help-13

[green mulch]

Btw I am in highschool so I don't have alot of knowledge

[jaunty pumice]

Analytically mean?

[jaunty pumice]

I will use section formula and area of triangle formula

[spiral fog]

Cord , trig , complex , etc

[green mulch]

Can u tell me the section formula

[jaunty pumice]

https://www.cuemath.com/geometry/section-formula/

[green mulch]

Oh I have read about this

But how can we use this here

And btw do u think the area of the small tri can do anything

[jaunty pumice]

Yeah

[green mulch]

Can u tell me how

[jaunty pumice]

Like you could find coordinate of P and Q and use area of triangle formula

But that will be a lot of calculations

[green mulch]

That would give us a variable ans right?

I believe there should be a simpler way to solve the question

[jaunty pumice]

Yes

[spiral fog]

If it helps i got that area of APQ = 1/8 area of ABC

[green mulch]

Can u show how please?

[spiral fog]

Notice that PQ is parallel to median from C

The rest follows

[green mulch]

Huh?

[spiral fog]

You know what median is?

[green mulch]

Yes

[spiral fog]

Draw it from C to BC

[green mulch]

C to bc?

[spiral fog]

Yes

[green mulch]

Shouldn't it be a to bc?

A*

[spiral fog]

No

For this one no

Basically call the point of intersection D

[green mulch]

How can u draw from c be the point lies on bc only?

[spiral fog]

Now notice that APQ an ACD are similar

[green mulch]

Hmm I am not understanding this

Sorry

[spiral fog]

AQ : QB = 1 : 3
AP : PC = 1 : 1
Therefore AQ : QD = AP : PC = 1

[green mulch]

I really appreciate your help but I think this isn't for me rn

Thanks alot

Sorry for taking your tim

Time*

[hot crag]

hmmm

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[shy spoke]

I had a math problem on today's regional Olympiad that I sadly didnt have time to solve 😦 Could you pls help me?
So I need to find natural number n that is bigger than 10^100 and so n^2 and (n+1)^2 are the same but with the numbers rearranged, for example 13^2 and 14^2 are 169 and 196
But I need n to be more than 10^100

[hot crag]

...

firstly

169 and 196 do not have the digits changed entirely

only the latter two

secondly, if you want "n" like that then the number of digits changed would be at the very least 100, off the top of my head

it appears you want a number that could be a palindrome with one outlying digit at the front, possibly

[cedar kiln]

@shy spoke Has your question been resolved?

[shy spoke]

Sorry, I mean they don't have to be changed entirely, just some of them

[shy spoke]

<@&286206848099549185> Help

[spiral fog]

You need to find it or prove it exists?

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[somber wing]

is the answ sqof6? , plscorrect me if im wrong

[flint plinth]

that is not correct

[somber wing]

oh

why so

[flint plinth]

show how you got it and i'll tell you what you did wrong

[flint plinth]

the square root should only be over the numerator

you took the square root of the whole thing

[somber wing]

ah

icic

how careless of me

anyways thx

[flint plinth]

it happens

[somber wing]

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[thorny tusk]

Let's say I have two infinite sets for example N and Z. How do I show they have the same size? well I have to come up with a function which is bijective for f:N--Z but it's not obvious to me how I can construct a function that ensures bijectivity

[flint plinth]

there's a standard idea you can use here, i'll just give a hint so as not to give it away

0, 1, -1, 2, -2...

[thorny tusk]

well that's just the set of integers

I actually have the answer to the problem above in my lecture notes but I'm more looking for a step by step guide to solve any problem of this kind, like the two infinite sets N and Q, how do we show they have the same size

[south tundra]

I don't think anything beyond "Provide a bijection" can be said given arbitrary infinite sets (nvm I thought the question is general)

[flint plinth]

it's not just the set of integers, it's in a particular order, intended to convey how you could define a bijection

f(0) = 0
f(1) = 1
f(2) = -1
f(3) = 2
f(4) = -2
...

(what happens after the ... ?)

[spiral fog]

You dont have to prove bijection

You can prove injection in both ways

[flint plinth]

^ assuming you know the schroeder bernstein theorem and are allowed to use it

[spiral fog]

I mean N -> Z is trivial and Z -> N you can always do some stuff with prime numbers

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[spare cloak]

How do u do this??

[misty flame]

Which one

[cedar kiln]

@spare cloak Has your question been resolved?

[spare cloak]

<@&286206848099549185>

Well this is the first question

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[muted bear]

it looks like you just have to isolate k

[latent torrent]

yea but how

[muted bear]

you only have k on one side

so what is the first thing you can get rid of

[latent torrent]

333

[muted bear]

with what operation

[latent torrent]

divide

[muted bear]

ok

the you are left with a 3^k on the right side

what can you use to isolate k?

[latent torrent]

log base3

[muted bear]

yeah

you have to use a calculator to find a numerical value for k

[latent torrent]

oh shoot ok thanks

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[marble hawk]

Graph the unit circle, and the angle in standard position that has measure π/4 radians.
What are the (x, y)-coordinates of the point that is the intersection of the terminal side of this angle with the unit circle?

