help-19

[shy smelt]

He meant that the eq of the tangent at x = 0 is y = x

[ornate inlet]

I mean, it's possible that c = 0 but you cannot infer that from the statement that y=x is a tangent

I get that

[shy smelt]

But the tangent goes through the point it's tangent to

So through the origin

[ornate inlet]

duh you guys are right

[shrewd tiger]

ok but i still do not understand why y'(0)=1

[ornate inlet]

y' gives the slope of the tangent at point x

since you are given that the tangent at the origin is y = 1 * x, it follows that y'(0) = 1

[shrewd tiger]

i see now thanks.

❤️

.close

[odd edge]

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[jovial bridge]

hello I need some help with this symbol manipulation problem

[jovial bridge]

I originially thought of distributing the m, but how would that work if there's another variable inside the parenthesis

or maybe I don't do that at all, I'm kind of confused

[leaden karma]

well, you can divide both sides by m

or distribute

it's up to you

[jovial bridge]

I'll go with dividing both sides, one sec

ok now I am left with this

[leaden karma]

yes

now can you isolate x?

[jovial bridge]

I could try to subtract 2, but how would that work? Would I just but it outside the fraction?

so something like this?

[thorn fractal]

and then make it positve x

[leaden karma]

yes

now just negate both sides

[jovial bridge]

I'm not sure if I was just supposed to make everything inside the fraction negative or just write a negative sign outside

[leaden karma]

no, you did it correctly

you could also negate the things inside the fraction, it would be the same

$-\frac{3-y}{m}=\frac{-(3-y)}{m}=\frac{y-3}{m}$

[clever fjord]

kheerii

[leaden karma]

notice how the - sign went to the numerator?

[jovial bridge]

yes how come it didn't also go to the denominator

[leaden karma]

it can

but that won't be constructive

[jovial bridge]

alright thanks for the help, I totally forgot you could just divide both sides instead of distributing

.close

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[oblique rapids]

Did I even do any right

[plain ruin]

First is incorrect

[oblique rapids]

What was the answer

[plain ruin]

I don't know but it can be only i dot

You should solve it

Multiply by 2/3

I guess

If previous steps were correct

Second is incorrect

3+1 is 4

4≤-4x

[oblique rapids]

Did I get everything wrong

[plain ruin]

third is almost correct

The answer is from -3 to 4

Fourth is incorrect

[oblique rapids]

wtf man

What was wrong with 4

[odd edge]

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.

[oblique rapids]

These interval notations don’t even make sense like 3 says 0 is the biggest so why does it go to 4

[plain ruin]

5x-3<7 5x-3>-7

[plain ruin]

One sec

Left side is -∞ and right side is +∞

Graph looks like this

We don't need 0 cause it's not solution

[oblique rapids]

It says 0 is biggest so it shouldn’t go past 0

[plain ruin]

Who says

[oblique rapids]

That’s what it says

[plain ruin]

Where

It's parabola

[oblique rapids]

What’s even the point of <0 then

[plain ruin]

nah

they dont say x<0

They say graph<0

so graph is less than zero when it's from -3 to 4

We only use roots as dots in graph

[oblique rapids]

So when am I suppose to use the negative infinite and positive infinite signs

[plain ruin]

When u need + ones

[plain ruin]

Left side is -∞ and right side is +∞

But we don't need +

Cause we need -

Cause graph<0

[plain ruin]

And the module you did wrong, this is the right way

Goddamn, you need to study polynomial inequalities...

[leaden karma]

why the judgement?

[plain ruin]

Cause he messed up everywhere

[leaden karma]

you have no idea about their situation. judgement is not appreciated on this server

[plain ruin]

Bro what a hell

[plain ruin]

He made a lot of mistakes, so he needed theory and practice more, isn't it obvious?

[odd edge]

@oblique rapids Has your question been resolved?

[odd edge]

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[vital birch]

where did i make a mistake?

[vital birch]

im trying to derive arctan'(x) by using the derivative of inverse function formula

[dim thicket]

whats the formula?

[vital birch]

its written on the left there

next to the 1/tan'(arctanx)

[dim thicket]

okok

but why are you multiplying by arctan'(x) on the denominator?

[vital birch]

because of the chain rule

[dim thicket]

its not the chain rule

its the inverse formula

[vital birch]

and im multiple with arctan'(x) not arctanx

[dim thicket]

i know

but the formula doesnt ask for that

you take f' and plug in arctan

[vital birch]

isnt that what i put? 1/tan'(arctanx)

[dim thicket]

yes and thats it

no chain rule

tan'(arctanx) = sec^2(arctanx)

its different from (tan(arctanx))'

maybe thats why youre confused

[vital birch]

is it?

[dim thicket]

it is

[dim thicket]

you would use the chain rule there

[dim thicket]

not there

[vital birch]

how would you write it with d/dx
(tan(arctanx))' is d/dx (tan(arctanx))

how would you do tan'(arctanx)

because i dont see any way to do it

[vital birch]

and also i simplified tan^2(arctanx) to x^2 since its equal to (tan(arctanx))^2

[dim thicket]

maybe $\frac{d}{dx}(\tan x) \circ \arctan(x)$

[clever fjord]

LF

[dim thicket]

where are you doing tan^2(arctanx)?

[vital birch]

in the second last line

[dim thicket]

oh sorry yeah thats good

[odd edge]

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[formal needle]

https://www.youtube.com/watch?v=wxOT2-Kjx78

[formal needle]

why can the numerator be equal to the power?

just watching the video, can someone explain why?

also, why does he divide by factors of 2?

