help-27

[sand thorn]

the x- and y-coordinates are equal

that's what's special about this line

[torpid locust]

Every point on the line with the x coordinate 1-9 inches and y coordinate 1-9 satisfies

the line

I think

[sand thorn]

wdym by "x coordinate 1-9 inches"

what's "1-9"

[torpid locust]

1 through 9

which is the distance of the paper horizontally

Well 8.5

[sand thorn]

so you're saying every point on the line has x-coordinate between 1 and 9 inches, and y-coordinate between 1 and 9 inches

[torpid locust]

Actually no

For example

the point (0,0) satisfies it

[sand thorn]

yes lol

modify what you said

[torpid locust]

There are specific points that can satisfy it

[sand thorn]

yeah

ok if I modify it, I would say:

all the points on the line have x-coordiante between 0 and 9 inches, and y-coordinate between 0 and 9 inches

[torpid locust]

sure

[sand thorn]

This is a wonderful observation, but it applies to every other point on the paper lmao

our goal was to describe the line uniquely and fully

in other words, we want to find a condition which forces a given point to lie on the line

[torpid locust]

so there are specific points that satisfy the line

[sand thorn]

Yes

My statement provides such a condition

namely, the x coordinate must equal the y-coordinate

[sand thorn]

the only points which satisfy this requirement are the ones on the line, right?

[torpid locust]

Yes

[sand thorn]

so i've found a complete description for the line

it's

the set of all points whose x- and y- coordinates are equal, and whose x- and y- coordinates lie between 0 and 9 inches

cool

[torpid locust]

Right so how does this relate to the question I initially had

[sand thorn]

I'm trying to explain how the equation of a line works

the way you would write such a description algebraically is:

y = x, where 0 ≤ x ≤ 9 and 0 ≤ y ≤ 9

[torpid locust]

Oh right

[sand thorn]

are you seeing it now?

[torpid locust]

Yeah

[sand thorn]

can you write an equation for this line?

this is slightly different

[torpid locust]

Um

Do you want a coordinate?

[sand thorn]

I want an equation, a description using coordinates

can you spot a pattern in the points

[torpid locust]

y = m x + b ?

[sand thorn]

what's m? what's b?

now you're going back to what you memorized

here

[torpid locust]

Well m is what we said earlier

[sand thorn]

does this help at all? the x-coordinate is 2 and the y-coordinate is 4

[torpid locust]

Yeah

well now you have a coordinate

you’d need a 2nd one to find the m

[sand thorn]

the "m"?

again, you're going back to the stuff you learned without understanding

[sand thorn]

just look at this picture: it's very obvious that the y-coordinate is exactly double the x-coordinate

so your equation is y = 2x

does that make sense?

you don't need to go back to your procedures

[torpid locust]

yeah

[torpid locust]

hey man im just trying to get my question done 😭

[sand thorn]

hahahaa

I know, I know

[torpid locust]

I just need this bit done cause my exam is tomorrow

[sand thorn]

ah I get you

okay, so I guess I'll just cut to the chase then

[torpid locust]

that would be amazing

[sand thorn]

this "m" you're looking for is the slope of the line

it describes the steepness

[torpid locust]

right

[sand thorn]

for this line, the steepness was 1

[sand thorn]

for this one, it was 2

in general, given a line, the steepness is given by y2-y1 / x2-x1

[torpid locust]

yes

[sand thorn]

it kinda makes sense that it should be this way

the "rise" divided by the "run"

anyway, okay, so that's how you calculate m

what if the line is off-centered? that is, it doesn't go through the origin?

[torpid locust]

What do you mean?

[sand thorn]

if I drew the line connecting these dots, it wouldn't go through (0,0)

so the equation isn't as simple as y = mx, where m is the steepness

[torpid locust]

right

[sand thorn]

how do we account for this?

it's pretty easy

the observation is simple: consider the same line, moved downward until it does cross through (0,0)

let me show you

this blue line has the same steepness as the black line

so their m's are the same

what's the m-value, by the way? @torpid locust

[torpid locust]

2/3

[sand thorn]

nice yeah

[torpid locust]

Top of my head

[sand thorn]

lol

what's the equation of the blue line?

(hint: y = mx)

[torpid locust]

2/3

[sand thorn]

that's a number, not an equation

what's the equation

[torpid locust]

Oh whoops

y = 2/3 x

[sand thorn]

yeah

ok nice

[sand thorn]

now look at this again

we seek the equation for the black line

the observation is just: every point on the black line is just a point on the blue line, except the y-coordinate goes up a little bit

[torpid locust]

Right

So the black line would be the same equation

[sand thorn]

not quite...

it's the blue equation, y = (2/3)x, except now we add the purple amount to each y-coordinate

do you see the purple line there

[torpid locust]

Yes

[sand thorn]

it turns out the black equation is y = (2/3)x + 1.3333...

how did I get this 1.333... thing?

obviously I'm cheating a little bit by using a graphing calculator

[torpid locust]

Uhhh

[sand thorn]

?

[torpid locust]

You take a y coordinate and isolate it with the 2/3 x?

