help-38

[carmine spade]

So since we know our constant term is 0, we know that x is a factor of function

But from the way it's written right now, that's not obvious

[lunar stirrup]

oh ok yes..are u trying to hint at differentiate and x=0?

[carmine spade]

We want to use row operations to make it obvious

No, no differenting

[lunar stirrup]

oh ok

thank god

[carmine spade]

Haha yeah

With the hint that we want to use a row operation, do you have any ideas?

[lunar stirrup]

let me just put the question here again one second

[carmine spade]

Sure thing

[lunar stirrup]

oh

the sum of each column or row is the same

so we can add then take them out

?

[carmine spade]

Not quite sure what you mean?

[lunar stirrup]

R1---> R1+R2+R3

then we will have 1+(1+x)^a + (1+2x)^b in the first row of all 3 columns

and if we have a thing common we can take it out

and we will get 1 in the first row

and then we can easily create 2 zero's

[carmine spade]

That is all true, and a nice spot, but I don't think that step really helps

[lunar stirrup]

oh

then i dont have any ideas 😔

[carmine spade]

Ok, see what happens when you do R1 -> R1 - R2

[lunar stirrup]

sum of elements in the first row will become 0?

[carmine spade]

Yes but we're not so bothered by the sun of elements, more the elements individually

Do you notice anything about them

Let's just look at the first element in the row (after the operation)

[lunar stirrup]

okay

(1+x)^a -1

[carmine spade]

Yep, what is its constant term?

[lunar stirrup]

-1

no wait

0

[carmine spade]

Yep

What about the other two entries

[lunar stirrup]

0 as well

[carmine spade]

Exactly

And if the constant term is 0 what do we know

[lunar stirrup]

not really sure...

[carmine spade]

(hint: its a fact we have already mentioned)

[lunar stirrup]

oh that f(x) is divisible by x?

[carmine spade]

Yes ! But in this case our entries are divisible by x

[lunar stirrup]

oh ok yeah

im a bit unclear on how to move forward

[carmine spade]

That's ok, we're nearing the final stretch

If all of the entries in our row have a common factor what can we do

[lunar stirrup]

take it out

[carmine spade]

Bingo

And so we have an expression for f(x)/x (after dividing)

[lunar stirrup]

yeah

[carmine spade]

We were interested in the coefficient of x in f(x), but this is the same as the constant term in f(x)/x

Does that make sense?

[lunar stirrup]

yes

so we put x=0 now?

[carmine spade]

Yep

[lunar stirrup]

Ohhhhh ok got it

yeah got the answer

[carmine spade]

Nice!

[lunar stirrup]

thank u so much man u were really patient

[carmine spade]

It's scary because those top rows are horrible expressions, but actually we don't care about it at all !

Yes nws, happy to help

[lunar stirrup]

yea bro how did u see that the constant term would be 0 on doing R1--->R1-R2

well anyway

[carmine spade]

Years of doing problems ig haha

[lunar stirrup]

XD

ill close this now

thx again

.close

[carmine spade]

Acc I have to close it lmao but all good

[lunar stirrup]

oh

its ur

😭

[carmine spade]

.close

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[raven plank]

Hello, I´m so bad in math so I need help with these problems

[empty orchid]

Do you need to simplify?

[raven plank]

yes

[empty orchid]

Okay, do you know exponent laws?

,tex .exponent laws

[solid kiln]

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

[empty orchid]

,tex .exponent rules

[solid kiln]

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

[empty orchid]

Bruh

[raven plank]

?

this?

[empty orchid]

Yeah

Try to use those now

[raven plank]

how?

can you explain pls

[empty orchid]

Try to use the first law on part a)

[raven plank]

and then?

[solid hemlock]

Factor in the roots: m^11 = m^9 x m^2 and n^17 = n^15 x n^2. Then, simplify the roots.

[raven plank]

thank you

[solid hemlock]

Note that, in all cases of simplification of roots, you will have to factor the roots according to it´s indices.

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[golden lintel]

.help

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[golden lintel]

If 1360.8 w/m^2 of solar energy reaches the top of Earth's atmosphere, how many fewer watts reach Earth when Mercury (diameter = 12 seconds) transits the Sun (diameter = 1909 seconds)?

[golden lintel]

<@&286206848099549185>

[sage mesa]

Mercury blocks (12/1909)² of 1360.8 W/m², so ~0.05 W/m² less energee reaches Earth

[golden lintel]

energy*

[golden lintel]

but whats the answer

[sage mesa]

typo

-0.05??????

💀💀

[golden lintel]

0.05 w/m^2?

.close?

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[minor mason]

Hi this is not exactly a math question but im just looking for insights on a mapping project im doing, in a 2 dimensional space I have a bunch of points which represent fire events. I have such a map of 4 different weeks across 5 years (ex: 2020 - w1,w2,w3,w4; 2021 -w1,w2 so on), I wish to categorise and classify these fire events in low moderate and high reccurance.

now in this 2d space a fire event may re-occur over time. So for instance at a single point a fire re-occurs 6 times. now this doesnt mean that the fire recurrs every year and one extra week in one of those years, since it can be that the fire recurrs every week of one year and 2 other weeks. And for such a recurrance there can be 20!/6!(20-6)! possible combinations (38760), and now there are 14 recurrance classes (ex: fire that recurs once, twice... upto 14 times)

I wish to filter out for this plane, fires that have recurred throughout combinations that are spread atleast accross 3 years. I can come up with a logic or something and programme it but I just wanted to know if these kinds of problems are encountered in math, and are there any statistical/geometrical concepts around this so I can be more informed I understand the temporality and the 2d plane part adds a very confusing dimension to it but yeah. Thanks!

