help-0

[ebon sparrow]

no number given

[primal flume]

hypotenuse

[ebon sparrow]

yeah i know but like i dont see any number

on this picture

[primal flume]

ok

both adjacent and hypotenuse

[ebon sparrow]

oh

you need to use cos

[primal flume]

so cos(x) = 5.5/4.5?

[ebon sparrow]

$cos(\theta) = \frac{4.5}{5.5}$

[ocean seal]

Akira

[primal flume]

oh

is that it?

[ebon sparrow]

yes

you need find the theta

[primal flume]

0.8181818181818182

[ebon sparrow]

now use cos shift

$cos^{-1}$

[ocean seal]

Akira

[ebon sparrow]

this ^

[primal flume]

i dont know what that is

[ebon sparrow]

thats why you need find the angle

you use shift

[primal flume]

im confused

[ebon sparrow]

do you have a calculator?

[primal flume]

yes

[ebon sparrow]

wait let me show you

those things are called arsin, arcos and artan

you need to use this to find the theta

[primal flume]

mhm

so what do i do?

cos to the power of negative 1?

[ebon sparrow]

can i see your calculator

there should be a shift button

[primal flume]

its a casio

yep

there is

[ebon sparrow]

yeah use this

like this

since it's 35.09 it should be 35.1

that works

[primal flume]

yep

i got that too

on my calculator

so thats the answer?

[ebon sparrow]

yes

[primal flume]

wait i got 38.99644581

[ebon sparrow]

did you write all this

[primal flume]

yes

[ebon sparrow]

you should write it just like this

if you're using a real calculator by hand then you gotta type it like this $cos^{-1}(0.8181818181818182)$

[ocean seal]

Akira

[ebon sparrow]

then it gives you 35.09

or 35.1

do you remember this?

[primal flume]

sorry im trying to find my phone so i can send you a photo

[ebon sparrow]

okay

you can use online calculator though

[primal flume]

nvm my calculator has been wrong before

[ebon sparrow]

oh I see

[primal flume]

so 35.1

ok

so

i have one left

one question left

then i can go to sleep

[ebon sparrow]

oki

[primal flume]

its this

but i need to use multiplication to figure it out

not division

[ebon sparrow]

I feel like this is the same question

you asked for the first place

[primal flume]

yes

but you told me how to figure it out using division

didnt you?

[ebon sparrow]

you because you were finding for H

[primal flume]

...

[ebon sparrow]

this

you said x was unknown

so

[primal flume]

my brain isnt working

is it okay if i ask you tomorrow?

[ebon sparrow]

on what

[ebon sparrow]

uh sure

[primal flume]

will you be on in 12 hours?

[ebon sparrow]

it's 10am for me

if 10 hrs more

then 10pm

yeah I can

you can dm me or ping

[primal flume]

k cuz its 12am for me and im tired

[ebon sparrow]

damn I feel you

I think it's time to rest

[primal flume]

yeah i was up earlier to do research on a battle for history

now my desk is covered in paper and books

mmm

ok so ill hopefully talk to you in like 12 hours

maybe earlier

goodnight

or goodday

idk

[ebon sparrow]

good night

[primal flume]

bye

[ebon sparrow]

it's morning for me

10am

[primal flume]

oh yeah

well have a good day

[ebon sparrow]

thanks

[lone heart]

@primal flume Has your question been resolved?

[lone heart]

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[tawdry urchin]

Assuming tan(a) - tan(b) = x, and cot(b) - cot(a) = y, find cot(a-b) in terms of x and y.

[tawdry urchin]

I know that cot(a-b) = 1+tan(a)tan(b)/tan(a)-tan(b)

so it is $\frac{1 + \tan a * \tan b}{x}$

[ocean seal]

dork9399

[slender gull]

You can find tan(a)tan(b)

[tawdry urchin]

but now what should I do?

how?

[slender gull]

rewrite cot(b)-cot(a) in terms of tan

[tawdry urchin]

thanks

.close

[lone heart]

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[shadow sparrow]

Back at it again with the basic mathematics, I'm not seeming to remember how to solve this, for some reason.

[shadow sparrow]

$\frac{-3}{4}+C=2$

[ocean seal]

huntifer

[mellow grail]

yo

[shadow sparrow]

Is it not multiply the reciprocal of the fraction by both sides?

[mellow grail]

add some thing to both sides, so that on the left side only the C remains

[shadow sparrow]

Ah, so its not using reciprocals here?

[mellow grail]

no

thats when the variable is multiplied by something

[sharp thorn]

solving for c?

[shadow sparrow]

Ahhh.

[shadow sparrow]

That makes sense

[sharp thorn]

try adding 3/4 on both sides

[mellow grail]

i wanted him to figure that out on his own, since he knows he just forgot

[shadow sparrow]

I see what you're saying, make it 0/4, which is 0.

