help-10

[fair pond]

say you have m and a point (x, y). Then, you calculate b = y - m*x

and you have the equation of the line

[violet compass]

you can't use y=mx+b in this situation.

[spiral maple]

you can.

[minor pilot]

I think he means immediately

[spiral maple]

Well obviously.

[violet compass]

but not the best.

nor ideal

[spiral maple]

Yes, that doesn't mean you cant use it.

[fallen mantle]

Equation of straight line: y = mx+c
now m is 3
x is -7
y is 2
c is unknown

Solve for c then you will have the equation of the line

[obtuse pebble]

@paper socket Has your question been resolved?

Channel closed

Closed by @paper socket

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[inland thicket]

x2 +5x +6 >= 0

[inland thicket]

I got results:

im confused to like

at the end the >= has to flip

she got -3 flipped

and other x>=-2 like normal

[minor pilot]

oh

It's cuz between -3 and -2, the value is a negative number

So even if the number was -4, it'd overall give a positive result

[inland thicket]

hmm

[minor pilot]

Here I put a graph for reference

Cuz u cant get the sqrt of a negative number right?

Do u see how it can be less than -3?

[inland thicket]

yeah but

i thought it would be a result of inequality

like

i thought you could calculate inequalities

and >= flips when something hapens

[minor pilot]

U can, it's just the domain

Like if it's -2.5 for example it doesnt work out

[minor pilot]

The inequality is just another way of writing what she wrong in the pink writing

The U means and

[inland thicket]

shes wrong?

ye i know the U

[minor pilot]

I dont think so

She's saying that x can be below -3 as well as above -2

[inland thicket]

like i just calculated the inequality and like

[minor pilot]

Im trying to be simple

lets see

So what's the confusing part

[inland thicket]

on test i would just write x >= -3 , x >= -2

[minor pilot]

im a bit lost

[inland thicket]

so i would fail

because thats what i got in calculating on paper

[minor pilot]

That one doesnt work tbh

Its the right idea buuut

[inland thicket]

wait

because its a polynomial

i have to conclude that

it has to be like

-3 can be lower or equal

is that the thing?

[minor pilot]

Yah

[inland thicket]

but yo

[minor pilot]

The inequality is the same thing as the (-inf., -3] U [-2, inf.)

[inland thicket]

here is also a polynomial

[minor pilot]

Yah this one's a bit easier to explain

U cant divide a number by 0, right?

[inland thicket]

ye

[minor pilot]

Basically it's just stating that u cant have x be -3 or -2 cuz that would make it 0 on the bottom

Cuz what u said, the polynomial would end up giving 0

Welp, im gonna head to bed it's really late for me

[inland thicket]

ait

fr fr

[minor pilot]

I wish I could help more thooo

[obtuse pebble]

@inland thicket Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lusty vigil]

idk

[obtuse pebble]

Channel closed

Closed due to the original message being deleted

[lusty vigil]

so

how will i calculate it?

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lusty vigil]

Two functions are given by the regulations f (x) = 2x^3−150x and g (x) = x^2 + kx + 25−500, where k is a constant. The function f has a local maximum at the same point as g has a local minimum. Based on this information, determine the value of k.

[lusty vigil]

I tried to find k, by differentiating f(x), g(x) -> f'(x), g'(x)
then finding an expression for x1,2 from: f'(x1) = 0, g'(x2) = 0
and solving the equation x1 = x2 to find k

but I couldnt get it right

who can help?

<@&286206848099549185>

[tardy epoch]

show your work and someone can review it

[lusty vigil]

uhm 2 sec

files too big

[hasty plinth]

take pictures w/ phone or something

[lusty vigil]

yeah ok gimme a minute

[tardy epoch]

use an app to downsize it too

[hasty plinth]

how do you know you couldn't get it right

is there like an answer key

or answer choice list

[lusty vigil]

i got 3

entered

it said wrong

[hasty plinth]

do you have multiple attempts

how did you get 3

[lusty vigil]

idk either

😭

that's why i need help

[lusty vigil]

1 attempt actually:/

so cant blow it

[hasty plinth]

ok so i don't take calc but i think i have an idea

So your idea of differentiating f(x) and g(x) is what I'm thinking of

and you know that the critical points of a function are when the derivatives are equal to 0 right

so taking the derivatives of your functions you get
f'(x) = 6x^2-150
g'(x) = 2x+k

So setting f'(x) = 0 we get 6x^2-150 = 0 so x = +/- 5 right

So if we plug in a value such as -6, we see that f'(x) will be > 0 so it will be increasing up until x=-5 so that means x=-5 is a local maximum...I think

So then plugging in x = -5 to g'(x)

We get 2(-5) + k = 0...so k must be 10 I think...?

This might be completely wrong idk

it might be k=-10

hm

that's weird though

because wouldn't both of those be a local min...

so it can't be k=-10

no what the hell it has to be k=10...am i stupid...?

i don't understand...

what am i doing wrong wtf

@lusty vigil do you see the error in my ways lmao

Ok let's review...

g(x) = x^2+kx+25-500 has a local minimum at
x = -b/2a
x = -k/2

We're given
f (x) = 2x^3−150x

the local maximum of f is the same as the local min of g

local max of f is x =- 5.

-5 = -k/2
-10 = -k
k = 10...?

wtf am i doing wrong

the points are different but they have the same x values...but i don't know what else I'm missing...?

ugh

pain.

someone else more qualified please ping me when you see what i did wrong lmao

[obtuse pebble]

@lusty vigil Has your question been resolved?

