help-28

[abstract anchor]

Okay

Just the range

[gritty patrol]

so the solution would be x>14

[abstract anchor]

I cant find the value of x if only 2 values are given

[gritty patrol]

any value of x larger than 14 fits

[gritty patrol]

correct

[abstract anchor]

Okay

Best i can get is the range

Okay!

I get it now

Yay!!!

[gritty patrol]

mhm. again its exterior INEQUALITY theorem

you cant get x=

hope that helped!

[abstract anchor]

Yes! It did

Thank you very much!

I'm grateful for your time and effort

Thank you!!!!

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[torn jolt]

Where is the similarity here??

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[rugged void]

@proven bane what is your question

@proven bane ?

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[golden shore]

How to solve it

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[unborn vessel]

hi i want to confirm if this is a valid way to answer the question.

[ripe horizon]

It is

but try to write the final thing as $y=\frac{3x+6}{7}$

[glossy valve]

Renz

[ripe horizon]

and $y=\frac{7x+6}{3}$

[glossy valve]

Renz

[unborn vessel]

thank you!!

will do

[unborn vessel]

so can i confirm that if they were inverses, i would have got the same final result for each??

[ripe horizon]

ye

when you inverse the left, you should get the right

while if you inverse the right, you should get the left

but they dont

so they are not inverses of each other

[unborn vessel]

thank you so much

i understand now

also can i confirm with someone that this is correct

[ripe horizon]

i think its vertical translation 2 units up

because when you put the -2 from the left to right it becomes positive 2

and $f(x)+2$ means a vertical shift upwards

[glossy valve]

Renz

[unborn vessel]

wow

how did i miss that

thank you so much

thank you thank you you actually saved me

🙏🏻

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[torn jolt]

The vectors u and v have the respective lengths 4 and 3, the angle between them is pi/4 rad.
What is this equal to?
(2u + 3v) ∙ (2u – 3v) (dot product)

[slate violet]

Can you try expanding the dot product?

So what's 2u dot 2u for example

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[frail karma]

simple question: are the poles of a function the same as the restrictions of the domain

[frail karma]

nvm I got it, it is the same as the vertical asymptote

correct?

<@&286206848099549185>

[dire osprey]

poles?

[frail karma]

in the context of the study of functions

roots

etc.

[dire osprey]

i assume you mean these

[frail karma]

is that the correct terminology? I study maths in spanish

thats the asymptote

[dire osprey]

I don't know I'm not great in english in educational terms

if it is the same as asymptotes then yes

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[tight whale]

I have proved (i), (ii) and (iv) but I'm a bit unsure about the proof of (iii)
Here's what I've got
A graph G is 2-connected hence $2 \le \kappa(G)$ and by (iv) $\kappa(G) \le \delta(G) \implies 2 \le \delta(G)$ so there exists a node v with degree greater than $2$ such that $G - v$ is connected now consider any 2 nodes in the neighborhood of v $v_1$ and $v_n$ there must exist a path between them $v_1 v_2 \cdots v_n$ so $v_1 v_2 \cdots v_n v$ must be a cycle in G

[glossy valve]

Frisk17

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[idle marsh]

Any degree level mathmaticians could help me with this, its related to number theory

[kindred grove]

alright, what issues are you facing right now ?

@idle marsh

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[idle marsh]

so i want to start with what the elements of F are

i got F = {[0], [1], [X], [X+1], [X^2], [X^2+1]}

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[kindred grove]

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✅

[idle marsh]

any ideas?

[kindred grove]

how did you get all these ? there are redundant classes

i.e. [X^2] = [X+1] and [X^2 + 1] = [X]

[idle marsh]

oh yeah youre right

its these then

[0], [1], [X], [X+1]

[kindred grove]

so all degree < 2 polynomials yes

in a way this works a lot like Z/2Z, where you have 2 = 0, and the classes correspond to the possible remainders when you divide an integer by 2, ie 0 and 1
here you have X^2 + X + 1 = 0 essentially, and the classes correspond to all possible remainders when you divide a polynomial by X^2 + X + 1, i.e. all degree <2 polynomials

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[idle marsh]

and for question 2 the element f has only 4 options right?

[0], [1], [X], [X+1]

if thats the case then i can prove the rest

[kindred grove]

yeah only 4 classes

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[worthy briar]

hello! i have to calculate the angle between these two vectors, but i don't understand why the final answer isn't 180-125.5 since the cos^1(/7/root170) > 90. does anyone know? thanks in advance

[worthy briar]

i apologise if this doesn't make alot of sense, i'm not english. if you need any more clarification on my question please say so^^

[limber forum]

Your angle between vectors can be more than 90 degrees, really it can be anything up to 360 degrees but generally it makes more sense to express it between 0 and 180

These are your two vectors

So you can see the 125 degree angle

You could do 360 minus that to get the bigger angle

[worthy briar]

ooh!! thank you!

in another exercise they required me to do 180 - an angle above 90 so i got a bit confused

but thanks alot!!

