help-10

[obtuse pebble]

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[obtuse pebble]

@willow acorn Has your question been resolved?

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[warm flame]

Looks like I made an error here, can anyone find it? I'm confused

[warm flame]

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[obtuse pebble]

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[obtuse pebble]

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[crystal horizon]

Your Mistake:
[(x-7)^2 = x^2 + 14x - 49]

[crystal horizon]

.close

[high lily]

.close

[obtuse pebble]

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[slate comet]

I saw this on bartelby and wanted to try and solve it. I just want to know: how do you get the total acceleration of point C?

This was my solution for it, but I’m pretty sure it’s wrong given that one of my friends told me the total acceleration isn’t purely horizontal.

[obtuse pebble]

@slate comet Has your question been resolved?

[obtuse pebble]

@slate comet Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[spice plinth]

Guys what is the key idea in a question like this

[spice plinth]

<@&286206848099549185>

[wooden cipher]

Have you drawn a diagram

[spice plinth]

Yes

Here

[wooden cipher]

I cant think of anything better than trig spam

Still thinking

[spice plinth]

Probably not intended

[wooden cipher]

Yeah

[spice plinth]

I thought to draw circles but that didn't work

Because it cuts the sides at 2 points

[wooden cipher]

I think the best is to set up an equation like this:
(XY/2)(1+sqrt3)=sqrt3

I got this by making a new point N that is the midpoint of XY

NM+NA=sqrt3

[spice plinth]

Hmm that could work

Let me try too

[sharp forum]

i got the answer its root 3 am i right ??

[rancid yacht]

does one use of law of sines not give you MX?

[spice plinth]

But you don't know that XY is parallel to BC

[spice plinth]

Idk

[wooden cipher]

You do

[rancid yacht]

use XM/sinBXM = BM/sinBXM right

everything except XM is known

find it and multiply by sqrt2

[sharp forum]

Since AXMY is an isosceles right triangle, we know that AX = AY and angle AXM = 90 degrees. Since ABC is an equilateral triangle, angle BAC is also 60 degrees.

[spice plinth]

Trig isn't Intended I think

[rancid yacht]

oh you want a non trig sol?

idk i think trig is really easy here

[wooden cipher]

AXM is not a right angle

[rancid yacht]

its like 3 calculations

[sharp forum]

i have marked diffrent

[spice plinth]

I don't know trig lol

[rancid yacht]

ok in that case you can do this

draw the height of MXY to XY from M

draw the height of AXY from A to XY

you know these 2 in terms of XY

and you know the height of the equilateral triangle

solve for XY

[wooden cipher]

Thats what i did

Pythag only

[rancid yacht]

ah nice

yeah

[spice plinth]

How do you know XY is parallel

To BC

[wooden cipher]

By construction

SAS on XMB, YMC

XB=YC

[rancid yacht]

yep

[wooden cipher]

Both are the same way up the triangle

[spice plinth]

Angle XMB = angle YMC is not given

I am definitely missing something

[rancid yacht]

i think what they mean is that like

XMY can be constructed in such a way that it satisfies the conditions given

and in that construction XY || BC

[wooden cipher]

Its the only way

[spice plinth]

I did some angle chasing and there is some contradicting because of this

Let me check again

That would mean XMB is 15

[wooden cipher]

How

[rancid yacht]

uhhh no

(180-90)/2

[spice plinth]

Take XMB as theta

BXM is 120 - theta

AXY is 60 -45 + theta

= 15 + theta

Now if XY is parallel :
15 + theta = 60

Ya it's 45

Kk

Thanks

[rancid yacht]

👍

[obtuse pebble]

@spice plinth Has your question been resolved?

[obtuse pebble]

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[timid silo]

this is a follow up from a previous question i asked but the channel got timed out: find a basis of P2 in which the vector x+3 has all coordinates equal to 1

[timid silo]

so i set up my equation

x+3 = b1 + b2 + b3

i need to find b1, b2 and b3

[twin sapphire]

whats x?

[timid silo]

idk

[twin sapphire]

and also whats the space

[timid silo]

P2

second polynomial degree

i dont know what x is

[twin sapphire]

oh its polynomials

my bad

[timid silo]

its some vector

[twin sapphire]

x is not some vector

its the x of polynomiald

[timid silo]

ah ok

so x+3 is the vector

so now i need to find a basis in which the vector x+3 relative to the basis has coordinates equal to 1

[twin sapphire]

ok so do you know a basis of P2 first?

[timid silo]

well the general basis for p2 is (1,x,x^2)

[twin sapphire]

idk wym by general but yes its a basis

usual maybe

[timid silo]

the standard basis

[twin sapphire]

ok so whats the issue with this basis?

[timid silo]

the coordinates relative to it for the vector x+3 is not 1,1,1

[twin sapphire]

yeh but more specifically

what are the cordinates

[timid silo]

let me see

(3,1,0)

sorry it took me long lol

i am slow

i calculated it

[twin sapphire]

ok so maybe you can modify the basis a little bit to make the first cordinate be 1 no?