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[dapper oxide]

We know that if you were to plot the set of all points (x, x^2, 0) where x is any real number, then you get the parabola. However, if we were to plot (x, x^2, 0) by itself where x is any real number, would you get a single, varying point? Or would you also get the parabola?

[obsidian zealot]

didn’t you ask this question before

[dapper oxide]

they said they don't know

that it's uncertain

[obsidian zealot]

its a varying point dependent on x, but when we graph it we graph all x to get a feel for the behavior of the function

but conventions are conventions and people use terms definitely so always confirm in your use case

[dapper oxide]

plotting the set of all points of the form (x, x^2, 0) gives you the parabola, but what if we plotted (x, x^2, 0) by itself? that's what's new to me here

[obsidian zealot]

when we say “graph (x, x^2, 0)” we implicitly mean graph all points

[dapper oxide]

I see

[obsidian zealot]

if we graph one point we have to know x

normally its just more useful to know how all points look which is why we graph it as it is

[dapper oxide]

when graphing (x, x^2, 0) by itself, let's say we pick x = 5, and then we pick x = 6, would we get two points on the same graph? or one point that moves from x = 5 to x = 6?

[obsidian zealot]

well

it depends if we’re doing like

(x, x^2, 0) x=5 / (x, x^2, 0) x=6

or just taking the first line and changing the value

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[obsidian zealot]

?

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[mortal kestrel]

yo

[vagrant prism]

yo

[full dock]

yo

[fallow bane]

yo

[mortal kestrel]

Express 1728 and 2025 in terms of prime factors hence evaluate. cube root 1728 / square 2025

[vagrant prism]

🪀

[mortal kestrel]

idk how to put symbols lol

[vagrant prism]

$\frac{\sqrt[3]{1728}}{\sqrt{2025}}$

[solid kiln]

artemetra

[vagrant prism]

like that?

[mortal kestrel]

thx yah

[vagrant prism]

what have you tried? have you started favtorizing them?

[mortal kestrel]

yea i got 2^6 x 3^3 for the numerator

and 3^4 x 5^2 for denominator

...

[obsidian zealot]

so

what does that tell you about each of the radicals

you can try converting them to fractional exponents

[mortal kestrel]

so 12/5^4x3^8

[obsidian zealot]

wait what

[mortal kestrel]

lets take a simpler approach my friend

[obsidian zealot]

how’d u get ur denominator

[mortal kestrel]

its square will be (5^2 x 3^4)2

thus use law of exponents

[obsidian zealot]

what

is it not easier to divide 4 by 2 and 2 by 2

[obsidian zealot]

its square will be notably not this

it will just be (5^2x3^4)

[mortal kestrel]

srry its actually 5^4 x 3^8

[obsidian zealot]

ummmm what

2025=25x81=5^2x3^4

[mortal kestrel]

oh yes

i was so wrong, srry

after that what do i do bro

[obsidian zealot]

what’s the fraction

[mortal kestrel]

i dont know how to get numerator

but i think its 12

[obsidian zealot]

fractional exponents…..

_whats the denominator??

[frail heron]

Correct

[mortal kestrel]

4,100,625

[obsidian zealot]

ummmm

its the square root of 2025 so it’ll be smaller

[obsidian zealot]

go from here

[mortal kestrel]

if we simplify it will be 15

[obsidian zealot]

yea

so

what’s the fraction

[mortal kestrel]

4/15

[obsidian zealot]

well

what’s the numerator

[mortal kestrel]

4(simp)

[obsidian zealot]

12 yes exactly

sooooooo what’s the fraction

[mortal kestrel]

4/5

[obsidian zealot]

yesssss

[mortal kestrel]

thx alot man

.close

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[regal hatch]

isn't there no combination of scalars that can make them collinear.. if they are originally noncollinear. the direction won't change

[regal hatch]

wait i misread the question

it wants me to choose

deff a and d right?

ok thanx

[frail heron]

Correct

[regal hatch]

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[simple panther]

Wait, how is it a and d (I didn't get it)

[frail heron]

Determine which of the above are colinear

Are you having problem with this one?

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[simple panther]

I mean, the vectors aren't given here

[trim jolt]

Please don't occupy multiple help channels.

[simple panther]

Closed the other one

[solemn parrot]

if you have any vector a

and you multiply it by a scalar factor

it will always be collinear (lie on the same line as the original)

because we're just stretching it

only options A) and D) have the same vectors

[simple panther]

Ohh, mb mb

[solemn parrot]

do you understan

[simple panther]

I didn't see the options properly

Lol

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[median sequoia]

hi so i have to find the diameter

[median sequoia]

i did like this bit

but im not sure how to move on with this

this is the question

[split chasm]

are you allowed to apply intersecting chords theorem (power of a point)?

