help-13

[cedar kiln]

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[earnest storm]

I need to find the x-interval, where the area M is halfed

[earnest storm]

Dm(f) is (2,8)

f(x) is 1/2x

g(x) is -3/4 * x^2 + 8*x - 12

[earnest storm]

I've calculated the area M to be 27

Only thing I wanna confirm is, whether I can approach the task like this?

[dense wing]

Idea's fine and correct

notation is poor

shouldn't have an integral bound and integration variable be the same

[cedar kiln]

@earnest storm Has your question been resolved?

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[cedar kiln]

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[indigo basin]

Hello

[indigo basin]

Nwm
.close

.close

[cedar kiln]

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[crimson sedge]

ok so ive come across a question and ive goten all the way down to 2^x = 1/2

[crimson sedge]

ive looked at the answer and its x = -1, how do i get to that

[mighty mortar]

Have you been taught logarithms?

Actually

[white hornet]

no need logarithms 😮

it is basic thing that x^-1 = 1/x

[tight herald]

Negative exponent rule

[crimson sedge]

yes i have

[mighty mortar]

Yeah that too
Logarithms is more helpful but, yeah, just the neg-

50/50 odds they're supposed to solve it with logs

[tight herald]

Do you know how to use logs where you change to base 10? That way you can use a calc

[crimson sedge]

ohhhhh

[mighty mortar]

This is true but logarithms are more helpful

[crimson sedge]

log2(o.5)

[mighty mortar]

Yeah

[crimson sedge]

= -1

[tight herald]

Yes

[crimson sedge]

thank you so much

[tight herald]

Or log(1/2)/log(2)

[drifting matrix]

No need for log 10, just do log_2(1/2).

[mighty mortar]

Change of base rule, but I vaguely remember being taught that a little late and not tested on it

[tight herald]

Yeah but I wouldn't usually just change base if its not so obvious

[glad kestrel]

change of base is by far the worst thing to ever happen to math

[tight herald]

Lmfao

I use it all the time

For logs

[mighty mortar]

Can you type ".close"

[crimson sedge]

.close

[cedar kiln]

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[tight herald]

Adios

[cedar kiln]

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[crimson sedge]

Hello

[crimson sedge]

I have to find the critical numbers of this function:

Now how to proceed furthure?

[glad kestrel]

do you know how to find critical numbers?

[crimson sedge]

Yes, take the derivative and evaluate at 0.

But I don't know how to do that with this function

[glad kestrel]

do you know the chain rule?

[crimson sedge]

I already did that.

[crimson sedge]

Please see this.

[glad kestrel]

that's not right

[crimson sedge]

I see. This is How I did:

f(x) = cos(x/2)
f'(x) = - sin(x/2)(1/2)
      = - sin(x/4)

[crimson sedge]

Please correct me where I did wrong.

[glad kestrel]

you're multiplying through the sin

can't do that

$2\sin(x)\neq\sin(2x)$

[wraith dagger]

a disappointing son

[crimson sedge]

Yes, I am sorry.

New function:

[glad kestrel]

looks better

so you know that $$\cos(\frac{x}{2})-\frac12\sin(\frac{x}{2})=0$$ is where critical points occur

[wraith dagger]

a disappointing son

[glad kestrel]

my tip: try getting that in terms of one trig function

[crimson sedge]

Would you please expand what you said?😀

[glad kestrel]

$$\cos(\frac{x}{2})-\frac12\sin(\frac{x}{2})=0 \implies \cos(\frac{x}{2})=\frac12\sin(\frac{x}{2})$$

[wraith dagger]

a disappointing son

[glad kestrel]

try getting that 1/2 isolated

[crimson sedge]

Got it.

$$ \cot{\frac{x}{2}} = 1/2 $$

[glad kestrel]

almost

you still have that 1/2 on the right side

[crimson sedge]

Will x = 0?

I am so sorry, I don't know how to solve this.

[glad kestrel]

so you have $$\cos(\frac{x}{2})=\frac12\sin(\frac{x}{2})$$ and you divided by $\sin(\frac{x}{2})$, which will leave you with $$\cot(\frac{x}{2})=\frac12$$, not equal to zero

[wraith dagger]

a disappointing son

Json

[crimson sedge]

Yes 🙂

[glad kestrel]

right, so are you familiar with how to get that x/2 out of the trig function?

using inverse trig?

[crimson sedge]

Yes,

[glad kestrel]

alright, so using inverse trig, what would you get x/2 equal to?

[crimson sedge]

$$\Arcoth(\frac{1}{2}) = \frac{x}{2} $$

I know only till here:

cot^-1 (1/2) = x/2

[glad kestrel]

right, now just solve for x

[crimson sedge]

pi/4

[glad kestrel]

not quite

it's not gonna be a pretty number

it's just isolating x now

[crimson sedge]

I don't know how to do that 🥺

[glad kestrel]

yeah you do

let $\cot^{-1}(\frac12)=u$,
$$u=\frac{x}{2}$$
solve for x and reverse substitution

[wraith dagger]

a disappointing son

[crimson sedge]

pi/2?

