help-17

[ornate ember]

What is the original area?

[wintry walrus]

What do you mean

[ornate ember]

The original vent is 30cm long and 10cm wide

what's that area?

[wintry walrus]

300cm

[ornate ember]

cm^2, but yeah you got it 🙂

ok so what is 75% of that area?

[wintry walrus]

225cm^2

[ornate ember]

yep! Now we add that to our original area

and that's our goal area

[wintry walrus]

So 525cm^2

[ornate ember]

yep!

that's our A

now solve for x

525 = x^2 + 40x + 300

that'll get you the dimension increase (aka we increased both sides by x). Once you have x add it to 30 and 10 respectively for their sides

[wintry walrus]

X= -45

[ornate ember]

well you should get 2 x values

since it's quadratic

[wintry walrus]

X = 5

And x = -45

[ornate ember]

bingo!

so since we're dealing with area

we can throw away the x < 0 value

(aka we dont' need negatives, it's erroneous)

so then we know that we increased each side proportionally by 5 cm, so what are our new dimensions?

[wintry walrus]

530cm^2 ?

[ornate ember]

no that's area

what was the original length?

[wintry walrus]

30 cm

And bridth was 10cm

[ornate ember]

and what did x represent?

[wintry walrus]

So the new dimensions are 35cm and 15cm?

[wintry walrus]

The dimensions increase

[ornate ember]

yep

[ornate ember]

bingo!

and just to verify

,calc 35 * 15

[twin meteor]

Result:

525

[ornate ember]

aaaye look at that

does that process make sense?

[wintry walrus]

Thanks ! 👍🏻

Yes very helpful

[ornate ember]

yep you're welcome! Best of luck

[wintry walrus]

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[strong jetty]

did I do some thing wrong here? I get a negative number in the square root

[vast shale]

google complex number

[strong jetty]

how would you solve this without a calculator ?

[vast shale]

the solutions are complex

there's a number assigned for $\sqrt{-1}$ which is $\iota$ such numbers are called complex numbers

[twin meteor]

deltaG

[vast shale]

@strong jetty

[strong jetty]

ok I just looked up the solution page and for now this means that there is no solution

at least not with real numbers like you said

[whole oasis]

4x^2 - 12x + 11 < 0
4x^2 - 12x + 9 + 2 < 0
(2x - 3)^2 + 2 < 0

[strong jetty]

I'll probably get to complex numbers in later classes

[whole oasis]

now you can deduce the solution

[vast shale]

write the complex solution and confuse your teacher

[whole otter]

get him to give you the GOOD homework

[vast shale]

yeah write the complex solutions and plot it on the imaginary graph

also write it in euler form

[whole otter]

in mirror writing

[vast shale]

no

[whole otter]

no.

ignore that.

[strong jetty]

btw how do I square root on a simple calculator

to the power of what number?

[trim cedar]

1/2

[vast shale]

1/2

[strong jetty]

is power 1/.2

[vast shale]

bro you have pre-uni role just saying

[strong jetty]

oh , actually I can't even do that on this calculator lol

[whole oasis]

first of all, that's inequality not an equation, using only the quadratic formula doesn't give you a solution

[trim cedar]

any epsilon delta masters in this lobby

[strong jetty]

what do you mean pre uni role

[vast shale]

just be patient for people to notice your help channel

the more you advertise it, the lesser the chances of people helping

[trim cedar]

sometimes I help other channels to get help faster

[strong jetty]

I haven't done mathematics in 10 years even if I'm pre uni role...

[trim cedar]

¯_(ツ)_/¯

[strong jetty]

so its kinda rude

[silent burrow]

did you get this ineq. like this originally or you arrived on this yourself?

[vast shale]

pre-uni is 11th and 12th grade

not done mathematics in 10 years implies the last time you did mathematics was when you were 7 or 6

[strong jetty]

no, not this case

[silent burrow]

cuz maybe the coefficients are wrong

[strong jetty]

then what did you imply by saying im in pre uni role , as in I'm too dumb for not knowing this huh

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[vast shale]

no, nvm

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[hollow karma]

Hello

Would I need to divide these 2 matrices to get A?

