help-10

[drowsy girder]

I see you can factor first equation

[fallen mantle]

what do you get?

[gritty dragon]

yeah I tried but I just didint get it

Can maybe some help me like step for step

<@&286206848099549185> can someone help me it due tomorrow and i don't want to get in any trouble

[obtuse pebble]

@gritty dragon Has your question been resolved?

[gritty dragon]

no

it hasn't

[trim oasis]

So, one of the methods to solve this is to solve one of the equations for either x or y and then substitute the variables of the other equation with your result.

I'd suggest solving 3y - 2x = 12 since it's the easier equation
Do you want to solve for x or for y?

[gritty dragon]

yeah but i don't know hoe

[trim oasis]

I'll help you, don't worry
Let's solve for x, it doesn't really matter too much

[gritty dragon]

oke

[trim oasis]

So you want to solve 3y - 2x = 12 for x, do you know how to do that?

[gritty dragon]

I think you have to trace out the x

[trim oasis]

Yep, so you'll want to move everything that isn't x to the other side of the equation (i.e. a variable, a number)

[gritty dragon]

okay

[trim oasis]

So what you want to do is subtract 3y from both sides, does that make sense?

[gritty dragon]

so that is 2x=-3y+12

yes

[trim oasis]

Yep 👍

Well, almost actually

[gritty dragon]

oh

okay

[trim oasis]

It's
-2x = -3y + 12

[gritty dragon]

oh yeah

[trim oasis]

Don't worry about that, it's an easy mistake to make
So now, you want the -2 to be gone as well. This means you have to divide both sides by -2

[gritty dragon]

yes i understand

[trim oasis]

And then, the equation is...?

[gritty dragon]

x= 1,5-6

1,5y

[trim oasis]

Exactly!

[gritty dragon]

oke

[trim oasis]

And now, you enter 1.5y - 6 for every x that appears in the other equation

[gritty dragon]

oh isee

yes

so now we got to do that for the y aswel

[trim oasis]

Then, the equation becomes ${(1.5y - 6)}^2+4y^2-8y+2*(1.5y-6)$
And the good thing about this is that you only have y's and no x's

[warm shale]

Asgard

[trim oasis]

No, then you just solve the resulting equation for y

[gritty dragon]

oh

wait let me see

[trim oasis]

For now, just try to get everything with a y or a y^2 on one side, you're going to bump into a problem you can't solve with eihgth grade maths

[gritty dragon]

okay so I can't answer that main question of 5x+5y

[trim oasis]

The problem is that you have to solve a quadratic equation and that's not possible without guessing (at least at the moment, you'll learn a formula for it later)

[gritty dragon]

you mean a^2+bx+c

[trim oasis]

Oh, do you know that already? :D

[gritty dragon]

yeah

[trim oasis]

Fantastic, then you can solve it

[gritty dragon]

GRADE

NICE

sorry caps

[trim oasis]

Dw :>

[gritty dragon]

so i am calculating the first question right now

[trim oasis]

Mhm
Do you know how to solve the equation?

[gritty dragon]

i am trying but if you could help me that would be great

[trim oasis]

Alright :D

[gritty dragon]

thx 🙂

[trim oasis]

So first, let's try to make the (1.5y - 6)^2 easier
Just multiply (1.5y - 6)*(1.5y - 6)

[gritty dragon]

so 1,5y*1,5y

is 2,25y^2

[trim oasis]

Not quite

[gritty dragon]

oh

[trim oasis]

When squaring this sort of expression, you have to sort of multiply everything with everything

[gritty dragon]

alr and how do we do that

[trim oasis]

Like this

[gritty dragon]

oh but I thought I was doing that

[trim oasis]

So the result is a*a + a*b + a*b + b*b

$a^2 + 2*ab + b^2$

[warm shale]

Asgard

[trim oasis]

You only did a*a

[gritty dragon]

oh yeah I see

[trim oasis]

That means our result is
$(1.5y)^2 + 3y*(-6) + (-6)^2$

[gritty dragon]

yeah okay

[trim oasis]

Not 21.5 lmao
Sorry, the bot seemed to skip the asterisk

[warm shale]

Asgard

[gritty dragon]

oh np

[trim oasis]

Do you think you can solve this yourself?

[gritty dragon]

yeah I am going to try

[trim oasis]

Alright, tell me if you're stuck

[gritty dragon]

i will

I got -15,75y+36

[trim oasis]

You seem to have dropped a y^2 somewhere 🤔

[gritty dragon]

oh

[trim oasis]

The 36 is correct, but it's -18y, not -15.75y

[gritty dragon]

oh

I at first had 2,25y-18y+36

[trim oasis]

Okay, well the correct result is 2.25 y^2 - 18y + 36

[trim oasis]

That was almost correct! You just seem to have forgotten a ^2

[gritty dragon]

yeah I think so

so what do we do with this answer

[trim oasis]

Alright, so now we're at 2.25y^2 - 18y + 36 + 4y^2 - 8y + 2*(1.5y - 6)

[gritty dragon]

yes

[trim oasis]

We just have to simplify one more thing

[gritty dragon]

the 4y^2-8y+2*(1,5y-6) right

[trim oasis]

Exactly
And we'll make our lives a whole lot easier if we solve 2*(1.5y - 6) first

[gritty dragon]

yeah so that is 3y-12

[trim oasis]

Yep

So now, it's
2.25y^2 - 18y + 36 + 4y^2 - 8y + 3y -12

[gritty dragon]

yes

[trim oasis]

And we can add things like 2.25y^2 and 4y^2 together to simplify everything even further

[gritty dragon]

yes so we get 6,25y^2-26y+24

[trim oasis]

Exactly!

