help-13

[sharp bay]

Pls can I have some help. I've done part ai), idk what aii) is even asking

[bold hinge]

What are you confused on?

A \ {a,b} = {c,d}

It just means "the set A, without the elements a and b"

[sharp bay]

and what does the x mean?

[bold hinge]

Cartesian product

Every ordered pair that contains one element from the first set and one element from the second set

[sharp bay]

ok, lemme try figuring it out

just to double check there are 16 different sets in ai)?

[bold hinge]

Yes

2^4 = 16

[sharp bay]

cool

[bold hinge]

2⁴ = 16

[sharp bay]

does this mean without both elements a and b individually or when a and b are in the same set?

[bold hinge]

It means this

[sharp bay]

ok

is this correct so far?

[bold hinge]

Yep

[sharp bay]

is this correct?

[bold hinge]

No

One element from the first set, one element from the second set

in order

[sharp bay]

this better?

[bold hinge]

Yes!

[sharp bay]

cool, I was debating whether or not to include {c, d, a} or {c, d, a, d} for example

why do we not include them here

[bold hinge]

Because that's not how the Cartesian product works. You form ordered pairs using one element from the first set, then one element from the second set

[sharp bay]

ah ok

also is my notation correct here using ({}) instead of {{}}

[bold hinge]

The Cartesian product is itself a set

[sharp bay]

oh ok

so {{}} then

theres a third part to the question, I'll try to answer it first tho

is this a reasonable first line of working?

because a and d are already in A

[cedar kiln]

@sharp bay Has your question been resolved?

[cedar kiln]

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[steel sun]

z = 1+2i

[steel sun]

w = 3-i

Solve (z+w)*

[hard vigil]

add the 1 and the 3, add the 2i and -i

[steel sun]

.close

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[wanton harness]

(a^m)^n = a^(mn)

[floral arrow]

(i)^3 = ((i)^2)^1
That's just wrong

I don't understand your question then

[cedar kiln]

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[worthy basalt]

hey help

[worthy basalt]

so

1024, -516, +256, -128, ....+1

it's a geometric sequence

I'm trying to find what term is 1

how do I DO THAT

I did

(a)(r^n-1)=1

I know

r is -1/2

(1024)(-1/2^(n-1)=1

I did that and when I did log I found some very strange results...

can someone help me

[night forum]

yes

So

You know that

[worthy basalt]

I don't know like

(-1/2^n-1)=1/1024

I did that

[worthy basalt]

log(-1/2^n-1)=log(1/1024)

(n-1)log(1-/2)=log(1/1024

n-1=log(1/1024)/log(-1/2)

I TRY DOING IT IT DOESN'T WORK

[night forum]

Wait wait

with this writing it's confusing

Plug the numbers you know into the first equation I wrote

You get this

If you solve it correctly you get a very normal looking number for n

Oh wait -0.5 not 0.5

[worthy basalt]

yes

then What do I do

[night forum]

Welp my solution still holds because n-1 is even

So 0.5 and -0.5 in the power of n-1 will give the same result

[worthy basalt]

so what should I do ?

[night forum]

Solve this

[worthy basalt]

I don't know how to solve this
I logged both

nothing happens

[night forum]

Because you tried to solve it with -0.5

Try with 0.5

[worthy basalt]

But wouldn't that change the equation

[night forum]

It won't because n-1 is even

[worthy basalt]

how is it even

2-1=1

4-1=3

[night forum]

So it doesn't matter if r is positive or negative

Because I solved it brute force and I see that n=11

Which means that n-1=10

and 10 is even

[worthy basalt]

I got 11

which is correct

but why is it even

[night forum]

Yes

Because I saw the answer

I don't know if you can prove that

And if you can't

[worthy basalt]

then how am I supposed to solve it

[night forum]

Brute force is still a solution

[worthy basalt]

what's brute force

[night forum]

Not a "nice" one

[night forum]

Calculate the numbers yourself

Multiply the numbers by -0.5 until you get 1

And count how many are there

[worthy basalt]

ah

[night forum]

Log isn't defined for non-positive numbers so you can't do it this way

Wait

[worthy basalt]

ok

[night forum]

You have this

You can say

[worthy basalt]

ye

[night forum]

The answer is positive

Which means n-1 is even

Because if it was odd then the answer would be negative

And then you can remove the negative sign

And solve it with logs

[worthy basalt]

ok

[night forum]

@worthy basalt when your question is solved you should write .close

[cedar kiln]

@worthy basalt Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[grim nimbus]

hi

[grim nimbus]

<@&286206848099549185> someone can help me here pls i got exam tomorrow

[gilded mesa]

Add both

[grim nimbus]

ok i have already done it

pqr is the multiplication of p,q and r

[gilded mesa]

Q+r=19,you can use 2,17

[grim nimbus]

what do you mean

by 2,17

[gilded mesa]

You need 2 prime number sum 19, you can use 2,17

Q=2,r=17

P=31

[grim nimbus]

aha

ok thank u very much samurai

.close

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[outer oxide]

So the question is :
Prove that there is one solution for g(x)=0 and that solution is a and 2.1 < a < 2.2
g(x) = x^3 - 3x - 4 Dg = R

[sage topaz]

k

wait

yeah

The factors are (x-1)(x^2+x+4)=0

@outer oxide

[outer oxide]

from where did the x-1 come from

[crimson sedge]

This works but there's a simpler way

[outer oxide]

and that makes it more than one solution bc 1 is not ...

