help-17

[viral copper]

You wrote it as sqrt(x^2)

But there should be a - outside

$x = -\sqrt{x^2}$ for $x < 0$

[twin meteor]

NEON

[safe nova]

Is the negative sign on -sqrt(x^2) there because we are approaching negative infinity?

[viral copper]

precisely

[safe nova]

That’s where I’m confused because I gotten some wrong when it says approaching positive infinity and the answer is negative

Let me show you another one

My internet slow rn

since it says approaching negative infinty i did -sqrt(x^4)

i figured it out just now

.close

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[fallen pilot]

Hi, I am having trouble proving that a certain set is connected in the complex plane

[fallen pilot]

let's say you ahve 2 distinct points, z1, and z2

define K = C - {z1,z2}

and now I am trying to prove that K is connected

the definition of connected is that if it is not disconnected, where disconnected means

$\exists A, B$ open sets such that $K\cap A \neq \emptyset, K\capB \neq \emptyset, K \subset A \cup B$ and $A\cap B = \emptyset$

[twin meteor]

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

[fallen pilot]

so I just ahve to prove that there does not exist such A, B open set

but I'm not sure where to go from here... I tried contradiction, but can't seem to find the way yet

any help/pointers would be very much appreciated

[brisk moss]

showing K is path connected might be worth looking into

[fallen pilot]

right, but how would I write it down formally?

i think i get the idea of that

there are infinitely many paths between 2 points in C, so just removing z1, z2 from C doesn't disconnect all paths

but uh

[brisk moss]

you could show any 2 points in C have 3 paths between them that don't overlap

[fallen pilot]

ah

and by removing 1 point

you are removing maximum 1 non-overlapping path

[brisk moss]

not sure wym, the point was that at least one of those paths must not go through z_1 or z_2

[fallen pilot]

yea that's what I mean

ok I can work with this

tysm for your help

[brisk moss]

sure thing 🙂

[fallen pilot]

!close

/close

uh

.close

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[solid laurel]

Ive been tryna do this problem for a really long time and I think it results in a no soluotion

Ive been at it for a long time, i just wanna make sure im right and that it has no sol

[quick estuary]

what have you tried

[solid laurel]

ive tried basically everything

[quick estuary]

thats a -1 right[

[vocal sleet]

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

[solid laurel]

yes

it is

dont mind the messy writing

[quick estuary]

im p sure thats supposed to be +1

if it is a -1

then yeah

[solid laurel]

but it is a negative right

on here

[quick estuary]

i suppose

[solid laurel]

alright

thanks

[quick estuary]

lol this comes 0

[solid laurel]

pardon?

[quick estuary]

the lhs

is 0

[solid laurel]

meaning..

that theres no sol obv

right

[quick estuary]

that the left hand side is always 0 whenever defined

yeah

[solid laurel]

alright

thanks for your help

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@solid laurel Has your question been resolved?

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[vast shale]

Need to find the perimeter of this shape in factorised form.

Any help?

[tidal umbra]

okay

so whers the problem at?

[vast shale]

I don't know how to factorise this shape I haven't done this type of question before

I need to find x

[zinc quail]

you can't find x from what is shown above

[obtuse sierra]

is the area or perimeter given?

[vast shale]

No

[obtuse sierra]

anything other than this data?

[silk ether]

no, i think it's asking you to find the perimeter with x in it, so find the unknown side first

[vast shale]

If I split the shape in to half it will become 2 rectangles

[silk ether]

you dont need to do that to find the perimeter

assuming these are are parallel, what do you think this side length is?

[vast shale]

side length x?

or the long top one

[silk ether]

yeah the long top one

because thats the only missing side length

[vast shale]

3? or 3x?

[silk ether]

no

[tidal umbra]

i dont think so you can find x

[zinc quail]

yea thought the same before, you shall not find it

just write the perimeter as a func

[silk ether]

its the sum of the sides opposite to it

which would be......

[vast shale]

1?

No

It would be

[zinc quail]

perimeter = 3+4+3+1+x+(3+x) @tidal umbra

[silk ether]

have a think just imagine x is any number

[vast shale]

the sum of the two opposite sides would be 7?

[silk ether]

no the sides opposite to the long one

[vast shale]

3+x

[silk ether]

yep

so now you know every side, what is the perimeter

[vast shale]

but what what was top sides length did we figure that out?

so its 3x on the top as well?

[silk ether]

no this is the top sides length

we dont need to figure out what x is

[vast shale]

Okay

[silk ether]

the side length on the top is just (3+x) because the opposite sides add to it

(assuming they're parallel)

[vast shale]

Ok so the total would be

13+x but if i right that as 13x its prob wrong

[silk ether]

ok so

13+x and 13x are different

13x implies it is $13*x$

[twin meteor]

jamar

[silk ether]

13x is multiplying both, whereas 13+x is adding these two

[vast shale]

Oh ok

[silk ether]

and 13+x is still wrong because you need to add every side

[vast shale]

13+x squared?

[obtuse sierra]

no

[tidal umbra]

yeah, i thought the same, but then he said you need to find x, so i got confused

[vast shale]

so in that case there is no need to find x?