[odd edge]

@formal needle Has your question been resolved?

[formal needle]

<@&286206848099549185>

[quasi sparrow]

Screenshot it outline the proof. Not many people are going to watch a whole video to help

[formal needle]

i literally did

[formal needle]

its here

[quasi sparrow]

You need to explain the assumptions

[formal needle]

^

[quasi sparrow]

Not the assumptions the proof makes

The assumptions in the problem

[formal needle]

no i just want to know like whats the rule that lets you set the numerator equal to the power of the denominator

[odd edge]

@formal needle Has your question been resolved?

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[tall veldt]

You can't pull that stuff involving x out of the integral

[clever fjord]

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

[nimble blaze]

x and t are related
you can't treat x as a constant

[clever fjord]

José

José

[nimble blaze]

don't think ibp ends well here

for ibp you'd also want the derivative of the other component to be get simpler. can't just focus on part of it.

for this question, if first approach this with partial fraction decomposition

takes practice to see what might help, how you should approach certain problems.

[odd edge]

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[royal steppe]

what is the solution set of sqrt root 4x+21 =x?

i got -3 and 7 but -3 isnt a solution, how?

[desert marlin]

plug it in

does sqrt(4(-3)+21)=-3?

[royal steppe]

yea thats what i did, but it doesnt make sense to me

why is that the case?

[desert marlin]

why is what the case

[royal steppe]

that -3 doesnt work?

because i formed a quadratic equation

and negative 3 should be an answer

[desert marlin]

immediately you can see that -3 is wrong because the square root can only output positive values

[royal steppe]

but is there any other way of seeing it?

[desert marlin]

performing the algebra correctly

[royal steppe]

i did tho, no?

[hallow pelican]

7 is right

[desert marlin]

I assume you squared both sides to begin

right?

[royal steppe]

yup

[desert marlin]

Okay so what did you get after that?

[royal steppe]

4x+21=x^2

[hallow pelican]

sqrt(4(7)+21) = sqrt(28 +21) = sqrt49 = 7

[desert marlin]

|4x+21|=x^2 *

[royal steppe]

why that sign?

[desert marlin]

sqrt(x)^2 is not equal to x

it is equal to |x|

[royal steppe]

what?

can u elucidate?

[desert marlin]

I have it backwards

apologies

sqrt(x^2)

=|x|

not = x

not the way I written it

[royal steppe]

huh?

i squared both sides

and got 4x+21=x^2

is that not correct?

[hallow pelican]

can't you also do it =±x

[desert marlin]

When you square both sides you pick up an extra solution because

[royal steppe]

im aware

[desert marlin]

if the left side was negative

then now it is positive

and you gain that solution

but the left side never could have been negative

[desert marlin]

no

[royal steppe]

?

bro im not understanding

[desert marlin]

|x| is not equal to +/- x

I was responding to the other person though

[royal steppe]

why are u making that symbol tho

|

[desert marlin]

that means absolute value

[royal steppe]

im aware

but why are u using it?

[desert marlin]

Okay I was making that symbol because earlier I thought your confusion came from that you did not realize that sqrt(x^2)=|x| and you had thought sqrt(x^2)=x

[royal steppe]

huh?

x^2 doesnt =x

[desert marlin]

do you not see the sqrt that I typed

[royal steppe]

yea

[desert marlin]

So what is confusing you then

[raven grotto]

the issue is that there is a context of the domain you're working in floating in the background that's easy to forget.

you have sqrt(4x + 21) = x. the square root only outputs (by definition) nonnegative numbers, so in order for x to be equal to sqrt(something), it's necessary that x >= 0.

now, you square both sides and get the parabola x^2 - 4x - 21 = 0. as a function from the real numbers to the real numbers, the parabola x^2 - 4x - 21 intersects the x-axis in two places: x = -3 and x = 7. however, remember, you're only looking for solutions that are greater than or equal to 0. so really, you're looking for where the graph of the parabola intersects the line {(x,0) : x >= 0}

[odd edge]

@royal steppe Has your question been resolved?

[royal steppe]

wait so square roots only output positive numbers?

but what about smth like square root of 4

its + - 2

[raven grotto]

yeah. you're pointing out that both 2^2 = 4 and (-2)^2 = 4. but sqrt(x) as a function is defined to output the nonnegative square root of x.

the reason for this is kind of clear. say you have 4, and you want to take the square root. you can only return one number (because functions only return one thing). thus, you have to pick one of the two options. the convention is to pick the one that's >= 0

that's why, as austin was saying, sqrt(x^2) = |x|. because sqrt(x^2) is the nonnegative number y such that y^2 = x^2. this number y must be equal to |x| (because its either x or -x, and you have to pick the one thats nonnegative. this is what |x| is defined to be)

its kind of obnoxious when you first learn this, since there's all these different meanings of square roots

[mystic saffron]

basicallly $\sqrt{4}=2$ but $x^2=4 \Rightarrow \sqrt{x}=\pm2$

[clever fjord]

Prime Minister

[mystic saffron]

hell yeah math

[odd edge]

@royal steppe Has your question been resolved?

[odd edge]

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[austere anvil]

Given A = [ 7 5 ] and B = [ 3 5 ], Find a matrix Q such that BQ = 2A
[ 8 9 ] [ 1 2 ]
The answer is supposed to be [ -52 -70 ]
[ 34 44 ]
but im getting [ -52 -70 ]
[ 64 68 ]

[manic sleet]

can you show your work?

