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[runic spruce]

Been trying to prove these vectors are linearly dependent

[gritty rose]

<@&268886789983436800>

[twilit leaf]

Thanks riemann :)

[runic spruce]

But whenever I try eliminating a symbol for the second time i go round in circles

It’s the only example I’m stuck on

[twilit leaf]

You can put them in a matrix and check the determinant

[runic spruce]

Uhh

[twilit leaf]

You can also use cross products

[runic spruce]

Idk if that’s what we’ve been taught

We’ve been taught to use alpha, beta gamma?

[twilit leaf]

Oh

[runic spruce]

And set up like eqns

I could use a calculator for this

But my professor wont be happy

[twilit leaf]

Multiply the vectors by alpha, beta, and gamma respectively then add them

[runic spruce]

Yes

[twilit leaf]

Then set the sum equal to 0 0 0

[runic spruce]

Yes

[twilit leaf]

Then its a set of linear equations

Or if you still want matrix form, its gaussian elimination

[runic spruce]

These are vectors though right so I dont need to use that surely

If I’m just proving something

[twilit leaf]

Yeah you can

[runic spruce]

When i eliminate alpha I get the following two equations

2beta + gamma = 0

-6beta -3gamma = 0

But when i try to use multiples to get rid of the next symbol i just get 0=0 lmao

Seems like that’s the method we have to use is multiples, we have covered Gaussian elimination later tho

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@runic spruce Has your question been resolved?

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@runic spruce Has your question been resolved?

[runic spruce]

Finally got it

.close

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[torn jolt]

The question goes like this: "We have 20 people in the class, 12 boys and 8 girls. There is 6 tickets being randomly distributed among the class members. Whats the probability that at least 3 girls get the tickets?"

First i calculated all possible distributions of 6 tickets among 20 people (pic1)

Then i reasoned like this: i will have to distribute 3 tickets among 8 girls to make sure at least 3 of them have tickets. And the other 3 tickets can be distributed among the rest of the class, so 3 tickets distributed among the other 17 class members (pic2)

Then i divided the pic2 possibilities that satisfy task criteria with all possibilities in pic1 and after simplifications i got 56/57 which is wrong according to the book. Where is the mistake?

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@torn jolt Has your question been resolved?

[wild sleet]

it counts too much

[torn jolt]

What do you mean?

[wild sleet]

if you distribute 3 tickets to g1,g2,g3 and then also g5, g6 get a ticket, it's the same distribution as if you give 3 to g2,g3,g6 and then g1,g5 get it on the second step

so that situation counts as one in pic1

but you count it many times for the numerator

[torn jolt]

Oh i see

Thank you

[wild sleet]

you want to count exactly 1 girl getting the ticket, then there's no issue with overcounting
then you do it again with exactly 2, and you do it again with no girls getting tickets

then it all adds up normally to "less than 3" and you can subtract

i don't know a shorter way

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[chrome idol]

To prove uniquness of any pde, do we just consider, say u1 and u2 to be solutions, then show that w = u1-u2, where w then must satisfy the initial consitions of the set pde?

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[covert sapphire]

what is the GCF of 15 and 81

[worn matrix]

what are the factors of 15 and 81?

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[rain idol]

I'm continually getting the (same) erroneous answer for this problem:

The correct answer is -sqrt(2)-sqrt(6) / 4. The answer that I keep coming up with is -sqrt(6)-sqrt(2)/4.

My solution uses cos(150+45), while the solution uses cos(135+60), though I'm fairly sure that shouldn't matter.

My solution is as follows: cos(150+45) = cos(150)cos(45) - sin(150)sin(45)

-> ((-sqrt(3)/2) x (sqrt(2)/2)) - ((1/2) x (sqrt(2)/2))

-> (-sqrt(6)/4) - (sqrt(2)/4)

-> (-sqrt(6) - sqrt(2)) / 4

Where am I going wrong here?

Thanks!

[rain idol]

For clarity, here's my answer:

[worn matrix]

it should be correct, what is the answer then?

[grim skiff]

It's the same

,w (-sqrt(6)-sqrt(2))/4

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[grim skiff]

,w (-sqrt(2)-sqrt(6))/4

[glossy valve]

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[grim skiff]

See the same

[worn matrix]

,calc cos(195deg)

[glossy valve]

Result:

-0.96592582628907

[worn matrix]

yup brother you got the right answer

[rain idol]

Ya know, that's what I thought too. Gotta love online math homework telling you your answer is wrong, haha.

[worn matrix]

ll

lol

[rain idol]

Thanks!

[worn matrix]

np

[rain idol]

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[hollow moth]

what's the fastest way to take the derivative of this vector

[hollow moth]

cuz the way i did it took me like a solid 10 mins

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@hollow moth Has your question been resolved?