And the equation that we get is
6,25y^2-26y+24 = 3, but we need the 3 to be a zero, so just subtract three on both sides

[gritty dragon]

oke

so 6,25y^2-26y+21

[trim oasis]

👍

[gritty dragon]

nice

[trim oasis]

Now, just enter the values into the formula for quadratic equations

[gritty dragon]

oke

don't we need to divide everything by 6,25 firs

first*

[trim oasis]

Not really

Just enter
6.25 for a
-26 for b
21 for c

[gritty dragon]

oh thanks

I am stuck man

[trim oasis]

Where exactly?

[gritty dragon]

My calculator says math error

[trim oasis]

Could you send what you entered?

[gritty dragon]

-26+ -26-4**-26* *21

[trim oasis]

So you entered +-?

[gritty dragon]

No only the +

and then afterward I would do the same but then for -

[trim oasis]

Mhm okay

Could you put parentheses around the (-26) and try again?

[gritty dragon]

i will try

yeah that worked

[trim oasis]

Awesome :D
What's the result?

[gritty dragon]

-2190.625

[trim oasis]

Hmmmm

[gritty dragon]

wait

i did something wrong

[trim oasis]

I think so...

[gritty dragon]

-1640,625

I still think its wrong

[trim oasis]

Give me a second

[gritty dragon]

and the other answer was -1803.125

[trim oasis]

The two blue points are the ones we're looking for
They're pretty close to 1 and 3

[gritty dragon]

oh then I was def. wrong

[trim oasis]

The solutions of 6,25y^2-26y+21 = 0 are
y_1 = 1.09
y_2 = 3.06

Oh wait hold up

[gritty dragon]

alr I will wait

[trim oasis]

There are two solutions, because it's a quadratic equation, but we need one value for y so we can continue to work with it

[gritty dragon]

so we have to go back to 3y-2x=12

[trim oasis]

I don't think we made a mistake there, the problem is somewhere else

[gritty dragon]

oh but if we want to know the value of y we have to trace back the y

[trim oasis]

Oh wait
Let's try both y for what we're going to do and see if one matches and the other one doesn't
Then we can rule it out

[gritty dragon]

smart idea

[trim oasis]

So we have these two values, we just have to plug them into 3y-2x=12 and then solve for x
(Two times because we have two values)

[gritty dragon]

yesh I get it

[trim oasis]

Awesome!

[gritty dragon]

so 3y-2x=1.09 and 3y-2x=3,06

[trim oasis]

Not really, you have to substitute the y in there with the value, not the 12

[gritty dragon]

so how would you write that down

[trim oasis]

3*1.09-2x=12
and
3*3.06-2x=12

Fricken Discord xD

[gritty dragon]

hah

3,27-2x=12

[trim oasis]

Actually, we can do it even simpler

[gritty dragon]

oe I like simple things

[trim oasis]

Just use the equation we got before:
x = 1.5y - 6

So
x_1 = 1.5*1.09 - 6
x_2 = 1.5*3.06 - 6

wait

[gritty dragon]

what is wrong

[trim oasis]

There isn't a condition we have to fulfill, they want you to calculate a value...

[gritty dragon]

what do you mean

[trim oasis]

The question was 5x + 5y = ?

[gritty dragon]

yas

yes

[trim oasis]

And since we have two y and we will have two x, there will be two values for "?"

There will be
?_1 = -16.375
?_2 = 22.35

Hmmmm

[gritty dragon]

we only have x and y what could it be

[trim oasis]

Well, we have two y, that means we can calculate two x, which will result in two equations

[gritty dragon]

2 coordination

[trim oasis]

(I) 5*x_1 + 5*y_1 = ?_1
(II) 5*x_2 + 5*y_2 = ?_2

[gritty dragon]

ahhh

so we have the 2

[trim oasis]

oooof I have no idea how to solve this tbh

[gritty dragon]

oh its okay

[trim oasis]

Maybe another person can help you, try pinging the Helper role after 15 minutes
It's getting kind of late where I live unfortunately

[gritty dragon]

yeah for me to

its almost 11pm

[trim oasis]

We're in the same timezone :D

[gritty dragon]

where do you live then

[trim oasis]

Germany 👀

[gritty dragon]

i am from the netherlands

found this on a site

I don't know if its right i an just going to turn this in

[trim oasis]

Nice :D
Well, good luck with your homework!

[trim oasis]

Not sure about that 🤷‍♂️

[gritty dragon]

thx man you too

I am to tired to go make math again so i am fine with it

well asgard thx for your help

.close

[obtuse pebble]

Channel closed

Closed by @gritty dragon

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timid silo]

(sin20)(tan10+cot10) without a calculator

[timid silo]

I just do not know how this is supposed to wkr

work

I converted into radians but this did not help my understanding

[obtuse pebble]

@timid silo Has your question been resolved?

Channel closed

Closed by @tranquil idol

Use .reopen if this was a mistake.

[timid silo]

.reopen

[obtuse pebble]

✅

[timid silo]

(sin20)(tan10+cot10) without a calculator

<@&286206848099549185>

[strong vale]

easy

[timid silo]

its easy I just forget how to do it

[strong vale]

2

[timid silo]

yeah I know the answer

[strong vale]

what is sin(20) in terms of cos(10) and sin(10)

[timid silo]

its just how to do it

[strong vale]

do u know?

