help-23

[knotty coyote]

I’m assuming were looking for only angle B correct?

[terse lichen]

CBE is 80 degs

[weary hull]

Let’s start here: If angle BA is 50°, then what is angle AED?

[hallow forge]

its 80 right?

[hallow forge]

thought so

found a corollary about this

maybe arc AD is half of measure AED?

[weary hull]

Well hold on, I’m not asking what CBE is, I’m asking what AED is. Remember that an inscribed angle is half the angle of the arc. So what would EDA’s angle be?

Notice how BDA and EDA have the same angles and that we know that the angle of arc BA is 50°

[hallow forge]

hmm

i dont get it sorry

[weary hull]

No worries. Whenever we have an inscribed angle in a circle, the angle of the arc is twice the inscribed angle. This is a theorem that you can use to help solve this particular problem and many geometric problems with inscribed angles

For example, if the arc AC of this inscribed angle was 80°, angle ABC would be 40°

So, if the arc of BA is 50°, then what is the inscribed angle BDA equal to?

[hallow forge]

25?

is it?

[weary hull]

Correct :)

So then what is the angle of DEA?

[hallow forge]

quite confused

40 maybe?

hmm wait is CED 160 degrees then?

[weary hull]

No

[hallow forge]

sad

[weary hull]

Think about the angles you have of triangle DEA so far. You have 2 angles so you can solve the third. What is the angle of DEA?

[hallow forge]

65??

sorry im very confused

[weary hull]

One sec

@hallow forge so we already solved for angle EDA (in red) by using the inscribed angle theorem. What’s the angle of DEA (the angle marked by the question mark)?

[safe radish]

@hallow forge Has your question been resolved?

[safe radish]

@hallow forge Has your question been resolved?

[hallow forge]

im not really sure man

what's the answer to it?

maybe i can figure it out

[weary hull]

Well, the interior angles of any triangle will always add up to 180°, so angle DEA must be 180-80-25

Which would be 75°

[hallow forge]

so if DEA were 75, that means the intercepted arc DA of it would be 150 degrees?

[weary hull]

Not quite, this is only true for inscribed angles which have to be on the circumference of the circle. Point E is not on the circumference of the circle

[hallow forge]

oh i see

if thats the case though how would you find the arc?

[weary hull]

We’ll get to that soon. For now I want to help you solve the inscribed angles :)

You already got the answers, butttt I think it’d serve you better to know how you got the answers

[hallow forge]

okay thank you for thattt

but now we got that side done isn't it time to move on to the next side?

preferably the CEB?

[weary hull]

Well, since chords AC and BD intersect at E, the opposite angles must be equal. So angle CEB is equal to angle DEA, right?

[hallow forge]

i think so yeah

if they're both the same, doesn't that mean its the arcs thats left?

that also means arc CD will be 160 if im not mistaken

because B and A are both 80 degrees

[safe radish]

@hallow forge Has your question been resolved?

[safe radish]

@hallow forge Has your question been resolved?

[flint cargo]

Why is this function not completely rational

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[peak estuary]

5 seems like a lot

what kind of magnitude are we talking here

rounding error maybe? how big are the numbers

weird. no clue

you could do it on your own ^^

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[mossy pilot]

how can i show a particle does not complete a full revolution

[mossy pilot]

in circular motion

[tacit idol]

it's angular displacement should be less than 2π

mostly

[mossy pilot]

could i also say that tension would be less than 0 at the top

because its a particle connected to a string

fixed at point O which is the origin

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[lean otter]

what have u tried

[keen drift]

gimme a sec writing numbers on discord pains my eyes

[lean otter]

lmao nw

[keen drift]

i understand u first expand bracket which gives u; 8x-u=hx+3h then u -hx which gives u 8v-u-hx then u +v which gives u 8x-u+3h+v

ok wait it was good on paper

[ebon cove]

first try leaving 8x on the one side only

maybe :0

[keen drift]

huh

oh

[lean otter]

where did v come from...

[keen drift]

like seperate it

v?

there is v there

oh.

its u

um

[ebon cove]

u gotta first leave whatever u wanna make subject on the one side and all the other thing on the other side

how can u do it?

first was right that u expanded

[keen drift]

8x = hx+3h+u

?

then -hx

8x-hx=3h+u

then divide h?

