help-10

[devout anvil]

in positive is the same basically, just inverted

[brave bramble]

,w graph 1/x^2

[warm shale]

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[brave bramble]

Odd, it connects the top like that

[devout anvil]

yeah here is limiting

[brave bramble]

Well, uh, log(x) might bend the way you want

[devout anvil]

this is the normal

im limiting myself

thats why

[brave bramble]

,w graph log(2(x + 0.5))

[devout anvil]

but yeah i think need to be with log

[warm shale]

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[brave bramble]

Hmm, that still increases too fast. You really want a hard turn

[devout anvil]

yeah

[brave bramble]

,w graph log(log(2(x + 0.5))) between -0.5 and 3

[devout anvil]

its very hard

Maybe need to be dis?

called "dose"

[warm shale]

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[devout anvil]

idk what that is

[brave bramble]

Oh duh. These are exponentials

[devout anvil]

semilog? hm

[brave bramble]

,w graph -e^(-x)

[warm shale]

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[devout anvil]

i got this

somehow lol

the thing is, i need to do that but the other way around

im not sure how to do that

[brazen viper]

Try -(-x)^(2+k)

@devout anvil

[devout anvil]

i got something with this lol

i guess that do the job

thx anyway, i didnt remember about exponentials

.close

[obtuse pebble]

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[cosmic ivy]

How long will it take for an investment of 1100 dollars to grow to 6900 dollars, if the effective rate of interest is 6.2 percent? (Assume compound interest at all times.)

I got 110 years for this one and I am pretty sure it's right as I've calculated it. However, I am unsure of the decimal points

[drifting wraith]

you won't tell how you're doing this?

[cosmic ivy]

I will

Ok so

I used the compound interest formula

For annual as it's continuous and I wanted to find t

Eventually I found that t equaled 110 but when I entered it on my homework

It was wrong

[tardy epoch]

show your work

[cosmic ivy]

82 and 110 are the answers I got

[tardy epoch]

log(a/b) is not equal to log(a) / log(b)

[cosmic ivy]

Ahh I see

Do I have the right formula?

.close

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[verbal cloud]

so I think I have the first multiplication correct but I'm kindof confused how to structure the product of the second set of multiplications

[verbal cloud]

((1, u),m) or (1, u, m) or something else?

[obtuse pebble]

@verbal cloud Has your question been resolved?

[rugged kite]

Technically it's the former, and what you did was correct

In practice, if you see someone insist on the distinction between the two outside of an introductory set theory class they deserve a slap to the face

@verbal cloud

[verbal cloud]

Okay cool, thanks! The first feels a bit better but It had me concerned

.close

[obtuse pebble]

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[sudden otter]

I know how to solve this given a radiant but i don’t know how to with a decimal

[tardy epoch]

too much glare

[wild dagger]

Man I had a doubt, I am really struggling with these "in a traingle APB, ab = pb ac = da etc etc, proove that 2ab = 1/2 dc"
I absolutely cannot solve These types of problems no matter how hard I think, my mind gets stuck onto so many possibilities, but I just can't and my mind goes blank.
Can anyone help? Please that would be appreciated

[tardy epoch]

this channel's occupied. read #❓how-to-get-help

[wild dagger]

;-;

[sudden otter]

Ah lemme take a second one

[tardy epoch]

,calc acos(0.04)

[warm shale]

Result:

1.5307856524409

[tardy epoch]

look at this show off converting decimals to fractions in their head

[tardy epoch]

do you mean radians?

why can't you do it with a decimal ? radians are usually given in terms of decimals

[sudden otter]

Yeah sorry

[sudden otter]

I just don’t know how

[balmy mortar]

calculator

it says round to 2 dp where accurate

[tardy epoch]

do you know how to calculate when it's not a decimal?

[sudden otter]

Yes

[tardy epoch]

what are your steps

[sudden otter]

Gimme a sec

You gotta use the unit circle

Then find whatever number coronates to the number your given

I’m forgetting my terminology

[sudden otter]

I just don’t know how to get that into the a rad

[tardy epoch]

this is radians already

[sudden otter]

Like the fraction

Cus I need to get the pi/something

[tardy epoch]

why?

[sudden otter]

It solve the problem

Wait so I just put in 1.5307856524409?

[obtuse pebble]

@sudden otter Has your question been resolved?

[obtuse pebble]

@sudden otter Has your question been resolved?

[obtuse pebble]

@sudden otter Has your question been resolved?

[brazen viper]

@sudden otter if I'm understanding correctly, it's round to two decimal places, so you'd put in 1.53

2πk + 1.53 that is

n.b. getting an exact answer in terms of pi will be quite tricky in this case

And definitely not expected

[obtuse pebble]

@sudden otter Has your question been resolved?

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[sudden otter]

.reopen

[obtuse pebble]

✅

[sudden otter]

yeah their asking for an exact

ive been stuck on this for like 2 days

[brazen viper]

If you need an exact answer then remember that arccos(0.04) is the angle of a right triangle with side lengths (1/25, b, 1), where b = sqrt (1-(1/25)^2)

@sudden otter

[sudden otter]

Yeah okie this is my first time doing this so im not exactly sure how.

I get the triangle im just not sure what to do with it

[obtuse pebble]

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[potent onyx]

Hi

[potent onyx]

hello guys could you confirm me the solution of this exercise pls.

[obtuse pebble]

@potent onyx Has your question been resolved?