[glad kestrel]

it's not gonna be a pretty number

it's not gonna be in terms of pi

just solve for x in u=x/2

you're just isolating x here

this is the very basics of algebra, you're just overcomplicating it because there's a trig function

[crimson sedge]

Do you mean this?

2 * cot^-1 (1/2) = x

[glad kestrel]

yes

[crimson sedge]

So, what will be my answer?

@glad kestrel

[glad kestrel]

do you have a specified range?

or is it just every critical number

[crimson sedge]

Every critical number.

[crimson sedge]

No range specified.

[glad kestrel]

then this

but $2\pi n +$ that

[wraith dagger]

a disappointing son

[crimson sedge]

I get it.

First we took the derivative and solved for the x.

Thank you for being patient and kind to me @glad kestrel 😊

I am very grateful to you 💐

.close

[cedar kiln]

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[cedar kiln]

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[final sonnet]

Hello I don’t understand this..// my work so far

[cedar kiln]

@final sonnet Has your question been resolved?

[cedar kiln]

@final sonnet Has your question been resolved?

[final sonnet]

<@&286206848099549185>

[cedar kiln]

@final sonnet Has your question been resolved?

[cedar kiln]

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[vague pond]

What is -5²? Is it 25 or -25?

[jovial ingot]

for (-5)² it is 25.
however for -5² it should be -25

[vague pond]

Why is that?

[obsidian coral]

Parentheses

Because $-5^2 = -1 \cdot (5^2)$

[wraith dagger]

dldh06

[jovial ingot]

you may use a calculator to check

[obsidian coral]

By order of operations

[vague pond]

I was answering this problem -5²-3(-5), then is the answer -10?

[jovial ingot]

-5²-3(-5)=-25+15=-10

[vague pond]

Thank you so much

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[jovial ingot]

wlc

[cedar kiln]

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[charred tide]

In my efforts to invert this 3x3 what am I getting wrong?

Apparently the answer is [[5 2 2], [5 3 2], [1 1 1]]

[obsidian coral]

How did you go from the second matrix to the third one?

Steps unclear

[charred tide]

Use row 1 to eliminate pivots below first pivot

I do columns at a time

[obsidian coral]

3R1 + R2 step is wrong

3(1 0 -2 1 0 0) + (-3 1 4 0 1 0)

Check that math again

[cedar kiln]

@charred tide Has your question been resolved?

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[cyan grotto]

dont get this part

[cyan grotto]

<@&286206848099549185>

[dire geode]

What's f(x)

[cyan grotto]

sinx

so basically f^3(x) = sin(sin(sinx)))

[cedar kiln]

@cyan grotto Has your question been resolved?

[cyan grotto]

could you help me out

<@&286206848099549185>

[cedar kiln]

@cyan grotto Has your question been resolved?

[cedar kiln]

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[honest tide]

i am struggling with the second part of the problem

[crimson sedge]

Did u try to multiply by -1

[honest tide]

wdym? why would i multipy by -1

for (5f(x)) -10)dx

[crimson sedge]

Oh

Just split them

Ull get it

U can do it

Did u get it?

[honest tide]

i think so

let me try

rq

wait @crimson sedge would i do

actualy idk

so in the #calculus channel it says i can switch the sign if i switch the bounds

[crimson sedge]

No need

[honest tide]

what do u mean by split?

[crimson sedge]

U can but its longer

[honest tide]

so i would do integral of the 5f(x) + the integral of -10?

[crimson sedge]

$\int_9^7.5 5f(x) - \int_9^7.5 10 dx$

[wraith dagger]

mohaaa

[crimson sedge]

$\int_9^7.5 \5f(x) - \int_9^7.5 \10 dx$

[wraith dagger]

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

[crimson sedge]

Omg

$\int_9^7.5 (5f(x)) - \int_9^7.5 (10) dx$

[wraith dagger]

mohaaa

[crimson sedge]

$\int_9^(7.5) 5f(x) - \int_9^(7.5) 10 dx$

[wraith dagger]

mohaaa

[crimson sedge]

Ok look

$\int_9^7 5f(x) - \int_9^7 10 dx$

[wraith dagger]

mohaaa

[crimson sedge]

See?

[honest tide]

yeah

[crimson sedge]

Dont worry about 7

Its 7.5

Imagine its 7.5 not 7

$\int_9^7.5 10 dx$ is easy

[wraith dagger]

mohaaa

[honest tide]

thanks i will try that

[crimson sedge]

$int_9^7 10 dx$

[wraith dagger]

mohaaa

[crimson sedge]

$\int_9^7 10 dx$

[wraith dagger]

mohaaa

[crimson sedge]

Its not 7 its 7.5

[honest tide]

okay thank u

.close

[cedar kiln]

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[trim quail]

how would i start 2.

[bold hinge]

10.15 - 10?

10.30 - 10.15?

10.45 - 10.30?

10.60 - 10.45?

[cedar kiln]

@trim quail Has your question been resolved?

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[steep bone]

I don't understand what to do in this question

[steep bone]

Like finding the domain and range of f(-4)

[glad kestrel]

it's not asking for that

it asks for the domain and range of the function, and the value of f(-4)

[steep bone]

The domain and function of y=f(x)?