[spiral fog]

!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 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

[dusky peak]

the full circle has 360° which is 2pi
so half a rotations would be pi
so a quarter rotation would be pi/2
and so on

that tells you how big the rotation would be

for the coordinates (x, y), we can get those with the trigonometric functions, since we know the angle and the length of the terminal side (1)

[marble hawk]

I know I need to draw the angle for pi/4 radians, but I am stuck on how to get those coords

[dusky peak]

kk

[marble hawk]

my professor gave us a equation of x^2 + y^2 = 1 but I don't know where to go with that

[dusky peak]

here we have a graph

notice that with an angle of pi/4, x=y

x^2+y^2=1 comes from the pythagorean theorem, since we are dealing with a right triangle
the 1 is the length of the terminal side
and since x=y

from this on, we can get the values of x and y

alternatively, we can use sin and cos

for example:
sin(pi/4)=y/h
cos(pi/4)=x/h
h here is the length of the terminal side
but since h=1, this gets easy

but if we wanna calculate the values, it is better to use x^2+y^2=1, since we cant really calculate sin or cos by hand

[marble hawk]

so to plug in x and y, would I use the x/y coords that are on that point of the unit circle?

[dusky peak]

well we are looking for this point right?

[marble hawk]

yeah

[dusky peak]

and in this triangle, we have x^2+y^2+1
x and y are the small sides of the triangle we see here

so the results for x and y will be the coordinates of that point

[marble hawk]

ohhh okay i see

one more question

how would i go about combining the x and y values? is it just side x of the triangle squared (in this case 1) plus the same thing for y? and I add those together?

[dusky peak]

combining x and y?

[marble hawk]

as in adding them

because its the x ^2 + y^2 = 1

[dusky peak]

well

we cant do x^2+y^2 if we dont know x and y

but

pi/4=45°
its like if f(x)=x

so x=y

therefore
x^2+y^2=x^2+x^2=2 * x^2

[marble hawk]

so in my professors example, she had the x ^2 + y^2 = 1 formula, and then in the next step just changed the y to an x. is that because they are equal to one another?

[dusky peak]

correct

we can also "see" that in the following way

what is the angle i left empty here?

[marble hawk]

45, because its a 45 45 90

[dusky peak]

good

[marble hawk]

and therefore x=y

[dusky peak]

this kidna shows us that the triangle is symmetrical around this red line

and thus x=y

[marble hawk]

alrighty, tysm for the help!

[dusky peak]

you're welcome^^

[marble hawk]

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[glacial steppe]

Yo what is this even asking me

[glacial steppe]

Is that not 45 degrees

[worldly chasm]

It is not

[glacial steppe]

??

[worldly chasm]

So let's take stock of what we know

We know A and B, but not R

right?

[glacial steppe]

No

[worldly chasm]

ok, you found R, but let's stick with we know A and B

because R looks horrible

[glacial steppe]

Yeah

[worldly chasm]

So you're learning about trig functions

do you remember SOHCAHTOA?

[glacial steppe]

No this is calc

[worldly chasm]

This is just a trig problem.

[glacial steppe]

Yup

[worldly chasm]

So we have A and B, which means we can find an angle using which trig function?

[glacial steppe]

Okay 15/26 and I’ll plug my result into inverse?

[worldly chasm]

inverse what?

[glacial steppe]

Sine

[worldly chasm]

why sine?

[glacial steppe]

I mean

I have the opp and hu

Wait no I don’t

Oops

[worldly chasm]

👍

which trig function do you use?

[glacial steppe]

Tan

[worldly chasm]

ok, and which side is opposite and which is adjacent?

[glacial steppe]

Side A is opposite

B is adjacent

[worldly chasm]

Good!

[glacial steppe]

About 30 degrees

[worldly chasm]

So tan(theta) = 15/26

so theta = arctan(15/26)

,w arctan(15/26)

[glacial steppe]

I actually gotta ask

[wraith dagger]

Results provided by WolframAlpha

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

Requested by omnipotententity

[glacial steppe]

What’s an arc

Like not an arc arc

[worldly chasm]

like arc in arctan?

[glacial steppe]

Ut arc as in the context of arc tan

Yeah never learned it in hs

Just another function?

[worldly chasm]

arc is short for arcus

I don't know what arcus means though

[glacial steppe]

Oh another problem

Given no context how does this work?

The worst part is I’m coming back from the lecture and I don’t even know what to do

[worldly chasm]

seems to be latin for arch,

so we're essentially back to where we're started

I guess someone in the 1600s-1800s decided that's a good name for it, and it stuck.

[glacial steppe]

Gotta love mathematicians……

[worldly chasm]

So it might be helpful to know a general fact about the pythagorean theorem

it doesn't matter how many dimensions you have, the pattern is the same.

so in 3d, we have a^2 + b^2 + c^2 = d^2

[glacial steppe]

Don’t know that

Ok ill plug the numbers in for it

[worldly chasm]

well,

before you just take my word for it

let's do some drawing.

imagine you have a point at (x, y, z)

Now consider two triangles.

(0, 0, 0), (x, 0, 0), (x, y, 0) for the first.