[ebon cove]

8x - u = hx + 3h than u get 8x = hx+3h +u yes

no look

8x-hx what to they have the same

x

so u factorize it

[keen drift]

oh

[ebon cove]

x(8-h)

and than you get x(8-h)=3h+u

what u gonna do next

[keen drift]

wait

can i write it from start rq

[ebon cove]

yeah

[keen drift]

so first factorise,

[ebon cove]

to make something subject, you want to get everything that contains that subject on the one side

and leave everything other on the other side

remember this

[ebon cove]

?

[keen drift]

sorry had to goto my meowing cat

[lean otter]

so you are solving for x?

[keen drift]

ye

[lean otter]

ok so step 1, you need to expand the brackets

[ebon cove]

yeah but remember what i told u, u need to leave everything that contains the subject (x in this case) to the one side

[keen drift]

ya

[keen drift]

ya

[ebon cove]

and hx contains x so get it on the left side as well

than youll have everything that contains x on the left side

[keen drift]

so its;

[lean otter]

ok ill leave this to ALOAp

[keen drift]

8x-hx=3h+u

[keen drift]

👌

[ebon cove]

yeah

[keen drift]

but if i divide h from each side

then h is gone completely

[ebon cove]

but the reason that youre getting everything that contains the subject to the one side is that you can factorize

[keen drift]

oh

[ebon cove]

since everything contains x you can just do 8x-hx = x(8-h)

[keen drift]

but i can /h?

[ebon cove]

why??

[keen drift]

nono im not going to

but is it possible

[ebon cove]

yeah why not

[keen drift]

but then h would completely be gone from the equation right?

[ebon cove]

no

if you divide both sides by h youll be left with h still

like in the denominator

[keen drift]

wait

what is 3h/h

[ebon cove]

3

[keen drift]

ye so

[ebon cove]

so?

[keen drift]

it'd be gone?

like fropm 3h+u /h

=u

[ebon cove]

=?

= 3 + u/h

youre still left with it

[keen drift]

oh.

ohhhh

okok

i did the other one

is this orrect

[ebon cove]

yeah

just follow what i told u and youre gonna do this typo things pretty easily

cya

[keen drift]

sorry thought someone was actually in my garden 💀 i think its a fox but anyway- thank you

[ebon cove]

no problem, close the channel

[keen drift]

.close

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[blazing solar]

ok

[glass carbon]

[blazing solar]

i tried to do 360-215 but nothing worked

[quasi bison]

by "nothing worked" you mean "that one thing i tried didn't work"?

[blazing solar]

ye

i tried to even do 215-ans

[quasi bison]

i think you should not try to do blind calculations.

this is the angle you are looking for

[blazing solar]

corrosponding angles add up to 180

k

ill give it a try again

thx

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[kindred rapids]

Hi im a bit confused as to how to calculate determinants as i seem to be getting some contradicting information. my textbook is saying to calculate like this, while keeping a row constant and going down a column but when i look it up people are saying the opposite thing, (seen in the screenshot with a black background) this is also what i thought it was but im confused as to why my textbook is saying the opposite. are these two equivilent, am i missing something? thanks

[kindred rapids]

(discord rendered the images weirdly but if you click on the black one you'll see the full thing)

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[paper knot]

Hello :

[safe radish]

@paper knot Has your question been resolved?

[paper knot]

<@&286206848099549185>

ah 😦 the ping helper don't work anymore

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[glad agate]

How do you turn the LHS to the RHS. What steps are involved in this?
I do not know where to start.

[glad agate]

<@&286206848099549185>

[rare tendon]

this is just division

[stuck crest]

dont ping helpers straight away

but u can put the 1 in front

[glad agate]

oh okay sorry i didnt know

[stuck crest]

then rewrite 1 as (2x+1)/(2x+1)

then work from there

[glad agate]

what

[stuck crest]

you want ur denominator to be the same so u can subtract

[glad agate]

cud u write it if its not too much of a hassle

i want to convert the LHS to the RHS

[stuck crest]

wait sorry LHS to RHS or RHS to LHS

oh mb

[glad agate]

like say i was just given the LHS

how do i get to the RHS

[rare tendon]

make the top (2x-1+2-2) which is equal to (2x+1-2) then split the fraction to 2 parts and sympify

[stuck crest]

i mean jus rewrite the numerator as (2x + 1) - 2

[rare tendon]

^

[glad agate]

2-2 because change to the eq shud be cancelled out?