[austere anvil]

uh i don't have a phone atm

but i can type it out

so

first i decided to find 2A

which gives us the 2x2 matrix [ 14 10 ]
[ 16 18 ]

[manic sleet]

ok now you have BQ=2A, how do you solve for Q?

[austere anvil]

then Q = B^-1 * 2A

i already found the inverse of B earlier

[manic sleet]

and that would be?

[austere anvil]

which was [ 2 -5 ]
[ -1 3 ]

but B^-1 * 2A gives me the answer i obtained above

which apparently is incorrect

[manic sleet]

I think the big numbers are messing you up a little just try doing B^(-1)*A

instead of B^(-1)*2A

I assuming you know how to matrix multiplication?

[austere anvil]

yes

i hope so

i could be wrong

[manic sleet]

do you think you can try to B^(-1)*A? If so, can you try to show your work :)

[austere anvil]

why don't i show u the what i did to obtain the product of B^-1 times 2A here?

[manic sleet]

sure that works

[austere anvil]

it's gonna take me some time to type it out

aight hold on

[manic sleet]

do you have it written down?

[austere anvil]

oh shit

i realized what i did wrnog

i made an arithmetic error

[manic sleet]

haha

happens to the best of us :)

[austere anvil]

i multiplied 1 with 14, instead of multiplying 14 with negative 1

[manic sleet]

always check your work!

[austere anvil]

the thing is

my handwriting on the paper was a bit too messy

so i interpreted it as positive 1 the whole time

[manic sleet]

I've said 3*4=7 a bit too many times then I'd like to say 😔

or 4*4=8

[austere anvil]

it was when i was typing out the matrix products i realized the error

[austere anvil]

lol

[manic sleet]

because you double checked always do that on tests :) especially when you got all those numbers

and trust me 3x3 and 4x4s only get way crazier with multiplication

[austere anvil]

yea

i have my math exam in 3 days

im so screwed

.close

[odd edge]

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[manic sleet]

good luck!

[odd edge]

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[timber atlas]

with this, we know that the triangle ABD is equivalent to the triangle EFD

[white kernel]

how

[white kernel]

similar or equal

[timber atlas]

the same occurs with the triangles CBD and EBF

[timber atlas]

equivalent

[white kernel]

how

[timber atlas]

they have the same ratio

[white kernel]

so EF= 12???

[median cedar]

He means same ratio not same length

[timber atlas]

if theres a line crossing a triangle 90° in relation to a face, it creates an equivalent triangle

[white kernel]

so they are similar

like if height is 12

base 2

then the other one would be

18 and 3

??

[white kernel]

pls continue

[timber atlas]

exactly

[white kernel]

ok

explain pls

@timber atlas

dude

[white kernel]

hello??

[timber atlas]

oh man it’s harder than i thought

[white kernel]

ye

[timber atlas]

i apologize as i think i can’t help you any further than this

[white kernel]

np

<@&286206848099549185> can someone help

.close

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[solar walrus]

i need help with this problem

[solar walrus]

im stuck i dont even know where to start

[mystic saffron]

do you know calculus

[solar walrus]

yes

but idk how to apply that here 😭

[mystic saffron]

find derivative of this function first

[solar walrus]

P'(x) = 3ax^2+2bx+c

help pls 😭

<@&286206848099549185>

[near compass]

4 equations will come up

[near compass]

2 for extremum and 2 for normal value of polynomial

[solar walrus]

what 💀

u lost me

[near compass]

Lol

[solar walrus]

like help me solve it i dont understand this terminology

[near compass]

Look at x=-2 p has the value of 3 so a.2³+b.2²+c.2+d =3

Similarly for the other point

Also a

[solar walrus]

yea i did that

and now im stuck

[near compass]

At those given point slope is zro

*zero

[solar walrus]

huh

slope is 0 at point R right

but Q is not a turning point so it wouldnt be

[near compass]

If is not Turning point then u can't solve it

Turning in the sense it's minima right?

[solar walrus]

i dont understand what you're saying im sorry

[near compass]

Gand maara

[solar walrus]

but there is a solution to this i just dont know how to get it

[odd edge]

@solar walrus Has your question been resolved?

[solar walrus]

.close

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[vague void]

do you guys have pascals triangle for high exponents

[vague void]

like 12-20

[tender lagoon]

Probly write a code or smth to generate mate!

[vague void]

uhh

is that the only wayt

[tender lagoon]

U can do it by hand or find it out in Internet

That's another way

[junior grotto]

Or you could calculate nCk

[tender lagoon]

By hand I meant that ^

[junior grotto]

you could use approximations

[vague void]

nvm I found internet

thx guys

[odd edge]

@vague void Has your question been resolved?

[odd edge]

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[clear flint]

Let ( ABC ) be a triangle with ( AB \neq AC ), let ( I ) be its incenter, and ( \gamma ) its inscribed circle, and ( D ) the midpoint of ( BC ). The tangent to ( \gamma ) through ( D ) different from ( BC ) touches ( \gamma ) at ( E ). Prove that ( AE ) and ( DI ) are parallel.

[clever fjord]

Alex

[clear flint]

any idea how to start?

[mystic saffron]

draw a diagram

visualise the problem

[fallen remnant]

Well
If the tangent is at D
How is it supposed to again touch the circle at E

[junior grotto]

Read carefully

it is written tangent through D

not tangent at D

[fallen remnant]

Oh

Mb

[clear flint]

I’m struggling to start with the problem. I understand that I need to prove that lines AE and DI are parallel, but I’m not sure how to use the fact that D is the midpoint of BC and how this relates to the tangent to the incircle

<@&286206848099549185> any hint?

