help-10

[cloud saffron]

yes

[brisk matrix]

so y^2

we can rewrite that as f(y)

where f(x) = x^2

[cloud saffron]

and f(g) = y^2>

?

[brisk matrix]

f(y) = y^2

if you want to use g

then we can write

[cloud saffron]

ok

[brisk matrix]

f(g(x))

where f(x) = x^2

g(x) = y(x)

[cloud saffron]

ok so we have

?

[brisk matrix]

well the derivative of f(g(x)) = f’(g(x)) * g’(x)

what is f’(x)

[cloud saffron]

2x

power rule

[brisk matrix]

so f’(g(x)) = 2g(x)

[cloud saffron]

2*y^2

in this case

[brisk matrix]

and what is g’(x)?

[brisk matrix]

no

[brisk matrix]

g is y not y^2

[cloud saffron]

what was the original term we were applying the chain rule to?

[brisk matrix]

y^2

[cloud saffron]

x^2*y^2 product rule

and because we have y^2

we have to then apply the chain rule?

correct?

[brisk matrix]

yes

[cloud saffron]

ok i finally see that

sorry

so with y^2 we have f(x) and g(x)

how do we break that down into f(x) and g(x) again?

y^2

[brisk matrix]

.

[cloud saffron]

f(x) = y?

[brisk matrix]

.

[cloud saffron]

ok

so once we applied the product rule to x^2*Y^2

we are still using that x^2 and y^2?

so the chain rule appears to be happening inside the product rule

this is how i understand it

.

.

<@&286206848099549185>

is this correct?

with x^2 * y^2 we have to apply both the product rule and the chain rule. I see the application of the product rule and the application of the chain rule within that. they're nested.

<@&286206848099549185>

@brisk matrix Thanks for your help and explanation. I was able to work the problem out and understand it. You cleared a lot of misconceptions up for me. I really appreciate it.

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[obtuse pebble]

@fluid snow Has your question been resolved?

[obtuse pebble]

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[echo scroll]

is the reason that for the upper half, that integral gives outside of circle, but for the half part integral gives inside of circle?

bad english sorry

what, i ask for if its the reason why that integral doesnt give area outside of circle

[echo scroll]

.

[obtuse pebble]

@echo scroll Has your question been resolved?

[timid silo]

Actually hmm I am not too sure sorry

[obtuse pebble]

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[upper yew]

how do i substitute if i have two y=mx+b equations.

[fierce lagoon]

Replace the y with the other equation

[timid silo]

Substitute?

Context?

[fierce lagoon]

So if y = A and y = B, then A = B

[upper yew]

like solving fo x and y

like for example

[timid silo]

Oh okay just do whAt umbra is saying

[upper yew]

y=3x + 4 and y = 5x + 9

i have to find x and y on a graph

but how do i substitue

[timid silo]

.

[ruby path]

Euclid's Axiom

[strange pagoda]

Find the 5th term in the expansion of (a+b)^8 in simplest form

[upper yew]

woh

im and 8th grader

[fierce lagoon]

#❓how-to-get-help

[ruby path]

Open your own channel

[strange pagoda]

oh

[fierce lagoon]

_ _

[upper yew]

alr

thanks

[obtuse pebble]

@upper yew Has your question been resolved?

[obtuse pebble]

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[exotic creek]

they divided the yellow highlighted answer with cos^3(x). why?

[civic zealot]

to get the tan^3 on the top and the 1 on the bottom

[exotic creek]

ah ok

thanks

.close

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[rough shard]

that's my work. i entered that the point (10/3, 10/3) is a saddlepoint, but it says that i'm wrong

trying to find my error

[latent walrus]

fxy isnt 0

[rough shard]

ah, crap, yeah. thanks! i was looking at the wrong equation when i solved that

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[obtuse pebble]

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[rough shard]

.reopen

[obtuse pebble]

✅

[rough shard]

but shouldn't it still a saddle point?

[latent walrus]

y=-x, not x

didnt see that last time

3rd line

[rough shard]

ah, yeah, that was it. thanks!