[rare tendon]

+2-2 cause it can cancell out yes. and it makes the fraction easier to solve

[glad agate]

okok i get it ty

.close

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[rare tendon]

np

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[dire escarp]

stuck

[dire escarp]

tried making u = cos2x but im gonna get sin2x anyway so i dont think thats the right step

please ping me if youre here

[dire escarp]

<@&286206848099549185>

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[manic heart]

how do I do this

[lean otter]

what's the equation of a circle ?

do you know the general form basically

[manic heart]

yeah

[lean otter]

what is it then

[manic heart]

(x-h)^2+(y-k)^2

[lean otter]

equal to?

[manic heart]

wdym

[lean otter]

what u wrote isn't an equation

[manic heart]

oh

[lean otter]

it's incomplete

[manic heart]

OHH WAIT

= r^2

mb

[lean otter]

yess

now what is (h,k)

[manic heart]

is this like x1 y1 x2 y2

or is this different

[lean otter]

(h,k) is the center of the circle

[manic heart]

0

[lean otter]

wot

[manic heart]

(0,0)

[lean otter]

yeah okay

the radius?

[manic heart]

it's either -4 or +4

[lean otter]

radius can be -4?

damn negative length just dropped /j

[manic heart]

so it can't be?

[lean otter]

no

[manic heart]

4^2?

[lean otter]

yes

[manic heart]

it can never be?

[manic heart]

whats the difference in b and c

same as a?

[lean otter]

okay as a VECTOR it can be

as a magnitude it cannog

you're dealing with magnitudes rn

[manic heart]

whats a vector

[lean otter]

don't worry about it it's not something you're studying right nkw

for now, always think of radius as positive

[manic heart]

okayy

what do I do b and c

same thing

[lean otter]

yes

[manic heart]

what does the (-3,5) outside mean

[thin bridge]

indicates a point on the circle which can be used in your calculations for the radius

[manic heart]

wdym

(x-h)^2+(y-k)^2

does it replace something from here?

<@&286206848099549185>

[molten apex]

Can’t help right now

But

(h,k) is the radius

In each problem

We see that the radius is 0,0

Also use the point (4,0) for question 1

The given point (-3,5) for question 2

And the distance formula for the point given in question 3

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[manic heart]

.close

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[quartz ridge]

help

[quartz ridge]

im having trouble on where to startt

[shy wing]

What type of problem is this

[minor harness]

Maybe find X?

[quartz ridge]

how

[minor harness]

Manipulate the equation

They are equal soo what you do to the other side is what you should do to the other side

[quartz ridge]

well 1 can sq rt

so it would be

x + 2 sqrt (x)

= 7

then

subtract x to isolate radical

then divide by 2 and square

?

hello

[minor harness]

Waaiit I'll tryyy xD

[quartz ridge]

.close

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[quartz ridge]

help

[quartz ridge]

how do u solve

find x

[stuck crest]

i mean u could try moving the x to the right side then squaring

to remove the root

then go from there

[dreamy heath]

I need alot of help with 8th grade math...

[cunning thistle]

!onechannel

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Please stick to your channel.

[cunning thistle]

!help

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[quartz ridge]

i got x^2 -18x + 45

= 0

can u factor

it

[stuck crest]

yes u can

split -18 into two terms that multiply to give 45

remember that because your b is a negative number but multiply to give 45 which is positive

that means both terms have to be negative

since negative * negative = positive

[quartz ridge]

15 and 3 right

[stuck crest]

-15 and -3 are the factors

so yes x = 3, 15 when solved

[quartz ridge]

so x = 15 and 3

then check

[stuck crest]

yep

check solutions

[quartz ridge]

for extranious

alr

thanks

[stuck crest]

yep perfect

no worries

[quartz ridge]

can u help with another

one

[stuck crest]

sure whats the question

[primal glen]

move the negative root 2x to the other sid

by adding it

to the left side

[stuck crest]

i mean i would immediately square both sides

but

[quartz ridge]

ya cus its easier to square rig

[stuck crest]

yh cuz u immediately get rid of the root on the left

youll be left w a root on the right side but you can square that later

[quartz ridge]

wait once u square the right side u have to get 4 terms right

[stuck crest]

well its a perfect square

so youll have four terms but you can combine two of them

this channel occupied go to #❓how-to-get-help

[quartz ridge]

i got on the right

9 - 6 sqrt (2x) - 2x

[stuck crest]

yep

and on the left?