[crimson sedge]

Do you know about the intermediate value theorem?

[outer oxide]

no

[crimson sedge]

it doesnt work

[outer oxide]

can you give an ex?

[crimson sedge]

g(1) isnt 0

[crimson sedge]

If f is a continuous function and f(0)=0 and f(1)=1 then there exists x such that 0 < x < 1 and f(x)=0.5

Basically since the function is continuous (meaning no jumps), in order to go from 0 to 1 the function has to go through every value in between

Now note that since g is a polynomial it's continuous and try to calculate g(2.1) and g(2.2)

[outer oxide]

didn't get any of that

[crimson sedge]

ah

[outer oxide]

how is that going to help?

[crimson sedge]

hold up, I'm trying to think of a way to make this visually intuitive

@outer oxide Okay, so here we have a continuous function but I only drew 2 of the points of the function: (0,0) and (1,1). I also drew a line at y=0.5. Assuming f is continuous, is there a way for it to go from (0,0) to (1,1) without crossing the y=0.5 line?

[sage topaz]

HI

[outer oxide]

no?

[sage topaz]

See

[crimson sedge]

Exactly, it's not possible

The function must go through the y=0.5 line

[sage topaz]

you get this

[crimson sedge]

Otherwise it's not continuous

[outer oxide]

so this problem has no posibility?

[crimson sedge]

@sage topaz Your method is great but it's far from the easiest one

[crimson sedge]

No, it's a good thing that it's not possible

[sage topaz]

k

[outer oxide]

why?

[crimson sedge]

Here I drew a bit of one possible continuous function going through (0,0) and (1,1)

Notice how there's a point where it crosses y=0.5

This means there's some value of x from 0 to 1 where f(x)=0.5

@outer oxide Does that make sense?

[outer oxide]

yes

so i need to find g(x) = 2.15?

[crimson sedge]

No

It's much simpler than this

[outer oxide]

ok

[crimson sedge]

First compute g(2.1) and g(2.2) and I'll show you how this can be useful

[outer oxide]

-1.039 0.048

i think

[crimson sedge]

Looks about right

Well the key here is that one of them is positive and one of them is negative

[outer oxide]

ok

[crimson sedge]

,w plot (2.1,-1.039) and (2.2,0.048) and y=0

dangit

,w graph x^3-3x-4 from x=2 to x=2.3

[wraith dagger]

Results provided by WolframAlpha

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Requested by labyrinthsplash

[crimson sedge]

There

[outer oxide]

ok how do i find a??

[crimson sedge]

You don't

You don't need to

[outer oxide]

why?

[crimson sedge]

The question only asks you to prove that there's only one solution and that it's between 2.1 and 2.2

It doesn't ask you to find the solution

[outer oxide]

but i need a to calculate other stuff

[outer oxide]

it does

[crimson sedge]

Prove that there is one solution

and that solution is a and 2.1 < a < 2.2

I don't see anything that requires you to find a

The first one is just proving that there's only one solution

And the second is proving that that solution is between 2.1 and 2.2

[outer oxide]

but in other questions there is like find f(a)

[crimson sedge]

Really?

well shit

[outer oxide]

like prove f(a)=3/2a+2

[crimson sedge]

@outer oxide Wait, what is f?

[outer oxide]

it comes later

[crimson sedge]

How is it defined?

[outer oxide]

f(x)=(x^3 + 2x^2)/(x^2-1)

Df = R-{-1;1}

[crimson sedge]

@outer oxide Okay so I'm gonna be honest with you
I'm still confident IVT would be the way to go for your question if the next questions weren't there but with the next questions I'm not sure what the best approach to the current one is

Basically what I'm saying is I am no longer sure I can help

I put a message for other helpers to see

[outer oxide]

thanks very much

[crimson sedge]

@outer oxide I think you'd get help faster by opening a new help channel. First, close this one with .close, then open a new one with the problem at hand and the next questions too so that other helpers don't make the same mistake as me

[outer oxide]

ok

.close

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[full wind]

Helping figuring out how they got step 3 on this problem

[full wind]

Went over this in class, and am now unsure how we turned 3Sin into 3/2

[drowsy sage]

it has to do with x/y for std angles

x = cos, y = sin

y val for pi/6 is 1/2, so 3 sin pi/6 = 3/2

plug in and then basic add/subtract

[full wind]

ah I see, thanks.

/close

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[plush wharf]

What I do wrong in 3

[clear steppe]

You might have mixed range with domain. The range is all the values the function outputs, and here it's 0 and all positive values. So, [0, +inf)

[plush wharf]

👍

.close

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[bold hazel]

I'm trying to figure out how to write this as a recurrence relation

"//" is floor division in python, but is there a symbol for that in math?

maybe I would just write it as normal division

oh, the prof clarified and said normal division operator is ok

[cedar kiln]

@bold hazel Has your question been resolved?