[unborn quarry]

Hello! Well you should practice more if it took you that time 🤔

There's no way to simplify the process, it's just product rule

You could replace the sum/subtraction of cos and sin as

cos - sin = -√2 sin(x - 7π/4)
sin + cos = √2 sin(x + π/4)

idk

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[rose grove]

Could anyone help me on this proof I'm very confused on what to do next

[rose grove]

nevermind i give up

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[modest wraith]

help me explain wat is forces and it meaning

[modest wraith]

<@&286206848099549185>

[gritty rose]

!15m

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[gritty rose]

Read this

What math question do you have

[modest wraith]

explaining what is forces

formula

and other things

[gritty rose]

#old-network for physics server

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[modest wraith]

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[rotund pulsar]

quick q

[rotund pulsar]

if i have two points $P_0$ and $P_1$ and I want to denote a vector going from $P_0$ to $P_1$, should i order it $\vec{P_0P_1}$ or $\vec{P_1P_0}$

[glossy valve]

omgatriple

[spiral vigil]

start -> end

[rotund pulsar]

tip to tail then

thanks

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[spiral vigil]

tail to tip

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[torn jolt]

Hi, I need to "sketch the region enclosed by the given curves and find its area"

[torn jolt]

Just wanted to make sure I shaded the right region

[spiral vigil]

yeah

[torn jolt]

Can you please help me with this one?

There's 3 functions so I'm confused

[spiral vigil]

there were three functions in the last one

graph all four of them and find the intersection

,w graph {x^2/4, 2x^2, 3 - x}

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@torn jolt Has your question been resolved?

[torn jolt]

Okay, I found the area I need to compute

The bounds are 0 to 2 right

The thing is, what's the difference of functions that I need to integrate?

[spiral vigil]

look at your plot

it might be multiple parts

[torn jolt]

Think I got it?

[spiral vigil]

your integrals look right

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[spiral vigil]

and the numbers are right

[torn jolt]

Yeah, just realized that I wrote (1, 0) instead of (1, 2)

For my plot

But fixed that

[torn jolt]

Thanks for your help

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[torn jolt]

,rotate

[torn jolt]

How do I do this?

Where would I begin

Idk where to start

[pseudo parcel]

F=ma, calculate the deceleration of the ball

it stops in 0.15m so u can calculate the time taken

[torn jolt]

Could you lead me through it by any chance ?

I’m not sure what’s my first step to get time

[pseudo parcel]

you could plot a v/t graph

acceleration is constant

[torn jolt]

I’m actually unsure what a vt graph is sorry

[pseudo parcel]

velocity time graph

[torn jolt]

Is there anything I can do with kinematicss

Since I didn’t learn about those

[pseudo parcel]

area under the graph is 0.15

or you could use

[torn jolt]

But isn’t it horizontal

[pseudo parcel]

v^2 = u^2 + 2as

same thing

just you can get a immediately

[torn jolt]

Hm

[pseudo parcel]

so what is your final and initial velocity?

and your displacement

[torn jolt]

Final is 41.1

Since that’s when he catches it

[pseudo parcel]

no

hes bringing it to rest

[torn jolt]

Oh

[pseudo parcel]

so what is the final velocity

[torn jolt]

0

[pseudo parcel]

yes

and initial?

[torn jolt]

41.1

[pseudo parcel]

do you know what to do now?

[torn jolt]

Hmm wait

Can I work this out then show you please?

[pseudo parcel]

sure

you should also write all your kinematics equations as well

[torn jolt]

Ok 👍

[pseudo parcel]

ping me when ur done

[torn jolt]

@pseudo parcel

Can it be negative?

[pseudo parcel]

yeah

Negative a is ok

Apply f=ma

[torn jolt]

M= f/a?

And mass will always be positive right?

[pseudo parcel]

yes

well

a is negative but you can take it as positive

since a is the deceleration

[torn jolt]

Ok

Thanks for the help I appreciate it

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[neon escarp]

Compute the volume of the solid lying above z=0, below the surface z=1-sqrt(x^2 + y^2), and inside the cylinder x^2 + y^2 = x

[neon escarp]

this is a triple integral problem

im confused on how to even start it

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@neon escarp Has your question been resolved?

[neon escarp]

update

so i said z=1-r

and r^2 = rcostheta

or r=costheta

and ofc z=0

from that i got 0 <= z <= 1-r

0 <= r <= 1

and 0 <= theta <= 2pi

alr i solved it

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[deft bramble]

Could someone explain to me how I derived 2^k + 1 = 2^(k+1)

These are some older notes I’ve returned to and just can’t quite figure out why

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@deft bramble Has your question been resolved?

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@deft bramble Has your question been resolved?

[pseudo frigate]

You didn't

You concluded that $2^k+1 \leq 2^k + 2^k$

[glossy valve]

daniGhost

[deft bramble]

I'm confused by this because I found this online which matches what I wrote exactly

But they also don't explain why

[deft bramble]

But that's where I'm stuck, why is 2^k + 1 <= 2^k + 2^k

[pseudo frigate]

subtract 2^k from both sides

what do you get

[deft bramble]

1 <= 2^k ?

[pseudo frigate]

yeah

do you believe this or would you like to justify it too

[deft bramble]

Think i've kinda got it now

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[cunning hatch]

i have no idea where to even start with this question

[glacial pasture]

diagram

[cunning hatch]

idek what to draw tbh

there would be a straight line as the shore

180m away perpendicular from the line at P

make another point which is P + 300

not sure from there

[glacial pasture]

yeah, draw that out

do you know about angular velocity?