[obtuse pebble]

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[fresh oriole]

chotam giác abc, gọi O là giao ba đường phân giác AN, CP, BM. chứng minh
AP/PB.BN/NC.CM/MA = ON/AN + OP/PC + OM/BM

[fresh oriole]

bruh 🙂

[fleet cliff]

does anyone know how to find the hypotenuse using radical form

[serene fossil]

your name is fucking trippy

[fresh oriole]

.close

[obtuse pebble]

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[serene fossil]

also cant read vietnamese

if thats what that is

[fleet cliff]

someone??

[serene fossil]

my friend

just use pythagorean theorem

a^2+b^2=c^2

[fleet cliff]

i know bht

[serene fossil]

you have a and b given

[fleet cliff]

i dont now how to use radical

[serene fossil]

$\sqrt{39}^2=39$

[warm shale]

llspacebarll

[serene fossil]

there ya go

[fleet cliff]

thank you

[serene fossil]

39+25=64

[fleet cliff]

and then i simply that??

[serene fossil]

so hypotenuse is 8

[fleet cliff]

using the swuare root

[serene fossil]

yea

[fleet cliff]

ok thank u

[serene fossil]

$\sqrt{64}=8$

[warm shale]

llspacebarll

[obtuse pebble]

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[timid silo]

Hey folks, I would like to know how to draw an equilateral triangle in a canvas in which I know it's Height and Width.

I don't really know how to calculate it's 3 points with that data. I only know how to calculate some of them.

I know that the X coord for the upper point is width / 2.
And I also know that points 2 and 3 share the same Y coord. Being height * cos(π/6)
Don't really know how to calculate Y for the first point and X for the second and third point.

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

No, but I just found out I'm dumb

I can use for coords

x1 = width / 2
y1 = 0
x2 = 0
y2 = height
x3 = width
y3 = height

[obtuse pebble]

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[timid silo]

How do i find the value of x and m < 1 in the figure In the figure below

[tidal pelican]

what do the numbers mean?

do you know the other angles?

[timid silo]

the numbers are values sorry

[tidal pelican]

so is 1, 1 degree???

what are the values?

[timid silo]

nope the values are missing

[tidal pelican]

all of them?!

[timid silo]

that, that's an angle name

[timid silo]

nope

[tidal pelican]

its kinda impossible if we dont have values?

isnt it?

[timid silo]

Not really

it's not missing all

[tidal pelican]

what are the given angles?

[timid silo]

< 1-7

m∠6= 108

[tidal pelican]

sorry i do not follow this

[timid silo]

Uhm, could someone else help...?

[tidal pelican]

<@&286206848099549185>

[timid silo]

could i try to help?

[tidal pelican]

please do, i'd like to know as well

[timid silo]

ok awesome

[timid silo]

so is there any other given information about this picture?

like are any of the lines parallel or perpendicular

[burnt scroll]

Are you trying to find angle 1 @timid silo

[tidal pelican]

.

finding x

[burnt scroll]

And the value of x

[timid silo]

yes and x

[burnt scroll]

Ok

Ok you know how the blue line is straight right @timid silo

[timid silo]

yup, 180 degs

[burnt scroll]

That means

That’s true right

[timid silo]

yes

[burnt scroll]

Also the red lines are parallel right

[timid silo]

Yup

[burnt scroll]

That means

Those angles are the same thing

Right @timid silo

[timid silo]

Yup, theyre both exterior. if so it would be, 3x =2x?

[burnt scroll]

Well

No

[tidal pelican]

oh 3x+2x=180

5x=180

x=36

right?

@timid silo whats the answer?

[timid silo]

Yeah yeah, I got it. Same answer to x

[tidal pelican]

ok problem solved then?

[timid silo]

Actually, the problem is m < 1. I forgot how to subsitute values to the angle

[burnt scroll]

You know 3x right @timid silo

[tidal pelican]

@burnt scroll you can take over =]

[timid silo]

ahh, 108 yep

[tidal pelican]

goodluck

[timid silo]

thank you still @tidal pelican

[tidal pelican]

no worries, i played a minor part haha

if yall need any help, @ me

[timid silo]

but, 3x is not m < 1 ?

[burnt scroll]

True

[tidal pelican]

no i think cause thats a straight line

[burnt scroll]

But you know that angle 1 + 3x is equal to 180

[tidal pelican]

180 - 3x = 1

so 180 - 108 = 72

1 = 72 ?

[timid silo]

yes but angle 1 still unknown

[burnt scroll]

But 3x is known right

[timid silo]

yup 108

[burnt scroll]

So angle 1 + 108 = 180

[timid silo]

ohh supplementary so m < 1 = 72

[burnt scroll]

Yes

[timid silo]

Thank you soo muchh!

Wow, I got my head stuck in here for a longg time

[burnt scroll]

👍

[timid silo]

.close

[obtuse pebble]

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[timid silo]

Hello

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[dry yarrow]

14

[dry yarrow]

I have no idea how to start this even

[royal basin]

give names to the lengths of the two pools

for example, something like:

Let x be the length of the smaller pool, in feet.
Let y be the length of the bigger pool, in feet.

write down equations in x and y based on the problem statement

@dry yarrow

[dry yarrow]

I’m really confused on the difference on the perimeter of 16 feet

[royal basin]

the perimeters differ by 16 ft

the bigger pool's perimeter is 16 ft more than the smaller pool's perimeter

does that clear things up for you?

[dry yarrow]

Kind of

Would the equation be something like (2x+16) + (2x-16) = 730?

[royal basin]

no, this equation is not correct

[dry yarrow]

X = y-16?

And y = x+16?

