help-10

[royal basin]

answer my question:

are you able to apply the formula that they give you

yes or no

[vital frigate]

yes

[royal basin]

ok

then you're able to do the problem yourself.

[vital frigate]

Ill answer it can you check it if its right?

[royal basin]

sure...

[vital frigate]

this is my answer

this process is what I've learned from my friend and im bothered if the steps I applied was wrong

[royal basin]

where did 0.166 come from?

you screwed up your arithmetic for sure.

you used a calculator, yes?

[vital frigate]

yea...

[royal basin]

ok

then...

why didn't you just calculate $2 \times 3.14 \times \sqrt{\frac{4.0}{9.8}}$?

[warm shale]

Ann

[royal basin]

there is no "process" to "learn" here. it is literally just plug and chug.

[obtuse pebble]

@vital frigate Has your question been resolved?

[vital frigate]

forgot to say, thank you ann

[obtuse pebble]

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[clear kraken]

how do you implicitly differentiate this?

[clear kraken]

$\frac d {dx} (2xyy'')$

[warm shale]

Kalgar

[clear kraken]

What is the solution?

I believe I need to use three-way chain rule

[dusky light]

is y given?

[clear kraken]

not

just abstractly

Do I need to "append" a y' term to EVERY term in the product rule?

I

'm so confused

[fallow plank]

Is y=y(x)

It is

So

Just start writing the product rule

[dense imp]

are you confused by how you would apply product rule to a 3 way multiplication first of all?

[dusky light]

(2xyy'')'=2(xyy'')'=2(x(yy'')'+yy'')=2(x(y'y''+yy''')+y'y'')

[fallow plank]

👽

[dusky light]

smth like that

[dense imp]

$=2yy'' + 2xy'y'' + 2xyy'''$

[warm shale]

Soosh

[dusky light]

o

oops i did y'y''

[dense imp]

@clear kraken still here / did you figure it out?

[clear kraken]

are you sure?

btw y is a function of x

we are differentiating in terms of x

so wouldn't I need to append y' terms?

i.e. $\frac{d}{dx}xy=y+xy'y'$ right?

[warm shale]

Kalgar

[clear kraken]

notice how the second term has a $y'$ term "appended" to it

[warm shale]

Kalgar

[dense imp]

so this part is supposed to be the derivative of y right?

just simple product rule

so what you are saying is that derivative of y is for some reason (y')^2?

i think what you are confusing yourself with is a case where you have a more complex expression in terms of y's that you implictly differentiate, for example:

$\frac{d}{dx}(3y^2 + 2y) = (6y + 2)\frac{dy}{dx}$

[warm shale]

Soosh

[dense imp]

where y is a function of x, here we had a more complex expression in terms of y that we used simple rules on then chain ruled the y, but in the problem you originally posted there is nothing to chain rule, derivative of y is simply y', derivative of y' is simply y'' etc.

[dense imp]

$\frac{d}{dx}(xy) = y\frac{d}{dx}(x) + x\frac{d}{dx}(y) = y + x\frac{dy}{dx} = y + xy'$

[warm shale]

Soosh

[dense imp]

does that clarify @clear kraken ?

[obtuse pebble]

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[fresh rune]

could someone explain how these complex numbers equations work

[obtuse pebble]

@fresh rune Has your question been resolved?

[fresh rune]

<@&286206848099549185>

[dim gulch]

Heyyy !!!!

[fresh rune]

hey @dim gulch

[obtuse pebble]

@fresh rune Has your question been resolved?

[fresh rune]

<@&286206848099549185>

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[unborn raven]

I need a concealer. Hello, I would like to know if I am ok and if not where and why. Thanks in advance !

[fossil crag]

For 2b) why did the real part become 4a-7?

And same for 2c)

[unborn raven]

because we have 3a+a before no ?

[fossil crag]

But why -7?

[unborn raven]

because I change sides the =2 so it becomes -2. -5-2=-7 Like this I have an equation =0

[fossil crag]

But that's only for 2a)

Why does -7 appear AGAIN in 2b) and 2c)

[unborn raven]

ohhhh ok thanks sorry i’am dumb

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[unborn raven]

I need a concealer. Hello, I would like to know if I am ok and if not where and why. Thanks in advance !

[unborn raven]

I hope now it’s good

[obtuse pebble]

@unborn raven Has your question been resolved?

[unborn raven]

<@&286206848099549185>

[primal iris]

Looks good

I might've read quickly but I think everything's good and you did understood what you had to do 🙂

[unborn raven]

ok thank you !

[obtuse pebble]

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[marble sand]

hey guys

[tepid mountain]

close this room as you don't need it and remain on the second one

[marble sand]

sorry check #help-23

[tepid mountain]

ok

.close

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[drowsy stirrup]

How to do this?

[royal basin]

@drowsy stirrup do you know vieta's formulas?

[drowsy stirrup]

wym, i havent heard of that before?

[royal basin]

sum and product of roots

and for cubics also the sum of twofold products

in terms of the coefficients

does that ring a bell?

[drowsy stirrup]

yh

[royal basin]

right

so you can get the values of α+β+γ, αβ + βγ + γα and αβγ

[drowsy stirrup]

what is α+β+γ, 1/2?