[sand thorn]

isolate it? wdym

[torpid locust]

Divide 2/3 from both sides

[sand thorn]

that's a good idea, but it doesn't really work

this is actually a rather difficult idea

but what I did is I switched out the variables, basically

I know the point (1,2) is on the line

so I switched out the variable "x" with "x-1" and the variable "y" with "y-2"

this way, the new equation reads:

(y - 2) = (2/3) (x - 1)

[torpid locust]

Oh right

[sand thorn]

this is called the point-slope form

anyway you don't have to fully understand why this works

but it helps to remember the formula

if the line crosses through (x0, y0), and the slope is M, then the equation reads:

(y - y0) = M(x - x0)

[sand thorn]

for our case, this is what we got

okay let's go to your question

should be easy now

[torpid locust]

I’ll try!

So the m is 1?

@sand thorn

Uh

<@&286206848099549185> been waiting for a bit for my question to answered now

[pseudo thorn]

do you need help with all 3?

[torpid locust]

Yes

[pseudo thorn]

ok so for the first one, we need to use the formula for slope of a line

so we need to find the y-values because they arent given

[torpid locust]

right

[pseudo thorn]

but all we need to do is plug in the given x-values into the given equation

so f(1) = 1

[torpid locust]

yeah

[pseudo thorn]

and f(2) = 8

so our two points are (1, 1) and (2, 8)

[torpid locust]

wait

where does the 8 come from?

[pseudo thorn]

when you plug in 2 into the original equation

x^3, (2)^3 = 8

make sense?

[torpid locust]

This is for a) right?

[pseudo thorn]

yes

[torpid locust]

Okay yeah

[pseudo thorn]

okay so we have our two points (1,1) and (2,8)

were gonna let the first set of points equal (x sub 1, y sub 1) and the second set of points equal (x sub 2, y sub 2)

for this equation

now all we have to do is plug in the values

[torpid locust]

Alright

So it’s 7

[pseudo thorn]

ok so now we have the slope of the equation of the secant line

but we need a y-intercept

how do you think we can find that

[torpid locust]

what do you mean y intercept?

[pseudo thorn]

so in any equation there is a y-intercept

[torpid locust]

ohhh

right right

[pseudo thorn]

sometimes it is 0, for example with x^3

remember y = mx + b

[torpid locust]

So you can pick either 1 or 8.

?

[pseudo thorn]

yes you can use either point to determine the y-intercept

so plug in known values into y = mx + b and solve for b

either point works for x and y but it has to be from the same point

[torpid locust]

so let’s say

1 = 7x1 + b

[pseudo thorn]

yes

[torpid locust]

so b would be -8

[pseudo thorn]

-6

[torpid locust]

-6

right

[pseudo thorn]

yes

and now we have all of our needed values to write the equation of the line

[torpid locust]

okayyyyyy

makes sense

[pseudo thorn]

so we have y = 7x - 6

if what you found for the slope originally was right i never checked

[torpid locust]

For the equation of the secant line

[pseudo thorn]

yes thats it

for a

question a

[torpid locust]

yeah

[pseudo thorn]

ok next question

[torpid locust]

This was what was confusing me

[pseudo thorn]

ok are you in calculus?

[torpid locust]

Yes

[pseudo thorn]

okay so you have learned to find a derivative right?

[torpid locust]

yeah

[pseudo thorn]

so by definition, the derivative is the slope at one single point

[torpid locust]

yes

[pseudo thorn]

which will kind of look like this]

shitty diagram but im sure you get the picture

[torpid locust]

yupp

[pseudo thorn]

so the line intersecting that point is known as the tangent line

so when they ask determine the tangent line im gonna assume they want an equation

[torpid locust]

yeah

[pseudo thorn]

because the question is vague they shouldve either said find slope of the tangent line or the equation

but im gonna assume equation

[torpid locust]

the equation

[pseudo thorn]

okay

[torpid locust]

Which is where you use the big limit formula

[pseudo thorn]

correct, but theres shortcuts

[torpid locust]

f(x + h) - f(x) / h

[pseudo thorn]

have you learned the power rule?

[torpid locust]

I think so but my teacher is evaluating me if I can use this correctly

[pseudo thorn]

ah i see

so for the equation of the tangent were going to use point-slope formula

which is similar its just rearranged

but again we need a point and a slope

[torpid locust]

What would the formula be ?

[pseudo thorn]

can you see that

[torpid locust]

Ehhh

[torpid locust]

I’d have to change my discord appearance

okay yeah

[pseudo thorn]

fortunately this is easier because we only need one point and the slope

[torpid locust]

But I don’t think I can use it in this exam

Well I don’t think we’ve learned it either

[pseudo thorn]

well you can use the other one as well its just a different format

as in y = mx+b

[torpid locust]

Yeah

[pseudo thorn]

but youll have to solve for b once again

the other one you doint

dont

but thats okay its not hard

[torpid locust]

Yeah

[pseudo thorn]

so we need the slope

to do this we are going to plug in the x-value into the equation of the derivative of f(x)

so this is where we will use the limit definition to find the equation of the derivative

what do you get when you do that

[torpid locust]

That would give us the m?

[pseudo thorn]

after we plug the x-value into the result

[torpid locust]

okay so the formula is f(x + h) - f(x) / h

[pseudo thorn]

yes

[torpid locust]

when the lim of x is close to 0

and we know that f(x) = x^3

and x=1

[pseudo thorn]

yes

[torpid locust]

So it would be this

[pseudo thorn]

yes

[torpid locust]

Well I forgot to add the f’s

[pseudo thorn]

not necessary

well i shouldnt say that

because the value of f(1) is 1 but thats there

[torpid locust]

right but knowing my teacher..