[vast shale]

how do you divide matrices

[hollow karma]

idk

[vast shale]

you dont need to divide matrices

that shit is cursed

treat them like constants, just write 5A=Matrix1-Matrix2

and subtract the matrices

also notice that one is in transpose

do that first

[hollow karma]

What have I done wrong here

[vast shale]

nope, i am not checking the calculations (cause im sleepy) the method is correct

[hollow karma]

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[quasi ivy]

Don't know how to do at all

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[cobalt raven]

do i just send a picture here

[hoary swift]

Could anyone help explain what to do in this question, I don’t get how the vector being parallel to the

[hoary swift]

Yeah you do

[cobalt raven]

okay, i just want to know what they want me to do here

so what now

[hoary swift]

Did the first 5 it just wants you to give another example of the proportion given with the objects

[cobalt raven]

so another equivalent

[hoary swift]

For example for the first one you would say 15 circles for every 24 triangle, 20 circles for every 32 triangles, 50 circles for every 80 triangles and 100 circles for every 160 triangles

[hoary swift]

Yeah

Btw once you don’t need anymore help do .close

And if you need help again just make another help request

[cobalt raven]

so, do i have to go by 3s?

and do i have to give 4 new ones or just two??

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[hoary swift]

4 new ones and you done have to go by 3’s, just something that you can multiply by both and get whole numbers, for example 2

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[boreal fable]

Is this in echelon form?

[late ingot]

yes

its also in reduced echelon form, matter of fact.

[boreal fable]

How so

@late ingot

What's the difference between not reduced and reduced echelon form?

[late ingot]

pivots are 1 and the other values in the column of the pivot are 0

[boreal fable]

What about the 3 and 8

[karmic sail]

my question is harder

[boreal fable]

Oh so reduced means that the leading entries are ALWAYS 1, and those leading entries MUST have 0's below and above it.

@late ingot

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[trim cedar]

Can I get some feedback on this proof

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[vast shale]

Can someone explain to me how I would do the first one

[shy willow]

do you know what displacement is?

its distance in a particular direction from a point

so horizontal displacement means the distance in the horizontal direction from a point

verticle displacement means the distance in the verticle direction from a point

here "from a point" would be "from the starting position" or at t = 0

so you need to see how far its moved vertically and horizontally from t=0 to t=5

the distance values are given on the horizontal and verticle axis

did this help?

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

Yea

It doesn’t really help with what you said

You just explained a concept

Not like formulas that I’d need or anything

[shy willow]

you dont need any formulas for the first part

all you need is to understand the concept and how its represented in the graph

the axis are verticle and horizontal, and give distance right?

so what would the position, relative to each axis, tell you about the ball at a given time?

for example, at t = 3, it appears to be at about position (100, 30) [assume the values are accurate, cause its not clear on the diagram). what woudl the values 100 and 30 tell you about the ball at time = 3?

@vast shale do you get it?

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

Sorry uhh

I’m confused how this will help me understand what the displacement vector would be

[shy willow]

theyre not actually asking what the displacement vector is, theyre asking what the horizontal and verticle displacement is

the graph gives its verticle and horiztonal positon

so the displacement is the final position - the initial positon on the graph

along the respective axis

[shy willow]

answer this, then youll get it

[shy willow]

the x axis is the distance in the x direction (horizontal, and the y axis in the distance in the y direction (verticle). the axis give distance in a particular direction (displacement). so using the values on the axis determine the displacement (distance traveled in a a particular direction) of the particle at the given time

its coordinates (x,y) give the horizontal and vertical displacement respectively.

[vast shale]

It would be 150, 55?

[shy willow]

mb, i misread the graph

but yes about 100,55

or actually

yeah either way

[vast shale]

I mean x over y

Or

Y over x

[vast shale]

And it would tell me the vertical distance and horizontal

Of that specific point

[shy willow]

yes

assuming it began at 0,0

but according to the graph it didnt, so it would be ∆x and ∆ y

it would be easier to consdier the positon of the top of the circle, as it begins at the origin, (0,0)

then at any point ∆x = x and ∆y = y

[vast shale]

So at 0 x and 0 y?

[shy willow]

basically you need to find how much its position changed across the x axis (horizontal displacement) and the y axis (vertical displacement)

[vast shale]

I see

I was absent for this day we did this so sorry

I’ll give you nitro if you help with the first two

[shy willow]

nah dont worry

[vast shale]

Is that fine with you?

[shy willow]

i dont need nitro dw

ill help you

[vast shale]

It’s ok I’ll give it to you anyways

[vast shale]

Ok so what do I do with this

[shy willow]

no 😭 its alr. i like helping people w math

[shy willow]

that’ll give you the horizontal and verticel displacement

[vast shale]

How do I use that to show the components of getting displacement vector

[shy willow]

im not sure its asking for the vector

its just asking for displacement

vertical displacement = how far its moved in the vertical direction
horizontal follows the same

[vast shale]

I mean it asks me to demonstrate understanding of getting displacement vector

I mean

Why do I keep saying vector

Nvm

[shy willow]

its asking to demonstrate your understanding of displacement vectors by finding the displacement

[vast shale]

Oh ok

[shy willow]

displacement is a vector since it has direction. its asking you to find the distance travelled in that direction

[vast shale]

So is it just

X -y

[shy willow]

no so x and y are in different directions so they give displacement in different directions

so since x is in the horizontal direction, you need to find the change in x to find the horizontal displacement

horizontal displacement = distance travelled in horizontal direction

[vast shale]

And change in x is calculated by..?