[cunning hatch]

its maths, cant use physics formuli

[glacial pasture]

... pretty sure you can

[cunning hatch]

syllabus in australia says you cant

we can do it that way

and ill just say i was confused

so angular velocity is change in theta/change in time right?

mate is that right?

@glacial pasture

<@&286206848099549185>

sorry thats displacement

no im right

This is my rough drawing

but idk what to do with it

not even sure its right

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[faint iron]

Could somebody teach me how to do partial fraction decomposition?

[faint iron]

for example $\dfrac{x+1}{(x-5)(x+7)}$

[glossy valve]

annyeong

[faint iron]

I know the process of making it into $\frac{A}{x-5}+\frac{B}{x+7}$ but i'm not sure how to find A and B

[glossy valve]

annyeong

[brave blaze]

either by identification
or you use this :

A = lim (x-->5) (x-5)F(x)

with F(x) the fraction

B = lim (x-->-7) (x+7)F(x)

[faint iron]

wait which fraction is F(x)

[compact zodiac]

Ooh Residue theorem in the wild

[brave blaze]

the fraction you just sent wtf

[faint iron]

oh

[brave blaze]

shhh

[faint iron]

so 6/12

1/2

[brave blaze]

Im too lazy to calculate
have fun

[faint iron]

alright thanks

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[torn jolt]

Hi. I'm stuck, need help solving for x

[jade herald]

!show

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

[torn jolt]

I'm quite unsure of what I should do to progress. I only have the idea of maybe substituting or smtn but idk

[torn jolt]

<@&286206848099549185>

[white smelt]

write log_9 x in terms of log_3 x

and then let log_3 x = t and you have a quadratic in t

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

I ended up with 8t^4 + t^6 + 8 = 0

Is that correct?

[white smelt]

no

show your work

[torn jolt]

Okay wait

[torn jolt]

It's t^4 - 27t^3 + 8 = 0

[white smelt]

this is correct

[torn jolt]

Question

[torn jolt]

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[torn jolt]

.reopen

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✅

[torn jolt]

@scenic osprey Sorry for the ping, but now that the t values are here, this means that log_3x = either t_1 ot t_2 right?
so log_3x = 26.999594 ---- x_1 = 7622196947932.53
log_3x = 0.672294 ---- x_2 = 2.09298326

when put into the equation,
x_1 = 0.00906249 which also is not equal to zero
x_2 = -0.00001345 which also is not equal to zero

How would I be able to know what the correct answer is for the value of x?

[white smelt]

the only reason you arent getting exactly 0 is because you have to take decimal approximations

if you were able to take exact value, those would be 0

cause they're very reasonably close to 0

[torn jolt]

so both x_1 and x_2 are correct answers?

[white smelt]

yes

[torn jolt]

okay makes sense

thanks a lot

[white smelt]

np

[torn jolt]

I have a last question, how would I be able to progress here?

quotient rule?

[white smelt]

yeah

[torn jolt]

I managed to get to 5log_5(2) - 5log_5(3x-4) = 5/6x-7

Am I on the right track?

<@&286206848099549185>

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[vivid berry]

excuse me

[vivid berry]

why is my graph wrong?

[woven sand]

You have an extra x-intercept, that (-5,0)

[vivid berry]

oh

[woven sand]

You only have four factors and none of them repeat so you should have exactly four x-ints

[vivid berry]

oh

its x=5

...

wait let me redo it

how come its mot like this

[wide sundial]

that still has 5 roots

you just didn't label one of them

the one on the left of 0

lol

[vivid berry]

oh

how am i supposed to know that in a test,

?

like bc

the ones with 3 roots

theyre either

[wide sundial]

there are 4 roots

[vivid berry]

up down up

or down up down

[wide sundial]

ok well

[vivid berry]

if its negative

[wide sundial]

this is a polynomial

these are easy

look at the power and sign of the highest degree

here it's 4

4 is even

[woven sand]

The number of factors in an equation (or highest power of x in standard form) is the number of x-intercepts/roots there are

[wide sundial]

that means either both ends go up or both ends go down

[woven sand]

Unless some factors repeat but idk if you need to worry about that

[wide sundial]

positive coefficient means it goes up

negative coefficient means it goes down

[woven sand]

[wide sundial]

for odd powers it has be 1 up 1 down

if you cant remember just remember the shape of x^3

x^3 is odd power so 1 up 1 down

left side down right side up

so positive coefficient means ^^

then negative coefficients mean left side up right side down

,w graph -2x^3

[glossy valve]

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[wide sundial]

,w graph -10x^5

[glossy valve]

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[vivid berry]

im ngl

those look the same....

