help-23

[manic radish]

By "R", do you mean the real numbers?

[soft matrix]

!15min

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[twin skiff]

yup

[rustic goblet]

in a metric space M, a subset E is totally bounded iff for all eps > 0, there exists a finite collection of open balls of radius eps whose centers lie in M and whose union contains M

[twin skiff]

oh okay, so for something like [0,1], for any eps > 0 you could just "place" the balls next to each other starting from 0 until 1, and theres a finite number of them because of the nature of [0,1]

or even something like [0,1] U [2,3]

[rustic goblet]

pretty much, yeah

[twin skiff]

i see, thanks

[rustic goblet]

of course, you can formalize the argument, but I'm not exactly in the mood to do that myself at the moment

the intuition is good though

[safe radish]

@twin skiff Has your question been resolved?

[wet wolf]

That’s precisely the defn of totally
Bounded

For any eps u can cover it with finite number of eps balls

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[twin skiff]

yeah sometimes i get confused when im supposed to take epsilon -> 0 like a limit (?) and when epsilon is just a normal (/colloquial) positive number. in the limit when epsilon approaches zero, the number of balls it takes to cover [0,1] will be a finite number that increases without limit as epsilon goes to 0. but in this case, epsilon is just a positive number of any value, so i was just overthinking it ;;;

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[gleaming bloom]

Can someone help. I've calculated what I think is the mean by adding all the variables and dividing by however many occurrences there are but I don't know how to calculate the standard deviation

[gleaming bloom]

Can you break that down for me, also I don't have a calculator that can calculate that

can someone else please provide more support. I have no way to input this equation

[young nexus]

well, you have your observations xi sum them up, divide by the number of observations (11) and you have the mean.
then take the difference to the mean of each observation, square them, sum them up, divide by number -1 and then take the root.

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[indigo terrace]

Hi there, I am playing a game which I have to socket in a "gem" into a weapon.
The weapon can have 5-7 gem slots, each time you socket a gem it's a 50% chance it successfully sockets inside and 50% chance to break ALL gems.
The first slot is always a 100% chance to socket successfully.
The gem costs 900g

Can anyone help me calculate how much it costs ON AVERAGE for each slot individually?
I wish to change the gem price accordingly / maybe use this formula in a spreadsheet.
Thank you!

I've tried these approaches (900g on gem cost)

First slot:  900 gold
Second slot:  1.8k gold (+900 gold) = 2700g
Third slot:  1.8k gold (+2.7k gold) = 4500g
Fourth slot: 1.8k gold (+4.5k gold) = 6300g
Fifth slot: 1.8k gold (+6.3k gold) = 8100g

[indigo terrace]

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[indigo terrace]

.

[indigo terrace]

bump

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[clever rose]

Hi, I just have a quick question about graphing in the complex plane the thing i want to graph is -3-i, the question I have is where do I go for the negative i, usually for these problems there is a number like (3i) and that is easy to graph. Im assuming all I need to do is count the i as 0 on a normal graph?

sorry for bad grammar

[thin bridge]

-i = -1i

[clever rose]

oh ok thank you

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[trail escarp]

hey i need help with this question, i know that FA || BE (opp sides of parm) i mostly need help with 1)

[manic radish]

Ok, so for 1. you need to show FB || EC and FE || BC

[trail escarp]

yes

[manic radish]

good, just checking

[trail escarp]

if i prove FE || BC then FB | | EC bcs of angle FBC and FEC (alternate)

but im not sure how i can prove that

[manic radish]

E, F, A, B, C form 5 corners of a regular hexagon

[trail escarp]

how do you prove that? I havent used pentagons in analytical geometry before so im guessing thats not what will be expected

[trail escarp]

*Euclidean geometry mb

[manic radish]

play around with this if you want.

https://www.desmos.com/geometry/ddre9apywm

I've made it in the desmos geometry tool.

[trail escarp]

thank you

[manic radish]

you can drag around A, B, and F

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[trail escarp]

i think i should get ready for school now, thanks for the help :)

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[dense wadi]

for a linear equation as such

[dense wadi]

how do i know if theres 0, 1 or infinitely many solutions?

[plucky elk]

There are lots of ways

What were you taught

https://www.wikihow.com/Algebraically-Find-the-Intersection-of-Two-Lines
This look familiar?

[dense wadi]

I wasnt taught thoroughly, but this linear equation gives = 8 and = 9 could that indicate that theres no intersection?

[dense wadi]

ty ill look into it

oh this ik

ohh

[plucky elk]

You mean 6x-4y=8

[dense wadi]

yeah i forgot i can just do that

[dense wadi]

yes

it = 8 and = 9

so no intersection if im not mistaken

[median vigil]

have you started on row reduction yet?

[dense wadi]

nope i legit just started

actually i probably know it but dont know the name

ty anyway

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[dense wadi]

how do i find the value of k here

[plucky elk]

Not enough information

[median vigil]

so in other words find out when the system does or does not have solutions

[dense wadi]

oh

[fast mason]

System moment

[median vigil]

"consistent" means "has a solution" (either 1 or infinite)

[dense wadi]

ahh okay

wait so how do i find if it is consistent in this

[fast mason]

It says so

[median vigil]

rewrite the augmented matrix as a system of linear equations and use the same method as any other

[dense wadi]

so multiple bottom row by -2 and (1x + ky = -8x -16y)?

[median vigil]

sure

[dense wadi]

💀 thing is im stuck here

idk hwo to continue with a k in the equation

[median vigil]

hmm. what techniques has this course covered so far for solving linear systems?