[royal basin]

are you sticking to my naming scheme or are you using your own names?

[dry yarrow]

I’m trying to go with your name schemes now

[royal basin]

ok, then y = x+16 is not correct.

[dry yarrow]

I’m sorry because I’m really slow on math

[royal basin]

i was not attempting to demean you in any way.

[dry yarrow]

I understand

[royal basin]

the bigger pool's perimeter is 16 ft more than the smaller pool's perimeter

[royal basin]

.

read these carefully

recall what the word perimeter means

or, if you don't remember what it means, tell me as much.

[dry yarrow]

I’m not sure but is it not the length of a boundary around something?

[royal basin]

it is the length of the boundary, yes.

[dry yarrow]

Kind of like a border to a painting

[royal basin]

in particular for the square, the boundary has a length that is easy to calculate.

if you know how many sides a square has.

[dry yarrow]

4

[royal basin]

okay, so would you like to try to make the equation on your own or would you like me to take you through it?

[dry yarrow]

Take me through it

[royal basin]

ok

here is our sentence again

the bigger pool's perimeter is 16 ft more than the smaller pool's perimeter

let's translate this into symbols piece by piece

what is the side length of the bigger pool?

(i'd like you to not overthink any of the questions i'm about to ask you. these are all really simple)

[dry yarrow]

4 ft?

[royal basin]

no

[royal basin]

.

we're sticking with the naming scheme that i gave at the beginning

and the answers to my questions will involve x and y. since, yknow, we're constructing an equation. of course x and y will appear in it.

[dry yarrow]

Y = 4(4)?

[royal basin]

no

you're overthinking it

the length of the bigger pool is just y.

that's it.

that's how we introduced the variable y after all

you must not forget the definitions of your variables - in fact, you should always write down what all of your variables mean at the beginning of your work.

so that both you and the teacher can refer to the definitions when needed.

should we continue?

[dry yarrow]

Yes

[royal basin]

the sidelength of the bigger pool is y. what is the perimeter of the bigger pool?

[dry yarrow]

Now is it Y(4)?

[royal basin]

lowercase y, not uppercase Y.

[dry yarrow]

y I meant yeah

[royal basin]

and typically when we have a raw number multiplied by a variable we write the number first.

not that (y)(4) is invalid, it's just weird.

[dry yarrow]

I understand

So (4)(y)

[royal basin]

no need for parentheses.

but ignoring these stylistic mistakes, yes, the perimeter of the bigger pool is 4y.

now, in the same token, what is the perimeter of the smaller pool?

[dry yarrow]

4x

[royal basin]

correct

now

the bigger pool's perimeter [4y] is 16 ft more than the smaller pool's perimeter [4x]

can you translate this into an equation now?

[dry yarrow]

Would the answer be 2 equations in parentheses?

[royal basin]

no, the answer is one equation

what we have been writing are expressions, not equations.

an equation has an equals sign in it.

i would like you to write down the equation that says "4y is 16 more than 4x".

[dry yarrow]

4y=4x+16?

[royal basin]

there we go.

ok, good.

we have this equation 4y = 4x + 16

let's set it aside for the time being

write it down on your paper

let me know when you've done that so we can continue

[dry yarrow]

Got it

Would the other equation be 4x=4y-16?

[royal basin]

no

4x = 4y-16 is exactly the same as the equation we wrote down previously.

it doesn't tell us anything new.

[dry yarrow]

Alright

[royal basin]

The sum of the areas of the two pools is 730 square feet.

this is the other sentence we need to turn into an equation

would you like to try it on your own or would you rather be taken through it?

[dry yarrow]

Taken through it again

[royal basin]

ok

what is the area of the smaller pool?

[dry yarrow]

I’m not too sure how to find that

[royal basin]

do you know how to find the area of a square?

[dry yarrow]

Length times width right?

[royal basin]

the area of a square

[dry yarrow]

Apparently I do not

I completely forgot about all of it

[royal basin]

okay, let me try it this way: what can you say about the length and width of a square?

[dry yarrow]

They are equal to one another?

[royal basin]

exactly, they are the same

so what is the area of the smaller pool, which is a square with side length x?

[dry yarrow]

4x?

[royal basin]

no, not the perimeter

the area

we're looking at area now

[dry yarrow]

Still a tad bit confused

[royal basin]

.

[royal basin]

.

[royal basin]

.

you said it already

the length and width of a square are the same

and the area of a rectangle is its length multiplied by its width

i don't know how else to say it

[dry yarrow]

I’m just not sure what the length or width is

[royal basin]

.

the smaller pool, which is a square with side length x

[dry yarrow]

xlw?

[royal basin]

??

what's l? what's w? where are these new letters coming from?

don't overthink it... please. seriously please don't overthink it, i'm begging you.

[dry yarrow]

x^2?

[royal basin]

yes!!!

i mean like, think about it! why do we call raising something to the second power "squaring"?

it's because x^2 is the area of a square with side x!

[dry yarrow]

Ohhh that makes more sense

[royal basin]

does the name not speak for itself?

okay so now that we've gotten past that little block

what's the area of the bigger pool?

[dry yarrow]

Y^2?

[royal basin]

lowercase

but yes, y^2

okay, now:

The sum of the areas of the two pools is 730 square feet.

can you now write this down as an equation?

[dry yarrow]

y^2+x^2 = 730

[royal basin]

bravo

okay, so now you have your two equations:

4y = 4x + 16
y^2 + x^2 = 730

are you now able to solve these equations for x and y?

[dry yarrow]

I would not be able to, would I have to use substitution?