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[glossy aurora]

i thought it was 9k 9k+1 or 9k+8

[glossy valve]

Akti

Akti

Akti

[glossy aurora]

yo sorry i cant really help too much with this i dont remeber much from descreet math i just remebered that all cubes of intergers where expressed as 9k 9k+1 or 9k+8

i could be wrong though

na you good i do stuff like that all the time

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[sour harbor]

how do I simplify this region 0 < t < s -t < infinity
or is this like the most simplified already?

[sour harbor]

I guess the infinity part can be deleted since everytthing is below infinity right?

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[sour harbor]

oh it seems like I can siimplify it to 0 < t < s/2
what is the correct way to see it?

is it like every inequality sign has it's own part so like I can split this: 0 < t < s -t < infinity
into :
0 < t and t < s -t and s -t < infinity

t < s-t can u simplify to t < s/2
0 < t can't be simplified and s -t < infinity can be left away?
so u are left with 0 < t and t < s/2
combined it gives the one

btw when I have the inequality < infinity I can always just leave it away right?

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[thorny cloak]

hi

[thorny cloak]

Cold

if tau divides both the sums above

why will it divide t

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[torn gust]

tau=c(n+t) and tau=d(n) so you also get tau=(c-d)t

as in you can add a bunch of taus to get to nij+nji, and then you can add some more to get to nji+t+nij, so the amount in between can be done too which is just t

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[regal timber]

Here I'm finding for the radius and interval of convergence

Where did they set the bounds between -1 and 1

I see they're using the ratio test but why did they add the -1

[vestal meadow]

it's $|x-2| < 1$, which is true when $-1<x-2<1$

[glossy valve]

lgkoo

[vestal meadow]

say we let a = x-2, then we're trying to find values of a such that |a| < 1

what values can that be? @regal timber

[regal timber]

ah I see, anything between 1 and 3

[vestal meadow]

?

I'm asking what values of a, such that |a| is less than 1

[regal timber]

isnt that just any value between 0 and 1

[vestal meadow]

what about -0.5

what is |-0.5|

[regal timber]

o

so anything between -1 and 1

[vestal meadow]

yeah

[regal timber]

o

[regal timber]

I see where -1 came from now is there a name for what this is called?

[vestal meadow]

modulus?

[regal timber]

ah I'll look into it, ty for the help

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[vestal meadow]

np

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[little owl]

for this do you just do the question as normal but without doing the square root part

so normaly you would suqare root 74 but since its exact value you just dont square root it so it isnt a infinate decimal?

[torn gust]

👍

[little owl]

!close

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[little owl]

ty

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[glossy aurora]

ayo i have another question im tryna google these but im too stupid to understand google

[glossy aurora]

can someone fact cheak this

would it be

1False

2false

3false

4true

5idk i need to learn about nullity still

6false

7false

htrue

[past ridge]

4 is false I think

since 3x4 matrix * 4x1 is equal to 3x1 , so range is comprised of vectors from R3

range is possible outputs

possible ouputs are 3x1 vectors

[glossy aurora]

ooo ok so you can only get vectors in R3

[past ridge]

I also think 8 is false

Do u want to know answer for 5?

[glossy aurora]

its ok i still need to learn about nuallity so i might not understand

thank you though

does this mean they are all false

[past ridge]

3 and 6 I have no clue, but everything else looks right

no

well maybe

[glossy aurora]

okok thank you

[past ridge]

8 is false cuz if u have the diagonal matrix [[1,0],[0,-1]]

[glossy aurora]

oh you right okok i think they are all false then

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[maiden pike]

im so lost what they want from me here

[proven mango]

that notation just means to find the second derivative of y with respect to x

[maiden pike]

i dont know how to get that

[proven mango]

do you know how to find the first derivative?

[maiden pike]

yes

[proven mango]

so the second derivative just means to differentiate the first derivative again

[maiden pike]

it would be 2x and -sinx?

[lime ether]

product rule

[proven mango]

mm you would need to use the product rule here

[maiden pike]

ohh right

[maiden pike]

ohhhh

its f''(x)?

i needa find?

[proven mango]

exactly

[maiden pike]

okayyy ill do it and come back

okay first derivative is =-x^2sinx + 2xcosx?

yayy

so second derivative now..

ill be backk

do i use product rule again?

[proven mango]

yep

[maiden pike]

do i use product rule to do product rule?

[proven mango]

wdym by that

just do the product rule the same way you did it before

but for each term

so differentiate -x^2sinx

and then 2xcosx

[maiden pike]

but for x^2sinx dont i use product rule to find derivative of that?

[proven mango]

yeah you use product rule for that

and the 2xcosx as well

[maiden pike]

thats alot of product rule

[proven mango]

yeah

[maiden pike]

where do i go from here?

[proven mango]

oh wait

all you needed to do was add the product rule you got for the -x^2sinx

with the product rule for 2xcosx

you didnt need to do the product rule again for those two

since they're not being multiplied together, but rather they are being added together

[maiden pike]

im a bit confused

[proven mango]

alright so we just need to do this:

$\frac{d}{dx}-x^2\sin(x)+\frac{d}{dx}2x\cos(x)$

[glossy valve]

y0shi

[proven mango]

d/dx means take the derivative of if you dont know

[maiden pike]

but dont we need to do the product rule to get the derivative of those 2?