[wide sundial]

,w graph -2x^101

[glossy valve]

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[wide sundial]

that's the point

[vivid berry]

wtf

[wide sundial]

they look the same

because odd power

and negative coefficient

left up right down

[vivid berry]

,w graph (x-5)(x-3)(x+1)(x+2)

[glossy valve]

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[wide sundial]

that's even power

and the coefficient of the highest power is positive

so you get both up

[vivid berry]

can u show me example of an odd power pls

[wide sundial]

,w graph x^3

[glossy valve]

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[wide sundial]

this is odd power

[vivid berry]

oh

[wide sundial]

the highest power is 3

[vivid berry]

oh..

[wide sundial]

and 3 is odd

,w graph x^5 - 2x^4 + x^2 - 3

[vivid berry]

down up

[glossy valve]

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[wide sundial]

this is odd power

coefficient is positive

[vivid berry]

ohh i see

[wide sundial]

left down right up

same as x^3

[wide sundial]

here you've drawn it left down right up

[vivid berry]

if the coefficient is negative it goes left up right down?

[wide sundial]

but that's not possible for even power polynomials

even power polynomials have the same on both sides

either both up or both down

both up for positive coefficent

both down for negative coefficient

[wide sundial]

for odd power polynomials yes

even/odd power polynomials sounds weird, i think it's "polynomial(s) of odd/even degree"

[vivid berry]

how about odd even polynomials

[wide sundial]

what odd even polynomials

[vivid berry]

i mean

I MEAN

OMG

NEGATIVE

😭😭😭

[wide sundial]

negative powers wont be polynomials

[vivid berry]

negative coefficient even polynomials

oh

OH

[wide sundial]

then they both go down

[vivid berry]

oh

i hate maths im ngl....

[wide sundial]

that's this case

[vivid berry]

ok thx :P

[wide sundial]

well you can even think about it

if you plug a very big number into your polynomial (positive)

you go like say for x^6 + x^2 - 3

you put x = 10000000

that's just gonna be really really positive

and even if you put -10000000

itll still be very positive becuase (-1000000)^6 ignores the - sign

cos ^6 is even

so whether you go to the left or right your output is really big positive

hence it goes up on both ends

similar deduction will tell you about the other cases

[vivid berry]

hmmm i see

okok thanks :D

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[white smelt]

can you square it and take the logarithm?

try that, idk if it works but i think it should

[stark hinge]

One is reciprocal of the other so might be able to use that.

[covert quarry]

No, I took this problem from my test on the topic "exponent function". We haven't gone through the logarithms yet, so there must be an alternative solution.

[covert quarry]

Yes, but a good question is "how?"

[stark hinge]

||1.) $m+/frac{1}{m}=10$||
||2.)Solve quadratic to get roots||
||3.)Take logarithm to find value(s) of x||

[glossy valve]

SirLancelotDuLac

[covert quarry]

I tried to do something about it and generally came to extremely dead ends.

And again, the problem must be solved without logarithms.

[stark hinge]

||m^2-10m+1=0 has roots (5+2sqrt(6)) and (5-2sqrt(6))||

||You can see that in this case there is no need for logarithms as one is (sqrt(5+2sqrt(6)))^2 and the other is its reciprocal so the values are 2 and -2||

[covert quarry]

No, wait, I don't understand

Where did you get this quadratic equation from, I missed this point

[full forum]

@covert quarry Has your question been resolved?

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@covert quarry Has your question been resolved?

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[scarlet lark]

Hello I am in grade 11 and I've encountered this problem that my teacher gave us to try and solve at home.
It is in french so I google translated it:
Let (E): x²+(m+3)x +6 +2m=0. 1) Study the existence of roots. 2) Find m so that the roots are distinct and negative. 3) Find m if (2x1-3)(2x2-3)=9 4) solve [x²+(m+3)x+6+2m]/(-x²-9)<0 for all real x. 5)ABC is a right triangle at A. Find m knowing that AB=|x1| AC=|x2| and BC = radical of 5
I have done part 1 and I'm mostly sure of it
As for part 2 I'm really hesitant
Part 3 felt somewhat easy
And part 4 completely lost me
I haven't tried part 5 yet I'm taking it step by step
So if anyone could please help me I'd appreciate it

[full forum]

@scarlet lark Has your question been resolved?

[scarlet lark]

<@&286206848099549185> so sorry for the ping^^

[novel sage]

It's okay

[scarlet lark]

hello are you a helper?

<@&286206848099549185> I’m really sorry but the assignment is due tomorrow and I would like to get it done in time and understand what I need to understand since I have an exam on these next week

[scarlet lark]

Should I just close this channel and open another one?

<@&286206848099549185>

[scarlet lark]

<@&286206848099549185> sorryyyy

[wicked seal]

Give me a bit and I’ll try to help

[scarlet lark]

Okayy thanks

[simple ridge]

ok so your solution of first 3 parts is correct

now for part 4

[scarlet lark]

Yes

[simple ridge]

$\frac{x^2+(m+3)x+6+2m}{-x^2-9}<0$

[glossy valve]

calculus is fun

[simple ridge]

you want to solve for x

or m?