[timid silo]

no

i forgot how to do lots of trig

its the identities i forgot

[strong vale]

u know sin(2a) = 2sin(a)cos(a)

[timid silo]

yes

[strong vale]

ok apply it and solve problem

u done?

[timid silo]

i did something stupid I think

(2sin10cos10)(sin10/cos10+cos10/sin10)

is this a valid step?

@strong vale

[strong vale]

no this is fine

[timid silo]

i dont know what to do next

[strong vale]

now carry out the multiplication

[timid silo]

I did it wrong

and got 4

after I add both terms

[strong vale]

how u get 4

(2sin10cos10)(sin10/cos10+cos10/sin10) start from here

and walk me through what u did

[timid silo]

no

i just dont know what im doing tbh

could you please walk me through the steps

[strong vale]

Wht is $2\sin{10}\cos{10} \times \frac{\sin{10}}{\cos{10}}$

[warm shale]

azeem321

[strong vale]

no im not solving it for you

[timid silo]

i dont want u to solve for me

if i wanted an answer i just use my calculator

this looks like sin10/2sin10 but its not

[strong vale]

carry out the multiplication and tell me your result

what is $\frac{\cos{10}}{\cos{10}}$

[warm shale]

azeem321

[timid silo]

1

omg i am cumb

dumb

ok

2sin^2(10)+2cos^2(10)

[strong vale]

yes

but it's fine keep studying and u will be less dumb soon

[timid silo]

its just

we did this topic months ago

[strong vale]

yes

[timid silo]

and my teacher randomly gives this assignment in the middle of studying logs

[strong vale]

trigonometry will follow you for years in your math classes so get good at it

when u get rusty watch youtube videos

[timid silo]

yes i do that sometimes

i just need review

ty ❤️

.close

[obtuse pebble]

Channel closed

Closed by @tranquil idol

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[copper lantern]

isnt this correct?

[haughty coyote]

,rotate 180

[haughty coyote]

wtf

what are these angles

[copper lantern]

sorry the pic is messed up

lol

[haughty coyote]

lmao

are you trying to integrate that fraction ?

[copper lantern]

you mean 0/x-3?

[haughty coyote]

no the rational fraction of "degree" 3/1

Cause then I'd say the result looks good to me

[copper lantern]

im supposed to use polynomial division on those 2 expressions and then integrate that

[haughty coyote]

,w integrate (x^3-3x²+2x-6)/(x-3)

[warm shale]

Results provided by WolframAlpha

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

Requested by themateo713#5303

[dreamy river]

@timid silo

Sup

[haughty coyote]

looks good to me yeah

[copper lantern]

but this is different

[haughty coyote]

no

[copper lantern]

how is there a x^2+6

[haughty coyote]

expand it

[copper lantern]

ohh yeah ok

[haughty coyote]

idk why it chose to write it that way

[copper lantern]

but look

[haughty coyote]

why did you integrate ?

[copper lantern]

i thought we were supposed to lol

[haughty coyote]

don't see it written anywhere

[copper lantern]

so do i just write x^2+2?

[haughty coyote]

what are you going to do if not try it

[copper lantern]

yeah you were right

thanks

can i ask u another question

[haughty coyote]

just post it

[copper lantern]

this is ibp tho

[haughty coyote]

yes

[copper lantern]

i've tried this multiple times and have gotten it wrong

i would show you my work but its all messy

[haughty coyote]

differentiate y

integrate e^-2y

[copper lantern]

only y?

[haughty coyote]

take the 9 out of the integral, don't bother with it

[copper lantern]

can i do this with tabular notation?

[haughty coyote]

what do you call tabular notation ?

[copper lantern]

u & dv

[haughty coyote]

if you want. I always forget how that looks but if that's what you're familiar with

[copper lantern]

ok i'll try

so i differentiate y until it's 0 right?

[haughty coyote]

Apply the formula, stop doing IBPs when you think it's not the easiest step anymore

[copper lantern]

?

oh yeah i put a negative in front of the 1

[haughty coyote]

should be -9(y/2+1/4)e^-2y

[haughty coyote]

if you write this into a sum properly, you should get that

[copper lantern]

does the 9 get distributed in those parenthesis?

[haughty coyote]

depends on how you want to write the final result. I put in factored form, you can expand if you want

[copper lantern]

i got -1/2ye^-2y-1/4e^-2y

but i haven't added the 9 in yet

[haughty coyote]

you'll notice it's the same

[copper lantern]

so do i put the 9 over the 2 & 4?

[haughty coyote]

you do whatever you want. There are many ways of writing this

[haughty coyote]

software like that should accept all equivalent answers as correct

[copper lantern]

oh ok so then i have to do the bounds now

that's the issue

i got this answer before

and it was incorrect

[haughty coyote]

looks wrong to me

[copper lantern]

its the same answer you got though

[haughty coyote]

you forgot to subtract the value at 0

[copper lantern]

oh yeahh thank you

do you mind helping with this

[haughty coyote]

I'd expect terms of all degrees from 3 to 0

not just 3

[copper lantern]

what do you mean?

[haughty coyote]

just integrate

by parts

then integrate by parts

then integrate by parts

[copper lantern]

u = t^3?