[proven mango]

yes

[maiden pike]

cause they seperate terms?

[proven mango]

but we just need to add them together

well they are being added

not multiplied together

so we just need to evaluate the derivative of each term

[maiden pike]

so how will i write that?

[proven mango]

if i plug in the derivatives you got into this:

$-2x\sin(x)-x^2\cos(x)+2\cos(x)-2x\sin(x)$

[glossy valve]

y0shi

[proven mango]

that'll be your final answer

[maiden pike]

oh those are the derivatives of what i found

[proven mango]

yep

we just add them together

[maiden pike]

why dont we multiply it by the normal g(x) and f(x)

[proven mango]

wdym?

[maiden pike]

cause the product rule

[proven mango]

what is g(x) and f(x)?

the -x^2sin(x) and 2xcos(x)?

[maiden pike]

yes

[proven mango]

we only do that when its a product of two functions

lets say if it was f(x)g(x), then the derivative would be f'(x)g(x)+f(x)g'(x)

[maiden pike]

yes

[proven mango]

but if it was just f(x)+g(x), the derivative would just simply be f'(x)+g'(x)

which is what we have up there

it is addition not multiplication

[maiden pike]

OHHH

thank u yoshi

[proven mango]

yw!

[maiden pike]

i will use what

i know no

w

i will be back

[proven mango]

alright

[maiden pike]

=-x^2 cosx -4xsin +2cosx

[proven mango]

thats correct

[maiden pike]

yayyy

thank u

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[honest ether]

Greetings.

There are four boxes:
In first there are 109 rocks
In second there are 110 rocks
In third there are 111 rocks
In fourth there are no rocks.

In one step you take 1 rock from each of any of 3 boxes (3 rocks in total) and place them in the other box.
(For example, I take 1 stone from 1st box, 1 from 2nd and 1 from 3rd and then place them in 4th box
109, 110, 111, 0 ->
108, 109, 110, 3)
You can do several steps

а)  Can there be 105 rocks in the 1st box, 110 in 2nd, 111 in 3rd and 4 in 4th?

B) Can it be 330 rocks in the 4th box?

C)  What is the biggest amount of rocks that can be in the 1st box?

All your solutions/thoughts to B and C should be mathematically proved

Thanks

[ancient folio]

"Can there be"/ "Can it be" doesnt really ask u to prove it mathemtically but rather just give ur teacher a solution to the problem thus, it could be done, thus satisfy the question

Trial and repeat then

[honest ether]

Good point, I will edit it

[ancient folio]

If u want to prove it, it probably requires number theory or some really advance theory

[honest ether]

It is, but I'm bad at number theory, so I'm here

[torn gust]

the biggest trick is parity, so rewrite all the numbers as even/odd and investigate how the game works

[honest ether]

Ok, so I solved A
109,110,111,0 ->
108,109,110,3 ->
107,108,109,6 ->
106,111,108,5 ->
105,110,111,4

But how to prove B and C?

[torn gust]

A) tells you that you can transfer 4 from any pile to another

that suggests looking at the piles mod 4, and seeing if something is invariant

the idea is that doing (+3 -1 -1 -1) keeps the differences between piles at the same number mod 4, since 3-(-1)=4

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[honest ether]

Nice, I've solved this thing using your idea

Thx

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[twilit trail]

3. The sides of a rectangle are 25 cm and 8 cm. What is the measure, to the nearest degree, of the angle formed by the short side and a diagonal of the rectangle? Express your answer into 2 decimal places.

I solved the hypotenuse by the use of sine and i got 26.29 as you can see rounded off but when i use pythagorean theorem i get 26.25? What should I use? sine or pythagorean I also need explanation pls and thank you

[ripe horizon]

its because you did some rounding

72.26 degrees was rounded to 72 degrees

so when you did sin(72)

the result became different by a bit

try using sin(72.26) or more decimal places

I used $\frac{25}{sin(72.26)}$ and i got $x=26.24812$

[glossy valve]

Renz

[ripe horizon]

which is close to the 26.25 you got from phythagorean theorem

so use atleast 2 or more decimal places

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[twilit trail]

Ahh I see

Thanks

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[agile trench]

my working out

[agile trench]

I'm doing part c

Is my working out correct?

[willow sun]

Missing a +tan²x

In place of the +1

[agile trench]

@willow sun Then what is my next step?

[viscid wren]

then direct formula

[willow sun]

This is now resolved to standard integrable forms

Just get the tan² in terms of sec² as well

Then use standard results

d/dx(secx)=secxtanx

d/dx(tanx)=sec²x

[agile trench]

Under which type of conditions do I need to use the double angle formula

[willow sun]

Why do you want to use the double angle formula

We generally use it when we have sin² or cos²

To convert them to integrable forms by getting them in terms of cos2x

[agile trench]

oh I see

[willow sun]

But for sec and tan we don't need that since we know the standard integration of sec²

[agile trench]

how about cosec and cot

@willow sun

[willow sun]

We treat them like tan and sec only

Since we know d/dx(cotx)=-cosec²x

[agile trench]

How show this

@willow sun

I got that two double angle equation

How do I combine them

What do they mean adding the above

Can someone help me please?