[royal basin]

you don't "have to" use substitution but it is a decent first step.

[dry yarrow]

If I would to use substitution would it be (x+4)^2+x^2=730?

[royal basin]

that is what equation you'd get after substitution, yes

do i understand correctly that you are now solving the equation (x+4)^2 + x^2 = 730?

[dry yarrow]

Yes

[royal basin]

okay

ping me once you have an answer or are stuck

[dry yarrow]

Would I be able to get to the next step if I square rooted both sides and getting 2x+4=27~ @royal basin

[royal basin]

...........

no, you fucked up.

$\sqrt{(x+4)^2 + x^2}$ is NOT the same as $(x+4) + x$.

[warm shale]

Ann

[royal basin]

sqrt(a+b) is not equal to sqrt(a)+sqrt(b).

[dry yarrow]

What would my next step be then?

[royal basin]

your next step would be to recognize that (x+4)^2 + x^2 = 730 is a quadratic equation in x.

and act accordingly.

namely, you should first clean it up

i.e. bring it into standard form

and then solve it in any way you like, such as factoring, completing the square, or the quadratic formula.

and if you do not remember how to solve quadratic equations at all then you really should go give that a review, because that roadblock is not something that can be overcome in a single discord chat.

[dry yarrow]

I remember those methods but for me either completing the square or the quadratic formula is the easiest

[royal basin]

i don't like repeating myself, but again, the exact method doesn't matter.

[dry yarrow]

Alright

The standard form of (x+4)^2 would be x^2+8x+16 correct?

[royal basin]

(x+4)^2 expands to x^2 + 8x + 16, yes.

[dry yarrow]

And then you add x^2 onto that which would give you 2x^2 + 8x + 16?

Or did I mess something up?

[royal basin]

no, nothing so far.

you have rewritten the equation as 2x^2 + 8x + 16 = 730. so far so good.

[dry yarrow]

Could I be able to divide both sides by 2? That would still equal the same thing right?

[royal basin]

"could i be able"...

[dry yarrow]

So I get something like x^2 + 4x + 8 = 365

[royal basin]

yes, you CAN always divide both sides by 2. that's not something you need special permission for.

not "something like"

you get exactly that, no more and no less

[dry yarrow]

Yes mam

From there I could do -8 from both sides and then take half of b and square it, then add that to both sides, factor the left side and then square root both sides

Is that correct or did I mess up something along the way

@royal basin

[royal basin]

would appreciate if you showed your work

[dry yarrow]

Alright let me do that, I’ll be right back

[devout solar]

quadratic formula should do the trick

[royal basin]

ping me once you're back.

[obtuse pebble]

@dry yarrow Has your question been resolved?

[dry yarrow]

So I’m back @royal basin

[royal basin]

uh huh

[dry yarrow]

The final equation I’ve come to is x = -2 + or - 19

[royal basin]

$x = -2 \pm 19$?

[warm shale]

Ann

[royal basin]

yes, that sounds right.

[dry yarrow]

Yeah

[royal basin]

now remember what x represented, and you should see that one of these solutions will have to be discarded and the other solution is your goal.

[dry yarrow]

It can’t be a negative if I’m correct

So x = 17 has to be the answer

[royal basin]

it's a length.

so yes, x cannot be negative.

and since you were asked for the length of the smaller pool, you need not do anything else.

but for the sake of completeness, i would like you to say now what the length of the bigger pool is too

[dry yarrow]

Alright

I will do that

Alright

I finished

I got that the length of the bigger pool is 21 ft @royal basin

[royal basin]

that is correct

[dry yarrow]

I just plugged in 17 for x in y^2 + x^3 = 730 and jsut solved it from there

[obtuse pebble]

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[royal basin]

you didn't really need to do that

[royal basin]

...oh

god damn it

but anyway yeah you did not need to do that

you could have just used the equation y = x + 4 that you already had

[dry yarrow]

Oooh right

I could’ve done that

I completely forgot about that equation

Well I appreciate the help ty very much @royal basin

[obtuse pebble]

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[torpid furnace]

Bob needs to bake the pizza in exactly 17 min. Bob have a sand clock that ends in 9 minute and 13 minute. Then how to bake the pizza in exact time? (math type: logical, grade level: grade 5)

[obtuse pebble]

@torpid furnace Has your question been resolved?

[torpid furnace]

<@&286206848099549185>

[quasi ferry]

you have 2 clocks, right ?

[torpid furnace]

yes

[quasi ferry]

i am not sure if thats the best way but probably you need to start the clocks first and later put the pizza in the oven

try to make one of the two clocks to end in a more useful time than the ones you have right now

[worthy comet]

think about which times from now you can get using only the first clock and which times from now you can get using only the second clock. Then notice that you can put the pizza in the oven at a time measured by the first (or second) clock and take it out at a time measured by the second (or first) clock

[timid silo]

start both clocks at the same time, when 9 runs out flip it and you know that 9 minutes passed, then when 13 runs out flip 9 again, it will have ran for 4 minutes and will run for another 4 so 9+4+4=17

[torpid furnace]

ive come up to an idea. you start the clocks at same time and when the 9 min clock runs out, put pizza in the oven. and immediately after 13 min clock runs out, flip it. and after 13 min clock runs out, take pizza out

.close

[obtuse pebble]

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[tawny compass]

how do i prove that 0.5 maximises x - x^2 ? i am aware this is probably the most stupid obvious insight ever and yet my idiot head has been banging against the wall

[high lily]

complete the square to get vertex form
or use calculus

[tawny compass]

i dont understand

complete what?

