help-13

[junior timber]

so

16 - x^2 + 25 - y^2

[runic garnet]

That’s not an equation

U have to make 1 equation

Equate the two values of h^2

[junior timber]

oh sorry

wait

16 - x^2 = 25 - y^2

i menat this

[runic garnet]

Nice

Now we have two equations with x and y in them

16 - x^2 = 25 - y^2

x + y = 5

Solve for x and y

[junior timber]

okk

y = 5 - x

so then

16 - x^2 = 25 - (5-x)^2

[runic garnet]

Keep going, ur on the right track 👍

[junior timber]

ohh

its a fraction tho

[runic garnet]

That’s fine

What did u get

[junior timber]

its 8/5

[runic garnet]

Nice!!!

Solve for y now

[junior timber]

or 1.6

[runic garnet]

x + y = 5

Yes that’s correct

[junior timber]

okkk

so then

5 - 1.6

[runic garnet]

=?

[junior timber]

3.4

[runic garnet]

https://tenor.com/view/happy-dance-way-to-go-gif-24438686

Ok so now we have x and y

[junior timber]

yes

then pythagorean

[runic garnet]

Yes

Solve for h

[junior timber]

okk

[runic garnet]

Then solve for area

[junior timber]

okok tytyy

[runic garnet]

See what happens when u trust the process

Np

[junior timber]

.close

[cedar kiln]

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[granite eagle]

[cedar kiln]

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[supple forum]

how is this solved? didnt come across anything similar yet

[red pumice]

taylor series most likely

[mighty drift]

it says deduce, show what's before it

[supple forum]

thats why

we didnt take taylor and maclaurin yet

.close

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[oak imp]

first degree inequalities with two unknowns

[oak imp]

if i do a inequality with 2 unknows

do the graphic

and check the inequation if it corresponds or not

how do i know if the solution is the upper part of the graphic or the other one

here i got confused about the solution

[granite eagle]

do u have a specific problem?

[oak imp]

yh idk why it isn't uploading

[granite eagle]

well assuming ur doing all this in a cartesian plane;

just take a point thats distinctly in one one those areas, check...

[granite eagle]

theres a more intuitive idea but cant really explain it without seeing the graph myself

[oak imp]

i got confused here

[granite eagle]

ok first off express that in terms of y

[oak imp]

i took of the x

and putted values to the y

[granite eagle]

ok

5x-6y less than 0

[oak imp]

and checked with 0,0 points

[granite eagle]

"thats distinctly in one one those areas,"

0,0 wont do

i'll try explaining once

[granite eagle]

lets just ignore the equal to part it just means the boundry line is a valid set of points too

if 5x-6y is less than 0

5x must be less than 6y

5x/6 must be less than y

now think of a extreme point

say 0,100

5*0/6 =0

its certainly less than 100

[oak imp]

so like

when putting the value

[granite eagle]

yes 0,100 and by extension the values above the line are valid

just exclude the line

[oak imp]

oh so this with every inequality?

[granite eagle]

sure

[oak imp]

i pick one value of the other line and one from other

and check

if it's true

oh kk

tysm

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[desert zephyr]

P (A) ∩ P (B) = P (A ∩ B)

how do i prove thi

[desert zephyr]

s

[grand shale]

what

[desert zephyr]

how do i prove that
P (A) ∩ P (B) = P (A ∩ B)

[foggy merlin]

what is P

[tropic oxide]

^

[grand shale]

your left hand side is a set and right hand side is a number , if p means probability

[foggy merlin]

it cant be a probability anyway

[grand shale]

p sure op was referring to probability

hence i said "what"

[desert zephyr]

parts

[foggy merlin]

what ?

[desert zephyr]

parts of a set

this

[tropic oxide]

this is called the power set.

but the "parts" you are talking about are properly called subsets.

[desert zephyr]

oh ok in my nat lang its called parts of a set

but yeah P(A) means subsets of A

[tropic oxide]

what is your native language?

[cedar kiln]

@desert zephyr Has your question been resolved?

[desert zephyr]

portuguese

[dusk willow]

What have u tried yet

[desert zephyr]

I tried this

what u think?

im not sure abt the first move

but i came up with this

@dusk willow

[cedar kiln]

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[rich haven]

How do I rewrite this using z

[cedar kiln]

@rich haven Has your question been resolved?

[cedar kiln]

@rich haven Has your question been resolved?

[cedar kiln]

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[solid latch]

Hey, how would I prove this?

[cerulean sail]

alt: complete the square

[livid hound]

*discriminant

[red pumice]

oh lord

[solid latch]

I’ve gotten here so far @red pumice

I got it

.close

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[river topaz]

how do i answer these two questions?

[gentle flower]

!status

[cedar kiln]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[river topaz]

1

for both of them

[zenith stag]

for the first one, what are the width, height and length

[river topaz]

ah, i think i understand what you mean

so 5 times 3 times 3

?

what abotu the second one tho

[gentle flower]

what is the surface area of a cuboid?

[river topaz]

height times width times length

or is that volume

[zenith stag]

that's volume

[gentle flower]

how many surfaces does a cuboid have?