[glad kestrel]

domain and range

do you know what domain and range are?

[steep bone]

I understand that domain is like the x or width and range is the y or height

I think

[glad kestrel]

yeah, domain is all the x values the function can take on, while range is all the y values the function can take on

so looking at your first graph, are there any x values the function can't take on?

[steep bone]

No

[glad kestrel]

so your domain is all real numbers

what about range, are there are y values your function can't take on?

[steep bone]

Also no

[glad kestrel]

are you sure?

[steep bone]

Ya

[glad kestrel]

can you find where f(x)=-1?

[steep bone]

Oh (-1,0)

[glad kestrel]

that's f(-1)

you told me f(-1)=0

but you also said your function can take on all y values

take a better look, are you sure?

[steep bone]

I'm not sure

[glad kestrel]

what is the minimum y value you see on this graph?

[steep bone]

0

[glad kestrel]

right

and those arrows mean it goes on forever up like that

so, what y values can your function not cover?

[steep bone]

It can't cover below 0?

[glad kestrel]

right

so the range of your function is all x greater than or equal to zero

[steep bone]

Oh I see

So the domain is all real number and the range is number that is greater or equal to 0

[glad kestrel]

yep

"number that is greater than or equal to 0" can be written in terms of x, though

$x\geq 0$

[wraith dagger]

a disappointing son

[steep bone]

So what would it be like in interval notation?

[toxic owl]

$x \in [0, \infty)$

[wraith dagger]

[TEB] darthlothins

[steep bone]

Thanks so much!

And for domain, would it be something like (-1,-inf) u (-1, inf]?

Since it can go both ways?

[dusty hazel]

Domain for this?

[steep bone]

Yep

[dusty hazel]

Wouldn't Domain have all real values?

[steep bone]

Ya

[dusty hazel]

So why this?

[steep bone]

Oh NVM found it

I thought of something else

[cedar kiln]

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[cedar kiln]

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[drifting path]

what am I doing wrong

.close

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[final sonnet]

this was never answered

[plush walrus]

Re ask your question.

[final sonnet]

ok

in this channel?

[plush walrus]

Yes here.

[final sonnet]

I dont understand this problem, having trouble figuring out b and c,

I did a bit of part a but it was not the answer the textbook gave

does anyone know how to go about it?

I can show a solved example similar to this one if needed

the absolute number is supposed to be less than 2.25

[plush walrus]

Yes show me that example.

[plush walrus]

Okay let me check it first. It will take little time.

[final sonnet]

Thank you, don’t worry

I’m going to be working on it too to see if I made a mistake

I made a mistake in the cross product, supposed to be 29•cos58 and -20•sin58. Just fixed that

[plush walrus]

This is part b or part c?

[final sonnet]

Still it gives me a result greater than 2.25

[final sonnet]

Not sure, it’s not the same format and it looks somewhat like a

[plush walrus]

Okay i am also finding your mistake. You continue. I will tell you in minutes

Hey. The angle that you took 58.
I think you should take -58.
Because it's measured from negative x axis.

[final sonnet]

Yes that works!!!! Thank you!!!!

[plush walrus]

What answer u got.

[final sonnet]

|-1.59k|<2.25

[plush walrus]

Okay. Cool. If you don't have any doubts you can close this by writing ".close"

[final sonnet]

I got one final question and then I’ll be good to go, just wondering for b, to find the angle I would do something like subtract 58 from 180?

[cedar kiln]

@final sonnet Has your question been resolved?

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[smoky stump]

D

[dense hornet]

What idea have you got for d?

[smoky stump]

@dense hornet not quite sure

[dense hornet]

Right, so basically, we have to use some kind of substitution here. And the best one would be u=sqrt(1-t)

Have you tried that

[smoky stump]

No

[dense hornet]

Well then go ahead with it, tell me if you are stuck

[smoky stump]

Okay thanks

.close

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[vagrant crescent]

Can someone check these for me?

[knotty moat]

second one must be 95.4

[vagrant crescent]

Thanks

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[west stag]

i solved 13 a and 13b but have no idea on c

any tips on how should i start

because f'(0) is 0

but f''(0) is 2

so im a little bit confused and i cant find any patterns also and its no point for me to solve higher derivatives

[fallen heath]

@west stagcouple hrs ago, I thought I asked you to frame an equation with fⁿ(0) using what you derived in part (b)

Do you not understand that, if, for every $x \in (-1, 1)$ you have proved that $$(1-x^2)f^{(n+2)}(x) - (2n+1)xf^{(n+1)}(x) - n^2 f^{(n)}(x) = 0$$, then it holds for $x = 0$ as well?

[fallen heath]

and as such, you have: $$f^{(n+2)}(0) - n^2 f^{(n)}(0) = 0$$

[west stag]

yes and i got confused again

[fallen heath]

and.. what's the confusion?

[west stag]

the confusion is how should i proceed with this

i could only get it in terms of f(n+2) and n

[fallen heath]

and you know f(1) as well as f(2)

can you not define f(2n) and f(2n + 1) seperately?

[dense wing]

$n\to n-2$ gives a recurrence relation for n

[wraith dagger]

Mosh

[west stag]

so f(n) (0) is 0 when n is odd

and when n is even

its 2^(n-1)

[fallen heath]

huh?