[quartz ridge]

just 2x + 4

where do u move the xs

now

[stuck crest]

mhm so you have

$2x + 4 = 9 - 6\sqrt{2x} + 2x$

[flat frigate]

taro

[quartz ridge]

ya

[stuck crest]

what do you have on the LHS and RHS

that are the same

[quartz ridge]

2x

[stuck crest]

right

so we can cancel that

[quartz ridge]

okok

[stuck crest]

$4 = 9 - 6\sqrt{2x}$

[flat frigate]

taro

[stuck crest]

you can bring the -6sqrt(2x) to the left side

to make it positive

[quartz ridge]

ya

[stuck crest]

so what u got now

isolate the x

[quartz ridge]

alr

lemme try

[stuck crest]

aight lemme know how it goes

[quartz ridge]

oohh ok

i got it

tysm

[stuck crest]

aight cheers

no worries

[quartz ridge]

.close

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[orchid canyon]

Yo this is incredibly random and might not even be related but what's the best way to estimate the number of candies in this jar

[slate thorn]

what information are you given?

do you have access to the internet?

[orchid canyon]

just these two pictures

i guess i could find the measurements of a candy

and compare it to the picture

[slate thorn]

you could also count how many are at the top

and count how many layers there are

you will overestimate when you do that though

so account for it be showing a range of candies that could be in there

[orchid canyon]

alright

thanks

.close

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[lavish berry]

Let F(x) be an antiderivative of (lnx)^3 / x . If F(1) = 0, then F(9)=?

[lavish berry]

I'm just simply confused on how to do the antiderivative of that

[thin bridge]

try usub

[lavish berry]

oh wait

im dumb

thank you

i forgot u sub exists

.close

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[dawn heron]

i dont know how to do this : (

[lean otter]

This formula should help

[dawn heron]

The equations are different i dont think it works forcit

[lean otter]

subtract y

to get 0

[dawn heron]

ohhhh ok

[lean otter]

and use those to get ur (a,b,c)

[dawn heron]

i am sorry i dont wanna bother but can you go through it step by step for me? i am pretty new to this

[lean otter]

Yeah sure

So look at the formula

Ax+by+c = 0

So first we need to rearrange the formula you have to make it ax+by+c=0

We are given y=2x+32

subtract y

we get

2x-y+32=0

im new

Now using the formula

whats happening

[lean otter]

Im helping someone right now, go to #discussion

sorry i leave

So using Ax+By+C=0

We can apply the terms

A = 2 B = -1 C = 32

@dawn heron are you understanding it so far?

[dawn heron]

yes

[lean otter]

And we know

x1 = 1 and y1 = 0

Since it gave us the point (1,0)

Now we use the other formula

d = (ax+by+c)/sqrt(a^2+b^2)

remember for the x plug in the x value given and for y the y value given

and you should get your answer

[dawn heron]

ok ok got it

thank you

[lean otter]

yw

[dawn heron]

would the answer be 6.8?

34/5

in the question 6.8 isnt one of the answers

<@&286206848099549185>

[safe radish]

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@dawn heron Has your question been resolved?

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[marsh walrus]

what has you stuck

[wanton hull]

um

well first

solving for the width

[marsh walrus]

whats the length of the interval?

[wanton hull]

its -1 to 2

i just got 3/8

i just want to make sure that right

[marsh walrus]

3/8 is what?

[wanton hull]

the width

[marsh walrus]

yea, this sounds right to me

[wanton hull]

okay

now

the hard

part

so basically its asking me to split the interval of [-1,2] into 8?

[marsh walrus]

yea, either left or right

[wanton hull]

well right

but

i just every example ive seen is the use of a graph

but like its hard to like write that graph out myself

and have the right numbers

[marsh walrus]

there are less moving pieces than youd think

so, its gonna be a sum of areas of rectangles, right?

[wanton hull]

yes

[marsh walrus]

so you need width, and height

width is easy

to get height, you just need the endpoints

do you get what i mean by that?