[silk ether]

you can't

the most you can find is the domain

but the perimeter can be found

[vast shale]

Which would be 13+x+x?

Thats all the sides added up

[silk ether]

yep

and 13+x+x=13+2x

[vast shale]

Ok

So how will i factorise 13+2x?

[silk ether]

that can't be factorised

do you have the actual question?

[vast shale]

Yes

The question says

"Find the perimeter of this shape in factorised form".

[silk ether]

13+2x is the simplest most factorised form

can i ask where you got this question from lol

[silk ether]

oh you missed a side btw

[vast shale]

How come that side is 1?

14+2x

[silk ether]

4-3=1

[vast shale]

Oh.

So 14+2x cannot be factorised?

[silk ether]

that can be factorised into 2(7+x)

because 14 and 2x have a common factor 2

[vast shale]

Mhmm

So the final answer will be 2(7+x)?

[silk ether]

yes

[vast shale]

maths assignment bro its a headache

[silk ether]

the question is written kind of poorly

cos you need to assume opposite sides of this shape are parallel for anything to work

[vast shale]

Ik the teachers are kinda retarted and don't make things work at some points

Write the equation of a line with a gradient of 7 and y intercept of -1

Would that be y= 7x-1?

What will be the equation of a line with a gradient of "m" and y-intercept of "c"?

y= m+c?

[obtuse sierra]

y = mx + c

you forgot the x

[vast shale]

Oh

This is Line F.

Can you help write down the equation in gradient intercept form?

[silk ether]

find the gradient and y-intercepts

and sub it in to y=mx+c

[vast shale]

y would be where the point intercepts the y axis?

1/3 would be gradient?

[silk ether]

y-intercept is when x=0

so yes when the point intercepts the y axis

[silk ether]

no

rise over run

take the section between the two black dots

[vast shale]

that section is a triangle

if you look at the rise its 1?

[silk ether]

no look at the scale again

what is the y intercept

[vast shale]

-4

[silk ether]

yrah

the rise is between y=-2 and y=-4

which is what

[vast shale]

2

[silk ether]

yep

and the run?

[vast shale]

3?

[silk ether]

yeah

so the gradient would be

[vast shale]

2/3

[silk ether]

yeah

sub the values into

[vast shale]

y=2/3x+c

c is 2?

[silk ether]

no

c is the y-intercept

[vast shale]

-4

y=2/3x-4?

[silk ether]

yes

[vast shale]

The bottom of the playground slide is 2.5m from the foot of the ladder. The gradient of the line which represents the slide is 0.3. How tall is the ladder?

Will I need to find X to find out how tall the ladder is?

<@&286206848099549185>

[signal drum]

the gradient is tangent which is relation between two sides of right triangle. You can use that to make up an equation and solve it

[signal drum]

yes

[vast shale]

Ok

Where do I start?

[silk ether]

gradient is rise over run

therefore x/2.5=0.3

rearrange and solve for x

[vast shale]

x= 0.75?

0.75/2.5=0.3

[silk ether]

yes

[vast shale]

.close

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[plain aurora]

For differentiabiliy at a point using the the definition of differentiation:

[plain aurora]

if either the left or right limit tends to infinity

does that automiatcally mean that the function is not differentiable at that point?

without even needing to calculate the other limit

i.e. suppose I were checking differentiability of y(x) at x=0. If $\lim_{h\rightarrow 0^-}\frac{y(h)-y(0)}{h}=-\infty$, do I even need to check the right limit

[twin meteor]

Kalgar

[dull maple]

It does.

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@plain aurora Has your question been resolved?

[dull maple]

Although, i can't think of such a function. Wouldn't it be discontinuous at that place already?

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[plain aurora]

.reopen

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✅

[plain aurora]

@dull maple what does it mean to shrink h from "below"?

i.e. $\lim_{h\rightarrow 0^-}$

[twin meteor]

Kalgar

[plain aurora]

because isn't $h=∆x$

[twin meteor]

Kalgar

[plain aurora]

how would I interpret shrinking h to zero from ... below?!? negative ∆x?

[vocal sleet]

@plain aurora Has your question been resolved?

[bleak oak]

$\Delta x = x_1 - x_2$

[twin meteor]

nadav

[plain aurora]

wat

[dull maple]

They (probably) mean that you are approaching zero from negative side of x-axis.
For example,
values are h are like: -0.000...5, -0.000...4, -0.000...3, -0.000...2, -0.000...1 and so on.

[vocal sleet]

@plain aurora Has your question been resolved?

[dull maple]

He asked the meaning of this. @paper depot What's the issue?

[paper depot]

you are talking about -0.000...1 as if it is a real number

[dull maple]

It's not?

[paper depot]

this sounds like you're talking about a number with infinitely many zeros after the decimal point

[dull maple]

Also, i know that number is infinitesimally small but i just tried to give an example.

[paper depot]

and to answer this question: something like x^(1/3) should work

[dull maple]

Yeah. What's the issue with that?

[paper depot]

it's a bad example!

that's not a real number!

you're inevitably gonna cause confusion!

[dull maple]

Yeah. Had gotten that now.