[haughty coyote]

just differentiate t^3

[copper lantern]

ok ill do that

[haughty coyote]

it's not hard when you got a polynomial, you take it's derivative until there's no polynomial anymore

guarantees the process ends

[copper lantern]

true i'll differentiate t^3 and integrate e^-2t

[haughty coyote]

,w integrate -8t^3 e^(-2t) dt

[warm shale]

Results provided by WolframAlpha

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

Requested by themateo713#5303

[copper lantern]

oh yeah let me add the 8

-8

[haughty coyote]

looks good to me

[copper lantern]

nvm i forgot the +c

[haughty coyote]

yes

[copper lantern]

mateo you are the literal goat

[haughty coyote]

I just told you to IBP and used wolfram calm down

and checked your answers

[copper lantern]

loll yeah i know but i ask some other people and they just tell me idk what im doing and to completely relearn ibp

[haughty coyote]

You look decently at easy with it yeah

[copper lantern]

so when i get someone that doesnt call me stupid im like enlightened lolll

[haughty coyote]

could get some more practice as these are very easy IBPs that you shouldn't struggle with, but for a help channel that's exceeding expectations for sure

[copper lantern]

nah i mean that even when i asked a teacher he said i have to relearn everything

[haughty coyote]

If you can do the exercises and you understand the method, the only thing left to "learn" is experience

[copper lantern]

yeah you're right

[haughty coyote]

That's often the difference on these problems between the help-seeker and the help-giver: we've seen these problems before, we know how to solve them, sometimes in our heads, and we can use methods that we're familiar with that you're not, it allows us to do these things more easily

experience makes one hell of a difference

[copper lantern]

for sure i agree

tbh i just like to make sure what the u & dv is for this because im not always sure

[haughty coyote]

It's like if you scroll up #help-6 with AuHasard, I had 4 methods for his limit

[haughty coyote]

it's whatever you want. Your job is to find the ones that work well

[copper lantern]

i put u=lnx

[haughty coyote]

That's experience with a type of problem, you end up having many solutions from different techniques

yes

indeed

first intuition as well

reduces to a nice IBP

you can do that one on your own I think

[copper lantern]

ill try

i think that lnx only differentiates to 1/x

[haughty coyote]

I don't like the "I think" here

[copper lantern]

i mean i know that it does

& it stops there or else it's too complicated

[haughty coyote]

Actually this one can also be done without u-sub

[copper lantern]

i got 1/3x^3lnx-1/3x^2

[haughty coyote]

,w integrate 8ln(x)/x²

[warm shale]

Results provided by WolframAlpha

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

Requested by themateo713#5303

[haughty coyote]

easy check

think it went the full IBP way

[copper lantern]

yeah thats different from what i got then

[haughty coyote]

and clearly not by just a constant

Fun fact: sometimes that constant difference is very hard to spot and 2 expressions are both correct even though they look significantly different

[copper lantern]

true

but im not sure what i did wrong

i could have done 1/x(1/3)x^3 instead

but x^3/x = x^2

[obtuse pebble]

@copper lantern Has your question been resolved?

Channel closed

Closed by @copper lantern

Use .reopen if this was a mistake.

[obtuse pebble]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[waxen musk]

anyone out there?

[warm pine]

what class is tghius

*this

[waxen musk]

Cal 1

[warm pine]

pls find someone else sorry

[waxen musk]

no probs thanks for replying atleast @warm pine

@analog python

@zinc flax

[obtuse pebble]

@waxen musk Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[serene hatch]

I don't know how to approach this

<@&286206848099549185>

[obtuse pebble]

@serene hatch Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[daring sparrow]

An integer is divisible by 11 if, and only if, the alternating sum of its digits is divisible by 11.

Does this prove both sides? or just the If n is divisible by 11 part?

[obtuse pebble]

@daring sparrow Has your question been resolved?

Channel closed

Closed by @daring sparrow

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[near burrow]

any idea how to solve this

[near burrow]

q 3

,rotate

[fallen mantle]

QPS + QRS = 180

[near burrow]

okay

so

b + d + c = 90

a = 60?

i need to know how to work it out

[fallen mantle]

b + d + c = 180
a + d = 180

d = 180 - a

[near burrow]

okay

[fallen mantle]

so: b + d + c = 180
b + (180 - a) + c = 90

[near burrow]

a + d = 180 because of angles on a line right

= 180

[fallen mantle]

not really

do you rmb the angle inside a four sided is always 360?

since there is already 2 angles with 90, so a+d would be 360-90-90

[near burrow]

yeah

angles of a quad = 360

wait

oh

so a + d = 180

im so confused

i hate geometry xd

[fallen mantle]

back to here

since b + (180 - a) + c = 180

[near burrow]

is 180 - a = 30

since 3 x 30 = 90

[fallen mantle]

so: b+c = a

the 180 can be cancelled out each other

[near burrow]

so what would i write

to prove b + c = a

ill probably get it if i write it out

[fallen mantle]

b + d + c = 180
a + d = 180
d = 180 - a

b + (180 - a) + c = 180
b + c = 180-180+a
b+c = a

done

[near burrow]

i still dont get it wtf

get the first line

why does a + d = 180

angles on a line or?

it cant be coint <'s because we arent using paralell lines

[fallen mantle]

rmb that all the 4 angles in a 4 sided polygon is always 360
since there is already got 2 angles which are 90, so a+d+90+90 = 360