Let $f^{2n}(0) = b_n$

[fallen heath]

then, $b_{n+1} = n^2b_n$

[fallen heath]

how da heck does your expression work here?

[west stag]

f'(0) is 0

and f''(0) is 2

[fallen heath]

so?

b_1 = 2

b_2 = 2

b_3 = 4 * 2

b_4 = 9 * 4 * 2

[west stag]

f(n+2)(0) - n^2f(n)(0) = 0 right

in the first case where n is 1

f(n+2)(0) = 0

so f''' (0)= 0

[fallen heath]

throw away the odd derivatives

they're all frickin 0

pay attention to even derivatives

[cedar kiln]

@west stag Has your question been resolved?

[west stag]

i mean like

i know b(n) is 2(n-1)!^2

but b(n) is f(2n)

[cedar kiln]

@west stag Has your question been resolved?

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[frank willow]

.reopen

i need help
if the slot here is done then my turn

x- 2y =8

-x+5y=-17
-5y -5y

-x=-5 -17

(-5 - 17)-2y=8

PLS HELP

[solemn torrent]

So this some system of equations?

I don't understand how you've formatted it

[frank willow]

uh yea i updated

[frank willow]

this is substitution method of solving a system of equation

i messed up and mixed someone's example sign in a video, brian mclogan to be exact and

[frank willow]

this is my equations that i need to solve
x- 2y =8
-x+5y=-17

[solemn torrent]

ok well all you have to do is really solve for the variable you want in either equation

then substitute that in

to the other

So let's just redo everything you've done

So solve for x in x-2y=8

[frank willow]

i chose -x instead of y like in the video

[solemn torrent]

It wouldn't matter

[frank willow]

that would be a good idea

so the equation is
x- 2y =8
-x+5y=-17 like i said

[solemn torrent]

you'll end up having to solve for the other one anyway, it's just you found -x instead of y

[frank willow]

yes

so after choosing my variable

[solemn torrent]

ok

[frank willow]

il do -x+5y=-17 by substracting 5y by -5y and -17 by -5y right?

[solemn torrent]

yeah so -x+5y=-17

transpose 5y

[frank willow]

yea its gonna be 0

[solemn torrent]

-x+5y-5y=-17-5y

-x=-17-5y

get x now

you have -x

we want x

[frank willow]

oh so the sign in the x doesn't really matter??

[solemn torrent]

what

[frank willow]

so our -x is -17-5y

right

[solemn torrent]

yes

-x=-17-5y

But we want x not -x

[frank willow]

oh ok i get it

after this

we can do x- 2y =8 right

[solemn torrent]

Because we're gonna have to substitute x in -x+5y=-17

in class atm be back in 1 min

[frank willow]

by substituting x by -17-5y

[frank willow]

ok

[frank willow]

so we are using the same equation instead of the other one

[solemn torrent]

wait yeah i got confused i was gonna solve for x in x-2y=8

and then sub that into -x+5y=-17

and i went along with that

[frank willow]

oh yea same

[solemn torrent]

but whatever you've done is fine

[frank willow]

so we gonna do this

subbing x for x-2y =8

[solemn torrent]

What you did was solve for -x (we want x as I said) in -x+5y=-17

[frank willow]

so what we gonna do now is gonna substitute it into x-2y=8 right?

[solemn torrent]

yes

but wait

you still haven't found x

you have -x but not x

[frank willow]

so what now

[solemn torrent]

-x=-17-5y

how would you get x

remember that the negative sign is -1

we just don't write the 1

[frank willow]

oh because x = 1

[solemn torrent]

So we have -1x, i.e. -1*x

no

wait lemme just do it

i'll explain after

(following what we've just done)

[frank willow]

oh ok sure

btw if its not too late here is the actual screenshot of the question

[minor heart]

I think

ya

I mean

wait

[frank willow]

why

[minor heart]

sorry

[obsidian coral]

Don't give out answers

[minor heart]

oh

I saw something wrong

[frank willow]

no its ok i want to learn something why is that answer

[minor heart]

it is not

I saw something wrong

sorry

[frank willow]

oh ok

[obsidian coral]

Then people will teach you the process on how to do it, and not give out answers

You should be able to apply the process yourself

[frank willow]

ok

[frank willow]

and even if he gave me the answer i won't put it on my homework cus i don't know on what grounds does it have

but i like doing the process more than receiving the answer

[solemn torrent]

my latex is so shit holy shit how'd i forget to latex

[frank willow]

yo why the font diff and message

nvm

so after we gonna do @solemn torrent which is putting -x into x-2y=8

[solemn torrent]

read this

[frank willow]

OHH

[obsidian coral]

So did you have to solve the problem for the person? You could have done a similar example, showed that then let the person apply to their own problem

[solemn torrent]

I condensed the steps because i felt like the other things aren't really necessary

because it's quite evident to what i've done

[solemn torrent]

i keep telling you we want x and not -x

[frank willow]

so the -x is turned in to x but -17 turns to 17 also

[frank willow]

oh yea

[obsidian coral]