@terse terrace

[neat aspen]

add the LHS of first and second lines, and add the RHS of the first and second liens

[agile trench]

So
2sin3xcos2x = 0

@neat aspen

right?

[neat aspen]

no, you added the LHS wrong

[agile trench]

How to do that ?

Can you explain to me please?

[neat aspen]

sin(3x+2x) + sin(3x-2x) = sin(5x) + sin(x)

3x + 2x = 5x

and 3x - 2x = x

[agile trench]

oh okay thank you

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[slate night]

Let there be a set A ={0,4, 8, 12}. Determine the amount of triplets (a,b,c) which can formed with the property that a<b<c

[slate night]

I get that I have to do C(4,3)

[sharp flame]

yeah

because for any 3 numbers you choose only one permutation will satisfy that property

[slate night]

But how do I make sure that the property is true?

Oh so

[vast fossil]

Integers are totally ordered

[slate night]

4 0 8

Will then be the same as 0 4 8

[sharp flame]

no no

[slate night]

And it'll count as a group

[sharp flame]

if you pick 4, 0, 8; there's only one way to order it such that a < b < c

its basically C(4, 3)*1!

[slate night]

Wait, so could you explain the from the start?

Don't really get this

So I need 3 numbers

[vast fossil]

Every subset with 3 elements will give you exactly one desired tuple because of this

Hence you only need to count the subsets with 3 elements

[slate night]

Alright I think I get this

But If it were for example, 4 elements instead of 3

Ehat would you do?

Or would it be the same thing?

[vast fossil]

Yes

[slate night]

I c

I think i get it now, I'll mess around with a bit to get the hang of it more

Thanks

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[tired blade]

A Ferris wheel at an amusement park completes one revolution every 40 seconds. The wheel has a diameter of 16 meters and its center is 12 meters above the ground.

a) Model the rider's height above the ground with a sine function
Need help with this.

[tired blade]

I did it as f(x) = 8sin(pi/20(x ))+12

im just not sure what the phase shift is

can someone help me find this?

[brisk obsidian]

There would be no phase shift.

[tired blade]

sin starts at the midline, so maybe its (40 / 2) = 20 sec?

oh

why?

[brisk obsidian]

Because the problem does not specify an initial height for a cart or rider on the ferris wheel. You can pick an arbitrary point on the "circle".

[tired blade]

ah

alright then, thanks 👍

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[tired blade]

Commercial bottling machines often use a circular drum as part of a mechanism to install tops on bottles. One such machine has a diameter of 120 cm, and makes a complete turn once every 5 s. A sensor at the left side of the drum monitors its movement. Take the sensor position as zero. (Assume starts at the far right)
I've got another question, what would be the phase shift for this one?

[tired blade]

the question:

Sketch the graph of the horizontal position of a point on the drum, h, in centimers, as a function of time, t, in seconds.

I'm just not sure if we start from the 120 or 60

i.e sin or cos

[lost dune]

60

Think about it

It’s a circle

[tired blade]

why not 120?

[lost dune]

Diameter of 120

Look at where the sensor is

To the left right

[tired blade]

right

[lost dune]

So it has to face the centre

[tired blade]

the question mentioned that?

oh ic

[lost dune]

But a circle doesn’t have uniform ‘faces’

So the only way it can be to the left is if it’s

How do I draw a circle

[tired blade]

wym lol

[lost dune]

Hold on

[lost dune]

The sensor has to be placed here

Halfway between the top and the bottom

So half the diameter

[tired blade]

ah got it now

ty

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[twin warren]

arent these the same?

[twin warren]

like if u distribute the negative sign on the calc answer

or am i wrong

[willow sun]

No the x-1 is moving

From the denominator and the numerator

[twin warren]

wait wdym by moving

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[ashen crypt]

hi

[ashen crypt]

how to solve for I

[wide sundial]

What have you tried

[ashen crypt]

I tried to convert it to polar coord

and i got this

nvm

thats wrong

Im really stuck

my professor gave an instruction

but he used this

rho, phi, theta, while all material + tutorial I found online use rho, theta, phi

so I just mess them up

I got these wrong already 😦

The E solid look like this

so i though both theta and phi goes from 0 to pi/2

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@ashen crypt Has your question been resolved?

[ashen crypt]

<@&286206848099549185> 😭

[radiant wadi]

stop pinging

[ashen crypt]

its been 15min?

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[torn jolt]

I'm having trouble determining wether functions are one-one algebraically

[nova sigil]

show an example you are unsure on

[vestal meadow]

the way to check if a function is one-to-one is usually to do a vertical line test. If any vertical line you draw doesn't intersect the graph more than once, than it's injective, i.e. one-to-one

for example, a circle graph fails a vertical line test, thus it is not injective

[torn jolt]

so I need to plot a graph everytime?