[median sequoia]

im not sure what that is, never learnt it

[split chasm]

its worth looking up, useful in many circle geometry problems
or from what you have here, you can consider ratio of sides of similar triangles

[median sequoia]

oh ok

yes

[solid hemlock]

Let P be the intersection point of AB and CD. So, we have: AP x PB = CP x PD (It´s called the power of a point P).
This will give you AP.
The diameter will be AP + PB

[median sequoia]

wait i can take this entire triangle as a right angle triangle right

ACB I mean

[median sequoia]

oh

[median sequoia]

this is the power of a point rule?

[blazing rune]

was it specifically given that AB is a diameter?

[solid hemlock]

Yep

[split chasm]

intersecting chords is one part of power of a point

[median sequoia]

oh i see

[solid hemlock]

Note that the center belngs to AB.

[blazing rune]

he's put the point there, was it mentioned in the qsn

[median sequoia]

I put the point actually

[blazing rune]

oh yes

[median sequoia]

wait so can this be right

[pseudo tusk]

u can use euclid

[median sequoia]

ok but how do i find AD?

[pseudo tusk]

it'll be 90 degrees if it looks at the diameter

u don't need AD

[median sequoia]

oh

[pseudo tusk]

euclid's theorem

4*x = 5^2

diameter will be 4 + x

[median sequoia]

i havent learnt that actually so ill need to learn that

[split chasm]

CB and BD isn't needed eitehr

[pseudo tusk]

oh

[median sequoia]

oh

[split chasm]

4*x = 5^2
diameter will be 4 + x
is the power of a point that Marcus and I've been referencing

[median sequoia]

ahh ok i see

[pseudo tusk]

ah

[split chasm]

applying similar triangles will result in the same equation

[solid hemlock]

Yes. You can also note that AB intersects the middle point of CD and BC is perpendicular with CD. That means, AB is the diameter.

[median sequoia]

ah ok so

i can apply this here with subsitution

right?

[solid hemlock]

The power of a point (or in this particular case, intersecting chords theorem) is a shortcut to using triangle similarity. So you can also find a good solution using triangles.

[median sequoia]

ok

[solid hemlock]

Nice image.

[median sequoia]

wait but then

so if a x b = c x d

how do i find a in this

actualyl nvm

so it would be a x 4 = 5 x 5

4a = 25

smth like that???

[solid hemlock]

Remember that the diameter is AB = AP + PB. Happy Easter.

.close

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[smoky kite]

,rotate

[smoky kite]

There’s no way these r equivalent right

[rare spear]

is this a h

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[ebon venture]

what does the question mean?

[marble wharf]

it means what it says

you need to be more explicit about which parts you do not understand

[trim lichen]

which part of the question is unclear to you?

[ebon venture]

what does almost squarefree and squarefree mean separately does it mean along with the conditions given it is also not a square number?

[marble wharf]

squarefree means that it has no square as a divisor

in particular, p^2 does not divide n for any prime p

[trim lichen]

"almost squarefree" is defined directly in the question

[ebon venture]

ok thanks

.close

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[wraith hinge]

No idea how to solve it

[mystic iris]

😭😭 incredibly possible with 7th grade methods

is it 2.5

just to check

[wraith hinge]

Yes but how did you get it

Show me solution

[mystic iris]

my solution involves a lot of circles

let me explain the process atleast

[wraith hinge]

Oh dear ok

[mystic iris]

so you place E at the origin

and you draw a circle with radius r

you dont know it yet

[wraith hinge]

Oh in graphs i mean it is one of the hardest way to solve this problem i am a 11th grade student

[mystic iris]

oh okay

im trying to figure out how to solve this without a method like this

[wraith hinge]

Same

[main sigil]

if anyone could prove that the green like goes through the center of AC, i could prove the rest easily

[wraith hinge]

Now way but how it is a median the one that you draw?

Okay what is the answer that you got

If we consider it as median

[main sigil]

2.5

[mystic iris]

yeah its true but how do you even know

[main sigil]

yeah, thats the question

[wraith hinge]

Intuition

[main sigil]

we could probably also use the angle bisector thm somehow

[wraith hinge]

What about it

Any idea

[mystic iris]

once you find BC its relatively easy and its sqrt(193)

just some sort of magic how you get there

I figured it out fully now

[wraith hinge]

Show it bro

[mystic iris]

This genuinely could be a book for 7th graders who just fully understand circles

[wraith hinge]

Aaa circles bruh

Btw is the numbers that you get i mean the ED is not an integer a number i guess?

[mystic iris]

yeah ED is not an integer

it is a rational

[wraith hinge]

No i mean i know 5/2 but like you did it via circles but does it have repeating decimals?

[mystic iris]

no repeating decimals

[wraith hinge]

No way it means you solved it via circles damn but it is time consuming

[mystic iris]

yeah

[main sigil]

Wait is angle A 90° or not?

[wraith hinge]

Nope

[main sigil]

that makes it wayyy more interesting

that means that ED must be constant

no matter what A is

[mystic iris]

yeah it is 90

[main sigil]

how do you knwo?