[high lily]

the square

look up completing the square

[tawny compass]

i am talking about only decimal inputs btw

[high lily]

what?

[tawny compass]

you mean the quadratic formula?

[high lily]

no

i mean what i said

[tawny compass]

ok

[high lily]

completing the square is a method that can be used to convert a quadratic expression from
general form to vertex form where you can easily identify extrema

[tawny compass]

wait i think i did it

but i didnt complete the square

and since im stupid it mnight be wrong can you verify it?

[high lily]

show what you did

[tawny compass]

alright so if were only considering decimal inputs

[high lily]

wdym by only considering decimal inputs

[tawny compass]

the function x - x^2 for only decimal inputs how do i prove that 0.5 is the max output

okay so i am trying to show that changing x any amount from that is going to result in something smaller

[high lily]

also max output of 0.5 and max output at 0.5 mean different things

[tawny compass]

shit yeah i meant at

[tawny compass]

we need to say that changing x any amount will yeild this:

x - y (y = amount of change) -(x - y)^2

and we cant forget adding y to x

x + y -(x + y)^2

so since i already hypothesise/know that 0.5 is it, i can substitute 0.5

0.5 + y -(0.5 + y)^2

since the output at 0.5 is 0.25, i can set it like so

0.5 + y -(0.5 + y)^2 < 0.25

if we evaluate the expression on the left side

0.25 - y^2 < 0.25

which makes sense

so now we should consider the case where we subtract y

0.5 - y -(0.5-y)^2 < 0.25

again evaluating we end up with 0.25 - y^2 < 0.25

which is correct since inputting 0.4 and 0.6 in x-x^2 yeilded the same result

so i can say that making any change to 0.5 as the input will result in something smaller than 0.25

is this all correct?

i am kind of worried thsi argument will still hold up for inputs that CAN be changed to output something larger

[high lily]

you didn't square properly

[tawny compass]

oopsies i didnt mean to do that

substitute x wherever it is with 0.5

ok wait lemme see if i did it correct

no i think its correct

lemme show the working

1/2 - y -(1/2-y)^2
1/2 - y -(1/4 +y^2 - y)
1/2 - y - 1/4 - y^2 + y
1/2 - 1/4 - y^2
1/4 - y^2

i may be missing something since i always make sillies but i think this is right

[high lily]

seems ok

i'd recommend using a variable other than y though

like h

[tawny compass]

i used epsilon

[high lily]

or Delta x

[tawny compass]

but i didnt want to say epsilon over and over

since idk the symbol for it

[tawny compass]

i saw someone use it for incremental changes when they were explaining derivatives so i just used it

but would my solution be correct?

[high lily]

it works

[tawny compass]

are you sure it doesnt just do the same thing for values other than 1/2 when in reality they could be changed to output something larger

[high lily]

well you could try and see what happens when you try another value

[tawny compass]

whati wanted was a smooth proof showing a linear change and a peak or something but i just went muddy

[tawny compass]

okay ill do that now

3/2 - y should be able to get larger so lemme try that

shit

i meant

2/3

ok so its working out for 2/3

its showing me that the inequality is false

lets fucking gooo

well actually my domain for the function x - x^2 is 0 to 2, but the real numbers between 1 and 2 would always return a negative number when inputted right(i am solving a bigger problem and this is one of the subparts i need to do[i think])

yeah i think they would since a number greater than one squared would have the numerator increasing much faster than the denominator even if they were consecutive

anyawys thanks for helping but do you have an elegant proof for it? you were talking about completing a square

[high lily]

look it up

[tawny compass]

yes yes i looked it up and understand it now, you complete the sides and add the amount you splitted by squared to the other side

but since im an idiot i dont understand how this would help me here...

how would figuring out the roots of that polynomial help me is what im trying to say

[high lily]

i didn't say anything about figuring out the roots

[tawny compass]

hmm

[high lily]

there's a difference between completing the square and solving an equation using completing the square

completing the square by itself refers to the part where you complete the square and balance the equation / maintain the value of the expression

[tawny compass]

i guess i could use the two equivalent expressions

[high lily]

$\cts$

[warm shale]

ℝamonov

[high lily]

since the coefficient of the leading term in x-x^2 is negative

you can consider factoring out -1 first

-(x^2 - x)

[tawny compass]

i actually did this

i did factoring out x aswell

but i didnt get anywhere

[high lily]

why factor out x

[tawny compass]

i dont know

[high lily]

-(x^2 - x)
and then use the approach outlined above to complete the square for the stuff within the parentheses

[tawny compass]

okay

thats great and all, but after doing that i have to maximise the value of the stuff in the parentheses while keeping track of minimising the value of the stuff in the last term we used to complete the square, which is its own challenge almost equivalent

oth are squared too

*both

[high lily]

what do you have after completing the square

[tawny compass]

i have x^2 - x + 0.25

[high lily]

no

[tawny compass]

wait but

(x+0.5)(x+0.5) = x^2 - x + 0.25

[high lily]

you should have an expression equivalent to what you started with

[tawny compass]

oh lol

[high lily]

show everything

[tawny compass]

(x + 0.5)(x + 0.5) - 0.5

[high lily]

don't cut out certain parts and ignore everything else

still not right

[tawny compass]

wait what did i cut out this time

hmmmmm

[high lily]

errors with signs too

are you doing this on paper?