[lusty vigil]

@hasty plinth

I looked

at it

and

seems better than what i did

but anybody qualified?

[obtuse pebble]

@lusty vigil Has your question been resolved?

[obtuse pebble]

@lusty vigil Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wanton dagger]

I just need help with part a)

[obtuse pebble]

@wanton dagger Has your question been resolved?

[wanton dagger]

<@&286206848099549185>

[nocturne minnow]

Did you attempt anything?

[wanton dagger]

I found the inverse of them

[nocturne minnow]

So for a, what's the inverse function?

[wanton dagger]

Y=x-7

[nocturne minnow]

Do you know what $f^{-1}[f(x)]$ means?

[wanton dagger]

Nope I’m confused on that

[nocturne minnow]

Do you know what composite functions are?

[wanton dagger]

I’ve heard of them

A function in a function?

[nocturne minnow]

Yes

[wanton dagger]

Oh ok

[nocturne minnow]

Have you seen (f o g)(x) before?

[wanton dagger]

I don’t think so

I might’ve seen it but I don’t know what it is

[nocturne minnow]

That is a different notation for composite function

[wanton dagger]

Ohh ok

[nocturne minnow]

So for instance, y = 5x, the inverse is y = (1/5)x, f^(-1)[f(x)] means plug in the function f(x) into f^-1

[wanton dagger]

Ok let me try understand

So 5x(-1)?

Or 5(-1)?

[nocturne minnow]

When I say something like f(3) = 3x - 4, do you know how to interpret that?

[wanton dagger]

Yh 3(3)-4

[nocturne minnow]

That's the same concept of f^(-1)(f(x))

[wanton dagger]

Ok

So is f(x) 1/5?

[nocturne minnow]

For my example, f(x) = 5x and f^(-1)(x) = (1/5)x

As you mentioned, f(5) means plug in 5 for x, in the function f

So same idea for f^(-1)[f(x)], it means plug in f(x) for x in the function f^(-1)

[wanton dagger]

So 1/5(-1)

[nocturne minnow]

Not exactly

f^(-1) is notation for inverse, right?

[wanton dagger]

Yes

So 5(-1)

[nocturne minnow]

No

Why is there a -1?

[wanton dagger]

Ok I’m confused in the notation

[nocturne minnow]

Here watch this
https://youtu.be/uiMNvOPqrSM?t=345

About 5 minute watch starting from the location I linked it at

[wanton dagger]

Ok sure

Ok I think I get it

Thanks for your help

.close

[obtuse pebble]

Channel closed

Closed by @wanton dagger

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[flat ravine]

May I get help with this, I am completely lost lol, prefferably a step by step help

[obtuse pebble]

@flat ravine Has your question been resolved?

Channel closed

Closed by @flat ravine

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stiff maple]

Hello! I'm having trouble with understanding how to use the half-angle formulas (particularly if the formula should be negative or positive). The question states: cos(x)=-4/5, 180 degrees < x < 270 degrees

[stiff maple]

I've worked it out and it seems to me that sin(x/2) should be negative, but other sources say otherwise without much explanation

Because since x is in quadrant 3, both sin and cos should be negative

[obtuse pebble]

@stiff maple Has your question been resolved?

[stiff maple]

<@&286206848099549185>

[tardy epoch]

Other sources are wrong

[stiff maple]

So the answer here is right?

@tardy epoch are you sure?

Because if both sine and cosine are negative, that would make tangent positive

[obtuse pebble]

@stiff maple Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[small bloom]

The sum of the first 10 terms of a geometric series with a common ratio of –1 is 0. What is the first term?

[small bloom]

I got 0, but this is multiple choice and that's not one of the answers

[deft nimbus]

hi

ping me when you are ready

[small bloom]

@deft nimbus hello! sorry ready for what?

[deft nimbus]

ok you got the answer choices?

[small bloom]

4, -4, 1

[deft nimbus]

those are the choices?

[small bloom]

mhm!

[deft nimbus]

hm

[deft nimbus]

they all should work

you can select multiple answers right?

[small bloom]

right yes I'd assume so? Idk how to get those though

[deft nimbus]

so

it says

the ratio is -1

which is also the rate

[small bloom]

right

[deft nimbus]

so the number itself should alternate between it positive and negitive forms

lets get 4 for example

so the first term is 4

next 9 terms would be

+4,-4,+4,-4,+4,-4,+4,-4,+4,-4

right?

[small bloom]

mhm

[deft nimbus]

well thats all 10

and if we were to add all of them together

what would you get?

[small bloom]

0

[deft nimbus]

yep

so since those numbers would cancel out

[small bloom]

but then aren't all numbers possible?

[deft nimbus]

it would work with any number

yes all numbers are posible

but since those are the answer choices they give you

[small bloom]

ohhh

[deft nimbus]

you would choose all of them

but

that only works for even terms

not odd

if it was 11 terms it would not add up to 0

[small bloom]

right yes makes sense

okay thank you!!

[deft nimbus]

is that it?

[small bloom]

mhm!

.close

[obtuse pebble]

Channel closed

Closed by @small bloom

Use .reopen if this was a mistake.

[deft nimbus]

have a great day

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[keen pecan]

how would you do this?

[tranquil rain]

Which question?

[keen pecan]

i need for number 61

[tranquil rain]

Ok what have u tried?