[pseudo thorn]

yes true

[torpid locust]

now what do I do next cause I was stuck here

[pseudo thorn]

but then again when you plug in, what you wrote is what you get and then you solve from there

okay

so we have to multiply out 1+h cubed

as in (1+h)(1+h)(1+h)

[torpid locust]

yes

soo:

(1+h)(1+h)(1+h) - 1 / h

[pseudo thorn]

should be correct

[torpid locust]

Okay I wrote it down

[pseudo thorn]

okay are you able to show me what you have?

[pseudo thorn]

okay now we have to continue to simplify

so you remember how to multiply (1+h)(1+h)?

you might know it as FOIL

[torpid locust]

(1+h)^2 ?

[pseudo thorn]

yes but were multiplying it out completely

[torpid locust]

Oh right

One sec so I can do it

[pseudo thorn]

okay

[pseudo thorn]

looks good

now you have to multiple (h^2 + 2h + 1)(1+h)

a little tedious but its necessary when using limit definition

[torpid locust]

I did this quickly so it might be wrong

[pseudo thorn]

lets assume its right if we end up getting the wrong answer we can go back

i know what right answer is so itll be okay

so we can take off the parenthesis now

and when this happens we see we have 1 - 1

which cancels

makes sense?

[torpid locust]

Yup

So we’d be left with the all the h’s

[pseudo thorn]

correct

and we have to somehow cancel the h so that we dont have 0 on the bottom

now we can factor out all of the h's from the top

[torpid locust]

So we’d be left with h^2 + 3h + 3

[pseudo thorn]

yes, is that after you canceled the h on top and on bottom?

[torpid locust]

Yup

You could check after me

[pseudo thorn]

okay great

now we can plug in 0 for h

what do you get

[torpid locust]

0

[pseudo thorn]

well remember you have + 3

(0)^2 + 3(0) + 3

0 + 0 + 3

[torpid locust]

Oh shit

You’re right

I didn’t write it down

[pseudo thorn]

all good but you now you have your slope

which is 3

so now we can plug in to find b

1 = 3(1) + b

1 = 3 + b

[torpid locust]

(1,1) or (2,8)

[pseudo thorn]

we can only use the point we used to find the slope

[torpid locust]

Oh true

[pseudo thorn]

remember we found the slope at (1,1) which has no relation to (2,8)

[torpid locust]

Okay yeah

Or else it would mess shit up

[pseudo thorn]

yes

[torpid locust]

OKAYYYY

Makes so much sense

[pseudo thorn]

good

whats ur equation

[torpid locust]

y = 3x - 2

[pseudo thorn]

should be correct

[torpid locust]

Yeah it’s the right answer

[pseudo thorn]

awesome

now the last one is also easy

[torpid locust]

so it’s the same but it’s just 1/m

[pseudo thorn]

correct

but then you have to solve for b again

[torpid locust]

so:

1 = 1/3x1 + b

[pseudo thorn]

yep

[torpid locust]

Oh it’s a negative number

-1/m

[pseudo thorn]

oh yes sorry i shouldve said

perpendicular slope is always the negative reciprocal

[torpid locust]

all good

Okay so it’s:

1 = -1/3x1 + b

so b would be 4/3

[pseudo thorn]

yes thats correct this time

yes

so your equation should be y = -1/3x + 4/3

[torpid locust]

yes

aaand it’s the right answer

thank you so much man

you’re amazing 🙏🙏

[pseudo thorn]

no problem

[torpid locust]

alright that’s what I needed

thank you again

[pseudo thorn]

of course its what im here for

[torpid locust]

.close

[devout snow]

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[devout snow]

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[woven edge]

Context:
If given a list of N points, and another point we will call A, How do you calculate the point furthest from A in the list of N points?

So I was thinking about it, and it seems as simple as calculating the distance each point is from A and seeing the largest one. But what got me curious is that I may be able to skip a step, but wanted to be sure it was safe to skip said step. When calculating the distance between two points, you can simply do Pythagorean theorem on the delta between each point. That would be this:((x1 - x2)^2 + (y1 - y2)^2)^(1/2). Couldn't I skip the last step, square rooting it if I only want a comparison for who ever is the furthest?

I don't need to store the actual length between each point, just the "squared" length.

[woven edge]

Lol, do i need to do a proof on this?

[strange arch]

Yes that's correct

in case this is a CS question

you can partially even skip the squaring

[woven edge]

wdym?

[strange arch]

But regarding why the root isn't required, it's because the function f(x) = sqrt(x) is strictly rising for positive x

which means if one point has a larger root distance then another

then it also has a large distance without using root than the other

the order of distance is maintained

[woven edge]

OMG. that's an excellent way of putting it. yes. why didn't i think of it like that. tysm

[strange arch]

Since you can first do a Manhattan distance check

which yields boundaries

but that's just for reducing the amortized time complexity of the algorithm

[woven edge]

Alrighty. I think i'm good here. thanks lunatic

[strange arch]

np

[woven edge]

.close

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[devout snow]

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[fast helm]

please help

[fast helm]

i tried doing proj(ab)
a=3,1
b=4,7

no work

[devout snow]

@fast helm Has your question been resolved?