[timid silo]

i found x^2, x-x^2, 3

is this a basis for p2?

don't i need to check?

if it spans and is linearly independent

[twin sapphire]

if its linearlt independant you are done

since you have 3 vectors

and p2 is of dim 3

[timid silo]

i see

ok

[timid silo]

so here is what i did:
$k_1x^2 + k_2(x-x^2) + 3k_3 = 0 \
(k_1 - k_2)x^2 + k_2x + 3k_3 = 0$

[warm shale]

Derivative

[timid silo]

here we get k_3 = 0

k_2 = 0

and k_1 = k_2 = 0

thus it is linearly independent

therefore it is a basis for P2

we are done!

.close

[obtuse pebble]

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[next wagon]

can someone explain why the absolute convergence of a series on an open space U => the series converges normaly on all compact spaces of U ?

[hexed gull]

a series or a power series?

[next wagon]

power series

[hexed gull]

hm I cant think of a proof directly, but you can look for it via weierstraß

weierstraß maj.-test

https://en.wikipedia.org/wiki/Weierstrass_M-test

[obtuse pebble]

@next wagon Has your question been resolved?

[obtuse pebble]

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[thin oxide]

Is it possible for anyone to explain to me how to use "voisinage" I think it is named Neighbourhood in english, in "continue" fonctions.

[thin oxide]

<@&286206848099549185>

[timid silo]

Do you know the definition to it?

[thin oxide]

Yeah but I found an issue using it.

[timid silo]

what issue

[thin oxide]

using it to extend the range of definition of a function "prolongement". so I can prove that it continues to a point.

[twin sapphire]

a neighborhood of a point M is a set such that M is in the interior of the set

state your problem then

[thin oxide]

you have the following:

f(x) = sin(x)sin(1/x) what can you do to see if f(x) can be extend to the point x=0.

[twin sapphire]

calculate the limit?

|sin(x)|< |x|

[thin oxide]

The steps he mentioned in the solution:
x is in v(positive 0): -sinx< f(x) < sinx => limx->+00 f(x)=0
Same with Negative 0.

[twin sapphire]

yes

[thin oxide]

why he used "voisinage"?

[twin sapphire]

tu peux ecrire en francais exactemznt ce que le prof a ecrit?

V(0+) ca veut pas dire grand chose

[thin oxide]

I know I could just calculate directly limx->0f(x).

• encaderement de f(x)

[twin sapphire]

they used the neighborhood to be rigorous about the fact that x is not =0 maybe because f is not defined at 0

but thats being overzealous

especially so if this isnt advanced topology

just stating x>0 is enough

or even x=/=0 and working with absolute values

[twin sapphire]

the proof written doesnt need to use the concept of neighborhood

just calculate the limit for x-->0, x=/=0

[thin oxide]

EXACTLY.

These Profs doing extra blah blah for nothing.

Anyway, thanks a lot.

.close

[obtuse pebble]

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[twin sapphire]

just write pour tout x>0 if you want to be rigorous about x

[obtuse pebble]

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[obtuse pebble]

@fallow pasture Has your question been resolved?

[timid silo]

non

no

no

n

o

[obtuse pebble]

@fallow pasture Has your question been resolved?

[obtuse pebble]

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[old grail]

Define a prism

[tardy epoch]

,w prism

[warm shale]

Results provided by WolframAlpha

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

Requested by rie.mann

[old grail]

,w pyramid

[warm shale]

Results provided by WolframAlpha

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Requested by anshul6368

[old grail]

,w Platonic solid

[warm shale]

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Requested by anshul6368

[old grail]

What is the shape of the side of a prism

lateral sides

[median nest]

Parallelogram

[old grail]

Thanks

[median nest]

.close

. close

[old grail]

.close

[obtuse pebble]

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[old grail]

.close

[obtuse pebble]

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[wintry viper]

What's a Jacobian?

[wintry viper]

I understand it's a row vector with the partials of every variable in the function. Is it like a change of basis for multivar variable substitutions

[worn yoke]

the derivative describes the best linear approximation of a function in the neighborhood of a given point (i.e., the tangent line). similarly, the jacobian matrix describes the best linear transformation that approximates a given transformation in multiple variables. This is useful in multivariable calculus since the determinant of the jacobian matrix (also called the jacobian) describes how much area (or volume, etc) is changed by the transformation/change of variables. see:
https://www.youtube.com/watch?v=bohL918kXQk
https://www.youtube.com/watch?v=wCZ1VEmVjVo&pp=ygUIamFjb2JpYW4%3D

[lone sky]

ik only jacobin club lol

is this some maths concept

[obtuse pebble]

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[wintry viper]

That was really helpful

Thank you

[obtuse pebble]

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[toxic elm]

Can someone help me with this question please?

[versed stratus]

Is this from a test??

a test you're doing now?

just want to be safe, didn't mean to insult you

[toxic elm]

No is not a test

I'm at home right now

Is a past paper

Is actually my homeowrk

[rocky shale]

you can factorize

180

and 80

[toxic elm]

What is the next step after the factorising?

[rocky shale]

take root5 common

from the numerator

its basically (6root5 - 4root5) / root5

which is 2

[toxic elm]

Okay thank you

[obtuse pebble]

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[obtuse pebble]

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[next wagon]

hello, could someone help prove that the poles of the compex function f(z) = Pi* cotan(PI*z) are simple ?