Gives the person a chance to do their own work without having to copy your steps word for word

[solemn torrent]

because we're gonna have to sub in x in x

[frank willow]

yea the x in x-2y=8

[obsidian coral]

There's a difference between doing the work for the person and letting them learn. Sure, you provided information on how to get it, but chances of people just copying and not caring about the process is still possible which is why you should provide a similar example

[frank willow]

man helped me, not gave me the answer and solution to it ok and i already did it myself there is just some little problem i really need help about

:)

[solemn torrent]

well yes as I do agree with your statement but

So anyways i just wanna go through everything

like litearlly everything

[obsidian coral]

The answer is embedded in that explanation so they did give you the answers

[frank willow]

thank you so much man about teaching me this

its like 11:14 pm right now so im kinda tired

but i don't want to sleep knowing my homework is due now

[solemn torrent]

I will digress from what we've spoken of but I wanted to get this out there:

A set of simultaneous equations, also known as a system of equations or an equation system, is a finite set of equations for which common solutions are sought.

i.e. there exists solutions to your set of equations given that they're related.

So there are three primary methods to solving systems of equations that is elimination, substitution, and graphing, you'll need to know how to do all three although you wouldn't really need graphing as you probably aren't given graphing paper, and it's the most time consuming assuming you don't have a graphing calculator or some other graphing device.

So when it comes to substitution just like elimination it doesn't matter what indeterminate you're solving, or what equation you're using, you'll just have to do the other one as if you done the preceding (this sentence probably doesn't make sense, I don't know how else to word it).

So in our case we decided to solve for x using the second equation, that means we have to substitute our x from our second equation into our first, then we solve for the single variable.

Now again I will solve it again just adding everything I've done and getting the solution set

[frank willow]

ok

[frank willow]

yes tysm

[solemn torrent]

I should colour code it so it's more clear

want me to do that?

probably should

I'll explain what i've done in each and every step

So here I've solved for x in the second equation

[frank willow]

lol i was about to ask if how u made the -x=-5- 17 but it looks like u got me

[frank willow]

how u made the equation go positive

[solemn torrent]

Because remember that

-x = -1x = -1*x

We just don't write the 1

So it's just positive x multiplied by -1

Which i divided to make x positive

[frank willow]

so i get it now

[solemn torrent]

The asterisk in text means multiplication

we just don't use x because it's definition may be ambiguous to the reader

Could be interpreted as a variable or the multiplication cross

[frank willow]

oh so just a symbol for the value

thus calling it a variable

[solemn torrent]

Yeah do you know what a variable is?

[frank willow]

yea

[solemn torrent]

yeah

So a variable is a quantity

[frank willow]

It is a symbol that represents the quantity in it

[solemn torrent]

A quantity (definition used in mathematics and physics) is a value or component that may be expressed in numbers.

the figure or symbol representing a quantity.

[frank willow]

that figure is the variable

[solemn torrent]

So a quantity is a letter that's used to express a number or something arbitrary

[frank willow]

yes

[solemn torrent]

so any number or anything used to express a number is a quantity

yada yada

[frank willow]

so anyways after finding x

[solemn torrent]

I genuinely hope anything that comes through this chat you understand

[frank willow]

we are just gonna do x=17+5(-3) right

[solemn torrent]

any queries lmk

[solemn torrent]

yes

[frank willow]

yes idk why i learn so much more in a discord server than an online class

[solemn torrent]

Since we found y=-3

We sub that into any of our equations

and solve for x

[frank willow]

so its gonna be x= 17+ -15

[solemn torrent]

yes

[frank willow]

and after that we are gonna add 17+-15 which is -2 right

or 2

i think its -2

[solemn torrent]

No

it's 2

+2

Recall your positive-negative rules

[frank willow]

oh yea

[solemn torrent]

search those up

[frank willow]

cus 17-15 makes it 2

[solemn torrent]

one way to put it

[frank willow]

like 17 being lowered by negative 15

[solemn torrent]

yes

[solemn torrent]

yeah 17 is subtracted by 15

[frank willow]

so for example -9/3 is -3 right

[solemn torrent]

yeah

[frank willow]

since their unlike signs

u don't mind if i screen shot some of our convo right

i want to keep if incase i forgot il just look at it and remember all of it

[solemn torrent]

I don't care lol

[frank willow]

nice

[solemn torrent]

you do you

[frank willow]

tysm again for the teaching

i don't wanna fall behind in school even im focusing in youtube

[solemn torrent]

lovely

[frank willow]

if i need help again i know where to go

[cedar kiln]

@frank willow Has your question been resolved?

[cedar kiln]

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[obsidian coral]

Khan academy

They have videos

Depends on the person

[hollow osprey]

Text is harder to read, and at trig there r still videos

Later on (check some of the advanced channels) there’s not nearly as many videos

So you should use them while you can

[cedar kiln]

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[final glade]

Hi everyone, i got a question
if x= 1 + sx - s^2 (s≠1) what is x
how do you solve this?

[bold vine]

You can move sx on the RHS to the LHS and solve for x that way

[final glade]

s^2+1 / s + 1 / R \ (1) / s - 1 / 1 - s

there are 5 answers possible

[bold vine]

Is this just an algebraic equation?