[river topaz]

6

[gentle flower]

what is the area of each face?

[river topaz]

legnth times width

length

is it length times width times 6?

[gentle flower]

what will be the formula for your surface area then

[river topaz]

LW times 6

?

[gentle flower]

is each face the same length and width?

[river topaz]

ok wait

length times width times 4 +

length and height times 2

because 4 rectangles and 2 cuboids

?

[gentle flower]

2lw + 2lh + 2hw

[river topaz]

NO WAIT

oh

[gentle flower]

what values are given to you?

[river topaz]

5 for width

3 for length and height

so it would be

[gentle flower]

plug those in

[river topaz]

30 + 18 + 30

78

right?

yes yes yess

[gentle flower]

length is 5 width and height are 3

[river topaz]

right

[gentle flower]

yes

[river topaz]

so its 78?

[gentle flower]

let me calculate it wair

[river topaz]

wait how is length 5

width is 5

i feel stupid

[river topaz]

alright hold on let me redo it then

[gentle flower]

2(5)(3) + 2(5)(3) + 2(3)(3)

[river topaz]

it ends up the same

78

am i right?

8?

78?

[gentle flower]

30 + 30 + 18 = 78

correct

[river topaz]

alr awesome thanks alot

do you by any chance know how to do the second one

i figured out how to do the surface area but not the volume

[lofty flame]

I can help just give me a sec

[gentle flower]

do you know the formula for the volume of a triangular prism?

[river topaz]

i forgot it

hold on

[lofty flame]

1/2 bh times L

(1/2) b x h x L

[river topaz]

what is bx supposed to be

oh

alr wait time to use my brain

i can figure it out give me a hot second

[gentle flower]

1/2 bh is the area of a ____?

[river topaz]

trianglle

but just wait give me a second

i have the question i just need to find the measurements

alr i have no clue

all i know is the square base at the bottom is 30.59 cm

not sure how that helps me tho

[zenith stag]

The volume of a prism = base area × height

The base area in this case would be the area of the triangle as there is another triangle parallel to it

and the height is CF

[river topaz]

hold on which triangle

is the base

there are multiple

[zenith stag]

any

they have the same dimensions

let's consider triangle ABC for now

how would you find its area

[river topaz]

12.1 times 13.3

divided by 2

[zenith stag]

not quite

13.3 is the hypotenuse

[river topaz]

so it equals 12.1 squared plus the other side squared

right?

[zenith stag]

yea

[spiral reef]

Prism volume (V) = B × h, where, B is the area of the base and h is the height of the prism.

[river topaz]

i tried doing that but

it ended up being a super specific decimal

[zenith stag]

oh wait yea, i messed up

[river topaz]

like 5.00000025162162612

[zenith stag]

it's 3 am here

[spiral reef]

No problem haha it happens to all of us

[zenith stag]

let's just say it's 5

so you can find the area now

[river topaz]

5 times 13.3 right?

[zenith stag]

base x height

/2

and 13.3 is the hypotenuse

[river topaz]

13.3 is the base

so its not the base?

man my brain is sort of turned off atm

[zenith stag]

the base and the height are perpendicular

[river topaz]

so

its

not

the base?

[zenith stag]

yea it's not

[river topaz]

OH WAIT IT CLICKED IN MY HEAD WHAT THAT MEANS

OK SO

THE BASE AND THE HEIGHT

HAVE TO JOIN TOGETHER

TO MAKE A 90 DEGREE ANGEL

ANGLE

?

right?

[zenith stag]

yea

YEA

[river topaz]

EYAAAA

IM SO SMART

[zenith stag]

yes you are

[river topaz]

ok ok wait wait

so it would be 12.1 times 5

[zenith stag]

ok so now u can find the ans using the values

[river topaz]

so 60.5

divided by 2

30.25

thats the area of the triangle

[zenith stag]

yew

[river topaz]

volume of prism is

[zenith stag]

yea*

[river topaz]

base times height

so 30.25 times

whats the height elt me check

2.3 cm

[zenith stag]

2.3

[river topaz]

69.575

THATS THE ANSWER

I FEEL THE BRAINPOWER SURGING THROUGH ME

thank you so much fo r the help

[zenith stag]

np

im gna get my prob much needed sleep now

have a good one

[runic garnet]

Wow

What a feeling

Synapses firing

[cedar kiln]

@river topaz Has your question been resolved?

[cedar kiln]

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[bold needle]

Am I wrong in believing the triangles in the illustration are not similar but congruent?

[crimson delta]

Well depends on which ones. But congruent implies similar

[bold needle]

DGE is congruent do DGF, no?

[crimson delta]

There is also DEF

[bold needle]

Well.. yes actually

[crimson delta]

But there is nothing wrong with calling triangles similar even if they are more than just similar

[bold needle]

They're similar because although they're different in size their angles are equal, right?

It's just DEF would be much larger than DEG

[crimson delta]

Similar means equal angles, yes

Congruent means similar and same size

[bold needle]

Right

And then, if segment EG = 2 and segment EF = 8, what steps would I have to take to find the length of segment ED

[crimson delta]

Wait what? Those aren't supposed to be right angles at G?