[tawny compass]

yes i am

i am a huge nimrod so expect it from me

[high lily]

try it again

[tawny compass]

sorry

alr

i meant (x - 0.5)(x - 0.5) - 0.25

i have no fucking clue why i typed what i did

[high lily]

missing the -( )

[tawny compass]

yeah i know but i thought that would come aftert

hmm alr

[high lily]

don't cut out certain parts and ignore everything else

[tawny compass]

0.25 - (x-0.5)(x-0.5)

[high lily]

it'll be clear if you express (x-0.5)(x-0.5) as(x-0.5)^2

$0.25 - (x-0.5)^2$

[warm shale]

ℝamonov

[tawny compass]

OHHHHH

THIS IS WAY BETTER

this is what i was looking for

so basically nothing positive can come out of that term so its best to make it 0 and 0.5 makes it 0 soooo

ahh this is much more satisfying

thank you very much

[high lily]

you should be able to identify the max and when it occurs in this form

[tawny compass]

yes yes i forever thank you, i only knew splitting the middle term and memorised the quadratic formula without understanding this properly, thanks for this dope tool

[high lily]

it seems like you're doing some sort of pre-calc here,
you should've already learned about this stuff

[tawny compass]

i knew about splitting middle term and quad formula but... yeah

i am doing precalc

for competition math

i mean this is competition math

thanks !

.close

[obtuse pebble]

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[steel steeple]

Hii somebody can help me understand binary search?

[steel steeple]

What I need if I have an even number of elements, where did i start?

The 5 element or the 6 element in the list

[nocturne minnow]

I believe that's your choice to make

[steel steeple]

i have a question of how many tries worst case i need to do for if i can only do > operation

For example for list with 10 elements and I can be right twice (2) the minimal number of tries is 4 worse case, and for 10 elements and 1 time to be right the you have to go 10 times worse case

Something like this

The question is completed and not in English

I know the answer have something tovdo sith log but i still dont sure

@nocturne minnow

[short spruce]

that's not binary search

if you're only using >

[short spruce]

this is incorrect as well

worst case scenario cannot be equal to the number of elements in your array

binary search would fail if you just used >

[steel steeple]

I want to do a Google search a minute for to see this question in English

This is the dropping egg question

[short spruce]

that question doesn't seem to only use >

it says if the egg survives, you move up a floor, if the egg breaks, you move down a floor

[steel steeple]

I know but i didn't know this question have a name when i tried to explain

But thx anyways

Good night

[obtuse pebble]

@steel steeple Has your question been resolved?

[obtuse pebble]

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[clear sinew]

This is not a homework lol. But just a question I was wondering about. I read in my scripture that for huge numbers $$(a+b)! \ge a!b!$$

[warm shale]

mrbrown

[clear sinew]

is there any way to prove that?

[obtuse pebble]

@clear sinew Has your question been resolved?

[tight oasis]

It is always true @clear sinew

[warm shale]

SkyTwX

[tight oasis]

@clear sinew sorry double ping

[clear sinew]

all fine, thank you very much but wait a bit pls!

in the first step you wrote out the faculty notation as far as i can see

i know $a! =1\cdot 2\cdot 3 \dots (a-3)\cdot (a-2) \cdot (a-1) \cdot a$

struggling a bit to grasp my head around (a+b)!

[warm shale]

mrbrown

[clear sinew]

$(a+b)! = 1 \cdot 2 \cdot 3 \dots (a+b-3) \cdot (a+b-2) \cdot (a+b-1) \cdot (a+b)$

[warm shale]

mrbrown

[tight oasis]

Factor out the first $a$ integers to make appear $a!$
Sorry for the ping again @clear sinew

[warm shale]

SkyTwX

[clear sinew]

Ohhhhhhhhhhhhhhhhhh I understand

$(a+b)!=1\cdot 2 \dots a \cdot (a+1) \cdot (a+2) \dots (a+b-2) \cdot (a+b-1) \cdot (a+b)$

[warm shale]

mrbrown

[clear sinew]

ah ok you dropped all the $a$'s except $a!$ on your next step leaving you with $a! b!$

alight!

thank you very much @tight oasis

[warm shale]

mrbrown

[tight oasis]

Np! Dont forget to .close ♡ @clear sinew

[clear sinew]

❤️

.close

[obtuse pebble]

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[crimson berry]

do you know what it means to divide

as in what it means to say x divides y

[analog karma]

like.. x is a factor of y?

[crimson berry]

right. there are two trivial cases of divisibility. Can you think of them?

Something along the lines of "everything divides ___" and "____ divides everything"

[analog karma]

1 divides everything, not so sure about the first one?

hmm

i mean i guess everything divides 0

too

[crimson berry]

ok great

now can you see how to use this information

[analog karma]

OHH YEAH

i may be dumb... is it t = 2 and t = 0?

[crimson berry]

yes

[analog karma]

THANKS a lot omgggg

[obtuse pebble]

@analog karma Has your question been resolved?

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[sly folio]

https://cdn.discordapp.com/attachments/896348563821436938/962126714710929508/image0.jpg

[sly folio]

I'm confused on how to continue with this problem. I need to find the arc length. When I plug in t=1 I get -1/3 - 1/2, and for t=0 I get undefined fractions.

[obtuse pebble]

@sly folio Has your question been resolved?

[sly folio]

<@&286206848099549185>

[tardy epoch]

mistake here

$\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$

[warm shale]

riemann

[sly folio]

Hmm okay, what would be used instead? @tardy epoch

[obtuse pebble]

@sly folio Has your question been resolved?

[sly folio]

<@&286206848099549185>

[tight oasis]

Whaz you need? @sly folio

[sly folio]

.