[final glade]

yes

[bold vine]

x = 1 + sx - s^2

x - sx = 1 - s^2

x(1-s) = 1 - s^2

x = 1 + s

s ≠ 1, so dividing by 0 isn't a problem.

[final glade]

okay first bring it on LHS and then look for x

i got it

What is the equation of the line passing through the point (4 , 1) and parallel to the line x + 4y = 1

if anyone can help me another question...

[bold vine]

You can use point-slope form

$y - y_1 = m(x - x_1)$

[wraith dagger]

Shen

[bold vine]

You have a point $(x_1, y_1)$ and a slope.

[wraith dagger]

Shen

[bold vine]

You can get slope from your 4y = 1

Another approach which may be simpler is that notice that a line parallel to y = 1/4 is a horizontal line, so you'll only need a y-value to describe this graph. This y-value is determined by the point (4, 1) given, so your answer will be y = 1.

[final glade]

horizontal with y = 1+x/4

[bold vine]

Oh I didn't see the x

So yeah only the first approach would be valid

You can use point-slope form

$y - y_1 = m(x - x_1)$

You have a point $(x_1, y_1)$ and a slope.

You can get slope from $x + 4y = 1$

[wraith dagger]

Shen

[final glade]

so which equation can be horizontal with y = 1+x/4

is y = 4x + 1 correct?

[cedar kiln]

@final glade Has your question been resolved?

[cedar kiln]

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[sterile bolt]

The body is in equilibrium, you have to find the values of T1,T2 and T3...

[sterile bolt]

I’ve done this much

Now what do I do

[obsidian coral]

Substitution

[frigid canyon]

is 100N it's weight?

[sterile bolt]

Yes

[frigid canyon]

well really i think we can't find it

we have two equations but three unknown

[sterile bolt]

I see...

[frigid canyon]

can you send the complete question

[sterile bolt]

Sure

[frigid canyon]

it can possibly be that you oversaw some info

[obsidian coral]

It gave the answer

Post that part

[sterile bolt]

Alright

They made it so complicated that I don’t even wanna look at their work.

[obsidian coral]

T2 is 100N

That's how they solved it

[sterile bolt]

Yeah but how did they find that out

[obsidian coral]

T2 isn't a separate force you need to find

T2 is the force given

[frigid canyon]

wth how?

[obsidian coral]

You have the weight hanging on a rope, that rope is T2

That's the FBD

The solution just decided to label an extra T2

[sterile bolt]

Shsksnaksndk

[obsidian coral]

As the 100 N and drew it there, which made it more confusing

[sterile bolt]

They really need to make their questions clearer dammit

[obsidian coral]

That's what I think though

I could be wrong

[sterile bolt]

Welp

I’ll try not to think too much on this

Thanks

.close

[cedar kiln]

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[hidden radish]

What did I do wrong?

[jaunty mural]

$\frac12\abs{4-\frac12x}\le5$

[wraith dagger]

Shuri2060

[jaunty mural]

Is this the original?

I'm going to guess this wasn't applied properly ---

You split into 2 cases: x >= 8 and x < 8

[cedar kiln]

@hidden radish Has your question been resolved?

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[green plover]

Hey, I just have a quick question

[green plover]

how can I evaluate summations

[sand ether]

Do you know what $\sum_{x=1}^{50} x$ represents?

[wraith dagger]

azeem321

[green plover]

like the sum of numbers from 1 to 50 ?

[sand ether]

Can you write out the first few terms here

[green plover]

wait which terms?

[sand ether]

of sum of numbers

from 1

to 50

[green plover]

woah

I used gauss's rule

n(n+1) / 2

[sand ether]

forget that for a second

[green plover]

aa okay

[sand ether]

So we would have 1+2+3+4...+50

You get that?

[green plover]

yeah

[sand ether]

And what does $\sum_{x=4}^{49} x$ represent?

[green plover]

aa I don't really know

[wraith dagger]

azeem321

[sand ether]

It represents the sum of all numbers starting from 4 to 49

[green plover]

aah yeah

[sand ether]

So (1+2+3+4...+50) - (4+5+6+7+...49)

[green plover]

yeah

[sand ether]

You need to calculate that

Don't get confused by the summation notation. It's just a shorthand way of writting long sums

[green plover]

yeah yeah

[cedar kiln]

@green plover Has your question been resolved?

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[cedar kiln]

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[sand ether]

What type of quadrilateral is ABCD supposed to be? Are we just to assume that it's quadrilateral and the type doesn't matter?

[cedar kiln]

@astral lava Has your question been resolved?

[astral lava]

yeah its just a quad

[sand ether]

Besides the vertical angles being equal, I don't know what over assumptions can be made without knowing what type of quadrilateral it is. If AD and BE were parallel we could do some things, but i dont think we are told that? someone smarter than me will probably come to help soon

[cedar kiln]

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[steel relic]

yo i can’t remember but i swear the area under a reciprocal graph between 2 points i something special

[steel relic]

saw it in a video but i’ve forgotten and it’s bugging me my bad

[trim sentinel]

,w integrate 1/x from 2 to 8

[wraith dagger]

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[steel relic]

oh is it log(range/2)

[trim sentinel]

Idk lets see

[steel relic]

,w integrate 1/x from 3 to 50

[wraith dagger]

Results provided by WolframAlpha

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Requested by elliott#0007

[trim sentinel]

Yep it is

[steel relic]

aight thank you

[trim sentinel]

to/from

[steel relic]

didn’t mean range

ye

thanks ☺️

[cedar kiln]

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[chilly warren]

Do you know integral by substitution?