[shy willow]

since the x axis gives the distance in the horizontal direction, find out how far its moved in that direction

[shy willow]

final x value - initial x value

or ∆x (change in x)

[vast shale]

Okay

And what do I do with delta x and y

Because sorry I’m not sure how to explain how the understanding of displacement vectors is applied by this knowledge yet

[shy willow]

this is the only understanding of displacement vectors you need

for this question

[shy willow]

delta x is the horizontal displacement, delta y is rhe verticel displacement

the displacement vector would be the displacement of the particlw

the red line would be its displacement, or displacement vector

the green line is the horizonal displacement

the blue line is the verticle displacement

the red line is related to the blue and green line and can be defined by it

[vast shale]

Oh ok

Wouldn’t this just be base * height * 1/2 then

Wait no

I’m thinking of base

Mb

[vast shale]

Uhh how would you get the hypotenuse then to find that angle, I think it’s tan-1(125/250)

Since it’s a vector I’d assume it wants me to get the angle as well

To tell the direction

And to get the length of that I’d do uh tan() of that angle?

Does that make sense

[shy willow]

yes for the bottom right angle thats correct

you would probably prefer the top left angle tho

since its relative to the origin

[vast shale]

Oh so I got it the opposite way

[shy willow]

yeah

you dont need that though

[vast shale]

Yeah that makes sense

[shy willow]

for this question

[vast shale]

I see

[shy willow]

hypotenuse would be pythagoras formula

[vast shale]

A plus b equal c?

All to the power of 2

Can I use that to find the length?

[shy willow]

yes

the total displacement from origin is c

but again, not needed for this so dw

[vast shale]

Ok thanks 🙏

I’ll write that all down

[shy willow]

np

lmk if u need help for part 2

ping me

[vast shale]

@shy willow

[shy willow]

sup

[vast shale]

Would I just use a kinematic equation for 2

Actually nvm

How do I get velocity in the y-axis

?

[shy willow]

the only force acting in the y (verticle/downwards) direction is gravity

use its acceleration to determine how much the velocity changes

did you get the x direction?

[vast shale]

No sorry

[shy willow]

dont be sorry dw, just asking

[vast shale]

So 9.81 at all times

?

[shy willow]

you can do it in any order

[shy willow]

yes

unless ur school uses 10

[vast shale]

No

That would be dumb

imo

[shy willow]

so if acceleration is 9.81 m/s^2, then velocity changes by 9.81 m/s every second

in the downward direction

[shy willow]

true

[vast shale]

Would I just be liking for final velocity for each

Second

Looking for

Or like how would I find vx

[shy willow]

yes, the velocity at that second

v = u + at if you need a formula

that goes for horizontal as well ( the formula i mean)

[vast shale]

U is uhh what exactly

Initial velocity?

Wouldn’t that be calculating the velocity at an angle though like

Not the x direction specifically

[shy willow]

yes

v if final velocity or velocity at t

[shy willow]

velocity is a vector, it can be in any direction. the formula uoholds as long as all vectors considered (initial velocity, final velocity, acceleration) are in the same direction

and that goes for any direction

vertical, horizontal, at an angle etc

[vast shale]

Ok thanks

So just use kinematic equation for that

Wow ok

You just helped me by a ten fold

I’ll send you nitro in DMs

Thank you

:)

[shy willow]

no no its okay really

i enjoy math

[vast shale]

Too late

Check your DMs

[blissful fable]

wtf

damn

[shy willow]

tysm

[vast shale]

@shy willow one more question

[shy willow]

yeah

[vast shale]

Would Vy be 9.8m/s squatted

Square

Or no

[shy willow]

just 9.8 m/s. velocity is m/s. acceleration is m/s^2

wait vy at what point?

vy changes

[vast shale]

At all points

[shy willow]

it would have 3 different values

its accelerating downward at 9.81 m/s every second

[vast shale]

Thought it was 9.8 all the way

[shy willow]

so its velocity changes by 9.81 m/s every second

[shy willow]

formula is V = U + aT

[vast shale]

So that’s what you meant

By it applies to Vy too

[shy willow]

as T changes, C changes

[vast shale]

I see

Thanks

I appreciate it

[shy willow]

apply the formula to every point

ofc

[vast shale]

I really got to go no sorry

Thank you

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[shy willow]

Vx is constant tho

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[elder gazelle]

ive been stuck on part a for a while Im unsure how to go about it

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[frozen carbon]

hi idk what to do 😭

[vast shale]

They've given expression for total number of oranges in n layers, you've to find number of layers containing 91 oranges

Does this click?