[royal basin]

do you mean like "what does that mean" or "what is its value"

[drowsy stirrup]

the value

[royal basin]

yes, it's 1/2

[drowsy stirrup]

so αβ + βγ + γα is 0?

[royal basin]

indeed it is

[drowsy stirrup]

how do u get alpha, beta, gamma?

[royal basin]

you don't.

[drowsy stirrup]

oh

[royal basin]

from these with some manipulation you can get the sum, sum of twofold products, and product for α^2, β^2 and γ^2

[drowsy stirrup]

so is one of the roots 1/4 for (α^2)?

[royal basin]

no

[drowsy stirrup]

oh

[royal basin]

typing out the manipulations in full will be difficult. you'd need to get some paper and like, expand out (α+β+γ)^2 for example

not just that

but as a starting point

[obtuse pebble]

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[gleaming rose]

Earth exerts a gravitational force on the Sun and the Sun exerts a gravitational force on Earth.
a) which exerts the larger force?
b) which exerts the greater acceleration?

[gleaming rose]

i believe that the sun exerts more force due to its mass?

and is the sun not stationary?

[leaden lion]

They both exert equal force

On each other

[gleaming rose]

always?

even when distance changes?

[gleaming rose]

and even though the mass is larger?

is this due to the first law?

[fallow plank]

G m1m2 / d²

Equivalent force on each other

[gleaming rose]

what about acceleration?

[alpine bison]

f=ma

so a=f/m

may i close this channel ?

atlas_ goes offline

i ll wait up to 18:20

[gleaming rose]

so the acceleration

is depedent on the mass

[alpine bison]

yes

are you fine?

[gleaming rose]

but not the force of gravity

?

[royal basin]

don't do that.

[gleaming rose]

i think that f= ma is alittle misleading

[alpine bison]

it depends on the distance and the product of the mass

[gleaming rose]

yes

so they would be equal

force of gravity

but is the acceleration different?

not equal correct?

[alpine bison]

you know that free channel are not many

[royal basin]

there are 3 open channels right now. and 4 that are hidden.

for a total of 7 more channels if need be.

[alpine bison]

this discord has 30k memebers

[royal basin]

don't close people's channels for them unless they're actually done and have said so.

[alpine bison]

🫡

[royal basin]

yes, the acceleration is different. the sun weighs about 300,000 earths if memory serves me right

so its orbital acceleration relative to the... center of mass of the solar system, i guess, is that many times smaller than that of earth

[gleaming rose]

so its accelerating much slower?

[royal basin]

indeed

i mean if you think about it

the center of mass is very close to the center of the sun itself anyway

simply bc the earth is, by comparison, a tiny pebble

[gleaming rose]

that makes sense

is there an equation to calculate orbital acceleration?

is that what it would be called?

[royal basin]

orbital acceleration from what data?

[tardy epoch]

Are you looking for something like
https://youtu.be/CCsbSq9wlyI?si=-MkjhJarwQbfrIu5

[royal basin]

yeah kepler's laws come to mind

if you're going for a circular orbit, you might also look at uniform circular motion, in which the centripetal acceleration is given by a = v^2/r [where v is the linear velocity and r is the radius]

[gleaming rose]

ill take a look into this thanks

[gleaming rose]

appreciate that ill try calculating it.

.close

[obtuse pebble]

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[formal echo]

How do I 12

[cobalt shuttle]

You can manually find the first 5 terms

u1 = 4, how do you find u2?

[obtuse pebble]

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[timid silo]

Can someone help me i just have no clue how to do these things

[timid silo]

like i just don

know how to get the answers for these

<@&286206848099549185>

[balmy wind]

put x=the given value in f

eg 13- put x= 0 in the equation

[timid silo]

so for a on 13 it would be 0^2 + 4(0)-12

@balmy wind

[balmy wind]

yes

correct

f(0) =-12

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

ok thank you

i think that helps

.close

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[nimble cloud]

$f(x) = x^{5} - 3x -1$

[warm shale]

.moka

[nimble cloud]

how would I solve this?

[true summit]

you want to factorise it?

[nimble cloud]

I wanna solve for x, as in find where y is 0

[balmy wind]

put x=0

you will ger f(x) or y

[nimble cloud]

wouldn't that just give me the intersection of the function at the y-axis?

[true summit]

ye idk tbh

[obtuse pebble]

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[nimble cloud]

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[ocean axle]

I was tyring to solve this proble

I don't understand why OYZ and OXZ are supplementary

https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_16

nvm it's cylic

[obtuse pebble]

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[formal python]

i got a small question about modular arithmetic
lets say i got an expression - (a^b modp)^c
why can i say its congruence to (a^b*c modp)?

[brazen gorge]

wait what?

wdym by (a^b mod p)^c?

[sinful plover]

so basically looking for proof of a = b mod p => (a^k) = (b^k) mod p where k is non negative integer and = refers to congruence?

[obtuse pebble]

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[timid silo]

i was doing a sequence questin and i used the geometric equation

[timid silo]

Ar^n-1

but the answer was using Ar^n

and i just am a bit confused

so i dud

Un=a x 0.85^N-1

U(3)=a x 0.85^2

[latent walrus]

saying 0.85^2 would mean only 2 years of depreciation

[timid silo]

isnt the formula Ar^n-1 though?