[vestal meadow]

in short yes. Although, on top of my head rn I can't think of any function that is not one-to-one but also doesn't have quadratic terms

[slate violet]

You mean a horizontal line test

The vertical line test is just to check if you have a function

[slate violet]

Well this is just the graph of 1/x translated right by 3 units

And 1/x is one-to-one from the shape of the graph (like use the horizontal line test)

[vestal meadow]

oh yh horizontal mb

[slate violet]

So 1/(x - 3) must be too

[torn jolt]

Do we get these typed of QNs on exams too? cuz I cannot think I would plot a graph asap and then write the answer on a paper

[slate violet]

So you don't need to draw anything as long as you know what the graphs of most common functions look like

[slate violet]

Depends on your exam but I would expect this to be on there

It tests if you understand what one-to-one means, firstly

And more importantly, it gets you to use it in a newish situation

[torn jolt]

I'm in CBSE

[slate violet]

So you can see if you truly understand the concept

[slate violet]

Yeah I wouldn't be surprised for like class 10

[viscid wren]

you are in cbse

one one is in 11th grade

so do you know to differentiate?

[torn jolt]

nah its in 12th this year

[viscid wren]

i see

so do you know to differentiate

there is a short trick to know if its one one or many one

[slate violet]

Yeah if the function were more complicated you should differentiate and check if f'(x) > 0 or f'(x) < 0 for all x

Constantly increasing or decreasing -> one-to-one

[viscid wren]

^^^

[torn jolt]

yea

[viscid wren]

alright then whatever function you get try to differentiate it

then if f(x)>0 or f(x)<0 for all x(i mean f'(x)

then its one one

[slate violet]

Ah so rational functions aren't in class 10, interesting
(I am not Indian)

But they are in class 11

[torn jolt]

ahhm

[slate violet]

And class 11 has calculus so yeah

[torn jolt]

we have basics like the intro of that in 11th

types of relations and functions are in 12th

[torn jolt]

yep we got limits and derivatives

[slate violet]

That just goes to show the gap between JEE and the normal CBSE syllabus is enormous

[viscid wren]

but i doubt your school teacher will accept derivative method though

[torn jolt]

I'm not really studying for any entrance (I'm NRI)

[viscid wren]

not really

[torn jolt]

idk either

[viscid wren]

its the recent times that both cbse and jee has reduced their syallabus a lot

[torn jolt]

CBSE is kinda prerequisite like precalc and JEE is a bit advanced

[slate violet]

Exactly

[slate violet]

Really?

[viscid wren]

i just gave my jee exam 4 days ago lol

[slate violet]

Wow congrats

[viscid wren]

tbh jee level has come down

[viscid wren]

ahh its not like i cracked it

CBSE concentrates more on the proofs and theoritcal knowledge

[torn jolt]

of is f'(x) is less than or greater than zero [except equal to zero] it is one-one

[viscid wren]

where as jee focuses on numericals and speed

[viscid wren]

yep if it satisfies for all x

[torn jolt]

I just derived this function

the 1st derivative is -1/(x-3)^2

[slate violet]

Exactly it's the speed that is the killer

I don't think the question difficulty is insane for JEE mains

[viscid wren]

yeee

[slate violet]

Maybe advanced but even still it's just a lot of content knowledge without much application

[viscid wren]

ahh i gtg now south check your dms i will ask you something later

plz help me out when i ask

[torn jolt]

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[slate violet]

Please just ask on the server instead

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[astral crag]

asking help on this one. Im already done with number 1 just the numbers 2 to 5

[somber hill]

Tried anything yet?

[astral crag]

hmm well tbh I tried 2 and 4 but Im quite not sure about the answer I got but on numbers 3 and 5 I really dont get it

[somber hill]

Well why don't tou try doing 3 and 5, see where it gets you?

If it's wrong I'll help you out but you should try once

[hollow summit]

!show

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

[astral crag]

ok ok

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@astral crag Has your question been resolved?

[astral crag]

so ughh

@somber hill I have my guesses now

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[restive bolt]

Can we say $|x|<y \implies |x+a|<y+a$ for all $x, a \in \mathbb{R},\quad y\in \mathbb{Z}^+$

[glossy valve]

babario

[keen vector]

no

.close

[twilit leaf]

if a is positive yes, if a is negative there are some tricky situations

[twilit leaf]

what

for example, x=-4, y=5 and a=-2

[restive bolt]

how about if x is negative

[twilit leaf]

it still depends on a

[restive bolt]

but if both x and a is positive it holds true for all right?

[twilit leaf]

yes

[restive bolt]

and then if x is negative it depends on a?