[bold hazel]

if I consider x a constant...

then I get T(n)=T(n/k^2)^(k^2)*c

[cedar kiln]

@bold hazel Has your question been resolved?

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[quick flume]

why does it change to -cos

pi/3

what does happen from the last bracket

[brave aspen]

They rewrite cos(2pi/3) as the negative of cos(pi/3)

These both have the same reference angle except one gives a positive answer for cosine while the other is negative

[quick flume]

But in this table it's not negative

[wraith dagger]

woods

[brave aspen]

It's not really necessary for you to arrive at the answer in this case

You could just directly evaluate cos(2pi/3) and get the same thing

[quick flume]

If I have the table at disposal

can't I just write -1/2

I wasn't at the lessons so idk.

But thanks for clearing things up.

alright

.close

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[quartz mauve]

for this problem I used the mean value theorem split into three integrals

I thought I did everything right but -7 is clearly wrong, did I use MVT incorrectly?

[sharp pike]

Your 3rd integral is wrong

It's not -x

[sand cradle]

h(x) is wrong

[quartz mauve]

wait what is it

is it -x - 8?

+8*

I did it with h(x) = -x + 8 but this still seems wrong, now the answer is 11

Shouldn’t the answer be around 3

[livid hound]

-x + 8 is still wrong

[quartz mauve]

damn

what is it

[livid hound]

use something like point slope form

to get your equation

[quartz mauve]

so (13 - 12) = -1(3-4)

wait that’s wrong

would it be (3-4) = -1(13-12)?

[livid hound]

wdym by "it"

[quartz mauve]

h(x)

[livid hound]

no

that equation is true,
but doesn't contain x,y and isn't representative of an equation of a line

[quartz mauve]

so how do I do that

[livid hound]

point slope formula or otherwise

[quartz mauve]

how would you use point slope formula here

[livid hound]

identify the slope of the line
identify a point on the line

plug those into the point slope formula

[quartz mauve]

oh right

so it’d be 3-y = -1(13-x)

which is y = 16-x?

[livid hound]

dunno why you didn't use a simpler point like the x-intercept (16,0)
but yes that works

[quartz mauve]

which makes the final answer 3

I’m so used to using point slope to find ordered pairs that I forgot how to use it to find functions

thanks for the help

[livid hound]

for that interval yes, but you can't just add up the averages for each section

[quartz mauve]

why not?

[livid hound]

that's not how averages of a total work

[quartz mauve]

so how can I apply mvt over this interval if there are 3 different functions

[livid hound]

(total area)/(total time difference)

[quartz mauve]

so I add the 3 integrals, then I divide by delta t?

isn’t that what I’m already doing

[livid hound]

mvt not really relevant

no that's not what you're doing

you're calculating the averages of each then adding which isn't the same thing

[quartz mauve]

well yeah but the averages of each are the integrals across their respective intervals and then divided by delta t

that’s the same thing as finding the total area and then dividing by delta t

[livid hound]

no

[quartz mauve]

why not

[livid hound]

consider travelling at an average speed of 20kph (for an unspecified amount of time
then 40kph (for an unspecified amount of time)
the average speed isn't 60kph

[quartz mauve]

I don’t see how that’s relevant to what I said above

[livid hound]

you can't add individual averages to get the total average

the area is just the integral, no division

[quartz mauve]

I don’t see how adding three integrals and then dividing by delta is different than dividing each integral by delta t before adding them

[livid hound]

you shouldn't be multipl lying those by 1/8, 1/4, 1/2

deltas are different for each one

and also you're first integral should be starting at 1 not 0

[quartz mauve]

oh right I misread that

so the answer would be 38/15?

[livid hound]

no

how are you getting that

[quartz mauve]

I multiplied each average by the reciprocal of delta t, added them, and divided by delta t

[livid hound]

can you show what youre doing

[quartz mauve]

how about this

[livid hound]

time interval isn't 15

the interval is from t=1 to 14

[quartz mauve]

it’s not 16-1?

oh right

I mixed it up with a different one

so it would be 37.75 / 13?

[livid hound]

yes

[quartz mauve]

okay

thanks

[cedar kiln]

@quartz mauve Has your question been resolved?

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[vast marsh]

Is this 0 or am I just overthinking this

[flint plinth]

0 is correct, what was your reasoning?

[vast marsh]

From my textbook

The question is an odd function

So I just went off this theorem

But that isnt much for an explanation and well, what if I were wrong

[white latch]

Well essentially if something is an odd function

[vast marsh]

Is it always 0?

[white latch]

then it is not symmetric about the y-axis.

[vast marsh]

Correct

[white latch]

It will behave like a linear function when we perform the odd test

f(-x) = -f(x)

[vast marsh]

Understandable

[white latch]

Are you still confused?

[vast marsh]

No

It makes sense

I appreciate it

[white latch]

No worries.

[vast marsh]

I just didnt wanna bulldoze through the question bc of what my textbook says without properly understanding it and potentially being wrong

[white latch]

Ofc, it is imperative one understands things in calculus as oppose to memorize them.

Your probs on applications of integrals rn I presume?

[vast marsh]

Yeah

[white latch]

Noice, next up for yah is differential equations right?