And in this instance its 2+3/4

which is actually 8/4+3/4

So its then 11/4

[mellow grail]

yep

[shadow sparrow]

That makes a lot more sense

I was sitting here like

"Reciprocal not comprende?"

So if addition, get rid of fraction with addition, if multiplication by the variable, get rid of fraction with multiplication

and if its division, use the fraction rules

Cool, thanks lots!

.close

[lone heart]

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[cosmic coral]

Here, in the circle, RS = SM = 6 units. Radius of the circle equals to 5 units. Find RM. Feel free to ask if anything's unclear, the drawing may be a bit shabby. Try not to use too advanced math(not above trig). Thanks.

[gray isle]

what have you tried?

[lone heart]

@cosmic coral Has your question been resolved?

[cosmic coral]

i drew a radius from O to R and O to M

but then i couldn't figure out anything

[gray isle]

draw a line from O to P

[cosmic coral]

yeah i did that too

[gray isle]

introduce a variable like x for OP for convenience

PS can be expressed in terms of x

OM is a radius, so also has length 5

an expression for PM can be obtained from both right triangles

[cosmic coral]

i see

then we can do 2PM and find RM

but is there a way to get a real value?

[gray isle]

an expression for PM can be obtained from both right triangles
set those two expressions equal to each other to determine x and hence PM

[cosmic coral]

ohh

then it's solved?

[gray isle]

if you do what's been outlined, you'll get the result

[lone heart]

@cosmic coral Has your question been resolved?

[lone heart]

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[autumn mountain]

is it possible to do smth to this using quadratic formula

[mortal trellis]

its a quadratic in h, so yes

[autumn mountain]

could i have some help doing that

[mortal trellis]

this is mostly just an exercise in keeping things organized

[autumn mountain]

idek which side i'm meant to do what to

[mortal trellis]

if you have it written down on a paper, it might help to eg mark all the terms that have h^2 in them in red, all those with just h in blue and the rest in green

or by underlining them in three different ways or something

[autumn mountain]

gotcha

[mortal trellis]

what you want to do in the end is to get (some long expression)*h^2 + (other long expression)*h + (third long expression) = 0

[autumn mountain]

i'm sorry it's rlly late and i don't rlly get how to use quad formula here

[mortal trellis]

do you know how to use the quadratic formula on eg 7h^2-13h+27=0 ?

[autumn mountain]

also it's meant to be h^2 at the very start i think

[autumn mountain]

yea

[mortal trellis]

then here its the same just that the numbers look uglier

[autumn mountain]

so do i have to include cot in it?

[mortal trellis]

cot(56) is just some number

you could calculate it with a calculator if it bothers you

(although that will introduce rounding errors)

[autumn mountain]

i see

[autumn mountain]

i frankly can't get it

[mortal trellis]

ok

what do you have currently

[autumn mountain]

i got to

100cot^2(56)+sqrt (-100cot^2(56))^2-4(cot^2(56))(2500cot^2(55))

all over 2cot^2(56)

[mortal trellis]

no

you ignored the right side

you have to bring everything to the left side before you apply the quadratic formula

[autumn mountain]

oh and have 0 on the other side?

[mortal trellis]

yes

[autumn mountain]

i see

[autumn mountain]

what do i do with the final part of the equation

the 2(61)(hcot55 thingy

[mortal trellis]

multiply out the bracket

then you have one term with h and one without

[pine grove]

boi what the

what the

[autumn mountain]

what do i do with the cos58 at the end 😭

wait nm

nvm

i just times it by both

[mortal trellis]

yes

[gilded mortar]

how much is 2+3?

it's 6 am i right?

[mortal trellis]

dont troll in help channels

[gilded mortar]

but I don't trolling, I really dont know

[ornate ginkgo]

How old are you?

[gilded mortar]

23

[vale wigeon]

at the very least open your own help channel

dont interrupt somebody elses

[autumn mountain]

wait so is it

gimme a sec

the top one

[mortal trellis]

,rotate

[autumn mountain]

oh and -61^2

[mortal trellis]

that looks good

unless I missed something

now you can factor h^2 out of the first two terms and h out of the next few

[autumn mountain]

ohh

[autumn mountain]

yup

got that

[mortal trellis]

and now it should look like this

[autumn mountain]

yup it does

i have -h at the start of the second group but is that ok

[mortal trellis]

-h*(expression) = h*(-expression)

[autumn mountain]

should i switch it

[mortal trellis]

your choice

not necessary as long as you treat it correctly in the next steps

[autumn mountain]

i will just in case

so once i've done that

then what do i do

[mortal trellis]

now you can apply the quadratic formula

although at this point it would probably be fine to first calculate these long expressions, shouldnt introduce too much rounding errors

[autumn mountain]

Tryna do that now

[mortal trellis]

I have to go now but you can definitely finish it from here

[autumn mountain]

Oki

Thx for the help

[lone heart]

@autumn mountain Has your question been resolved?