[bold needle]

what?

[crimson delta]

Those lengths don't fit at all with the sketch. I hate sketches that are drawn as if certain angles are right angles even if they aren't necessarily

[bold needle]

Yeah the lengths are not accurate

It's just like filler numbers

That's how most of it is layered the inaccuracy is tricky

[bold needle]

But even then, how would I go about finding the length of the segment ED

[crimson delta]

Without more information this is then impossible

[bold needle]

That's where I get lost

[crimson delta]

I think

[bold needle]

Well D is a right angle

90 degrees

[crimson delta]

Not enough

Do you know thales?

[bold needle]

Do I know thales?? 🤣 no..

[crimson delta]

The theorem

[bold needle]

In geometry, Thales's theorem states that if A, B, and C are distinct points on a circle where the line AC is a diameter, the angle ∠ ABC is a right angle.

[crimson delta]

Yes google

[bold needle]

But that's when you're working with a circle?

LOL

[crimson delta]

Imagine a circle with diameter EF

D could be anywhere on the circle

[bold needle]

ok Im imagining

[crimson delta]

And it would give a right angle

And the length of ED could be essentially anything

[cedar kiln]

@bold needle Has your question been resolved?

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[dreamy zenith]

$-2x=\sqrt{{\frac{48-20x}{2}}$ solve

[dreamy zenith]

of so i square both sides to remove the root

$4x^2+10x-28=0$

[wraith dagger]

putridplanet

[dreamy zenith]

$x=-10+- \frac{\sqrt{10^2-4\cdot4\cdot-28}}{2\cdot4}$

[wraith dagger]

putridplanet

[dreamy zenith]

$x=-10+- \frac{\sqrt{100+448}}{8}$

[wraith dagger]

putridplanet

[dreamy zenith]

$x=-10+- \frac{\sqrt{548}}{8}$

now what

[wraith dagger]

putridplanet

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

[dreamy zenith]

<@&286206848099549185>

[lofty flame]

Alright let me solve

[lofty flame]

This one?

[cedar kiln]

@dreamy zenith Has your question been resolved?

[dreamy zenith]

yes

[lofty flame]

X = aprox -12.925

[cedar kiln]

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[crimson sedge]

so, i got this fun question

[crimson sedge]

<enumitem>
Let $(X,\AA)$ be a measurable space. Let $\eta: \AA \longrightarrow [0,\infty)$ be a set function such that
\begin{enumerate}[i.]
\ii $\map \eta{\emptyset} = 0$
\ii $\map \eta{X} < \infty$
\ii for all $U \subseteq V$ in $\AA$ we have $\map \eta U \le \map \eta V \le \map \eta U +\map \eta{V\setminus U}$
\end{enumerate}
Let use define [
\map{\eta^*}Z =\inf\set{\sum_i \map \eta {U_i}: \bigcup_{i=0}^\infty U_i \supsete Z}
]
for every $Z \subseteq X$

\vs{3 mm}
Show that $\map \eta U = \map {\eta^*}U$ for every $u \in \AA$

[crimson sedge]

I managed to prove that $\eta$ is finitely subadditive, but i am unsure on how to proceed

[crimson sedge]

Any ideas?

[crimson sedge]

Let us define*

[cedar kiln]

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[charred beacon]

hi! can someone help me to check if this is right? thanku!

[flat harbor]

hey mate

not sure if there is enough information here

are there other parts we should be aware of

what is h(x) and k(x)

[cerulean sail]

Assuming the context is #help-27 message

[charred beacon]

yes sorry

@flat harbor

[flat harbor]

id copy it here mate

[charred beacon]

so h x = what i said and k x is just h x dialated by 5/4

thats all the information just dialating the h (x) by 5/4

i just dont know if i did it right

[flat harbor]

ill be honest im not the right person here

bit above my paygrade in undrestanding

thought i could help

sorry mate

[charred beacon]

allg

[crimson sedge]

You just take the reciprocal of it and do $\map h {\f 45 x}$ I believe
So like

[crimson sedge]

,w simplify h(x) = ((4x/5)/800)(4/5 x-20)(4/5 x +20)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by ♡Lixera♡#5050

[crimson sedge]

So what u did seems to be right!

Although I haven't quite seen someone do the steps you did before

[cedar kiln]

@charred beacon Has your question been resolved?

[charred beacon]

OH OK THANKU!

[crimson sedge]

No problemm

[cedar kiln]

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[royal light]

can anyone help me find this as a definite integral?

[cedar kiln]

@royal light Has your question been resolved?

[drifting marlin]

have you rewritten it in the requested notation?

[royal light]

no thats the part i want help with

i can solve the integral

[drifting marlin]

alright, what does a riemann sum look like in general?