[dense wadi]

for instance the exercise above is about single elementary row operation

i really havent gotten far

[median vigil]

well using elementary row operations, can you make the first coefficient the same?

[dense wadi]

if you mean the same without accounting for the last column

https://cdn.discordapp.com/attachments/903481106819612714/1278948930951974933/image.png?ex=66d2a982&is=66d15802&hm=79cd6ccfc512228ee624caac94810fe668ebc12992476bfe79f1ae59979c70f1&

then k = 2?

[median vigil]

what would happen if k = 2?

[dense wadi]

uhm

this would be the same except

with different

therefore k = all solution except 2?

im taking like a wild guess here

💀

[median vigil]

yes

[dense wadi]

oh its actually this? wow

tysm!

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[median vigil]

we can interpret it geometrically as changing the slope of a line, and no solutions being when the lines are parallel

[dense wadi]

.reopen

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✅

[dense wadi]

ah yes

but

this

is this the smae as the first one i sent?

is it again all values of k except 2

[median vigil]

well consider what happens if we do this on the new matrix

[dense wadi]

i would assume k=2 as well?

oh but if k=2

then its all -4

so k = 2 works as well so just k = N?

ah yeah okay tysm!

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[median vigil]

well k can be any real number

and we can interpret this geometrically the same way, except instead of just being parallel they become the same line (so the solution is their intersection, the entire line)

[dense wadi]

ah yes tysm

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[zinc epoch]

when solving quadratic equation, do the value of variable required to be a whole number?

[thin bridge]

no

[lean otter]

i dont think so

iirc even a decimal is ok

[exotic cypress]

even surds are fine

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[dense wadi]

how many solutions could i expect from this?

[dense wadi]

my guess is infinite

[exotic cypress]

what's ur thinking

[dense wadi]

if a b c comes poitns from a parabola

a parabola has infinite amount of points

therefore a b c have an infinite solutions

[quick crater]

hint: plug in the x and y's

[exotic cypress]

it's asking for the solutions of values for a,b,c to get the parabola to pass thru the 3 points

not how many points are in the parabola

[rich stratus]

Find (but do not solve) 😭😭😭

[dense wadi]

when i replace x^2 with -1

is it (-1)^2

or -1^2

[rich stratus]

you are squaring x

which is -1

so you are squaring -1

[rich stratus]

So it has to be this

[dense wadi]

kk

actually one solution?

since theres a parabola that passes through all the points

[exotic cypress]

mm

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[lean otter]

,rptate

,rotate

[lean otter]

i was just here to rotate so if someone wanted to help

[fathom jewel]

Which one of these do you struggle with?

[sly creek]

both of them

couldnt get how the lesson works earlier in our discussion

[fathom jewel]

Ok so for 35. you can use the vertex form formula but we would have to do it in terms of y, because the axis of symmetry is horizontal.

[flat frigate]

bacc

[fathom jewel]

You are given a second point too, that way you can calculate out a.

[lean otter]

whats the question

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[leaden swift]

asecX = 1 - btanX
a²sec²X = 5 + b²tan²X
a²b² + 4a² = kb²
k = ?
<@&286206848099549185>

[grim kraken]

What is sec again ? We don’t see it intentionally in France

[leaden swift]

secant

[coral jay]

asecx + btanx = 1
So, asecx - btanx = 5
Solve for each term

[leaden swift]

one of the trigonometric ratios

[grim kraken]

Yeah but what trigonometric function is associated to it ?

[leaden swift]

Check the square 🙂

[leaden swift]

😭 wdym?

[coral jay]

Yep u derive it from the second equation given then u manipulate it with given relation to get the ans

[leaden swift]

I tried several times! It'd be best if you show me the method PLEASEEE

[sinful hedge]

secant is inverse trigo function of cosine

[coral jay]

a=3cosx
b=-2/tanx
Then manipulate it with the given relation

[coral jay]

Ok wai

a² (b²+4) = kb²
9cos²x(4(1+cot²x)) = k*4cot²x
9cos²x(4cosec²x) = k * 4 cos²x/sin²x
9(4/sin²x) = 4k/sin²x
36 = 4k
k = 9

[leaden swift]

🤯🤯🤯

WOW

@coral jay THANKS ALOT MAN

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[leaden swift]

Yeey

[coral jay]

Np

[tight rapids]

Yes

[leaden swift]

Prove that
| 1+x²-y² 2xy 2y |
| 2xy 1-x²+y² -2x | = (1+x²+y²)³
| -2y 2x 1-x²-y² |

<@&286206848099549185>

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[leaden swift]

Prove that
| 1+x²-y² 2xy 2y |
| 2xy 1-x²+y² -2x | = (1+x²+y²)³
| -2y 2x 1-x²-y² |

[leaden swift]

<@&286206848099549185>

[lean otter]

hmm i dont think we can create 0 anywhere

maybe we can do R1=R1-yR3

@leaden swift

[leaden swift]

okay next

I got 1 zero

[lean otter]

what expression do you have now

can you send a pic

[leaden swift]

hol up

[lean otter]

okay

ping

[leaden swift]

@lean otter

[lean otter]

that is so wrong

[leaden swift]

I plused

[leaden swift]

whyy

[lean otter]

u applied the operation wrongly

[leaden swift]

If I minus, It doesn't become zero 🙂

so I plused

[lean otter]

why do you think it isnt becoming 0?

[leaden swift]

Yeah

I got only 1 zero

[lean otter]

the horizontal lines are the rows

and the vertical ones are the columns

[leaden swift]

@lean otter will you please solve the whole math??

