \equiv 4 mod 5?

with n = 1/2

[hardy lion]

Im a little shaky when it comes to involving fractions

[small mortar]

it states at the end for any n E Z

[hardy lion]

But for integer cases, the statements are definitely true

[solar hazel]

I know I was just commenting on this

[hardy lion]

I know that

[small mortar]

fractions become too case by case to deal with. it's annoying

[solar hazel]

I would just... not think about non-integers

[hardy lion]

Fair enough

[small mortar]

yeah exactly

so is everything okay then?

[hardy lion]

Think so

[small mortar]

alright, thanks

[solar hazel]

uhh no

[small mortar]

hmm?

[solar hazel]

a ≡ b (mod c) <=> a + n ≡ b + n (mod c) <=> an ≡ bn (mod c) for any n E Z

[small mortar]

yeah what's wrong with that

[solar hazel]

an ≡ bn (mod c) => a ≡ b (mod c) isn't true in general

[small mortar]

give me an example

[solar hazel]

$2\times 3\equiv 2\times 2\pmod{2}$ but $3\not\equiv 2\pmod{2}$

[flat frigate]

layla💜

[small mortar]

ah right, it only works when n is coprime with the modulus

[solar hazel]

yes

[small mortar]

thankss

[solar hazel]

or at least, it works when n is coprime with the modulus

[small mortar]

yeah, i'll take note of that

thank you very much

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[buoyant path]

Hi so ignore the writing underneath but I’ve tried working this out and I didn’t get 1296/(x+2)^3

[buoyant path]

I did this

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[lean otter]

how do i solve this

[safe radish]

@lean otter Has your question been resolved?

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[vapid knoll]

5 = Square root of n^2

[vapid knoll]

would this mean n = 10?

[foggy salmon]

n=5

[tacit fog]

sqrt(n^2)

or sqrt(n)^2

[oblique cloak]

• or - 5

[vapid knoll]

Oh okay

How can I solve this

one sec

11 B

How do I solve this

[dire crypt]

yh

[foggy salmon]

um i dont think its ever -5

[royal ivy]

could be

[tacit fog]

right

[vapid knoll]

^

11 B please

[dire crypt]

why? sqrtx^2 is |x|

[vapid knoll]

...

[dire crypt]

use distance formula
its not ur channel

#❓how-to-get-help

[tacit fog]

it is

his channel

[royal ivy]

can you express the distance between (-2,2) and (4,n)

[dire crypt]

oh

lmao

[foggy salmon]

oh mb

[vapid knoll]

Perhaps using 10 = (-2,2)^2 + (4,n)^2?

[foggy salmon]

i meant in the case the .. person who deleted the msg said

[tacit fog]

me?

[royal ivy]

? what's the square of a vector?

[foggy salmon]

idk, it was (sqrtn)^2

[vapid knoll]

Idk

I am dead lost at point 0

[royal ivy]

Well somewhere in your lesson there should be a formula for the distance between 2 points

[vapid knoll]

d=√((x_2-x_1)²+(y_2-y_1)²)

this is right?

[royal ivy]

Yes

so go for it, input (-2,2) and (4,n) in there

[dire crypt]

he wrote Square root of n^2

[foggy salmon]

zz doesnt matter go back to the Q

[vapid knoll]

(-2,2) (4,n)

10 = (4+2)^2 + (n+2)^2

[royal ivy]

don't forget the square root

[vapid knoll]

oh yea

my bad

[royal ivy]

also, it should be n-2

not n+2

But we're getting there

10=√((4+2)²+(n-2)²)

[vapid knoll]

okay thanks

do I then evaluate

[royal ivy]

Yeah do some simplification

develop (n-2)², etc.

[vapid knoll]

okay ty

wait

I am lost

I evaluated

and got

10 = √ 16 + 4 + n^2 - 4

[royal ivy]

Not quite

[vapid knoll]

why?

[royal ivy]

You're at least missing an n

[vapid knoll]

an N?

I wrote n^2 tho

[royal ivy]

yeah, you should have a term in n² and a term in n

(4+2)²=36

[vapid knoll]

OHHH

[royal ivy]

(n-2)²=n²-4n+4

[vapid knoll]

why -4n?

[royal ivy]

What's (a-b)²?

[vapid knoll]

a^2-b^2

[royal ivy]

not quite

[vapid knoll]

bruh whats so funny

[royal ivy]

let's see

what's (2-1)²?

Well it's (1)² so it's 1

And what's 2²-1²?

Well it's 4-1=3

[vapid knoll]

yes

[royal ivy]

So (a-b)² is NOT a²-b²

It's a²-2ab+b²

you can prove it by writing (a-b)(a-b)=a(a-b)-b(a-b)=a²-ab-ba+b²

[vapid knoll]

how can you evaluate n^2- 4n?