[royal light]

like this

well the LHS

[drifting marlin]

alright, now identify your f(x) and your delta x

[royal light]

i understand 2/n is my delta x

but i dont understand the f(x)

ik that xi = a + i(delta x)

but theres no i values

@drifting marlin how would i know f(xi)

[drifting marlin]

there are "i values," they just don't have i in this form

[royal light]

i understand in xi = a + i(delta x) the a is 1

[drifting marlin]

remember i is nothing more than the index of summation

[royal light]

ok got it

how am i supposed to get f(x)

[drifting marlin]

take a look at the sum

your i*delta x is (hopefully obviously) present

[royal light]

i see it ya

2/n + 1 is the first thing

[drifting marlin]

alright, so what do you think is your f(x_i)

[royal light]

f(2/n + 1)?

[drifting marlin]

ok but what's f?

[royal light]

that i dont know

[drifting marlin]

when i = 1, f(x_i) = 2/n + 1
when 1 = 2, f(x_i) = 4/n + 1

figure out the pattern

[royal light]

the numerator of 2/n gets mutiplied by whatever is

i is

[drifting marlin]

so what's f(x_i) for a general i?

[royal light]

2i/n + 1?

[drifting marlin]

there we go

[royal light]

so f(x) = 2i/n + 1

or thats f(xi)?

[drifting marlin]

f(x_i)

[royal light]

so then whats f(x)

[drifting marlin]

something that won't contain i

[royal light]

so just 2/n + 1

[drifting marlin]

no

there will be no n either

recall what the definition of x_i is, and see if you can rewrite f(x_i) in terms of x_i instead of i and n

[royal light]

how do i find f(x) from f(xi)

[drifting marlin]

by replacing x_i with x

[royal light]

so f(x) = 2i/n + 1

[drifting marlin]

again, no

[royal light]

how do i get a fucntion that doesnt have n or i in it tho

[drifting marlin]

why haven't you done this?

[royal light]

so 2xi/xi + 1

[drifting marlin]

no

what is x_i?

[royal light]

xi = 1 + 2i/n?

or 2/n + 1

[drifting marlin]

there needs to be an i

so the first one

[royal light]

ok

[drifting marlin]

so then what is f(x_i), in terms of x_i and not i and n?

[royal light]

what do i replace i and n with

[drifting marlin]

what do you mean replace?

you're aggressively overthinking this

you've previously found f(x_i) = 2i/n+1

you've also identified that x_i = 2i/n + 1

so then what is f(x_i)? (do not say 2i/n+1, say something that contains x_i)

[royal light]

oh

(1 + xi)?

[drifting marlin]

no

look a little more closely

[royal light]

i dont understand what the function surronding xi is

[drifting marlin]

it's not that deep

f(x_i) = 2i/n+1
x_i = 2i/n+1

[royal light]

f is 1

[drifting marlin]

what is f(x_i) in terms of x_i?

f is not 1

[royal light]

there the same

[drifting marlin]

they are the same

so what is f(x_i)?

[royal light]

xi

[drifting marlin]

so what is f(x)?

[royal light]

the same as x?

[drifting marlin]

so f(x) =

[royal light]

x?

[drifting marlin]

yes

you've identified a already, now identify b and you can evaluate the integral

[royal light]

ok

[royal light]

@drifting marlin is this how i would evaluate the integral?

[drifting marlin]

yes

[royal light]

ok thank you

.close

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[vagrant hamlet]

how do i find c

[tropic oxide]

your graph depicts 1 period of the cosine wave, and c is the length of said period.

[vagrant hamlet]

so i sub y = 4 into the equation

but i cant get the answer

i only got 0

i mean

0

[livid hound]

wdym you only got 0

oh

your graph depicts 1 period of the cosine wave, and c is the length of said period.
try identifying the period of the cosine wave from the equation

[cedar kiln]

@vagrant hamlet Has your question been resolved?

[vagrant hamlet]

okay

the period is 1/2

so 2pi x 1/2 is pi

[livid hound]

the period is 1/2
no

[vagrant hamlet]

waht

uh

HU

1/2 x

[livid hound]

the coefficient of x isn't the period

[vagrant hamlet]

huh

then what

[livid hound]

but you can use that to get the period

i'd recommend looking up period of a trig function

[vagrant hamlet]

waht

cos 1/2

period?

[livid hound]

no

stop guessing

i'd recommend looking up period of a trig function

[vagrant hamlet]

4pi

the answer

4pi is the period

[livid hound]

yes

[vagrant hamlet]

yes

now what

[livid hound]

your graph depicts 1 period of the cosine wave, and c is the length of said period.
you've found the period,
which means you've just found the value of c

[vagrant hamlet]

hu

wha

Wow

anazing

thank you

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[sick ledge]

Hello! I wanted to ask a question regarding this problem

[sick ledge]

Does the x=0 case make the whole statement invalid ?

[south tundra]

1. x here can be 0
2. Why are you considering a choice of x though?

Oh, wait, yeah

[sick ledge]

I want to consider a choice of x here because we have two cases

[south tundra]

But you should say it's the y = 0 case

[sick ledge]

The non zero case and the zero case

What should I put in the final answer then ? Sorry I think I might still be confused

Because I’m one case the statement is false and the other is true

In*

[south tundra]

The statement is false because for y = 0 there is x such that y = x^2 (it's x = 0), but y would not be positive in that case

[sick ledge]

They makes more sense because if you write the truth table out P is true and Q is false which means p-> q is false

Yeah

One last thing, to write the inverse of the statement, would I need to invert everything, including the = sign to not equal to? I know to write an inverse statement you have to write the negation of both sides of the statement

[cedar kiln]

@sick ledge Has your question been resolved?