[fast helm]

@dry robin please help guys

[devout snow]

Channel closed

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[fast helm]

.reopen

[devout snow]

✅

@fast helm Has your question been resolved?

[devout snow]

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[urban matrix]

I dont understand how to name the quadrant in which the terminal side of each angles lies
98 deg, 350 deg, 450 deg, 735 deg, -240deg, -315deg

[urban matrix]

I forgot what the terminal side is can u explain it real quck

[urban matrix]

like terminal vs inital

thanks btw

[soft umbra]

The initial side starts at the positive direction of x axis

[urban matrix]

oh

and terminal is awp?

[soft umbra]

The terminal side is determined by how much degree you rotate

[urban matrix]

ic

[soft umbra]

Any questions?

[urban matrix]

so if I have a large angle like 800 deg how would I determine what quad it would come in?

[soft umbra]

800 deg represents you rotate 360 and 360

After that

You rotate 80 deg

So the terminal side will be on I quadrant

[urban matrix]

so it would come in quad 1

[soft umbra]

ye

[urban matrix]

ic

[soft umbra]

If you get a degree greater than 360deg

You can simply deduct it with 360n n is an integer

The result remains the same

[urban matrix]

for my questions I got quadrants 2, 4, 1, 3, 3, 4

is that right?

[soft umbra]

I dont understand how to name the quadrant in which the terminal side of each angles lies
98 deg, 350 deg, 450 deg, 735 deg, -240deg, -315deg

Lemme check

[urban matrix]

alr

[soft umbra]

False

[urban matrix]

shit

[soft umbra]

Try again

[urban matrix]

🫡

2, 4, 1, 4, 3, 4

[soft umbra]

2 4 y axis 1 2 1

[urban matrix]

if this is wrong than third time is the charm

?

[soft umbra]

Incorrect

[urban matrix]

idk man

[soft umbra]

Could you tell me how did you figure them out?

[urban matrix]

aight

[soft umbra]

The third to the last

[urban matrix]

for an example the angle 735 I did 735/360 which is equal to 2.041 menaing that it would take to revolurtions and 10 degrees(735-360*2 = 10) hence itsa quadrant 4

i think

[soft umbra]

no

[urban matrix]

LOL

[soft umbra]

..

[urban matrix]

example

[soft umbra]

deduct

E.g. 1280 deg

[urban matrix]

okay...

[soft umbra]

360*3=1080

[urban matrix]

yess

[soft umbra]

1280-1080=200

deg

[urban matrix]

yes

ohhh

[soft umbra]

III quadrant

[urban matrix]

so 200

[soft umbra]

yes, try again

[urban matrix]

so diff would be the quadrant

[soft umbra]

yes…

[urban matrix]

https://tenor.com/view/you-didnt-have-to-cut-me-off-meme-you-didnt-have-to-cut-me-off-big-brain-big-brain-meme-gif-26965490

which ones are wrong?

[soft umbra]

The third one to the last one

All of them

[soft umbra]

Here is the correct answer

[urban matrix]

2, 1, 1, 1, 3, 1 my answers

y axis?

btw what does coterminal mean

I did not pay attention in class so sry

wait

how is number 2 quad 4

360-350=10

and 10 is in quad 1

where is bro gone

<@&286206848099549185>

Thanks for the help ig man real good for u to leave

.close

[devout snow]

Channel closed

Closed by @urban matrix

Use .reopen if this was a mistake.

[soft umbra]

yeah

The angle is 90°

[urban matrix]

damn hes back

.repoen

[soft umbra]

Open

Typo

[urban matrix]

.reopen

[devout snow]

✅

[soft umbra]

The second one?

Lemme check

[urban matrix]

yes

[soft umbra]

I dont understand how to name the quadrant in which the terminal side of each angles lies
98 deg, 350 deg, 450 deg, 735 deg, -240deg, -315deg

[urban matrix]

oh ic

I did it the other way around

lol

and what about number 3, 4, 5, 6?

lol

[soft umbra]

The y axis one?

[urban matrix]

ye

whats that

[soft umbra]

The terminal side lies on y axis

Have you calculated its degree?

[urban matrix]

yes

[soft umbra]

What is it?

[urban matrix]

,calc(360-450)

[woven radish]

Result:

-90

[urban matrix]

wait other way

[soft umbra]

90

[urban matrix]

90

yes

[soft umbra]

yeah

[urban matrix]

so its 1

[soft umbra]

Where is 90° at?

[urban matrix]

quad 1

[soft umbra]

Are you sure?

[urban matrix]

100 percent

[soft umbra]

Could you draw it?

[urban matrix]

sure

see

[soft umbra]

Draw me the terminal side

[urban matrix]

bruh

[soft umbra]

Like this

[urban matrix]

man its where I drew the 90 angle

just show it pls

[soft umbra]

that’s right

[urban matrix]

yes

so its 1

[soft umbra]

Its terminal side is stick to the y axis

[urban matrix]

ohh

[soft umbra]

That means it doesn’t belong to any quadrant

[urban matrix]

so its denoted as y axis?

[soft umbra]

yes

[urban matrix]

alright are the last three good?

[soft umbra]

negative means the direction they rotate is different

[urban matrix]

Yes

they are

[urban matrix]

yes

[soft umbra]

yeah, so you’ve gotten it

[urban matrix]

Thanks for the help sry for being a dick earlier

.close

[devout snow]

Channel closed

Closed by @urban matrix

Use .reopen if this was a mistake.