[odd edge]

@clear flint Has your question been resolved?

[clear flint]

feel free to ping me

[odd edge]

@clear flint Has your question been resolved?

[odd edge]

@clear flint Has your question been resolved?

[odd edge]

@clear flint Has your question been resolved?

[odd edge]

@clear flint Has your question been resolved?

[supple mural]

Do you know projective geometry?

If you do, use this lemma:
If ABCD is a harmonic quadrilateral inscribed of (O) and l is the tangent line at D, the intersections of the tangents at A,B,C to (O) and l form a harmonic bundle

After that it’s easy

@clear flint

[clear flint]

that mean that the cross ratio of the four points is -1 right? if I apply this to our problem, I should look at the tangents to the circle at points A, B, C, and the tangent line at D. But I’m not quite sure how to construct this harmonic bundle from the tangents

ok i got it, thx @supple mural

.close

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[supple hound]

hi i need help asap

[supple hound]

not sure how to solve this

inegral of (sin^3nx)dx = ?

[tough mesa]

Is it sin³(nx)

[odd edge]

@supple hound Has your question been resolved?

[supple hound]

yea

but i figured it out

[odd edge]

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[restive hare]

I need help with multivariable Taylor series specifically with sin(x+y)

[restive hare]

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[echo knot]

Find two polynomials p(x) and q(x) that satisfies the given condition and such that p(x) -> +Infinity and q(x) -> +Infinity as x -> +Infinity

[clever fjord]