[fossil crag]

If for you the initial value is $a_1$ then it corresponds to after 0 years

[warm shale]

rafilou2003

[timid silo]

oh if you take A=1 to be year 0 year 3 would be N4 and so it would still be ^3

[fossil crag]

Yes

[timid silo]

thank you very much!!

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[rancid hedge]

I need to compute the expectation of X

[rancid hedge]

This was the answer

And i dont understand where three comes from

[sinful plover]

yep true

[rancid hedge]

true?

[sinful plover]

2 * P(X=2) won't be equal to 2 * 3 * p^2 * (1-p)

3 is wrongly multiplied above

[rancid hedge]

thats I thought too, i think its a mistake with the calculation

anyways i was just checking

thank u

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[worthy rivet]

Middle school students were taking a test, with the max score being 15 and the avg. score rounded up to the nearest decimal is 10,4. Max figured out that he had a few mistakes. If he corrected the mistakes, he would gain 4 more points and the avg. score rounded up to the nearest decimal would be 10,6. What is the max. and min. number of students taking the test?

[worthy rivet]

I got answers 14 and 39

as min and max

through the use of the form 10,4-Q and 10,6-R where Q and R some numbers on the interval [-0.5 , 0.5)

I just really need someone to verify this as soon as possible

Help would be greatly appreciated

[obtuse pebble]

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[dapper wagon]

how can I do part A if I dont know f(z) ?

[brazen gorge]

image on the way?

[dapper wagon]

sorry

I know that z = x+iy (or I assume)

@brazen gorge any idea of how to start it?

[brazen gorge]

didn't study conplex integration so no

[dapper wagon]

ok ty anyway

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.close

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[frigid dove]

How do you solve this?

[obtuse pebble]

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[uneven horizon]

i know how to do partial derivative questions but i have completely no idea whats going on here, can someone explain?

[brave bramble]

fx(x,y) ≈ [f(x+h,y) - f(x,y)] / h

[uneven horizon]

what do the graphs mean??

[brave bramble]

You'll use h = 1 in this case

The left graph is the value of f(x,y) at some integer values of x,y

It looks like the right graph is the values of fx

[uneven horizon]

can i assume h to be 1 in these type of question all the time

[brave bramble]

No lol

[uneven horizon]

then how do i know what h is

[brave bramble]

But they're giving you integers so that's the best you can do

Smaller h is better

[uneven horizon]

i can choose?

[brave bramble]

Sure, but as h gets larger fx gets less accurate

So they're expecting you to pick the smallest h you can

[uneven horizon]

is h supposed to be delta x

[brave bramble]

Oh they're using Δx yeah

[uneven horizon]

so for yy

do i have to make anither graph similar to fx

why is it different from mine

[obtuse pebble]

@uneven horizon Has your question been resolved?

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<@&286206848099549185>

.close

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[kind hollow]

hi

[kind hollow]

help please

<@&286206848099549185>

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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.

@kind hollow Has your question been resolved?

[wispy wagon]

basically unit circle except dialate it

so it would be (20 times cos(195), 20 times sin(195))

[wispy wagon]

20*cos(195 degrees) is -19.319

20*sin(195 degrees) is -5.176

[obtuse pebble]

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[kind hollow]

oh

mb

thanks man

[wispy wagon]

np

[kind hollow]

9 more to do in 20mins

doubt ill finish

[obtuse pebble]

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[kind hollow]

hw due in 20 mins

[kind hollow]

jus to clarify incase u think its a test

[dark stirrup]

@kind hollow did you still need this open?

[obtuse pebble]

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[timid silo]

So I handed this exercise in and got graded and comments and I'm trying to fix it
I feel like I've mostly fixed it but I can't come up with ideas on how to seal the deal on completeness of Y
any ideas?

[obtuse pebble]

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[stuck willow]

Someone please help me solve this
Find m so that the equation has infinitely many solutions
a=5
b=3

[analog steppe]

Can someone factorize this completely everytime i do it i get diffrent answers: 6p^2+7p-20

[stuck willow]

can u solve mine

[analog steppe]

bro what i aint no math helper

[stuck willow]

😦

<@&286206848099549185>

[stuck willow]

please help me this

[obtuse pebble]

@stuck willow Has your question been resolved?

[leaden lion]

Have u tried something?

[stuck willow]

try what

[leaden lion]

Evaluate lhs first

[stuck willow]

how to?

when the equation has infinitely solutions

I don't know that 😢

[leaden lion]

Forget about infinitely solutions

First multiply the matrices

[stuck willow]

ya

so then...

[obtuse pebble]

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[warm shale]

cookie2

[cosmic veldt]

.close

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[gleaming rose]

Earth exerts a gravitational force on the Sun and the Sun exerts a gravitational force on Earth.
Which has the greater acceleration? Explain your choice.

[gleaming rose]

all that i can figure working with are these two formulas

Newton’s Law of Universal Gravitation

or

Kepler’s Third Law

[daring rock]

certainly the gravitational force exerted by the earth onto the sun is equal to the gravitational force exerted by the sun onto the earth, right?

equal in magnitude, anyway

How does the force applied on an object relate to the acceleration it causes?

[obtuse pebble]

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[tired ingot]

solve the third one pls

does it converge or diverge

bruh

[unreal musk]

what do you think?