[wraith hinge]

If it is 90 your median is gonna be valid i mean true

[wraith hinge]

This one

[main sigil]

Yeah

[mystic iris]

i found the whole side to be sqrt(193(

[main sigil]

How?

there is no way you couldve solved for that side

[mystic iris]

coordinate geometry

[main sigil]

no way

[wraith hinge]

Bruh but there is too much decimals

[main sigil]

if you dont make any extra assumption, then the only things you know about the triangle are its 2 sides

and nothing more

[wraith hinge]

Mhm indeed

This kind of problems are more like you can solve it only by improvising and checking if your answer is right

[main sigil]

This assumes A is 90°

[wraith hinge]

Bro is masochist

[main sigil]

the thales circle above BC

and yes, its quite masochist

for the special case where A = 90°, what I did works perfectly fine in much less work

[mystic iris]

it simply must be

[main sigil]

Why?

it doesnt have to be

[mystic iris]

i cant figure it out but the math is always right

[main sigil]

your math assumed it's 90°

otherwise the circle above BC wouldnt necessarily pass through A as you drew it

[wraith hinge]

Indeed

[mystic iris]

the way i wouldve tried to find the radius of the rightmost circle it was numerical in the start

but i found a way to get it closed form

[main sigil]

take an arbitrary triangle with sides 7 and 12 (not necessarily 90° at A). Now halve BC to get E, make thales above AC and draw bisector of angle A. Where they intersect, mark point D and you got exactly the same setup as in the question with non-90° angle at A

[mystic iris]

and coincedentally it manages to be 90 degrees

[main sigil]

it can be anything

there is nothing in the question what could force it to be 90°

[wraith hinge]

MathIsAlwaysRight ironic nickname

[mystic iris]

can you find another solution to solve the triangle?

i think this is the only triangle which fulfills all the criteria

[main sigil]

i mean sure

i can show you a counterexample

lemme open up geogebra

here you go

[mystic iris]

is AB 12

is AC 7

is EB = CE

[wraith hinge]

Well okay but bro what about your solution here can you show me how did you get the 2.5?

[main sigil]

ánd even better

[main sigil]

Yes

[main sigil]

yes

[main sigil]

yes

[main sigil]

geogebra calculated that it's exactly 2.5, i still dont know how to prove it

aside from specific cases such as A = 90°

[mystic iris]

so its apparently the same in all cases

[main sigil]

perhaps the angle bisector theorem

[main sigil]

Yeah

pretty surprising result tbh

[wraith hinge]

Yeah okay lets consider it 90 degree i am curious how you went further

[main sigil]

I just calculated all necessary angles

and proved that AB || ED

if you can prove that, you won

if AB is parallel to ED, then the extended ED must join midpoints of AC and BC, and so its length must be half of AB = 6

[wraith hinge]

Oh ok

[main sigil]

then by thales circle above AC, we will have the length from the midpoint of AC to D = 3.5 (half of 7)

and 6 - 3.5 is 2.5

what remains to be proved is that ED is parallel to AB

[wraith hinge]

Ohh dang

[main sigil]

lets do some angle chasing I guess

after chasing every angle i could, i still didnt figure out anything

[mystic iris]

what is this book called again

[wraith hinge]

I can send you guys a pdf

[mystic iris]

please do

[wraith hinge]

Here you go

[mystic iris]

thanks

[wraith hinge]

This task is in page 111

15 16 17 are the ones that i am struggling with

Well 17 seems easy but 16 is not defo not

[main sigil]

angle bisector theorem

[wraith hinge]

Bro how like the angle bisector is not touching the BC

It intersects with DC

[main sigil]

i mean you can always extend it, though it might not be what they want here

[wraith hinge]

We don’t have the BE and EC

And like 12/7 is not equal to 1

You know

12/7 is not equal to BE/EC

Even if we extend it we don’t have values that’s the thing

We are just gonna loose the BE=EC

[main sigil]

i was thinking of imitating the proof but it seems to only intrdocue another useless line

ddamn

[wraith hinge]

Yeah i have looked through every topic but no one helps to solve this problem dang

[main sigil]

I feel like we're extremely close

the whole problem is reduced to proving that DE is parallel to AB

and we can even ignore all the side lengths, because its a statement which should hold in general

did all the previous problems include the angle bisector theorem?

from that part of the book

[wraith hinge]

Yeah all of them easy as hell

But these idk man

I feel like the author was drunk when he was writing these problems

[main sigil]

Done

i think i did it

I proved AB || ED

[wraith hinge]

No way show it

[main sigil]

oh nvm i didnt 😢

[wraith hinge]

Bruh you scared me dude

[main sigil]

i got lost in my assumptions and accidentally assumed that they already are parallel

[wraith hinge]

I mean there is one thing we can use or we can not idk do we still consider the angle A as 90 degree?

[main sigil]

if we do that, then its simple

or maybe it isnt

but its certainly doable with coordinate

[wraith hinge]

But it is defo should not be solved with coordinates

[main sigil]

im too interested at this point

ill repost the new question

that is, prove that ED is parallel to AB

[wraith hinge]

Okay

[main sigil]

What if we reflected it somehow

wait no

nvm

or maybe

no

actually yes

it worked

[trim jolt]

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[real spindle]

Got this question that I'm not sure whether it's solvable. It says: "Two parabolas intersect on the x-axis. Parabola A has a vertex of (0, -49.5), and Parabola B has a vertex of (0, 29.7). Write an equation in standard form to represent Parabola A." Could someone let me know if this is solvable and how to do it? Thank you! (P.S. I'm pretty sure this equation is based off of Pluto's orbit).

[pseudo goblet]

Pluto's orbit is an ellipse; parabolas are... not ellipses

Do you know what the standard form of a parabola is?