[near burrow]

oh

so

why does b + c = a

[fallen mantle]

you have to substitute and simplify to get b +c = a

[near burrow]

hmph

i dont know why i dont get it

so

which 2 angles are 9

90

[fallen mantle]

PQR and PSR are both 90

[near burrow]

why

[fallen mantle]

rmb that: PQR + QRS + RSP + SPQ = 360

[fallen mantle]

this was stated in the question and diagram

[near burrow]

oh

OH

so

left side

and bottomleft side

and right side, bottom right side are 90

you

are in

a taken channel

[sick plank]

oh sorry

[near burrow]

#help-1

go there

and try

okay so

anonymous

[fallen mantle]

yes

[near burrow]

uhhh

idk where would i learn this

just written and explained simply in text form

i just cant understand it and ive got an even harder next question

[obtuse pebble]

@near burrow Has your question been resolved?

Channel closed

Closed by @near burrow

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tepid timber]

Heyo

[tepid timber]

Just wanted to clarify minor details for finding the determinant of a matrix

To find the determinant of a matrix, we just use laplace expansion on the top row of the matrix right?

And then that's all?

[mental solstice]

or any row or column, really

[tepid timber]

Wait wat rlly

So I can apply it to any single row or column

And it would find the determinant of the matrix

[mental solstice]

yes, so it's always a good idea to pick a row or column with a lot of zeroes

[tepid timber]

Oh

I see, thank you so much :))

[mental solstice]

no prob 🙂

[tepid timber]

.close

[obtuse pebble]

Channel closed

Closed by @tepid timber

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sleek sinew]

Are these triangles similar? I rotated it two different ways

One was similar, one was not

Just not sure which line segment UG corresponds to

If we make it correspond to FE, then SAS says the triangles are similar

But is it possible UG corresponds to BE, resulting in not similar?

[tepid timber]

I believe it is similar

angle UGA = angle BEF

UG/EF = GA/BE

The ratios of two lines and a corresponding angle is similar

And I think that's one of the proofs for similar triangles? I really don't trust myself

[sleek sinew]

Yes that would be correct if we did it that way

Just wondering how we know for sure UG corresponds to EF

Because if we did UG/BE = other ratio, it will not work

[tepid timber]

Yeah since UG/BE doesn't work and UG/EF works, then UG corresponds to EF

Cus if u said triangle ABE is similar to triangle PLO for example, AB/PL = BE/LO

And AB/LO wouldn't be equals to PL/BE

[sleek sinew]

Okay thanks

.close

[obtuse pebble]

Channel closed

Closed by @sleek sinew

Use .reopen if this was a mistake.

[obtuse pebble]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rapid sentinel]

need help solving 1 and 3 please, asap

[rapid sentinel]

i looked up the first one but didnt understand anything too

[fallen mantle]

,rotate

[river zephyr]

Oh lol very similar question to mine

If it’s above x axis, discriminant is <0

So arrange the equation to b^2-4ac <0

[rapid sentinel]

a and c are both gonna be 1 yes?

[river zephyr]

No

C is k

[rapid sentinel]

ah i see

i get k<9

[river zephyr]

Hmm

It should be k>9

Do you have the mark scheme with you by any chance

[rapid sentinel]

yes

[river zephyr]

What’s the answer?

[rapid sentinel]

its k>9

[river zephyr]

Ok one sec

Ok so when u multiply negative on both side of inequality

The inequality change sides

[rapid sentinel]

ohhh

i get it now

how do i do the second part

[river zephyr]

Is the answer k=11 ?

[rapid sentinel]

yes

[river zephyr]

Ok since the line intersects with curve

At one point bc tangent

U can equate the two equation

The line and curve equation, combine it into one

[rapid sentinel]

mhm

[river zephyr]

after than u should get a quad equation

Then u can use discriminant on it

Since it’s a tangent and touches at one point only

B^2-4ac=0

U should get k=11 after that

[rapid sentinel]

yea got it

tysm

[river zephyr]

Np

[rapid sentinel]

in the 1st part of the 3rd question i do the same?

equal the both equations and solve them?

no. 3

[obtuse pebble]

@rapid sentinel Has your question been resolved?

[rapid sentinel]

<@&286206848099549185>

[obtuse pebble]

@rapid sentinel Has your question been resolved?

[obtuse pebble]

@rapid sentinel Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[spiral yoke]

so i got -30 in this question but it was incorect 5(a –b) if a = −4 and b = −2

[rocky goblet]

be careful with negatives

[spiral yoke]

?

[rocky goblet]

you're subtracting negative two

[spiral yoke]

so -6

[rocky goblet]

no

this is why you have to be careful

think about it: if you take 2 and add negative 2, what do you get

[spiral yoke]

?

[rocky goblet]

what's 2 + -2

[spiral yoke]

0

[rocky goblet]

yep, it's 0

2 + -2 is the same as 2 - 2

[balmy mortar]

$a-b = a+(-b)$

[spiral yoke]

ok

so 4+(-2)??

[rocky goblet]

so if 2 + -2 is 2 - 2 which is 0
what should 0 - -2, subtracting negative two from zero, be?