[crimson sedge]

wait i got you bro

i have it almost done

ill upload a picture

thats what i have so far

the triangle part is confusing

[gentle lintel]

wdym triangle part

[crimson sedge]

so i have to convert it back to X

right now its theta

[dense wing]

I don't think trig sub is even needed

[crimson sedge]

probably not but i have to use it

worksheet explicitly says to only use Trig Sub

[chilly warren]

I think you are overthinking it.

There is no need to use trigonometric substitution here.

[crimson sedge]

ok, but i have to use it

[gentle lintel]

forced?

[crimson sedge]

yes

whole worksheet was designed to make us practice trig sub

[chilly warren]

Your teacher is a dum dum. The whole point of mathematics are creatively solving the problem, not how to hammer in a technique to solve a problem even though it may make the problem more complicated.

[crimson sedge]

I get you, but she just wants us to practice trig sub, because we are learning it

i could've used u-sub

but im being forced to trig sub

[gentle lintel]

this is dumb 😦

[crimson sedge]

gosh dammit

lol

im litterally at the ending

just dont know how to do it

any suggestions?

,rotate

[chilly warren]

Also, never drop the $d\theta$'s

[wraith dagger]

Xwtek

[chilly warren]

Integral without that symbol is an illformed expression

[crimson sedge]

ah ok

[chilly warren]

Also, what is $\tan \theta + \frac{\tan^3 \theta}{3}$?

[wraith dagger]

Xwtek

[crimson sedge]

thats the final answer

now that its in Theta, i have to convert it back to X

my bad it goes from the LAST expresion

Up to the next

Ran out of space on the bottom

i just drew and arrow from the bottom,up to that expression letting it be known that the following expression

[chilly warren]

You use the sec2−1=tan2x identity

[crimson sedge]

yes

tan^2(x) +1 = Sec^2(x)

[chilly warren]

Also, you need to be careful about $\sqrt{(\tan x)^2}$

[wraith dagger]

Xwtek

[chilly warren]

It's not simply $\tan x$

[wraith dagger]

Xwtek

[crimson sedge]

In problem its , $\sqrt{(\tan ^2x)}$

[wraith dagger]

Rush Fan Boi

[crimson sedge]

,$(tan(x))(tan(x))=tan^2(x)

[chilly warren]

Yeah, but $\sqrt{y^2}=|y|$, not simply $y$

[wraith dagger]

Xwtek

[chilly warren]

Anyway, reorder this equation so you get just tan(x) on the left.

[crimson sedge]

yea same

[chilly warren]

Sorry, I mean just sec(x)

then you plug in the sec(x) to the $x= \frac{\sec x}{3}$

[wraith dagger]

Xwtek

[chilly warren]

And then you reorder the new equation to get just $\tan x$

[wraith dagger]

Xwtek

[crimson sedge]

?

My u was tan

after integration, i plug back in tan(theta)

<@&286206848099549185>

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[cedar kiln]

@crimson sedge Has your question been resolved?

[green belfry]

Srill stuck?

[crimson sedge]

i finished it

im disappointed tho, over an hour

no

one

[green belfry]

What was the question

[crimson sedge]

just had to convert the theta answer in to terms of x

easiest part of the whole problem

,close

.close

[cedar kiln]

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[desert creek]

<@&286206848099549185>

[desert creek]

could someone just give me the answer?

X^2 + 7x - 10?

and then factor giving (x+2) (x-5)

?

[gentle fog]

nopes this is not the right factorization

[wraith dagger]

yatharth

yatharth

[desert creek]

listen this isnt my work or assignment i was asking because this guy wanted the answer,

but thank you 😛

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[north solstice]

hey, if i have this function

[north solstice]

how can i find

[gentle lintel]

just differentiate it as usual

[north solstice]

the function doesnt hv the variable t

[gentle lintel]

dh/dt is obtained

[gentle lintel]

doesn't matter

you have dh/dt

[north solstice]

so i differentiate normally?

[gentle lintel]

yea

[north solstice]

i get

but the ans is

<@&286206848099549185>

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[pure geode]

Going to assume this is as straight forward as saying that since H \leq G then for any h \in H we know \phi (h) \in \bar{G}

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[worthy snow]

how do i calculate the areas separately?

[worthy snow]

i tried part a

$x=\int^{6}_{0} y-5 \mathop{dy}$

[wraith dagger]

breadiculous

[worthy snow]

which resulted me with -12

[solid quarry]

they want positive area

when y = 0, x = -5

[worthy snow]

so would it be positive 12?