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[obtuse pebble]

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[plush fractal]

hi

[novel trench]

hi

[plush fractal]

can u help me plz

[nocturne minnow]

.close

[obtuse pebble]

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[thorn rock]

hello

[thorn rock]

how do i transform this function

i have point (4,6)

the y wont be changed since its 1f

but the x

shouldnt i divide by 2 and add 4?

[thick gyro]

yes

[thorn rock]

so it turns out to be (6,6)?

my friend told me he somehow got 4,6 so im not sure

maybe hes wrong

[thick gyro]

wait

i think its the other way round

yh

add 4

[thorn rock]

wdym

[thick gyro]

divide by 2

[thorn rock]

uh wait fr?

[thick gyro]

yh

if 2x - 4 = a

then

x = (a+4)/2

[thorn rock]

isnt by left to right?

like in y

[thick gyro]

no

y and x

have opposite

reactions

[thorn rock]

but y is in the left and right order right?

[thick gyro]

yh

if it was 2f(x) - 4

we'd times y by 2 and - 4

[thorn rock]

and x stays the same

[thick gyro]

yh

[thorn rock]

since its divided by 1

[thick gyro]

yh

[thorn rock]

okay

so when x is in the parentheses its right to left?

or should i make up an equation

like u did

[thick gyro]

yh

so theres no ambiguity

just say that f(2x - 4) is f(z)

so 2x - 4 = z

[thorn rock]

2x= z+4

x= 1/2z +2?

[thick gyro]

yh

[thorn rock]

what is z tho

the old x?

[thick gyro]

z is the original

yh

[thorn rock]

so if the old x (z) was 4

we would multiply by 1/2

and add 2?

[thick gyro]

yh

thats why your friend got 4

[thorn rock]

ohh

i see

and if i want to remember it for the y, its just in left to right?

or are there anyways to make up an equation for that too

[thick gyro]

for y if you want

you can just say y = 2w - 4

where w is the old y

but it doesnt matter, since its less ambiguous

for y

[thorn rock]

oh btw

what do i do here

when i have these

doesnt that make it positive

[thick gyro]

yes

[thorn rock]

so how do i solve it?

[thick gyro]

whats it asking you for?

[thorn rock]

transform 4,6 in this situation

should i make the whole expression positive to get rid of them?

[thick gyro]

no its fine

keep it as it is

since its just f(x)

the x coordinate doesnt change

but for the y coordinate

[thorn rock]

what about the y

yea

[thick gyro]

put in f(x) = 6

and what is g(x)

[thorn rock]

g(x) is the new function of f(x) being transformed

[thick gyro]

yh

its no different than writing y = |-f(x) + 32|

[thorn rock]

yeah

but what about the y here tho

in 4,6

[thick gyro]

put in the old y value of 6

into f(x)

and get out the new y value

[thorn rock]

so y= | -(6) +32|

[thick gyro]

yh

[thorn rock]

y= 26?

[thick gyro]

yh

[thorn rock]

isnt it just solving without the Absolute value and then making everything positive

like
y= -f(x) +32

and making every y and x positive

because of the Absolute value

[thick gyro]

yh

no not the x positive

only the y becomes positive

[thorn rock]

why

the x is also in Absolute value

[thick gyro]

the x

only cares

about whats inside the f( )

[thorn rock]

oh

[thick gyro]

it could give less of a shit about everything outside

[thorn rock]

okay

so only y gets positive

if i had

g(x) = | f(x+1) +32 |

originally its minus 1 since its to the left once

the Absolute value
doesnt change it?

[thick gyro]

no

[thorn rock]

okay

[thick gyro]

what do you think (4,6) becomes

under this

[thorn rock]

well it will become
(4,6) ==> (3,38)

?

[thick gyro]

yh

[thorn rock]

okayy

y in left to right order

right?

[thick gyro]

for y, its just the normal bidmas order

so yh

left to right order

[thorn rock]

okay

last question

what within the ( )

say for example

y= (3x)

its divide by 3 right?

since 3x=a

x= a/3

[thick gyro]

y = (3x) or y = f(3x) ?

[thorn rock]

does it matter?