[sick ledge]

Done

.close

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[crimson sedge]

can line of reflection vary depending on an object and an image or line of reflection is always in b/w obejct and image?

[short blade]

well you could have the image be on the line

[crimson sedge]

im talking about line of reflection

[short blade]

can you clarify your question

[crimson sedge]

i am not able to understand what is line of reflection

[plain plover]

Well, it's maths so yh ...
If we we're to talk bout image formed in Physics in lens or mirror then the image might even be on the side of the object

[short blade]

well no, the line doesn't have to be between the image and object

the image and object could be on the line

[plain plover]

I mean yh that's true

But i think he meant as in both being on same side

[crimson sedge]

im trying to understand line of reflection

[short blade]

if you reflect a line , it is possible to have part of the image on the same side as part of the original line

[short blade]

it's what we reflect over

take the original point, and draw a line directly to the line of reflection

it should make a right angle with the line of reflection

then continue that line over the line of reflection to be the same distance as the first half (i'll draw an image in a sec)

that's where the reflection lies

[crimson sedge]

right

[short blade]

the red line is the line of reflection

think of the left as the original object

we draw a line directly to the red line from the object (this is the green line)

then we continue and draw the blue line until it is as long as the green line

that is where the image is

(the right dot)

think of it as mirroring the original object

[crimson sedge]

i understood it

thank you.

.close

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[queen horizon]

What is 5^2022 + 11^2023 mod 64

[crimson sedge]

Use mathway calculator... It will answer it

[queen horizon]

Uh i need the way not the result

[short blade]

have you learn about the totient function?

[queen horizon]

Euler?

[crimson sedge]

It also gives steps bro

[short blade]

yes

[short blade]

pls don't outright suggest using a calculator

it's not helpful

[crimson sedge]

See bro you get whatever you want in every way possible.... Saying it's not useful is something people say who have not tried it or lack the knowledge of finding and understanding the answer...

[queen horizon]

no

[short blade]

that's why people are here, to get help in finding the answer

[short blade]

so you haven't heard of the euler totient function?

[royal loom]

helpers are here to help people learn math, understand math, and know how to do their problems. Not to hand out solutions and teach people how to use calculators

[queen horizon]

yes

[crimson sedge]

I am pretty much familiar with that ... But look at its question bro ... It's lot easier to understand there .... And it's not just a calculator it's more like a solution book

[short blade]

this is the general idea

use this to compute 5^2022 mod 64 and 11^2023 mod 64 separately, then add their remainders and mod it by 64

$\phi(n)$ here is the euler totient function

[wraith dagger]

maximo

[short blade]

no

a = 2022 mod 32

b = 2023 mod 32

then compute 5^a mod 64

and 11^b mod 64

[queen horizon]

So its 5^6 mod 64 + 11^7 mod64?

[short blade]

yes

mod 64

[queen horizon]

the answer is 70?

[cedar kiln]

@queen horizon Has your question been resolved?

[cedar kiln]

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[clever basalt]

i need help with this grade 8 geometry problem

[hushed spoke]

!status

[cedar kiln]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[spare fiber]

you should approach this by focusing on triangles ABC and XAY

[clever basalt]

what next do i do?

[hushed spoke]

You have to first show us what you’ve done so far?

[clever basalt]

i dont understand the correlation between them

[hushed spoke]

Drawing it out like that, is unlikely to be very useful

Okay look at the AD/DB part and locate it on your diagram

[clever basalt]

yes?

[hushed spoke]

Now are you aware of the Basic proportionality theorem?

[clever basalt]

i am not

[hushed spoke]

It say that if DE is parallel to BC, then AD/DB = AE/EC

Does that look familiar?

[clever basalt]

oh yea that makes sense

[hushed spoke]

You have another name for this theorem?

[clever basalt]

no but we learnt something very similar

[hushed spoke]

Okay now

Can you apply this in this problem for AD/DB part

What is AD/DB equal to?

( DE|| BC )

[clever basalt]

i dont understand

ad/db is the same length as ae/ec?

[hushed spoke]

This ratio is same

Length might not be the correct term

Okay so now you have AD/DB = AE/EC, yes?

[clever basalt]

yes

[civic eagle]

maybe this helps? I found it in a pdf and there's a problem that is similar to yours in the same pdf.

[clever basalt]

thankk youu but i still have no idea how to prove it😭

[civic eagle]

through the properties of complementary angles and knowing which lines are parallel to each other, you can prove these relationships.

[clever basalt]

how?

[civic eagle]

corresponding angles and parallel lines will proportionately dissect triangles. look at triangle ABC, and see how angle ADE and angle ABC will be the same. i tried to mark what angles are the same similarly.

the double hash mark one is wrong, my b. but i found a khan video to help explain https://www.youtube.com/watch?v=L1c0SP4W4ro

[cedar kiln]

@clever basalt Has your question been resolved?