[frozen carbon]

yeah i know so am i supposed to do smth like y=91?

[vast shale]

Yes.

You'll get a cubic equation

Solve it and then you'll find "n"

This to b clear

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[rigid kernel]

How to do -5^2+7

[soft walrus]

wdym by "how to do"?

simplify?

[rigid kernel]

What do I do

Per

Order of operation

With exponent

[soft walrus]

that doesn't really clear things up 😭 were you given any instruction

[rigid kernel]

Is it -25+7

[raven owl]

Yes

Google calculator

[rigid kernel]

Ohh ur right

I put all of ut in calculator

.close

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[high heart]

How do ppl divide one matrix into two matrices? Seems pretty complicated to me

[rugged orchid]

How do you mean divide

[high heart]

Just a sec

This he made it into

[rugged orchid]

That’s not a matrix

[high heart]

what

[rugged orchid]

That’s the determinant of a matrix

Which is just a number

They are using properties of the determinant function here

[high heart]

oh

How'd they split into two though

[rugged orchid]

They aren’t splitting the matrix itself

[high heart]

like how do they know u gotta add Cos alpha and Sin Alpha and 0 and all

[high heart]

oh

[rugged orchid]

You do a lot of questions

[high heart]

💀 this appearantly is the second question so-

[rugged orchid]

Well you just have to know

Ideally since it’s 0 you just want to find a column/row of 0’s

[high heart]

until I solve the question I wouldn't know if it's 0 right

[rugged orchid]

No

It says show it’s 0

So you know it’s 0

So you’re aiming to find a row/column of 0’s

[high heart]

well in a test they gave it as an mcq 💀

welp thanks ig

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[vagrant basin]

I'm trying to prove that McCarthy 91 function is equal to 91 for all positive integers less than or equal to 101. I found the following pattern:
M(101) = 101 - 10,
M(100) = 100 + 11 - 10 - 10,
M(99) = 99 + (11 - 10) + (11 - 10) - 10,
M(98) = 98 + (11 - 10) + (11 - 10) + (11 - 10) - 10,
...
M(n) = n + (11 -10) * (101 - n) - 10

In effect, the result of recursive call is that the function is continuously adding 1, 101 - n times that makes the input equal 101, and last subtractions takes away 10 giving 91. I also see that {n + adding 1} - 10 is basically the base case def of the mccarthy with n = {n + adding 1} for n > 100. Thus, after 101 the function outputs 92, 93, 94 ....

However, I am not sure how to begin writing the formal proof

[vagrant basin]

Def of McCarthy 91 function

<@&286206848099549185>

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[hearty dragon]

i got part a but idk how to do part b

[grizzled halo]

solve for gradient = 1, -1, 5, -5, 0

[hearty dragon]

oh nvm its just working backwards i got it now thx

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[hard bay]

Hi, in Algorithm Design by Kleinberg and Tardos page 237, on the use of FFT to compute the convolution of two vectors a = (a0, a1, ..., an-1) and b = (b0, b1, ..., bn-1) in O(n log n) time, the procedure is as below:

1. Define A(x) = a0 + a1x + a2x^2 +...+ an-1x^(n - 1), B(x) = b0 + b1x + b2x^2 +...+ bn-1x^(n - 1)
2. Define C(x) = A(x)B(x) = a0b0 + (a1b0 + a0b1)x +...+ an-1bn-1x^(2n - 2), a degree (2n - 2) polynomial. Note that the coefficient vector of C is the desired convolution.
3. Use Divide and Conquer to evaluate A and B on the (2n)th roots of unity in O(n log n) time
4. Use the results of evaluation to compute the evaluation of C on the (2n)th roots of unity in O(n) time
5. Use the 2n values of C to perform polynomial interpolation and obtain the coefficient vector of C, and thus the convolution, in O(n log n) time.

I understand that as long as we have > 2n - 2 values for a polynomial of degree (2n - 2), then we can perform polynomial interpolation to obtain the polynomial. This book used 2n values to interpolate a degree (2n - 2) polynomial, one more value than necessary.

In page 241, the book (wrongly?) assumed that C is a degree (2n - 1) polynomial rather than a (2n - 2) polynomial. The rest of the interpolation follows naturally under this assumption.