Riemann called out a mistake but I'm not sure what to use instead of what I did

@tight oasis

[tight oasis]

Think about it

What does square root mean

[sly folio]

Right, so I ended up doing integration by parts and it was messy, but I have an answer close to the actual but I'm not sure what happened.

The actual answer is 1/3(2sqrt(2) - 1)

[tardy epoch]

,w int (1+x^2)^(1/2) / x^4

[warm shale]

Results provided by WolframAlpha

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

Requested by riemann#2529

[sly folio]

Yeet, thats what I got

[tardy epoch]

Why didn't you evaluate the antiderivative at 0?

[sly folio]

Isn't that division by 0?

@tardy epoch

[tardy epoch]

look up the definition of improper integral

[dusk ruin]

It’s when your integrals are lewd

[obtuse pebble]

@sly folio Has your question been resolved?

[obtuse pebble]

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[plush anchor]

In 2nd question how do we convert from binary to division

[plush anchor]

Decimal*

[short spruce]

you can just do binary division

[plush anchor]

Someone please explain how to find the base in decimal number

[short spruce]

that sentence doesn't make sense

[plush anchor]

I meant

[short spruce]

if you need help converting from binary to decimal, google it

[plush anchor]

How did we get base 10

[short spruce]

by knowing how to convert from base 2 to base 10

[plush anchor]

Yes that's what I am asking

[short spruce]

google

[plush anchor]

How do we do that

[short spruce]

look up a video

[tardy epoch]

reading's faster
https://www.purplemath.com/modules/numbbase.htm

[obtuse pebble]

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[brisk arrow]

Ask for sin theta

Those dotted lines are the bisector of corresponding angle

AC=3*(sqrt7)

BD=3(sqrt19)

I need some help

[real horizon]

law of sines

you can use law of sines on triangle ABC

since you know all its 3 sides and 1 angle

[brisk arrow]

I don’t get it

[real horizon]

here's an image from google

[brisk arrow]

I see

[brisk arrow]

The answer is wrong

[real horizon]

your work is wrong

show me exactly how you did it

[brisk arrow]

I split the whole thing into a smaller triangle

Then I make a equation that red circled using laws of sine

[real horizon]

oh okay
how do you know there is a 60deg angle in that triangle though?

[brisk arrow]

Cuz it’s a bisector

[real horizon]

it's not?

it's the diagonal of the parallelogram

[brisk arrow]

Ohh

[real horizon]

it's best if you use law of sines on triangles ABC and ABD

[brisk arrow]

So Diagonal didn’t split the main angle into two equivalent angles

[real horizon]

you can see why it's not true:

[brisk arrow]

I see

I see the problem

[brisk arrow]

Then I think it’s impossible to apply laws of sine into this question

[real horizon]

hmm???

for ABC you know all 3 sides and 1 angle
in fact, you only need SSA to apply it

[brisk arrow]

but those three angles of triangle ABC doesnt involve theta

[real horizon]

yes

however

you will find these two angles (actually one can be found from the other, but....)

[brisk arrow]

what if its not a special angle

[real horizon]

from which (after finding AADB as well) you can find theta

[real horizon]

it's fine?

[brisk arrow]

let me try

[brisk arrow]

i found sin P= 3(sqrt3)/2

that mean angle P isnt a special angle

that mean i cant recognise its degree

[obtuse pebble]

@brisk arrow Has your question been resolved?

[obtuse pebble]

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[wind hemlock]

if I am given a pair of points (x,y) and I am asked to find the lagrange interpolation, do the interpolation points of type x0 need to be the x values given in the pair of points when I calculate the lagrange polynomials or can I choose any interpolation points?

[obtuse pebble]

@wind hemlock Has your question been resolved?

[light sand]

<@&286206848099549185>

[obtuse pebble]

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[obtuse pebble]

@frozen willow Has your question been resolved?

[obtuse pebble]

@frozen willow Has your question been resolved?

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[gloomy valve]

What's the problem?

[glossy ibex]

69

[ocean topaz]

don't troll in help channels

[teal prawn]

for domain it's just x + 4 ≠ 0, every other real value of x

[wanton dagger]

I want help in range

Ok let me see

[teal prawn]

if you equate the fraction to any real number, you can cross multiply and find x

but for 0, that x + 4 vanishes and you get 3 = 0

[royal basin]

@teal prawn do not just give out answers.

[teal prawn]

ig thatll help

[wanton dagger]

Wait I don’t understand the cross multiply part

[teal prawn]

so if 3/(x + 4) = a
3 = a(x + 4)
for any constant value of a other than 0, you just get a simple linear equation

[wanton dagger]

Oh ok

[teal prawn]

but if a = 0

then 0(x + 4) = 0 ≠ 3

[wanton dagger]

Yh I see

Ok I’ll give other questions a go, thanks

.close

[obtuse pebble]

Channel closed

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Use .reopen if this was a mistake.

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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.

[gray creek]

I’m confused with these 2 questions

[zenith spade]

What have you tried

[gray creek]

That’s the thing I don’t know where to start

[long sinew]

Is this from class?

[gray creek]

Yeah I’m getting practice for a state test next week

[long sinew]

Then review the notes and study material?

That would be a good start

[gray creek]

We never covered this

[long sinew]

I doubt that, what grade?

Or

What math class?

[gray creek]

Algebra 2

[long sinew]

Yes they definitely covered both of these in alg 2

The second one is algebra 1

[gray creek]

But they never did for my class

[long sinew]

Do you remember exponent properties?