[clever basalt]

so how do i prove that ex/ey = ad/db?

because segment ex and ey and ad db do not make a triangle so how do i prove that theyre the same?

[civic eagle]

ΔAYE ~ ΔCXE. does this help?

ΔABC ~ ΔADE

[clever basalt]

but how about the ad/db?

[civic eagle]

AD/DB = AE/EC

theres still more to the problem, but do you understand the reasoning yet?

[clever basalt]

OHHH yea

[civic eagle]

theres still algebra you must do

gotta prove AD/DB = AE/EC too

[clever basalt]

how do you prove so?

[civic eagle]

ΔABC ~ ΔADE can give us more ratios to prove it.

can you figure which

[clever basalt]

no i dont know

[civic eagle]

two triangles, with angles ∠ADE = ∠ABC, ∠AED = ∠ACB. because they are "similar," in that they share at least two angles, proven by properties of parallel lines producing similar angles with triangles.

you can draw conclusions such as AD/AB = AE/AC = DE/BC

you can even break these up into more segments

AD + DB = AB and so on

we need to use algebraic reasoning to prove AD/DB = AE/EC

this isn't even the original ratio you were tasked to prove. but knowing how to prove one allows you to prove all of them.

[clever basalt]

so then how do i prove my original ratio knowing this

[civic eagle]

you take all the ratios you can prove from parallel lines and similar triangles, and you should have enough evidence to algebraically manipulate ratios to get the result you want.

EX/EY = AD/DB

AY || XC means ΔAYE ~ ΔCXE. DE || BC means ΔABC ~ ΔADE

knowing these facts, you can produce a lot of ratios

also like i said, you can break larger segments into its smaller components within these ratios

AD/AB = AD / (AD + DB)

AY || XC means ΔAYE ~ ΔCXE this fact is super important for proving the ratio you want

ΔAYE ~ ΔCXE means we can derive 3 ratios. these three ratios and their manipulation can bring you to your answer.

[clever basalt]

what are the ratios?

[civic eagle]

can you guess

[clever basalt]

no?

[civic eagle]

why not

[clever basalt]

i know that theyre similar bc they have the same paralel side nd a fixed angle but idk the ratios we can derive out of them

[civic eagle]

similar triangles are when they share two angles, which guarantees the third angle is also similar. since no matter how long the sides of a triangle are, they are governed by ratios of these angles, we can produce ratios without knowing the true lengths of the sides.

when a segment of one triangle has the same angles as a segment of another, they are similar.

similarity provides us reasoning to derive ratios. in this case DE || BC means ΔABC ~ ΔADE, which would mean AD/AB=AE/AC=DE/BC. three ratios of each corresponding segment of the triangles.

now you can try applying the same logic to the other set of similar triangles and derive their ratios.

[clever basalt]

so if AY||XC then it means that ΔAYE ~ ΔCXE and means that CX/AY = CE/AE = XE/YE?

[civic eagle]

good

now take a look at the answer we need again

how do we get there?

starting from CX/AY = CE/AE = XE/YE and AD/AB=AE/AC=DE/BC

how do we manipulate these ratios to make EX/EY = AD/DB

[clever basalt]

i dont know

[civic eagle]

CX/AY = CE/AE = XE/YE this already has one piece of the puzzle

XE/YE is there already

can we manipulate the other two ratios to be AD/DB

[clever basalt]

yes? although i dont know how

[civic eagle]

think about the identities we have here

AC = AE + EC, AB = AD + DB, XY = EY + EX

[clever basalt]

so AD = AC +CD?

[civic eagle]

CD?

[clever basalt]

oh sorry

i dont know which ones to add to make ad or db

[civic eagle]

we already know the ratio of ad and db

[clever basalt]

AB = AD+DB?

[clever basalt]

@civic eagle ?

[civic eagle]

i feel like your original problem might've been copied wrong? should be EY/EX = AD/DB not EX/EY = AD/DB

im going to bed, but look at this. i tried explaining it to you. idk what else to do. good luck.

[clever basalt]

oh thank you so much.

[cedar kiln]

@clever basalt Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[steel mica]

In a coordinate system where the origin is at the top left. how would i translate it into a coordinate system where the origin is the center.

[steel mica]

.close

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[crimson sedge]

total (T) = 20 choose 3
right triangles (R) = (20 choose 1) times 2 times 1

but the probability is not coming out to be 40/1140

[fair geyser]

i don't know how you did the right triangles, but it looks too small

[crimson sedge]

choose a vertex, pick its neighbour, connect to its opposite

[fair geyser]

why a neighbor

[crimson sedge]

im actually using an octagon as a reference which could be why its wrong

[fair geyser]

as long as you connect to the opposite, you get a right triangle

what you actually do is pick a diameter, and then any third point works

[crimson sedge]

it doesnt seem to work on an octagon

[fair geyser]

paint doesn't have an octagon shape 😦

[crimson sedge]

yeah

[crimson sedge]

oh nvm im dum

[fair geyser]

basically this is also right triangle

they all are

[crimson sedge]

yeah my drawing was terrible that i thought it was wrong lmaoo

i realized that it will be right triangle because of the umm central angle theorem

i forgot the name

subtended arc theorem?

something like that

thanks so much :)

now im getting 6/19 as answer

,w (20 * 18)/(1140)

[fair geyser]

there's 10 diameters, not 20

[wraith dagger]

Results provided by WolframAlpha

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

Requested by ! LaV.#7677

[crimson sedge]

oh yeah

.close

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[crimson sedge]

.close

[lavish plank]

question and answer. unsure on acceleration denomiantor

[cedar kiln]

@lavish plank Has your question been resolved?