[keen pecan]

well since i got stuck on part a i did substitution for part b

i got 1/2* [1/(x^2+3)]0,-2

however

i dont know how they got 3, 7

[tranquil rain]

When ur substituting u

U have to substitute the limits as well

Ex when $x=0$
$u=x^2+3$

Plug in x and u will get it

[warm shale]

Glory

[keen pecan]

oh

dang

thank you

how do you do part 1 tho?

[tranquil rain]

By finding the indefinite integral first?

[keen pecan]

how would you do that with this one?

[tranquil rain]

Same method

Substitution

But ur doing it separately

Ur treating it as if ur disregarding the limits

[keen pecan]

so you just substitute again??

[tranquil rain]

Yeah then swap out u

So it doesn't affect the calculations of the limits later on

[keen pecan]

isnt that the same as part 2?

[tranquil rain]

No for part 2 u have to change the limits

In part 1 u don't have to

[keen pecan]

wait wont that get you a different answer since the equation is the same but the limits are different?

[tranquil rain]

Won't bc for part two pay attention to 4th line where the u wasn't swapped out

[keen pecan]

so how do they get 4/21?

[tranquil rain]

By applying basic integral rules

[-1/3+1/7]

[keen pecan]

...

thank you so much

i got the answer

[tranquil rain]

Np

[keen pecan]

.close

[obtuse pebble]

Channel closed

Closed by @keen pecan

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[marsh trail]

I need help with this math problem thats due in 2 hours

[marsh trail]

and this one too if that's okay

I was sick for a week, I told my teacher but he refuses to give me more time to finish this or explain how to do it

[obtuse pebble]

@marsh trail Has your question been resolved?

[marsh trail]

<@&286206848099549185>

[obtuse pebble]

@marsh trail Has your question been resolved?

[obtuse pebble]

@marsh trail Has your question been resolved?

[hasty plinth]

well first, you need to know how your teacher defines bisector and altitude

euclidean geometry is a fucking PAIN

my teacher personally made us write everything in the form of conditionals which made a single proof a page long...which isn't much in comparison to more complex proofs but considering how many problems he gave us it was a lot

first you state your givens

then you use the definition of your givens to derive something

(in english)

then you write that derivation in english

so 1 goes to 1
2 goes to 3
4 goes to 5
6 goes to 6...?

and then 8 goes to 8...?

that's probably wrong but

yeah

you need to know how your teacher defined each term as he should've gone over it in class

get notes from a classmate

these proofs are always so dumb because teachers always nitpick every little thing to ingrain the definitions into your head

its probably too late anyways rip

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timid silo]

given: S is the area in the first quadrant within the circle x^2+y^2 ≤ a^2
and under the line y =sqrt(3)x
where a ∈ R

[timid silo]

how do i find the borders of Theta

or how do I find the angle itself?

nvm i found it

[obtuse pebble]

@timid silo Has your question been resolved?

Channel closed

Closed by @low bane

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[primal niche]

is it possible to prove that the vertical line is perpendicular WITHOUT using pythagoras' theorem? i think i need to prove that it is perpendicular so that i can then prove that pythagoras' theorem can be used on it

[vernal wren]

You can use trigonometry to prove it?

[primal niche]

we havent learnt trigonometry yet so idk if thats how i should solve it

[deft hazel]

generally if the angle formed between two lines is 90degs then its a right angle, thus perpendicular/orthogonal with respect to something else

[vernal wren]

Nvm then

[primal niche]

i dont know if we are allowed to assume this, is there any other way to prove it apart from this or using trigonometry?

because if not then i guess i have to assume it then

[deft hazel]

You're tryna show that the line segment of length one not on the x-axis is perpendicular to the x axis?

So you don't know your trig ratios?

Like sin, cos, tan

"sohcahtoa"

no?

[primal niche]

our school program starts trig a bit after this, i have a bit of understanding of soh cah toa, but it hasnt been properly taught

i just know a bit from watching videos online, etc

[deft hazel]

hm okay how about using cartesian geometry?

[primal niche]

ok sure lets try that

[deft hazel]

hm just thinking about it don't think you can do it with cartesian geo

Since i was thinking to show that

[deft hazel]

But since the y-axis has a gradient of 0? idk to word this

wait y'kno you just do your things

[primal niche]

?

[deft hazel]

nevermind

god i'm confusing myself

[primal niche]

do you think it would be fine if i just assume that its a 90 degree angle?

this is just a high school topic test

[deft hazel]

Yes it is orthogonal

[primal niche]

ok

[deft hazel]

you can verify that by looking at the triangle formed

it's an orthogonal triangle

[primal niche]

ah ok

[deft hazel]

Hm but if we set the vertices are coordinates

we can say that, that vertical line formed

is perpendicular to the x-axis

Because we can see that the coordinates are collinear

and the y-axis goes up/down in increments of integers

no

uh

ye ig all you have to say that the points are collinear

or something

[obtuse pebble]

@primal niche Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timid silo]

hi can i please get help

im struggling to find v

i intergrated but it seems to be wrong

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185> please help me :((

.close

[obtuse pebble]

Channel closed

Closed by @untold juniper

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wanton dagger]

I need help with finding the volume of this shape.

[wanton dagger]

I’m having trouble with finding the area of the front shape

[vernal wren]

Can you find the area of the front if there was no hole?