[real spindle]

ax^2 +bx +c

I was guessing that they took the aphelion and the perihelion and used that as the two vertexs to make a rudimentary graph of pluto's orbit

[pseudo goblet]

Oh that could make sense

We'll try using a different standard form, only because it makes the maths hella easier

If I had a parabola and I completed the square, we'd be able to find the vertex of that parabola, right?

[real spindle]

Yeah

[pseudo goblet]

What would the completed-square form of the parabola look like?

[real spindle]

which one?

[pseudo goblet]

Well let's answer the questions in order so

Parabola A

[real spindle]

Ok

y = ax^2 - 49.5

[pseudo goblet]

yee

You can probably do the same for B

[real spindle]

y = ax^2+29.7

[pseudo goblet]

(though don't use "a" - since that now has a meaning: it's part of Parabola A's definition)

[real spindle]

good point

m?

[pseudo goblet]

Sure

(tbf I'd've gone with "b" because that's the next letter in the alphabet)

(and it goes with B the parabola's name)

[real spindle]

so y=mx^2 -49.5

Then we just need to solve for m

which we would do by finding the x-intercepts

[pseudo goblet]

...

I meant
A has eqn y = ax^2 -49.5
B has eqn y = bx^2 +29.7

[real spindle]

oh

sorry

[pseudo goblet]

So you can easily tell which coefficient of x^2 belongs to which equation

[real spindle]

got it, that makes more sense

how do we find the x-ints

[pseudo goblet]

If any graph is on the x-axis, which value's 0? x or y?

[real spindle]

y

[pseudo goblet]

yee

So solve for y = 0

[real spindle]

0=ax^2+49.5

-49.5=ax^2

-49.5/a = x^2

radical -49.5/a = x

[pseudo goblet]

+- that radical, yh

And similarly for B?

[real spindle]

yeah

oh wait, +-√49.5/a

no negative, my bad

[pseudo goblet]

You need the negative

ik you're thinking you can't square root a negative

But we know there's a solution

[real spindle]

but the original is ax^2-49.5

you would add to get rid of it

[pseudo goblet]

So if -49.5/a can be square-rooted, what does that tell you about a?

oh wait yh mb

I misread that

[real spindle]

Its ok, I mistyped it

[pseudo goblet]

that one's right yh

[real spindle]

Parabola b has that

[pseudo goblet]

btw what does the rest of the question ask?

[real spindle]

Write an equation in standard form for Parabola A

[pseudo goblet]

I meant the rest of the whole question

That's got to only be part of it?

[real spindle]

That's it

[pseudo goblet]

Otherwise why would they be giving you a parabola B

[real spindle]

Maybe to find the x-intercepts?

can you solve for a without it?

[pseudo goblet]

There's a whole family of solutions is the ]thing

Without any extra information all you can really tell is how much a/b is

(basically you similarly solve for the x-ints of B, and you argue that since those are the same as from A, the bits in the square-roots are the same)

[real spindle]

huh

ok

so 49.5/a = 29.7/b, which gives you a/b?

or am i crazy?

[pseudo goblet]

= -29.7/b *

[real spindle]

right

sorry

[pseudo goblet]

but yh

[real spindle]

similarly, you could get 49.5/-29.7, right?

[pseudo goblet]

to get b/a?

yh

wait

er

a/b

one or the other

I'm typing this shet on a computer and playing with a graphing calculator lol

[real spindle]

lol

[pseudo goblet]

My point being that you can get any number of solutions:

[real spindle]

looks like +-13

[pseudo goblet]

(actually this is a better representation)

[pseudo goblet]

Note the slider there

[real spindle]

oh

[pseudo goblet]

You can pick any (negative) value for a, and there'll be corresponding value for b, is my point

[real spindle]

yeah

[pseudo goblet]

The question may as well have been "Parabola A has a vertex of (0, -49.5); write an equation in standard form to represent Parabola A."

[pseudo goblet]

To which this would be the right answer

[real spindle]

but don't we still need a?

[pseudo goblet]

Because without any more information you can't really say what a is

[real spindle]

oh

ok

wait

does the graphing calculator tell you the intercepts when you hover over it?

[pseudo goblet]

It's a parabola, it's in standard form, and it has that vertex

It does - I'm using Desmos, you can try it on any browser

[real spindle]

alright

if we have the intercepts, can we use intercept form and convert it to standard form for a?

y=(x+11.1243)(x-11.1243)

[pseudo goblet]

What I mean by this is that where it crosses the x-axis depends on a

If I'm changing a though, then those intercepts will change

[real spindle]

oh

[pseudo goblet]

and I can always find a b to match

[real spindle]

sorry

yeah

that is useless

[pseudo goblet]

So the order of operations doesn't really reveal anything here

yh

Hence my asking if there's anything else to this question

[real spindle]

I'll tell the person who gave me this

I had a feeling they just wrote some algebra 1 stuff down expecting for it to be hard, without checking if the answer could even be narrowed down

Thanks for your help

(PS your avatar is really cool)

[pseudo goblet]

hehe 🍋

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[stone condor]

I've been struggling with this problem here. The picture is a semicircle with diameter AB and center C.

[stone condor]

I tried some values, but no cigars.

[wet lintel]

the hell is a cigar doing?

[wet lintel]

is C the center?

[stone condor]

Hahaha just an expression.

[stone condor]

Yes.