[soft umbra]

np

[urban matrix]

yep

clockwise is negative and vice versa for CCW

[soft umbra]

Have a good one

[urban matrix]

u too man

[devout snow]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[river tapir]

yo

[river tapir]

<@&286206848099549185>

[twilit comet]

!15m

[devout snow]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[twilit comet]

please state your problem

and do NOT ping helpers before 15 minutes have passed

[river tapir]

that was by mistake i was trying to ping my friend?

you can come back later lmao

[twilit comet]

well, you can ping them individually or in a separate channel

you're just occupying a channel for no reason

@river tapir please close the channel, if you don't have a question

[weak cove]

.close

[devout snow]

Channel closed

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[twilit comet]

thanks lmao

[devout snow]

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[tawdry venture]

can someone explain this to me

[tawdry venture]

is there like a rule or something to it

[stone stump]

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

if you expand it

you can view this as an instance of difference of cubes. a^3-b^3=(a-b)(a^2+ab+b^2)

[gray flax]

Real

Try distributing it @tawdry venture

As in multiply the 2 factors

[devout snow]

@tawdry venture Has your question been resolved?

[tawdry venture]

isnt it a^2 + 2ab + b^2

[eager nova]

No

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

[woven radish]

Samuel

[tawdry venture]

oh and shouldnt it be negative x

i mean -1

[eager nova]

$x^3 - 1^3 = (x - 1)(x^2 + x\cdot1 + 1^2)$

[woven radish]

Samuel

[tawdry venture]

or are the a and b variables not considering whether its a + or - in front of it

cuz i remember for difference in squares it was considering the - or + in front of it

[eager nova]

You have a problem of focus

Let me fix it for you

$(x)^3 (-) (1)^3 = (x - 1)(x^2 + x\cdot1 + 1^2)$

[woven radish]

Samuel

[eager nova]

Can you see now?

[tawdry venture]

ohh yeah

is that just a rule i have to remember

[eager nova]

It doesnt matter if its negative or positive, look

$x^3 + (-1)^3 = (x + (-1))(x^2 -(-x\cdot1) + 1^2)$

[woven radish]

Samuel

[eager nova]

If yoi solve the inside inside parenthesis u get the same

This would be the sum of cubes

Using negative numbers to prove it doesnt matter

[devout snow]

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[tawdry venture]

ohh okay

i get it now

[devout snow]

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[chrome elk]

Hello, I've got two questions. Does the rational zeros theorem always guarantee a factor? The second question is when am I allowed to say plus or minus a square root? I've seen some people say by definition square roots have to have a positive answer, in other words you have to take the absolute value of the answer of the root. I've also seen some people just say plus or minus a square root without any care. Follow up question, does it change whether it's root rather then specifically a square root?

[stone stump]

the rational roots theorem does not guarantee that the polynomial has a rational root. it only that if it has a rational root, then it can only be one of a few options

when you solve something like x^2=5, then x can be both x=sqrt5 and x=-sqrt5. but sqrt5 itself only means the positive one

[chrome elk]

But if it did have a rational root, it HAS to be one of the options or it CAN be one of the options

[stone stump]

if it has a rational root then it HAS to be one of the options

[chrome elk]

So if the question had a square root itself, then it has to be positive? But if I take the sqrt it can be both positive and negative.

[eager nova]

The square root of a number has only one solution, because the square root of a number is a number

And a number cannot be equal to two different numbers

x^2 is not a number

Its a function

So when u equal it to a number

U might have two different numbers thay match with your solution

[chrome elk]

I've seen some people try to explain it to me like this. Sqrt has to definitions, if you take it by the normal definition then it has to be positive, but if you take it by the second definition, it can be both negative and positive. Is that correct?

[eager nova]

What is sqrt(4)?

[chrome elk]

Idk, that's why I'm asking. It's confusing, it can be 2, or it can be both 2 and -2

[eager nova]

You are saying sqrt(4) is 2 and sqrt(4) is -2

In other words

2=-2

Which is false

[chrome elk]

I still don't get it. Which is the real answer. 2 or -2?

[eager nova]

2

[chrome elk]

Why not negative 2?

[eager nova]

Sqrt(a) = |a^(1/2)|

So its always positive

[chrome elk]

I see, but then why do some people say plus or minus sqrt if it's always positive?

[worthy tundra]

okay... it's both. What's (-2)^2?

[chrome elk]

I'm more confused now...

[eager nova]

No its not both

Stop confusing him if you dont know this

[worthy tundra]

very well

[eager nova]

You are confusing the equation x^2=4

Where when u take the sqrt both sides u get

Sqrt(x^2) = sqrt(4)

Now lets solve this

Whats is sqrt(x^2)?

[chrome elk]

x...

[eager nova]

No

It is |x|

Now you get an equation with absolute value

|x| = 2

[smoky wraith]

Hi

[eager nova]

Do you know how to solve this?

[smoky wraith]

Someone French here ?

[eager nova]

This channel is occupied

[smoky wraith]

x=2 ?