hecker

$$
\[ \lim_{x\to\infty} [p(x) - q(x)] = 3\]
$$
```Compilation error:```! LaTeX Error: Bad math environment delimiter.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.53 \[
        \lim_{x\to\infty} [p(x) - q(x)] = 3\]
Your command was ignored.
Type  I <command> <return>  to replace it with another command,
or  <return>  to continue without it.```

[echo knot]

There were 3 similar questions before and I solved them

But I don't understand how a polynomial that goes to infinity subtracted by a polynomial of same behaviour can equal 3

[pastel dew]

maybe all is need is p=q+3

[echo knot]

Ohh yes that works, didnt think of that

I am having some trouble with limits you see

But thanks

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[pastel dew]

👍

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[warm crater]

dont know how im supposed to get the equation of the quadratic.

[tough mesa]

They gave you vertex

And a factor

[echo knot]

Well based on the fact that the turning point is in the middle of roots

another root = 4 + (4 - (-2))

= 10

With the formulas of

Product of roots = c/a
Sum of roots = -b/a

[clever fjord]

hecker

[echo knot]

Now you have to find out what a is such that the turning point is (4, 24)

Based on some eye balling in desmos I can say that a= -2/3

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[cosmic marten]

If you have a line defined by:
5x - 3y - 8z + 42 = 0
x - 9y +4z +28 = 0
How do you find the numbers for the direction vector of this line?

[cosmic marten]

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[left seal]

36/64

[left seal]

Cant I do like

6^2 / 8^2

Square root

[elfin zodiac]

Why do you want to square root

[south plume]

what?

are you supposed to simplify?

[left seal]

Yh

[elfin zodiac]

Then why square root

[south plume]

no, you can't do square root

[left seal]

I know answer is 9/16

[lavish mural]

U can't square root

[left seal]

With divison 2

[elfin zodiac]

You can simplify in different ways, depending what do you want to achieve

[lavish mural]

U have to divide by same number

[left seal]

Im asking if i can do square root

[lavish mural]

Yeah u cant

[left seal]

To simplfy

[lavish mural]

Cuz 1/4 isn't equal to 1/2

[left seal]

ok but if i see a big numbers

[south plume]

simplification is the process of transforming what you have while preserving the value

[left seal]

Can i do something else than divison 2

[elfin zodiac]

Square roots has nothing to do with common factors

[south plume]

square root does not preserve the value

[lavish mural]

U can divide by any number

Just divide or multiply

[elfin zodiac]

The proportion when u divide both, numerator and denominator are the same when u divide with the same number

This doesn’t happen when u apply square root to two different numbers

[lavish mural]

Like 1/2 isn't equal to 1/4

But yes u can divide by numbers that aren't 2

So 3/6 can become 1/2 (divide top and bottom by 3)

[left seal]

Ok

.clode

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[frigid canopy]

how would I find the number of triplets satisfying 2^x+2^y+2^z=2336

[frigid canopy]

can I have. a hint

is there any standard way to solve such dopiantine equations?

,w prime factrorise 2336

[clever fjord]

Results provided by WolframAlpha

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

Requested by physicsrocks

[frigid canopy]

ok, so I'm guessing I have to use the prime factorisation somewhere

but where

<@&286206848099549185>

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[graceful sand]

Lets say I have a polar curve:
$r=1-2sin(3\theta)$

[clever fjord]

rainy

[graceful sand]

How would you know that it corresponds with D

I know 1-2sin(theta) has this kind of shape

i just dont know how to handle the 3 theta

[south plume]

1. it has two have 3 "petals" (due to 3 theta)
2. substituting 0 should yield 1

[toxic yew]

you can check where the curve is equal to one

oh mb

you go ahead

[south plume]

so the only option is D

lol np

[graceful sand]

so based upon your 1), my only options are A & D
and then subsituting should give me D

lets see

[south plume]

ye

[graceful sand]

im not sure what it means when its 1

basically r=1 ??

[south plume]

yes

[graceful sand]

Ah so, subsituting 0 as an angle (theta) gives back where r is at that angle

okkkkkkk

i get it

thx

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[mystic saffron]

Is this a valid solution to the problem?

[pastel orbit]

Looks valid to me

[mystic saffron]

Ok thanks!

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[night rapids]

right?

[brittle beacon]

Is y(0) = C given to you anywhere? with exception of that, everything else seems fine to me

[odd edge]

@night rapids Has your question been resolved?

[night rapids]

yes

[brittle beacon]

do you have the original question in picture form? seems strange how there's reference to the integration constant in the question

[night rapids]

Consider the two differential equations
y′(x) = 4py(x)x (1)
y′(x) = y(x)2. (2)
(a) For all C > 0, find a continuous function y : [0, ∞) → [0, ∞)
with y(0) = C which is differentiable on (0, ∞) and satisfies the diffe-
rential equation (1) here

@brittle beacon

[brittle beacon]

Not entirely sure whether that C in the question is supposed to match up with the integration constant you find

[night rapids]

yeah same

[brittle beacon]

Think that the final answer you get should be in terms of that C

[night rapids]

i am not sure whether it is the same C

bit odd to not call it C_1 or somethingh then

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[brittle beacon]

But like say when you get to like the point $2\sqrt{y} = 2x^2 + c$ (little $c$) then you make use of the fact that $y(0) = C$ to get that $2\sqrt{C} = c$

[clever fjord]

@brittle beacon

[brittle beacon]

.reopen

[odd edge]

✅

[night rapids]

yeah it doesnt change too much but i also thought of that

[brittle beacon]

At which point $\sqrt{y} = x^2 + \sqrt{C}$ and stuff

[clever fjord]

@brittle beacon

[night rapids]

@brittle beacon

[brittle beacon]

Yep of course you can simplify that down by cancelling down the 2's

[night rapids]

"Consider the two differential equations
y′(x) = 4py(x)x (1)
y′(x) = y(x)2. (2)
(a) For all C > 0, find a continuous function y : [0, ∞) → [0, ∞)
with y(0) = C that is differentiable on (0, ∞) and satisfies the differential equation (1) here."

Do you think in this i have to prove it it continuous and differentiable?

[brittle beacon]

I mean I doubt you do, but showing it satisfying the differential equation pretty much does all three at the same time (any differentiable function must be continuous!)