[lean otter]

-_-

[leaden swift]

😭

I forgot

Sorry

[lean otter]

try again

[leaden swift]

😭

[lean otter]

why u apolozing bro 😭

[leaden swift]

Sometime I mess row and column

[lean otter]

yeah its ok, now that you know, try again

[cloud jetty]

you can brute force it. just expand the entire determinant and slowly simplify. it is the same amount of work. and I think it might be easier than trying to get a zero.

[lean otter]

i mean making a 0 might simplify some of our calculations already

and then we can expand

[cloud jetty]

yeah on the face of it yes. but in this case with the squares and all - I dont see an easy way to do that.

the simplification should be more fun.

[leaden swift]

what's next

@lean otter

[lean otter]

it is still wrong 😭

[leaden swift]

Why

[lean otter]

the calculations are wrong

[leaden swift]

Please do the whole math😭

I suck at math

[lean otter]

😭

[leaden swift]

@lean otter

[lean otter]

man its just basic subtraction

WHAT

X+Y=1 X-Y=0.5 X,Y=?

<@&286206848099549185>

[spiral fern]

add up the equations to find x and then put the value of x in either equation to find y

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[next crater]

A function f(x) = cos(nx), x in (0,2] have 2019 intersection point with g(x) = x2 - 2x + 1. Find the value of n/phi

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[safe radish]

@lean otter Has your question been resolved?

[dawn sequoia]

Opposite angles in a quadrilateral are congruent. Parallel lines have congruent corresponding angles.

[young nexus]

asking a solved question again?

you are asking A again

what was not clear?

which other 2 angles?

ade=cbf comes from the paralleogram ABCD, cfb=dea is a result of the sum of angles in a triangle. if 2 angels are the same then the third has also be tme same since the sum up to 180.

extend AE and draw a line where both extensions cut AB and CD.

[faint belfry]

You guys are very smart i must say!

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[green sequoia]

Does a common bisector mean it's angle is divided in half or smt

[hollow isle]

yes, the angle gets bisected into two equal angles

so angle ABC gets bisected into ABC and ADC (as said in the question) and BD is the common bisector as the given line segment bisects the thing

[thin bridge]

the word "common" wasn't really needed here

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[frail ridge]

A is the set of all ordered pairs of real numbers;
(a, b) R (c, d) if and only if a = c. Is R transitive?

[frail ridge]

@manic radish I know the definition as well as the answer. I just want to check if the answer that I have is correct

[manic radish]

Ok.

[frail ridge]

(a,b), (c,d), (w,z) in A. (a,b) R (w,z) if and only if c = a and d = b. But d doesn't necessarily equal to b. So it is not

Assuming that (a,b) R (c,d) and (c,d) R (w,z)

[manic radish]

pretend this picture is marked as a spoiler (oops)

Is that what you've got?

[frail ridge]

#help-23 message

[manic radish]

that doesn't follow the definition of transitivity. You need to show what it means for (a,b) R (c,d), and (c,d) R (w,z), and from that deduce that (a,b) R (w,z).

It's a one-way implication, so you can't necessarily go from (a,b) R (w,z) to showing (a,b) R (c,d) and (c,d) R (w,z).

make sense?

[frail ridge]

I'm not using (a,b) R (w,z) to show (a,b) R (c,d) and (c,d) R (w,z). I'm doing the opposite. For (a,b), (c,d), (w,z) in A. By the def. of transitivity we show that if (a,b) R (c,d) and (c,d) R (w,z), then (a,b) R (w,z).

[manic radish]

where did "and d = b. But d doesn't necessarily equal to b" come from?
I thought (a,b) R (c,d) <=> a = c

[frail ridge]

I thought to prove the transitivity I needed to use the assumption that (c,d) R (w,z) to show that (a,d) R (w,z) since c = a but d != b so it's not transitive

[manic radish]

Does the relation care anything about b and d?

* b and d

[frail ridge]

It only applies the restriction to the first terms of the ordered pair

[manic radish]

Ok, so you don't need to use b and d in your proof of transitivity.

[frail ridge]

Ohh

[manic radish]

Just follow the definitions, and what they mean.

• (a,b) R (c,d) means a = c
• (c,d) R (w,z) means c = w
• a = c and c = w implies a = w (notice this is one-way)
• a = w means (a,b) R (w,z)
  done

"means" and "if and only if" are two ways of expressing the same idea

[frail ridge]

I've been struggling with applying definitions to problems. Thanks

I know the definitions but can't correctly use them sometimes

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[manic radish]

lots of practice and you'll get good at them 🙂

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[shrewd topaz]

are they using chain rule or product rule first

[orchid atlas]

product

[red delta]

Both

[shrewd topaz]

yes theyre using both

but which one comes first

[orchid atlas]

product rule

[shrewd topaz]

kk ty

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[cosmic juniper]

I'm having trouble finding the answers to these questions and just knowing how to get there.

[subtle yew]

r u in the same class as aadi or sum lmao

[cosmic juniper]

Wait i think i got it mb

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[cosmic juniper]

For this i was working on question b and don't understand how to get the answer.

[devout shale]

If you say, find the solutions to sin(x)=0 in [-3pi, 3pi]

and then divide each of these solutions by 3

The new numbers will satisfy sin(3x)=0

And what interval will they lie in?