[next wagon]

the poles are reached when z is an integer

[obtuse pebble]

@next wagon Has your question been resolved?

[next wagon]

<@&286206848099549185>

[obtuse pebble]

@next wagon Has your question been resolved?

[next wagon]

.close

[obtuse pebble]

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[obtuse pebble]

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[toxic elm]

Can someone help me with this question please?

[toxic elm]

Do I need to use substitution?

[royal basin]

can we see part a

given the word HENCE at the beginning, we're gonna need to use something from part a.

[royal basin]

ok so you have reduced the thing to an equation in x only

2x^4 - 15x^2 - 50 = 0

with this in mind do you see what to do?

[toxic elm]

Is my part a correct?

@royal basin

[royal basin]

looks fine except for missing =0's in the two lines above the last

but i believe if this is cambridge or igcse you wouldn't get marked down for missing these

[obtuse pebble]

@toxic elm Has your question been resolved?

[toxic elm]

okay thank you

[toxic elm]

Binomial expansion?

or factorising?

[royal basin]

idk what you were going to expand there when it's already fully expanded.

but i would make this into a quadratic by substitutiong t := x^2

then solve the quadratic by any means available

[toxic elm]

But don't I need to solve the simultaneous equation?

[lone sky]

wat is surd

is it like some number in a fractional power?

[royal basin]

yes, but reducing to a single-variable equation is the first step in doing that

or i guess the first goal

like almost no matter what you do -- substitution, elimination, whatever else -- your intermediate goal is to get an equation that has only 1 variable in it

then solve it

then get the value of that variable (in your case x)

and plug it in everywhere and in doing so get the other variable

[royal basin]

"surd" means a number with some square roots in it, to say it vaguely

[lone sky]

yeah precisely

[obtuse pebble]

@toxic elm Has your question been resolved?

[obtuse pebble]

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[waxen forge]

If two zeroes of a cubic polynomial px³+3x²-qx-6 are-1 and-2, find the third zero and also the values of p and q.

[rocky shale]

using these two roots

find the value of p and q

then find the 3rd root

[waxen forge]

Can you my solution for finding the value of p and q.

<@&286206848099549185>

[rocky shale]

Cmon you can do this on your own

[waxen forge]

I just need if this answer is correct because my friends got different answer

[lucid flare]

in the last step

-2+q=3

you accidentally wrote q=3/(-2)

while it would be q=3+2

[waxen forge]

Yeah

I know

q= 5

[waxen forge]

@lucid flare So -1 + -2 + Gama = -3/2

[waxen forge]

But b is 3x²

How did you got -(-3)/2

[meager laurel]

( ax + b = 0 \implies x=\frac {-b}{a} )

[warm shale]

Natural7

[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.

[waxen forge]

So it should be -3/2

Not -(-3)/2

[meager laurel]

If found b=-3

[waxen forge]

The answer is 3/2

Right??

[meager laurel]

If you found p=2 and q=5 , then be sure that the third zero is 3/2

[waxen forge]

Yeah

.close

[obtuse pebble]

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[obtuse pebble]

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[nimble gulch]

shouldn't be terms in descending order when doing long division?

[royal basin]

ordinarily yes

but they do it differently

[nimble gulch]

so both are correct?

[royal basin]

wdym

[nimble gulch]

?

[royal basin]

what do you mean when you say "both are correct"?

[nimble gulch]

so when i do long division i can sort terms in ascending order as well as descending?

[scarlet gale]

No. That's why they say that they use a procedure like long division rather than that they use long division itself.

[nimble gulch]

i see

[scarlet gale]

They're not dealing with polynomials, so there's no way to do regular long division.

You can't start with the term with the highest degree because there is none.

[nimble gulch]

it's weird though because they have not provided what it means to divide two series but provide procedure to do that

what is the definition of division of two series?

i know the definition of division of two polynomials as well as the procedure

but here only procedure is provided

[scarlet gale]

It's the inverse of multiplication.

So, it's what you multiply the denominator by to get the numerator.

So, you can use the definition of multiplication to get the definition of division.

[nimble gulch]

ok tnx

.close

[obtuse pebble]

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[whole dew]

topic: Antidifferentiation and U-Substitution
problem: why do I need to substitute -u with v, can i not operate on e raised to -u?

[ripe venture]

Yeah I don’t know on that one, the v sub seems unnecessary

Just a “here’s e^x instead of e^-x”

[whole dew]

if I were to convert it into 1/e^u. That is still operable right

here is the previous substitution, might help

[ripe venture]

Sure or just treat as e^-x, whichever

But if you’re wanting to flip e^-u as 1/e^u, then maybe that’s why they did the u=-v

[whole dew]

the final answer also:

[ripe venture]

So you don’t do silly fraction stuff

[whole dew]

https://www.youtube.com/watch?app=desktop&v=oMkQWXNxEVI

oh I think e^-x doesn't satisfy this:

i think this identity states that the exponent must be positive

I may be wrong, hoping someone can confirm or un-confirm

[ripe venture]

Just saying \int e^x dx -> e^x is when it’s just ^x

But there’s related identities you can get pretty quickly from this substitution

Like e^ax dx -> (e^ax)/a

[whole dew]

oh wait wait

so

e^-1x = (e^-1x)/-1

woahhhh ive never seen that identity

[ripe venture]

Yep

You can show it pretty quick

Using exactly what was shown

Or seeing that the derivative lines up

[whole dew]

is a^xdx=1/lna a^x + c applicable for e, considering e is a constant

although i would think this leads to nowhere

[teal depot]

ln e is 1

so yes u can put it for e

you'll just get e^x

[whole dew]

so right back where it began

okay so i skipped ahead to the next module, apparently this identity does pop up. So the example given was used just to show U-Substitution despite having already operable terms. thank you @ripe venture and @teal depot

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[obtuse pebble]

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[worn cove]

could anyone help find my mistake

picture on the right should be the solution

i got 0 🫣

[inland matrix]

What's the question?