[dull maple]

How is it not a real number? Not talking about infinitely many zeroes before 1. Ellipsis was just meant to show that it's a very small number.

[paper depot]

if you want a very small number then write a very small number that doesn't look like an infinitesimal.

[dull maple]

Infinitesimal means very small. 😐

Indefinitely small*.

So, that's what i tried to do.

[paper depot]

infinitesimal does NOT mean merely "very small".

[dull maple]

I corrected in next line, literally. 😶

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[urban scarab]

Excuse me, I got stuck at my task.
In the space rectangular coordinate system in the image, a straight line intersect z-axis and x-axis at point A and point B.Point C lies on straight line AB.The centre of a circle shown in the image is Point C, and the normal vector of Circle C is (vector)AC

$$Known:$$
$$A(x_A,y_A,z_A)$$
$$AC=d$$
$$R_{\odot C}=r$$

[twin meteor]

wrmdcxy

[urban scarab]

I have several problems:(please use the parameter given as constants)
How to get vector AC?
How about the parametric equation of Circle C?(Please use θ{limit:[0,360)} to express it.)

[vocal sleet]

@urban scarab Has your question been resolved?

[urban scarab]

<@&286206848099549185> Can anyone help me?

[vocal sleet]

@urban scarab Has your question been resolved?

[vocal sleet]

@urban scarab Has your question been resolved?

[bleak oak]

do you know B coordinates? or the line?

[urban scarab]

I think B coordinates can be calculated with A coordinates and the angle alpha

[bleak oak]

yes thats true

you have the angle?

[urban scarab]

Exactly in my project it can be known by some calculation (this is part of my project)

I think we can just use alpha as constant to calculate

[bleak oak]

the angle between the line and Z axis or line and X axis @urban scarab

[urban scarab]

the line with X axis

[bleak oak]

tan(phi) = zA/xB

B(zAtan(phi),0,0)

now you can calculate the vector AB by doing B-A

and a general form for the line would be A+tAB

you cna solve for coordinates of the point C as C(x(t),y(t),z(t))

express the distance from C to A in terms of t

and set it equal to d

you can solve for t and get 2 points

one between A and B and the other after A like C A B

[urban scarab]

OK, I see. And what about the circle, please?

the parametric equation of Circle C?(Please use θ{limit:[0,360)} to express it.)

[bleak oak]

one sec

[urban scarab]

Excuse me, how are the things going? Are the given values not enough?

[bleak oak]

nah its enough its just harder

i looked it up and found this, @urban scarab https://i.imgur.com/6aj3WSO.png

id be happy to explain it if you need

in this case OA is AC

[urban scarab]

hmm i'm sorry that because my country's internet limit , i can't open this image...

[bleak oak]

oh uh

[urban scarab]

i'm not in America sorry

[bleak oak]

its ok me neither

do you know how to find vector perpendicular to AC?

[urban scarab]

sorry i don't . im just enter high middle school

[bleak oak]

you can a vector (x,y,z) and do dot product with AC and set it to 0

that is $x x_AC + y y_AC + z z_AC =0$

[twin meteor]

nadav

[bleak oak]

yeah sure

[urban scarab]

oh i see

the next step is?

[bleak oak]

and then that vector v_0

and w is v_0/len(v_0)

then you use the formula v(theta) = cos(theta)v_0 + sin(theta)omega x v_0

where x is cross multiplication

make v_0 have lenght of the radius

bit busy so cant give more detailed solution

srry

[urban scarab]

thats ok. it's my pleasure to receive your help.
My problem is solved. thanks a lot!
(In addition is theta radian or degree?)

[bleak oak]

wouldnt matter itd just loop around but if since its 0,360 itd be degrees

[urban scarab]

OK i see thank you so much!

.close

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[dim silo]

Top one is correct?

[distant eagle]

Yes

[dim silo]

.close

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[dim silo]

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✅

[dim silo]

How to solve

For n?

[dull maple]

You can divide both sides with a term of form $(2k)^m$ when k is non-zero.

[twin meteor]

Enemagneto

[dull maple]

Choose such m so that you can get rid of (2k)^p term on one side at least.

[dim silo]

Divide by (2k)^n-2?

[dull maple]

(2k)^(n-3) would be a better choice but both work.

[dim silo]

But how will i divide it after

[dull maple]

Use the exponent rules.

Are you not familiar with computing $\frac{a^{m}}{a^{n}}$?

[twin meteor]

Enemagneto

[dull maple]

If not, as a reminder:
$\frac{a^{m}}{a^{n}} = a^{m-n}$

[twin meteor]

Enemagneto

[dim silo]

Im supoosed to get n=6k+2 what did i do wrong?

[mild flower]

well for one thing you didn't multiply by 3 correctly, but are you sure that's the right answer? it doesn't seem to match your original problem

[dim silo]

Im so close whats the problem?

[mild flower]

3(n x 2k) ≠ 3n x 6k

3(n x 2k) = n x 6k or 3n x 2k

[dim silo]

I dont understand why it is wrong?

This is what i did

[mild flower]

2(3*4) is not 6*8 that's not how the distributive property works

distributive property applies specifically to a*(b+c)

what you have there is 3 x (n x 2k) which is also known as 3 x n x 2k

[dim silo]

So it equals 6kn ?