[autumn mountain]

Yes

[lone heart]

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[junior warren]

can anyone let me know if this statement is somehow true, pulled this out on a midterm and got full marks

but it was a complete guess as i was stuck

so just interested in knowing if this step specifically has any merit

[trim sorrel]

What is the statement

?

[junior warren]

the ms paint writing

[trim sorrel]

What are A and B

[junior warren]

ah

i knew how to find the inverse but couldnt come up with anything quick

to prove it was invertable

and i just dont see my statement having any correlation to invertability

[trim sorrel]

Ngl i dont see how it connects either

Let me think some more

[junior warren]

im thinking it mustve been a marker error

we got these back in a day so its likely they just let it slide lol

[lone heart]

@junior warren Has your question been resolved?

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[fierce narwhal]

yo can anyone help me with this

[fierce narwhal]

don't know how to start

[fossil latch]

I don't like how the denominator 1-sin²x part

Let's change it to cos²x first, shall we, since x -> π/2 and sinx becomes 1, not 0

[fierce narwhal]

alright

[fossil latch]

I also hate how it's sin²2x instead of being with sinx and cos x

Have you learnt the trig identity sin2x= 2sinxcosx?

[fierce narwhal]

i learned that like 20 mins ago

[fossil latch]

Okay then you know it :D

So finally the limit equation becomes 4cos²x sin²x / cos²x

This limit is now much more solvable

That's it

[fierce narwhal]

yeah i can see it now

wait can you explain how it becomes 4 cos^2x sin^2x

[fossil latch]

(2cosxsinx)²

[fierce narwhal]

👍

thanks

[fossil latch]

General tip: if you see + or - signs on fractional limits you generally want to consider factorising or changing into more favorable terms

Especially for trigs.

[fierce narwhal]

.close

[lone heart]

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[frosty sparrow]

how can i prove that if x² is a multiple of 3, then x is a multiple of 3

[frosty sparrow]

i thought about doing it using a contradiction, but how can i express that a number can't be divided by 3 mathematically?

[echo socket]

Considering the cases x = 0, 1, 2 (mod 3) is one way

[frosty sparrow]

yep

but how can i demonstrate this?

[echo socket]

Just show that x^2 = 0 (mod 3) happens only in the first case, meaning x = 0 (mod 3) must be true

Meaning x is a multiple of 3

[frosty sparrow]

ohh got it

thx

.close

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[barren portal]

I didn't get why they said r cong to r-9 (mod 9)

[worn fox]

You don't get why it's true or you don't know where they've used it?

[barren portal]

I couldn't understand the solution, my reasoning is that, set of possible least residues modulo 9 are {0,1,...8}

so x^2, has to be square congruent to square of each of them

resulting in x^2 cong to {1,4,7,9} (mod 9)

[worn fox]

Where did you get 3 from?

Right, that's the same as they're doing

Just that they've used for example that 7 = -2 mod 9

To make calculations easier (not that they're overly difficult without)

[barren portal]

why is it +- ?

[worn fox]

.

[barren portal]

they didn't do the sepearetly right?

as in they didn't compute 5,6, 7, 8, but rather used that fact 5 mod 9 is cong to -4 mod 9

similarly for others

[worn fox]

Yes

[barren portal]

thank you

.close

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[night lodge]

how do we get 1/2 here?

[alpine sable]

divide both sides by A_0

as said on the image

well, the original 1/2 gets here

because you are finding the half life

(A_0)/2 is half of the original amount

[night lodge]

ohhh okay

ty that rlly clears it up

[alpine sable]

np

[night lodge]

.done

.close

[lone heart]

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[weary wyvern]

for the covariance of 2 rvs to be defined, do they need to be on the same probability space?

[lone heart]

@weary wyvern Has your question been resolved?

[marsh rapids]

surely, they ought to be otherwise you can't take the expected value of their product

E((X-E(X))(Y-E(Y))

E(XY) - E(X)E(Y)

either way you need XY to be defined and to be able to take the expected value of that, so you need a common probability space

[weary wyvern]

makes sense

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[patent swallow]

hint plz 😄

I can share what I did but I'm getting into problems

I need to find fx

[vale crag]

sure

hint comes after

[patent swallow]

ok

uploading

its tricky

I tried this but it wont help I guess

since dt is dividing

and then to the right you can see my last attempt

and it doesnt look right at all

[vale crag]

alright yea you seem kinda lost

[patent swallow]

indeed

[vale crag]

this one actually seperates quite nicely

[patent swallow]

I guess I need to get f'x/fx and integrate both sides

but yeah no luck yet

[vale crag]

try factoring all the f'(x) together

at the start

[patent swallow]

so 5f'x to the other side first

right?