[wanton hull]

yes

[marsh walrus]

and, the endpoints aren't really that hard to get

you know how they're spaced

and how many you'll have

so if you can find the first one, you can get the rest pretty easily

so say for example, the left endpoints

wheres the first one? what x value?

on -1 to 2

[wanton hull]

-1

[marsh walrus]

right

and then the next one is -1 + 3/8

and then the next next one is -1 + 3/8 + 3/8

[wanton hull]

and then we dont use 2 right

[marsh walrus]

and then -1 + 3/8 + 3/8 + 3/8

yea, youll end up missing 2

[wanton hull]

okay

[marsh walrus]

but

you can actually write this now

$x_i = -1 + i \frac{3}{8}$ for $i = 0, 1, \dots, 7$

[flat frigate]

jan Niku

[marsh walrus]

8 left end points

and your approximation is the sum of the areas of the rectangles, so

$\sum _{i=0} ^7 \frac{3}{8} \cdot f(x_i)$

[flat frigate]

jan Niku

[marsh walrus]

i forgot what f was

[wanton hull]

lol

[marsh walrus]

but, hard part done

its not really that bad

i hope

[wanton hull]

well today like just started doing it

but it doesnt seem that hard

[marsh walrus]

no, it seems like theres a lot of moving pieces but once you place one piece the rest kind of fall into place

if you understand it conceptually its very fast

[marsh walrus]

you can adapt this very quickly to the right end points, too

[wanton hull]

so like

now what

[marsh walrus]

what was f, x^3?

[wanton hull]

do i use this?

oh its uh

[marsh walrus]

sorry, 1/(x+3)

[wanton hull]

yeah

[marsh walrus]

yea, its easier if you have a calculator that can take sigma inputs

[wanton hull]

how do i use that

[marsh walrus]

what calculator do you have

or i guess a better question

what are you allowed to use

[wanton hull]

ti-84 plus and im pretty sure

we can use it

for exam

[marsh walrus]

oh

alright

dang i dont have one on hand

i cant remember how to type sum in

one sec

[wanton hull]

okay

[marsh walrus]

if you hit math then up do you see summation?

[wanton hull]

for 0 yeah

[marsh walrus]

so you can type this very easily

just plug in the definition we used for x_i

and the problem gives you f

[wanton hull]

also what does i mean ehre

here*

[marsh walrus]

i is the variable of summation

like uhh

$\sum _{i=0}^5 = 0 + 1 + \dots + 5$

[flat frigate]

jan Niku

[marsh walrus]

https://www.youtube.com/watch?v=AA-GMzNGRb0

heres the video im referencing

[wanton hull]

thank you

[marsh walrus]

you may need to change i to x

[wanton hull]

yeah i was gonna say i is not workin

[marsh walrus]

you can always go old school and make a table, too

or use desmos

probably not bad to learn to use the calc tho

[wanton hull]

this is gonna sound stupid

bit

but

how do i put what u showed me in the calc

like

i cant put xi ofc

but idk im just confused

[marsh walrus]

so, yea, youll have to use the definition

i guess you can start with f so it comes out as

$\sum _{i=0}^7 \frac 3 8 \cdot \frac{1}{x_i + 3}$

[flat frigate]

jan Niku

[marsh walrus]

and $x_i = -1 + i \frac{3}{8}$ for $i = 0, 1, \dots, 7$

[flat frigate]

jan Niku

[marsh walrus]

so what do you get when you substitute this definition in?

[wanton hull]

uhh

i got .97499

like

when i put everything in

[marsh walrus]

whats the actual value

of the integral

[wanton hull]

what u mean

[marsh walrus]

i mean, actually calculate the integral

see if its close

,w integral from -1 to 2 of 1/(x+3)

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by janniku

[marsh walrus]

so, yea, not too far

since your left sum over estimated

[wanton hull]

what the freak is that

[marsh walrus]

youd probably expect that your right is slightly under

[marsh walrus]

the integral?

[wanton hull]

yeah

[marsh walrus]

oh, do you not know calculus

[wanton hull]

well

im taking like brief calc right now

and this is like my first lesson

on this stuff

my last math class was pre calc

[marsh walrus]

ah okay

well an integral just gives area under the curve

[wanton hull]

okay

[marsh walrus]

if you cant do that then

idk how youd know if youre close or not

maybe use triangles

[wanton hull]

LOL

i mean

[marsh walrus]

itd be hard

[wanton hull]

yeah that my fault i think there are more videos to watch

its even harder cause im online

[marsh walrus]

its normal i guess to do this kind of thing before integration

[wanton hull]

thats how he did it

[marsh walrus]

yea seems about right

[wanton hull]

is the way we did it easier

[marsh walrus]

theyre the same

[wanton hull]

oh

well

so the answer for the right endpoints

is .9750?