[cedar kiln]

@lavish plank Has your question been resolved?

[cedar kiln]

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[tender imp]

can someone help with this question 😅

So, the idea I thought for it is that the rock has to cover 8 squares in 2 dimensions in 4 moves

so there are 3 possible cases for that

Case 1: 1 Vertical move and 3 horizontal moves
Case 2: 2 Vertical moves and 2 horizontal moves
Case 3: 3 Vertical moves and 1 Horizontal moves

[frozen valve]

I was going to say to look at cases, so 1 move up then 3 sideways, or 2 up 2 sideways, or 3 up 1 sideways

but i feel like theres another way to do it

oh you had the same idea as me

[tender imp]

yeah

[frozen valve]

well 3 horizontal and 1 are equal no?

[tender imp]

but the permutations for move was going harder

[tender imp]

yes

21

[frozen valve]

actually no...

[tender imp]

to be exact

[frozen valve]

you may have cases where one horizontal move reaches the corner

in a 3 horizontal

case

you have to exclude those

[tender imp]

like 3 horizontal and 1 vertical will have same as 1 horizontal and 3 vertical

[tender imp]

no I solved for those

[frozen valve]

well if its one vertical, you need that vertical to reach the top, but what if one of the horizontal reaches the edge, that will be a dq case

thats what i was saying

[tender imp]

like the equation will be

x + y + z = 8

[tender imp]

yes so only 1 case

[frozen valve]

are you sure?

say rook reaches a8 first move

[tender imp]

I will write my idea down on a notebook so that you will be understand it clearly

[frozen valve]

then you have x + y <= 7

21 cases that are right

for 3 horizontal

x2 for 1 horizontal

so solve for 2 horizontal 2 vertical

there are 49 cases for 2 horizontal 2 vertical

i think

[tender imp]

yes

[mighty drift]

just arrived, but how is that not an even number when you have symmetry between the directions

[tender imp]

that's where I arrived

[frozen valve]

ok well lets look how many possible ways you can cover 8 squares in 2 moves

although now that i think about it, its not 49 but 56

but ill get to that later

for horizontal you can have x and y 1 to 8

[tender imp]

we can't have zero though

[frozen valve]

yeah right

[tender imp]

yeah

[frozen valve]

but i thought about vertical first

and thought we could not have 8

actually no we cant have 8

man tricked me

right so x and y 1 to 7 right?

[tender imp]

yeah

[mighty drift]

|{(a, 8-a), a in {1, 7}}| = 7

[frozen valve]

and we need their sum to be 8, so by x, y is fixed

[frozen valve]

yeah so horizontal 7, so is vertical 7, so their product is the # of ways no?

[mighty drift]

order matters

[frozen valve]

OH nevermind

[mighty drift]

(h h v v), (h v h v) etc

[mighty drift]

that's where my symmetry argument comes in

it has to be even because of that factor

[frozen valve]

right lol, this was a brainfuck

forgive me for misleading you

I feel like we can still solve this problem by looking at cases

[mighty drift]

which is 4C2 = 6
h h v v, h v h v, h v v h and the symmetrics

[frozen valve]

3 cases include: 1 horizontal move, 2 horizontal moves, 3 horizontal moves
And by simetry, the #of ways in 1 horizontal move and 3 horizontal moves should be the same

[mighty drift]

of course you can

just better do it in a smart way that minimizes actual case counting

[frozen valve]

Right, idk what happened with me that caused me to mess up so bad

[mighty drift]

idk because I barely read anything before correcting you

[frozen valve]

Lol good

Anyways

[fair geyser]

it's just 6×6 yes?

[mighty drift]

what is "it" ?

[fair geyser]

it is the answer

[mighty drift]

I already had 7*7*6 with a specific case

[fair geyser]

yeah, you divide a move of 7

[frozen valve]

So for 2 horizontal moves, we get 7 possibilities, as well as vertical moves
So shouldnt the # of cases for 2 horizntal moves be 49*42?