[vast marsh]

I believe so

[white latch]

Goodluck 🙂

[vast marsh]

Calc 2 stuff

So fun

:')

[white latch]

uhhh thats calc 1 stuff?

i'm currently taking a calculus course and am on differential equations rn.

[vast marsh]

Ah I may have misunderstood

My courses just started again, so this is reviewing calc 1 stuff going into/mixing in calc 2 topics

A refresher

[white latch]

Oooooh nice. Calc 2 I believe is mostly integrals from what I hear. Make sure your all up to par on that.

Integration by parts my fav.

u-substitution pretty lit too. Also jus ensure u understand Riemann sums.

[vast marsh]

Yeah thats whats being done rn, riemann sums and usub

Which is still partly calc 1 but

[white latch]

Noice lmk if u got any questions abt that stuff.

[vast marsh]

Dealing more with it

[white latch]

Best way to learn

is to struggle urself tho.

Can't emphasize that enough.

Gl to u grind hard brother.

[vast marsh]

Thanks, I appreciate your help and words of encouragement

Good luck to you as well

.close

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[crimson sedge]

6x-9=10x+7

[crimson sedge]

Im trying to get x on one said, so first i want to move 10x to the left

but

do i divide EVERYTHING by 10? or just the 6 and -9?

or the 7 aswell?

[livid hound]

same operations to both sides of the equation

divide/multiply the whole side, not just individual terms

[crimson sedge]

so not 7?

[livid hound]

also its recommended that you first separate the variables and constants

take this one step at a time

what do you have after the first step

[crimson sedge]

its annoying how i forget to do simple equations even though im currently factoring polynomials and then solving systems via elimination

like i forget the basics of alg 1

but can do advanced stuff

6x-9=10x=7

so first i divide 10, 6, and -9 by 10, correct?

[livid hound]

you "can" divide each individual term by 10,
but that won't really help you

its recommended that you first separate the variables and constants

[crimson sedge]

thats what im doing

to re arrange i have to do inverse

[livid hound]

dividing by 10 doesn't separate constants from variables

[crimson sedge]

oh

the equation was wrong

6x-9=10x+7

is right

i wrote it wrtong

couldnt i just subtract 7 from the -9 on the left side, and the subtract 6x from 10x

[livid hound]

you'd subtract 7 from both sides
and subtract 6x from both sides

that would be a more efficient approach, yes

[crimson sedge]

yes im aware

i dont even think of that as a step to "cancel out"

i just think of it more as using that operation

so one i use that 7 to subtract on the left side its gone

i DO know youd do whats on both sides to cancel out and BALANCE it

but it takes more thought process for me that way

makes it take longer

[livid hound]

doesn't take that much longer

its helpful when you have more complex problems

[cedar kiln]

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[plush wharf]

Decreasing is 0,4??

[unborn verge]

there are two distinct intervals where f is decreasing

[plush wharf]

So I have to use U?

I thought u can combine them together?

[dusk goblet]

it’s not 0,4 because of the middle piece

the middle piece is increasing

not decreasing

[plush wharf]

Oh right

But if that wasn’t there then my answer would be correct?

[dusk goblet]

u have to break it up

yea if it were decreasing in the middle too

[plush wharf]

I mean like no graph there

[dusk goblet]

if nothing was in the middle then no

because it’s not in the domain

so how could it be decreasing

[plush wharf]

👍

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[full vault]

how i get the average rate of change

[livid hound]

slope formula

[full vault]

(0,6) and 5,4) so 4-6 and 5-0 which is -2/5?

aight thx

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[crimson sedge]

can u please help me solve this

[sharp lotus]

do you notice anything about the number of cubes in each layer?

[sharp lotus]

?

[cedar kiln]

@crimson sedge Has your question been resolved?

[buoyant igloo]

it should be adding 4 each time to the answer

cuz from 1 to 1+4, u have to add 4. Same thing on the third figure where from 1+4 to 1+4+9, u have to add 4 to 5

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[crimson sedge]

nope

[sharp lotus]

1 + 4 = 5 not 4...that isn't the pattern 😐

[sharp lotus]

each layer is a SQUARE pile of cubes, each one differs in number by 1 for example the pile of 9 is a 3x3 pile of cubes

the next one will be a 4x4 pile of cubes, so each layer is just the next square number

so if you have 8 layers you would add up:
1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 =
1x1 + 2x2 + 3x3 + 4x4 + 5x5 + 6+6 + 7x7 +8x8

so that adds up to 204, answer D

[cedar kiln]

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[buoyant igloo]

Idk

That how my brain functions

Srry abt that

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[crimson sedge]

woahh, thats so cool

[crimson sedge]

all that brain work 😭

omg

thank you though!

it made a lot of sense

[sharp lotus]

there is an interesting additive pattern between each consecutive square number though....
0 + 1 = 1
1 + 3 = 4
4 + 5 = 9
9 + 7 = 16
16 + 9 = 25
25 + 11 = 36
...etc.
or in other words:
1 + 3 + 5 + ... + (2n-1) = n^2
aka:
if you add up the first n odd numbers you get n^2 : )

[buoyant igloo]

For a sec, i thought you add odd number from 1 to 4 which is 3 and from 4 to 7 is , which are odd numbers

Makes sense

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[dire flame]

Triangle ABC is acute. On the outside of ABC is 2 right isosceles triangle ABD and ACE at A.
a. Proves that DC=BE
b. proves that DC is perpendicular to BE
c. AH is perpendicular to BC at H and intersect DE at M. Proves that M is the midpoint of DE.
d. I is the midpoint of BC. Proves that AI is perpendicular to DE.