1. Can I confirm that the book is actually wrong, and C is actually a degree (2n - 2) polynomial?
2. Assuming the book is wrong, I ignored C(w_2n-1,2n) and tried to mimic what the book did. I defined D to be a degree (2n - 2) polynomial instead. I realised that by doing this, I can no longer simplify D. Maybe I should use (2n - 1)st roots of unity instead? Maybe I should drop C(1)? What is correct thing to do here?

https://github.com/davie890/CS102-Algorithm-Analysis/blob/master/Algorithm Design by Jon Kleinberg%2C Eva Tardos.pdf
pdf of the book

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[hard atlas]

well it's maybe a bit imprecise to call a cubic poly as degree 4, but we can still treat a cubic poly as if it had degree 4, and then just find out that actually the coefficient of x^4 is zero

I don't wanna check the pdf but I assume the arguments would go through if instead of "poly of degree bla" they wrote "poly of degree <= bla"

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[short matrix]

I don't even know where to begin. How would i solve this?

[sudden compass]

what do 2nd,3rd,4th etc. represent?

[silk comet]

probably first difference, second difference (difference of differences), etc

[short matrix]

yeah difference

9-0=9 etc

[sudden compass]

ah k

[short matrix]

<@&286206848099549185>

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[shy willow]

sup

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@short matrix Has your question been resolved?

[short matrix]

help

[shy willow]

with?

i gtg soon tho

[short matrix]

.

oh okay

[shy willow]

its differences

the bottom number - the top number

[short matrix]

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[night radish]

does the connection point between the three medians in a two equal sides triangle is at the middle of its main height?

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[woeful hull]

180-x = 120?

x=?

do you know special triangles?

[regal bane]

You'll want a unit circle open nearby, if you're not used to your unit circle yet

[woeful hull]

i mean if he is given sin(180-x) = sinx he doesnt really need the unit circle

yeah you cant

you need to give an exact value

i mean i would say use special triangles but i dont know if you learned that

prob did

do you have figure 25?

you see how they are getting 1/sqrt(2)

they are using special triangles

can you send a picture of figure 25

and also make sure to review ur notes

cause this is basic trig that should be taught to you

oh that

you know soh cah toa?

ok draw a triangle with a right angle and 60 degrees or pi/3

and draw it in the first quadrant where everything is positive

this

whats the angle on top

-60 but yeah

its 30 degrees

this is a special triangle

1,2,sqrt(3)

then you just do soh cah toa

https://math.libretexts.org/Courses/Rio_Hondo/Math_175%3A_Plane_Trigonometry/01%3A_Right_Triangles_and_an_Introduction_to_Trigonometry/1.02%3A_Special_Right_Triangles#:~:text=Hypotenuse equals twice the smallest,x√3%3A2x.

i suggest you read this

no matter what value

you put in the shortest side

there will be a multiple of 2 on the hyp, and nsqrt(3) for some int n on the longest side

which is what we have dont we?

sin60

sin120 is identical to sin60 but is in a different quadrant

yeah

you can be asked sin270

but the thing is

that all the angles are the same

i mean its pretty easy to memorise

2 special triangles

thats it

you use that for everything

cause 270 degrees is the same thing as 90 degrees

225 degrees is the same as 45 degrees

its just in a different quadrant

then you use a calculator 👍

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[vast shale]

Corresponding angles are equal

Equate both of them

You're welcome

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[hazy kelp]

What is the proportionality constant of the equation y=3/x ?

[near totem]

I dont think there is one

[hazy kelp]

yea i can't find it either but there is no option for "no solution"

<@&286206848099549185>

i looked at the answer and they said it was 3/x but its just wrong

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[leaden berry]

i got the max horizontal range as 118m

Vf as 38.6

and max height as 27.5m

[timber cypress]

What's the question

[leaden berry]

can someone confirm pls

asking for the 3 values ahead

above

final distance, max height and final velocity

[vast shale]

Uhhhhhh

,w solve -17 = 14.28t - (1/2)(9.8)t^2

[leaden berry]

??

[twin meteor]

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Requested by eragon._._.

[vast shale]

Wait

Yeah

118 metre is correct

,w solve v^2 = (14.28)^2 +2(9.8)(17)

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[vast shale]

,w solve v = √((30.81)^2 + (23.17)^2)

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[vast shale]

Your Vf is correct too

,w solve s = ((14.36)^2)/((2)(9.8))

[twin meteor]

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[vast shale]

I got 17+11.5 when i took g = 9.8

So i mean your answer is correct too

@leaden berry all three correct

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[leaden berry]

how do I do this

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[cinder oyster]

Hi so they ask for domain and range

i did vertical line test and found out it was function

but is domain and range look right?

or it could be [-3,-1] for rnage omgg

[leaden berry]

its not a function boss

all is wrong

[cinder oyster]

watttttttt

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[dusky wasp]

Is this correct?