[mild flower]

yeah

but all of this is mostly immaterial since i think your original equation isn't right

can you show me how you got to that equation?

[dim silo]

Sure

Here

[mild flower]

you crossed out n on one side but not the other

[dim silo]

Oh

That just becomes 0?

Or 1?

[mild flower]

1

[dim silo]

Ill try again now

Finallt i got it🙏🙏

Thank you

[mild flower]

wooo

[dim silo]

.close

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[vast sparrow]

I'm stuck on both of these ;/ only ones i have left

the second one im way farther like I know dy/dx = 1 but how does that equate to anything

[peak matrix]

I think in the exercise they wanted you to find 2 different representations of the point

[vast sparrow]

I know like HOW does that even work? If it's X AND Y there can't be like two different

like how can their be two different points for that since it's given as X,Y

[peak matrix]

There can be 2 different represantions, one of them with negative r

the problem is that one of your answers is correct and it didnt get accepted

[vast sparrow]

oohh.. I see

[peak matrix]

oh wait it isnt correct

[vast sparrow]

welp nvm

[peak matrix]

Yeah, your answers aren't correct

[vast sparrow]

but 10 is the circle that the point is on

[peak matrix]

how did you even get sqrt(2)/2?

[vast sparrow]

because the point

the x,y given are -5sqrt(2)

which is the same as -sqrt2/2

same number

[peak matrix]

but you need to find the angle they make

supposedly in radians

full circle is 2pi, right

[vast sparrow]

why is this even on the homework it isn't explained in any way

yeah?..

so r is in radians???...

[peak matrix]

do you know polar coordinates?

Polar coordinate represents point as (r, theta)

where r is distance from origin and theta is angle between point and positive x axis

like this angle

[vast sparrow]

yeah definitely

but r is 10 right?.. like it has to be

what else can it be

so like 5pi/4 ???

cuz the unit circle

so that's theta ?..

[peak matrix]

Yeah that's theta with positive r

[vast sparrow]

but you'ce gotten multiply is by 10

[peak matrix]

so (10, 5pi/4) is first solution

[vast sparrow]

oh wait no cause r = 10

so what about the second one like? positive radius?

could you do 10, -pi/4

well no

mane I really just don't know

[peak matrix]

it's the r that would be negative

[vast sparrow]

okay so I got the larger one correct (10, 5pi4)

but the negative which I put as (-10, 5pi/4) is still wrong

[peak matrix]

great, for the second one I will try finding video

[vast sparrow]

I don't see how theta could be negative

[peak matrix]

it's hard to explain how does negative r work

[vast sparrow]

hmm

yeah they don't go over this at all

[peak matrix]

so basically, negative radius flips the point around

[vast sparrow]

OHH

YEAH

okay yeah

[peak matrix]

So do you know the second solution?

[vast sparrow]

makes sense I got it

[vast sparrow]

yep it's -10 pi/4

so this one

the second

I feel like im super close like

the bottom is 0 like always

like dy/dx = 1

at theta = 0

so how do I like go aobut

nvm I got it!

[peak matrix]

Are you sure?

[vast sparrow]

.close

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[marsh coyote]

Im trying to go about showing that row equivalence is an equivalence relation and am having trouble showing that symmetry holds.

If A is row equivalent to B, then theres a sequence of elementary row operations that turns A into B. Intuitively, I can just reverse this sequence of elementary row operations to get from B to A but im not sure how to word this rigorously. Am i suppose to be using induction here to show that any n sequence of elementary row operations can be reversed to get from B to A?

[bleak oak]

have you seen that each row operation can be expressed as a matrix?

[marsh coyote]

you mean like R_2->R_2+3R_1 for example to get a new matrix?

[bleak oak]

no

whatever

[marsh coyote]

actually i dont think so, i dont know that row operations can be expressed as matrix

[bleak oak]

you could use induction yeah

show that each matrix operation is invertible

then if you get from A to B by the sequence of operators a1 a2 ...

then you get from B to A by an^-1 an-1 ^-1 ... a1^-1

[marsh coyote]

alright i think i get it, just to confirm, in the definition of row equivalence, the sequence of elementary row operations is taken to always be finite right?

[bleak oak]

yeah

i mean theres no real meaning to infinite sequence here

[marsh coyote]

ye my notes just say sequence so i wanted to confirm

thanks alot for the help, i think i can do it now

.close

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[dim silo]

how to simplify 30276 into form K root 3 ?

[dim silo]

where k is an integer

[paper depot]

$30276$ cannot be simplified into $k \sqrt{3}$ for $k \in \bZ$.

[twin meteor]

Ann

[paper depot]

k sqrt(3) is never an integer (except for k=0), but 30276 is.

[cedar kernel]

hm

ah right

integer

nvm

[dim silo]

what did i do wrong here then

[vocal sleet]

@dim silo Has your question been resolved?

[dim silo]

Better quality image

[vocal sleet]

@dim silo Has your question been resolved?

[dim silo]

<@&286206848099549185>

[hard plaza]

second line of solving
i think for the coefficient 60. the power is 4. did you typo? or err?