[twilit leaf]

at that point it would just be x<y => x+a<y+a

[restive bolt]

yea

[twilit leaf]

if x is negative and a is positive, it is still true

[restive bolt]

wonder if u can show it algebraically tho

like $-y<-x<y \implies a-y<a-x<a+y$ which is always true because y is bigger than -x in the first place

yep

[glossy valve]

babario

[restive bolt]

nvm

[remote basin]

i might be stupid, but I threw this into desmos and the graphs look different when a isnt 0

[restive bolt]

yea but im just asking if LHS is true than will RHS always be true

[restive bolt]

wait tiz doesnt make sense tho on second thought

[remote basin]

sorry i was talking about this

[restive bolt]

ya i was talking about tat too

[restive bolt]

any help?

oh nvm |x|+a<y+a becomes |x+a|<y+a only if a is positive no matter x

yea now its obvious

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[runic spruce]

Don’t get part 2

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[rare dragon]

is e^x surjective?

[rare dragon]

I'm getting a mixed reply on the internet

i think it's not surjective because we can have -ve x but not -f(x)

Can someone confirm it?

[hot wadi]

it's not surjective on R because it can never be negative yeah

[rare dragon]

so it's not surjective right?

[spice knot]

🤔

it's not surjective

[keen vector]

depends on what you define the codomain lmao

[rare dragon]

Oh ok

[rare dragon]

How?

Ok then I'll close the channel

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[keen vector]

if codomain (0, inf) then surjective

[rare dragon]

Id say codomain is R

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[native matrix]

X^2+(m+n-12)x+(m-2n+3=0 If the sum of the roots and multiplication of the roots is prime and reverse multiplication sum of the roots is 4/5( not sure what this is in english reverse multiplication of x is 1/x)

[native matrix]

In this question by using vieta i saw that the sum of the roots is -m-n+12 and multiplication of roots m-2n+3

(1/x1)+(1/x2)=4/5 i turned that to x1+x2=5/4

And since -m-n+12 is the sum of the roots i found 12-5/4=m+n which is wrong

Where did i go wrong

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[sturdy fulcrum]

what am i doing wrong here?
my procedure:
-1<n 3>(x)^2n-15
2n-15 = 0
n = 15/2

[sturdy fulcrum]

<@&286206848099549185>

[limber forum]

I can't really make sense of what your workings are suggesting you're trying to do, but how do you work out the nth term in a binomial expansion?

[sturdy fulcrum]

for (a+b)^n; the nth term would be <n n>(b)^n

since the rth term is <n r>(a)^n-r. (b)^r

[limber forum]

True, but I think you're counting one too high

[sturdy fulcrum]

oh yeah

[limber forum]

Term k means r = k-1

[sturdy fulcrum]

that'd be the r+1 th term

right on

[limber forum]

Ye, so your binomial for this one is just nC2 * a^(n-2) * b^2

[sturdy fulcrum]

shoot you're right

aight yeah i got it now

thanks a bunch

[limber forum]

np

[sturdy fulcrum]

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[grand hawk]

need help, I'm trying to find how to solve for the standard deviation

[shrewd hamlet]

for number 3 on the first one: how many multiples of our st.dev (150) can we fit between our mean(2000) and 1550

[grand hawk]

the equation I've been given is 1- (1/(k^2))

[shrewd hamlet]

idk what that means

but just do my method

its basically just (1550-2000)/150

[grand hawk]

oh ok

[shrewd hamlet]

if ur familiar with the concept of z-score, thats what it is

[grand hawk]

so, 3

[shrewd hamlet]

yea

[grand hawk]

how do I solve for 4?

[shrewd hamlet]

for 2450?

[grand hawk]

I'm so bad at z scores, my school basically told me to figure it out

tips on finding the z score?

[shrewd hamlet]

z score just tells u how many standard deviations above or below the mean your data point is

its just

$\frac {data point - average}{st. dev}$

[glossy valve]

Stephen

[grand hawk]

data point? so the datapoint would be 3?

(3-2000)/150 ?

[shrewd hamlet]

an example of a data point would be 1550, the number of hours

[grand hawk]

oh ok

so 1550 and 2450 are the data points?

[shrewd hamlet]

3 is the number of standard deviations from the mean(1550 is 3 standard deviations from the mean of 2000)

[shrewd hamlet]

yes

those are examples of data points

[grand hawk]

do I just add the points together?

[shrewd hamlet]

no

[grand hawk]

so then how do I find the between portion?

[shrewd hamlet]

i think ur confused about the essence of the z-score

[grand hawk]

maybe a bit

[shrewd hamlet]

https://youtu.be/2tuBREK_mgE?si=HSj6YDKUuhoiUnnS

maybe this might help

[grand hawk]

I'll watch it

[grand hawk]

so, to find z, I'd use (1550-2000)/150 ?

and (2450-2000)/150

[shrewd hamlet]

Perfect

[grand hawk]

omg thank you, ok lemme do that rq and I'll be back for the second part

[grand hawk]

so I got -6

but that isn't right

I'm running into an issue with the standard normal table, as I don't have one on hand, is that something universal or do I have to make one with the info I have?

<@&286206848099549185>

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@grand hawk Has your question been resolved?