[scarlet lark]

No I need to solve for m I think because they said to solve for all values of x

So I think I need m

[simple ridge]

ok so lets do it for m

so first of all you need to check the sign of the denominator

to do that you first check roots

does it have any real root

[scarlet lark]

Yes

[simple ridge]

what is it

[scarlet lark]

Isn’t it 3 and -3

[simple ridge]

you just got the roots of x^2-9

the denominator is -x^2-9

[scarlet lark]

I don’t really know how to do it any other way

[simple ridge]

do you know quadratic formula ?

yes

[scarlet lark]

Yes do I use it here?

[simple ridge]

but no dont

there is an easier way for this

the roots of -x^2-9 satisfy the equation -x^2-9=0 right

[scarlet lark]

Yes

[simple ridge]

ok now how to solve this

just isolate x

[scarlet lark]

So no roots? Since x^2 is never equal to -9?

[simple ridge]

exactly

now you know table of sign of second degree polynomial

if it doesnt have a real root then how is its sign

[scarlet lark]

-?

[simple ridge]

are you talking in general or for this particular polynomial

[scarlet lark]

Are you asking about here?

[simple ridge]

yes

[scarlet lark]

If so then I’m talking in this polynomial

[simple ridge]

ok

correct

[scarlet lark]

In general it’s the sign of a

[simple ridge]

exactly

this is for a 2nd deg polynomial with no real roots

so we know that the denominator is always negative and cant be 0

[scarlet lark]

Yes

[simple ridge]

now we wish to get rid of the denominator this will make it easier

how do we do that

does anything in the inequality change?

if yes then why and what changes

if no then why

[scarlet lark]

I don’t know? It stays <

I’m not sure how I can even get rid of the denominator here

Could you explain it to me?

This is the only thing that comes to mind

@simple ridge sorry for the ping but you still here?

[simple ridge]

i am here

this is partially correct

[simple ridge]

dw np

it is correct to multiply both sides by -x^2-9

but the mistake is here

when you multiply something will change

[scarlet lark]

Ohh I know

[simple ridge]

why something should change

[scarlet lark]

When you multiply by a negative number < becomes >

[simple ridge]

because we are multiplying by a negative expression

so for example 1<2

multiply both sides by -1

using what you did you get -1<-2 which is wrong

[scarlet lark]

-1 > -2

[simple ridge]

so you should flip the inequality sign

[scarlet lark]

I understand

[simple ridge]

ok what happens if we multiply by a positive expression instead ,say x^2+1.

[scarlet lark]

It stays the same

Right?

1<2 multiplied by 2 is 2<4

[simple ridge]

correct

so now lets return to oir problem

after multiplying both sides by -x^2-9 what do we get

[scarlet lark]

x^2 +(m+3)x +6 +2m>0

[simple ridge]

ok nice

[scarlet lark]

Now I need the roots?

[simple ridge]

so now you already know when there are roots and how many of them there are

[scarlet lark]

Yes it was in part 1 I think

These?

[simple ridge]

you already know the sign of the roots over the domain of the polynomial

from 2 you got when the roots are distinct and negative

[scarlet lark]

Lebanese

[scarlet lark]

Yess

[simple ridge]

i suspected that

[scarlet lark]

Yep

[simple ridge]

i am lebanese too nice to meet you

[scarlet lark]

Nice to meet you what a coincidence 😂

[simple ridge]

because it is the first time i see someone use table of sign in this server

thats why i suspected that you are lebanese

[scarlet lark]

Oh wow

[simple ridge]

also the name

[scarlet lark]

Ye it gives it away

[simple ridge]

a nice one 😂

lets continue and we will talk when we are done

[scarlet lark]

Sure

[simple ridge]

so when are the roots distinct and negative

[scarlet lark]

When m is a part of ]-3; +inf[

[simple ridge]

from this and from the part that your polynomial has 2 distinct roots when m is in ]-inf,-3[U]-3,+inf[ when does this polynomial have 2 positive distinct roots

[scarlet lark]

m is in ]-inf;-3[ ?

[simple ridge]

exactly

[scarlet lark]

Is the part done?

[simple ridge]

no wait a sec

let me rethink about something

[scarlet lark]

Okayy

[simple ridge]

wait we need values of x not m

[scarlet lark]

We do?

[simple ridge]

yes

[scarlet lark]

:/

[simple ridge]

it is easy

we are halfway

we are going from x^2+(m+3)x+6+2m>0

[scarlet lark]

okay

[simple ridge]

so from here

we have 3 cases

when this has 2 distinct roots

when it has 1 double root

and when it has no real roots

let me check somethig fast

[scarlet lark]

Did we do something wrong?

[simple ridge]

lets continue from here

i liested the 3 cases

do you understand why we have 3 cases

[scarlet lark]

Yes it depends on delta

[simple ridge]

ok

we will start with the first case

when we have 2 distinct roots

we dont care about the sign of the roots we just care that they are distinct

for what values of m does the original polynomial have 2 distinct roots

[scarlet lark]

]-inf;-3[U]5;+inf[

[simple ridge]

correct

what do you like to call the 2 distinct roots

[scarlet lark]

Do you mean x1 and x2?