[marsh walrus]

thats what i get

[wanton hull]

um

[marsh walrus]

youve entered the left sum into the right box

[wanton hull]

oh

[marsh walrus]

R_n, for Right

[wanton hull]

so we did the left?

[marsh walrus]

we did the left because the first endpoint is at the far left end of the interval

and the right end point is not included

[wanton hull]

okay

oopsie

so now how do i do the right

[marsh walrus]

all you have to change is x_i

[wanton hull]

this is messing with my brain

so 8 would not be at the top

[marsh walrus]

it depends how you write it

you know where the rectangles end

the last endpoint is at 2

[wanton hull]

why does i = 0

could i = 1 and the top be 8?

im just confused cause we put 7 on top but 8 is the n

[marsh walrus]

yea

then youd start at -1 + 3/8

which you should notice is the left endpoint of the second rectangle

so its the right endpoint of the first rectangle

i gotta go to bed but good luck okay

[wanton hull]

okay thank you for all ur help

[marsh walrus]

this is the whole thing

to get the right endpoints, you just start one index later

and go one index further

nothing else changes

[wanton hull]

okay

[safe radish]

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[amber kraken]

I need 2. Solved

actually i have the answer

but could someone explain the 3rd row? i dont understand it.

<@&286206848099549185>

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@amber kraken Has your question been resolved?

[lean otter]

the average of both of chris's term test and final is 68

so the avg (Test 1 + Test 2 + Final exam )/3 =68

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[lean otter]

!help

[safe radish]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[lean otter]

Wait wrong one

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[karmic ore]

1

i cant get trhough it

i dont even know where a(2) would come in at

[ornate dome]

according to the function a(r), a(2) means that r = 2 (but also I've never learned of the polynomial remainder theorem so I could be interpreting it wrong, sorry)

[quiet juniper]

Divide poly by x - 2

Remainder is a(2)

[karmic ore]

alright 😎

can you check if something a binomial rq?

f (x) (x+2) (x+1)(x-2)

[safe radish]

@karmic ore Has your question been resolved?

[karmic ore]

.close 😔

[safe radish]

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[coarse kelp]

how is the answer this:

I can't see how they did it

[quasi bison]

well

let's see

[coarse kelp]

I guess we can start from inclusion exclusion of 3 sets

[quasi bison]

there are three types of regions we are considering

outer (part of exactly one circle), middle (part of exactly two circles) and inner (the threefold intersection)

the outer regions are counted once each, and only in the terms P(A)+P(B)+P(C)

with me so far?

[coarse kelp]

yeah

[quasi bison]

ok

the middle regions are each counted twice by P(A)+P(B)+P(C), and once with a coefficient of -2 by the terms -2P(A&B)-2P(A&C)-2P(B&C), for a total coefficient of 0

do you follow?

[coarse kelp]

okay

[quasi bison]

now the inner region

the first three terms count it thrice

and the next three terms count it -6 times

for a total coefficient of -3

but we need its coefficient to be 1

[coarse kelp]

ohh

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[misty holly]

idk how to say this, but is there a way to work out the surface area of a cube you can see by looking at it from above knowing the rotation of the cube

[misty holly]

or rather % of the entire surface area

like if this is the cube

<@&286206848099549185>

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[misty holly]

ohh

ooo I didn't realise that, I see some people do it instantly

[minor goblet]

you mean if you know the side area can you figureo ut the % surface area?

1. what are you trying to achieve

2. yes just do the area/surface area lol

[misty holly]

ok I'm prob going to explain this horrible

like knowing what angle this cube is rotated at

[minor goblet]

why?

what are you trying to achieve?

[misty holly]

and how far away you are from the cube

what is the area of the shape

[minor goblet]

but why?

[misty holly]

like if you look at this as a 2d image

[misty holly]

because it's intresting to know

[minor goblet]

ah i thought you were trying to make something

[misty holly]

prob would at some point

[minor goblet]

sounds possible yeah

[misty holly]

for starters then ig it would be useful to know the max possible area

or I'll just close this

.close

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[cobalt lynx]

In an isosceles triangle, AB=AC=4, BC=6, point D is on the base BC, ∠DAC=∠ACD, fold the triangle ACD along AC so that point C falls to point E, connecting BE , find the length of BE/The picture is here but I dunno how to prove ∠1=∠2！

[cobalt lynx]

<@&286206848099549185>

[tiny wraith]

I think you might be able to get away by angle chasing

Starting from the left angle that's marked

[cobalt lynx]

∠1？

[tiny wraith]

I can't read which angle is which, but I meant the one with AD in it

[cobalt lynx]

Ok I know

[tiny wraith]

It might not be the quickest way

[cobalt lynx]

HAD right?