[vale crag]

ya

[patent swallow]

going

ok done

ah

ok lemme see if I can do it now

finishing

1 moment

doesnt look good either 🤣

is this right? 🤔

[vale crag]

how did you integrate the left hand side ?

seems very wrong

[patent swallow]

wow

why

where

🤔

[vale crag]

it sounds like you split your fraction by the denominator to integrate your function

but fractions don't work like that

you're not an algebra 1 student in disguise aren't you?

[patent swallow]

bruh

right

yeah fucked up mistake just saw it lmao

hahahaha

thanks I'll re-do it 😄

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[lone heart]

@patent swallow Has your question been resolved?

[lone heart]

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[dapper lynx]

Hey all

[dapper lynx]

The original text of the question

Is it asking this, paraphrased?

[worn fox]

that is what it is asking yes

[dapper lynx]

Awesome

XD

Math with teachers for whom English is a second language can be tricky

I've lost more than a few marks over the terms

.close

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[proven leaf]

Khan Academy and YouTube are good online places to independently study mathematics :) that's how I started but there are many ways to approach

[foggy current]

I'd suggest the basic algebra section on Khan Academy

[alpine sable]

what have you studied so far/

[foggy current]

Or if you want to get more basic, pre-algebra might be good

Yes

There's also the section 7th grade math, which also may help

Because math is not just algebra

[alpine sable]

Does anyone know how to do this if so can u show me the details how it's done plz

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[icy jewel]

idk what i did wrong for the domain

[crisp cargo]

can x be less than -2?

[icy jewel]

oh

im blind

i had brainfart

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[icy jewel]

.close

[alpine sable]

Wait what

Shouldn't the domain be (-infinity,1 ]n[1,infinite)?

[crisp cargo]

nah

[alpine sable]

what is the answer?

[crisp cargo]

[-2,infinity)

weird that the chat doesn't close

[alpine sable]

How though how about the other line and the closed circle on 1

[crisp cargo]

look at the piecewise definition

[alpine sable]

Oh yeah sorry my bad 😆

The graph was confusing

[lone heart]

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[quick gust]

Hello could someone check this for me : )

[shy dove]

yeah they look fine

[quick gust]

Coolio thankyou

.close

[lone heart]

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[snow salmon]

anyone have all formulars for y11 GCSE

[lone heart]

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[grave eagle]

whats the difference between

[grave eagle]

-f(x)

and

f(-x)

i just realized how stujpid the question is nvm

[long axle]

Lol

U got it?

[grave eagle]

i mean yeah

but im stil stuck

i get what the difference is but im lost here

i just need some more thinking give me a sec

i have no idea

[torn elk]

which one u don't know?

[grave eagle]

all below f(x-1)

[torn elk]

read the first f(x) till the end

and if you know what's f(-x) you'll be able to solve it

here's an example

[grave eagle]

y axis flip

i see that

so (x-1) is supposed to change the x value

by +1

doesnt that mean that y is unaffected??

[trim sorrel]

You want to find f(x - 1) when x = -1

so whats x - 1

[grave eagle]

-2?

ohhhhh

[trim sorrel]

Right

So you want f(-2)

For that one

Its just a matter of plugging things in

[grave eagle]

bruh\

OH

F(-2)

not 02

-2

[torn elk]

for -1 it's f(-2)

[grave eagle]

so just type f(-2) in

[torn elk]

which is

[grave eagle]

the key says the answer was 1?

[torn elk]

you can see here

[grave eagle]

O

OHHH

do they all go back

had to get a new one since i got the other one wrong

[trim sorrel]

Yeah you have to reference the original function values

[grave eagle]

so here

-4-1

=-5

thres no -5

[trim sorrel]

Hm?

[grave eagle]

on x's

[trim sorrel]

f(x - 1) when x = -1

f(-2)

[grave eagle]

bruh

ok

[trim sorrel]

Wheres -4 coming from>

?

Did you confuse it with this box

[grave eagle]

absolute brainfart

my bad

[trim sorrel]

Lolll

You got it now tho

[grave eagle]

ye

im a math god

thanks

[trim sorrel]

W

[lone heart]

@grave eagle Has your question been resolved?

[lone heart]

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[lone heart]

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[kind wind]

For question 4

[kind wind]

i cannot seem to solve it for a reason

quadratic formula and man does not work

caculate the angle that a line with a slope of 4 makes with the x-axis

another question i do not understand

[trim sorrel]

Do you know about the discriminant

Oh wai

It asks for the actual zeros not number of zeros

Sorry misread

Why doesnt quadratic formula work?

Oh does it mean to say x^2 - 2x + 24 it might be cut off

[lone heart]

@kind wind Has your question been resolved?

[kind wind]

the

_4ac part

-4ac

it is a negative under a root so my calculator keeps saying math error

[trim sorrel]

Right

Can you take the square root of a negative number?