[spiral yoke]

2

[rocky goblet]

yep

subtracting -2 is the opposite of adding -2, and adding -2 is subtracting 2, so subtracting -2 is the same thing as adding 2

[balmy mortar]

-(-a) = -1 * -1 * a = 1 * a = a

[rocky goblet]

so now if we go back to -4 - -2

subtracting -2 is the same as adding 2, so this is -4 + 2

[spiral yoke]

-2

[rocky goblet]

yep

[spiral yoke]

soo -10

[rocky goblet]

-4 + 2 = 2 + -4 = 2 - 4 = -2

and then yeah multiply that by five

[spiral yoke]

ok thx for the help

.close

[obtuse pebble]

Channel closed

Closed by @spiral yoke

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rich holly]

was it a mistake adding all 4 equations?

[empty cypress]

what's the question?

[rich holly]

find the value of P(5)

[balmy mortar]

consider a polynomial Q

which has something to do with P

is my hint...

Q = P + ???

[rich holly]

uhh

no idea

[balmy mortar]

Let Q(x) = P(x) + R(x)

where R(x) is a polynomial to find

there is a sensible choice

[rich holly]

so P(5)

R(x)=P(5)

[balmy mortar]

how is that a polynomial

well i guess it is

P(5) is just a constant though

[rich holly]

hmm yeah

[balmy mortar]

Q(x) = P(x) - P(5)

how will this help (it wont)

You try it out and see if u get anywhere.

[balmy mortar]

The polynomial R to choose is not complicated, I can assure you

[rich holly]

wait isn't $P(5)=(5)^4+a(5)^3+b(5)^2+c(5)+d$?

[warm shale]

hyperlix26

[balmy mortar]

It is.

a b c d are constants

[rich holly]

yeah

and how would Q(x) = P(x) + R(x) help me?

what's R(x)?

[balmy mortar]

Read what I said

[rich holly]

I know what P(x) is

[balmy mortar]

I said you have to find a suitable R(x)

that is helpful

That is my hint

[rich holly]

idk what the suitable R(x) is, wanted to make it equal to $(5)^4+a(5)^3+b(5)^2+c(5)+d$ but you said it wasn't equal to P(5) so that's not it

[warm shale]

hyperlix26

[balmy mortar]

skdjfnfkiffifofofofkfkfifo

Dont give up before you've even tried

[rich holly]

what

[balmy mortar]

oh nvm misread

I told you though

to find a suitable polynomial R(x)

And I also said it isnt complicated, it is simple

Write down a few guesses, see if they could help you any way.

If youre stuck, then say what R have you tried

[rich holly]

yeah nope

nothing else makes sense to me except for $(5)^4+a(5)^3+b(5)^2+c(5)+d$

[warm shale]

hyperlix26

[rich holly]

other than that I'm just making stuff up

if R(x) is a polynomial to find, why not make R(x)=Q(x)-P(x)?

[balmy mortar]

that isnt even a choice

of R

Do you understand what a polynomial is?

[rich holly]

yeah

[balmy mortar]

I hinted a polynomial, so please realise it probably isnt a constant

So what example of R could you try

just name one

any

[rich holly]

the same as P(x) but it doesn't make sense

[balmy mortar]

...

Tell me an example

of ANY polynomial

[rich holly]

P(x) is a polynomial

[balmy mortar]

omg

can you keep it simple?

[rich holly]

ax+b?

[balmy mortar]

if i randomly asked you for an example of a polynomial you wouldnt give me that

why the letters???

[rich holly]

it still counts right?

[balmy mortar]

An explicit example please?

[rich holly]

2x+7?

[balmy mortar]

Yes. Good.

Now I asked you to consider the new polynomial Q(1) = P(x) + 2x+7

What can we say using the given information?

It must still be of degree 4

But in particular what is Q(1) and so on

[rich holly]

1+2+7

that's Q(1)

which is 10

[balmy mortar]

indeed

and you can compute Q 1 2 3 4

yes?

[rich holly]

yeah

[balmy mortar]

Now my hint was to pick a nice R(x)

what 'nice' means, try to figure it out

I will say now that Q(1) = 10 isn't helpful, so we need a better R

[rich holly]

yeah I'm figuring out how to find a better R

[balmy mortar]

Q(1) = ?
Q(2) = ?
Q(3) = ?
Q(4) = ?

First maybe think what we want these to be

for some useful information

Then you can find a good R

[rich holly]

ok if I said we want those to be 0, how would it help?

[balmy mortar]

If you know Q(1) = 0

What does this tell you about Q?

Is there a theorem you can use perhaps?

[rich holly]

no idea

[balmy mortar]

(x - 1)

[rich holly]

oh

1 is a zero of the function?

[balmy mortar]

well yes

so the theorem you can use is?

[rich holly]

no idea

[balmy mortar]

Factor theorem

Remainder theorem

surely you have done either?

[rich holly]

maybe

[balmy mortar]

Review it then

otherwise you'll be stuck

[rich holly]

I don't know what they are called here

[balmy mortar]

For a polynomial f
f(a) = 0 <=> (x-a) is a factor of f(x)

For a polynomial f
f(a) = k <=> f(x) = (x-a)g(x) + k for some g

Factor and remainder theorems

[rich holly]

yeah I've seen them but I don't understand them enough

if you can divide a polynomial by (x-a) and the result is 0, then (x-a) is a factor right?

[balmy mortar]

no...

[rich holly]

what

why

[balmy mortar]

if i divide a polynomial by another polynomial I won't get 0

unless the original polynomial is 0

???

[rich holly]

I mean the remainder

[balmy mortar]

yh ok but thats just restating what i said

Try applying the theorem to this problem...