[worn cove]

just to solve rhe differential equation

[inland matrix]

Oh ok

[tender stratus]

are you differentiating wrt to alpha in (2)

[worn cove]

ialso messed up by replacing the x to y, because im used to using y and x instead of x and t

but it shouldnt change my calculations

[worn cove]

to t

oh

wait

i differentiated to alpha , thats my mistake, but wouldnt it =0 if i differentiate to t?

then i wouldnt have any solution

[tender stratus]

something seems off about the assumptions you made here. what does y_p represent here

[worn cove]

particular solution

ithink

i indeed messed up my yp

[tender stratus]

I seem to have forgotten most of my DE knowledge, but I think you should be assuming x(t) to be a quadratic polynomial here since the second derivative has no t left

[worn cove]

youre right i think i got it now

thank you sm!

[tender stratus]

no problem

[worn cove]

.close

[obtuse pebble]

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[hasty pine]

How do I solve this trig question?

[hasty pine]

I tried solving it but when it got to putting it in my calculator I got imaginary roots

[royal basin]

$\cot^2(\theta) = \csc^2(\theta) - 1$

[warm shale]

Ann

[hasty pine]

Oh I see now

Thanks!

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[reef merlin]

Why has my lecturer put the third bracket for the Jt=1/3 formula different as from the others?

[reef merlin]

original question

[plucky condor]

you are missing only a single bracket

1/3((0.25

do it like this see it will all make sense

[reef merlin]

I see, just confused as why the value of t was changed to 0.01

[obtuse pebble]

@reef merlin Has your question been resolved?

[obtuse pebble]

@reef merlin Has your question been resolved?

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[haughty swallow]

Wait I realized my mistake

MY BAD

.close

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[obtuse pine]

can someone help me with please

[obtuse pine]

specifically 2b

the topic is roots of polynomials

<@&286206848099549185>

[carmine pasture]

vieta's formulas?

[obtuse pine]

yeah

i have managed to get up to the part where i expand the bracket, but subsituting k im getting stuck on

and finding the actual value of k

<@&286206848099549185>

<@&286206848099549185>

[civic sapphire]

Could you show us your expansion? What you expanded and the result you got?

[obtuse pine]

yeah one sec

this is the question

[meager laurel]

The expansion is not correct .

[obtuse pine]

ok let me redo it, thanks, but regarding the q

question, i dont know what to do after the expansion

[meager laurel]

After that , you can use the identification method .

[obtuse pebble]

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[placid rose]

Another day

Another set of problems

Here’s the first one

Wants me to find the domain and solve

Usually domain is all real so let’s just work on solution

How do I find the solution

[hot harness]

I think you should solve for x first

[placid rose]

Ok

How do I do that

Well I know how to solve in general

For x

But in this situation

What do I do

[timber fox]

inspection ?

[placid rose]

?

Bro

I don’t get it

[timber fox]

I mean just looking at the equation the solution is fairly obvious, if you want to solve for it algebraically I guess moving everything to one side and writing everything with a common base could work

[placid rose]

Ok

Well

I know I can write 9 as 3^2

Would that make it 3^(2(x+1))

[timber fox]

yep

[placid rose]

Hm

Ok

[timber fox]

I would try to isolate the finicky bits

so your 3^x

[placid rose]

But I can’t really get rid of anything can I

[hot harness]

Algebraicly solving for x is a little tricky. You can get an integer solution by trying numbers tho

[timber fox]

you can rewrite this in a helpful form

[hot harness]

But I do agree with this process to make it easier to solve

[placid rose]

I just realized how dumb I am lol 🪦

I’m only in 10th grade and this isn’t even school I’m doing work for

But anyway

[placid rose]

You mean like testing numbers?

[hot harness]

Yes

[placid rose]

Wait

That brings me an idea

[timber fox]

its quite a bit quicker in this case :D

[placid rose]

Can I graph this?