[wet lintel]

okay, so you konw the inscribed and central angle theorems?

[stone condor]

Yep. Tried to play around with it.

But couldn't get to ∠BCF.

[wet lintel]

Hmm

[stone condor]

I identified some of the arcs with their respective values.

[wet lintel]

yeah, but this feels really obvious you know what i mean

[stone condor]

Tried completing the circle to find maybe some symmetry.

[stone condor]

Sometimes stating the obvious help, I think.

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@stone condor Has your question been resolved?

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@stone condor Has your question been resolved?

[stone condor]

Maybe triangle CFI is equilateral somehow. For that, smaller arc DF should be 120º, though....

(sorry for the whole useless measurements on the pic, 'been trying for a while now)

[nova glade]

I feel like DCF and DGF both have to be right triangles based on the fact that the two triangles have two congruent sides and one congruent angle, but it's been long enough since I've worked with this that I don't remember exactly why. If true, the answer should be 50 degrees.

Maybe they aren't right triangles from looking at the pic you've drawn, but there should be some relationship between those two triangles that should help you calculate BCF. Probably a good place to start.

It's also noteworthy to say that those are the only two triangles that have two congruent sides and one congruent angle in that exact way, so you should in theory be able to solve for both triangles completely based on that fact alone.

[stone condor]

Interesting. Usually this helps transfering some angles around.

Oh.

[nova glade]

Review your SSA methods for solving triangles

[stone condor]

I think I got it.

Quadrilateral DCGF is a cyclic quadrilateral, since both ∠CDG and CFG are "looking" to side CG (and both CF and DG are diagonals for this quadrilateral). So, there's a circle that passes through D, C, G and F. Since ∠FCG is an inscribed angle on arc GF of this circle, and ∠GDF is another inscribed angle on the same arc (GF), we have that ∠GDF = ∠FCG = θ. Thus:

2θ = θ + 20°
θ = 20º.

Here's a better visual solution.

Thanks for the attention, guys!

[vagrant kraken]

remember to do .close after you are done

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[mystic iris]

.close

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[west wharf]

Can anyone help me go through this integral? I dont know how to aproach this

[vagrant imp]

have you tried substitution?

[west wharf]

I was thinking on letting u = x but that doesnt seem right

[vagrant imp]

yeah that doesn't help since we get the same thing again

[solemn parrot]

the first thing you should be looking at is if any derivatives occur inside of the integral

ln(x) -> 1/x

[west wharf]

oh

wait dont solve it yet, i wanna try

so without simplifying it, (1/5)*(1/x)^5 + C

right?

which basically boils down to 1/5x^5 + C

right?

[vagrant imp]

nope im pretty sure thas wrong

[west wharf]

hmm

[vagrant imp]

can you show what you substituted?

[west wharf]

ok so u = ln x, du = 1/x dx. Since there is an 1/x in the integral and a dx as well, I can just replace it with du

[vagrant imp]

but we don't have a lnx

[west wharf]

we dont?

(lnx^4)/x dx isnt it?

[vagrant imp]

is that power for ln or for the x?

[west wharf]

good question, my friend just wrote it down but didnt tell me where

so if it were on the ln, would it be done how I did it?

yeah he just confirmed, on the ln

[vagrant imp]

alr then yeah proceed with how u did

i have to check the ans tho give me a min

[west wharf]

sure

[vagrant imp]

you'll get int u^4du yeah

[west wharf]

cool

[west wharf]

so this is it?

[vagrant imp]

wouldn't that become u^5/5

and u=lnx

so...

[west wharf]

oops

I applied it wrong

nice, so it would be (lnx)^5 all over 5

• C ofc

[vagrant imp]

corrct

[west wharf]

aahhh thanks dude

I appreciate the help very much

[vagrant imp]

your welcome

[west wharf]

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[midnight pebble]

help

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[drifting whale]

how exactly do i do this question

[drifting whale]

it asks me to find the equation of this line

so the thing is

ai and my answers have completely different things

this is my working

so since theta is 0 < pi/3 < pi/2 so that means theta is in quadrant 1

in quadrant 1 tan is positive

so i find the ref angle of that so theta ref = pi/2 - pi/3

which gives me pi/6 for my ref angle

then i put my angle into tan theta

so tan pi/6 = 1/ square root 3

and then i sub x int and find c which is -2

so my answer is y= 1/ square root 3 x -2

but ai says i dont need to find ref angle since the angle is acute and in quadrant 1 so its giving me a complete diff anser

wait so my question is

do i need to find the ref angle?

since its a acute angle and also lies in quadrant 1

[sharp heart]

It's unclear what's going on in the image (what does pi/3 represent there; what does 2 sqrt(3) represent).

Is pi/3 the angle between the y axis and the blue line? Is 2 sqrt(3) the x intercept?

,calc tan(pi/6)

[solid kiln]

Result:

0.57735026918963

[sharp heart]

,calc sqrt(3)/3

[solid kiln]

Result:

0.57735026918963

[sharp heart]

For the last part, you can use the point-slope form instead of the slope-intercept form.

Slope-intercept is y = mx + b. Point-slope is y - k = m(x - h).

So, y - 0 = 1/sqrt(3) (x - 2 sqrt(3)). That simplifies to y = 1/sqrt(3) (x - 2 sqrt(3)) and then you're done.