[rich holly]

yeah basically that's all I know, I don't understand much from the remainder theorem

[rich holly]

@balmy mortar anyways using this method, was it a mistake adding all 4 equations?

[balmy mortar]

did you see me suggesting such a method?

[rich holly]

no

[balmy mortar]

Why would you end up doing such a thing in the direction I've been hinting at

[balmy mortar]

also there arent even 4 polynomials

in your working

theres just 1

you mean 4 equations i suppose...

[rich holly]

oh equations

yeah

[obtuse pebble]

@rich holly Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[brittle swan]

how do we solve
$a_n = 3 a_{n - 1} + 10?$

[warm shale]

Chromium

[spiral maple]

what's the initial condition?

just a_0?

[brittle swan]

oh uh

nth

.close

[obtuse pebble]

Channel closed

Closed by @brittle swan

Use .reopen if this was a mistake.

[balmy mortar]

rifififififfg you hibihgigog ficicciciocovo

[spiral maple]

Ok then.

[obtuse pebble]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[compact shadow]

a(n)+5=3(a(n-1)+5)

[compact shadow]

.close

[obtuse pebble]

Channel closed

Closed by @compact shadow

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[late pond]

Hey everyone,
I need to calculate the radius and height of a cylinder, where the materiale usage is $20m^2$
The material usage can be calculated using this formula:
$O(r)=10\cdot r^{-1}+\pi+r^{2}$

I would assume that I could put 20 instead of $O(r)$, then isolate the radius $r$ but that gives me 3 solutions:
$20=10\cdot r^{-1}+\pi+r^{2} = $
r=2.221046 or r=−2.743439 or r=0.522393

[warm shale]

Juia

[obtuse pebble]

@late pond Has your question been resolved?

[potent dagger]

hello

by material usage does it mean the material used to cover the whole cylinder

[obtuse pebble]

@late pond Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[late pond]

Yes, it does 🙂

[late pond]

@potent dagger

<@&286206848099549185>

[potent dagger]

if that's the case then u could try using the formula for total surface area of cylinder

and equate it to 20

then you'd get values for the height as well

[late pond]

Is it this one?

$V=\pi \cdot r^{2}\cdot h$

[warm shale]

Juia

[late pond]

@potent dagger

[potent dagger]

no no

not the volume, i meant the surface area

volume has units of m^3

not m^2

[late pond]

Yeah 🙂

[potent dagger]

the formula should be 2pir(r+h)

[late pond]

Yeah, but that gives me 2 unknowns (radius and height)

[potent dagger]

you do have two radius values tho

from what you solved earlier

tryin using each value and see what value of height u get

by using the equation we just discussed rn

[late pond]

Okay.

I have these two radius: r=2.221046 or r=0.522393

$20 = 2\cdot \pi \cdot 2.221046\cdot (2.221046+ h)$

[warm shale]

Juia

[late pond]

and

$20 = 2\cdot \pi \cdot 0.522393\cdot (0.522393+ h)$

[carmine bear]

Was the equation in the question given by the question or is it something you deduced

[warm shale]

Juia

[late pond]

This one was given in the question:

[carmine bear]

well you could create 2 simultaneous equations

one using O(r) = 20 and the other using surface area = 20

unless O(r) comes from the surface area equation in which case that probably won't work

essentially what im trying to say (or hope will work) is if you try to solve $$20 = 10r^{-1} + \pi r^2$$ $$20 = 2\pi r(r+h)$$

[warm shale]

I can't believe you've done this

[carmine bear]

simultaneously

not sure if it will work but worth trying

like perhaps try finding an equation for $r$ using the first equation and then sub replace both $r$ in the second equation with the one you got from the first one

[warm shale]

I can't believe you've done this

[obtuse pebble]

@late pond Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rocky dove]

Hello everyone! I am just struggling on how to find critical points with e problems 🥲 not sure if I am doing it right. Is it truly possible to not have critical points? If so what to do?

[obtuse pebble]

@rocky dove Has your question been resolved?

[rocky dove]

<@&286206848099549185>

[obtuse pebble]

@rocky dove Has your question been resolved?

[compact shadow]

Use x=rcos(t), y=2rsin(t)

That becomes e^{-r^2sin(2t)}

Maximum when r=1 and sin(2t)=-1,minimal when r is 1 and sin(2t)=1

[rocky dove]

okay and I assume that helps me find the critical points

ooof okay i see what you are getting at

my question is how did u know that y=2rsin(t)? it just is? or its coming from a formula

just curious

[compact shadow]

?

You don’t know sin(2t)=2sin(t)cos(t)?

This is all I used

[rocky dove]

its been a while ;-;

[compact shadow]

sin(x+y)=sin(x)cos(y)+cos(x)sin(y) then let x=y=t

You have sin(2t)=2sin(t)cos(t)

[rocky dove]

ohhh

thank you so much!

[compact shadow]

Np

[rocky dove]

.close

[obtuse pebble]

Channel closed

Closed by @rocky dove

Use .reopen if this was a mistake.

[obtuse pebble]

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[ashen sierra]

The question: Solve tan²θ = 2 * 1/cosθ for 0° ≤ θ ≤ 360°.

The answer in the book was: 64.3°, 295.7°, 219.9° and 140.1°.