[tired ingot]

i tried root test

but it failed

now im not sure if this is in my syllabus so if u can pls tell me the test ill verify and stop wasting my time on this q

ive been stuck for a while

pls

[unreal musk]

Well there are [probably] multiple steps for this one: comparison

[tired ingot]

how?

like what do u compare it to

[unreal musk]

Think about rewriting $\sqrt{n^3} - \sqrt{n^3 - 1}$ and then from that point it should become a bit easier to see what to compare to, hopefully...

[warm shale]

@unreal musk

[tired ingot]

rewriting it to what?

[unreal musk]

Well you have two square roots, if there was a way to get rid of them somehow...

[tired ingot]

ive tried everything i can think of

squaring them?

like a sort of conjugate?

[unreal musk]

[tired ingot]

so i square this entire expression?

because if i do square it how do i balance the substitution

[unreal musk]

Multiply by a conjugate

[tired ingot]

oh so multiply the num and denom and try?

let me see

[unreal musk]

[tired ingot]

ok i rewrote it and got 1/sqrt(n^3) + sqrt(n^3-1)

now what?

do i compare this to 1/sqrt(n^3)?

[unreal musk]

That'll do

[tired ingot]

ok tqqqqqq

it worked

[unreal musk]

There ya go well done

[obtuse pebble]

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[manic atlas]

They ask about the necessary/sufficient conditions for (a to b), (b to c), and (c to a). Tbh, this is a new thing to me (I'm self-taught), I've tried searching related material from google but most of them are word based problems. I still have no idea how to tackle this

[obtuse pebble]

@manic atlas Has your question been resolved?

[unreal musk]

Starting with (i):
If you know a), does b) follow from that? [if so, then a) is sufficient for b) - a) being true is enough to make b) true - if not then it isn’t]
If you know b) to be true, then does a) have to be true? [if so, then a) is necessary for b) - you can’t have b) true but a) not]

Maybe starting from there, assume that x + y = 5 and xy = 3. Are you then able to find the value of x^2 + y^2 with that information?

[manic atlas]

Yes

Then at least a is necessary for b?

[unreal musk]

Justify why?

[unreal musk]

This question is about sufficiency

[manic atlas]

Oh, I mean sufficient, right?

So, then the necessary is the reverse? Like using b information to find a?

[unreal musk]

Yep, that’s it: if b is true, then does a have to be true(?) that’s the question for necessity of a for b to be true

[unreal musk]

And if you were able to show that x^2 + y^2 has to be 19, yep

[manic atlas]

I see, I understand the necessary and sufficient now. But to prove the ability to find the truth is still vague. Does that depend on my algebraic manipulation? Or maybe any tips?

[unreal musk]

Pretty much in this case, it’s how you do manipulations really, for example for some of these you won’t get one implying that the other must be true

Did you find whether a was necessary for b or not yet?

[manic atlas]

Yes, that's just swapping the role. But yeah, I wonder if the equations were a bit more complicated, then it'll take time to figure it out

@unreal musk anyway, thanks

.close

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[tawdry vortex]

hello

why is it -ln exactly

I know how to do the solving until there

[sage geode]

a = -b -> b = -a
ln(11.5) = -x -> x = -ln(11.5)

[tawdry vortex]

no difference?

between those two

or

oh

i see

[obtuse pebble]

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[tawdry vortex]

.close

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[tepid crag]

hi

[tepid crag]

ik what the product rule is

it’s just that i don’t under stand what’s dy/dx and how do they know it’s x^2-1^4

[royal basin]

x^2-1^4
do not drop the parentheses oh my god

i don’t under stand what’s dy/dx
dy/dx is the derivative of y, your function, with respect to x

applying the product rule leaves one derivative not yet evaluated: d/dx [(x^2-1)^4]

[tepid crag]

because u cant evalute it?

[royal basin]

no, because it takes another step to do that.

[tepid crag]

yeah

ngl i tried to find the derivative my head hurt

okay so

it’s dy/dx because it’s harder to find the derivative compared to others?

[royal basin]

...

no you're overthinking it

this is not about harder or easier

this is about not doing 1000 things at once

[tepid crag]

okay thank u

.close

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[proven olive]

HEllOOO

[proven olive]

Is this a legal chain rule?

product rule

the top is already in terms of f'' but it simplifies down to this just want to make sure its legal algebra?

[obtuse pebble]

@proven olive Has your question been resolved?

[proven olive]

nope

[obtuse pebble]

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[plush mesa]

Hi, I’m just not sure how to start this problem. If anyone can give me a clue that would be great thanks!

[lyric crane]

i'd start by finding the sum of the interior angles of the whole outer shape PQRST...

[plush mesa]

Ok I’ve got that

[lyric crane]

alright

so there are 8 rectangles around the edge right

meaning that there are how many right angles around the edge

[plush mesa]

16

[lyric crane]

right

so whats the sum of the measures of those right angles

[plush mesa]

Well idk but it leaves me with 1080 degrees, and 16 ‘x’ angles. So 1080/16 equals 67.5, would that be the answer?

[lyric crane]

sounds right

[plush mesa]

Ok thank you!!

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[cyan copper]

Do I have to add a constant if a polynomial doesn't have one?

[pseudo swift]

wdym?

[cyan copper]

Like 4x²+2x³-4x⁴+6x⁵

Do I have to add a constant if I want to turn it into standard form?