[cosmic juniper]

I thought since sin(3x)=0 then 3x = 0 and when finding what quadrants sin(0) is in i thought it would be 0, 2pi, -2pi

for the range you specified as i understand how to get that

and then i divided by 3 to get 0, 6pi, -6pi where only 0 is in the range [-pi,pi] but then after when i checked the answer

the textbook says it is

I have no clue how they got this as their answer.

<@&286206848099549185>

[young nexus]

just do what austin said

for which x in [-3pi, 3pi] is sin x = 0?

[cosmic juniper]

-3, -2, -1, 0, 1, 2, 3

[young nexus]

sin(3)=0?

[cosmic juniper]

No it would be 3pi

Would it not since 3pi is refrance angle of pi no?

Like if i think of it in degrees sin(540) equals 0

[cosmic juniper]

For this i meant -3pi, -2pi -pi, 0, pi, 2pi, 3pi

Which when divided by 3 as the above line only make 3x

would make x

[cosmic juniper]

Which is the same as this

BOOM i solved it.

anyways thank you.

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[dry mural]

Hi! I’m having issues with this physics question and I need some help.

[dry mural]

This is physics so if anybody is good at physics help would be appreciated :)

(please ping if you're able to help)

[junior smelt]

,rccw

[junior smelt]

Remember you want their positions, rather than their velocities, to match (the whole “you can’t say I’m speeding because you had to get up to [speed] to catch me, if you did the speed limit and I did, you’d never catch me!”)

[dry mural]

So you want me to draw a pvt graph?

[junior smelt]

You could, if you wanted to alternatively you could e.g. do it “algebraically” by writing their positions both in terms of when the police car spots the speeder, or you could remember the area under the velocity graphs will represent displacement and try and use that…

[dry mural]

Uh

Okay I'll try that I gues

[junior smelt]

Whichever you find is easier, up to you those are effectively all equivalent methods really anyway, the “area under the graphs” one potentially the easiest to work with maybe…?

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[true wing]

I really need help with this exercise, Anyone can help me?

[kind forge]

!status

[safe radish]

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

@true wing Has your question been resolved?

[true wing]

1

[safe radish]

@true wing Has your question been resolved?

[safe radish]

@true wing Has your question been resolved?

[safe radish]

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[tardy copper]

Hi, just curious did I do this simplification right?

[tardy copper]

The answer key says the one before simplyfying is correct so im just trying to check if i did the next part properly too

[vale oriole]

looks fine

[tardy copper]

Ok bet thank you

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[indigo seal]

pls help

[indigo seal]

is this

(4x+1)(x+1) = 0

I need to solve the x

anyone pls help

[thin bridge]

the factorisation is incorrect, how did you get that?

[indigo seal]

I just guessed

please help

[thin bridge]

how much have you been taught on factorising quadratics

[indigo seal]

2days

[thin bridge]

what quadratics have you been able to factorise

[indigo seal]

x^2 + x + a

this

[thin bridge]

what's a

[indigo seal]

numbers

1

2

3

4

5

[thin bridge]

that doesn't have a nice factorisation for the values you've listed

can you show an example of something you've actually factorised

and how you did it

[thin bridge]

,rotate

[thin bridge]

ok

have you been taught something like
ac method

[indigo seal]

r u impressed

no

[thin bridge]

how did you get
3x+1,
x-4
for 3a)?

[indigo seal]

I asked my friend

but he is a terrible explainer

[thin bridge]

normally when factoring a monic quadratic:
x^2 + bx + c
you'd consider the pair of values that
multiply to c and
sum to b

when your quadratic isn't monic,
you'd want to first consider the pair that
multiplies to ac
and sum to b
this will be used to split the bx term,
and then continue with factor by grouping

[indigo seal]

oh

[thin bridge]

and for 3b) you can also
consider whether you have something in the form
$$(ax+k)^2 = \blue{a^2x^2 + 2akx + k^2}$$
(a perfect square trinomial)

[flat frigate]

ℝαμOmeganato5

[indigo seal]

i see

btw

was my question equatoin

(2x+1)^2

I used the a^2 + 2ab + b^2 = (a+b)^2 equation

[thin bridge]

yeh, the factorisation for b) will be
(2x+1)^2
the equation would be
(2x+1)^2 = 0

[indigo seal]

O

thank you

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[loud cedar]

how do I graph this

below 2 is 8

[sturdy rune]

u can factor out the

x

x(x+2)

and that has zeros at x=0,-2

from there draw a parabola that crosses x at 0 and -2

[thin bridge]

each pair of corresponding values can be represented by a point

[thin bridge]

start by plotting them

[loud cedar]

Idk on how make cartesian plane for quadratic functions..

like whats the limit/end for each quadrants

what. No

@thin bridge

[thin bridge]

forget about what type of function you have

you have a table of values

start by plotting that

[sturdy rune]

it's not important as long as the given points are visible

[loud cedar]

I do not know how to make cartesian plane for this...

[sturdy rune]

so try like 10 in each axis

[thin bridge]

do you know how to plot points?

[loud cedar]

But ik how to plot the points

[thin bridge]

show what you've done

[loud cedar]

IDK HOWW TO MAKE CARTESIAN PLANE

But wai4 ill make

[sturdy rune]

print one lol

[humble lantern]

just draw two perpendicular lines (x and y axis)

[sturdy rune]

it's just a cross

[humble lantern]

yes

[loud cedar]

yes but idk the numbers on it

What numbers to do I put

[sturdy rune]

try 1 to 10

[thin bridge]

you said you knew how to plot points

[humble lantern]

just write equidistant numbers, for example -3 to 2

[humble lantern]

on x and y axes

[loud cedar]

Wait lemme write it

uhm

Helps

[humble lantern]

?