[wanton dagger]

Yes

[vernal wren]

And can you find the area of the hole? Its a semi circle

[wanton dagger]

Yh ok let me try again

Ok got it

Thanks

.close

[obtuse pebble]

Channel closed

Closed by @wanton dagger

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wooden bay]

why does this limit equal infinity, ignore first equal sign

[obtuse pebble]

@wooden bay Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rocky cosmos]

if\ $$tan(x)=2$$\ what\ is\ $$sin(2x)$$?
$$\frac{sin(x)}{cos(x)} =2;\ 2cos(x)\ =\ sin(x)$$

[warm shale]

AuHasard

[rocky cosmos]

how do i proceed from here

$$\left( 2cos(x)-sin(x)\right)^{2} =(2cos(x)-sin(x))(2cos(x)-sin(x))\ =\ 0$$

[warm shale]

AuHasard

[rocky cosmos]

i dont know haha

hmm

2cos(x) = sin(x)

if i multiply by 2 on both sides

nah

i know what it is

2sinxcosx

but we have sin x on the right

so what are you trying to tell me

i didnt know that

[spiral maple]

??

[rocky cosmos]

??

yes, i didnt know that

[spiral maple]

what was literally anything you were saying then?

[rocky cosmos]

$$sin(2x)=2sin(x)cos(x)$$
$$\frac{sin(2x)}{sin(x)} =\frac{1}{sin(x)} \times 2sin(x)cos(x)$$
$$

can you chill

mate

sin(2x)/sin(x) = 2cos(x)

right?

[spiral maple]

yes.

[rocky cosmos]

that's what i'm trying to prove up there

ok

[spiral maple]

provided x != n*pi blahblahblah

[rocky cosmos]

ok so we have 2cos(x) = sin(x)
and we know that sin(2x) = 2sin(x)cos(x)

so 2cos(x) is basically just 2sin(x)cos(x)/sin(x)

what then?

let me check what you wrote

yes

$2sinxcosx=(sin(x))(2cos(x))=sin(2x)$

[warm shale]

AuHasard

[rocky cosmos]

yes?

[spiral maple]

pointless symbol pushing

but yes

It is.

[rocky cosmos]

i would've if it was that easy

here let me show you the alternatives

”what's sin (2x)”

[spiral maple]

@rocky cosmos if you had y=3x and I asked you what y was when x=2, would you be able to do that?

[rocky cosmos]

yes

[spiral maple]

Ok, what would it be, and why?

[rocky cosmos]

y would be 6 because the value of the function is 6 at x = 2

[spiral maple]

yes

so If I had $\sin(2x)=2\sin(x)\cos(x)$ and I asked you what it became when $\sin(x)=2\cos(x)$

[warm shale]

Mosh

[spiral maple]

you should be able to do this on the back of what you just said.

I disagree that it'd be easier

[rocky cosmos]

I don't know tbh

and I feel bad

I've studied trig identities for a while

[spiral maple]

why is 3x 6 when x=2?

ELI5

[rocky cosmos]

because of the value

of the function

[spiral maple]

so 3x=2?

[rocky cosmos]

no, x is 2

[spiral maple]

yeah but you just said cause of the value

[rocky cosmos]

you probably know what i meant

what's your point

[spiral maple]

Cause you're failing to recognize that this problem too is just a substitution

[rocky cosmos]

$2sinxcosx=(sin(x))(2cos(x))=sin(2x)$

[warm shale]

AuHasard

[spiral maple]

No

[rocky cosmos]

nobody said if this was wrong

[spiral maple]

Yeah, cause it's not

it's just pointless

$\sin(2x)=2\sin(x)\cos(x)=2(2\cos(x))\cos(x)$

[warm shale]

Mosh

[spiral maple]

cause sin(x)=2cos(x)

[rocky cosmos]

can we treat them as variables?

[spiral maple]

yeah?

[rocky cosmos]

no i mean the trig functions

[spiral maple]

Yeah

algebra doesn't magically break as soon as trig appears

[rocky cosmos]

ok so a = sin(x), b = cos(x)
a = 2b

so i need to prove that a is equal 2b

[spiral maple]

NO

That is the assumption tan(x)=2

[rocky cosmos]

okay type it out

$a=sin(2x)=2\sin \left( x\right) \cos \left( x\right) =\sin \left( x\right) (2\cos \left( x\right) )=bc$

[warm shale]

AuHasard

[rocky cosmos]

hmm

okay so sin(x) = 2cos(x)

yes

$b=\sin \left( x\right) \wedge c=2\cos \left( x\right) \Longrightarrow b=c\Longrightarrow \sin \left( x\right) \times \sin \left( x\right) =\sin^{2} \left( x\right)$

[warm shale]

AuHasard

[rocky cosmos]

So $\sin \left(2x\right) =\sin^{2} \left( x\right)$

[spiral maple]

omg

clearly not?

[rocky cosmos]

let me check

no sin(2x)

[warm shale]

AuHasard

[rocky cosmos]

okay so which one of these are our answer

[spiral maple]

you're nowhere near done

you need to know what sin(x)= now

[rocky cosmos]

lol

[spiral maple]

what lol?

[rocky cosmos]

too many steps

what do you suggest

[spiral maple]

Pythagorean identity...

[spiral maple]

$\tan^2(x)+1=\sec^2(x)$

[warm shale]

Mosh

[rocky cosmos]

$\sin \left( 2x\right) =1-\cos^{2} \left( x\right)$

[spiral maple]

no

well, yes cause finding cos^2(x) is much easier

[rocky cosmos]

i've not worked with sec yet, only tan cos and sin

[rocky cosmos]

so should i post the last equation?