Or just leave it as it is after turning it into standard form

[pseudo swift]

it's fine

[cyan copper]

Ahh okay

[pseudo swift]

there's no x coefficient in your thing anyway if we're going that way

would be useless to write

and longer to read

[cyan copper]

So 6x⁵-4x⁴+2x³+4x² and that's it?

Thanks

[pseudo swift]

yea

[timber spoke]

Somebody please help

I am losing my mind

[pseudo swift]

go to #help-13 I'll come there @timber spoke

.close

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[ancient iris]

if the vertical dilation for a hyperbola is negative, does that mean it reflects about the x-axis?

[hexed gull]

ys

[ancient iris]

ohh okay

so regarding the gradient for the hyperbola

i just wanted to make this sure

if the gradient > 1, the hyperbola compresses
if the gradient < 0, it reflects about the y-axis
but what occurs when 0 < gradient < 1?

[obtuse pebble]

@ancient iris Has your question been resolved?

[obtuse pebble]

@ancient iris Has your question been resolved?

[obtuse pebble]

@ancient iris Has your question been resolved?

[obtuse pebble]

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[flat geyser]

it's right

[unreal musk]

You can simplify that sqrt{x}/x to make it easier to work with, if you’re wondering…

[flat geyser]

someone else is probably wrong

oh then he's right

[unreal musk]

Same thing

[flat geyser]

sqrt(x)/x is the same as 1/sqrt(x)

$\frac{x^{\frac{1}{2}}}{x}=x^{\left(\frac{1}{2}-1\right)}$

[warm shale]

Combustion

[flat geyser]

wuh

[obtuse pebble]

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[west ocean]

Hello! In properties of exponential function is there always no zero and x intercepts?

[west ocean]

And how do i get its aymptotes?

[obtuse pebble]

@west ocean Has your question been resolved?

[west ocean]

<@&286206848099549185>

[past sand]

A zero is the same as an x-intercept, isn't it?

On the reals, no f_a(x) = a^x has a root (a f_a(x) = 0)

Except 0^x I suppose

[west ocean]

How do you get its horizontal and vertical asymptotes? In exponential function

[past sand]

There is no vertical asymptote, it's continuous and defined on all of R

The horizontal asymptote is always y=0

Of course it will change if you add stuff to the function

[obtuse pebble]

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[obtuse pebble]

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[pulsar quarry]

,w 16000(9.8+((900)^2/2))

[warm shale]

Results provided by WolframAlpha

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

Requested by yajatjamal

[pulsar quarry]

this is whaat i am getting, theres a lot of difference in my answer and thier's

instead of converting velocity in m/s i converted radius in km

i believe it is something related to units of acc due to gravity

,w 9.8 m/s^2 in km/hr^2

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by yajatjamal

[pulsar quarry]

,w 16000(127000+((900)^2/2))

[warm shale]

Results provided by WolframAlpha

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

Requested by yajatjamal

[pulsar quarry]

but this is wrong too

ik the motions dynamics, i just need help with this unit part

[tardy epoch]

can you just show your work without all the numbers plugged in

should just do all the algebra and plug in numbers at the end

[pulsar quarry]

ok i give me 5 mins

here @tardy epoch

[subtle sinew]

It's mainly because you're mixing units, a lot. Newtons is kgm/s^2

[subtle sinew]

This has units of kg(m/s^2 + (km/hr)^2/km)

[pulsar quarry]

yup yup right

[subtle sinew]

This has units of kg(km/hr^2 + (km/hr)^2/km)

[pulsar quarry]

so i need to simplify it in every in MKS right?

everything*

[subtle sinew]

MKS = meters, kg, seconds?

[pulsar quarry]

yes

[subtle sinew]

Yes, you should ideally convert everything to kg, m, s first

So that speed of 900 km/hr should be in m/s

[pulsar quarry]

hm yea OK, thankyou

[subtle sinew]

That's why this didn't work, because you had m/s^2 + (km/hr)^2/km

[pulsar quarry]

yea right

thanks

.close

[obtuse pebble]

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[forest sorrel]

PLEASE PLEASE HELP ME I AM REALLY STRUGGLING

[balmy wind]

what's the answer?

104?

[forest sorrel]

Thank you very much

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[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

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[severe raft]

Hello, I would like someone to check my answer. The question states that there is a line y = kx +1 which is tangent to a circle. K is a constant. This circle has a center at (5,1) and a radius of 3. The question is asking what the value of K is

[subtle sinew]

,rotate

[severe raft]

If needed I can explain part of my process

[obtuse pebble]

@severe raft Has your question been resolved?

[wispy wagon]

Sorry i'm not sure that is correct

(x-5)^2+(y-1)^2 = 9 is correct

Substituting y = kx+1

Expand and simplify

The reduced equation should be a quadratic in terms of x

Since it is tangent, the discriminant of that quadratic should be 0

[obtuse pebble]

@severe raft Has your question been resolved?

[obtuse pebble]

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[manic wolf]

how do I factor x^3 + 4x^2 +x -6 without rational root theorem

[thin star]

i thought you got it

[manic wolf]

tried x^3 - x + (4x+6)(x-1) and continuing but it doesnt work

[manic wolf]

its an other example

U helped me so well with the previus one

it was easy after i added that x XD

or maybe im doing something wrong

but this ones diffrent the method of adding x doesnt work

[dark stirrup]

There are no rational solutions. You would have to use cubic formula

[dark stirrup]

!xy

[obtuse pebble]

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.