Like it should be easy to see that $y = \pqty{x^2 + \sqrt{C}}^2$ is differentiable of course

[clever fjord]

@brittle beacon

[night rapids]

but my intergration constant i more than 0

if C=0

[brittle beacon]

well you're given that C > 0 already, in the question ("For all C > 0...")

Which is a good thing because you can take the square root of it

[night rapids]

https://tenor.com/view/cute-cat-cat-saying-ok-gif-27677526

[night rapids]

i read it as for C>0 (which in turn you have to prove it is)

[brittle beacon]

I more read it that they give you that C is (strictly) greater than zero, which you can take freely, and you then work with that

[night rapids]

do you like cats

[night rapids]

ok nice ill just hope that is what it is

[brittle beacon]

I do

[brittle beacon]

Yep pretty sure that they probably intend it that way (it's the way that is most likely )

[night rapids]

b)
"Find two different continuous functions y : [0, ∞) → [0, ∞) with y(0) = 0 that are differentiable on (0, ∞) and satisfy the differential equation (1) here.

(c) For all C > 0, find a number K > 0 and a continuous function y : [0, K) → [0, ∞) with y(0) = C that is differentiable on (0, K) and satisfies the differential equation (2) here. For the given C, what is the largest possible value of K?"

[frozen parrot]

anyone please help me with this question-Consider that a, b, c, d are positive real numbers satisfying (a + c)(b + d) = ac + bd.
Find the smallest possible value of S=a/b+b/c+c/d+d/a

[brittle beacon]

!occupied

[odd edge]

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).

[brittle beacon]

The differential equation $y'(x) = 4 \sqrt{ y(x)} \cdot x$?

[night rapids]

yes

no, with the root

over y(x)

[clever fjord]

@brittle beacon

[brittle beacon]

So like that?

[night rapids]

$$ y' = 4\sqrt{y} \cdot x $$

here

[clever fjord]

// mav

[night rapids]

yes

[brittle beacon]

Wait for part b or for part c?

[night rapids]

b)
"Find two different continuous functions y : [0, ∞) → [0, ∞) with y(0) = 0 that are differentiable on (0, ∞) and satisfy the differential equation (1) here.

with the equation we have just shown with TeXit ab ove

[night rapids]

b

[mental jetty]

What is it?

y'

[night rapids]

y' = dy/dx

[brittle beacon]

Then we've basically done that already all the stuff above where we got that y = (x^2 + sqrt{C})^2

[night rapids]

yh

[night rapids]

now what

how do i find two constants

when C = 0 ?

since y(0)=0

x=0 , y=0

[brittle beacon]

Basically pick two different [strictly] positive C's really, say C = 4 and C = 9 or whatever

Then they're different functions but as per what you found must satisfy the DE

[night rapids]

isnt that a bit too easy

[brittle beacon]

Well I mean it doesn't need to be that challenging, as it follows from all the hard work you did before

[night rapids]

but how do i just pick C = 4

[night rapids]

C has to be 0, no?

[brittle beacon]

Oh yea, didn't read they do say y(0) = 0, nvm

Well I mean in that case, the one you found satisfies the DE, but also rather trivially the constant function y = 0 also meets all the requirements stated too

[odd edge]

@night rapids Has your question been resolved?

[night rapids]

how is y=0 a value of C

and none of the are above 0

nvm it doesnt have to be thatg here

[brittle beacon]

The function that just outputs zero, is what I meant haha

[night rapids]

what

i guess y=0 would be correct

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@night rapids Has your question been resolved?

[odd edge]

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[mystic saffron]

how would you theoretically do these questions with "Guess and check" method

for the first one

[desert marlin]

4^(what)=32

try 4^2 and 4^3

[mystic saffron]

oh

[desert marlin]

16 and 64

too small too big

somewhere in between

[mystic saffron]

2.5

ah I see

okay then, thank you

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[ember shell]

suppose R is (3,5) and S is (8,-3). Find each point on the line through R and S that is three times as far from R as it is from S.

[ember shell]

i finished finding hte first solution

which was 6.75,-1

however idk how i should find the second point

i have somehwat of an idea

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[versed obsidian]

hello i was trying to solve a question and i wonder if i got the integral correct: two identical conical piles of sand with a height of 1m and radius of 1m overlap each other. the distance between the two vertices is also 1m. find the volume of the combined pile.

[versed obsidian]

i just set up this integral $2\int_{-1}^{\frac{1}{2}}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\left(-\sqrt{x^{2}+y^{2}}+1\right)dydx$

[clever fjord]

Jill ♡

[versed obsidian]

@mystic saffron help

<@&286206848099549185>

[odd edge]

@versed obsidian Has your question been resolved?

[versed obsidian]

<@&286206848099549185>

[night rapids]

I came back now, but how am i supposed to write it? y(0) is the function that just gets 0

ik you cant write here, but noone was helping so i just did it anyways

[versed obsidian]

this isnt your channel anymore...

[brittle beacon]

(You’ll need to open a new channel but tl;dr just have y(x) = 0)

[versed obsidian]

use another one

[night rapids]

$$ V = \iint\limits_{D} \left(1 - \frac{1}{h} \sqrt{x^2 + y^2}\right)^2 - \left(1 - \frac{1}{h} \sqrt{(x-1)^2 + y^2}\right)^2 , dx , dy $$

[clever fjord]

// mav

[night rapids]

could that be an answer?

[versed obsidian]

i dont understand what the 1/h is for

[versed obsidian]

i guess but the cones are identical in volume so we dont need the 2nd root

[night rapids]

idk, i am not really sure on this one tbh

but i think you're in the right direction at least

what class are you doing?

[versed obsidian]

none i found this online

the cones intersect each other on a straight line

so i just assumed i should decrease the integral boundaries

<@&286206848099549185>

[odd edge]

@versed obsidian Has your question been resolved?

[night rapids]

do u have a pciture

[versed obsidian]

i just plotted the thing

they intersect at x = 1/2

[odd edge]

@versed obsidian Has your question been resolved?

[versed obsidian]

.close

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[mystic saffron]

What am I doing wrong here?

1. I found the circumference of the wheel doing $r=0.5(20)$ and then $c=2 \pi \cdot 10$.

2. Then I took the number of revolutions per minute and multiplied it by the circumfrence of the wheel (in inches) which should've gave me the inch/min. $(2 \pi \cdot 10)195=12252.211349$

[clever fjord]

Matt

[mystic saffron]