[loud cedar]

the prob is idk what numbers to write

[humble lantern]

do what mwb117 said

[humble lantern]

.

[loud cedar]

whats axis again

[humble lantern]

?

[thin bridge]

doesn't matter too much as long as your scale is consistent
and its appropriate for the values you're trying to graph

[loud cedar]

what is an axis

[thin bridge]

the vertical and horizontal lines you just drew are the y and x axis

[humble lantern]

you're supposed to write numbers like these

[loud cedar]

why -4 only

And not 1-10

[humble lantern]

i give u an example

[thin bridge]

that's just a cropped image

[humble lantern]

you can choose any range

[thin bridge]

doesn't matter too much as long as your scale is consistent
and its appropriate for the values you're trying to graph

[loud cedar]

if I choose any range itll give the same ans right?

[humble lantern]

you aren't finding any answer

[loud cedar]

Whats a domain and range?

[thin bridge]

make sure its big enough to be able to represent what you want

[humble lantern]

domain: range of values a function can take
range: range of values a function can output

[sturdy rune]

just do the biggest x value + 2 as the right-most point on the x axis

[loud cedar]

Uh guyss

I think I did it wrong

Am I gonna plot the 0, 0??

[humble lantern]

yeah

lower point should be (-1, -1)

[thin bridge]

is (0,0) in your table?

[loud cedar]

yes

[humble lantern]

yes

[loud cedar]

What I did not saw it

[thin bridge]

then yes, if its part of your table you plot it
(0,-2) however, is not

[humble lantern]

looks correct

[safe radish]

@loud cedar Has your question been resolved?

[safe radish]

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[analog heath]

how to find length of a function between 2 points?

[median vigil]

https://tutorial.math.lamar.edu/Classes/CalcII/ArcLength.aspx

[analog heath]

(some russian error)

[median vigil]

to find the arc length of the curve $y = f(x)$ between $x = a$ and $x = b$ we use the formula [ L = \int_a^b \sqrt{1 + [f'(x)]^2} \odif x ]

[flat frigate]

cloud

[plucky elk]

VPN is your friend

[analog heath]

thanks

!done

[safe radish]

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[analog heath]

.close

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[analog heath]

.reopen

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✅

[analog heath]

is this is correct?

[median vigil]

the derivative was not squared correctly

[analog heath]

what is this weird brackets?

[median vigil]

the brackets are square brackets [] here, the screenshot is just low quality

[analog heath]

oh

how square this correctly?

[median vigil]

we know the derivative is (2x) so to simplify ((2x)^2) we need to apply the square to the entire thing

[flat frigate]

cloud

[analog heath]

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[verbal falcon]

ok so f(x+h) is (x+h)^2 -7x + 6

so I have (x+h)^2 -7x +6 -x^2-7x+6 / h

im not sure what to do next

[terse lichen]

-7(x+h)

[terse lichen]

use brackets, the signs for -f(x) are wrong

[verbal falcon]

there is no -f

[terse lichen]

you are doing f(x+h) - f(x)

and you are not using brackets when doing -f(x)

could cause sign problems if you are not careful

[verbal falcon]

oh so I have to destribute the negative

[terse lichen]

yes

[frozen marlin]

lmao

[junior smelt]

remove the $ after \to

[flat frigate]

ren

[frozen marlin]

thanks chart

[brave wolf]

Also there is a small error in here, it should be -7(x+h) instead of -7x

[verbal falcon]

so I get (x+h)^2 - x^2 -7h

[frozen marlin]

yes

divided by h ofc

[verbal falcon]

yea yeah

i simplfiied and got h+2x-7 , is that it

[brave wolf]

that's correct, but there is still a limit in front of that

[visual pebble]

h tends to zero

[verbal falcon]

is that the final answer? or am I supposed to plug in 0

[visual pebble]

plug in the zero

[verbal falcon]

so it's just 2x-7

[flat frigate]

MæthIsAlwaysRight

[verbal falcon]

Thanks yall

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[pliant jolt]

I'm tryna prove the gaussian integral value using Feynman's trick with the parameter I(t)=integral of e^-tx^2

[pliant jolt]

And I'm ending up with an ode in terms of I' and I

But I'm getting I(1).= 1

If only the first integration by parts bit didn't evaluate to 0 we might be able to trig sub

😔😔😔

[faint seal]

this looks like a pretty interesting idea honestly

the second integral won't converge though

$I'(t)=\int_0^\infty -x^2e^{-tx^2}\dd{x}$

[flat frigate]

kheerii

[faint seal]

you can't really differentiate like that when the limits are infinity

notice that condition -inf < a(x), b(x) < inf

[pliant jolt]

How come this trick works for stuff like sinx/x then

Where limits are 0 to infinity

[pliant jolt]

Why not surely the exponential will outgrow the x^2

[safe radish]

@pliant jolt Has your question been resolved?

[quick plinth]

Regardless
Firstly u need a +C
Secondly defining I(t) that way makes no progress, cuz obviously I(t)=I(1)/sqrt(t) from a simple u-sub
U need to know what I(t0) is for some value t0, which ends up back where u start in evaluating the gaussian integral

[pliant jolt]

I(0) diverges

I guess it's because they're both definite integrals

Why would I need a constant of integration

Both I' and I are definite

[safe radish]

@pliant jolt Has your question been resolved?