[dim silo]

which part are you saying is wrong?

[hard plaza]

the power?

[dim silo]

oh i was looking at the first image i sent

so i was confused

[hard plaza]

i just used calculator instead of writing it all down and your k value should be 1560
subtract those two binomials. then divide by radical3

[dim silo]

yes the power for that should be 4

looks like i missed it

[dim silo]

what is a radical?

[hard plaza]

fixing that gives u the answer i hope? because the working out is pretty fine. maybe a calculation error if u have one

radical is another word for square root forgive me

[dim silo]

dont worry ill try this again with the correct power hopefully it works out

[hard plaza]

sure

[dim silo]

i got the same answer tf

[hard plaza]

lmao did u work it out? attach a pic if u did

[dim silo]

wdym

i did the one with x as -root 3 and it gave me -2573.99963

[hard plaza]

one mo

[dim silo]

I fixed the power and got that

[hard plaza]

are u writing it down on a calculator? you're supposed to get decimal values

[dim silo]

yeah i put it into my calculator

[hard plaza]

think ur entering it wrong then. mind sending a pic of your equation?

[dim silo]

This?

[hard plaza]

this is fine tho-....cant see how youre getting tht answer

[dim silo]

Oh wait i know why

I put it like this

60(-root3 ^4) and it made it negative

[hard plaza]

yeah thats what lol calculator entry error

you're fine

i got this working it out. this is just for (2-sqrt(3)) ^6

[dim silo]

Now i get 3.7 x10^-4 is that right

[hard plaza]

yass

slay

its right

[dim silo]

Finallly😭

[hard plaza]

after doing the algebra of those two binomials. ull prolly have a decimal right?

[dim silo]

Yed

Yes

[hard plaza]

divide it by sqrt(3)

the value u get is k

hope uk why

if not lmk

actually lmao this is easier. working it out

[dim silo]

I got it🥳🥳

[hard plaza]

yes good

[dim silo]

Ty man

[hard plaza]

np

[dim silo]

Can you take a look at another question for me?

[hard plaza]

keep tab open i was in a hurry to leave out mb

or else u can dm me ill check it when im back

[dim silo]

Alr

[dim silo]

.close

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[vast shale]

Is it possible to build a recurrence relation for the sum of the first n^2 numbers?

[vast shale]

I have been thinking about it but nothing really clicks yet

[worthy citrus]

you mean sum of the first n square numbers?

[vast shale]

Yes

[somber yew]

Like $a_n = \sum_{i=0}^{n^2} i$ ?

[twin meteor]

DAILI

[vast shale]

Oh no

What sigma wrote

[worthy citrus]

sure, sum of the first n+1 is the sum of the first n, plus (n+1)^2

[vast shale]

What I said was just worded badly

[urban edge]

https://youtu.be/aXbT37IlyZQ

[vast shale]

You know no joke that was my line of thinking

But I realised

[urban edge]

You can find the first few terms and then hope that some nth order difference is constant

[vast shale]

[
a_n = a_{n-1} + n^2
]
Doesn't quite....make sense

[urban edge]

Wtf

[worthy citrus]

it would be a_{n-1}

[urban edge]

Yeah

[vast shale]

Yeah

Okay fair enough

[urban edge]

Explain

[vast shale]

I mean would that be satisfactory

[urban edge]

Oh yeah that is recursion

$a_n-n^2=a_{n-1}$ clearly shows removing the highest term

[twin meteor]

GarlicBredFri

[vast shale]

Removing the highest term?

[vast shale]

Oh so that wouldn't be valid be a_(n-1) would just be like

Linear?

[urban edge]

Wdym

[vast shale]

What's your point with what you said

[somber yew]

It's the same formula

[urban edge]

That is a recurrence relation

[vast shale]

Yeppie okay that works

Tyy

. close

.clowe

.closef

.close

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[final sierra]

Can someone just walk me through the process? I know sin^2 + cos^ 2 = 1

[civic otter]

Nice, that is the correct equation to use

[final sierra]

And 1 + tan^2 equals sec^2

I just get hung up I guess with the 1/2 exponent?

[civic otter]

Now, since in the question you have cos² and in all the answers there is a sin², you'd better explicitate cos² in terms of sin² from the fundamental trig equation

[civic otter]

You don't need this

[final sierra]

Gotcha

[civic otter]

In other words, you have $\sin^2(x) + \cos^2(x) = 1$. Therefore $\cos^2(x) =$?

[final sierra]

Yeah this is where I get hung up I think, in the order of operations bit

[soft walrus]

\sin and \cos make them look better btw :)

[twin meteor]

Alberto Z.

[civic otter]

Thanks

[soft walrus]

[final sierra]

I don’t know what I can touch first

[civic otter]

You can, if it helps, substitute sin²(x) with S and cos² with C @final sierra

[final sierra]

Since it’s all under the 1/2 exp

What can I actually do

Cos would equal 1-sin

[civic otter]

2 not 1/2

Or you are thinking of another equation??