[thick gyro]

well not really,

but

in the case of

y = ax

you can treat it as

either

y = af(x)

or y = f(ax)

where f(x) = x

[thorn rock]

yeah

it doesnt matter tho

abc= acb

[thick gyro]

you can either divide the x value by 3

or multiply the y value by 3

in that specific case

[thorn rock]

since if y= (3x)

its also equal to
y= 3(x)

[thick gyro]

yh

[thorn rock]

thats what u saying?

okay

so either way if i divide the x value by 3 or multiply the y value by 3 it will be correct both ways?

[thick gyro]

yh

[thorn rock]

g(x)= f(3x)

g(x) = 3f(x)?

[thick gyro]

no

that was only for the specific case of

f(x) = x

[thorn rock]

how is different

it

[thick gyro]

because

if we substitute 3x into f(x)

we get

f(3x) = 3x

if we have 3f(x) we get 3f(x) = 3x

no other equation can do this

[thorn rock]

okay so these are two different equations

[thick gyro]

we cant have a general rule of g(x) = f(3x) = 3f(x)

if f(x) = x then that rule works fine

but its a specific rule

not a general one

[thorn rock]

what about f(x) =2x

divide by 2?

the x

[thick gyro]

yh

[thorn rock]

or multiply by 2 the y?

[thick gyro]

yh

both work

[thorn rock]

okay

i will do the x to not get confused as i got alr used to it

being x in the ( ) and y is what outside

[thick gyro]

sure

[thorn rock]

thanks for ur time and help

[thick gyro]

also f(3x + 2) is the same as 3f(x) + 2 if f(x) = x

[thorn rock]

for this specific thing i will do the way you taught me by using 3x+2= z

[thick gyro]

yh

[thorn rock]

z being old value of x

[thick gyro]

yh

[thorn rock]

thanks again

.close

[obtuse pebble]

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[tacit jacinth]

hello, this is random but is there any people here who's an architect or an engineer kindly dm me for school purposes hehe tenks (or just an archi or engineering student)

[obtuse pebble]

@tacit jacinth Has your question been resolved?

[obtuse pebble]

@tacit jacinth Has your question been resolved?

[tardy epoch]

Just ask your question here

[tacit jacinth]

it's like an interview

but aight

[tardy epoch]

Help channels are for math problems. Ask in #discussion

[obtuse pebble]

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[split lintel]

Why does v come first instead of t?

[ruby path]

Eh doesn't really matter they probably chose v because the coefficient of v^2 is 1

[split lintel]

ok thanks;

.close

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[brisk matrix]

chain rule

oh lol

antiderivative my bad

lemme see

,rccw

[brisk matrix]

you integrated with respect to x at the end

you need to integrate with respect to u

then sub back the x’s

they are both with respect to u

[sudden wadi]

1 just become u if you integrate with respect to u

[brisk matrix]

it would benefit you to write it as $\int (1 + \cos (2u)) , du$

[warm shale]

maximo

[brisk matrix]

your integral should become u +sin(2u)/2

then you should substitute back in for u

[obtuse pebble]

@radiant hornet Has your question been resolved?

[brisk matrix]

i’m confused by your questions

can you be specific

[brisk matrix]

what you wrote is incorrect

if you integrate with respect to u the function that comes out should be in terms of u

would you be able to tell me what $\int 1, du$ is?

[warm shale]

maximo

[brisk matrix]

u + C yes

yes, but what you wrote doesn’t make sense

you start with cosu, then we suddenly get cos(u/6), and the dx changed to a du

can you ask your question again in a different way

i’m not understanding what you want to ask

this is not correct

du/dx = 2x -> du = 2xdx

so we can put du in the integral

and integrate with respect to u

i’m saying what you wrote is wrong

the textbook did it correctly as far as i can tell

[brisk matrix]

you’re missing an x

du/dx = 2x -> dx * 2x = du

-> dx = du/2x

[brisk matrix]

yeah

[obtuse pebble]

@radiant hornet Has your question been resolved?

[obtuse pebble]

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[molten cosmos]

Can someone help me with a mistake I made with an exact ODE question

[molten cosmos]

I have the solutions to a and b which are correct but apparently my c is wrong

Solution to a:

solution to b:

My incorrect solution to c:

feedback:

I dont rlly understand why its wrong though

<@&286206848099549185>

[obtuse pebble]

@molten cosmos Has your question been resolved?

[obtuse pebble]

@molten cosmos Has your question been resolved?