[dire flame]

I'm stuck on d. Help!

<@&286206848099549185>

[sage field]

you are in what grade?

[tame zodiac]

what is this?

[dire flame]

It's the shape of it

[tame zodiac]

oh you changed pic?

[sage field]

can you share your working for a,b,c?

[dire flame]

i hide some stuff of a and b bc i thought it was irrelevant but ok

[sage field]

did you drew all this in question or was it given?

[dire flame]

some of it wasn't given

so this is the stuff before d.

So a) is proving triangle ADC = triangle ABE

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[steady elk]

am i right?

[tropic oxide]

well your handwriting sucks a little bit

but you are right, yes

$\int 0 \dd{x} = C$ and $\int_a^b 0 \dd{x} = 0$

[wraith dagger]

Ann

[steady elk]

really can u rate it bqz actually my schoolteachers say the same im actually worried but i dont pay that much attention

[steady elk]

ok

[tropic oxide]

rate it on what scale

[steady elk]

some standard handwritting

[tropic oxide]

uhh

i don't know of any such thing as a "standard handwriting"

[steady elk]

ok how to write elegantly

[tropic oxide]

personally i think it looks like cursive gone bad

and i think you would benefit from adopting a style closer to print

like ok this is my personal opinion but i think the best handwriting lies somewhere between full print and full cursive

here's a sample of mine

[steady elk]

ohhh i see , i understand. Is this ok for long hours writings

[steady elk]

this

[tropic oxide]

i don't see why not

[steady elk]

ohk ,

thanks

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[boreal radish]

A thin transparent boiling tube of mass m length l is sealed at both ends. It is
suspended from a string so that the bottom of the tube is l above the table. A fly of mass m is sitting at
the bottom of the tube. The string is cut, and while the tube is falling, the fly moves from the bottom to
the top of the tube. Find the time that elapses between cutting the rope and the bottom of the tube hitting
the table

[boreal radish]

this is a question where i'm supposed to use centre of mass to find the answer to the problem, but i'm not entirely certain as to how to go about it

[mighty shuttle]

What have you tried?

Tried drawing a diagram?

Also, how high from the table is the tube suspended

[boreal radish]

honestly, i drew a diagram and that's about as far as i got. I've been thinking about different methods but i'm pretty stumped.

the same length as the boiling tube, l.

the mass of the boiling tube is mass m, and the fly inside of the tube is also mass m

[mighty shuttle]

Hm, so the angular displacement is 90deg right?

Of the tube

[boreal radish]

yeah

[mighty shuttle]

Find the torque and find everything else from that

[mighty shuttle]

For this you should know how much time the fly takes to travel through

Or alternatively use the fact that the centre of mass won't move due to changes in internal forces acting on the system

[boreal radish]

right, i didn't know that

that actually helps a lot just from that little piece of information

thank you

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[short stratus]

$$\lim_{x \to \frac{\pi}{2}} \frac{\sin x - (\sin x)^{\sin x}}{1-\sin x + \ln (\sin x)}$$

[wraith dagger]

xd_senBugha

[short stratus]

i applied L'Hopital's rule and ended up with

$$\lim_{x \to \frac{\pi}{2}} \frac{\cos x - (\sin x)^{\sin x}(\cos x \times \ln( \sin x) + \cos x)}{-\cos x + \cot x}$$

[wraith dagger]

xd_senBugha

[short stratus]

which is basically a dead end since another derivative would be more complex (considering this is still in 0/0 form)

is there any other way to solve this limit?

[eager sapphire]

this looks really awful but maybe you could try a taylor approximation?

$\sin(x) = x - \frac{x^3}{6}$

[wraith dagger]

AlphaNull

[short stratus]

I'd still end up with the same problem

[mighty shuttle]

well this is the taylor approximation at x=0 not π/2

,w taylor series of (sin(x))^(sin(x)) at x=pi/2

[wraith dagger]

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[mighty shuttle]

try finding the taylor series of sin(x)^sin(x) at x=π/2

[short stratus]

Alr

[raven shard]

in my opinion this is actually not that bad to expand

you would multiply the numerator and denominator by (sinx)/(cosx)

[cedar kiln]

@short stratus Has your question been resolved?

[short stratus]

,w taylor expansion of cosx at x = pi/2

[wraith dagger]

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[short stratus]

,w taylor expansion of ln(sinx) at pi/2

[wraith dagger]

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[short stratus]

i tried putting taylor series of all the terms at pi/2 then substituting, im getting -1

The answer given is 2

[mighty shuttle]

Hmm, not too sure in that case , sorry

[short stratus]

Nvm i found a way using substitution

I can take sinx = u

And then apply L'Hopital's rule twice

Thank you tho

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[crimson sedge]

how to solve (a+b+c)^2

[scenic brook]

$(a+b+c)^2$

[wraith dagger]

Zemitrix

[scenic brook]

?