[spiral maple]

sure

[rocky cosmos]

$\sin \left( 2x\right) =1-\cos^{2} \left( x\right)$

[warm shale]

AuHasard

[rocky cosmos]

hmm.

so i use that

$\sin^{2} \left( x\right) +\cos^{2} \left( x\right) =1$

[warm shale]

AuHasard

[rocky cosmos]

and solve for x?

[spiral maple]

Rule of Thumb: You never need to find x explicitly in these problems.

for reference, this is also wrong

[rocky cosmos]

they wanted to know what sin(2x) was, so true, not x per se

maybe i should've left it like this

[spiral maple]

and cause finding x explicitly is just a shit method

$\sin^2(x)+\cos^2(x)=1\overset{\sin(x)=2\cos(x)}{\implies}4\cos^2(x)+\cos^2(x)=1$

[warm shale]

Mosh

[spiral maple]

didnt sqrt properly.

no

I mean you just didnt square root properly

$x^2=a\implies x=\pm\sqrt{a}$.

[warm shale]

Mosh

[spiral maple]

k.

[rocky cosmos]

$$5\cos^{2} \left( x\right) =1$$
$$\cos^{2} \left( x\right) =\frac{1}{5}$$

[warm shale]

AuHasard

[spiral maple]

yes.

[rocky cosmos]

aaaand we are still not done

[spiral maple]

its one more step...

[rocky cosmos]

hold on

$$1-\sin^{2} \left( x\right) =\frac{1}{5}$$
$$1-\frac{1}{5} =\sin^{2} \left( x\right)$$
$$\sin \left( 2x\right) =\sin^{2} \left( x\right)$$
$$\sin \left( 2x\right) =\frac{4}{5}$$

[spiral maple]

sure

though you had 1-cos^2(x)=1-1/5 available to you, but I guess we can't avoid pointless symbol pushing

[warm shale]

AuHasard

[rocky cosmos]

lol

it helps if i write it all down

memorization i guess

thanks for the help

.close

[obtuse pebble]

Channel closed

Closed by @rocky cosmos

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wind bluff]

Pls help me with this problem

I am in 9tg grade and I can't do this

[obtuse pebble]

@wind bluff Has your question been resolved?

[wind bluff]

<@&286206848099549185>

[wind bluff]

<@&286206848099549185>

[obtuse pebble]

@wind bluff Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lapis smelt]

Can someone explain this to me

[civic zealot]

You pick a door. There's a 1/3 chance you're correct and a 2/3 chance you're wrong.
That doesn't change by showing you a door with a nothing behind it, you knew one of the other doors had nothing behind it. All this does is show you which door to switch to if you choose to switch.

[gilded needle]

It's called a "paradox" because many people intuitively don't think it could possibly make a difference whether you switch doors or not, after all, aren't they all equally likely to have the prize? To help with the intuition regarding why it could make a difference, suppose that instead of 3 doors there are one million doors. You choose one, then the host reveals 999,998 empty doors. There are only two remaining that haven't been open. Which one do you think is more likely to contain the prize? Yours or the other one that the host didn't open?

[scarlet gale]

@lapis smelt On that, you don't know whether to switch or not. They could have chosen to reveal a door only when you're right, and then switching never works.

The proper statement is that the host always reveals a door with nothing behind it.

Just seeing the host reveal a door once doesn't tell you the host's strategy.

If they always reveal a door with nothing behind it regardless of what you picked, then you have a 2/3 chance of winning by switching.

[lapis smelt]

I don’t understand still. If you chose one out of the 2 doors that are not revealed. Why is it favorable to switch? When the third one was revealed empty?

[scarlet gale]

Because you have a 1/3 chance of getting it the first time and a 2/3 chance of not getting it.

[gilded needle]

yes, you know there is 2/3 probability that it's behind one of the doors you did not choose. the host is helping you by ruling out one of those unchosen doors, so now you know that the other unchosen door has 2/3 probability to have the prize

[scarlet gale]

If you switch, you lose 1/3 of the time (when your first guess was right) and win 2/3 of the time (when you didn't pick right first time).

There is 1 way to pick correctly on your first guess and 2 ways to pick wrongly. 1 way will make switching fail and 2 ways will make switching work.

That's why it's 2/3 chance of winning if the host always reveals a door.

[lapis smelt]

Now I understand thank you

[scarlet gale]

No problem.

[lapis smelt]

.close

[obtuse pebble]

Channel closed

Closed by @lapis smelt

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sweet jay]

How can I determine its conv/div ergence?

[sweet jay]

By which test?

[gloomy valve]

Integral test may work

$k^2 e^{-k^2} \leq k^2 e^{-k}$ for $k \geq 1$

[warm shale]

Navix

[sweet jay]

ah tq

.close

[obtuse pebble]

Channel closed

Closed by @sweet jay

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[opaque sandal]

.reopen

[agile musk]

$cos^4(at)$

[warm shale]

Zoro

[agile musk]

can someone help me expand this?

the answer i get using two power reduction methods is not the same as what my prof got

and ive been at it for a while and i still keep getting the same answer i got

what i get is 3/8 + 1/2cos(2at) + 1/8cos(4at)

but im supposed to have 5/16 + 5/8cos(2at) + 1/16cos(4at)

[civic zealot]

what did you do?

[agile musk]

so i first expanded it to: [1/2 + 1/2cos(2at)]^2

squared this expression

then got 1/4 + 1/2cos(2at) + 1/4cos^2(2at)

then expanded the cos^2 again

[civic zealot]

and got?