[timber fox]

u can

[placid rose]

Nice

That’s gonna save so much time

[timber fox]

u would find the solution in the process of doing so

[hot harness]

But it's kinda hard to see initially. If you simplify you can figure it out quicker

What did you get

[placid rose]

Some very weird numbers

I wonder if they want it written differently

[timber fox]

ah I thought you meant graph on paper by computing numbers

[placid rose]

No

Not rlly

[timber fox]

I would consider 9^(x+1) + 3^(x+2) as one 'thing'

and 18 as the other

[hot harness]

No u got to graph 9^x+1 + 3^x+2

Yea

[placid rose]

Oops my bad

Ok so I got 0

As my answer

Wait

[hot harness]

Yes

[placid rose]

I feel stupid

Why didn’t I think of this

Well keep in mind

If you see me do any stupid stuff

It’s cos I’m stupid at math

But you won’t remember me so that doesn’t matter

Anyway I got it rig it

[hot harness]

So how can you find the domain now

[placid rose]

Well

I got it right by typing x=0 for the solution

Usually on every problem with a domain and soliton

The domain is always All real

And solution is whatever I got

[hot harness]

Is the domain not 0

The domain is all possible x values

And 0 is the only possible x value

@timber fox Am i dum

[placid rose]

You’re probably not

But this website says All real

The solution is 0 tho

(Ur probably smarter than me)

Ok

So I know how to logarithm

With numbers

Like I know that

Log5^25 is 2

But like

In a situation like this

Where it’s 5^3

[hot harness]

just know a^log base a of x =x

[hot harness]

ok lmao sorry

[placid rose]

Yeah I mean

Ur like pre university

And helpers

So that’s good

(I’m only a high schooler that is doing this math way too early)

Anyway

Can I put 5 in

Just wondering

Like there’s a 5 outside the log

On the first part

Could I slot that in

[hot harness]

You could but I think it's easier to cancel the 5 and 5^2

Wait sorry the spacing is confusing

[placid rose]

That would leave me with this

Whoops wrong photo

[hot harness]

is the second log_5(3) in the exponent of 5^2?

[placid rose]

It’s 5^2log_5(3)

[hot harness]

like this?

Ok

[placid rose]

That would give me

5^-log_5(3)

[hot harness]

I would cancel the 5^(2*log_5(3))

[placid rose]

Ok hold on

I have everything in common

Except for the 2

[hot harness]

Do you get this step

[placid rose]

Not sure what’s going on here

[hot harness]

The 5^log base 5 cancel

[placid rose]

So it’s just the same thing but without the 5

Then what

[hot harness]

Then simplify it

so the3^2 is 9

[placid rose]

From 5 to 3

Oh

Wait but

How did you get 3^2

[hot harness]

Look up log laws

sorry i gtg

[placid rose]

I understand

Soooo

[obtuse pebble]

@placid rose Has your question been resolved?

[obtuse pebble]

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[toxic elm]

I still don't really understand how to do part b can someone help me please?

[native inlet]

this is substituion

do you know what that is?

[obtuse pebble]

@toxic elm Has your question been resolved?

[toxic elm]

okay got it

thank you

[obtuse pebble]

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[proven oar]

how has the first line been simplified into the second

here is entire working/question

[tardy epoch]

Binomial theorem

[proven oar]

wdym

[tardy epoch]

What

[proven oar]

like can you explain a bit

[tardy epoch]

Do you know binomial theorem

[proven oar]

yes

[tardy epoch]

apply that to (1+3x)^n

[proven oar]

oh yeah ok

i get it

ty

.clos

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[zealous parcel]

hey! i was a little stuck on this one and was wondering if i could get help with starting

[zealous parcel]

im assuming for example g(2) is gonna be 4 based on the graph but im stuck algebraically getting there and explaining my work for the next problems

[dull rune]

Ok so, $$g(0) = \int_{0}^{0} f(x) dx.$$ So what does that mean?

[warm shale]

Zander

[zealous parcel]

wait would it be f(x)dx or f(t)dt still

if i follow the same logic wouldn't that be the area under the curve on the graph from 0 to 0, but that's just a line?

or i guess a point rather

[dull rune]

Yea and what's the area of a line

Or what would it make sense to be?

[zealous parcel]

undefined? i thought a line didn't have area

[dull rune]

Well let's think about the limit as x goes to 0 from both sides

Of g(x)

What number will that approach

[zealous parcel]

2?

or am i confused

[dull rune]

Well, we are taking the area of something smaller and smaller each time

Like g(0.000001) is close to what

[zealous parcel]

oh wait does it just approach 0

[dull rune]

Yea

[zealous parcel]

so then g'(0) is also 0?

[dull rune]

So we like to have $\int_a^a f(t) dt = 0$

[warm shale]

Zander

[dull rune]

What tells us about the relationship between derivatives and integrals

$g'(x) = \frac{d}{dx}\int_{0}^x f(t)dt $

[zealous parcel]

they're the inverse operations of one another

[dull rune]

Yea, so what'd g'(x) be

[zealous parcel]

i guess i'm not totally sure

my assumption was g'(0) would be zero since the derivative of a constant is always 0

and if g(0) = 0 would it not be the same either way?

[dull rune]

Well g(x) is not a constant function right

[zealous parcel]

oh i see

[dull rune]

We have that $g'(x) = \frac{d}{dx}\int_{0}^x f(t)dt.$ I will input the fundamental theorem of calculus below: $$\frac{d}{dx}\int_a^x f(t) dt = f(x),$$ for nice enough functions

[warm shale]

Zander

[zealous parcel]

hmm

i'm sorry i'm still a little stuck since my instinct is that if g'(x) = f(x) then it would still be 0?

sorry i think i'm missing something here

[dull rune]

Ok so if g'(x) = f(x), then g'(0) = f(0) = (what)

[zealous parcel]

ohhhhh

2

[dull rune]

Yea

[zealous parcel]

okay great thank you :) !!

now im stuck on g(-2)

for this, do i just find the slope and guess based on that

[dull rune]

Ok let's go back to how integrals are defined

Did you cover this in your class?