[eager nova]

No

[chrome elk]

It can be -2 too

[worthy tundra]

Right... this would mean x = [+-]2

[eager nova]

K so

|x| = +-sqrt(4)

As u can see

[chrome elk]

So it's both plus and minus then

[eager nova]

The +- is outside

The sqrt

[worthy tundra]

Ah. Gotcha

[eager nova]

So when u have a quadratric equation u get two values but its not becuase of the sqrt

[chrome elk]

I still don't get it. You said it has to be positive and now your saying it's both positive and negative?

[eager nova]

Its because of sqrt a function raised to 2

No

I said it is always positive and i never changed that

What i said its both its the equation u had left

With an absolute value

But the sqrt is always positive

[worthy tundra]

so I'm assuming the answer is: it depends on the context. If equation, both. If expression, positive

[eager nova]

Here

What is sqrt(4)

Where did u put the + and the -

[chrome elk]

Idk, can you try explaining this. It may help me

[eager nova]

Can you tell me what are the rwo definitions they gave u?

[chrome elk]

Wait a sec

Principle root and just left out the other ones name

[eager nova]

The negative square root

Lets solve this equation

y^2=x

Give me all steps

[chrome elk]

Wait but what happened to this

[eager nova]

You will see with this equation

[chrome elk]

Oh I think your referring to the positive and negative value of R when you graph it

Since it makes two lines

[eager nova]

Maybe its a problem of focus

Let me fix this for u one second

$\sqrt{x}$

[woven radish]

Samuel

[eager nova]

$-\sqrt{x}$

[woven radish]

Samuel

[eager nova]

Now let me add something

$-((((\sqrt{x}))))$

[woven radish]

Samuel

[eager nova]

The value of sqrt is always positive

The negative comes from outside

Because of a quadratic equation

But the sqrt itself if u solve it its always positive

[chrome elk]

Alr, is this the first definition of sqrt then?

[eager nova]

This is the only definition

[faint zinc]

Well... There is one annoying exception where you run into conflicting usages.

Odd roots of negative numbers

[eager nova]

Right

But i think his problem is about even roots

[chrome elk]

Let's just understand normal sqrts

Then move to the other ones

So a sqrt is always positive, cool. We can get a negative answer by placing a minus OR plus or minus before it. For example on the quadratic formula.

[eager nova]

Correct

[chrome elk]

Now when am I allowed to say plus or minus sqrt when solving things

[eager nova]

When you are facing a quadratic equation because sqrt(x^2)=|x|

And |x| will have negative and positive solutions

[chrome elk]

Idk remember any examples but I've seen people solve things where they take the sqrt of both sides and then put a plus or minus next to the sqrt. What's that all about?

[eager nova]

They are skipping steps

Its like i say this

x-1=2
x=3

Instead of

x-1+1=2+1

X=3

In your case

They do this

Sqrt(x^2)=sqrt4

x=+-2

Instead of

|x|=2

As you can see the step they skipped

I said sqrt(4)=2

And sqr(x^2)=|x|

[chrome elk]

Nono, it's more like this. .....=plus or minus sqrt(x)

[eager nova]

If they say

x=+-sqrt(4)

Thats also good, they just skipped the step

Of |x|=sqrt(4)

But sqrt(4) will be always 2

Whats +-2 is the |x|

[chrome elk]

So when solving, when can I say plus or minus sqrt?

[eager nova]

When u are facing an absolute value

Never skip this step and u will get used to it

Remember always that sqrt(x^2) =|x|

[chrome elk]

What if the equation doesn't have an absolute value? For example, z^2=x, plus or minus sqrt(x)= z

[eager nova]

And dont confuse it with (sqrt(x))^2

This one is actually x

[eager nova]

No

You made a mistake there

Because u sqrt both sides

And sqrt(z^2) =|z|

$|anything| = \sqrt{(anything)^2}$

[woven radish]

Samuel

[eager nova]

You can use this as a definition of absolute value

[chrome elk]

I am on a another level of confusion. It's not important, I'll close this and come back at another time. One last question. I asked my teacher about the fundamental theorem of algebra and he said each equation to the nth degree has n solutions, BUT those solutions can be repeated. For example (x+2)(x+2)(x-3). By definition it has 3 solutions but the x=-2 repeated twice.

Is that true or not?

[eager nova]

Yes, the solutions can repeat

[chrome elk]

In another words, it doesn't always have n unique solutions

Is that true?

[faint zinc]

It can have only 1 solution repeated n times.

x^n = 0 is a good example

[eager nova]

Correct, for example x^100=0

[eager nova]

[chrome elk]

Does this apply to non real solutions too? For example, (x+2)(x+2)(x-3) doesn't necessarily have to have an unreal solution to fit the criteria of having 3 solutions. Is that correct?

[faint zinc]

Yes, it applies to complex solutions. Without those you could have situations like x^2 + 1 = 0, which have 0 solutions under the reals.

But 2 if you include complex numbers.

[chrome elk]

But is it correct to say (x+2)(x+2)(x-3) doesn't necessarily have to have a complex solution to meet the 3 solutions criteria?

[faint zinc]

Going farther, it definitely does not, because if it did it would have more than 3 solutions

[eager nova]

Well u can clearly see in that example its already fsctored in 3 products =0

So either multiplier 1 = 0

Or multiplicsnd 2 =0

Or multiplicsnd 3=0

Three solutions

[chrome elk]

Idk it feels like cheating to count repeated solutions as different solutions

Is there no catch?