[kind wind]

no

[trim sorrel]

Right

[kind wind]

gives me like -59 or someting

[trim sorrel]

(You actually can, but you'll learn that later)

Anyways,

[kind wind]

so how do i get the zeros

[trim sorrel]

What that means is that the quadratic equation produces invalid values

This means that there actually are no zeroes

[kind wind]

u sure

u tried it?

[trim sorrel]

x^2 - 2x + 24?

Right?

[kind wind]

yes

[trim sorrel]

Assuming you havn't learned about complex numbers, then yes there are no zeroes

[kind wind]

yeha

yeah

ok thats makes sense

[trim sorrel]

👍

[kind wind]

idk how to do this question still tho

[trim sorrel]

You gotta use trig

[kind wind]

i dont know how to apply slope tho

like if it gives me a side length or something maybe i can get the angle

it confuses me

this is what i have

[trim sorrel]

What's an equation of a line with slope 4

[kind wind]

4/1

[trim sorrel]

Equation of a line

[trim sorrel]

Actually, this works

Nvm

So slope = rise/run right

4/1

[kind wind]

yes

[trim sorrel]

That means that for every increase of x by 1, y increases by 4

Think about how that fact can give you a triangle

[kind wind]

i dnot understand

[kind wind]

i thought abput it

and its not giving me triangle

[trim sorrel]

Youre like almost there

Take one of those points

Like (1, 4) for example

[kind wind]

ya

[trim sorrel]

Take the origin

And take (1, 0)

You get a triangle

[kind wind]

ohhhhhh

[trim sorrel]

A right triangle

With side lengths you know

Or at least you know two of them easily

[lone heart]

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[trim sorrel]

And then use trig with the angle

[kind wind]

just tan

right

[trim sorrel]

Right

[lone heart]

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[kind wind]

i get it now, thanks alot dude

[kind wind]

.closed

.close

[lone heart]

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[glass onyx]

Is possible to evaluate value of trig functions like sin(76.2) etc without calculator

[glass onyx]

And arctan arcsin arccos?

[vapid shuttle]

it is possible

why are you asking though?

[glass onyx]

Is it difficult?

[vapid shuttle]

depends on the angle

[glass onyx]

With that quadrant trick right?

It can't get values like 67 72.4 etc

[median oar]

you can do series expansion

[vapid shuttle]

there are ways to get those values approximately without using a calculator, but it is difficult. Why are you asking?

[median oar]

you can solve it numerically like calculators do

[glass onyx]

Taylor one right?

[vapid shuttle]

of course angles that are on the unit circle, or near angles on the unit circle, that is not so hard

[glass onyx]

Just in case

[vapid shuttle]

in case...??

in case all calculators and tech in the world get fried and you must know the exact value of arcsin(72.4)?

[median oar]

nah it's a valid question to ask how to numerically approximate something like that

[sour dove]

just wait until robot overlords take over all technology. Then we'll be happy we know how to find sin(76.2) by hand

[vapid shuttle]

(I'd say if it was a valid question, the reason would have to be more valid than "just in case", but I digress lol)

[glass onyx]

Yes

[lone heart]

@glass onyx Has your question been resolved?

[lone heart]

@glass onyx Has your question been resolved?

[lone heart]

@glass onyx Has your question been resolved?

[lone heart]

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[scenic wing]

how did they solve it when inside the root is negative

[outer kite]

uh

[scenic wing]

49-196=-147

[outer kite]

so it's root(-147)

which

using index laws

is root(-1) times root(147)

root -1 is i

[scenic wing]

root of 147 is 12.124

what did they simplify it to

wait

[outer kite]

firstly do you get the i part

[scenic wing]

yes

[outer kite]

if so then 147=7x7x3

so you take make it root(49) times root(3)

root 49 is 7

I can write out the working out if you want

[scenic wing]

ok so root of 49 and root of 3

root of 49 simplifies to 7

how did it disappear

[outer kite]

it didnt

its ontop of the fraction

[scenic wing]

now i see

[outer kite]

7/2 root 3

[scenic wing]

they simplified the denom

[outer kite]

yeah ok

[lone heart]

@scenic wing Has your question been resolved?

[lone heart]

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[queen lion]

Hey

[queen lion]

Just need a quick help

How are we able to turn 5 into 1/5 those are not the same in this instance

[mortal trellis]

dividing by 5 is the same as multiplying by 1/5

[queen lion]

Okay nice

Does it go other was around too

Dividing 5 = multiply 1/5

[lone heart]

@queen lion Has your question been resolved?

[mortal trellis]

dividing 1/5 = multiplying 5

[lone heart]

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[rare bay]

can anyone help or refer me to where I can learn how to graph/plot simple linear equations on my HP prime G2 calculator?

[rare bay]

there aren't really any good tutorials out there

[tardy stag]

you might just have to explore

i found the manual http://h10032.www1.hp.com/ctg/Manual/c05332710.pdf plotting seems to be starting on page 72 or so

[lone heart]

@rare bay Has your question been resolved?