[final sierra]

Wait why it’s a square root right

[civic otter]

No

[civic otter]

Do you agree on this equation? @final sierra

[final sierra]

3-2cos^2x under the square root

[final sierra]

Yes

[civic otter]

Forget about the exercise for a moment

[civic otter]

Nice

Can you answer that question?

[final sierra]

It should equal 1-sin

[civic otter]

No... is it sin + cos = 1 or sin² + cos² = 1 ? Which one is correct?

[final sierra]

Both of them squared

It should be 1-sin^2

[civic otter]

So now we know that $$\cos^2(x) = 1 - \sin^2(x)$$

[twin meteor]

Alberto Z.

[final sierra]

Yes

[civic otter]

Therefore, instead of cos^2(x) in the expression under the square root we can replace itwith 1 - sin^2(x)

[final sierra]

Yea

Kind of twists me up but I get it

[civic otter]

So, what do you get under the square root?

You had $$\sqrt{3 - 2\cos^2(x)}$$

[twin meteor]

Alberto Z.

[final sierra]

I’m only getting wrong answers

It’s not 2-2sin^2 right

[civic otter]

Ant this is equivalent to $$\sqrt{3 - 2 \cdot \left( 1 - \sin^2(x) \right) }$$

[twin meteor]

Alberto Z.

[final sierra]

Oh yeah looking like that is much better

[civic otter]

Which equals $$\sqrt{3 - 2 + 2\sin^2(x) }$$

[twin meteor]

Alberto Z.

[final sierra]

Oh wait

You don’t distribute 1 into 1-sin^2?

[civic otter]

Distribution means multiplication and addition... but there's no 1 multiplied by something

[final sierra]

The 3-2 times 1-sin part

[civic otter]

$a \cdot \left(b + c\right) = ab + ac$

[twin meteor]

Alberto Z.

[final sierra]

Here

[civic otter]

This is distributing the a over the sum of b and c

[final sierra]

Yeah I understand

But you have 3-2 times 1-sin^2 there

[civic otter]

I've never written this 😅

[final sierra]

^^^^

[civic otter]

Ah there ok

What didn't you understand of that step?

[final sierra]

3 minus 2

Outside of 1-sin squared parentheses

Why don’t we multiply -2 into 1-sin

?

[final sierra]

But it’s still multiplication binding the -2

Not even foiling it

Order of operations states id have to multiply before doing anything with the 3

?

[crimson jetty]

That's what has happened here

[final sierra]

Yeah

Yeah idk what that person was talking about

?

?

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@final sierra Has your question been resolved?

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[vocal sleet]

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[vast shale]

Please tell how to get the answer 😢 I don't know what went wrong lol

[pastel ruin]

its correct, till step 3

[pastel ruin]

i think, u didnt take the root, its √100

not just 100

[ornate ember]

Expand out the first few terms and you'll see a pattern you can eliminate (called a telescoping series)

[soft walrus]

hey there Mellow!

[ornate ember]

awww hello Fancy!

[vocal sleet]

@vast shale Has your question been resolved?

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[elfin moon]

5×mean deviation =4 ×standard deviation

[elfin moon]

Is this correct?

[dull maple]

Only for moderately symmetric distribution.

[elfin moon]

Please elaborate

[vocal sleet]

@elfin moon Has your question been resolved?

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[elfin moon]

The event of getting 'H' on first coin and the event of getting 'T' on the second coin in a simultaneous
toss of two coins are independent events

<@&286206848099549185>

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[elfin moon]

The event of getting 'H' on first coin and the event of getting 'T' on the second coin in a simultaneous
toss of two coins are independent events

[dreamy viper]

agree

[elfin moon]

why?

[dreamy viper]

how would a coin toss somewhere else affect anything

[elfin moon]

but outcomes probability can?

[frozen bobcat]

P(First is heads | second is tails) = P(Heads then tails) / P(second is tails) = (.25)/(.5) = .5 = P(First is heads) --> independent.

[elfin moon]

i did not get it?

please write with latex

[frozen bobcat]

$P(First\ is\ heads \mid second\ is\ tails) = \frac{P(Heads\ then\ tails)}{P(second\ is\ tails)} = \frac{.25}{.5} = .5 = P(First\ is\ heads) \implies independent$

[twin meteor]

Zybikron

[frozen bobcat]

two events are independent if $P(A\mid B) = P(A)$.

[twin meteor]

Zybikron

[elfin moon]

i guess you wrote it wrong

head probablity should be denomenator

4 total outcomes

HT is favourable outcome

[elfin moon]

1/4 probability

[frozen bobcat]

nope

[elfin moon]

we put downside p(B) which happened first

[elfin moon]

now see here

[frozen bobcat]

ok, what are you calling events A and B in your problem then?

[elfin moon]

0.25 how?

[frozen bobcat]

how many outcomes are there when flipping two coins? How many of those outcomes are HT? Whats the probability of getting HT?

[elfin moon]

as I have wrote above

4

HT=1

[frozen bobcat]

So P(first heads and second tails) = 1/4

[elfin moon]

1/4/2/4

this is correct

i got it

you should not have written it on decimal

[frozen bobcat]

ok then. Bye

[elfin moon]

.close

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[elfin moon]

I got u =+-2

Question 4

[untold surge]

b

[elfin moon]

I am not asking answer

The solution is incorrect

U^2 =4

But they wrote u^2 =2

[untold surge]

If u you mean P(X=k)=u^k e^-u/k!. Then yes u=2. Standard deviation of a Poisson distribution is sqrt(u).