[tiny wraith]

Yes, those points are nice

[cobalt lynx]

Then?

[tiny wraith]

HAD = x

You can figure out ADC and thus AED

AED and ACD are congruent by construction

[cobalt lynx]

Err…so?

[tiny wraith]

Do you know how to find ADC and AED?

It doesn't follow immediately from angle HAD

[cobalt lynx]

What I know is ABO equals DIO

So IDB equals BAE

[tiny wraith]

So you've draw the line segment ID as well?

[cobalt lynx]

And 1 is the half of BAE

[tiny wraith]

that might be overcomplicating things

[cobalt lynx]

ID?

[tiny wraith]

Yeah, you talked about angle IDB, or did you mean EDB?

[cobalt lynx]

Oh yes

Sorry

[tiny wraith]

You've proven this?

where angle 1 = HAD

[cobalt lynx]

Yes

And have no idea of HAD

[tiny wraith]

Then it's not too difficult, since angle 2 (IAE) is half BAE

Oh, I'm confusing angle 1 and 2

[cobalt lynx]

🤣

I think I must apologize for my poor English

2 is HAD

IAE is 1

[tiny wraith]

angle HAD is defined to be angle 2 right?

okay

[cobalt lynx]

Yep

[tiny wraith]

let angle 2 = x

What would ADC be in terms of x?

[cobalt lynx]

X+90

[tiny wraith]

Yeah, so you can figure out AED as well

with the help of congruency

[cobalt lynx]

HDA = 90-x

So EDB =2x?Am I right?

[tiny wraith]

So x + 90 = ADC = ADE = EDB + HDA = EDB + 90 - x

Yeah

[cobalt lynx]

Ok I can make it!thanks!

.close

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[gritty cradle]

hey, this ques want real x,y

[safe radish]

@gritty cradle Has your question been resolved?

[gritty cradle]

<@&286206848099549185>

[radiant bough]

that's i right?

[gritty cradle]

yes

[steep bramble]

so for the top of the first part you can factor that with difference for squares

[gritty cradle]

(x + 2y)(x + 2y)?

[steep bramble]

wait no lemme think about this for a sec

[radiant bough]

difference of squares, not sums of squares

[steep bramble]

I don't think thats the way to do it tho

[gritty cradle]

BTW

the answer is y=2
and x=3

[gritty cradle]

oh

[steep bramble]

one way might be to isolate the real parts and the imaginary parts

[gritty cradle]

how is that?

How can we separate the real parts from the imaginary?

[steep bramble]

im not sure really

[gritty cradle]

hmmm

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[somber thorn]

Does anyone know if there's a theorem that states that if f is continuous on (a, b), and the limit as f approaches a from right side exists, and limit as f approaches b from left side exists, then (a, b) has a maximum? I feel like the EVT can apply to an open interval in the case where the limits exist.

[crimson field]

Define f(x) outside (a, b) to be f(a) for x < a and f(b) for x > b so that f(x) is continuous for all x. You can do this by defining f(a) and f(b) to be those limits you assume to exist. Now either f(a) and/or f(b) are where it obtains the minimum/maximum or it's in the interior. You can prove that using the Heine-Borel theorem to prove it has a maximum/minimum on a <= x <= b (which requires a closed interval). Then use boundedness to show it has a least upper bound (or greatest lower bound) and that it attains it.

In short, you can define the points f(a) and f(b) so it's continuous on a closed interval since the limits exist and use the regular theorem.

You just have two cases though then:
If the maximum was to occur at these extreme points it might never attain it. For example, f(x) = x, 0 < x < 1 never reaches a minimum or maximum.

[merry sleet]

a and b finite values?

[crimson field]

You could make a theorem then that says if f(x) > f(a) and f(b), for any x, a < x < b then it attains a maximum on a < x < b.

[somber thorn]

Thanks

.close

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[hallow anchor]

I'm supposed to solve for that
The solutions are:

shouldn't the second part include k*pi/2?