You subtract the y coordinate of a point on the line from y. You subtract the x coordinate of that point from x.

[drifting whale]

well the chapter was about

finding the gradient of a straight line using tan theta

[sharp heart]

Oh, OK.

[drifting whale]

ye

so pi/3 represents the angle and 2 sqrt(3) is the x int

so equation was find equation of the line

[drifting whale]

i mean i think so

[sharp heart]

OK, the idea above was to skip finding the y intercept.

[drifting whale]

this is the solution

[sharp heart]

Like you found out that the y intercept was -2.

[drifting whale]

yep

[sharp heart]

But you don't need to if it just asks for an equation.

[drifting whale]

oh

[sharp heart]

It's like you have a point on the line that's (2 sqrt(3), 0).

[drifting whale]

ye

[sharp heart]

And you have a slope of 1/sqrt(3).

So, you can use this point-slope form:

y - __ = __(x - __)

[drifting whale]

oh that

y-y1=m(x-x1)

[sharp heart]

Right.

If you have the slope and a point, it's quicker to do it that way.

[drifting whale]

ohh

alr then

ima conitnue doing questions

[sharp heart]

OK.

[drifting whale]

ty for help and cya

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[sharp heart]

No problem.

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[regal hatch]

c)

[regal hatch]

my answer was:

∀x (∃n∈N, x=n^2+n+1) → (∃m∈N, x=2m+1)

is it right?

[trim lichen]

seems so yeah

[regal hatch]

so this is the answer in my solutions manual, they are "obviously" equal except that its not so obvious to me, is this a lack of intution on my part

I just used the definition of the "is a subset of", and expanded from there

[trim lichen]

well they basically rephrased it so as not to include x

n^2+n+1 represents a number in A

[regal hatch]

A?

[trim lichen]

err sorry

the first set from the original

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[regal hatch]

.reopen

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✅

[regal hatch]

For all 𝑥, if 𝑥 = 𝑛^2 + 𝑛 + 1 for some 𝑛, then 𝑥 = 2𝑚 + 1 for some 𝑚 (my version)
Every number produced by 𝑛^2 + 𝑛 + 1 is also of the form 2 𝑚 + 1 (book solution)

I guess I don't understand how "for all x, if x = ..... for some n" is the same as "for all n", like the change from the existential to universal n

[summer haven]

idk the notation too well but isnt it saying that n^2 + n + 1 is always odd for natural n?

[regal hatch]

i think so

[marble wharf]

your statement is of the form "forall x, if x=something then bla"

if x!=something then you dont care what happens

so you can just directly start with "forall x=something"

and at that point you can rewrite it to get rid of the letter x

[regal hatch]

i think this is the reason why I think the two are different, the second version omits this check

what is the universe of discourse of x?

i guess just numbers

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[inner yarrow]

Someone please dm me if they are willing to help me with my pre cal hw it’s just 2 questions

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Please don't occupy multiple help channels.

[frail heron]

#help-27 message

.close

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[wraith hinge]

How do i do q4

[wraith hinge]

I got i which = to 90°

How to i get k anf j

[trim lichen]

look at triangles ADO and DOC

see if you can say something special about them

[wraith hinge]

Ok

Idk both are isosc triangles?

[native shuttle]

keep going

[wraith hinge]

Mm

So angle ADO will be 40

[native shuttle]

You have the answer key?

[wraith hinge]

Yep

[native shuttle]

Ok ado is 40

[wraith hinge]

j=80
K=50

But udk how they got that

[native shuttle]

ok so do you know cyclic quadrilateral property

[wraith hinge]

Yep opposite angle in a cyclic quadrilateral are supplementary

[native shuttle]

ok

u know ado=40

use that

i=90 u found

Or you can even tell adc=90

[wraith hinge]

40+90+k=180

K=50

Ohhh

Ok i got it

[native shuttle]

:)

[wraith hinge]

Thank u very much

Hv a nice day!

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[kindred jungle]

hi chat

[kindred jungle]

can someone pls explain this

its proving the limit of the product is the product of the limit

my lecturer's proof i dont get it lol

[zinc ginkgo]

which line don't you understand first

[kindred jungle]

i understand up to here

past that i dont rlly get

[zinc ginkgo]

Do you know why this M>0 with -M < m < M exists

[kindred jungle]

like

why did we do that

i dont get why we did that

[zinc ginkgo]

it's used later

proofs should be read and understood in its entirety

[kindred jungle]

okk

[zinc ginkgo]

and a lot after that

[kindred jungle]

wait first

when we write this

is this the same as saying "for all x E R \ {x_0}, |x - x_0| < d_1 ->"?

[zinc ginkgo]

A isn't defined anywhere, so unclear

[kindred jungle]

oh

uh 😭

domain of the function i suppose

ah yes he said it here

that "f(x)" at the end should say "g(x)"

but yeah A is the domain of functions f and g

[zinc ginkgo]

just need to update x is also in A then

[kindred jungle]

oh yeah

so its the same thing as that

if x E A

okay so

we defined some M>0 so |m| < M and so in the neighbourhood of x_0 |g(x)| < M

cool

i understand this but thats js simple algebra

now where does this come from?

[kindred jungle]

why eta?