[safe radish]

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[devout palm]

how do i begin conditioning

[jaunty owl]

wsg tqd man

[devout palm]

wsg

i need to nail conditional prob

[jaunty owl]

which problem is this

i want to submit a sol

before i start yapping

[devout palm]

quantable

[jaunty owl]

yes

do u know the title

[devout palm]

Red Cube III

do u have sol

[jaunty owl]

i do have it

but i didn’t check

[devout palm]

u wanna know answer

[jaunty owl]

nah it’s ok let me think for a sec

[devout palm]

ok

[jaunty owl]

wait you wanted to know what to condition on?

[devout palm]

yes

how do i think about this

[jaunty owl]

well we just want P(6th face is white | 5 of the known faces are white)

did you get that part

[devout palm]

i just set up an lotp

how did u write formula for smaller cube

[jaunty owl]

smaller cube has (n-2)^3

unit cubes

[devout palm]

how do u know

[jaunty owl]

i know from comp math but u can visualize it with the 3x3 cube

smaller cases is the way

[devout palm]

so is that removing the exterior

and leaving interior cube

[jaunty owl]

yeah

[devout palm]

okay

keep going

[jaunty owl]

yeah so basically i just do P(All 6 are white)/(At least 5 faces are white)

or something like that because you have to consider for the cubes with 5 white faces, there is only a 1/6 chance that the red face is down

i’m some what using expectation

Like i counted outcomes tbh

i didn’t calculate probability

so like there’s (n-2)^3 cubes that satisfy the numerator

do you see that part

[devout palm]

no

[jaunty owl]

oh ok

like it’s the thing we were talking about earlier

removing the exterior

we only want the inner cube

[devout palm]

can u draw

it

[jaunty owl]

I TRIED ASKING GOT

GPT

DOES THIS HELP I DOUBT IT

issue is i’m not artist

but like if you want to reason it then

let’s take the height for example, the inner cube would remove one block from the top and one block from the bottom making the new height (n-2) and the same reasoning applies for the other dimensions

[devout palm]

n-2 is half i see

keep going

[jaunty owl]

is half?

or you somewhat get it

[devout palm]

is that what n-2 is is

like its just removing the faces

like first layer

[jaunty owl]

yeah

do u get it yet

[devout palm]

sort of

go on

[jaunty owl]

yeah ok

so actually let’s do it the conditional way

so we want P(all 6 faces are white) in the numerator

and there are (n-2)^3 unit cubes that satisfy that

and there are n^3 total cubes

so the probability is (n-2)^3 / n^3

[devout palm]

okay

[jaunty owl]

ok

so

you get all this so far right

if not lmk

[safe radish]

@devout palm Has your question been resolved?

[safe radish]

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[warped epoch]

I don't know why the math problem wants 4 as the answer when it only has a slope of -1/2 and it's y is 2

[waxen inlet]

You want to get 0 when you put in 8

-1/2x + 2 gives you -2 when you put in 8

Its y value is 2 when x=4, the y value we add to x's coefficient is the line's y value at x=0
If you continue that line all the way till x=0, you see that it's 4 at x=0

[warped epoch]

what? I didn't put in 8. I put in 4 since the y=2 I would need 2 to get to 1/2 since it goes down 1 and right 2 every 1 in the formula and I only need to go down 2 times to het to 0

[waxen inlet]

By that I meant if you put in 8 in place of x you're gonna have to get 0, like the diagram

Or is simpler terms f(8) = 0

[warped epoch]

I'm not getting something here since just by putting in 1 I already get to 6 here

[waxen inlet]

6(-1/2) + 2 = -1

[warped epoch]

+4 is the right answer but if I had plus 4 the problem would star up here

[waxen inlet]

(4, 4) is not on the line though?

[warped epoch]

yeah that's why I'm confused since 1/2(x) is the slope and +4 is the y

[waxen inlet]

Oh I see where you're confusing it

The +4 is the line's y value at x=0

Basically f(0) = 4

The line doesn't start at 0 in your diagram, but if we continue drawing it with the same slope, at x=0 we'll get 4

[warped epoch]

ahhh, I think I'm getting it now maybe let me see here

[waxen inlet]

Another example would be the other line in your diagram, it's y= -2x + 4 because f(0) = -2(0) + 4 = 4 (ik there's a hollow hole there but you get the idea)

[warped epoch]

Evidently I still don't understand

[waxen inlet]

But if we cut your line and just draw the part after let's say 1, its formula won't change

[warped epoch]

I'll reread this here

[waxen inlet]

Take ur time, let's do it step by step?

There are 2 lines in the diagram, one starts at 0 and ends at 3, the second one starts at 3 and ends at 8

Right?

[warped epoch]

sure, so I have my cords at (3,3) and the line ends at (0,8) so that means the slope has to be-3/5 since the x has a difference of 5 and it does down 3

[waxen inlet]

My internet went out sorry lmao

[warped epoch]

(don't sweat it)

so then I need to find my + #

[waxen inlet]

Yea u got the slope right

[warped epoch]

Now I think it is 3 since the y is (3,3) as it states there

, but it is not it

[waxen inlet]

Yes, that + # is the function's y value at 0, so imagine if we draw that line all the way to x=0

At x=0 we're definitely gonna get a y value higher than 3, right?

[warped epoch]

indeed

[waxen inlet]

Yup

Now we already know (3, 3) is a point on that line

So let's do something, we'll call the + # we don't know something like b

So we have -3/5x + b

Does it make sense so far?