(0, 0, 0), (x, y, 0), (x, y, z) for the second

take a moment and convince yourself that these are right triangles.

[glacial steppe]

What am I loooking at

[worldly chasm]

whatever you've drawn to represent this situation

presumably

I would draw it an take a picture, but I have a giant jerk cat sitting directly on my desk.

[glacial steppe]

Well I have no idea what that draws

[worldly chasm]

here, let me make a drawing on the computer.

[glacial steppe]

You want a (0,0,0) (x,y,0) (x,y,z)

[worldly chasm]

that's a triangle, with three corners located at those points

[glacial steppe]

It’s all 0 and variables

[worldly chasm]

it's variables because this is true no matter what x, y, and z are. here, let me draw it.

So you have some point (x, y, z)

it's sitting directly above the point (x, y, 0)

So this form a triangle when you consider also (0, 0, 0)

If we call the origin O, and the point (x, y, z) P, and the point (x, y, 0) A. Then the triangle OAP is right, because the angle OAP is 90 degrees.

This means the OP^2 = OA^2 + AP^2

just by regular pythagorean theorem

We know AP, that's just z, because it goes from (x, y, 0) to (x, y, z)

OP is the value we want to find, let's call it w, I guess.

then OA we can find with a second triangle.

Let's call the point B at (x, 0, 0)

now OBA is right, because angle OBA is 90 degrees,

so we have OA^2 = OA^2 + AB^2, which is OA^2 = x^2 + y^2

substituting back we just have w^2 = OA^2 + z^2 = (x^2 + y^2) + z^2

so

w^2 = x^2 + y^2 + z^2

[glacial steppe]

Yeah I’m guessing my hs left that out….

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[lost spade]

,, 8/(8*(1/3))

[wraith dagger]

Ayk

[lost spade]

algebraically, cant we cancel 8's?

apparently we cant

[flint plinth]

what is the *, multiplication?

[lost spade]

yea

[flint plinth]

sure, you can cancel the 8's

what did you get when you tried?

[lost spade]

1/3

calculator shows 3

[flint plinth]

no

you must have typed it wrong

wait

[azure horizon]

$$\frac{8}{8 \cdot \frac{1}{3}} = \frac{1}{\frac{1}{3}} = 3$$

[flint plinth]

calculator shows 3, that is correct

1/3 is incorrect

[wraith dagger]

Kaisheng21

[lost spade]

oh

1 comes

i see

.close

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[low fern]

I just started this topic and done a few basic ones but I don’t know where to start for this question anyone can help? Thanks.

[low fern]

.close

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[wraith trellis]

hello

[wraith trellis]

pls help

can u just write the equation and breifly explain the wording

[zenith sail]

do you know what "supplement" and "complement" mean?

@wraith trellis

[wraith trellis]

ya

[zenith sail]

so if I gave you an angle with a measure of x, what would its supplement and complement be?

[thorn phoenix]

Hello

@wraith trellis

[wraith trellis]

hi

[thorn phoenix]

Okay so

[wraith trellis]

um

[thorn phoenix]

I will write the solution here

Let an angle be x

We have been given that its supplement is three times its complement

Supplement is 180 degrees

and Complement is 90 Degrees.

Its complement angle will be 90 Degrees - x.

And its supplement angle will be 180 Degrees - x.

Supplement angle = 3(Complement angle)

180 Degrees - x = 3(90 Degrees - x)

180 Degrees - x = 270 Degrees - 3x.

180 Degrees - 180 Degrees + 3x - x = 270 Degrees - 180 Degrees - 3x + 3x. (Equals added to equals are equals).

2x = 90 Degrees.

2x/2 = 90 Degrees/2.

x = 45 Degrees.

Hence proved.//

@wraith trellis

[wraith trellis]

ooohhh ok

can u write the equation

[thorn phoenix]

What equation?

Supplement angle = 3(Complement angle)

180 Degrees - x = 3(90 Degrees - x).

Those equations?

This helped?

[wraith trellis]

ohh ok

thanks

[thorn phoenix]

Okay

do .close

pow

.close

@wraith trellis do .close

[wraith trellis]

.lcose

.close

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[candid comet]

can someone help me with this question. I dont understand and my teachers answer key isnt always correct

[potent wave]

I think you must find the formula of f(x) depends on g(x+3)

[candid comet]

I got the domain as 6,6

But I feel like that makes sense

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[covert tinsel]

Hey

[covert tinsel]

This problem is just for fun

Idk how to do it tho

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[stark knot]

i dont know how to do this

[cedar kiln]

@stark knot Has your question been resolved?

[stark knot]

<@&286206848099549185>

[bitter finch]

For question 5, measure the angles and put them in the table

[stark knot]

HOW

[bitter finch]

Hmm maybe you have a tool for it? Although it looks like News and commercials are both about 60 degrees

[stark knot]

ok

[bitter finch]

You can maybe figure out the rest

[stark knot]

how tho

[bitter finch]

For example DJ Talk and Commercials together look like they make a 90 degree angle right

Commercials is about 60

(60 is the angle in an equilateral triangle)

[stark knot]

what that

[bitter finch]

That’s a triangle where all the side lengths are equal

[stark knot]

so a triangle

[bitter finch]

https://tenor.com/view/triangulo-equilatero-triangulo-triangle-equilateral-triangle-shape-gif-14615939

[bitter finch]

Not every triangle has equal side lengths

Most of them don’t

[stark knot]

how was i supposed to know that like

[bitter finch]

I dont think you were

But do you perhaps have a tool for measuring angles?