[zealous parcel]

i guess so, as the area under the curve

[dull rune]

Did you see like, Riemann sums or whatever

[zealous parcel]

yeah

[dull rune]

Ok so let's just say f is continuous for simplicity sake. If $[a,b]$ is an interval, I wanna talk about the area from $a$ to $b$ of $f$ and in doing so, I divide into a bunch of intervals and take a limit:

$$\int_{a}^{b} f(t) dt := \lim \sum f(x*)\Delta x.$$

[warm shale]

Zander

[dull rune]

If I wanna go the other way, and say something like ``what is the accumulation going from $b$ to $a$," then it is just the negative of "what is the accumulation from a to b."

[warm shale]

Zander

[dull rune]

So like I take $$\int_{b}^{a} f = -\int_{a}^{b}f.$$

[warm shale]

Zander

[dull rune]

(And using this kinda of what of defining the integral I can also prove the Fundamental theorem of Calculus in like two lines:

Let's work with $f'(x)$ over $[a,b]$ then for each rectangle, $\Delta y \approx f'(x^)\Delta x,$ and thus $f(b) - f(a) = \lim\sum \Delta y = \lim\sum f'(x^)\Delta x = \int_a^b f'(x)dx.$ Which is a digression, but the intuition is: the change in y is f(b)-f(a) but that's exactly the change in y collectively over each rectangle (which cancel out in a way). But that is nothing more than the derivative at some point times $\Delta x,$ which is exactly your integral)

[warm shale]

Zander

[dull rune]

This is the main thing youd be interested in though

[zealous parcel]

okay i'm getting a tiny bit lost hold on

[zealous parcel]

but for this, i'm a bit confused bc if ur summing on the interval a to b for, say, x+y+z then summing from b to a would give z+y+x? why would they be inverses

and i think i got a bit lost on the relation to g(-2) sorry

[dull rune]

Well the idea is the things we summed we do like $x_{i+1} - x_i$

[warm shale]

Zander

[dull rune]

And now we wanna do $x_i - x_{i+1}$

[warm shale]

Zander

[zealous parcel]

okay this is starting to get a little more familiar.. what is $x_i$?

[warm shale]

newchucks

[zealous parcel]

is that the initial

[dull rune]

It's where each of the rectangles start:

[zealous parcel]

wait huh

wdym where each of them start

[dull rune]

Like we cut the interval [a,b] up into a bunch of small pieces

[zenith raft]

why is this necessary

[dull rune]

To understand why we do things the way we do

[zealous parcel]

i mean i get the idea of reimann sums but i'm confused on what exactly x_i is representing and its relation to g(-2)

i'm sorry i might not be totally understanding

[dull rune]

Im motivating I guess how we would wanna do this

Cause what is $g(-2)$, well it's just $g(-2) = \int_{0}^{-2} f(t)dt.$

[warm shale]

Zander

[dull rune]

So how do I deal with that expression

[zealous parcel]

well i think what i'm trying to figure out is how to do it graphically or algebraically – like from the graph i would guess i would need to find the slope but algebraically i'm not sure since i don't have an equation to plug in for f(t)

[dull rune]

Ok so do you agree that $\int_{0}^{-2} f(t)dt = -\int_{-2}^0 f(t)dt?$

[warm shale]

Zander

[zealous parcel]

i'll take it as a given for now since i'm a little shaky on exactly why but i guess i wanna see where it's going

[dull rune]

Ok

So $\int_{-2}^0 f(t)dt$ is saying "what is the area under $f$ between $-2$ and $0$, right?

[warm shale]

Zander

[zealous parcel]

right

[dull rune]

So let's look the graph, what is the area from -2 to 0

[zealous parcel]

well the slope is 1/2 so i'd find the area under y = 1/2x on the interval [-2, 0]?

[dull rune]

Yea, you can also just do some basic geomerty if that's easier

You have a rectangle that's 1 by 2 and a triangle that's also 1 by 2

[zealous parcel]

true true although maybe it'd be useful for future problems if i knew how to do it algebraically, do you think you could help explain that method?

like is there a formula im meant to plug into

for reimann sums or something

[dull rune]

Of "what is the area under 1/2x from -2 to 0"?

[zealous parcel]

yeah

well

i mean ik thats what the integral is

but in terms of prior math

like summation notation or something

which is what the reimann is right?

[dull rune]

You could set up an integral: $$\int_{-2}^0 \frac{x}{2} dx$$ and evaluate it

[warm shale]

Zander

[dull rune]

This is probably the next easiest; we don't really like using Riemann sums unless we wanna understand why something is the way it is

[zealous parcel]

so would that just be $\frac{-2}{2} - \frac{0}{2}$?

[warm shale]

newchucks

[zealous parcel]

(which is -1 - 0 = -1?)

[dull rune]

x^2/4 from -2 to 0

[zealous parcel]

wait huh

[dull rune]

You need to take the antiderivative of x/2

[zealous parcel]

ohhh and then plug and subtract from there?