[manic wolf]

there is

[gentle sphinx]

add 3x and substract 3x

[manic wolf]

yea tried that

[dark stirrup]

Oops I switched a + and -

[thin star]

you see that 4 and 6?

you can turn the 4 into 3+1

[manic wolf]

its (x-1)(x+2)(x+3) btw

[manic wolf]

yea

[thin star]

3 and 6 can be used together

that's all the hint i'm giving

[manic wolf]

aight wait it worked last time XD

there are so many combinations holy moly this gonna take me a min

[thin star]

factoring can be an artform in some cases

[manic wolf]

I give up

how am I meant to chose the right path

[thin star]

group x^3 and 3x^2

[manic wolf]

x^2(x+3)

[thin star]

now how about the rest of the terms

[manic wolf]

left with -x^2-x-6

[thin star]

+x^2

[manic wolf]

oh yea myb

left with this

oh wait

I see

[thin star]

i think you can see rest from here

[manic wolf]

yea

but like how tf

am I meant to chose the right path

[thin star]

lol practice

trial and error

you start seeing patterns

[manic wolf]

like i went for factoring 3x^2 -x-6 and its just a mess

[spice swift]

Hi, idk how to solve 15 if you can help me, thanks a lot.
Statement : In this part, we look for functions f defined and derivable on I (with I an open interval of R containing 0) that verify: f(x) + f(y) = f((x+y)/(1-xy))
Let f be a function that satisfies these hypotheses.
14.calculate f(0) (I found f(0)=0)
15. let, a∈I with a≠0
determine λ>0 as for all h∈[-λ;λ], the equation
a+h = (a+y)/(1-ay), (of unknown y) admits exactly one solution. This solution is denoted y(h). Then show that h->y(h) is derivable on [-λ;λ]. Calculate y(0) and y'(0)

I have several leads but nothing comes to fruition.
I first thought about solving this equation :
(a+h)(1-ay)=(a+y) which brought me back to a 2nd-degree equation, but I don't think that was very helpful (or idk how to deal with it)
Then, as the lim of (a+y)/(1-ay) is -1/a
So I thought it was sufficient to choose lambda small enough so that [a - lambda, a + lambda] does not contain -1/a. But what lambda?
Or I thought as [-λ;λ] is '' symmetrical '' it contains 0 so we can choose h=0 but with this method I can't conclude about the possible value of lambda...
I'm stuck here, so if someone could help me out 🙂
(the goal of the problem is to find the arctan function so lambda should be equal to pi/2??)
Thanks

[manic wolf]

ty ig

leme just finish it to make sure im not a donkey

yea its correct

.close

[obtuse pebble]

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[fervent gate]

Consider the following sums: 3 + 5 = 8 and 3 + 3 + 3 = 9. Show that any integer greater than or equal to 8 can be written as a sum of threes and fives using induction proof and recursive formula

[signal ledge]

if you have an integer i
i > 8 means that i has the minimum sum of 3 3s
f(i) = either sum of all 3s or all 5s
if f(i) does not satisfy that, replace 2 3s with with 1 5. this proves it for f(i-1)

I don't know if it makes sense

but its just substituting 3 3s with 2 5s for i+1 and 2 3s with 1 5 for i-1

[fervent gate]

kinda confusing to me

[signal ledge]

okay consider you're proving this for a number k, where f(k) = k

if you have 5 as a number which is a part of the sum, then substitute the 5 with 3*2. This would give you f(k+1)
if you do not have 5, then subtitute 3*3 with 5*2. This would give you f(k+1) too

I don't know if this makes much sense, I'm not too familiar with the concept of induction but this is how it should go

[fervent gate]

i wrote out like this

n>=8, n=3k+5m

p=3k₁+5m₁

p+1=3k₂+5m₂

and here is where i got stuck

[signal ledge]

or, you can write it p+1 as 3(k₁+2) + 5(m₁-1)

its the same thing

[fervent gate]

why +2 and why -1

how do u know its +2 and -1

[signal ledge]

open the bracket

p+1 = 3(k₁+2) + 5(m₁-1) = 3k₁ + 6 + 5m₁ - 5 = 3k₁ + 5m₁ + 1 = p + 1

[fervent gate]

ya

hold on

for just one second

im still stuck on the

3(k₁+2) + 5(m₁-1)

[signal ledge]

you could also write it as 3(k₁-3) + 5(m₁+2)

[fervent gate]

why k1 + 2

yea but how

[signal ledge]

you have to express 1 in terms of 3s and 5s

[fervent gate]

am i supposed to figure out these numbers

[fervent gate]

but there is a 2

[signal ledge]

think about possible combinations that could give you 1 and you're only using + or - with 3 and 5

[signal ledge]

yes but that is 3*2

1 can be written as 3*2 - 5*1

[fervent gate]

ah

ur actually

so

[signal ledge]

it can also be written as 5*2 - 3*3

[fervent gate]

just down it down for me a lil, why are we expressing 1 ?

is it cus p+1=?