[ashen sierra]

Here's my working out. Idk what I did wrong

<@&286206848099549185>

oh well .close

.close

[obtuse pebble]

Channel closed

Closed by @ashen sierra

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cosmic moon]

I am stuck on part E. I am doing past paper questions and I am confused on how I am supposed to do this

[short spruce]

what have you tried

[cosmic moon]

Honestly, looking at the question hade made me not try but from the rules I know of, I can't think of a way to do it

[short spruce]

i assume you know exponent laws? if not, google them

hint: rewrite 1/(9^x) with a base of 3

[cosmic moon]

I got 1/3^2+x

[short spruce]

?

how so

[quartz mica]

how

[cosmic moon]

Well, I learned exponent laws recently so there might be a flaw in my method but what I basically did was I assumed that since I can write 9 as 3^2, and when you multiply indices together you add the power so that's what I did

[quartz mica]

the 1/9 means 9^-1

[short spruce]

refer back to your exponent laws

[quartz mica]

ye[p

good luck aqua man

[short spruce]

actually i think you just interpreted it incorrectly

you rewrote $9$
as $3^2$, yeah, but that gets you $(3^2)^x$

[warm shale]

a disappointing son

[short spruce]

the law you tried to apply states $a^b\cdot a^c=a^{b+c}$

[warm shale]

a disappointing son

[short spruce]

but you're raising an exponent to an exponent here, not multiplying

we've already established that. this isn't your problem

[quartz mica]

sorry again

that was inapropriate of me

@cosmic moon do you know how to convert 1/9x to an exponent

[cosmic moon]

I did some working out, and got $3^2x-a+2^x$ = $3^2y-x$

[warm shale]

AquaDrix

[cosmic moon]

This basically

[quartz mica]

now you can cancel the 3s

[short spruce]

that ain't right

[quartz mica]

so it becomes 2x - 1 + 2x = 2y

[quartz mica]

he canceld the x

i thought the same thing

[short spruce]

don't do people's work for them

also that's wrong

[quartz mica]

i didnt

[short spruce]

yes, you did

that's not up for debate, i am looking at the work you did for him

[quartz mica]

i guess

i did some

mb

[obtuse pebble]

@cosmic moon Has your question been resolved?

Channel closed

Closed by @cosmic moon

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sturdy ledge]

(Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2𝜋). (Enter your answers as a comma-separated list.)
tan(𝜃) + cot(𝜃) = 4 sin(2𝜃)

[strong vale]

Let's start by getting all the trig functions in terms of theta

How do you get sin(2theta) in terms of just theta

[sturdy ledge]

u change it to sin(theta)cos(theta)

[strong vale]

yh

but $\sin{2\theta} \not= \sin{\theta}\cos{\theta}$

[warm shale]

azeem321

[sturdy ledge]

i meant 2sin(theta)cos(theta)

[strong vale]

good

now we have

$\tan{\theta} + \cot{\theta} = 8\sin{\theta}\cos{\theta}$

[warm shale]

azeem321

[sturdy ledge]

yep

[strong vale]

now what

[sturdy ledge]

now u change tan and cot to sin(theta)/cos(theta) and cot to cos(theta)/sin(theta)

[strong vale]

good and after doing that how about multiplying both sides by $\sin{\theta}\cos{\theta}$

[warm shale]

azeem321

[strong vale]

Giving $\sin^2{\theta} + \cos^2{\theta} =8(\sin{\theta}\cos{\theta})^2$

now what

[sturdy ledge]

distribute the 8?

[warm shale]

azeem321

[strong vale]

sorry i made a typo

now what

[sturdy ledge]

square root both sides

?

[strong vale]

what is s^2+c^2

[sturdy ledge]

1

[strong vale]

Can you solve it now?

[sturdy ledge]

umm tbh no thats the part im stuck on

I turned that into 1/2=(sin(2theta))^2

[strong vale]

good

So now you must solve $\sin{2\theta} = \pm \frac{1}{\sqrt{2}}$

[warm shale]

azeem321

[strong vale]

u can do this now?

[sturdy ledge]

which I get pi/8 and -pi/8

[strong vale]

notice the domain

and remember sin(2x) is periodic

with period pi

which means u can add/subtract pi

and it will be a solution

so if x=a is a solution

x=a+pi is a solution

x=a+pi+pi is a solution

x=a-pi is a soluition

but the question only wants the solutions between 0,2pi

[sturdy ledge]

we just add pi to both of them until we get all the possible solutions in the domain?

[strong vale]

let's put this on a back burner

let me show u something important

this is the graph of sin(x)

Suppose we want to find all values of x, for which sin(x) = a

we know there is a solution at the red line

How do you get the second solution?

[sturdy ledge]

the next point where they intersect

[strong vale]

yh

but most calculators won't give you that solution

so what algebra can u do to get that solution

look at the symmetry of the curve

$\sin{x} = \sin({\pi-x})$

[warm shale]

azeem321

[sturdy ledge]

origin symmetry

?

so its odd

[strong vale]

Notice that the curve is the same either side of the red line

[sturdy ledge]

ohhhh

I take pi minus what I got?

[strong vale]

That would be for just sin(x)

[sturdy ledge]

to get the other answer

[strong vale]

But since since we have sin(2x)

You do $\frac{\pi}{2}- \frac{\pi}{8}$

[warm shale]

azeem321

[sturdy ledge]

because the period was half of what it was before?

[sturdy ledge]

3pi/8

[strong vale]

So for $\sin{2x} = \frac{1}{\sqrt{2}} \implies x = \frac{\pi}{8}, \frac{3\pi}{8}$