[stark knot]

no

[bitter finch]

Hmm alright

So yea I guess we can go with the estimate that Commercials, and News are both 60 each

What do you think the angle of DJ talk is

This is kind of a weird question ngl

[stark knot]

120?

[stark knot]

ik

[bitter finch]

Do you think dj talk has a bigger angle than commercials?

[stark knot]

huh

idk

[stark knot]

no ig

[bitter finch]

What would you say is its proportion to commercials

Like is dj talk 2 times smaller than Commercials? 3 times? 4 times?

[stark knot]

2

[bitter finch]

I agree

So if we said commercials is 60 then what would you say is the angle of Dj

[stark knot]

30

?

[bitter finch]

Yes

Ok and knowing all this what would the angle of music be

[stark knot]

210

[bitter finch]

Yep

[stark knot]

i am so smart frfr

likeee

[bitter finch]

Facts

[stark knot]

anyyways

.close

[cedar kiln]

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[lapis arrow]

compute the scalar-surface integral f Ds:
double integral g x^2zdS where G is the cylinder (including the top and bottom) x^2 + y^2 = 4, 0 <= z <= 3

[tepid ginkgo]

!original

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[lapis arrow]

thats the original

compute the scalar-surface integral f Ds:
double integral g x^2zdS where G is the cylinder (including the top and bottom) x^2 + y^2 = 4, 0 <= z <= 3

[tepid ginkgo]

nevermind

[lapis arrow]

<@&286206848099549185>

<@&286206848099549185>

[cedar kiln]

@lapis arrow Has your question been resolved?

[lapis arrow]

<@&286206848099549185>

@tidal ravine

<@&286206848099549185>

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[versed elm]

How do I find the expression

[cedar kiln]

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[versed elm]

<@&286206848099549185>

Someone help

I’m desperate

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[novel seal]

Probability need help with problem E

[spiral fog]

!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 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

[novel seal]

I don’t know where to begin

1

[spiral fog]

It seems to me that your prev solutions can help alot

[cedar kiln]

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[azure galleon]

Im having trouble with this improper integral question. Should i use the partial fraction method?

[cedar kiln]

@azure galleon Has your question been resolved?

[gloomy crater]

partial fractions wont work since the degree of the numerator is less than the degree of the denominator

[gloomy crater]

isnt this just integral test backwards

[gloomy crater]

since x^3/(x^2+3)^2 is positive, continuous, and decreasing on x>=5

see if $\sum^{\infty}\frac{n^3}{(n^2+3)^2}$ converges

[wraith dagger]

mXd

[azure galleon]

oh ok

[azure galleon]

well its already know to be improper

but im not exactly sure how to solve it cuz of the fraction in the bottom

[azure galleon]

see if $\sum^{\infty}\frac{n^3}{(n^2+3)^2}$ converges

[wraith dagger]

ItzCJ

[cedar kiln]

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[azure galleon]

.close

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[deep cave]

need some help setting up this equation

[deep cave]

i think it is 4r+8r=108 right

[crimson sedge]

yes

right

b is 18

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[prime island]

O

[prime island]

I've been stuck on this for hours

[prime island]

I'm not sure what i'm doing wrong but I'm substituting in the function and changing around the variables, then i get a common denominator for the top 2 variables

[livid hound]

can you take a pic of your work

[prime island]

but ya idk what the answer or steps are

its a mess i might have to rewrite it in a readable way

ill do that one sec

im not sure what to do after that^

or if its right up till that point

(also we aren't supposed to use derivatives yet, just derivative definition)

[livid hound]

you messed up these products

[prime island]

oh you can't multiply them together and take it out?

oh wait nevermind i forgot the sqrt?

[livid hound]

trying to forcibly, combine, cancel stuff

[prime island]

^ this was also wrong

[livid hound]

trying to forcibly, combine, cancel stuff

[prime island]

its over h which is 0

[livid hound]

$\sqrt{a + b} \cdot \sqrt{a} \redneq \sqrt{a}\cdot\sqrt{a} + \sqrt{b}$

[wraith dagger]

ℝαμΩℕωⅤ

[prime island]

ooh okay

[livid hound]

so many math crimes there

here you should just leave that as
$$\sqrt{11}\sqrt{11 + 2h}$$

[wraith dagger]

ℝαμΩℕωⅤ

[livid hound]

or combine to get
$$\sqrt{11(11+2h)}$$

[wraith dagger]

ℝαμΩℕωⅤ

[prime island]

so im assuming you can't subract $$\sqrt{11} from $$\sqrt{11+2j}?

[wraith dagger]

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

[livid hound]

no, you can't forcibly cancel stuff if that's what you're trying

can you show what you have now?

[livid hound]

try simplifying a bit so that you don't have multiple fractions

[prime island]

like this?

oops that +2h is under that sqrt but ignore that

[livid hound]

simplify further, there stuff multiply fractions there

[prime island]

like this?