[signal ledge]

induction basically means reproving f(k) for f(k+1) if I put it blandly

[signal ledge]

yes

[fervent gate]

ah

[signal ledge]

f(k) is defined as k

so f(k+1) is defined as k+1

[fervent gate]

yea my teacher hasn't gone through induction as functions

so what you're saying is not being registered in my head unfortunately

[signal ledge]

okay hear me out

forget the f(k) part

ill start with where you left it off

[fervent gate]

alright bro

im listening

[signal ledge]

n>=8, n=3k+5m
p=3k₁+5m₁
p+1=3k₂+5m₂

this is what you understood by yourself and its well and good

now understand that we're only dealing with 5s and 3s
in induction if you basically prove that if a function/identity is true for a value k, then it should be true for k+1 too
suppose you represented p as 3k₁ + 5m₁
then you can represent p+1 as 3k₁ + 5m₁ + 1
now try rearranging the 1 in terms of 3s and 5s. This can be 3 + 3 - 5, 5 + 5 - 3 - 3 - 3 and so on
Consider 1 = 3 + 3 - 5
then p + 1 = 3k₁ + 5m₁ + 3 + 3 - 5 = 3k₁ + 3 + 3 + 5m₁ - 5 = 3(k₁ + 1 + 1) + 5(m₁ - 1) = 3(k₁ + 2) + 5(m₁ - 1)

so here in the last line you proved that p + 1 is also represented in the form of 3s and 5s

basically what was needed

[fervent gate]

3k₁ + 5m₁ + 3 + 3 - 5 = 3k₁ + 3 + 3 + 5m₁ - 5 = 3(k₁ + 1 + 1) + 5(m₁ - 1)

this part

[signal ledge]

yes

what about it

[fervent gate]

nvm

hold on

[signal ledge]

its alright don't press yourself

[fervent gate]

let me think about this for like

2 minutes

[signal ledge]

I actually gotta hit the bed since its 2 already 😭

[fervent gate]

so we have 3k₁ + 6

[signal ledge]

yes

[fervent gate]

we factor out the 3

[signal ledge]

yes

[fervent gate]

then factor out the 5m1-5

[signal ledge]

yes

[fervent gate]

factor out 5

[signal ledge]

yes

[fervent gate]

alright

so after this

can i take another example

with different numbers

that would leave 1

[signal ledge]

and then the k2 and m2 you used earlier are basically now k1 + 2 and m1-1

[fervent gate]

and if i get the same expression

we prove it

so like

[signal ledge]

like?

[fervent gate]

3(k-3)+5(m+2)=

the first expression

if they equal eachother

[signal ledge]

yes

[fervent gate]

we've proven

i

it

ya?

[signal ledge]

yes

we have

[fervent gate]

alr ty bro

so this is universal?

like

[signal ledge]

no worries

[fervent gate]

if i get another problem

like this

i want to express +1

in the terms

of the other problem

[signal ledge]

for this problem? yes. for another problem? depends on what terms you have to represent it in

[fervent gate]

lets say

n>12

4k+3m

[signal ledge]

okay

yes

[fervent gate]

4(k+)+3(m-)

and after the + and -

something that would

equal

1

in terms of 3 and 4

but probably wouldn't work in this case

[signal ledge]

yes, so if you have to do it using 4s and 3s for n>12, you can simply do + 4 - 3

[fervent gate]

ah

yea

true

[signal ledge]

there are n number of combinations with 4s and 3s

[fervent gate]

4(k+1)+3(m-1)

like this

4k+4+3m-3

[signal ledge]

yes

[fervent gate]

4k+3m+1

=p+1

alr

[signal ledge]

example, if you had to do it with 7s and 8s
you could do 8(k+8) + 7(m-9) and so on if p = 8k + 7m

[fervent gate]

yea

makes

[signal ledge]

its just making a given sum with 2 given numbers

[fervent gate]

immensive sense

alright

tyty

[signal ledge]

no worries

[fervent gate]

!close

.close

[obtuse pebble]

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[bitter wraith]

Would N -> N : f(x) -> x be injective?

[warm canopy]

do two elements get mapped to the same thing?

[bitter wraith]

no

[warm canopy]

then it is injective

what you've written down is the identity map and that in general is bijective

[bitter wraith]

is it the identity

ah it is true

f : R -> R , f(x) -> x^2

wouldnt be injective

but
f: N -> N, f(x) -> x^2 , this would be injective again

[warm canopy]

correct

[bitter wraith]

where N stands for natural Numbers

[warm canopy]

yes

[bitter wraith]

thanks ❤️

but not every element in the image must be hit right?

for injective

[final thunder]

No that’s surjectivr

[bitter wraith]

true

[warm canopy]

not every element in the codomain

[bitter wraith]

not i said

"not every

[warm canopy]

the image is precisely the things that are hit

[bitter wraith]

A -> B

injective

[final thunder]

Oh yeah missed that

[bitter wraith]

so basically not every element in B must be mapped

[warm canopy]

yes injectivity makes no claims about what elements are hit

[bitter wraith]

dumb question but

how can i read the ":"

here

[final thunder]

Such that

Oh wait

Can you show the whole thing

It’s probably something like for every positive real number y, there exists a real number x such that blah blah

[final thunder]

Yeah It’s probably something like 'because for every nonnegative real number y, there exists a real number x such that x^2 = y, namely x = sqrt(y)’ but tbh idk German conventions

[obtuse pebble]

@bitter wraith Has your question been resolved?