[simple ridge]

yes

ok we will go with this

so they are x_1 and x_2

to not get in troubles we will consider x_1<x_2

you can let x_2<x_1 no problem

we just want to assign one less than the other to know their order in the table of sign

[scarlet lark]

Okayy

[simple ridge]

lets go with this

from here draw the table of sign of x^2+(m+3)x+6+2m

[scarlet lark]

Can I call it f(x)?

[simple ridge]

call it what you want

as long as it is valid notation

[scarlet lark]

Like this?

[simple ridge]

i am asking you to draw table of sign because it is faster than typing the sign for each interval

[simple ridge]

ok great

remember we want x^2+(m+3)x+6+2m>0

so from this table this occurs when x lies in the interval?

[scarlet lark]

]-inf; x_1[ U]x_2; +inf[

[simple ridge]

exactly

now we are done from case 1

so case 1: $m\in ]-\infty,3[\cup ]5,\infty[ \ x\in ]-\infty,x_1[\cup]x_2,\infty[$ where $x_1,x_2$ are the roots of $f(x)$ with $x_1<x_2$

[glossy valve]

calculus is fun

[simple ridge]

any questions till here

[scarlet lark]

Nope

[simple ridge]

ok

now case 2

1 double root

let this root be x_1

this happens when m=?

[scarlet lark]

-3 or 5?

[simple ridge]

yes

now what is the sign of f in this case

[scarlet lark]

Positive

It follows the sign of a

In the table of signs, before and after the double root

[simple ridge]

nice

so case 2 :$m\in{-3,5} \x\in\mathbb{R}$

[scarlet lark]

Isn’t it -3?

[simple ridge]

mb tysm for the correction

[scarlet lark]

no worries

[simple ridge]

now case 3 when x has no real roots

this happens when m lies in ?

[scarlet lark]

m lies in ]-3;5[

[simple ridge]

yes

[scarlet lark]

f(x) is positive

[simple ridge]

yes

[glossy valve]

calculus is fun

[simple ridge]

case 3 (last case): $m\in ]-3,5[ \x\in\mathbb{R}$

[glossy valve]

calculus is fun

[simple ridge]

case 1: $m\in ]-\infty,3[\cup ]5,\infty[ \ x\in ]-\infty,x_1[\cup]x_2,\infty[$ where $x_1,x_2$ are the roots of $f(x)$ with $x_1<x_2$\case 2 :$m\in{-3,5} \x\in\mathbb{R}-{x_1}$\case 3 (last case): $m\in ]-3,5[ \x\in\mathbb{R}$

[glossy valve]

calculus is fun

[simple ridge]

part 4

[scarlet lark]

Nicee I only have part 5 I think I know how to do it

I’ll try it rn and I’ll send it here give me a sec

[simple ridge]

take your time

[scarlet lark]

Is this right?

[simple ridge]

,rotate cw

[simple ridge]

you have a mistake

(-m-3)^2=(-1)^2(m+3)^2=(m+3)^2

correct your mistake

[scarlet lark]

a is -1?

In this identity I mean

Oh nvm I got it

[uncut cove]

sorry to interrupt, but in the 2nd case shouldn't we exclude the root from our solution?

[simple ridge]

which root

[uncut cove]

the 1 double root

x_1

[simple ridge]

i excluded it

case 2 :$m\in{-3,5} \x\in\mathbb{R}$

[glossy valve]

calculus is fun

[uncut cove]

shouldn't it be R\ {x_1}

[simple ridge]

ohh i get what you mean

you are right

mb tysm for pointing out the mistake

[scarlet lark]

Did we miss smth in part4?

[simple ridge]

yes there was a mistake

case 1: $m\in ]-\infty,3[\cup ]5,\infty[ \ x\in ]-\infty,x_1[\cup]x_2,\infty[$ where $x_1,x_2$ are the roots of $f(x)$ with $x_1<x_2$ \case 2 :$m\in{-3,5} \x\in\mathbb{R}-{x_1}$\case 3 (last case): $m\in ]-3,5[ \x\in\mathbb{R}$

[glossy valve]

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

[scarlet lark]

Okay I see what we missed
We can’t have x= x_1 because then it would be f(x)=0 and we have f(x)>0 only right?

[simple ridge]

exactly

[scarlet lark]

And did you check this?

[simple ridge]

the work is correct

[scarlet lark]

Nicee

Thanks a lot

[simple ridge]

np you did all the work

ill send you a friend request

[scarlet lark]

Yeah but you guided me through it

[simple ridge]

where in lebanon are you from

[scarlet lark]

Bchelleh it’s near Jbeil

wbu

[simple ridge]

i am in south lebanon

a village called houla

[scarlet lark]

Nice to meet ya

[simple ridge]

nearest city is bint jbeil it is about 20 km away

nearest to my village

[simple ridge]

you too

[scarlet lark]

Okay I see

[simple ridge]

so you are going to school tomorrow ?