[crimson sedge]

yup

[scenic brook]

I was confused why ^^2

[crimson sedge]

ok

[dull oxide]

What's there to solve?

[scenic brook]

Maybe expand

[crimson sedge]

it"s value

[scenic brook]

What...

[dull oxide]

!xy

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[crimson sedge]

.close

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[split pike]

wtf?

[cedar kiln]

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[crimson sedge]

hey

[crimson sedge]

Give an example of a composite number whose decimal representation contains exactly 1994 digits "1" and one digit "7" (and no other digits)

How to proceed ?

[empty berry]

hey

can anyone help me

[mighty shuttle]

!occupied

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[raven shard]

what divisibility rules can you easily apply here?

[crimson sedge]

wdym by divisibility rules?

[raven shard]

a number is composite if it's divisible by an integer other than 1 so all we have to do is show an example of a number that fits the constraints

you mentioned 3, 7, and 11

i'm not sure how we'd apply 7 here, but 3 and 11 make sense

[crimson sedge]

yeah

[raven shard]

so try

using the fact that the number is made up of 1994 1s and a single 7

[crimson sedge]

okay lemme try

It's divisible by 3

ig

[raven shard]

ye

no matter how you set up the number, it's always divisible by 3

[crimson sedge]

so what would be an example ?
1111111111111117 ? (1994 1s)

[raven shard]

yeah pretty much

i'll try to find an example of the notation

[crimson sedge]

I'm really thankful

In the mean time I'll think about the case of 1993 1s and one 7

[raven shard]

it's not uncommon to write like this

but i guess you could do it however you wanted to

[crimson sedge]

$\underbrace{99 \ldots 99}_{321 \text { digits }}$

[wraith dagger]

Mellow

[raven shard]

ah nice

[crimson sedge]

hmm okayy

luv u

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[smoky niche]

ok whats the best method for simplifying this

[smoky niche]

(sorry for the awful image btw)

[tropic oxide]

$\frac{1}{(\sqrt{3} + i \sqrt{3})^2} - \frac{1}{(\sqrt{3} - i \sqrt{3})^2}$

[wraith dagger]

Ann

[tropic oxide]

made it better for you

[smoky niche]

ty

[tropic oxide]

anyway i would notice that $(\sqrt{3} + i \sqrt{3})^2 = 3(1+i)^2$ first and foremost

[wraith dagger]

Ann

[smoky niche]

hmm

sorry I don't see how that helps

you're obviously right but idk I'm a bit stuck

[brittle stag]

what can (1+i)^2 be rewritten as?

[smoky niche]

2i

ah that makses much more sense

[smoky niche]

but how do you get from whats shown here

[brittle stag]

?

[smoky niche]

like from (root3 + iroot3)^2 to 3(1+i)^2

[brittle stag]

take common sqrt3 and power distribution over multiplication

[smoky niche]

ty

[cedar kiln]

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[glacial summit]

how would you ideally look to solve this

[glacial summit]

i can brute force it but i feel its much simpler than im making it out to be

[drifting marlin]

What do you mean solve it? What's the question?

[glacial summit]

to equate it

[drifting marlin]

It's an expression

What are you equating it to?

[glacial summit]

it should simplify to zero

[drifting marlin]

Try rewriting the denominator of the middle term

[glacial summit]

(5a+b)(5a-b)

[drifting marlin]

Indeed

That should speed up the process of creating a common denominator across your other two terms

[glacial summit]

let me work on that

oh got it

huge thanks

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[eager flicker]

I've got this set of point

And I need to function that goes closest to them

How do I do this? (Computer science, first sem)

[brittle stag]

there is an infinite way of reaching these points approximately

you want the function to be linear quadratic or whatever else?

[eager flicker]

Um, this probably should be circle

Or elipsę

*elipse

[eager flicker]

I mean function that is closest to all these points, like the exact value

Not approximate values

Does not have to go straight trough point, but closes to them

To keep function as circle or elipse

<@&286206848099549185>

Let's do circle for simplicity

[brittle stag]

you can maybe rewrite your data to be in polar coordinates, rho and theta, and you'll have a linear regression of rho = atheta? or rho = a

[eager flicker]

I don't even understand what you mean, never had it on lectures

Only rho, but it was physics class

[brittle stag]

rho as in length of the segment

polar coordinates uses rho for length of segment and theta for angle of the segment with the x axis

[eager flicker]

Circle function
(x - X)^2 + (y - Y)^2 = r^2

Find X,Y

[eager flicker]

We didn't had polar coordinates at all

( @brittle stag I do have the answear for this, but idk how to do it in case I was presented with one on test)

It's
||(x-2.97767620117582022)² + (y-2.11515086283465292)² = 24.86432053||

[brittle stag]

i don't really know anything about prediction modelling, just wanted to see if i can provide anything

[eager flicker]

I thought it's simple statistics

...

[brittle stag]

haven't taken statistics yet, next semester

[eager flicker]

I didn't as well

I had this in algebra class, in computer science or engineering major

First sem

[cedar kiln]

@eager flicker Has your question been resolved?

[gilded mesa]

You can get closets polynomial

[eager flicker]

How do I get it ?