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[hoary sonnet]

für alle y aus den positiven Reelen Zahlen existiert ein x aus dne Reellen Zahlen
also basically kannst du den : als "gilt" übersetzen

[obtuse pebble]

Please don't occupy multiple help channels.

[hoary sonnet]

.close

[obtuse pebble]

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[snow sparrow]

Can someone explain the thought process for this problem? I dont understand how my teacher got this solution.

[obtuse pebble]

@snow sparrow Has your question been resolved?

[obtuse pebble]

@snow sparrow Has your question been resolved?

[steep flax]

Hm. First of all, there is no thought process, there is no problem, and there is no solution. The image simply shows three different ways of expressing the same information.

It's like writing a/b = a*(1/b). There's no thought process of going from one to the other, there is no problem, there is no solution.

What abt the image confuses you exactly?

[obtuse pebble]

@snow sparrow Has your question been resolved?

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[bitter wraith]

why can i define a tuple like this

[tardy epoch]

you can define whatever you want

[bitter wraith]

should i just accept it tha i can write (1,2) -> { {1}, {1,2}}

[drifting wraith]

there's no simpler way if you want a set

[tardy epoch]

.close

[obtuse pebble]

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[obtuse pebble]

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[sage elbow]

How do I solve this? (have to find a, b)

[sage elbow]

I was thinking of equalizing the y's but it doesn't make sense, as the line and the exponential could meet in 0, 1 or 2 points...

<@&286206848099549185>

[obtuse pebble]

@sage elbow Has your question been resolved?

[deft merlin]

wdym by equalizing the y's

[sage elbow]

e^x + a = x + b

[deft merlin]

oh

wait can you show me the entire qn

[sage elbow]

OHH ahaha

the question acutally says "The function is represented in the following graph, find a, b".
But in the picture i got the graph is not there

that exaplains all

indeed, i was wondering how could i find a,b without any other information

[deft merlin]

and what topic is this on?

is this on contiunity?

[sage elbow]

Limits but this part 'find a,b' doesn't matter with limits.
After that questions, it asks for limits.
In italian, this function's type is called "Piecewise Function"

(i am translating literally)

[deft merlin]

k have you learned the definition of continuity

[sage elbow]

Yes

why?

[deft merlin]

my best bet is that you can use that

the parameters of the qn dont really make sense

but you can make a system of equations with the defn of continuity

[sage elbow]

For example?

[deft merlin]

you find f(1)
find left and right limits for x -> 1
solve by system of equations using f(1) and the lim as x -> 1

[sage elbow]

Never did this in exercise.
Could you show me this way of solving it?

[deft merlin]

sure

[sage elbow]

Thx

[deft merlin]

do you have a solution by chance?

[sage elbow]

No, this ex was taken from an old school test

[deft merlin]

ah

[sage elbow]

Use google lens if you want but it won't help

[deft merlin]

yea no nvm defn of continuity wouldnt work

[sage elbow]

Ok but how would it work?

[deft merlin]

like the process?

if this problem could actually be solved using defn of continuity

[sage elbow]

Yes

Or, if you have a video tutorial on yt

i could watch it

[deft merlin]

gimme a sec

[sage elbow]

Ok

[deft merlin]

so first i checked if f(1) is defined

• f(1) = e + a

then i checked the one sided limits for x - > 1

and we know that the one sided limits must be equal to each other

but then after that theres nowhere to go

which is why it wouldnt work

oh wait i messed up

[sage elbow]

Pratically you are looking for an intersection of the 2 functions, right?

[deft merlin]

it should be
f(1) = 1 + b
but even still it wouldnt work

[deft merlin]

kinda?

this is really just solving for the constants
but using this method doesnt work with this problem

[sage elbow]

That's what i came up with initially but i've tought that the intersection could lead to 0, 1 or 2 points.
For example :

[sage elbow]

I see

So, without any other information given from the exercise, this ex is not solvable, right?

[deft merlin]

no theres definitely a solution out there

[sage elbow]

Indeed the exercise says to look at the graph (which i do not have)

[deft merlin]

oh

yea theres that too
but im sure theres a metthod that you could use

[sage elbow]

Where to calculate a, b

[deft merlin]

without needing the graph

oh

ok nvm then if it says you need to look at the graph then uh

probably the graph is really important then lol

[sage elbow]

I don't think so because if you have 1 equation with 2 unknown params (x and a) you won't be able to find the value of a without a second information

[deft merlin]

yea you right

[sage elbow]

However, thank you for your time.
I appreciate it

[deft merlin]

yea sry i couldnt help you find an answer

[sage elbow]

You did, i just read bad the question, it said from start to look at the graph but i did not see it in the picture because it's not there

i'll close the post then.
Thank you again

[deft merlin]

alright np

[sage elbow]

.close

[obtuse pebble]

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[quasi oasis]

i know that X Unif(5,8) but i dont know what to do with that

[quasi oasis]

im very lost on 3 and 4 i dont know where to start

[daring sorrel]

In a uniform distribution all values have the same probability

[obtuse pebble]

@quasi oasis Has your question been resolved?

[obtuse pebble]

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[river kindle]

hey i just have a question about numerical differention