[livid hound]

no

still mutliple fractions, and the length of your fraction lines aren't ideal

how would you simplify
$$\frac{\br{\frac ab}}{c}$$

[wraith dagger]

ℝαμΩℕωⅤ

[prime island]

this?

[livid hound]

yes,

as for what you were attempting before
$$\sqrt{a+b} \redneq \sqrt{a} + \sqrt{b}$$

[wraith dagger]

ℝαμΩℕωⅤ

[prime island]

ooh got it

[livid hound]

from here, rationalise the numerator using conjugates

(do not unnecessarily expand the denominator)

[prime island]

how do I go from here?

[livid hound]

you didn;t multiply the numeratirs correctky

[prime island]

oh im special

so you said i dont have to expand the denominator by multiplying the denominator of the conjugate right?

[livid hound]

you still didn't multiply the numeartor properly

you should have 11-(11+ 2h) there

[prime island]

oh okay

[livid hound]

for the denominator, just leave the product as is.

[livid hound]

after fixing that result cancel the common factor of h

you'll no longer have the component responsible for the 0/0 form and from there you can plug in h=0

[prime island]

i did something wrong again

[livid hound]

you didn't fix the numerator completely

simplfiy the 11-(11+2h) properly

[prime island]

OH

I understand now

[livid hound]

and you missed a factor of sqrt(11)

[prime island]

i did something wrong again

[livid hound]

you have a common factor in the numerator and denominator

simplify

[prime island]

... oh

It still shows it as incorrect

[livid hound]

what are you entering into the system

[prime island]

when i do -1/11(sqrt(11))

[livid hound]

should be correct

oh wait

you're just asked for the difference quotient

just after cancelling the h, just leave that as is

don't plug in h=0

[prime island]

i can simplify further right?

[livid hound]

where'd the - sign go

the factor of h can still be cancelled

just don't do anything with the other hs

[prime island]

liiike this?

i forgot the negative on the top but other than that*

[livid hound]

there should've also been a - sign in that top line,
but yes

[prime island]

ok ill try it

got it thanks!!!

[cedar kiln]

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[tender cipher]

idk what this question means

[mighty shuttle]

They have given you the path it moves along

it's initial position

and angular velocity

based on that parametrise the motion

also , this isn't from a test right?

[tender cipher]

nope

[civic coral]

Do you know the parametric expression for a circle

[tender cipher]

x^2 + y^2 = r^2 ??

[civic coral]

That’s the standard form

[tender cipher]

oh

[civic coral]

You isolate the x and y and, making them a system

That’s my interpretation of parametric representation

[civic coral]

So the system shown in the picture is, probably, the final answer of the question

However, I just want to show you a general concept of parametric representation

Also would you like to know how to derive the parametric representation from the given standard form of a circle

[tender cipher]

yeah sure

[ripe yew]

Since in real life nobody will tell which is the right angle of a triangle, what if the theta is placed on the right angle?. How to identify the hypotenuse, opposite and adjacent?

[hollow trail]

!occupied

[cedar kiln]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[civic coral]

@tender cipher

That is the algebraic approach to acquire the parametric representation of the given circle in your question

I think a geometric interpretation of it would be much better

Have you learned about trig function yet? @tender cipher

[tender cipher]

oh ok

i used trig identities to solve trignometric parametrics before but that was to find the cartesian equation

i havent done any questions where you do it backwards

[civic coral]

Good, then I think you’re capable of deriving the parametric representation of the circle when its graph is presenting in front of you

P is a movable point on the circle

Could you use trig functions to indicate the coordinates of P

Just manipulating the information (theta, radius) on the graph

[lucid pecan]

can you help me with my math

its Fractions

[civic coral]

Sir you have to open a new channel for your question, since this one has already been occupied.

[lucid pecan]

ohhhh

can you help me itts really quick

SUper Qucik

[tender cipher]

x = 2costheta, y = 2sintheta

[civic coral]

Terrific

That’s how we derive the parametric representation of a circle through geometry

[civic coral]

However, there’s some conditions of your just obtained parametric representation

You have to modify it in order to fit all of these conditions given by the question

[tender cipher]

yeah i dont understand the conditions

[cedar kiln]

@tender cipher Has your question been resolved?

[cedar kiln]

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[civic coral]

Help

Solved

.close

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[hasty flame]

can a normalised mean square error value be negative?

[fading summit]

i dont think so

[dire geode]

All squared real numbers are nonnegative

x^2 >= 0 for all real numbers x

[hasty flame]

like for this

[dire geode]

,w plot x^2

[wraith dagger]

Results provided by WolframAlpha

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

Requested by rie.mann

[hasty flame]

o its in decibels

is that supposed to help?

like decibels can be negative right?

[fading summit]

I know decibels is a logarithmic scale

so negative means that the actual loudness is less than 1, but not negative

[hasty flame]

can you explain why nmse is negative in this chart?

or why you think it is

cuz genuinely i just wanna know

[hollow trail]

where does it say that it's negative?