[zinc ginkgo]

what other letter would you propose

[kindred jungle]

because we've already used epsilon

right

[kindred jungle]

how come they're both eta tho

like why isnt it eta_1 > 0 for f(x) and eta_2 > 0 for g(x)

how come we cant use epsilon again cuz we've used it already before but they use eta twice here

[zinc ginkgo]

delta2 and delta3 depend on eta

[molten venture]

hi

[kindred jungle]

hi

[kindred jungle]

why tho

like why are we allowed to do that

[zinc ginkgo]

keep reading to find out

[zinc ginkgo]

because

[zinc ginkgo]

if you do this separately, you can just think of eta as the minimum of eta1 and eta2

[molten venture]

what line you dont understand

[kindred jungle]

what does this mean?

[kindred jungle]

this line now

[zinc ginkgo]

which part of that is confusing

[kindred jungle]

im not sure what minimum of eta1 and eta2 means

[zinc ginkgo]

min(a, b) = a if a < b and b if b < a

min = minimum

[kindred jungle]

okay

[kindred jungle]

so why do we take eta = min{eta1, eta2}

[zinc ginkgo]

keep reading to find out

[zinc ginkgo]

.

[kindred jungle]

im still confused lol

like ive read the whole proof but it js doesnt make sense to me

[molten venture]

thats just unpacking the statement

[kindred jungle]

the limits of f(x) and g(x) equalling l and m right

[molten venture]

yes

[zinc ginkgo]

try doing the proof without eta and just use eta1 and eta2

[molten venture]

take the smallest of the three positive numbers

[kindred jungle]

my lecturer said you cant use epsilon because it's already been used in |f(x)g(x) - lm| < e
so why can we use eta in both f(x) and g(x) being l and m

[kindred jungle]

but i dont get the proof 😭

wait let me step back a little

[molten venture]

bound variables can be renamed

[zinc ginkgo]

yes so do it the way that doesn't get you the right answer so you can learn why the current proof works

[kindred jungle]

okay

so if i continued on

what does it mean to take delta as the minimum of those three

picking the smallest delta out of d1 d2 and d3

why?

and shouldnt delta be in terms of epsilon

[zinc ginkgo]

min(a, b, c) = min(min(a, b), c)

[molten venture]

it just means you had three safe zones

[kindred jungle]

three safe zones?

[molten venture]

one to keep |g(x)| in control and the other one to make |f(x)–l| small

also |g(x)–m| small

[kindred jungle]

i mean like

in all the epsilon delta proofs ive seen

its always like "choose d = min{sqrt(e)/2, cbrt(e/3)}" or smth

where the min{} only is in terms of epsilon inside the brackets

we're trying to choose delta in terms of epsilon

so how can we choose delta to be the smallest of hte three deltas

when we wanna choose it in terms of epsilon?

[molten venture]

if u make every one of them true at once you simply pick the smallest δ out of all

[kindred jungle]

whats d_1 in terms of epsilon, whats d_2 in terms of epsilon, whats d_3 in terms of epsilon?

[molten venture]

cant answer

[kindred jungle]

i think im thinking of this completely wrong

[molten venture]

nosols

[kindred jungle]

can u pls eli5 from the start?

im like hella confused

[molten venture]

each of those three δs was itself picked in response to your original ε or to η which you defined as ε / (M+|l|) they all functions of ε

[kindred jungle]

huh 😭

sorry i dont understnad

[trim jolt]

@kindred jungle Has your question been resolved?

[kindred jungle]

.close

[trim jolt]

Channel closed

Closed by @kindred jungle

Use .reopen if this was a mistake.

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[dusky thunder]

Does anyone know when is root test used over ratio test? I havent encountered a question where i couldnt prove for convergence with the ratio test and could with the root test. Is there even a case where root test is used because the ratio test fails or is hard to evaluate the limit?

[ionic pendant]

i think this is a good example of when you'd want to use root test over ratio test

the ratio seems more annoying to evaluate

[dusky thunder]

let me try it

ah i see

the denominator has (n+1)^(n+1)

thanks

.close

[trim jolt]

Channel closed

Closed by @dusky thunder

Use .reopen if this was a mistake.

[bright quarry]

yes

root test is stronger i’ll give you an example

[pallid lake]

.

[bright quarry]

no one cares bro

[dusky thunder]

"Stronger" in what sense?

more rigorous?

[pallid lake]

Sorry I thought this was closed

apparently not

[dusky thunder]

.reopen

[trim jolt]

✅

[bright quarry]

if the ratio test works then so does the root test

but not the other way around

[dusky thunder]

oh really

damn

is there an example where root test works and ratio test fails

[bright quarry]

$\sum_{n = 0}^{\infty} 2^{(-1)^n - n}$

[solid kiln]

knief

[trim lichen]

[bright quarry]

also are you using the formal statements or like the ones from paul online math notes

[empty orchid]

What the

[trim lichen]

this one's sneaky

[dusky thunder]

stewart

[bright quarry]

ok so like paul’s online math notes

[trim lichen]

innit like 1 + 1/4 + 1/4 + 1/16 + 1/16 + 1/64 + 1/64 + ...

[dusky thunder]

yes

[trim lichen]

[bright quarry]

the formal statement uses limsup and liminf are you familiar with these

[dusky thunder]

is it the upper and lower bound of the limit when it oscillates or doesnt settle down

[bright quarry]

you mean expanding the sum?

it’s

2 + 1/4 + 1/2 + 1/16 + 1/8 + 1/64 + …