[agile musk]

3/8 + 1/2cos(2at) + 1/8cos(4at)

[civic zealot]

,w expand 1/4(1/2+1/2cos(4at))^2

[agile musk]

how come there is a square to that expression?

[civic zealot]

well that didn't give me what I wanted

ah, my bad, looked at the wrong thing

[agile musk]

im honestly not sure if my prof made a mistake

or if i keep making the same mistake that doesnt make our answers match up

[civic zealot]

yeah, Seems like they made a mistake

[high lily]

,w cos^4(at) = 3/8 + 1/2cos(2at) + 1/8cos(4at)

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ramonov#0921

[agile musk]

ok I think this clears it up more, thanks!

.close

[obtuse pebble]

Channel closed

Closed by @agile musk

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[drowsy cosmos]

i need some help with differential forms, could someone check what step is wrong here?

[drowsy cosmos]

I know the last one is wrong because it doesnt hold $$\varphi^* (d \sigma)=d(\varphi^*\sigma)$$

[warm shale]

Enoo58

[drowsy cosmos]

but i dont know where my mistake is

sorry this is already taken

[little crow]

Mb

[compact shadow]

Under 2

[drowsy cosmos]

did i do the wedge product wrong?

[compact shadow]

x_2 sin(x_2) not x_2

[compact shadow]

Just you forgot sin(x_2)

[drowsy cosmos]

in the last term right?

[compact shadow]

Yeah the coefficient or the second term of the right hand side under 2

[warm shale]

Enoo58

Enoo58

Enoo58

[drowsy cosmos]

so there is again something wrong

@compact shadow

[compact shadow]

I calculated φ*σ and I got

$cos(x_{3}x_{4})dx_{1}Λdx_{2}Λdx_{4}-x_{2}sin(x_{2})(x_{4}dx_{1}Λdx_{2}Λdx_{3}+x_{3}dx_{1}Λdx_{2}Λdx_{4})$

[warm shale]

Cogwheels of the mind

[compact shadow]

So you are right

[drowsy cosmos]

how do the $x_3,x_4$ get there in the last term?

[warm shale]

Enoo58

[compact shadow]

I got same d(φ*σ) with you

Just the Jacobi matrix is $\begin{pmatrix}1&0&0&0\0&1&0&0\0&0&x_{4}&x_{3}\0&0&0&1\end{pmatrix}$

[warm shale]

Cogwheels of the mind

[compact shadow]

They appear in the third row

[drowsy cosmos]

ah i see

yeah but still we should have this equality

$$\varphi^* (d \sigma)=d(\varphi^*\sigma)$$

[warm shale]

Enoo58

[drowsy cosmos]

i see it now i think

we get the x4 term just like before if we do the LHS

[compact shadow]

I got this

I calculated mine , they equal

Your dσ is correct

Now φ* it

[drowsy cosmos]

alright i think im having trouble with properly using $\varphi^*$

[warm shale]

Enoo58

[drowsy cosmos]

i thougt i just have to exchange the coordinates

[compact shadow]

No it’s

$ω=f(y_{1},…,y_{n})dy_{j_{1}}Λ…Λdy_{j_{r}}$, then $φ*ω=\sum_{i_{1},…,i_{r}} f(φ(x_{1},…,x_{n}))\frac{\partial (y_{j_{1}},…,y_{j_{r}})}{\partial (x_{i_{1}},…,x_{i_{r}})} dx_{i_{1}}Λ…Λdx_{i_{r}}$

[drowsy cosmos]

ah so you calculated the jacobian to get the $\frac{\partial (y{j{1}},…,y{j{r}})}{\partial (x{i{1}},…,x{i{r}})}$ term

[warm shale]

Enoo58

Cogwheels of the mind

[compact shadow]

Yeah jacobian term

[drowsy cosmos]

and only of those where we have the corresponding $dx_n$ term right?

[warm shale]

Enoo58

[compact shadow]

Idk, just calculate by definition

[drowsy cosmos]

for example here the ferst term uses derivatives $x_1,x_2,x_4$ but they are all 1 so nothing happens

[warm shale]

Enoo58

[compact shadow]

Just the first and second the fourth rows and choose three columns

Calculate the determinant

Only none zero terms are we choose columns 1,2,4 or columns 1,2,3

[drowsy cosmos]

alright got it

thank you very much

do you know a source where i can read more about it ?

[obtuse pebble]

@drowsy cosmos Has your question been resolved?

Channel closed

Closed by @drowsy cosmos

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hybrid owl]

Hey can someone help me?

[hybrid owl]

solve in R the equation, where a is the real parameter

[compact shadow]

It’s a quadratic function of 6^|x|

[hybrid owl]

Yes

you can note 6 ^ (x) = t

[compact shadow]

Yeah then you just solve it for different cases, a<0,0<=a<1,a>=1

[hybrid owl]

And thats all?

[compact shadow]

Yeah

[hybrid owl]

Ok thank you

But how to solve equation for this cases?

the teacher didn't teach us that

I dont know how

[obtuse pebble]

@hybrid owl Has your question been resolved?

[compact shadow]

?

What ?

You don’t know how to solve quadratic function?