[rare bay]

thanks

[lone heart]

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[grand berry]

Solo and James are standing at the seashore 35 miles apart. The coastline is a straight line between them. Both can see the same ship in the water. The angle between the coastline and the line between the ship and Solo is 40°. The angle between the coastline and the line between the ship and James is 62°. How far is the ship from Solo?

Here's my take on the problem. Though, I'm not sure if it is correct. Please point out my mistakes if there are any

[languid bolt]

uhh how is the angle in the middle 62 and 40? @grand berry

tbh the angles are very wrong

the angle that the line from the ship to the coastline makes would sum up to 180 since the coastline is a straight line

hello?

you here?

[languid bolt]

you don't need it tho

but the angles you wrote down was wrong

[lone heart]

@grand berry Has your question been resolved?

[grand berry]

Hello, I'm here

My bad

[languid bolt]

are you familiar with the sine formula?

[grand berry]

Yes

[languid bolt]

try using it

[grand berry]

Uhh use it how exactly?

[languid bolt]

are we on the same page here... wait

what is the sine formula?

[grand berry]

sinA/a = sinB/b = sinC/c

[languid bolt]

yes

you have the angles and you have one of the sides (the coastline)

try inserting it into the formula

[grand berry]

Well wouldn't the triangle split into two triangles? I'm just puzzled as to how I should 'insert' it

[lone heart]

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[shadow sparrow]

Not completely sure what I'm missing here... What is f(x)? Its not explicitly shown?

[wary stream]

f(x) is that plot

[shadow sparrow]

OH. I'm dumb.

[wary stream]

You're using that to find the area from 5 to 9

[shadow sparrow]

I was looking at the x axis

Its asking for what the height is on the y axis

🤦‍♂️

Sorry lol, silly question, thank you for pointing that out

.close

[lone heart]

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[patent swallow]

guys my professor corrected something for me and I've few mistakes, I understand them all except from the integral

to start with, why does he write that everywhere?

I'm integrating by parts.

it's like if idk how to solve a simple integral :/

am I missing something

no

[late token]

yes

[mellow grail]

that straight line means you do F(a) - F(b) where F(x) is the integral of f(x)

for eg. if you have the integral of cos(x) (indefinite)

we have sin(x)

if you consider the definite integral it must have some bounds

say 0 (lower) and pi/2 (upper)

we would write it as $\int_{0}^{\frac{\pi}{2}} cos(x) dx$

[ocean seal]

itzkraken.

[mellow grail]

which we can write as

$sin(x) |_{0}^{\frac{\pi}{2}}$

[ocean seal]

itzkraken.

[mellow grail]

now you do F(b) - F(a) (a=0, b=pi/2)

[mellow grail]

that straight line is standard notation for evaluating an integral

[lone heart]

@patent swallow Has your question been resolved?

[lone heart]

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[safe tartan]

how would you do this ?

[molten pivot]

Try differentiating under the integral sign

[safe tartan]

i did u substitution of u=y+1 but really didnt do anything

[echo socket]

Geometry or trig sub, which path do you want to go with?

[safe tartan]

umm trig sub because i could do it geometrically

[echo socket]

Then consider either one of y = sin(theta) or y = cos(theta)

[safe tartan]

yep

considered

[echo socket]

Well, let us pick y = sin(theta)

[safe tartan]

oh the root3/2 is a giveaway i guess to do trig sub

[echo socket]

No

[safe tartan]

oh is it not

cuz like exact value

[echo socket]

The sqrt(1 - y^2) hints at trib sub

[safe tartan]

hmm yea

[echo socket]

Then dy = cos(theta) dtheta and the integral becomes

$\int_0^{\frac{\pi}3}\sqrt{1 - \sin^2\theta}(\cos\theta\dd\theta)$

[ocean seal]

alonelybean

[echo socket]

Or simply $\int_0^{\frac{\pi}3}\cos^2\theta\dd\theta$

[ocean seal]

alonelybean

[echo socket]

Ever integrated cos^2 before?

[safe tartan]

umm yea

what was it again like

actually na i dont think i have

[echo socket]

The double angle identity for cosine states that [ \cos2\theta = 2\cos^2\theta - 1]
Rearranging the terms here, we get [ \cos^2\theta = \frac{\cos2\theta + 1}2]

[ocean seal]

alonelybean

[safe tartan]

ah yep i know that one

and now cos becomes

1/a sin

right

so here it would be 1/4 sin theta +1/2theta?

2theta ofc

soz

[echo socket]

Yes

[safe tartan]

thats genius

i always struggled with squared ones

thanks

[wanton nova]

https://tenor.com/view/the-universe-tim-and-eric-mind-blown-mind-blown-meme-mind-explosion-mind-explosion-meme-gif-18002878

[safe tartan]

me rn

i got an answer which was 3 solutions but

its not an option so

[naive crystal]

Divide both side by 2??