[scarlet lark]

I am

[simple ridge]

nice

i am on vacation now for a while 😂

because of whats going on

[scarlet lark]

Yes I was meaning to ask about it but I didn’t want to be insensitive

[simple ridge]

np dw

what are you planning to do in uni

[scarlet lark]

Chemical engineering ^^

[simple ridge]

nice

i am planning to do pure mathematics

this is my last year

i am in grade 12

GS

[scarlet lark]

Ooh nicee

[scarlet lark]

That'll be me next year if everything goes well

[simple ridge]

thats nice

i will give you this advice

study well for this and the next year get the highest grades

in order to get into universities like AUB with full tuition

[scarlet lark]

Yeah uni is very expensive

You planning to go to aub?

[simple ridge]

do some activities outside school like reading some uni topics or to do some projects this will be a boost too

yes i need to get one of first 10 in lebanon

then i will get full tuition

[scarlet lark]

I actually have no clue what to do

[scarlet lark]

Oh wow that's a challenge isn't it

[simple ridge]

i may get up to 80% if i dont get this high because of tests like SAT ,early admissions ,financial aid ,high grades and so on

[simple ridge]

it is but enrolling there deserves this

because from there you can get scholarships to known universities world wide

of course if you show your skills

[scarlet lark]

I need to do the Sat

[scarlet lark]

Do you have any advice

[simple ridge]

so you like chemistry the most ?

more than physics math and bio ?

[scarlet lark]

Yep

[simple ridge]

ok

what comes next

[scarlet lark]

I don't mind them both tbh ig physics

[simple ridge]

ok so i suggest that you read books/watch videos about chem where you study chem

not just learn some scattered facts

[scarlet lark]

So to try and get ahead of everyone?

[simple ridge]

check the order chemistry is studied in using google or by searching for chemistry server on discord where you can ask

and ask about good books for self studying these topics

then start the journey

[simple ridge]

yes

i am doing the same thing

i am far ahead from my class in math

[scarlet lark]

Should I do it with maths too?

How much do you get on the "livret"

[simple ridge]

i asked you what you like the most because you have to enjoy this

not to study this just to be ahead

i like math so i dont consider this like studying in class

because i enjoy studying math

[scarlet lark]

Oh yeah maths comes before physics that's for sure but chemistry first so I'm gonna focus on it like you said

[simple ridge]

and math is important of course it is found in every STEM major

[simple ridge]

whats that

[scarlet lark]

Your final grades

[simple ridge]

idk how much math is involved in chemistry you can check this too

[scarlet lark]

The average of all your grades at the end of the year

[simple ridge]

ahaa

last year it was around 18.5 eventhough i dont study much

[scarlet lark]

Nicee

[simple ridge]

what about yours

[scarlet lark]

Last year so grade 10 I got 19.42

[simple ridge]

nicee

which school

i will not know it probably

but just asking

[scarlet lark]

College jeune fille du Liban beit hebbak

u?

[simple ridge]

does it have some university classes or stops at grade 12

[scarlet lark]

It stops at 12

[simple ridge]

Al Mahdi high school(the one in bint jbeil)

[simple ridge]

same for mine

check #old-network

you can find a chemistry discord server there

[scarlet lark]

Oh wow thanks

[simple ridge]

np

how did you find this server btw

were you searching for a math server and found this ?

[scarlet lark]

Yep I needed help

Wbu

[simple ridge]

i was searching for a math server to get answers to my questions (questions from the self study) and because it is useful to be in a server where many have similar goals to you

for example many are here to discuss math

so one can learn alot from conversations like this

also whenever i want to ask about books or what topics to study before others i can ask

same goes for you when you enter the chem server

[scarlet lark]

Yess I'm gonna do that

[simple ridge]

you will benefit from experienced people in the field of chemistry

also if you have close friends who are interested in math ,physics ,chemistry ,biology present the idea of self study to them

[scarlet lark]

They don't even like to study for school I've tried convincing them to let me explain to them what they don't get, but they don't care about stuff like this
I'll ask anyways

[simple ridge]

you will see that what you take in school is nothing but memorizing and studying fractions of larger concepts

[simple ridge]

by friends i dont mean all classmates because it is clear that most of them wont do this

[scarlet lark]

Yeah I meant like my close friends

[simple ridge]

i mean friends who understand this or who are interested in a specific subject

[simple ridge]

ohh

doesnt any of them like physics or math or something specifically ?

[scarlet lark]

They just like sleep

😂

[simple ridge]

😂

no problem give them the idea

[scarlet lark]

They are not highachievers like me

[scarlet lark]

I will for sure

[simple ridge]

grades dont matter

you should believe that

[scarlet lark]

Not talking bout grades but I like to grow my knowledge when they don't care about it all

[simple ridge]

grades only matter to get a scholarship or so

in my opinion school wastes alot of time

[scarlet lark]

It does because they teach us stuff we won't use
That time could be spent studying what we want and need
And what we're interested in

At least that's my opinion

We should be able to chose what classes to take

[simple ridge]

yes thats way better

lets move to private chat if you want to continue to let others use this channel

[scarlet lark]

Oh yes

[simple ridge]

type .close

to close the channel

[scarlet lark]

.close

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@mint ore Has your question been resolved?

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@mint ore Has your question been resolved?

[full forum]

@mint ore Has your question been resolved?