[hybrid owl]

I know that

Never mind

[obtuse pebble]

@hybrid owl Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[true willow]

yo yo

[true willow]

ok so this probably shouldnt be too long

i have x^3 - 1 = f(x)

or f(x) = x^3 - 1

and i need to find the complex zeros of it

i tried using synthetic division

and got (x-1)(x^2 - 1)

[fickle lantern]

well you can already say x=1 is a solution

[true willow]

ye

but i wanna find the complex zeros too

[fickle lantern]

mmm hmm

ok

[true willow]

and to do that i gotta get like

factor it down

however

whenever i turned it into (x-1)(x^2 -1)

[fickle lantern]

Whenever you are solving an equation for complex zeros, you can always set x=a+bi

[true willow]

i factored it back out and got x^3 - x^2 + 1

which isnt x^3 -1

[fickle lantern]

oh yeah I think u did that wrong

[true willow]

how so

[fickle lantern]

I think maybe its (x-1)(x^2+x+1)

try expanding that

[true willow]

x^3 + 2x -1 -x^2 -x -1

[fickle lantern]

I'm getting x^3-x^2+x^2-x+x-1

[true willow]

wh

?

wait no that makes sense

wait but then

where does the x-1 thing work at the end

it should be -2 right

and even then, factoring out again gives us not x^3 -1

[fickle lantern]

I think you're doing something wrong

[true willow]

ok then lets back up a bit

[fickle lantern]

cos x^2(x-1)=x^3-x^2

x(x-1)=x^2-x

1(x-1)=x-1

[true willow]

O.o wha

im sorry i dont think i follow

ok so

cos x^2(x-1)?

[fickle lantern]

thats how u expand things?

[true willow]

how did you get that from x^3 - 1

[fickle lantern]

im expanding (x-1)(x^2+x+1)

[true willow]

so then how did we get that?

[fickle lantern]

you could use polynomial division by x-1

[true willow]

this i mean

[true willow]

i used synthetic division

and i got (x-1)(x^2 -x -1)

[fickle lantern]

oh

i think ik what u did

[true willow]

._.

[fickle lantern]

actually

ugh

i forgot synthetic division

:/

[true willow]

ok so basically

[fickle lantern]

I just did it the old way

[true willow]

you take like all the numbers

and put it on a thing

and then like multiple and add and stuff

its weird but it works but i guess not this time lol

[fickle lantern]

no no i think u probs tried to subtract (x-1), but forgot that the minus carries to the -1

which means u subtracted 1 instead of adding 1

that was very badly said

[true willow]

yeah i think i died

my brain i mean

ok so like subtracting x-1?

what?

[fickle lantern]

yeah just redo it but make sure you don't make any silly sign mistakes

[true willow]

ok

if i do it correctly i should get x^3 -1 if i factor it back out right

cus i feel like (x-1)(x^2 -x -1) should become x^3 -1 right

due to synthetic division

ok it turned into x^3 - x^2 +2x +1

but that doesnt work since the intercept is supposed to be -1

[fickle lantern]

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

[true willow]

ok

[fickle lantern]

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

[true willow]

oh!!! so i did mess up on the whole synthetic division stuff

[fickle lantern]

yeah

anyway

so now complex numbers (yay)

so basically just solve x-1=0 and x^2+x+1=0 and ur done

[true willow]

yeah complext numbers shouldnt be too hard

obv 1 is a zero, then

wait actually

x^2 +x + 1

ok so i need to use the quadratic formula

[fickle lantern]

ye

[true willow]

x = -1 +- square root of 1 - 4

over 2

-1/2 + or - 3i?

not sure if thats right lol

wait

it stays in root

[fickle lantern]

$\frac{-1\pm\sqrt{3}i}{2}$

[true willow]

ye

and then the 3 is an i

[warm shale]

LTHMath

[fickle lantern]

fixed it

[true willow]

so then the + or - inverts

so -SR3i/2 or SR3i/2

[obtuse pebble]

@true willow Has your question been resolved?

Channel closed

Closed by @true willow

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[bleak dragon]

So I have a relatively practical problem and I would like to get some advice on how to approach this problem, as I lack the skill/experience to properly formulate it.

I want to form a Catmull-Rom like spline-curve based on 4 quaternions (expressing rotations from a fixed axis) using spherical interpolation, such that the resulting curve is closed and lies fully on the surface of a unit sphere. I can do this in a practical sense, but I dont know how I could find a formula for that curve, if that is even possible.

Now given an arbitrary but fixed point on the sphere I want to find the closest point to the given point on that curve.

Does anyone have an idea how I should approach this problem? I had no luck searching for answers anywhere so far...

[obtuse pebble]

@bleak dragon Has your question been resolved?

[obtuse pebble]

@bleak dragon Has your question been resolved?

[obtuse pebble]

@bleak dragon Has your question been resolved?

[obtuse pebble]

@bleak dragon Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hybrid creek]

I am trying to interpret dU/dS(inf, v, t)=1

[hybrid creek]

These are the conditions being specified for pricing an option in a market

I am unsure on how to interpret the infty here. U is the value of the option contract

And my interpretation is that this is saying for any value S, the change in the value of the options contract wrt S is 1. That is, they change at the same rate.

The other interpretation is that as the price of the underlying stock S approaches infinity, the value of the option itself changes closer and closer to the change in S itself

Now that I have typed it out

I understand haha

That is it

.solved

.close

[obtuse pebble]

Channel closed

Closed by @hybrid creek

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[full relic]

Prove that if a and b are different integers, then there exist infinitely many positive integers n such that a+n and b+n are relatively prime

[full relic]

I tried let d= gcd(a+n, b+n) . WTS d=1. Had difficulty in setting n in terms of a and b

[compact shadow]

We assume a<b.you just consider prime numbers p greater than b-a

For such p, let n=p-a

Then a+n=p, b+n=p+b-a

They are coprime

[obtuse pebble]

@full relic Has your question been resolved?