[gray creek]

Yeah

[long sinew]

Then apply some of them to simplify those expressions

[gray creek]

I’m kinda confused with applying them into the equation

[long sinew]

Simplify option D, tell me what you get :)

It's just distributive property pretty much

[gray creek]

25XY + 25X^1/2

[long sinew]

I don't think that's right...

[gray creek]

Oh 5XY + 25X^1/2

[long sinew]

$5\sqrt{x} * \sqrt{y} + 5\sqrt{x}$ * 5$

[warm shale]

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

[long sinew]

So this can be further simplified, because x and y are both in radicals, we can say that

$5\sqrt{xy} + 5\sqrt{x} * 5$

[warm shale]

Breeziboi

[long sinew]

Now finally

The 5\sqrt{x} * 5

$5\sqrt{xy} + 25\sqrt{x}$

[warm shale]

Breeziboi

[gray creek]

So what properties you did use?

[long sinew]

Distributive property, 1/2 power is square root, and sqrt(a) * sqrt(b) = sqrt(ab)

So D is the answer, but try simplifying C, for practice

[gray creek]

Would you sqrt first or after distributing it?

[long sinew]

$\sqrt{x} * (25\sqrt{y} + 5)$

[warm shale]

Breeziboi

[long sinew]

This is C

[long sinew]

I would just change 1/2 power to sqrt first, because the original expression is in sqrt

Also I think it looks nicer :)

I already did so for you in the TeXit bot

[gray creek]

Would it b 25 sqrt xy + 5 sqrt x?

[long sinew]

$25\sqrt{xy} + 5\sqrt{x}$

[warm shale]

Breeziboi

[long sinew]

Like this?

[gray creek]

Yeah

[long sinew]

Yes I believe so

Good job

[gray creek]

I thought we had to add the 2 1/2 when distributing for option D

25XY + 25X^1/2

That’s why I thought I got this

[long sinew]

Oh

Wanna try B, or do you think you got it?

[gray creek]

do you have any good websites that have those properties

[long sinew]

Khan Academy

I spent a lot of time on there for exponent properties

Because I am in geometry currently, and taking precalc next year so I covered a lot of alg 2 on my own

[gray creek]

Gosh I’m taking Calc next year

[long sinew]

Are you in 9th?

[gray creek]

Oh no 10th

[long sinew]

Oh ok

I'm 8th grade

[gray creek]

Woah you’re pretty good

[devout solar]

Oh damn

[long sinew]

Anyways I think the second question is a piece of cake for you?

[long sinew]

Thank u :)

Dunno what I'm gonna do in 11th and 12th after calc

:'(

[gray creek]

Ap calc I guess

[long sinew]

But thats what im doing in 10th

[gray creek]

Statistics most likely

[long sinew]

Oh ya that would be useful

[devout solar]

Everything has a 50% chance of happening

[gray creek]

God I forgot about non real solutions

[devout solar]

on the first one?

[gray creek]

Second one

[long sinew]

When the square root is negative in the quadratic formula

[devout solar]

yeah

[long sinew]

Non real or complex is when you have a negative square root

a + bi

[gray creek]

Oh so we’re just doing the quadratic formula that’s easy

[long sinew]

Yeah

[devout solar]

so if d<0?

[gray creek]

D < 0?

[devout solar]

yes

[long sinew]

Wait what is going on here lol

[zenith spade]

Have you heard of the discriminant?

[devout solar]

Me

[long sinew]

oh

I see now

[zenith spade]

Essentially it’s the stuff under the square root in the quadratic formula

[gray creek]

I think so but I need quick explanation

[long sinew]

Yeah that would make any quadratic have no roots haha

Or complex roots :)

You could say

[devout solar]

discriminant is b^2-4ac

[gray creek]

Oh

[devout solar]

if my memory isn't wrong

[long sinew]

Yes

[gray creek]

Yeah you’re right

But is D the disciminant?

[devout solar]

yes

I guess

[gray creek]

Btw is it C?

[long sinew]

Should be yea

It's definitely not A or D

I just did it in my head the discriminant is negative so yeah you're right

[gray creek]

It’s not b either

[devout solar]

yeah it should be c

[long sinew]

Got any more practice questions?

[gray creek]

Yeah

[gray creek]

So how would we do this?

[devout solar]

should probably try to get rid of the square root

[long sinew]

You can factor the first two terms and the second two terms

[long sinew]

Oh

I thought you were talking about 7

[gray creek]

Nah it was 6

[devout solar]

(x-2)^2 is not x^2-4

do you remember the things I dont have the words for in english

like what is (a-b)^2

[gray creek]

Oh right

[devout solar]

You should memorize what those are

you will need them

[gray creek]

So would the expression be X^2 - 4x + 4 = X^2 - 4x + 4?

[devout solar]

wdym

[gray creek]

After disputing the (x-2)^2

[devout solar]

oh yeah

think so

But I usually struggle with square roots

[gray creek]

So what would we do after?

[gray creek]

How come?

[devout solar]

Idk

Imma check before I ruin your problem

Im too dumb to be helping

[gray creek]

It’s okay

[devout solar]

Am I allowed to give the solution that a calculator gave or not?

[gray creek]

If it’s right I think so

[devout solar]

hmm imma check the rules

probably not

ask the smart guy

[gray creek]

Are you still there?

[long sinew]

Now

I am

So you get x^2 - 4x + 4 = x^2 - 4x + 4

Hahaha, infinite solutions

[devout solar]

But it doesnt have infinite

[long sinew]

Why not?