[shrewd hamlet]

howd u get -6

[grand hawk]

3+-3

[shrewd hamlet]

no u dont add them

also, 3+-3 isnt -6, but that isnt relevant

1550 and 2450 are individual data points

they get individual z scores

[grand hawk]

then I converted that into this

[grand hawk]

I gathered that part

[grand hawk]

I don't remember how I did it, that was like 40 minutes ago

[shrewd hamlet]

1550 is (1550-2000)/150 standard deviations below the mean

[grand hawk]

yep

[shrewd hamlet]

2450 is (2450-2000)/150 standard deviations above the mean

[grand hawk]

yes

3 and -3

[shrewd hamlet]

-3 and 3 yes, thats all ur problem is asking for

[grand hawk]

which I then put into 1- (1/3)

[grand hawk]

it's asking for a %

[shrewd hamlet]

number 3?

i dont see that

[grand hawk]

number 4

[shrewd hamlet]

i cant see number 4

[shrewd hamlet]

Yeah i couldnt see that problem before

so for this we want

P((1550-2000)/150 < Z < (2450-2000)/150))

which is P((-3 < z < 3))

[grand hawk]

yeah

[shrewd hamlet]

oh wait ur using chebyshevs

uh lets see

[grand hawk]

1-(1/k)

I think

[shrewd hamlet]

ok heres an excellent image

x bar means the mean

k is our z score, number of standard deviations above and below the mean

s is our standard deviation

[grand hawk]

so 1.1111111-0.8888889

?

[shrewd hamlet]

Howd u get 1.11111

its 1 - 1/k^2

k = z score

[grand hawk]

the 3 and -3

[shrewd hamlet]

When u plug in 3 and -3, they yield the same thing

its $1- \frac 1{(-3)^2}$ and $1 - \frac 1{(3)^2}$

[glossy valve]

Stephen

[shrewd hamlet]

Same thing

[grand hawk]

the negative changes it

[shrewd hamlet]

No

the negative is inside the parentheses

[grand hawk]

oh ok

[shrewd hamlet]

anways, what answer did u get

[grand hawk]

0.888889

[shrewd hamlet]

yea 8/9

[grand hawk]

88.9%

[shrewd hamlet]

yep

[grand hawk]

in terms of this, how would I find a specific percentage like this

I understand how to get the standard deviation, but getting a specific % like that is confusing me

[shrewd hamlet]

1 - 1/sd^2 = percentage/100

this is the format we followed for the previous question

Use that to find the sd

what did u get @grand hawk

[grand hawk]

sorry, I was eating dinner

[grand hawk]

.75 = 1-1sd^2

sqrt.75= 0.8660254038

[grand hawk]

.5

[shrewd hamlet]

no

show me how u did it

[grand hawk]

1+1(sqrt75)

[shrewd hamlet]

$0.75 = 1 - \frac 1{k^2}$

[glossy valve]

Stephen

[shrewd hamlet]

solve for k

[shrewd hamlet]

not sure how u got this

[grand hawk]

2?

[shrewd hamlet]

yes

[grand hawk]

thank you !! You really don't understand how much this has helped me

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[shrewd hamlet]

sure np lol

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[gritty stratus]

I dont know how to solve this, can anyone help me..?

[glacial pasture]

had any ideas about it?

[grand hawk]

104 I think are the amount of outcomes, as there are 2 outcomes for each draw

[thick hedge]

!nosols

[full forum]

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

[hard veldt]

ty lol

[gritty stratus]

I only need the letter b

[grand hawk]

sorry

[hard veldt]

in how many ways can 2 cards be chosen from the 13 diamonds

@gritty stratus

[buoyant yew]

why did i get pinged?

[hard veldt]

mb

typo

[gritty stratus]

with repitition?

[hard veldt]

question says without replacement

[gritty stratus]

6?

and one is left

[hard veldt]

how so?

[gritty stratus]

13/2 is 6.5

[hard veldt]

Given four distinct balls

How many ways can you choose 2 of them

(without replacement)

[gritty stratus]

what does w/o replacement means?

I dont really get it, and thats the part I dont understand

the replace or w/o it

[hard veldt]

it basically means that after you draw the first ball, you don't put it back with the rest of the balls

[gritty stratus]

I know... but I dont get how it changes the answer

can you give me an example?

[hard veldt]

let's work through this problem

in how many ways can you choose your first ball

[gritty stratus]

4

2nd ball is 3

3rd ball is 2

4th ball is 1

am I correct?

I think his phone died lmao

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[primal warren]

how do you find the exact value of tan(-pi/12)

[austere cove]

You can observe that -pi/12 = -pi/4 + pi/6

And use the tan angle addition formula

[primal warren]

it should be the same if i use the subtraction formula right?

[austere cove]

For pi/6 - pi/4, yes

[primal warren]

then there must be an error in my working

[austere cove]

I honestly never bothered memorizing the subtraction formulas myself though

[primal warren]

ok

[austere cove]

Because you can get them from the addition formula

[primal warren]

ok that will help in the exam

[austere cove]

tan(a+b) = (tan(a) + tan(b))/(1 - tan(a)tan(b)) right?