But for some reason thats wrong, not sure why as multiplying the circumfrence by the amount of revolutions/min should give me inch/min

at least i thought so lol

[odd edge]

@mystic saffron Has your question been resolved?

[hasty dome]

Think about how a standard bicycle works. You pedal, which moves the front sprocket attached to the pedals. The front sprocket moving causes the chain to move. The chain is attached to the front sprocket and the rear sprocket, so the chain moves the rear sprocket. The rear wheel is attached to the rear sprocket, so the rear sprocket moves the rear wheel.

If the cyclist pedals at 195 RPM on the front sprocket with a 6 inch radius, how far in inches does that move the chain along the front sprocket per minute?

@mystic saffron

Then, the chain will also move that far along the rear sprocket. That sprocket has a 4 inch radius, so how many RPM is the rear sprocket doing?

Then, the rear wheel, which has a radius of 20 inches will move that many RPM, so how far does it go in a minute?

[odd edge]

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[mystic saffron]

the answer to this is apparently vertical strech by a factor of 3 (i agree with) and a vertical shift 3 units up????

should it not be horizontal comp by a factor of 1/8?

[manic sleet]

$8=2^3\implies 8x^3=2^3\cdot x^3$

[devout dove]

Try using some of the log laws to separate it

[clever fjord]

PajamaMamaLlama

[mystic saffron]

yeah I get this

but if it were a vertical shift

it would have to be

$(x^2 \pm something)$

[devout dove]

$log_28x = log_28 + log_2x$

[clever fjord]

peteozzy2018

mj

[devout dove]

This is why

[devout dove]

^

[mystic saffron]

okay

but still

where is the shif

t

[devout dove]

Do you mean vertical shift up 3?

Vertical shift right isn’t possible

[mystic saffron]

yeah up sorry

log2 8 is 3

is that why

okay then

[devout dove]

Yep!

[mystic saffron]

thank you

.close

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[devout dove]

Np

[odd edge]

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[fallow thunder]

so when it says distance from origin, surely it means shortest distance right?

because a plane would have infinite number of distances from origin?

[manic sleet]

yes that's what that would mean

[fallow thunder]

well i dont really understand how this works tho

what does root 14 even mean

hmm ig it doesnt have to mean anything, its just for finding equation

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[acoustic lynx]

I mean I wouldn't consider it easy

You could try Cramer's rule(very tedious) and the normal way to find the matrix determinant and compare those

But I don't know if there is an easy way to check

[quasi sparrow]

Just use any online calculator that supports determinants like wolfram

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[merry grotto]

i need help

[merry grotto]

when this is asked do i just have to plug in 4 years where the t is at?

because when i plug that in i get 500^125./e^109/e

[royal herald]

yea find A(4) which is the amount of radioactive lead remaining after 4 years

@merry grotto

and round to the nearest integer

[merry grotto]

okay thank you

(:

[odd edge]

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[tough plinth]

How do I make $k^2(k+1)^2+4k+4/4$ be $(k+1)^2 (k+2)^2/4$

[clever fjord]

ICANTCOMEUPWITHANAME

[tough plinth]

Is it allowed to add k² to the 4k + 4 so things would be much more simpler? Even though k² is attached to (k+1)²

[analog basin]

$\frac{k^2(k+1)^2+4k+4}{4}$ be $\frac{(k+1)^2 (k+2)^2}{4}$

[clever fjord]

Wither

[tough plinth]

Ah so thats how you fraction it

<@&286206848099549185>

[chilly egret]

I don’t think u can, cause it’s stick with the first expression

Priority to multiplication

[tough plinth]

Might have a made the induction wrong

[chilly egret]

Maybe

[odd edge]

@tough plinth Has your question been resolved?

[tough plinth]

Hell no

<@&286206848099549185>

[tough plinth]

There's somthing wrong with this induction and Idk what

[old whale]

,rotate

[old whale]

What are you trying to solve?

[tough plinth]

I cant make that expression from the left equal to the right

[old whale]

Ahh yes

[tough plinth]

Prove that the left side is equal to the right

[old whale]

The k squared on the left is connected to the (k + 1)^2

[tough plinth]

Yes

I cant just move it to 4k + 4 thats bending laws

[old whale]

So it’s actually (k^2 + 4)(k + 1)^2 / 4 being correct

Wait no

I don’t think you can simplify this

Not really

[tough plinth]

;-;

[old whale]

Yeah you can’t simplify it

Best you can do is

(k + 1)(k^2(k + 1) + 4)

[tough plinth]

,rotate

[old whale]

Yes

[tough plinth]

$(k + 1)(k^2(k + 1) + 4)$

[clever fjord]

ICANTCOMEUPWITHANAME

[tough plinth]

Hmmm

It would still be equal no?

[old whale]

Yes

[old whale]

This is equal to

The left side

The right and left are not equal

[tough plinth]

It is though

[tough plinth]

Reference to those at step 1 , I used 3 examples

And it all came out equal

1, 9, and 36

[old whale]

That’s for (k^2][k + 1]^2

/4

[tough plinth]

.close

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[mystic saffron]

question: find the sum of all possible 4 digit numbers that can be made using the digits 2,3,4 and 5 exactly once. I found a way to approach this problem but i think there's a flaw in it. Break the 4 digit number in the following manner: 10³×a +10²×b +10¹×c +d. 'a' can take any of the four values (2,3,4,5) and that would repeat 3! times. Similarly, for b, c and d. So the answer would be 10³×(3! ×14)+ 10²×(3! ×14)+ 10¹×(3! ×14)+ 10⁰×(3! ×14). But i suppose the cases would be repeated. For example, consider the number 2345. It would appear once when a=2, once when b=3, once when c=4 and once when d=5, which means it has been repeated 4 times.

[wooden python]

wym?

yeah of course it would appear 4 times

you're counting each of its four places separately

the thousands, hundreds, tens and ones

[mystic saffron]

but the answer is correct

2345+2354+2435+2453+2543+2534+3245+3254+3425+3452+3524+3542+4325+4352+4235+4253+4532+4523+5342+5324+5432+5423+5234+5243 = 93324= 10³×(3! ×14)+ 10²×(3! ×14)+ 10¹×(3! ×14)+ 10⁰×(3! ×14)

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

<@&286206848099549185>

[mystic saffron]

<@&286206848099549185>

[tepid shore]

the answer is 93 thousand 300 and 24

[mystic saffron]

i know that

[mystic saffron]

AI answer

your answer is wrong

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@mystic saffron Has your question been resolved?

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[sharp bay]

I need help solving this equation system, I see that it is a vandermonde determinant but I'm not sure if that's helpful or not