[glass carbon]

it's equivalent to

[flat frigate]

Modus

[glass carbon]

x + π/4 = π/2 + kπ
x = π/4 + kπ

not kπ/2

(solving when cos(x + π/4) = 0)

[hallow anchor]

how'd you arrive at that top equation?

[glass carbon]

multiply by cos(x) both sides

then sin(3x) = cos(x)

now sin(π/2 - x) = cos(x)

it gives

sin(3x) - sin(π/2 - x) = 0

and finally it's of form you're asking about

[hallow anchor]

ah I see, Imma redo that one.
Thanks for the help!

.close

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[hollow brook]

Can someone help with this I can up with this solution but it doesn’t accept it

[sharp tiger]

You can use da basic trigo triangle

[hollow brook]

What do you mean by that

[faint seal]

$\cos^{-1}(x)=\arccos(x) \ne \frac1{\cos(x)}$

[sharp tiger]

So as we know, cos is adjacent divided by hypotenuse

[flat frigate]

kheerii

[hollow brook]

How would I use this in the problem?

I’m sorry my teacher didn’t explain this and this hw is hard rn

[faint seal]

...

do you even know what $cos^{-1}(x)$ means or represents?

[flat frigate]

kheerii

[hollow brook]

Yes

[sharp tiger]

O he wrote it 1/cos

Nvm

[faint seal]

can you explain?

[hollow brook]

The adjacent side

Based on notes

[faint seal]

i think u should revise

[hollow brook]

Okay I guess I will

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[oak nest]

Hi guys I am having trouble with grasping the topic of domain and range

Im assuing the domain is x^2+6x+8>0

Or is there more to it

And y is equal to real number

[faint seal]

indeed

[faint seal]

how?

[oak nest]

Id that it

[oak nest]

Idk thats what my teacher says

They dont really explain how to find it

[faint seal]

it is..

[oak nest]

So is rhe domain just this

How to find range?

[faint seal]

no, you have to use some method to find out the range of x

[oak nest]

Wdym

Can you give an example

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@oak nest Has your question been resolved?

[oak nest]

<@&286206848099549185>

[hot matrix]

Yes the sqrt can't be 0 or negative

Then ur gonns get the domain and with the domain u can get the range

[oak nest]

Ok thx

[hot matrix]

👍

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[dull plank]

anyone know how to simplify?

[dull plank]

I am stuck

[lean otter]

[
\f ab : \f cd = \f ab \by \f dc ]

oh bruh they changed it

[lean otter]

start with this

[steep bramble]

since when?!?!

I thought it was like ad=bc

I could be wrong but

[lean otter]

multiplication is the inverse of division so...

[steep bramble]

why is a ratio equal to a forgot the name

[dull plank]

back

[dull plank]

I know that

but that shit dont work

it does, but

i did wrong after

i tried m/3n

and multiply i got still wrong answer

@lean otter

[safe radish]

@dull plank Has your question been resolved?

[safe radish]

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[safe radish]

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.

[crimson idol]

can someone check my answer, or tell me whats wrong, i got 50.067

[thin bridge]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[crimson idol]

get that

by taking derivative

then plugin values

[thin bridge]

and you haven't done anything with h'

[crimson idol]

yah i need to find hprime

i put dv and then started using prime for r and h

i wasnt sure how to find r prime though

like what it plugs in for

[thin bridge]

oh, you were trying to implicit differentiate this

[crimson idol]

yah i think so

dv/dt

[thin bridge]

it gets a bit messy like this

note that you know the relation between h and r

[crimson idol]

is there an easier way

[thin bridge]

which allows you to express V in terms of a single variable

[crimson idol]

what is the relation between them

im confused

[thin bridge]

what does the question say about height of the cone

[crimson idol]

its 24 feet

[thin bridge]

before that

[crimson idol]

diameter and height are the same

?

[thin bridge]

yes

and what is diameter

[crimson idol]

2radius

[thin bridge]

yes

so what does that tell you about height and radius

[crimson idol]

r = 1/2h

[thin bridge]

and you can sub that into your volume equation

to get an expression for V in terms of r or h depending on personal preference

[crimson idol]

like that

or before i take deriv

[thin bridge]

before

[crimson idol]

like that?

[thin bridge]

no

again making things more complicated for yourself

simplify that expression first

before differentiating

[crimson idol]

h cubed over 2

[thin bridge]

the whole thing

[crimson idol]

im confused