[primal warren]

yeah thats about right

[austere cove]

So tan(-pi/4) is what?

And tan(pi/6) is what?

[primal warren]

-1

[primal warren]

1/sqrt 3

[austere cove]

Exactly

So we have (-1 + 1/√3)/(1 - (-1)(1/√3))

(-3/3 + √3/3)/(3/3 + √3/3) = (-3 + √3)/(3 + √3)

What answer did you arrive at?

[primal warren]

textbook says this

[austere cove]

These are equivalent

Divide out √3 from the top and bottom from my answer to get the LHS of the image

And to go from the LHS to RHS in the image multiply top and bottom by (1 - √3)

[primal warren]

bruh i just realised my error

[austere cove]

Excellent!

[primal warren]

since its tan(-pi/12)

it = tan(pi/6 -pi/4)

[austere cove]

Yup

[primal warren]

when i put it in the formula it is tan(pi/6) -tan (-pi/4)

right?

i think before i was doing tan(pi/4) without the negative

[austere cove]

If you use the tan addition formula tan(a-b) is (tan(a) + tan(-b))/(1 - tan(a) tan(-b))

Then you can use the fact that tan is an odd function to get the difference formula

(tan(a) - tan(b))/(1 + tan(a) tan(b))

So it's "- tan(pi/4)" without the negative.

You probably just have a sign error somewhere

[primal warren]

ok

[austere cove]

But without seeing your work it's difficult to tell

[primal warren]

ill send a pic of my work in a second

[austere cove]

Well, I need to go afk actually

[primal warren]

ok np

can anyone help me find my error:

[full forum]

@primal warren Has your question been resolved?

[primal warren]

<@&286206848099549185>

[robust slate]

,w tan(-pi/12)=\frac{\sqrt{3}-3}{\sqrt{3}+3}

[glossy valve]

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Requested by civil_service_pigeon

[robust slate]

it's fine tbh

but you can simplify it

from here, if you multiply the numerator and denominator by sqrt(3), you get

$\frac{1-\sqrt 3}{\sqrt 3+1}$

[glossy valve]

Civil Service Pigeon

[robust slate]

then you can multiply the numerator and denominator by (sqrt 3 - 1)

[primal warren]

in the last step i did could i also multiply it by the conjugate to get the answer?

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[mystic reef]

Help

[full forum]

Please don't occupy multiple help channels.

[primal warren]

what do you need help with

!status

[full forum]

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

[primal warren]

@mystic reef

[mystic reef]

Yeah

[primal warren]

what do you need help with

[primal warren]

to solve the volume you can substitute the values into the formula

[mystic reef]

Do what that means

[primal warren]

?

[mystic reef]

Idk*

[primal warren]

k ill show you then

V=L * W * H

[mystic reef]

K

[primal warren]

H=12, L=8, W=16

so but these values into the formula

so V=8 * 16 * 12

V=1536 cubic units

does that make sense

[mystic reef]

A lil

10

[primal warren]

do u know how to do surface area

[mystic reef]

Not at the moment

[primal warren]

ok so surface area is all the area on the block ( front, back, left, right, top and bottom) all added together to find the surface area

[mystic reef]

Ok

[primal warren]

so you have to find the area for all of those then add them up

so Surface area = 2(8* 12) + 2(12 * 16) + 2(8 * 16)

=2(96) + 2(192) + 2(128)

=192+384+256

SA=832

is that all good for you

[mystic reef]

Yeah I think I can do it maybe

[primal warren]

try a couple more to get the hang of it

[mystic reef]

Well there’s 2 more questions

[primal warren]

k if u have any trouble you can ask me

[mystic reef]

It says when all side lengths are cut in half

Do I just cut answer in half

[primal warren]

no

you need to do the calculations again but with different numbers

[mystic reef]

Okay

And then it’s 25%

I got 192 and 208 @primal warren

[primal warren]

192 is right

let me check other one

yeah they're both right

good job

[mystic reef]

Nice

Then I got 24 52 @primal warren

[primal warren]

24 is right

nice their both right

[mystic reef]

Alright

[primal warren]

👍

[mystic reef]

Close

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[wet delta]

Write the formula N=80 x 1,28^4t-3 in the form t= a + b x (1,28)log(cN)

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[fossil wedge]

Why do I get this weird result? Is it wrong?

[torn jolt]

Is that the modulus

[fossil wedge]

Yes

[pallid coral]

is this supposed to be a proof of the limit

[fossil wedge]

Yes

Where delta is that weird number i get at the end

[pallid coral]

I think the usual way is to simplify the
|f(x)-L| part on its own first

then try representing it in terms of delta

at the very end, find out what delta should be in terms of epsilon in order for that expression to give epsilon, and go back to the beginning of the proof and let delta to be that

[fossil wedge]

Yeah

[fossil wedge]

So like this and delta = epsilon?

Then we also have to check the other way of course

[pallid coral]

Yeah i think so

yeah i think that works

[fossil wedge]

Ok thanks very much all clear now

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[idle scroll]

have to find the derivate of this using product rule