[hasty flame]

the y axis its down to -30 dB

[hollow trail]

the nmse itself isn't negative, it was plotted on a decibel scale

[hasty flame]

o

[hollow trail]

so 0 on the decibel scale would be some reference value (maybe 1, or some other "baseline" quantity), then 10 on the decibel scale would be 10x that quantity, and 20 on the scale would be 100x that quantity, and so on. Similarly, -10 on the scale would be 1/10th of the quantity, and -20 would be 1/100th of that quantitity

[hasty flame]

ohhh

that makes more sense

thanks a lot!

have a good one yall

!close

.close

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[mental quail]

can someone help walk me through this?

[cerulean sail]

What's f(-1) and f'(-1), in terms of a and b, given the fact you have f(x) and f'(x)?

[mental quail]

im not sure what youre asking ...but like f(-1) = 12 so should i be setting 12 = the function?

[cerulean sail]

Basically yep but replacing x = -1

[mental quail]

i did this but, im not sure where to go

also

but not sure if thats on the right path?

[cerulean sail]

That's all good, keep that in mind, now do similar with the derivative while keeping that in mind

[mental quail]

okay give me a minute to write it out

what should i be doing with these equations?

[clear umbra]

simultaneous equations, done them before?

[mental quail]

im very bad at them 😦

[cerulean sail]

Well at least you have b equalling two things

[mental quail]

am i supposed to stick them back into the original equation and solve for the other variable?

thats what i vageuly remember

[cerulean sail]

why -2 here though?

[mental quail]

unnnn

[clear umbra]

you have two equations

[mental quail]

wait ill recheck

[clear umbra]

in one equation, you isolate 1 variable

then put that variable into the other one

get an answer for 1 variable, then put that into any of the equations to get the second one

[cerulean sail]

Actually I think that one is fine, it's the one above it

[mental quail]

so its the a thats wrong?

[clear umbra]

by the other one, its the second equation

[cerulean sail]

Yea this one I think, but anyways, the b one is more convenient and fine

You have b = -(3a + 8)/2 and also b = a - 2

[mental quail]

how is that wrong 😭 ug

one is for f'(x) other is for f(x) ?

im still getting a = (-b-8)/3

oh 2b

is that right @cerulean sail ?

[cerulean sail]

That's it

In any case, whichever ones you have, it's all good, what you want is to express one of the variables, either a or b, in the two different forms...

[mental quail]

i dont get what this part means

[clear umbra]

(honestly using the elimination method rather than substitution would be much easier)

[cerulean sail]

Get either "a = ", one from the f(-1) equation and one from the f'(-1) equation, or "b = "

[mental quail]

so f(-1) = a(something + b(something)
and substitute a into it?

[cerulean sail]

As in either

[cerulean sail]

Get to here

[cerulean sail]

or use that, and the fact that the f(-1) = 12 gets you that a = b + 2

[mental quail]

so this is the work i have for f(x) side. should i be just substituting one of the b=a-2 into b for the original f(x) with x = -1 to solve for a?

[cerulean sail]

Not into the original, but into what you get from the f'(-1) thing

[clear umbra]

you got a = something from here

[clear umbra]

now put that a into here

[mental quail]

so im putting a=b+2 into f'(x) ?

[clear umbra]

you see that a = b+2 you wrote?

[mental quail]

oh, so the a= something from the derivative equation into the base equation?

[cerulean sail]

one of these

[clear umbra]

yes

into the a = b+2 you got

so you will get that something = b+2

[mental quail]

i feel like im still a bit confused but ill write down what i think you guys are saying...

[cerulean sail]

Alright, a slightly different question:

If I say that y = 2x + 3, and that x = 5, what would you think to do with that?

[mental quail]

am i doing this?

[mental quail]

plug 5 into y=2x+3

[cerulean sail]

Cool, we're basically saying similar here(!)

that you have, for example, ${\color{green} b} = a - 2$ and that ${\color{green} b} = \frac{-3a - 8}2$

[wraith dagger]

@cerulean sail

[mental quail]

wait so im setting the two b equations = to each other?

then to clarify, this is wrong and gets me nowhere 😬

[cerulean sail]

Well, it's much easier to set equations equal to each other than to put those into the original equations, as you'd basically end up doing the same thing anyway

[mental quail]

i dont really understand why im doing anything 😦 but maybe ill see it in the end

[cerulean sail]

Setting the b equations equal (or solving the simultaneous equations in any other valid way) will let you find what a is, then b

[cerulean sail]

Yep, we wanted the value of a, and that's it!

[mental quail]

i can use this value of -4/5 and plug into any of the two original equations to find b?

or can i just set the two b equations equal to each other to also find b?

o imean a equations

[cerulean sail]

Quicker to use the "b =" equations but this works, just much more pain

[mental quail]

on the bottom right...