[elfin moon]

.close

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[elfin moon]

Question 15 again

[elfin moon]

We can feel they are independent events

But we cant check by p(A/B)

[untold surge]

P(A fails or B fails)=1-P(A succeeds and B succeeds)

=1-(1-P(A fails))(1-P(B fails))

[vocal sleet]

@elfin moon Has your question been resolved?

[elfin moon]

My question is independent checking

@untold surge

[untold surge]

That’s probably a given condition without saying

[elfin moon]

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[celest crystal]

i still dont understand this, can anyone pls explain this to me in easier and simplier term?

[celest crystal]

i read it like 6 times already

[ebon rapids]

Ok, so first off you’re given the mean of the 9 numbers

If your tenth number is also 42, the mean will still be 42, which is clearly not greater than 42

Well, first, do you know what the mean is?

[ebon rapids]

what did you try anyway?

[vocal sleet]

@celest crystal Has your question been resolved?

[rugged orchid]

At what point in the explanation did you stop following?

[vocal sleet]

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[celest crystal]

.reoipen

.reopen

[vocal sleet]

✅

[celest crystal]

i went to toilet sryr

[celest crystal]

i do erything add up, which result is 378 right?

and i use that + x all divided by 10 = 42\

basiclaly i make smthing like an equation

[celest crystal]

that last part

the this means part

[rugged orchid]

Do you understand why it’s divisible by 10?

[celest crystal]

because theres 10 number in this means?

[untold surge]

10| (9 times 42 +x), 60>=x>42, it’s not hard to find the unique x satisfying both

[rugged orchid]

So where does the explanation lose you

Which specific sentence

[celest crystal]

the thus 378+x must have a ones digit of 0? i guess

and x has a ones digit of 2, the only possible value of x is 52.

like

how they even get 52

i got 42 because i work out the equation

i tried redo the equation thing that i make

[untold surge]

x>42

[celest crystal]

ik but like

[untold surge]

So can only be 52

[celest crystal]

its 42<x<60

[celest crystal]

how do i get to that number

[untold surge]

10|(378+x), x=10k+2

42<x<60, k=5

[celest crystal]

x=10k+2?

what?

whats that

[untold surge]

?

378+x=-2+x mod 10

x=2 mod 10

[celest crystal]

where does -2+x mod 10 came from?

[untold surge]

?

378=-2 mod 10

[celest crystal]

whats mod

[untold surge]

a=b mod c means that c divides (a-b)

[celest crystal]

c/(a-b)?

[untold surge]

(a-b)/c being an integer

[celest crystal]

can u pls reexplain it without using the term mod pls, its a new word for me. it make me confuse

[untold surge]

10| 378+x therefore

10 divides 378-340+x=x-2

So x-2=10k, x=10k+2

[celest crystal]

where does 340 came from?

[untold surge]

c|a, c|b, then c|(a+b).
Here c=10, b=378+x, a=-340

[celest crystal]

i give up, its fine. theres no way im understanding this

thank you for helping

.close

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[untold surge]

Np

I guess you just need to read it again. I have explained everything

[celest crystal]

i just dont understand the part you explain

[vocal sleet]

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[clever forge]

I need help so apparently i can hard maths but i am forgetting basic math.. so i can't do basic math i need help with basic before this test comes in, it's call minimum standard and it requires you to get 3 or 4 lvl and i kept getting 2.

[next sleet]

[clever forge]

so i need someone to teach me and help me out.

[next sleet]

uh all of them?

[clever forge]

yes.

[next sleet]

ok i will try

[clever forge]

mhmm thank you

[next sleet]

q16. 280 passengers, 3/4 of them is how many?

its 280*3/4 right

uh you there?

[clever forge]

so would be 210?

[next sleet]

yeah

ok but what couldn't you get about this question

is it inability to do the calculation, or inability to get that it's the fraction multiplied by the no. of passengers

i'm not criticising, i just need to know where you are going wrong

[clever forge]

i was confused by what is . Three-quarters of the passengers was

[next sleet]

what do you mean

it's a fraction

or 75% if you prefer it in percentage form

or 0.75

fractions, decimals, percentages are all the same thing actually

[clever forge]

ohhh

[next sleet]

it means that 3 out of 4 passengers are adults

so in every group of 4 passengers 3 are adults

[paper depot]

also btw

[paper depot]

.

y'all need to get a new channel

this one will close imminently

like, in less than 5 minutes.

[next sleet]

why is what i said wrong

[next sleet]

ah ok

bruh what

why is it wrong

ok i see

[paper depot]

in a group of 1000 people where 3/4 of them are adults, it's very easy to pick out a quartet where there aren't 3 adults

so you're misleading op

[next sleet]

yes yes we both know that wasn't what i meant though...