@solid quarry

[solid quarry]

probably not

$x=\int^{6}_{0} |y-5| \mathop{dy}$

[worthy snow]

would i have to like split the equations into 2 integrals of 0-5 and 5-6

[wraith dagger]

DETOX

[solid quarry]

yup

[worthy snow]

then i would absolute the negative area

[solid quarry]

ye just negate it

[worthy snow]

but will the question always as for the positive area

how would i know if the question is not asking for the net area instead of the sum of area

[solid quarry]

in this case it says to break it up

theres no other reason to do that

[worthy snow]

i havnt seen a question that asks for that and idk if there will be since im new to calc

[solid quarry]

its kinda common

youve for sure gotta know how to do it

[cedar kiln]

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[trail pulsar]

is this correct

[frigid canyon]

i can't see the use of doing this

this problem can be reduced to solving a simple quadratic equation

[trail pulsar]

fractions confuse me so im trying to simplify it

[dusty hazel]

You're actually increasing the work.

[frigid canyon]

yeah

just do the normal way

[trail pulsar]

hm but is my answer above correct?

[cedar kiln]

@trail pulsar Has your question been resolved?

[upper abyss]

How did you get that second line?

Equation 1 does not look correct, no.

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[copper peak]

what is a trivial solution? Is it when a matrix can be equal to zero?

[copper peak]

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[leaden crater]

why is it {19 +n \over 4} shouldnt it be 20 not 19 since it starts with 5

wait nvm it makes sense

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[supple comet]

so im trying to compute the convergence radius but i dont really understand how it is supposed to be sqr(5) and not just 5?

[supple comet]

im using this formula if that helps

[upper garnet]

never seen that

just do ratio at first

this might help

https://tutorial.math.lamar.edu/classes/calcii/PowerSeries.aspx

when completing ratio test

take the terms with x outside of the limit

then the one with limit will normally be constant

taking the reciperate of it it would become the radius

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[crimson sedge]

There are 33 rooks on a regular (8 × 8) chessboard. Show that you can find 5 among them such that any pair among them does not attack each other.

[crimson sedge]

This is what i had so far:
We can start by figuring the minimum amount of rooks for each row. Each row has a capacity of 8, and 33 can be written as 4*8+1, which means at least 5 rows must have at least 1 rook. The same goes for columns since they have a capacity of 8 as well.

i dont know how to continue at all

[fallen heath]

Shuri got the answer to your question.

[crimson sedge]

yeah i saw

[fallen heath]

Did you understand it?

[crimson sedge]

i didnt understand the mod 8 part

[crimson sedge]

tbh not really

[fallen heath]

Lol, it was a little complicated so

Anyways, here's what I meant when I hinted you the first time

Consider any random arrangement, okay?

[crimson sedge]

ok

[fallen heath]

so, by pigeonhole principle, you have at least 5 columns with rooks in them, yes?

[crimson sedge]

yes

[fallen heath]

pick one column from among them. subsequently, pick one rook from among them.

[fallen heath]

like I did here

kay?

by PHP, there's always gonna be a 5th column with a rook in it

[crimson sedge]

k

[fallen heath]

now,

[crimson sedge]

yes

[fallen heath]

as you pick this rook

you cannot pick any rook in the row / column same as this one picked rook, say R_1

as per the requirements of the question

[crimson sedge]

but there are two more rooks in that column

[fallen heath]

"cannot have any of the 5 rook attacking each other"

[crimson sedge]

not any, just a pair of the 5 rook

[fallen heath]

Yes there are.. exactly why we cannot pick one of these two for our 5 rook pair

in this case, I cannot pick any rook from the 3rd rank, except for rook at f3

or it'd be attacking my rook ;-;

right?

[crimson sedge]

right

[fallen heath]

now,

your concern is, how to pick the remaining 4 rooks such that they don't attack this one or each other

Again, by PHP you can argue that, at least 5 rows have rooks in them?

[crimson sedge]

yes

[fallen heath]

so, starting with your first pick R_1, you cancel out this row, and this column.

Try to accommodate as many rooks as you can in these row and column, because, later it boils down to "at least this many rooks exist such that they're not attacking R_1"

[crimson sedge]

row and 4 columns

[fallen heath]

nope, only this row. and this column. containing R_1

rearrange such that this is your situation

[crimson sedge]

right

[crimson sedge]

but we cant choose any in that row or column

[fallen heath]

Yep

[crimson sedge]

arent there many many options we can choose?

[fallen heath]

Yeah there are. But you only need to show that in every possible situation, there exists an algorithm to help you pick 1 such pair

we started with showing that:

no matter how the arrangement is prepared, we have at max 3 active files, i.e., not every rook is suffocated within 4 files [ because if they are, again by PHP, one can argue that any two might definitely be attacking each other ]

[crimson sedge]

ok

[fallen heath]

hence, it's possible that one such pair of 5 rooks can be picked such that they're not on the same file [ which was based on there being 33 rooks total ]

ok?

[crimson sedge]

ok

[fallen heath]

now, we're interested in showing that... for any such arrangement of rooks within these 5, 6, 7, 8 files, there exists at least 5 ranks with rooks in them, such that they're not on the same file

ok?

[crimson sedge]

wdym by files?

columns?

[fallen heath]

files mean the columns, yes

and ranks mean rows for a chessboard