[warped epoch]

yes, since b is the y in this case

[waxen inlet]

Yup

So if we put 3 in this function, which is -3/5x + b, we'll get 3

So -3/5(3) + b = 3

[warped epoch]

does it? wouldn't it be -9/5

[waxen inlet]

We're adding b here too, we agreed to write the function as -3/5x + b

So putting in 3 will give us -9/5 + b, and that equals 3

[warped epoch]

I do not understand because we are multiplying by 3 and that mean anything on the other side of the quation will equal 0 since -3/5(3) + b = 3(0)

okay

so

[waxen inlet]

The equation isn't = 0 though,
We write a line like y = ax + b
When we have a set of coordinates, like (3, 3), we can put in these numbers in place of x and y

[thorn pier]

b= 24/5

plug in slope, x coordinate, and y coordinate to y=mx+b and you'll have
3= (-3/5)(3)+b

3= (-3/5)(3)+b
goes into
3=(-9/5)+b

3 is 15/5

[waxen inlet]

We're trying to explain how we get to that man, that's the idea lmao

[thorn pier]

(15/5)+(9/5)=24/5

[thorn pier]

wdym?

[warped epoch]

hmm?

oaky

I'm trying to understand how to got the answer

[thorn pier]

alright see the graphed line on the right?

[warped epoch]

so I can do feature problems, but I'm not seeing it still

[thorn pier]

tell me the coordinate provided on the graph on the right

[waxen inlet]

Let's do an example, yk the line on the left is -x + 3, right?

[warped epoch]

indeed

[thorn pier]

scp do you know how to calculate slope?

[warped epoch]

it does down one for every right 1 is goes

y=mx+b?

[waxen inlet]

So -1 is the slope, and 3 is the line's y coordinate the line is at x = 0

[thorn pier]

that's not the slope calculation, but it involves it

calculating any slope requires two points on the linear graph

aka you need two plotted points on 1 straight line

[waxen inlet]

They know how to find the slope, they're struggling with the +b part

[thorn pier]

(im using the line on the right) clearly on the right you can see (3,3) and (0,8)
(sorry nima for any interruption)

alright let me define each of the letters

[warped epoch]

yes it starts at (0,3)

[thorn pier]

y is the y-cord
x is the x-cord
m is slope
y is always y-intercept

[thorn pier]

yes and another point on the line that's on the line is (0,8)

[waxen inlet]

Yup, in other words f(0) = 3, or like nick said, 3 is the y intercept

[thorn pier]

it doesn't matter which point you dub as point 1 or point 2

but the slope calculation is (y2-y1)/(x2-x1)

if we pretend (3,3) is point 1 then x1 is 3 and y1 is 3

then (8,0) is point 2

plugging that in, you get: (0-3)/(8-3)

simplifying into -3/5

(for the line on the right)

[warped epoch]

yeah I got that already

[thorn pier]

alright perfect

now from here

we have m.

[waxen inlet]

The slope is -3/5...

[thorn pier]

wait did i type the numbers in the wrong places

[warped epoch]

aaa too many cooks in the kitchen

okay

[thorn pier]

what the sigma

[waxen inlet]

Lmaoo yeah sorry bout that

[thorn pier]

oh wait whenever i said (0,8) i meant (8,0) 😭

dont worry scp

let me fix that

[thorn pier]

i edited these messages

alright

now we understand how to calculate m

and we can use any point for y and x in the equation y=mx+b

we have 3 of the 4 variables, and now we can solve for b, which like I said earlier: is the y-intercept

you following me?

[warped epoch]

yyyes

[thorn pier]

alright from HERE... if you plug any x and y coordinate (im gonna use (3,3)) in then youll get the same answer

3= (-3/5)(3)+b

[warped epoch]

why is the Y = 3 when you just put 3 in the x variable?

[thorn pier]

we need to solve for b before leaving y as a letter

[warped epoch]

where did y go then?

[thorn pier]

y=mx+b
3= (-3/5)(3)+b

y is 3

im borrowing any coordinate on the line

in this case im using (3,3)

[warped epoch]

(3,3) isn't on the line, it is empty

never mind

wait no

[thorn pier]

then ill use (8,0)

[thorn pier]

youll get the same answer regardless

0=(-3/5)(8)+b

using (8,0)

[waxen inlet]

(it is empty, but realistically you'll be fine, u can also use (8, 0) )

[warped epoch]

-24/5

[thorn pier]

that goes into -24/5 as well

yup

now you know where that line would "hypothetically" cross the y-axis

i say hypothetically bc clearly the line doesn't continue

[warped epoch]

so like this?

[thorn pier]

not over 25

[warped epoch]

sorry typo

just -24/5

[waxen inlet]

And +24/5

[thorn pier]

+??

nima wdym

[waxen inlet]

b - 24/5 = 0
b = 24/5

[warped epoch]

ah right

[thorn pier]

oh yes

+++++++

[warped epoch]

since it is the first quatrent

[thorn pier]

^^

[warped epoch]

so I still do not understand. 24/5 is close to just 5

are we finding the y intercept?

[thorn pier]

yeah

math doesnt care how close 0.999999 is to 1

[warped epoch]

no I get that

[thorn pier]

if the computer asks for a fraction just put the fraction yk?