[safe tartan]

maybe i added solutions after multiplying by x somewher

[naive crystal]

,w a^2-2a+2=0

[ocean seal]

Results provided by WolframAlpha

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

Requested by _basudev#0

[safe tartan]

oh

[echo socket]

$(2^{\sin^2{x}})^2 \ne 2^{\sin^4{x}}$ btw

[ocean seal]

alonelybean

[safe tartan]

oh yeah oops

[echo socket]

And A^ - 2A + 2 =/= (A - 1)^2

[safe tartan]

yea

i still get the same quadratic though

for some reason

[echo socket]

Instead it should be (A - 1)^2 = -1, hence no real solutions

[safe tartan]

oh that makes sense

quick question, when doing solids of revolution and rotating about the y axis

since it becomes pi integral x^2

do i just rewrite f(x) in terms of x

[echo socket]

No, it's x^2 dy

[safe tartan]

oh ye

so i just write x in terms of y by rearranging

[echo socket]

Yup

[safe tartan]

ive identified it to be B or D, but im not too sure on the bounds

do i find the y intercepts of f(x) to find the bounds?

[echo socket]

Should be from y = 0 to y = 0.6pi

[safe tartan]

hmm not sure how you would derive that

[echo socket]

No, it's just that cos^-1(x - 1) goes from 0 to pi by the definition of its range

[safe tartan]

ah i c

[echo socket]

So y goes from 0.6 * 0 to 0.6 * pi

[safe tartan]

thanks

for this integral, for what bounds do i integrate the positive case and what bounds do i integrate the negative case?

[echo socket]

You can immedialy rewrite that as 2 times the same integral but going from 0 to 1

Because the integrand is an even function

Generally $\int_{-a}^af(x)\dd{x} = 2\int_0^af(x)\dd{x}$ if $f$ is even

[ocean seal]

alonelybean

[safe tartan]

yep

[echo socket]

And, |x^2 - 1| = |x - 1| * |x + 1|, but now we are taking 0 < x < 1, so this is equal to (1 - x) * (x + 1)

[safe tartan]

is this the case when x^2-1 >_0 since we are only looking at the right side

wait no

[echo socket]

x^2 - 1 is negative in this case

Either way you look at it

[safe tartan]

oh i see just -x^2+1

[echo socket]

Yes

[safe tartan]

oh yeah makes sense cuz from 0<x<1 we take the positive case and theres a minus in front

is that somewhat right

[echo socket]

Yes, that is right

Even if we were not to half the interval

We would have -1 < x < 1

Meaning 0 <= x^2 < 1 and -1 <= x^2 - 1 < 0

[safe tartan]

mmhm

so now im simply integrating -x^2+2

[echo socket]

|x^2 - 1| = 1 - x^2
-|x^2 - 1| = x^2 - 1

[safe tartan]

huh

[echo socket]

There is a negative in front of the absolute value

|x^2 - 1| = 1 - x^2

So

-|x^2 - 1| + 1 = -(1 - x^2) + 1 = x^2 - 1 + 1 = x^2

[safe tartan]

sorry why does the thing inside the absolute value become switched

[echo socket]

|z| = -z when z is negative

We showed that x^2 - 1 is negative

Meaning |x^2 - 1| = -(x^2 - 1)

[safe tartan]

oh r u saying that x^2-1 has negative y value for the given bounds

that makes sense

sorry im quite sloq

[echo socket]

It is fine, take your time

[safe tartan]

yep i got the question, and the answer. Thanks!

.close

[lone heart]

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[rough sequoia]

Find the value of
$cos 20 + 2 sin^2 55 - \sqrt 2 sin 65$

[ocean seal]

omniscientzenith

[vale wigeon]

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[alpine sable]

Who turned Ann into an NPC /j

[ornate ginkgo]

That's an mc in disguise of an npc

[quiet vector]

[lone heart]

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[gilded marten]

You guys do science physics here?

[gilded marten]

I kind of forgot what this formula meant

[grim berry]

I think there's another server dedicated to that

but idk maybe

someone here probably knows

[quiet vector]

in #old-network

[grim berry]

yep

[gilded marten]

ahh ty ty sorry for the inconvience

.closew

.close

[lone heart]

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[steady moon]

hello, I must find this function's primitive:

[steady moon]

I've done that

but, why on the answer, they added (1/2)???

[gray isle]

your u' is wrong

d/dx of (1+x^2)
isn't x

[proven leaf]

also btw is the primitive just the antiderivative?

[steady moon]

omg it's 2x!!