[gilded mesa]

But it is a lengthy process

I cant explain

What you need this for

There is a langrage interpolation formula

And netwon forward diffrence formula

[cedar kiln]

@eager flicker Has your question been resolved?

[eager flicker]

Can you do like list of steps

[floral arrow]

It's clearly not a polynomial

Can't you just compute the average of all these points to get the center of your circle?

[valid egret]

Got a quick general question

[floral arrow]

And then the average distance between the center and each of the points to get the radius

[valid egret]

Can I ask?

[floral arrow]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[eager flicker]

I did for x, it's slightly of compared to given answer, but pretty close

Average is 2.994
Answer is 2.97767620117582022

Profesor was using some method with squares or sth like that

[floral arrow]

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

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[strong pawn]

I just had a quick question about the general transformation formula (x/k + d, ay+c)

[strong pawn]

is this actually the correct formula

[gilded mesa]

Ya correct

[strong pawn]

My translations end up wrong because of the fact that a negative D value = going into the x positive side of the graph.

So do I take the D value and flip the sign when plugging it into the formula?

[gilded mesa]

Flip the sign

[strong pawn]

Alright, thanks, I was getting confused.

[gilded mesa]

Take Positive it goes to negative side

[strong pawn]

Yeah ik.

Ok thanks, so just flip it when plugging into formula.

[gilded mesa]

Yeah

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[tiny socket]

Let $a$ be constant, and a series with $n$ elements, every element equals to $\frac{a}{i}$. $a = ?$ if the sum of the series is $1$

[wraith dagger]

Shachar

[tiny socket]

does someone knows how to solve this?

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[mossy night]

Can anyone help me with dot product. I cant figure out how to determine if they are parallel, purpendicular or neither.

[vagrant elbow]

do you understand geometrically what the dot product is

or if you don't, do you know the general formula

[mossy night]

yeah i have my answers but i cant figure out how to know which is which. the first one i got 0 second one i got -9 third i got 15 and last got -34

[vagrant elbow]

well

[vagrant elbow]

.

if i just drew two vectors on some paper and gave you a ruler and pencil would you be able to find the dot product?

[mossy night]

im sorry no. what i have been taught is formulas. and the only thing that is mentioned in my book is if the angel is perpendicular then the angle is pi/2

but it doesnt say anything about what the angle is if it is parallel

[vagrant elbow]

ah

$\vb a \cdot \vb b = |\vb a||\vb b| \cos \theta$

[wraith dagger]

NEON

[vagrant elbow]

i assume this is familiar then!

[mossy night]

YES

[vagrant elbow]

it was meant to be a question mark

but i appreciate the enthusiasm

in any case

θ here represents the angle between the vectors a and b

do you know when the cosine of an angle is 0?

[mossy night]

yes at pi/2 or 90 degrees

[vagrant elbow]

yeah

there you go then

two non-zero vectors are perpendicular iff their dot product is 0

[mossy night]

but how do i tell if the angle is parallel or neither.

[vagrant elbow]

right

so

can you draw two parallel vectors and tell me the angle between them

[mossy night]

no im sorry i dont think they have taught that yet

u dot v/ |u| |v|=cos theta?

[vagrant elbow]

hmm

well

do you know that vectors are basically arrows in space

[mossy night]

yes

[vagrant elbow]

so can you draw two parallel arrows out

or think about them

if you join the tails of these parallel arrows

what would the angle between them be

[mossy night]

see they only taught if the dot product is negative then its obtuse and if its positive then it's acute.

[vagrant elbow]

well

i'm trying to teach you something new

just try to keep an open mind

[mossy night]

will do

[vagrant elbow]

do you agree that these vectors are parallel

[mossy night]

ues

yes

[vagrant elbow]

and what's the angle between them?

[mossy night]

0

[vagrant elbow]

excellent

[vagrant elbow]

using this formula

what would the dot product of two parallel vectors be

you just need to put 0 for θ

[mossy night]

0

[vagrant elbow]

uh

what's the cosine of 0

[mossy night]

90

[vagrant elbow]

uh

hmm

this is a problem

[mossy night]

nvm it's cool ill find a video on Mathispower4u. thank you for trying.

[vagrant elbow]

or maybe not

i didn't ask the cosine of what gives 0

i asked for the cosine of 0

[vagrant elbow]

whatever floats your boat :/

[mossy night]

really thank you!

[crimson sedge]

you should refer to the unit circles for instances like this

[waxen plume]

is this dot product related?

when dot product or scalar product is zero you know the angle is perpendicular

cos(pi/2) = 0

[waxen plume]

unit circles is an excellent approach

[mossy night]

OK i got it if you have vector a <4,1,0> they are parallel if the next vector is the same or can be factored to look the same as vector a. so vector b is <-8,-2,0> so factoring out a -2 you have vector b= -2<4,1,0> so they are parallel

[waxen plume]

exactly they are parallel

[vagrant elbow]

i'd just like to point out one thing

if you find it to be a negative factor of the other vector

then the vectors are called antiparallel

if it's positive then they're considered parallel

[mossy night]

OH OK. im writing that down. thank you!

idk i just do what my Webwork tell me to do lol and its only options were perpendicular, parallel or neither. but i will def keep that in mind as im sure it will come up as i progress through Cal 3. thank yall!