[cerulean sail]

Those are the values for a and b

[mental quail]

im going to plug everything into f(-1) and check

[cerulean sail]

Yep now you can put the a and b into the original then work with that

[mental quail]

😭 its magic

so, what i did was....solve a and b for their respective equations first...

find a and b for f(x)
find a and b for f'(x)

then...set them equal to each other so that the two equations would euqal with the same values

idk, how to put it into words concisely

[mental quail]

any chance you know of a resource where i can get practice problems like these? regular systems of equations seem to simple but then when derivatives and all the extra stuff are involved i get super confused

[cerulean sail]

Not any I have stored and ready, sorry

[mental quail]

thank you for helping me through the problem though in anycase. @clear umbra u2

.close

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[mortal bone]

i have to determine wether this series converges or not

and if yes i have to find the value to which it conerges

i tried liebnitz test since this looks like an alternating series

[flint plinth]

notice that 3 - pi < 0

[mortal bone]

but then in the solution i saw this statement

which i didn't understant

how can we omit (-1)^n

[flint plinth]

^

[livid hound]

it wasn't omitted

[flint plinth]

rewrite 3 - pi as (-1)(pi - 3)

[mortal bone]

then I understod that its s geometric series with ration <1 which converges

[livid hound]

the combined them into single exponent since the powers were both n

[mortal bone]

oh ok

i got it

[livid hound]

a^n * b^n = (ab)^n

[mortal bone]

yes

thank you guys

i'm closing

.close

.close

[cedar kiln]

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[smoky falcon]

guys this is the def of limit of a sequence ok, I need to know how he estimates in the last steps

[smoky falcon]

from number 1

he makes it n over n^2

how ?

<@&286206848099549185>

[dire thorn]

!15m

[cedar kiln]

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.

[dire thorn]

Keep in mind they used a less than symbol, not equal to

Look at which terms in the denominator are greater than 0. This means we can remove them since a bigger denominator means a smaller number

[smoky falcon]

ow so less terms

mean smaller deno

so i can make it as n over n power 2

right ?

I guess so thank u

.close

[cedar kiln]

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[dire thorn]

Yeah we basically just deleted the things that are annotated with >= 0

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[glad dirge]

good evening fellas

[glad dirge]

for the a) im guessing m = f(x)/x is always a fraction

so df/dx has to be less than one

then for b i don't know what the difference is

[azure horizon]

i don't understand what you mean for a)

df/dx isn't just f(x)/x

it's differentiation, right?

[glad dirge]

yeah i think its completely wrong my logic

[dull oxide]

b) is the converse statement of a)
If-then statements are thought of as directional statements. For example "if a number is a perfect square, then it is not prime" is a true statement. But the statements inverse, "if a number is not prime, then it is a perfect square," is not true for every number. Thus, a conditional statement and it's inverse can (and often are) different

[glad dirge]

i see

if f(x) is less than x

it means the x is increasing at faster rate than the y

or staying the same as y

meaning the gradient should be less than or equal to 1

thats my guess for a)

[warm bone]

So your guess is T or F?

[glad dirge]

T

[warm bone]

Prove it

[upper laurel]

try coming up for a function that's got a slope of 1 but is bigger than x

and if you get stuck, consider what values f(0) can be, given that "function thats got a slope of 1" is your only requirement

[glad dirge]

if i use x for example

it holds

[warm bone]

I don't think you're supposed to be considering f(x)=x

I'll try a little hint: can you consider piecewise-linear functions for your examples as well?

Actually nvm, I'm assuming differentiation is defined for every value x

[upper laurel]

just finding that it can work for a f(x) isnt enough, they want this to work for every f(x)

[warm bone]

Instead of piecewise-linear functions. Consider piecewise functions with a derivative defined for every point

[cedar kiln]

@glad dirge Has your question been resolved?

[glad dirge]

isnt it technically only for all x and atleast one f(x)

[warm bone]

For both to be true they have to be true for every f, which should have a value for every x.

[glad dirge]

but its not true for every f

only select f

[warm bone]

A statement is either true or false

Then if you find out if it is false for some f
If it is false for some f, it is false.

[glad dirge]

obviouslly the only functions that would work would be the ones where f(x) is less than x so that would rule out many

[warm bone]

That's not what I care about

For both to be true, they have to be true for every f satisfying the conditions provided

Then you either prove this, starting from the conditions

Or show a function satisfying the conditions that does not imply what is stated

[glad dirge]

i see

[warm bone]

If you're really 'into math' then you can even show that some statements are logically independent of each other.
The question you have shown does not have such properties.

[glad dirge]

could the answer be found using the differention identity what ever its called

[warm bone]

Well here's a thought: I think both are false

[glad dirge]

so to prove false you only need one example where it doesnt hold

[warm bone]

Yeah

[cedar kiln]

@glad dirge Has your question been resolved?

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[silver arrow]

I need some help guys

[silver arrow]

Task

The following information is given about a function f
Dm(f) = {x E Rix -::f:.-61\x-::/:- 6}
f(-1 1) = 0, f(-7) = 0 and f(-3) = 0
f(O) = 2
f(x) > 0 in the intervals ]-11; -7[, ]-3; 6[ and ]6; oo[
f'(-9) = 0, f'(0) = 0 and f'(9) = 0
fer growing in the intervals ]-oo;-9], ]-6; 6 [ and ] 9; oo [

a) Write down the coordinates of the curves of the graph with the coordinate axes.
b) Sketch the graph of a function that fits the given information

Like this:

So how do we do it? It can be done over castool, we just have to explain what we do?