[royal ivy]

You want to solve for n

[vapid knoll]

yes

[royal ivy]

So, what do we have now

[vapid knoll]

10 = / 36 + N^2 - 4n + 4

10 = / 40 + N^2 - 4n

[royal ivy]

yes

[vapid knoll]

then what

[royal ivy]

ok, let's square both sides, this square root is annoying

[vapid knoll]

ight bet cuh

100 = 40 + n^2 - 4n

[royal ivy]

yes

[vapid knoll]

60 = n^2 - 4n

then what

[royal ivy]

rearrange a bit, we get 0=n²-4n-60

that's a degree 2 equation

[vapid knoll]

what the hell

alright what next

[royal ivy]

well, you solve this equation

[vapid knoll]

but idk what n^2 - 4n =

[royal ivy]

You want to find n

[vapid knoll]

ye

[royal ivy]

do you know how to solve equation of the form ax²+bx+c=0?

[vapid knoll]

no that is quadratics

[royal ivy]

Yes

You should have somewhere in your lesson on quadratics a way to solve this kind of equations

[vapid knoll]

can u help me solve it

[vapid knoll]

im not doing this for school

[royal ivy]

You've never solved a quadratic in your life?

[vapid knoll]

never

[royal ivy]

Well I'll have to direct you to a class on quadratics then

I mean, there's a wikipedia page

https://en.wikipedia.org/wiki/Quadratic_formula

[vapid knoll]

😭

alright

this looks so complex

[bold ferry]

it’s not too bad once you learn it

[royal ivy]

There's a way to bypass this, but you should learn about all that anyway. I'll tell you a cheat method, that only works in this specific problem, not in general.

[vapid knoll]

Alright

[royal ivy]

You're looking for points which are at a distance 10 from (2,-2)

and which have first coordinate 4

Trace a circle of radius 10 around (-2,2) and a vertical line going through (4,0)

You can read the coordinate on the drawing.

You find (4,10) and (4,-6).

HOWEVER

It could be that the actual answer is something like (4,9.99999) and we're reading it slightly wrong

So, you need to input those two solutions into the distance formula (evaluate for n=10 and n=-6) to see whether they are the actual solution.

This works because in our case, they are nice integer solutions.

But if the solution was something like sqrt(2), you would read 1.4 or so on the drawing, and it will never be quite accurate.

So: if you think you know the solutions of a quadratic, you can just check that they work easily. If you have no clue what the solution is, you need the general formula.

[safe radish]

@vapid knoll Has your question been resolved?

[safe radish]

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[daring jay]

@peak estuary Hi!
I was wondering if we could pick up where we left off yesterday, like we agreed

[peak estuary]

sry, not now

later

[daring jay]

No problem!

.close

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[pseudo bridge]

is tan(x) an onto function from the reals to the reals?
becuase every element of the output set has a corresponding input

[pseudo bridge]

but the inputs arent unique

and also tan(x) is not defined at some points in the real input space ?

[thick sparrow]

No. $\tan(x)$ is onto from $(-\frac{\pi}{2}, \frac{\pi}{2})$ to $\mathbb R$, but NOT from $\mathbb R$ to $\mathbb R$.

Like you said, we need the function well-defined on its domain b4 we can even start talking about whether or not it is onto.

[flat frigate]

jimmy1234

[lean otter]

wtf

is this

[pseudo bridge]

ohh really

[lean otter]

im scared what grade is this...sorry for asking

[pseudo bridge]

what about from the set:
[-pi, pi] excluding pi/2 and -pi/2
to the reals ?

[thick sparrow]

Trigonometry (in radian) is probably high school pre-calculus algebra topic. Don’t worry if you have not reached to this level yet.

[thick sparrow]

I think it is still onto, but not 1-to-1 in this case.

[pseudo bridge]

alright, thankyou :)

[lean otter]

god damm im in high school first year

the more i look the more terrified it looks

[pseudo bridge]

.close

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[humble oracle]

Hello. What would the closed form for this function be?

[humble oracle]

n > 1 by the way

[unique bison]

the last line is not true

[humble oracle]

Can you elaborate?

[unique bison]

should be f(n-1) on the lhs

[lean otter]

seems correct to me

[humble oracle]

Can you explain how?

[unique bison]

$f(n) = \frac{f(1) + f(2) + \dotsb + f(n)}{n^2}$

[flat frigate]

giannis_money

[lean otter]

he subtracted f(n)

[unique bison]

um

[humble oracle]

This is fine but it’s recursive