[steady moon]

I'm not an expert at all but yes, the primitive is the function that's not derivated🤔 I've the derivated function, I have to find the non-derivated function called primitive

[proven leaf]

I'll just stick to my derivative and antiderivative lol

[steady moon]

.close

[lone heart]

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[zinc lion]

i have a question abt solving complex fractions

(both)

[lone heart]

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[steep marten]

Hi, could anyone please explain why the bounded area integral trick top function - bottom function doesn't work for this question, and when in general it can be used. Thanks in advance

[vale wigeon]

how do you intend to use it

[steep marten]

I tried doing integral from 0 to 1(x^r-tangent line)dx but it didn't work

[vale wigeon]

well there's a simple reason why it didn't work. you calculated the area of the wrong region!

[steep marten]

I know we could do the integral from 0 to 1 (x^r)dx and subtract a triangle area but I am confused as to when the bounded area thing can be used

[vale wigeon]

this is the region whose area you calculated.

[steep marten]

Oh wait what

[vale wigeon]

this is the region bounded by y=x^r and the tangent line between x=0 and x=1.

[steep marten]

Okay wait so if I draw a line where the area starts and it hits both curves and then I do it again for where the area ends. If these 2 conditions are met and then I do top - bottom integral it gives me the total area between 2 curves

Could this be a way to think about it?

[vale wigeon]

nyeh

sure

[steep marten]

But there has to be a clear top and bottom curve. For example it wouldn't work here because there is no top and bottom curve and the lines at the start and end of the area don't hit both curves?

[vale wigeon]

But there has to be a clear top and bottom curve
yes

[steep marten]

Alright thank you so much Ann I think I understand when I can use it now, when the lines hit both curves and there is a top and bottom curve. Have a great day!

Would anyone know how I close this channel?

[lone heart]

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[gentle shadow]

can someone explaine this step to me? Idont know where the +4x² at the end come from.

[sharp thorn]

$(2x)^2$

[ocean seal]

bettimsucks

[mellow grail]

use $(a-b)^2 = a^2 - 2ab + b^2$ with $a=x^2, b=2x$

[ocean seal]

itzkraken.

[gentle shadow]

k thx a lot

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[mystic comet]

can anyone confirm if it's -1/2 or 0

[tardy stag]

I'm not sure how it could be negative or zero because all of the terms are positive

[gray isle]

as this is a series of positive terms both values are wrong

[mystic comet]

I made it into sum (1/k) - sum (1/(k+1))

[tardy stag]

and what did you get for sum(1/k)?

[noble sinew]

Telescoping series

[gray isle]

what did you do after that

[mystic comet]

1/n which I feel is wrong

[mystic comet]

I did this

telescoping

[gray isle]

can you show your work

[mystic comet]

sure

[noble sinew]

Then write it as sum from k=1 to n

[noble sinew]

You only have 2 terms that don’t get cancelled

Then take limit

What is $$\sum_{k=1}^5 \frac{1}{k}-\frac{1}{k+1}$$ for example?

[ocean seal]

scapeprof

[mystic comet]

(1/5) -(1/6) ??

[gray isle]

no

[noble sinew]

No

[gray isle]

do you understand summation notation

[mystic comet]

yes i do but it isn't geometric or arithemtic

[gray isle]

can you write what
$$\sum_{k=1}^5 \frac{1}{k}$$
represents, leaving that as a sum, don't bother simplifying

[ocean seal]

ℝamonov

[mystic comet]

1 + (1/2 )+ (1/3) +(1/4) + (1/5)

[gray isle]

yes

same question for
$$\sum_{k=1}^5 \frac{1}{k+1}$$

[ocean seal]

ℝamonov

[gray isle]

hence what do you get for:
$$\sum_{k=1}^5 \br{\frac{1}{k} - \frac{1}{k+1}} = \sum_{k=1}^5 \frac{1}{k}- \sum_{k=1}^5 \frac{1}{k+1}$$

[ocean seal]

ℝamonov

[mystic comet]

1+(1/6) ?

[gray isle]

no

can you show your work

[mystic comet]

is this correct ?

[gray isle]

for what i was asing for, yes

[mystic comet]

oh

1-(1/6) ?

[gray isle]

yes

[mystic comet]

forgot the sign

[noble sinew]

So can you say what $$\sum_{k=1}^n \frac{1}{k}-\frac{1}{k+1}$$ is now?

[ocean seal]

scapeprof

[mystic comet]

1-(1/(n+1))??

[noble sinew]

Yes

[mystic comet]

it's the same rule isn't it?

[noble sinew]

Yes

[mystic comet]

i got it thank you

[gray isle]

helps to write out a few terms and align stuff

$\begin{aligned} 1 &+ \frac12 + \frac13 + \dots + \frac1n \
&- \frac12 - \frac 13 - \dots -\frac1n - \frac{1}{n+1} \end{aligned}$

[ocean seal]

ℝamonov

[mystic comet]

(-1)^n is undefined right?

[gray isle]

there are real value outputs for integer n