[next sleet]

ok fine if you wish to be pedantic

it is possible to split the passengers into groups of 4 such that every group has 3 adults

or something like that

[paper depot]

yes that's better

anyway this channel is on the chopping block

@clever forge are you still here

[next sleet]

indeed

[clever forge]

alrighty thank you but i think i understand it now i won't be needing help

[next sleet]

birdy open a new one asap

before the channel is consumed in the void

[clever forge]

[next sleet]

ah well

[clever forge]

thanks tho

[vocal sleet]

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[soft wren]

solving for x in logarithmic equation problems

[viral copper]

log of what is 0

[viral copper]

Also you cannot simply split logs like that

[soft wren]

Yeah can see that now

[soft wren]

Not sure what you mean

[peak matrix]

log(x) = 0, are you able to solve this?

[soft wren]

Apologies but I still don't understand the question, doesn't log(x) = 0 need a base?

[peak matrix]

it's actually same for all bases, but let's say it's base 5

[soft wren]

Oh well okay yeah, if its log5(x) = 0 then 5^0 = x and x=1 right?

[peak matrix]

yeah, why not do the same here?

you will get a quadratic equation

[soft wren]

Ohh so would that mean 1 = x^2 - 9x + 21?

Then just solve it from there

[peak matrix]

yeah, subtract one from both side and it's a completely normal quadratic equation

[soft wren]

Ahh, cheers didn't see it like that appreciate it

.close

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[bleak prawn]

https://i.imgur.com/Xg2VkXO.png

[bleak prawn]

I dont really know how to start this one

We dont know the location of point P so we cant find point Q and vice versa

And we dont know the location of x = p so we cant get the equation of the line x = p

[round plover]

you know the difference between f(p) and g(p) is exactly ln(4)

[bleak prawn]

Yeye

Wait

So do I get √((ln(9) - 2x)^2 - ln(3 - x)) and set that equal to ln(4)?

[round plover]

it's a vertical line

it's just ln(9-2x) - ln(3-x) = ln4

[bleak prawn]

Ah oke

So
ln(9-2x/3-x) = ln4
(9 - 2x)/(3 - x) = 4
9 - 2x = 12 - 4x
2x = 3
x = 3/2

[round plover]

yep

[bleak prawn]

And thats it?

[round plover]

yeah

[bleak prawn]

Ah nice haha

These questions are always really easy but I never spot how to do them

Like I always understand the maths but im afraid I will do bad in my exam just cause I dont spot the trick

Its annoying haha

But yeah thx again

❤️

.close

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[hollow thorn]

Can someone teach me to prove this sin(A+B) = sin C

I can't figure out how sin C from ?

[hybrid flicker]

alright, since A,B,C are angles of a triangle, what is the relationship between them ?

[hollow thorn]

A + B + C = 180

[merry python]

And you can write sin(180-x) as?

@hollow thorn

[hollow thorn]

Let A + B = 180 - x ?

[hybrid flicker]

well

A+B+C = 180

so what is x in this context ?

[hollow thorn]

uhh idk why

[hybrid flicker]

Using this, A+B = ...?

[hollow thorn]

180 - C

[hybrid flicker]

yes

and sin(180-C) = ?

[hollow thorn]

huh ?

should i to expand that ?

[merry python]

sin(180-x)=sin(x)

[hollow thorn]

aight get it

how bout part (b) ?

same issue ?

[hybrid flicker]

well this time, cos(180-x) = ?

[hollow thorn]

ahh aight

get that tqtq

Am I draw right?

@hybrid flicker

[hybrid flicker]

yeah that's a pretty good drawing, but the proportions may be off

Oh there's a problem it think, BC is supposed to be a side of the rectangle

[hollow thorn]

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[river kettle]

not sure how to begin

[rugged orchid]

What’s the definition of independence?

[river kettle]

an event not affected by any other event?

[rugged orchid]

No

There’s a specific formula

That is the definition of independence

[river kettle]

oh

[rugged orchid]

In terms of probability

[river kettle]

wait

P(A) * P(B) = P(AnB)

that right

[paper depot]

you meant $P(A) \cdot P(B) \overset?= P(A \cap B)$

[twin meteor]

Ann

[paper depot]

also better to stick to the names of the events you're given, so X and Y instead of A and B.

[river kettle]

yes

well P(X) * P(Y) then

[paper depot]

$P(X) \cdot P(Y) \overset?= P(X \cap Y)$

[twin meteor]

Ann

[paper depot]

this is what you need to verify

[river kettle]

yes

how do i do that

[paper depot]

find $P(X \cap Y)$ (through means OTHER than this unverified equality!)

[twin meteor]

Ann

[paper depot]

you might be interested in the inclusion-exclusion formula for 2 events: $$P(X \cap Y) + P(X \cup Y) = P(X) + P(Y)$$

[twin meteor]

Ann

[river kettle]

usually its written as P(XUY) = P(X) + P(Y) - P(XnY) right

[paper depot]

it can be written in a variety of forms.

if you're interested in finding P(X ∪ Y) then the form you put will help you.

[river kettle]

so im interested in XnY then

[paper depot]

P(X ∩ Y)*

but yes.

[river kettle]

P(XnY) = P(X) + P(Y) - P(XUY)

= 0.3 + 0.42 - 0.594