[thorn pier]

yeah 24/5 is the y intercept

[warped epoch]

I don't understand why I would need to find the y intercept here

[thorn pier]

because the equation of a linear (straight line) graph is y=mx+b

[warped epoch]

since the line only goes a cut length

[thorn pier]

the whole +b part helps us know where to place the slope at

[waxen inlet]

You can think of functions as machines, you give them a x value and they'll modify it and give you a y value

In your case, f(x) = mx + b
To find out what this function actually does to your x value, u gotta find m and b

[warped epoch]

yeah and the slope starts at (3,3)

I mean I can clearly see that I am wrong

, but just don't know how

[waxen inlet]

Here you know the slope is -3/5, and if you give this function 3, it'll give you 3

So f(3) = 3

[warped epoch]

oh yeah

[thorn pier]

look at this graph scp

yellow line KNOWS where to cross the y-axis and therefore knows where to be placed

blue and red have no clue where to cross the y-axis even though they were given a slope

so they dont even know where to be drawn

[warped epoch]

wa

[thorn pier]

when you have your slope and y-intercept you know exactly how to draw a line

[waxen inlet]

So f(x) = -3/5x + b
So f(3) = -3/5(3) + b = 3
This is how we find b

And b just happens to be the y intercept in every line

[thorn pier]

when you only have your slope and no y-intercept b then you cant pinpoint where to draw the sloped line

think of m(slope) as "how steep"
think of b(y-int) as "where from" or "where at" or literally just where

[warped epoch]

I'm just going to do the next problem and see if I get it

[thorn pier]

alright

remember how to calculate slope @ me if you need me

[warped epoch]

@waxen inlet Okay I have gotten the right answer the first time

thank you for helping me!

[safe radish]

@warped epoch Has your question been resolved?

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[solid thorn]

!status

[safe radish]

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

[solid thorn]

1

[solid thorn]

<@&286206848099549185> can someone help

I got 2 more questions that i want help in too

so if someone can stay here please help

[safe radish]

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.

@solid thorn Has your question been resolved?

[solid thorn]

<@&286206848099549185> can yall help

[neat yew]

1. The triangles are similar (try to prove why, and find the ratio)

[solid thorn]

<EFD=<CFB

<FED=<FBC

Done

Ratio is 1/2

[solid thorn]

but wtf am i supposed to do with that now

[neat yew]

Ratio of perimeters is the same...

[solid thorn]

Wait really?

Damn i didnt know that

[neat yew]

Check by definition of similarity and by definition of perimeter

[solid thorn]

Thanks, ill try to do 2 and if i cant ill tag you are you ok with that?

[neat yew]

Ok

[solid thorn]

So @neat yew im not sure but i think i solved 2, so basically we know ratio DF/FC is 1/2 ill say DF is x FC is 2x and EF is y BF is 2y. So they tell us in 2 there are new midpoints which make them be cut in half so that means FN and FE are y and DF and FM are x

so to prove a paralelogram you have to prove that the diagonals are half of each other

[solid thorn]

Did I do it correctly?

[neat yew]

Yes I think

[solid thorn]

Ok, so can you help in another question?

[neat yew]

Yes

[solid thorn]

Ok im sending

[neat yew]

ABE and GCE are similar with coefficient a/b

So...

[solid thorn]

Wait how are they similar

I see GEC and BEA

But what else?

BC is parallel to AD sure but that doesnt help in GCE or does it?

[neat yew]

AB parallel to CD

[solid thorn]

Yes and what angles are similar?

From that

[neat yew]

A = G

[solid thorn]

Right

[solid thorn]

So hold on ill write it down rq

[solid thorn]

So it should be a²/b²?

Oh no it should be

a²*s/b²

since the triangles are similar

Right?

[solid thorn]

Or am i wrong

[neat yew]

Yes , squared

[solid thorn]

So in ADG do we have to make trianlges GEC and GAD similar?

[neat yew]

Yes

[solid thorn]

Sure ill do it and then tell you what i got

[solid thorn]

Ok, so i cant really do much with these ratios so i thought maybe we can prove ADG is similar to ABE?

ill try to do it and tell you what i got

Ok so @neat yew i dont really get it since i tired to do similarities and didnt get anything, can you help?

[neat yew]

AD = BE + EC

[solid thorn]

Ok yeah

Now what do we do

[neat yew]

AD/EC = 1 + a/b

[solid thorn]

AD/EC?

Oh i get it

yeah

[solid thorn]

Why the 1 tho?

[neat yew]

Divided this by EC

[solid thorn]

Wdym by that

[solid thorn]

Shouldnt AD/EC be a+b/b?

oh

i get it

Wait so the area of ADG is (a+b)²*s/b²?

[solid thorn]

Yes thats what it says in answers

[solid thorn]

Ok so i will try to do 3 and ill tag you incase i dont know

Ok @neat yew so ive done 3, and now im sending the last problem

Here @neat yew

Yo you here?

[neat yew]

Let AD=x, KN = y

Then area of AKL is xy

Area of ABC is 6h

They're similar with ratio 6/(2y) = 3/y

So 6h = 9/y^2 xy = 9x/y

[solid thorn]

Can you explain about how you got to the areas?

[neat yew]

1/2 base × height

[solid thorn]

0.5y*x?

[neat yew]

Base is KL = 2KN = 2y

[solid thorn]

Right

Area of abc should be

[neat yew]

3h yes

[solid thorn]

BC*H/2?

[neat yew]

Yes

[solid thorn]

Okay and how did you get to 6h?

[neat yew]

BC = 6

Area is 3h

[solid thorn]

ohhhh yeah

Yep

AKL and ABC are similar

[solid thorn]

Wait can you explain how are they similar?

[neat yew]

KL BC parallel

[solid thorn]

Why?

Oh

because

KL parallel to bc

to NM

Sure

[neat yew]

Actually there's a simpler solution..

Since triangles are similar, ...

[solid thorn]

yeah

[neat yew]

KL/BC = AD/AE

[solid thorn]

2Y/6

x/h