[timber cypress]

No

[dusky wasp]

Damn.

Where did I go wrong?

[timber cypress]

(xcosx)'≠x(cosx)'

Use the product rule

[timber cypress]

Nahhh💀

[dusky wasp]

Fuckkkk

[timber cypress]

Why you going on an infinite series bro

[dusky wasp]

That’s what I’m sayin

[timber cypress]

You solved it on the first line

[dusky wasp]

I see the motion

I’m a fucking wizard bro

Thx

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[upbeat lotus]

hi

[vocal sleet]

Please don't occupy multiple help channels.

[vocal sleet]

Please stick to your channel.

[upbeat lotus]

k

[karmic imp]

.close

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[crimson grove]

Could someone help me with everything past b?

For b, I'm saying since the graph of the function is concaving upward, the derivative is likely to be positive

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<@&286206848099549185>

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[echo smelt]

sup

[trail notch]

hi

[echo smelt]

aw man

[trail notch]

lmao sorry

[echo smelt]

wasnt quick enough

its aight

[trail notch]

ik how u feel

[trail notch]

idk how to know if a equation is linear or not

[simple mason]

!show

[vocal sleet]

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

[simple mason]

!statues

[trail notch]

idk where to start

cuz idk what is a linear quesiton

bro 😭😭

[simple mason]

What is the fromula of the liner equation?

[trail notch]

idk

[simple mason]

Ah

You need to watch a video

About liner equations and its properties

https://youtu.be/bAerID24QJ0?si=nxP_BNkt6LTvK3wk

[trail notch]

can u tell how to know if an equation is a linear one?

[simple mason]

https://youtu.be/Ft2_QtXAnh8?si=d5hiZbIjVN8zx7kd

All of those will help

[simple mason]

If it looks like a liner equation

[trail notch]

that’s not helpful to me

[karmic imp]

If x^1 is the highest power

[trail notch]

so no exponents?

an equation w no exponents higher than 1 is a linear equation?

[karmic imp]

Pretty much, but y = 1/x, even though that's 1/x^1, applying exponent rules, it's actually x^(-1) hence not linear because it's not x^1

Like it truly has to only contain x^1

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[woeful hull]

Can someone help me with this

[woeful hull]

I have no clue what to do

I got some help earlier but don't know how to finish the proof

<@&286206848099549185>

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[woeful hull]

<@&286206848099549185>

[harsh valley]

Where's the question bro

[woeful hull]

i have to prove it by contradiction

?

its implied in the img i sent

i have a claim that i have to prove

[harsh valley]

A is included in C so A-C=ø

[woeful hull]

i have to prove it

that doesnt help me

i need to do by contradiction

this is what i have done so far

[harsh valley]

Just say that first inclusion implies that a is included in C

In your final implication you said A is not included in C

Contradiction therefore A-C=ø

[woeful hull]

what

i started off with A-C has elements

bwoc

so i need to get to AxB is not a subset of BxC

so i can have a contradiction

thus prove the claim

i know what i need to do but don't know how

[harsh valley]

You want to prove that A-C=ø yeah?

[woeful hull]

no

thats not what i am doing

[harsh valley]

you're proving it by contradiction

[woeful hull]

well yes in an implication

if AxB is a subset of BxC

[harsh valley]

Then A is subset of C

[harsh valley]

The negation of this is a is an element A and a is not an element of C

[woeful hull]

yeah but how to i get to the point that A is a subset of B which is a subset of C

[woeful hull]

yes

i cant have this in cases btw just 1 whole proof

I know its similar to this

[harsh valley]

(a;b) € A,B => (a;b) € B,C

a € A => a € B

And so on

You get A is a subset of C

In the end

[woeful hull]

wouldnt my final statement have to be AxB is not a subset of BxC

pre sure thats the case

[woeful hull]

like this

[harsh valley]

No, the implication of A being a subset of C is that A-C=ø but our premiss was A-C≠ø

[harsh valley]

Yeah or like that

same thing tbh

you just have to find the contradiction

[woeful hull]

wdym

cause this claim is almost the same as the img i sent

[harsh valley]

Yes just say A€C and A-C≠ø => A not included in C

[woeful hull]

i cant just do that though

i have to show every step

and every step has to have a certain reason

i don't even know what you are saying

i get what ur saying

but i dont think it applies in what im trying to do

.close

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[tardy belfry]

three letters are selected from the 9 letter of the words CAMERAMAN. find the number of selections of the three letters include at least 1 one M

[tardy belfry]

ans is 16 but i got 15

i did 5C2 + 5C1

[dark kiln]

you're missing aam

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[vast shale]

I have 0 clue what Im doing and ppl keep trying to explain me through it but Im still not understanding

I need someone to pen paper it so I can read the math

If possible

[rustic portal]

I tried so many things and have nothing. I’m so sorry

Maybe any of this makes any sense or helps you think of anything

[vast shale]

does the altitute cut the

hypotenus

in 2

?

or is that an assumption]

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[fiery wave]

I thought we just add these two vectors together right????