[dull rune]

Recall: $\int_{a}^{b} f'(x)dx = f(b) - f(a)$

[warm shale]

Zander

[dull rune]

yea

and then to get the answer you want, take the negative

[zealous parcel]

i see

which is the same as just doing this method for from 0 to -2 which is what they ask originally right

[dull rune]

and i guess to remember that is "we are now going in the other direction, so we should be negative"

[zealous parcel]

bc that would be 0^2/4 - 2^2/4

which would give -1

or did i mix something up

[dull rune]

Hmm; your intial function seems to be wrong

It's not x/2, but rather x/2 + 2

[zealous parcel]

wait huh

[dull rune]

The function from -2 to 0

What's equation of that line

[zealous parcel]

from -2 to 0 it's y = 1/2 x and from 0 to 2 it's = -1/2 x

[zealous parcel]

ohhhh lmao right sorry

obviously

it crosses 2

whoops

[dull rune]

yea no worries, but if you use that instead youll get like 3

Or, I guess -3

Which one may verify geometrically

[zealous parcel]

oh okok i see

so then i take the antiderivative of x/2 + 2

which is $\frac{1}{4}x^2 + 2x$?

[warm shale]

newchucks

[zealous parcel]

or i guess x^2/4 + 2x

[dull rune]

Yea

[zealous parcel]

i see and so now i plug in -2 to that since plugging in 0 and subtracting would just be subtracting 0

[dull rune]

yea

[zealous parcel]

ok perfect thank u

and then finally what does it mean for the antiderivative of f(x)

since the equation is f(t)

[dull rune]

Well those are just burner variables

Remember we said that $g'(x) = f(x).$

[warm shale]

Zander

[zealous parcel]

hmm

wait sorry 😭 ik this is annoying i sort of forgot why

oh wait

nvm

[dull rune]

fundamental theorem of calc

[zealous parcel]

i see

but wdym by burner variables

like does it just mean f(t)?

[dull rune]

I could say that :) ranges from [-4, 4] and ask you "what is $$\int_{0}^{2} f(:)) d:)"$$

The variables name I choose doesn't matter

[warm shale]

Zander

[zealous parcel]

oh so like it doesnt matter the variables are inconsistent?

they'll mean the same thing? f(x) = f(t)?

i mean i get the idea of variables i was just stuck on the consistency

[dull rune]

The variable on top, namely the x hops down

It would be bad notation to write $\int_{a}^{x} f(x)dx$, for example

[warm shale]

Zander

[zealous parcel]

oh okay i see

wait so if the top is the derivative of the integral is f(x) then isn't the antiderivative of f(x) the integral?

[dull rune]

yes

$\int_0^x f(t) dt = \int_{a}^{x} f(t)dx + C$

[zealous parcel]

i see

so it's not a number, it's just the equation

[warm shale]

Zander

[dull rune]

yea, the answer would be "g is an antiderivative"

In general, if something is an antiderivative it better be a function

I can then evalute that function at a point, say

[zealous parcel]

got it i see

and the reasoning is ftc

[dull rune]

Yea, I don't know how your class introduced it (usually you have FTC1 and FTC2 and the one we used is the first)

[zealous parcel]

alright perfect thank u sm!! :)

[dull rune]

When I last taught Calc 1, this is a question that showed up on my final if you wanted some practice with it

[zealous parcel]

i haven't gotten to trig yet since we just introduced integrals but thank u, i'll keep it in mind

[dull rune]

I don't know what kind of class youre in but if its AP just doing P1 is enough for the exam

[zealous parcel]

okay great, tysm :)

[obtuse pebble]

@zealous parcel Has your question been resolved?

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[obtuse pebble]

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[winter pewter]

randomly choose a natural number k with k <= 2018. The probability that k^1009 has a remainder of 2 when divided by 2018 is...

[obtuse pebble]

@winter pewter Has your question been resolved?

[timid silo]

@winter pewter you can show that k = 2 (mod 2018)

[obtuse pebble]

@winter pewter Has your question been resolved?

[obtuse pebble]

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[obtuse pebble]

@brave wigeon Has your question been resolved?

[brave wigeon]

<@&286206848099549185>

[timid silo]

did you solve equation?

@brave wigeon

[brave wigeon]

No

I tried but no clue

[timid silo]

use this:

[brave wigeon]

🫣🫣

Integration?

[timid silo]

integration has nothing to do with this

[brave wigeon]

Bro please delete it

[timid silo]

@brave wigeon did you try this?

[brave wigeon]

10 (1-cos2x)^2/4+15 (1+cos2x)^2/4

Yes i am trying it

{10+10cos^2 2x-20cos2x+15+15cos^22x+30cos2x}/4

25+10cos2x+25cos^2x/4=6

[timid silo]

evaluate cos2x

[brave wigeon]

Means?

@timid silo

[timid silo]

find the value of cos2x

@brave wigeon

[brave wigeon]

10 cos 2x + 25 cos^2 2x =-1

No idea further

@timid silo

[timid silo]

it's quadratic equation

[obtuse pebble]

@brave wigeon Has your question been resolved?