[eager nova]

No

[faint zinc]

Repeated solutions behave differently from simple zeroes.

[chrome elk]

What are simple zeroes?

[faint zinc]

Non-repeated zeroes

[chrome elk]

How do they "behave" differently?

[faint zinc]

You'll learn more about it when you get to complex analysis.

[chrome elk]

But I want to know now

[faint zinc]

Sure

[chrome elk]

Thanks lol :)

[faint zinc]

When you walk around a simple zero in the complex plane your arg of your output will make one full revolution.

When you walk around a double zero, the arg will make two revolutions

And so on

[chrome elk]

You mean the imaginary y axes and real x axes, right?

[faint zinc]

Yeah

[chrome elk]

How would a simple zeroe(assuming its real) form a line? Isn't it just a plot on the x axes?

[faint zinc]

What do you mean by form a line?

[eager nova]

This is why you should wait a bit to understand this

[chrome elk]

Your wording made it sound like a line was being formed

[eager nova]

You have to learn other things before

[chrome elk]

I just want tem to clarify there wording. Do they mean solutions in the form of w=x+iy?

[faint zinc]

Not really a line, but a path. Imagine taking a walk around the complex plane, and you end where you start, and while you're doing that you have a compass that points in the "direction" of the complex number (the arg). If, in your wandering, which doesn't cross itself, your compass makes no revolution then you didn't encircle a zero. If it winds once you encircled one zero, twice two, and so on.

The zeros are just somewhere in the interior of the path you walked.

And this is a set of complex numbers, yeah.

z = x + iy

But this is pretty conceptually dense, so don't worry about not understanding it immediately.

[chrome elk]

Idk, I'll have to do more research. Thanks everyone! :)

.close

[devout snow]

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[devout snow]

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[wary ruin]

Hello,
I've been delving into a problem related to rational functions and their vertical asymptotes. Specifically, I'm examining the function (f(x) = \frac{x + 10}{(x - 2a)(x + a - 1)}) and trying to determine the values of the constant (a) that result in the graph of the function having exactly one vertical asymptote.
Through initial exploration, we derive the conditions for the vertical asymptotes from the denominator as follows:
[
\begin{cases}
x-2a=0\
x+a-1=0
\end{cases} \Rightarrow
\begin{cases}
\frac{x}{2} =a\
x-1=-a\Leftrightarrow 1-x=a
\end{cases}
]
Simplifying further:
[
\begin{cases}
\frac{x}{2} =a\
x-1=-a\Leftrightarrow 1-x=a
\end{cases} \rightarrow 1-x=\frac{x}{2} \Rightarrow 2-2x=x\Rightarrow 2=3x\Rightarrow \frac{2}{3} =x
]
By setting (a = a) and replacing (a)'s with the expressions derived from the equations, we initially find:
In the first equation, with (x = \frac{2}{3}), we get (a = \frac{2}{6}) as one potential value.
However, this only provides us with one value of (a), and I'm on the hunt for three distinct values that satisfy the condition of the function having exactly one vertical asymptote. How do we deduce the other two values of (a) given the constraints of the problem?

[woven radish]

dgh

[river gorge]

how do you know that there are 3 values of a for such a scenario? I tried this using my own method and I also got only a = 1/3

[wary ruin]

I don't know how to find the other 2.

[river gorge]

no but how do you know that you are supposed to have 3

what if there's only 1 solution

[wary ruin]

The question says so

Find the 3

[river gorge]

oh

[wary ruin]

"tre olika" = three different

[solar goblet]

the denominator should also share 1 root with the numerator

[river gorge]

ah right

[wary ruin]

why's that

[solar goblet]

vertical asymptote happens at, x=a, for example because at a, the value approaches inf/-inf

thats when numerator is not equal to 0, whereas the denominator = 0

[wary ruin]

what about a implies we geet a inf/-inf situuation?

[solar goblet]

but if both =0, its limit can be evaluable

[solar goblet]

as per l'hopital

[wary ruin]

i've not learnt that

[solar goblet]

this only works reliably when both numer and denom are polynomials though

[wary ruin]

If x-2a = x+a-1, you have an asymptote of multiplicity 2

If x-2a = x+10, you have a hole instead of an asymptote, and x+a-1 will be the only asymptote

If x+a-1 = x+10, you have a hole instead of an asymptote, and x-2a will be the only asymptote

This was a soln. from my friend.

,w solve x-2a=x+a-1, x-2a=x+10, x+a-1=x+10 for a

[woven radish]

Results provided by WolframAlpha

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

Requested by dghf

[wary ruin]

,w solve (x + 10) / ((x - 2a)(x + a - 1)) for a

[woven radish]

Results provided by WolframAlpha

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

Requested by dghf

[devout snow]

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[fair storm]

You solve problems like these with the Lambert W function\
\
$a^x = bx + c$\
\
What about problems like these?\
\
$x^x = ax + b$\
\
What about these too?\
\
$x^x = a^x + bx + c$

[woven radish]

Roman_Garland

[fair storm]

<@&286206848099549185>

[devout snow]

@fair storm Has your question been resolved?

[devout snow]

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[random river]

can someone help me find the two points that 4x^2-5xy-3y^2=19
and y=x-5 intersect?
im not sure how to simplify 4x^2-5xy-3y^2

[green kelp]

solve both the equations

you can substitute y as (x-5) in the curve's equation and solve the quadratic in x

looks like a hyperbola eq

[devout snow]

@random river Has your question been resolved?