Someone can help me? 🙂

<@&286206848099549185>

[cedar kiln]

@silver arrow Has your question been resolved?

[silver arrow]

Hello?

[silver arrow]

I hope someone can help here.

[tame zodiac]

May I ask you about the meaning of the intervals you wrote? I never saw those kind of notations yet

[silver arrow]

My teacher wrote it that way.

I do not know anything

I'm just asking for help

[cedar kiln]

@silver arrow Has your question been resolved?

[silver arrow]

Help.

I still need help

[silver arrow]

hello

[silver arrow]

<@&286206848099549185>

???

[nimble veldt]

i am not sure, what is exactly your question?

[cedar kiln]

@silver arrow Has your question been resolved?

[west wyvern]

I believe ]x; y[ means {z: x<z<y}

[cedar kiln]

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[formal pasture]

how do i do #5

i dont get

how to find the lim

[vernal lantern]

id first off graph it

its a piece wise function right

[formal pasture]

yes

[vernal lantern]

so id graph those points first, putting an open dot where x =/ 4, and putting a closed dot where x=4

youd see then that theres two differeny points on that specific part x = 4

so since at x =4 theres two points, you have a jump discontinuity

[formal pasture]

yea

[vernal lantern]

if im remembering correctly, the lim should actually be 1, since thats where the graph will actually tend do when x does equal 4

thats the closed dot right, so thats where the actual limit should ve

youd have 2 solutions, but 1 limit

[formal pasture]

i dont get it

hows the limit 1

[vernal lantern]

wait - 1

i misread apologies

[formal pasture]

no

the answer is 1

but i duno how

😭

[vernal lantern]

ohhhhh right i get it now

hang on

[crimson sedge]

a limit is defined as something that approaches a particular point, but never quite reaches there

[formal pasture]

yea

[crimson sedge]

so a limit at x = 4 would be akin to approaching it from values such as x = 3.9999 and x = 4.000001

so from the left and right directions

[formal pasture]

yes

[crimson sedge]

but the catch is that it is never quite x = 4

so what do u interpert from that?

[formal pasture]

a hole

[vernal lantern]

that should be what your graph looks like. f(x) is definied as 1 when its suuuuuper close to x = 4, but it suddenly jumps down to -1

just because it is -1 at x=4, that doesnt mean thats what the limit is

[formal pasture]

yea

[vernal lantern]

so as far as limits themselves are concerned, youre really looking for what are the values when x is realllllly super close to 4

thats why its 1

[formal pasture]

i see

okkk

tyy

[vernal lantern]

ofc!

[formal pasture]

.clsoe

.close

[cedar kiln]

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[cedar kiln]

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[knotty saffron]

How does 4w^3/2 equal 3/2w^1/2

3/2 minus 1/2 equals one so i dont know how they got 1/2 as an exponent answer

[cedar kiln]

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[dull oxide]

Open a new channel

[dull oxide]

!msgdel

[cedar kiln]

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[vocal sun]

Can someone lend a hand at knowing what this is/solving for?

[vocal sun]

just like, the basics tho

[crimson delta]

https://www.symmetrymagazine.org/article/the-deconstructed-standard-model-equation?language_content_entity=und

[twin tinsel]

why did you put a doughnut with numbers on it

[vocal sun]

gotcha gotcha

many thanks!

[dull oxide]

No trolling in help channels. Go to #discussion

.close

[cedar kiln]

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[lucid topaz]

$=\int_2^{x^2}cos(t^3)dt+x\cdot\dfrac{d}{dx}\int_2^{x^2}cos(t^3)dt\=\int_2^{x^2}cos(t^3)dt+x\cdot\cos(x^6)$

[wraith dagger]

ryan

[lucid topaz]

did i go wrong somewhere?

[dusty hazel]

yes.

[hollow trail]

need to apply the chain rule when doing the fundamental theorem of calculus

[dusty hazel]

yeah

[lucid topaz]

ohh so $=\int_2^{x^2}cos(t^3)dt+2{x^2}\cdot\cos(x^6)$

[wraith dagger]

ryan

[lucid topaz]

right thank you :')

.close

[cedar kiln]

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[short crow]

how to solve this?

[flint plinth]

first thing to check, what do the numerator and denominator do as x -> pi/2?

[short crow]

hmmm

I didnt get any answer

do you think this will give an answer?

[flint plinth]

it will help

[short crow]

well

tell now what to do?

[short crow]

0/0

[flint plinth]

that's not what i get

check the numerator again

[short crow]

ah sorry 1/0

.close

[cedar kiln]

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[cedar kiln]

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[quick flume]

What is that stuff in the bubbles

And how do you get it

[dire geode]

They solved for x first then plugged them into equation for y in terms of x to get y

x^2 = 9 implies x=3 or -3

[quick flume]

What

I don’t get it

[dire geode]

Which part

[quick flume]

By x you mean what

The 9 or -4

[quick flume]

.

[dire geode]

Do you know where that equation came from?

[quick flume]

I did it like this

With discrimimant

[quick flume]

.

[dire geode]

You solved for a, not x

[quick flume]

Ok so what do I do now when I solved for a