[brave wigeon]

5 cos2x+1=0

Here cos2x will be
-1/5

[timid silo]

yes

now you can describe thing you have to evaluate in terms of cos2x

[brave wigeon]

81/sin^2x+8 tan^6x

@timid silo

[timid silo]

make it in terms of cos2x

[brave wigeon]

How?

@timid silo

[obtuse pebble]

@brave wigeon Has your question been resolved?

[obtuse pebble]

@brave wigeon Has your question been resolved?

[tardy epoch]

Use half angle formula

,tex .half angle

[warm shale]

riemann

[obtuse pebble]

@brave wigeon Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[lilac pebble]

Translation: In triangle ABC, BD=DE=EF=FC, AP=PC, BP makes AD,AE,AF,AC and M,N,O,P, the two shaded regions have a difference in area of 10cm^2. Find the area of triangle ABC

so corrently all the information i have:

BD=DE=EF=FC

AP=PC

Triangle ABD - POFC = 10cm^2

Triangle AOP = 10cm^2

Triangle APB = Triangle PBC

[nova pawn]

Hi, I’m struggling with these exercises, could someone please explain how I would go about working out the solution?

[lilac pebble]

!occupied

[obtuse pebble]

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).

@lilac pebble Has your question been resolved?

[obtuse pebble]

@lilac pebble Has your question been resolved?

[lilac pebble]

<@&286206848099549185>

[obtuse pebble]

@lilac pebble Has your question been resolved?

[obtuse pebble]

@lilac pebble Has your question been resolved?

[lost tree]

do we know that the triangle is isosceles

[waxen cove]

yes it is isosceles

triangle abd and afc also have the same height

hey @lilac pebble do bp perpendicular with ac?

btw

[obtuse pebble]

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[hasty snow]

hello what is the Domain and range of this y^2=x+5

[outer moat]

Do you know what domain and range are?

[slow granite]

y= +- sqrt(x+5) -> domain: x+5>0 -> x>-5

[outer moat]

You arent really supposed to give the solution but anyway

[hasty snow]

i got confused because y=sqrt(x+5) and y=-sqrt(x+5)

[slow granite]

oh, first time in the server

sorry

[hasty snow]

i want to put them as one but idk how

[woeful cove]

You want to what

[outer moat]

You dont need to put them as one. Just look at both cases. We have that x+5 is under square root, right?

From here we can tell the domain

[woeful cove]

Also this is not a function

[outer moat]

Kinda sus problem tho

[hasty snow]

i get domain (-5,-infinity)

[outer moat]

Nah it is. We just have y instead of f(x)

I assume its some school stuff

[outer moat]

Almost right. We can have 0 under the square root

[woeful cove]

[-5,inf)

[hasty snow]

and for positive one (-5,+infinity)

[woeful cove]

Not -inf

[outer moat]

Nah, look

We are considering square root of x+5

What is the domain?

[woeful cove]

For both ±

[outer moat]

You dont need to worry about the stuff outside of the square root right now

And we cant have something negative under the root

What do you think the domain is then?

[hasty snow]

-5 to infinity ?

[outer moat]

(-5, infinity) like this you mean?

[hasty snow]

y

[outer moat]

Almost right

We can have 0 under the square root

[hasty snow]

but we cant have negative numbers like -6 and -7 and so on

[outer moat]

Do you know the difference between [] and ()?

[outer moat]

Yeah thats right

[hasty snow]

yes ( doesnt include and [ includes

[outer moat]

So should we include -5 or no?

[hasty snow]

yes

oh i get it now

[outer moat]

Then what will the domain be

[hasty snow]

[-5,+inf)

[outer moat]

Thats right

Now the range

We have that $y^{2} = x + 5$

[warm shale]

Fen

[outer moat]

And we found the domain

So can we just tell the range without solving some stuff? (We can)

[hasty snow]

how

[outer moat]

If we look at the domain we can see that it allows only for numbers that are positive or 0

x+5 where x belongs to [-5, +infinity)

So if x+5 is positive or 0

And y^2 equals to x+5

What can we assume?

[hasty snow]

idk

[outer moat]

Lets replace y^2 with 'a'

And keep the domain

We get that a=x+5

And according to the domain x+5 is either 0 or positive

And if a = x+5

[outer moat]

Then can we put 'a' instead of x+5 in this message?

if you dont understand why x+5 is positive or 0 lmk

[hasty snow]

yea i dont get why is it pos or 0

[outer moat]

Look at the domain

Its [-5, +infinity)

[hasty snow]

y u mean domain

[outer moat]

Which we calculated from the statement that x+5 was under the root

[outer moat]

Oh yeah my bad

[outer moat]

So this means that x+5 in our domain must be positive or 0

[hasty snow]

yea yea

[outer moat]

Any number you choose from [-5, +infinity) and put instead of x you will get a positive number or 0

So back to 'a'

a=x+5

If x+5 is positive or 0 (which is basically the range of x+5)

What can we tell about 'a'?

Do you want me to word it in a different way?

[hasty snow]

yes

[outer moat]

Okay lets take an apple

I say that apple equals x

Then I say that x equals 5

So we get that apple = x and x = 5

So in other words apple = x = 5

[hasty snow]

and apple=5 too ?

[outer moat]

Yes!

So lets say that apple = x+5

And the domain is [-5, +infinity)

Which means that x+5 is either positive or 0