[fiery wave]

What am I missing

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[vast shale]

Can someone help explain this to me? I don't see Alice and Bob can compute x without sharing x_A and x_B?

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[barren folio]

can someone explain this

I dont know my infinite series

[timber cypress]

What part of it do you not understand

[barren folio]

hi i do not get the expression in the denominator

how does the sereis become that

are we using this formula:

if so what exactly is r in this example

[flat whale]

e^t=r

[barren folio]

do I need to show the series itself

like can i just say by inspection that r = e^t

[flat whale]

Yes this is fine

[barren folio]

for this one, since it begins at x = 1 do I need to define r in terms of r^k-1

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[honest heath]

How would I go about the second half of part 4?

[honest heath]

Like, is there anything to be said about parallel lines other than that they have the same slope? How could I incorporate that into a computation for a proof?

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[gentle solar]

I get stuck at:
$$∫2π(4sinhθ)[(2sinh2θ)^{2}(4coshθ)^{2}]^{\frac{1}{2}}dθ$$

[twin meteor]

KnightOnF1 ♘

[gentle solar]

Using cosh²x + sinh²x isn't helpful here due to the doubled argument of the sinh

[vocal sleet]

@gentle solar Has your question been resolved?

[gentle solar]

<@&286206848099549185>

[grizzled halo]

$\sinh2\theta=2\sinh\theta\cosh\theta$ innit

[twin meteor]

chlamydia

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[pallid cloak]

Prove that var(𝑋 + 𝑌) = var(𝑋) + var(𝑌) + 2 cov(𝑋, 𝑌) using the definition
cov(𝑋, 𝑌) = E[(𝑋 – E(𝑋))(𝑌 − E(𝑌))].

[pallid cloak]

Im stuck at how to begin this Problem

Is this the Right direction

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[pallid cloak]

<@&286206848099549185>

[pallid cloak]

<@&286206848099549185>

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[pallid cloak]

.close

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[river kettle]

Finding the area bound by the curves y = sinx and y = cosx for pi/4 to 5pi/4, did I do this right?

[river kettle]

Also I’m not sure if the labelling of the area is correct t but

[blissful fable]

labelling is incorrect but integral is correct

[river kettle]

Rip again

What part of the area am I finding?

[blissful fable]

A doesnt include these parts underneath cosx

remember, it's between the 2 curves

[river kettle]

Oh so like this?

I see so it has to be completely enclosed by the two curves

[blissful fable]

yes

[blissful fable]

exactly

[river kettle]

That makes sense

Okay thanks

.close

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[vague prairie]

Hey, what's the mistake in this? The answer is that it doesn't exist

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@vague prairie Has your question been resolved?

[chrome drum]

U din multiply t in the numerator

In 3rd step

Oh wait
My bad
Din see the whole root

Mayb when you take the whole function out of root, u get a positive and a negative value
So does it hav anything to do with it ?

[vague prairie]

whatever, I know that my solution is wrong, but I think if I've done valid operations on a true statement then I can't get to a false answer

I don't know which step is wrong

<@&286206848099549185>

[fathom minnow]

t approaching zero

then the denominator would become zero

[vague prairie]

yes it's a 0/0 form

[fathom minnow]

yes

[vague prairie]

why is my answer wrong?

[fathom minnow]

can you send the question

[vague prairie]

is at the top of my solution

[fathom minnow]

directly

putting x to 2 in the limit

[vague prairie]

I thought it is 0/0

is it?

[fathom minnow]

yes

do you have the solution

?

[vague prairie]

yes

it's some another way

[fathom minnow]

can you show me

[fathom minnow]

the assumptions are contradicting the answer

thats why it doesnt exist

[vague prairie]

wdym?

[chrome drum]

Bro like i said, its the function when coming out of the root will need a mod
So if u consider the limits there, the answer will be doesn't exist

[fathom minnow]

true

[vague prairie]

nothing is coming out of the root in my solution

can you point out the line

[chrome drum]

Umm ok a sec

Isn't it how u solved it ?

[vague prairie]

that's not my solution, my sol has no sin

my sol is the first image

not the latter

[chrome drum]

How did u arrive from step 3 to step 4 then?

Yeah
So how did u get the 1/2 ?

[vague prairie]

the cos stuff * 4 is under the root, the cos stuff is a proven result, lim x->0 (1-cosx)/x^2 = 1/2

got any answers?