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[lost gazelle]

q57, i got -2sinycosy but ans was -2cos^3(y)siny

[twilit comet]

,rccw

[lost gazelle]

,rccw

[twilit comet]

LMAO

TIMING

[lost gazelle]

😄

[twilit comet]

arctan (x-2) = y?

[lost gazelle]

i differentiate both sides with respect to y

so i gey dx/dy

then i take the reciprocal and its dy/dx

[twilit comet]

mhm

what'd you get?

[lost gazelle]

crab

crap

[twilit comet]

nah

[lost gazelle]

i forgot to put dy/dx

[twilit comet]

ur method works

[lost gazelle]

i forgot chain rule

so i need to multiply by dy/dx when differentiating cos^2y

[twilit comet]

what??

[lost gazelle]

okay i know why now, thanks anyways

[twilit comet]

ok i ese

yea lmao

[lost gazelle]

okay cya

,close

[twilit comet]

*see

[lost gazelle]

.close

[devout snow]

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[devout snow]

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• Show your work, and if possible, explain where you are stuck.
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[hard knot]

This

[hard knot]

Into this

2sinAcosA = sin2A,
can i make something out of the 2sin^2cos^2 = ? sin^2 2A ?, i guess not

Ik the 2nd image is 1 - (sin^2 A/2)

[twin thunder]

you cannot put the square like that in the formula

[hard knot]

makes sense, can you turn it into that tho? if so how

[twin thunder]

trying

[twin thunder]

rather, it would be sin(2A) sinA cosA

[dusk briar]

just multiply 2 and divide by 2, then (2sinacosa)^2 becomes sin^2(a)

[twin thunder]

i got sin²(2A) inside,

instead of sin²A

[hard knot]

i dont think so? u sure

Oh shit

It is 2A

Yeah my bad on that

[hard knot]

no thats correct sry

my bad

[hard knot]

i cant follow sry

[hard knot]

how

[sand dove]

wut

[hard knot]

yeah this is it

[sand dove]

$2sin^2Acos^2A = \frac 12(2sinAcosA)^2 = \frac 12sin^2(2A)$

[woven radish]

rafilou2003

[hard knot]

oh

you guys are amazing

truly

thank you

its like ur 2nd language

.close

[devout snow]

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[plucky dome]

why is num 2 discontinuous?

[restive river]

right, so the interval is (0,3), the bottom part of 2, 2x-4, at what number for x would this equal 0

[plucky dome]

4 for 2x and 2 for x

[restive river]

think of it like setting equal the bottom part to zero, 2x-4=0, and solving that for x

And since you cant divide by zero, this function will not exist at that point

[plucky dome]

2x -4 ≠ 0
2x ≠4
x ≠ 2

what

[restive river]

exactly

x = 2

which falls in between (0,3)

[plucky dome]

ah

icic thx

.close

[devout snow]

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[vague vortex]

Can someone please help me

[devout snow]

@vague vortex Has your question been resolved?

[vague vortex]

<@&286206848099549185>

[lost laurel]

show the complete question

[vague vortex]

That’s it

That’s the question

[lost laurel]

the top right isn't there

the information isn't legible

increasing at a rate of what?

[lost laurel]

why have you scratched out the top

[vague vortex]

Explain your work

[lost laurel]

I'm thinking, just a minute please

[vague vortex]

Ok

[lost laurel]

this isn't from a test, is it?

[vague vortex]

No hw

[lost laurel]

ok

[vague vortex]

L * w * h

V= a^3

V =pie r^2h

[lost laurel]

*cube

[vague vortex]

V=a^3

[woven radish]

Why am. I here

[vague vortex]

Oh ok

Can someone please help me with this

<@&286206848099549185>

[wary stone]

It will depend how much of the container is full

Also whether the tip of the container pointing up or down

[devout snow]

@vague vortex Has your question been resolved?

[devout snow]

@vague vortex Has your question been resolved?

[devout snow]

@vague vortex Has your question been resolved?

[devout snow]

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[restive river]

the fn is continuous, b = pi/2

how do I find a?

[scarlet sequoia]

limit from the right of the top expression

when x approaches to 0

[solar goblet]

evaluate the first equations limit as its approaching 0+

solve the limit = b

[restive river]

i understand that, i don't know how to evaluate

[scarlet sequoia]

use sinx/x limit

rule

[restive river]

using squeeze theorem or the property sin x /x = 1

[scarlet sequoia]

So you're struggling with applying that property?

[restive river]

i'm just confused

cus of the 2x

[scarlet sequoia]

if sinx/x -> 1

then sin(2x)/x -> 2

[restive river]

oo

so 4/a?

[scarlet sequoia]

yes

[restive river]

so a is 8/pi?

thank you

.close

[devout snow]

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[devout snow]

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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>.
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• 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.

[severe pasture]

yo quick question is this correct or would the negative also distribute to the 4

[severe pasture]

,rotate

[devout snow]

@severe pasture Has your question been resolved?

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[severe pasture]

.reopen

[devout snow]

✅

[severe pasture]

^^^

[soft umbra]

Still need help?

[severe pasture]

yeah

[soft umbra]

Okay

[severe pasture]

im just checking