[odd edge]

@sharp bay Has your question been resolved?

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@sharp bay Has your question been resolved?

[sharp bay]

can someone help me?...

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@sharp bay Has your question been resolved?

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[hidden igloo]

so pi/3 is the only answer and I don’t understand why

[fervent hawk]

hint: when is cos(theta)=1 in 0≤theta≤2π

[hidden igloo]

in the first and 4th quadrant?

[fervent hawk]

not really, it's just exactly when theta=0 or 2pi

[hidden igloo]

yeah that makes sense

i was just overthinking it

[fervent hawk]

Cheers!

[hidden igloo]

thanks

.close

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[winged kernel]

is it possible/allowed to

[winged kernel]

use l'hospital

here

Yes or?

[brittle beacon]

I mean it does have the required form, though it is much more pain to do it with lopital's

You may want to multiply and divide $\sqrt{x^2 + 1} - x$ by its conjugate

[clever fjord]

@brittle beacon

[winged kernel]

i see

ty

.close

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[vapid elbow]

,w integrate (2x+1)/(x^2-1)

[clever fjord]

Results provided by WolframAlpha

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

Requested by __eminem__

[vapid elbow]

Chat can I not

$\int(\frac{2x}{x^2+1} + \frac{1}{x^2+1})$

[clever fjord]

EMINEM

[vapid elbow]

oh shit my bad

.close

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[modest badge]

Need help solving this solved the quadratic but idk what to do with the modul bit

[noble forge]

|x|+7 is always >7 for all real x

so you need to only consider the num

the inequality reduces to x^2-5x+6<=0

can you factorize the quadratic?

[modest badge]

Mmm so |x|+7 >0?

[noble forge]

yes

[modest badge]

Can it be equal?

[noble forge]

no

the logic is pretty simple

|x|>=0 for all real x

[modest badge]

If I square it would it work?

[noble forge]

so, |x|+7>=7 for all real x

[noble forge]

what would you square?

[modest badge]

This

[noble forge]

why would you want to square it?

[modest badge]

Ok so what would i do?

[noble forge]

just focus on this

[modest badge]

Yea i solved that

@noble forge

[noble forge]

you should have something like (x-2)(x-3)<=0

[modest badge]

Ok but where dose 7+x go?

@noble forge

.close

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[mystic saffron]

what can you say about the order of Q

[hearty cape]

wdym by that

this is the solution, im curious as to why the inverse R is on the right of I on the rhs instead of being to the left of the rhs

[pastel orbit]

because R is on the right of Q on the LHS

[hearty cape]

So XA = B
= XA*A' = B A'

?

A' is just A inverse idk how else to write it

[pastel orbit]

yes

[hearty cape]

and if B was a 2x1 matrix there would be no solution?

[hearty cape]

is this scenario

[pastel orbit]

if a matrix isnt square, it doesnt have a 2 sided inverse

[pastel orbit]

what are X and A in that case?

2x2?

[hearty cape]

X is 2x1 A is 2x2

that isnt possible right?

a question like that isnt possible

[pastel orbit]

well you cant even multiply the matricies X and A in that case

[hearty cape]

yeah

[pastel orbit]

since the codomain of one isnt the domain of the other

so that is impossible, yes

[hearty cape]

so if it was ABC = X, where they are all 2x2 matrices
A'ABC = A'X
A'AB'BC = B'A'X?

if i were solving for c?

or would it be
A'AB'BC = A'B'X

[pastel orbit]

let's see

ABC = X

so (AB)'(AB)C = (AB)'X

so C = B'A'X i believe

[pastel orbit]

this is the correct answer, but slightly wrong process

A'ABC = A'X yes

but it's B'A'ABC = B'A'X

B' needs to go to the left of A'

[hearty cape]

why?

[pastel orbit]

because B'A'ABC = B'(A'A)BX = B'(I)BC = B'BC = C

if you prefer to think about matricies as linear transformations

then it's because we first apply C, then apply B, then apply A to get the matrix ABC

to "undo" this

[hearty cape]

ahh

[pastel orbit]

we must apply the inverse linear transformations backwards

[hearty cape]

like 'followed by' is reversed

so b followed by a

is AB

not BA

[pastel orbit]

so A' comes first, then B' comes after

in order to get back to C

[hearty cape]

in this solution why do they not start from R

[pastel orbit]

yeah it's the same as for function composition

[pastel orbit]

because Q is in the middle, so it doesn't really matter which we do first

you get the same result in the end either way

[hearty cape]

i see

thanks man

[pastel orbit]

[f o g o h]^-1 = h^-1 o g^-1 o f^-1

[pastel orbit]

👍

[hearty cape]

anyone know much about physics, specifically how to find critical angles and refractive indices

[spring pebble]

Which grade is that in if I may ask?

[hearty cape]

not in usa but i am 16

[pastel orbit]

for refractive indicies, if you know the speed of light in a material, then divide c by that speed to get the refractive index of the material

[hearty cape]

the question is a weird one

[pastel orbit]

show?

to answer about the critical angle

the critical angle is the angle of incidence such that the resulting angle of refraction is 90 degrees

any angle of incidence greater than the critical angle will result in total internal reflection

[hearty cape]

part iv, part iii is backround info i guess

i know the basics like how to find crit angle and all that

but this question is a weird one

[pastel orbit]

what's your issue with it?

if you can calculate critical angle, then this is pretty straightfowards

[hearty cape]

show pls

[pastel orbit]

what

i want to know where your struggle lies

[hearty cape]

Its asking for the critical angle of the core/cladding boundary

[pastel orbit]

yes, it is

[hearty cape]

but we only know the refractive index of air to core and air to cladding

[pastel orbit]

why is that an issue?

[hearty cape]

because we need the critical angle from core to cladding im pretty sure

[pastel orbit]

yes, but we have the refractive indicies

[hearty cape]

how would i use them to find refractive index or critical angle of the core / cladding boundary

refractive index = 1/sin C

btw

[pastel orbit]

i'm a bit confused

the refractive indicies of the two materials are all you need

what else is there to calculate?

theres no 3rd material

as for critical angle, you said that you know how to calculate it i think?

so it's the same as always if you've done it before

[hearty cape]

how would you do it? im unsure as to which figure to use

[pastel orbit]

okay

you know Snell's Law, yes?

this thing

[hearty cape]

yes

[pastel orbit]

observe that we want to solve for θ_1

we know n_1, n_2, and θ_2

so it's just a matter of rearranging