[chrome drum]

But applying it in this particular case, u need to think of the square root

[vague prairie]

wait

it approaches 1/2 and is under the root so applying root to 1/2 gives the 1/root2 and we multiply by 2 to get root2 as the answer

[chrome drum]

Here if u remove the mod
U get two answers
+1 And -1

[vague prairie]

that's not the point, I'm asking why is my answer wrong?

what have I done wrong?

[chrome drum]

Do u hav the proof or derivation for the cos thing ?

[vague prairie]

google it bruh

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[mortal kite]

Yoonie is a personnel manager in a large corporation. Each
month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately
four hours each to do with a population standard deviation of 1.2 hours. Let Χ be the random variable representing the time
it takes her to complete one review. Assume Χ is normally distributed. Let X
¯
be the random variable representing the mean
time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.
2. Complete the distributions.
a. X ~ (,)
b. X
¯
~ (,)

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[tough blade]

I do not see why the theorem requires q be greater than p and not just any q that is not p.

I referred to some other text and I think it said something like,"say q>p ,swapping p and q if necessary"..but I don't see how you can swap p and q? But still, Why not just let q≠p?

Moreover, considering the example below, then you can't find the primes, that have certain prime p in it's quadratic residues list, which are lesser than p.

[worthy citrus]

Because they're talking about dividing q by p with remainder, this doesn't make much sense if q < p

[tough blade]

by the division algorithm, you still get a remainder that is between 0 and p?

[worthy citrus]

Yeah you're right that is true

[tough blade]

is it just a typo..?

I'm not sure

But i saw some other text that also did the same

[tough blade]

atleast i think so

[worthy citrus]

Ah no if q < p then r = p

5 = 0*13 + 5

Hm

[tough blade]

so?

wait that's not even true

[worthy citrus]

Yeah I mean q

[tough blade]

then r=q

[worthy citrus]

Which is fine

But you can see that the theorem isn't going to be true right bc it would say that p/q = q/p

[tough blade]

yeah then..

oh shit wait..

did i really miss that

[worthy citrus]

You would have to hunt down in the proof where condition q > p is used

[tough blade]

okay..idk how high i was but thx for the help

[worthy citrus]

No worries

[tough blade]

yeah, i'll look into that

.close

.close

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[mortal kite]

Previously, De Anza statistics students estimated that the amount of change daytime statistics students carry is
exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students.
a. In words, Χ = __
b. Χ ~ (,_)

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Please don't occupy multiple help channels.

[mortal kite]

/close

.close

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[vast shale]

hi

[vast shale]

why we actually write sup

not not max

?

[paper depot]

max might fail to exist in an infinite dimensional vector space

so it's a little bit more general this way

[vast shale]

@paper depot thanks

.close

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[ebon pond]

Heyo! Can I get an explanation to this? my answer is (x-2)^2/9 - (y+3)^2/4 = 1 but the correct answer is (x-2)^2/9 - (y-3)^2/4 = 1

Can someone tell me where I went wrong in my solution?

[vast shale]

wait a guy will join

[ebon pond]

Thankies!

[split wind]

-(9y²-54y)

so (y-3) not +

[ebon pond]

if i bring out the 9, wont the -54y turn to +6y?

so the square of 9 is +3

[split wind]

you bring a +9 out, so you don't have to change signs

[ebon pond]

oh wait

oohhhhhhhhhh

so it stays -6y?

9(y^2-6y+9)?

[split wind]

done~

[ebon pond]

wait need more clarification lol

so lets say if i bring the factors out

[split wind]

yes?

[ebon pond]

(4)(x^2-4x) "-" (+9)(y^2-6y+9), the negative sign is kinda just its own thing?

[split wind]

+/- signs are stick to the expression right side of the sign

so it's -9(y²-6y+9)

[ebon pond]

now that makes sense

thank you!!

how do i close lol

[split wind]

type .close

[ebon pond]

ahh thank you again :))

.close

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[thin root]

I don't know why it's meaningless

[thin root]

can someone show me why

{BB, BG, GB, GG}

left hand side 1/3

right hand side, idk what that means

but I think it's still 1/3

so why is it meaning less?

🤔

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[sharp sinew]

so i got the function f:[-1,∞)->[1,∞) and i have to tell if its possible to do f○f, i know thats not possible but i want to know why

[river minnow]

The composition f○g exists if and only if the codomain of g and the domain of f are the same

[sharp sinew]

oh also f(x)=x^2+1

[river minnow]

So f○f should exist when the domain and codomain of f are the same

So, are the sets [-1, ∞) and [1, ∞) the same?

[sharp sinew]

no