[fair geyser]

and there's 6 spots where 7 can be divided

[frozen valve]

42x6x6 variations for 2 horizontal

[mighty drift]

there's this thing in chess where pieces can go backwards

[fair geyser]

oh that assumes you only go towards the goal, but you don;t have to

got it

[mighty drift]

say k in [1, 8] is the first step
Then you want (a, 8-k-a) with a in [-k, 7-k]
right

giving sum for k in [1, 8] of (8) = 64 ?

no

should be 7*6

[mighty drift]

say k in [1, 7] is the first step
Then you want (a, 7-k-a) with a in [-k, 7-k] \ {0}

giving 7 * 6

alright

hence 7*6 * 8 + 7²*6 = 630

[tender imp]

the answer is between 500 and 600

[mighty drift]

damn

[tender imp]

yeah

I have been trying to solve it but no luck

[mighty drift]

it's 7*6 * 8 + 6²*6

552

made the same mistake on the 1st one as the 2nd

noticed on the 2nd one, didn't correct the 1st one

[tender imp]

ohh

can you explain that in a bit simpler form,

[mighty drift]

6^3: 2 horizonal 2 vertical
for each direction, 6 intermediate stops
6 permutations of the directions

7*6*8: 3 and 1
for the direction that has 3 moves:
choose a 1st stop, then a 2nd, then it's forced
then there's 8 permutations and symmetries

the details of the counting are left as an exercise to the reader

[tender imp]

ohh

thanks

.close

[cedar kiln]

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[prime lily]

what does the nCr mean in the binomial distribution formula

[tender imp]

sorry, I didn't get this one

can you explain that?

[mighty drift]

the ways of choosing which trials are successes, in r among n

which is by definition, nCr options

[mighty drift]

that's the complicated version

actually enumerate and detail which values are allowed at each step

which is any but the first for k
any but the current one and the last for a+k

[tender imp]

ohh

[prime lily]

hey

why is the probability of event A occuring then event B P(A) * P(B)

why is * and not some other operation

[mighty drift]

only if they're independent

which is a hypothesis for the binomial distribution

[mighty drift]

intuitively or rigorously ?

[prime lily]

intuitively

[mighty drift]

look at examples

it's like making a table of outcomes

what are the odds of the die being 4 on the first throw then 3 ? make a table

it has 6*6 = 36 entries

only 1 matches that, the entry (4, 3)

so it's naturally 1/36

which is (1/6) * (1/6)

P(4) * P(3)

[prime lily]

what about the rigorous one

[mighty drift]

you're not satisfied ?

[prime lily]

well

i am

but i want the rigorous version

[mighty drift]

I suppose you're a high schooler ?

[prime lily]

yeah

[mighty drift]

I'm going to try and get rid of some of the unnecessary rigor to make this a bit more approachable
Since they're different events, we're looking at the space of outcomes S², set of all pairs of events of S (where A, B are sets of events in S)
then we take an event of this space, (w1, w2), representing a pair of individual events
We want P((w1, w2) in A * B) = P((w1 in A) and (w2 in B))

Then we use conditional probabilities: given 2 events E and F (to avoid reusing A and B)
P(E | F) = P(E and F) / P(F)
Hence P(E and F) = P(E | F) P(E), which is just P(E) P(F) since we said our events are independant

Hence, taking E = (w1 in A) and F = (w2 in B), we get
P((w1, w2) in A * B) = P(w1 in A) P(w2 in B)

therefore the question has been transferred to the understanding of Bayes's formula and conditional probabilities

[cedar kiln]

@prime lily Has your question been resolved?

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[warm tree]

why is this wrong

actually i see what i did

so same answer but not negative

2 + sqrt3

i used 60 - 45 instead

i guess it was 2 - sqrt3

<@&286206848099549185>

i can get -2 + sqrt3 and 2 + sqrt3 but i dont know how the answer is 2 - sqrt3

.close

[cedar kiln]

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[marsh pond]

I can’t figure out what step I did wrong

Using e^x as an example of how to differentiate an exponential

I messed up something to do with tanx right?

[drifting marlin]

comparing against e^x is probably not great here since your base is also a function of x

though logarithmic differentiation seems to be the way to go, on the topic(-ish) of e

[dull oxide]

$f(x)=e^{\tan{x} \ln(1+x^2)}$

[wraith dagger]

SWR

[violet night]

pretty ugly final answer

[drifting marlin]

such is life

[dull oxide]

The final answer to life is indeed quite ugly

[marsh pond]

.close

[cedar kiln]

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[lament linden]

Exercise: Find the following derivative:

[cedar kiln]

Please don't occupy multiple help channels.

[bright bridge]

uh

[dense nacelle]

Can you translate in english plz?

[gentle flower]

he did already

[lament linden]

For the following functions, calculate the derivative indicated

[dense nacelle]

Ok

[dense nacelle]

Didn't see that

[lament linden]

Whats the approach here? @mighty drift

[dull oxide]

Induction may be the way to go

Look for a pattern after applying derivative a few times

[cedar kiln]

@lament linden Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[granite crystal]

stuck on part b

i have absolutely no clue where to start

part a was fine

c = 6, b=3, a =2

[drifting marlin]

X1 = X2 if they both equal a, or they both equal b, or they both equal c

[cedar kiln]

@granite crystal Has your question been resolved?

[granite crystal]

ok

i alr knwe that

thats their distribution

im just confused on what to do next

sorry i wasnt clear

thats my fault

[cedar kiln]

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[hollow garnet]

So here if we take the limit it's 1/1 or just 1 because the powers are the same