[obsidian isle]

If C divides A and B then it also must divide the difference A-B

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[static karma]

Can the sides and the perimeter of a triangle be an arithmetic progression

[pine sail]

I do not think so.

[haughty sable]

Amxela communityshi

Kartveli

[pine sail]

?

[haughty sable]

Nothing

[pine sail]

Bruh.

[haughty sable]

Why cant it be

[warm shale]

What the hell am I doing here?

[pine sail]

Someone will help you, do not interrupt other's channel.

[static karma]

But does it have to be the next term of AP?

[pine sail]

That doesn't matter, these are general terms.

Besides if we say, d>0

Then it must be the next term.

If d<0 then it should be the first term.

[static karma]

Let’s say sides are:4,6,8, P =18

[pine sail]

Sure.

[static karma]

In this case P is the eighth term of the progression

[pine sail]

Oh you mean 18 is some term in the AP somewhere later on.

Got it.

That is correct, I thought you meant they should be consecutive terms.

Of course it is possible like that.

You already have the example.

[static karma]

The answer in my book says that it is impossible

I guess they mean consecutive terms

[pine sail]

Right.

I mean you won't say

4,6,8,18 are in AP.

Would you?

[static karma]

They are in AP, aren’t they?

But they’re not consecutive

[pine sail]

They are not.

[static karma]

Why?

[pine sail]

4,6,8,18 being in AP in itself suggests that
6-4=8-6=18-8 which is not true.

[static karma]

Yeah, they aren’t in that order

[pine sail]

Yes, if they exclusively ask that they can be any term in some AP then and only then you'd give your example.

Otherwise, what I said remains.

[static karma]

Okay, thanks a lot

.close

[obtuse pebble]

Channel closed

Closed by @static karma

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tall badge]

.rotate

,rotate

[tall badge]

Hi, could someone let me know if I have completed 5a right?

[obtuse pebble]

@tall badge Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stoic anchor]

<@&286206848099549185>

[civic zealot]

• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping @Helpers.```

,rotate

[civic zealot]

what are you asking here?

[stoic anchor]

wtf does this question mean

i got exam tommorow, please stay on and help me fast

i got tons to revise

@civic zealot wtf is this question saying

maximum of what

[civic zealot]

If you take x from the domain of that expression, then it has a maximum output, what is that maximum output?

[stoic anchor]

what does that mean

well square root cnt be negative

so x has to be 0

oh yeah so negative 104

[civic zealot]

if x=0 then you have root(-3)

[stoic anchor]

@civic zealot i mean 3

3-3 = 9

0

[civic zealot]

right. since you're multiplying by -4, the larger that term in parenthese is, the more negative the expression

so the maximum value is when $(\sqrt{x-3}+26)$ is at a minimum. or, in other words, when $x=3$

[warm shale]

Zybikron

[stoic anchor]

@civic zealot how bout this

[civic zealot]

what is the inverse of y=mx+2?

[stoic anchor]

x = my + 2

x - 2 = my

y = x-2/m

y = x/m - 2/m

where m < 0

im confiued

@civic zealot elp

help

[civic zealot]

that's correct

so a = 1/m and b = -2/m

[obtuse pebble]

@stoic anchor Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stoic anchor]

@civic zealot what is the inequality about

[civic zealot]

if m<0, what do you know about a=1/m?

[stoic anchor]

oh

is it that simple

bruh

i done that first

stupid questions

@civic zealot i think like some sort of m sign change lol

like i thought after we do this, m sign changes opposite xd

[civic zealot]

nope, you know the sign of m, so you can find the sign of a and b

[stoic anchor]

15 last question

i don't realise how easy

these questions are

[civic zealot]

for 15, b is in the domain, so 5-b is in the range. So you need a bracket

[stoic anchor]

@civic zealot we substitue in right

[civic zealot]

yeah

[stoic anchor]

is square bracket because we are using b in the domain with square bracket right @civic zealot

[civic zealot]

yeah

[stoic anchor]

how do we know

[civic zealot]

square bracket means that the end point is included in the interval

[stoic anchor]

if we write 5a or 5b first

it got something to do with a<b righ

t

[civic zealot]

parenthesis means the endpoint is excluded from the interval

[civic zealot]

yes, if a<b, the 5-b<5-a

[stoic anchor]

how though

that what confuses me lol @civic zealot

[civic zealot]

which one takes more away from 5, b or a?

or, a<b means -b<-a and then add 5

[civic zealot]

plug in some numbers to convince yourself it's true

[stoic anchor]

hmm

@civic zealot SINCE a < b, we write 5-b first because it is a smaller number one?

[civic zealot]

b is the larger number, so when we subtract it we are removing a larger number

which results in a smaller number, yes

[stoic anchor]

@civic zealot would it be correct

othe way around

with same circle and square

?

[obtuse pebble]

@stoic anchor Has your question been resolved?

Channel closed

Closed by @stoic anchor

Use .reopen if this was a mistake.

[stoic anchor]

.reopen

[obtuse pebble]

✅

[fervent comet]

What do you mean?

Do you mean if [5-a,5-b) would be correct?

That wouldn’t be correct

a<b is what’s stated

If b is the larger number, when it’s subtracted from 5, it should become the smallest number

If a is the smallest number, when it’s subtracted from 5, it should become the largest number

Thus, [5-b,5-a)

@stoic anchor

Idk if I answered ur question

Think of it like this

Let’s substitute a=3 and b=4

3<4