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[thorn crescent]

The probability distribution of all possible values of the sample proportion is the
a. sampling distribution of x̄
or
b. sampling distribution of p bar

[thorn crescent]

i think its b

not 100% sure

ping me if u can help

[full forum]

@thorn crescent Has your question been resolved?

[thorn crescent]

.close

[full forum]

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[torn jolt]

how to tpye this in latex

[hollow wharf]

$\sqrt[k]{a}$ and $b^c$

[glossy valve]

CST (please ping when replying)

[hollow wharf]

Put them together

[torn jolt]

oh

[hollow wharf]

yeah this is the only way I can present this such that it helps you but also doesn’t give the answer away

[torn jolt]

@hollow wharf how can i make

the font bigger

this is not for a test or anything lol

i thinkj its fine for u to tell me

im just writing a note for myself

[hollow wharf]

the font bigger?

just…increase the font size in your word processor?

If you mean make A uppercase, you can type A

[torn jolt]

what is word processor

[hollow wharf]

Microsoft Word, Pages, Google Docs, etc.

[torn jolt]

oh

[left bone]

i mean they are using latex so its probably overleaf

[torn jolt]

can i increase the font in overleaf hto

yeh

[left bone]

something something \documentclass[Xpt]{something something}

maybe you can look it up I dont remember the exact command sorry

its something very similar to that though

[torn jolt]

ok sure

.close

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[desert cypress]

hhow is the third

not a polynomial

is it smthn with division

[astral steppe]

yes

[desert cypress]

like what

how is it not polynomial

i cant figure

rational

[astral steppe]

polynomials can only addition subtraction multiplication and integer powers

[desert cypress]

never division

[astral steppe]

like the general form for polynomials is ax^n + bx^(n-1) ...

[astral steppe]

nope

[desert cypress]

ikokkol

tyty

[astral steppe]

np

[desert cypress]

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[granite olive]

a * (b / c) = a * b / c

Why do these two equations evaluate to be the same value? In the first one the division is being evaluated first and then it's quotient is being multiplied by a. In the second, a is being multiplied by b and it's product is being divided by c. They are being solved in a different order of operations, and it's confusing because I just always thought that if multiplication and division wasn't done left to right you'd get a different answer; meaning the first expression would evaluate to a different value. However, that's not true here.

[brazen moth]

a = a/1

$\frac{a}{1} * \frac{b}{c} = \frac{a*b}{c}$

[glossy valve]

890s

[granite olive]

well i figured that too, i was like ok well if use fractions it makes sense

[brazen moth]

you should usually use fractions when it comes to division

[granite olive]

I'll explain what I was thinking real quick.

[brazen moth]

it makes the algebra a lot easier

[granite olive]

in the second expression, since multiplication and division is left to right - (a *b) / c - is what it is. You can't turn b and c into a fraction then

[brazen moth]

technically you can, but not in a way the is representative of the order of operations used here

actually

you could do it like how i did it above

[granite olive]

oh ya you can

[brazen moth]

$\frac{a*b}{c}$

[glossy valve]

890s

[granite olive]

ok that makes sense

🤦‍♂️

[brazen moth]

dont worry dude

we all have those moments

[granite olive]

well thanks see ya

[fathom cradle]

they both become ab/c when you do the simplifying

[granite olive]

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[twin wolf]

how can I solve x - 1 < (1 - x)(3 + x) ?
Dunno how to start

[viral jasper]

distributing might be a good first step

[twin wolf]

I get x - 1 < -x^2 - 2x - 3

[hollow wharf]

You want to put everything on one side and 0 on the other

[twin wolf]

-x^2 - 3x -2 > 0

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[tacit siren]

Factor again

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[proven jay]

Why do we have principal value of a complex number from interval (-pi,pi]?

[proven jay]

my natural notion is, between (0,2pi] the values of angles are unique

then adding integer multiple of 2pi will give me equivalent angles

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[slate oriole]

Hey all

[slate oriole]

I'm trying to get help.. trying to make the Numbers 3, 7, 12, and 2 in that order, using each number once equal 517 using pemdas

And I'm allowed to use any exponent on any of the numbers

I've tried so many things for the last few hours and can't get anywhere

Just the bottom one

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[slate oriole]

I got 517.5... that's the best I could do

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[woeful bay]

I need help with part e, I dont understand what the question wants from me when its asking for savings

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[teal whale]

So i got a^2=0 and ab=0 but how should I get the y?

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[sharp flame]

Maybe write $z = re^{i\theta}$

[glossy valve]

jan Nejon

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[crimson jasper]

Calculate the nth order covariate using the recurrent relationship method

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[cunning hatch]

no idea where to start here

[vestal acorn]

can u use calculator?

bc u can just use sin^-1 to get x if u can use calculator

[cunning hatch]

exact value

[raven quail]

Find the sign of cosx from the quadrant given and find the value of cos x

Then use sin2x = 2sinx cosx

[raven quail]

You don't need a calculator for this question

[vestal acorn]

ik, too lazy to think lol

[cunning hatch]

wdym by this