[vagrant elbow]

yeah it's better to follow what your course tells you atm

[cedar kiln]

@mossy night Has your question been resolved?

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[winter rampart]

pls help

[winter rampart]

<@&286206848099549185>

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[winter rampart]

<@&286206848099549185>

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[winter rampart]

<@&286206848099549185>

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[plucky owl]

Please stop repeatedly pinging helpers

[winter rampart]

ok

[plucky owl]

You are supposed to ping once after 15 minutes like that bot message says

[winter rampart]

can i have help

.close

[cedar kiln]

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[ivory finch]

basically asking if 2 groups are isomorphic

[ivory finch]

Z_5^X is {1,2,3,4} multiplication mod 5 and WTS it's an isomorphism

I'm at a roadblock where I need to test f(a~_1 b)=f(a) ~_2 f(b)

so i think f is {1,2,3,4}->{0,1,2,3}

so then f(x)=x-1

which satisfy bijection

but afterwards i need to test the f(a~_1 b)=f(a) ~_2 f(b) for 16 values?

11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44

f(a multiply mod 5 b) = f(a) add mod 4 f(b)

f(x)=x-1
1,2,3,4->0,1,2,3

and at 2,2 it's wrong

it seems like the problem is implying it's an isomorphism tho

[hasty crown]

Try to add Earth's diameter

[muted bear]

dont troll

[hasty crown]

Okay mister

[cedar kiln]

@ivory finch Has your question been resolved?

[ivory finch]

<@&286206848099549185>

[worn briar]

can you put your question in TeX?

[ivory finch]

$$\text{Recall that } \mathbb{Z}_4 \text{ denotes {0,1,2,3} together with addition mod 4. Show that } \mathbb{Z}_5^\times \cong \mathbb Z_4.$$

[wraith dagger]

cookie2

[worn briar]

i mean this. i do not get the notations

[ivory finch]

f: G_1 -> G_2

o = G_1's operation

o' = G_2's operation

a and b are in G1

[cedar kiln]

@ivory finch Has your question been resolved?

[cedar kiln]

@ivory finch Has your question been resolved?

[cedar kiln]

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[rare lintel]

what am i doing wrong?

it looks like a loop

[dire geode]

yea l'hopital alone won't work

try uhhh t = x^(1/4) or x^4 = t

then lim x-> inf means t-> inf

[rare lintel]

isn't it the same thing?

substituting t for x

[opaque root]

What grows faster

x^(1/4) or e^(2x)

As e^(2x) grows faster, the limit tends to 0

[rare lintel]

could you expand?

[opaque root]

Unless I’m reading wrong this is solved in one step

$\lim_{{x \to \infty}} \left( \frac{x^{1/4}}{e^{2x}} \right)$

[wraith dagger]

Samuel

[opaque root]

I have to go now, I’ll check later your answer but if I read correctly this is solved like this

[rare lintel]

I got stuck on this

[opaque root]

denominator grows faster

[rare lintel]

so its just 0

is this correct?

[cedar kiln]

@rare lintel Has your question been resolved?

[dire geode]

yea

samuel was right

[rare lintel]

alright

thank you

[cedar kiln]

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[quartz mauve]

At first I thought I should just subtract the y value at t=3 from t=15, but that was wrong

I tried integrating the two functions, finding the two y values for each one, and subtracting that but that was also wrong

[muted bear]

what does the y axis represent?

[quartz mauve]

this is my work so far

[quartz mauve]

velocity

[muted bear]

label your intervals

[quartz mauve]

the first integral is from 0 to 8 and the second is from 8 to 16

is there something wrong with my calculation or am I trying to do this the wrong way?

[muted bear]

youre not integrating properly

[quartz mauve]

wdym

[muted bear]

do you know how to integrate

[quartz mauve]

yes

[muted bear]

write the expression properly please

[quartz mauve]

which expression

[quartz mauve]

which integral did I mess up?

[cedar kiln]

@quartz mauve Has your question been resolved?

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[edgy mason]

how do i find the shared point

its a skew line

i have the direction vector but idk where to get the point from

[x,y,z]=[2,5,3]+t[1,2,2]

dk where 2,5,3 came from

[cedar kiln]

@edgy mason Has your question been resolved?

[edgy mason]

<@&286206848099549185>

[cedar kiln]

@edgy mason Has your question been resolved?

[cedar kiln]

@edgy mason Has your question been resolved?

[gilded mesa]

Which points you need

[cedar kiln]

@edgy mason Has your question been resolved?

[cedar kiln]

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[crimson sedge]

i need help with this please

[sharp lotus]

use the idea of ratios of corresponding sides being equal

so the ratio of 28/35 = ?
what is ? for the smaller triangle?

[crimson sedge]

idkk

thats all they gave me

[sharp lotus]

so similar triangles will have a relationship that ratios of corresponding sides will be equal

[crimson sedge]

wait, how do i find the perimeter of triangles?

[sharp lotus]

let's label the unknown side of the small triangle x, and the unknown side of the big triangle y

[sharp lotus]

well we need to find the length of all the sides first