[obtuse pebble]

@molten cosmos Has your question been resolved?

[obtuse pebble]

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[celest whale]

i got the result

-6/-5

do i keep it like that

or do i make it positive

6/5

[light flume]

make it positive

[celest whale]

ok but i dont have to flip them

right

i just have to take the negative sign away

.close

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[nova bramble]

Hey, I'm not even doing math homework at the moment. Just making a program that needs to error-check impossible circumradii of regular polygons, what circumradii are impossible?

[fierce lagoon]

Not sure if radii can be a complex number in magnitude

[nova bramble]

if it's just a double value, it is possible though?

[fierce lagoon]

Wdym double value

[nova bramble]

A real number basically

[fierce lagoon]

Well your radius can't be -2 in magnitude

[nova bramble]

Yeah, so just anything positive?

[fierce lagoon]

It's any real number inclusive 0 if your polygon is degenerate

Although your program might make negative radii just shoot out in the antiparallel direction

Assuming that your regular polygon isn't degenerate, then 0 will never work

[nova bramble]

Ok, thank you very much!

.close

[obtuse pebble]

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[native steppe]

How do I convert an XYZ euler angle into a XYZ coordinate representing the endpoint of a line 1 unit long in the direction of my angle

[obtuse pebble]

@native steppe Has your question been resolved?

[obtuse pebble]

@native steppe Has your question been resolved?

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@uncut fulcrum Has your question been resolved?

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[timid silo]

This is a super quick question, what does my professor mean by " show your intermediate calculations" (this is for a homework assignment)

[timid silo]

I'm assuming he means show you're steps?

[nocturne minnow]

Basically show all your work

[timid silo]

ok cool

.close

[obtuse pebble]

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[lost tree]

,rotate

[floral rock]

Need help with solving for X i tried it but I got the answer wrong

Can we not divide the angle by 2?

[lost tree]

Not without knowing that the triangle is isosceles

[obtuse pebble]

@floral rock Has your question been resolved?

[covert dust]

You can get two equations in x and y where y is the left side of the base
sqrt(x²-169) = y
And, x² + 14.32² - 2x * 14.32* cos(104) = y²

Solve this I guess

Idk

[obtuse pebble]

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[hollow narwhal]

how should i start with this question?

[timid silo]

It's a little vague... There lacks a definition for which point is considered to come from which other point on the other graph

[weary aspen]

i think we need to see the graph

[timid silo]

Hopefully it makes it clearer

Although on seconde thought I doubt there is a graph

I think you're just meant to work with any arbitrary function f

[weary aspen]

A.y=f(2x) is what hes given tho?

[timid silo]

||(she)|| Yes...?

[weary aspen]

my bad i wasn't paying attention to that

my apologies

[hollow narwhal]

no graph unfortunately

[timid silo]

I would assume (2a,b) is the answer, but (a,b/2) seems equally valid... Hmmm...

Do you have any way of asking your teacher for clarification? (I know we're supposed to be the ones to clarify things but the question is really vague here)

[hollow narwhal]

nopee

its the weekends

would it help if i showed another question

[timid silo]

Sure

[timid silo]

Yeah I think we're just meant to go with intuition

[timid silo]

Also I just realised it would be (a/2,b) and (a,2b)

[hollow narwhal]

oh

[timid silo]

So uh hmmm

Yeah intuition is probably what they expected us to do

[hollow narwhal]

okk

ima save that question for last

i have another burning question

[timid silo]

For example, given y=f(2x), what would you intuitively change about (a,b) to make sure that equation correct?

[timid silo]

I'm all ears... Or all eyes?

[hollow narwhal]

hmmm

[hollow narwhal]

im not sure i understand the way you worded that question

but

for this question

the inverse is kinda throwing me off

after reflecting about y=x should i be applying a vertical stretch?

[timid silo]

I'm trying to think about this intuitively but it takes a lot of brainpower... I think rewriting it as y=f^-1(x)/2 makes it simpler

Because then you can easily see that you should flip the function and squish it down vertically by 2

[hollow narwhal]

sooo it is a vertical stretch

[timid silo]

Yeah

[hollow narwhal]

it squishes down to the x axis?

[timid silo]

If that's your way of saying it gets squished vertically, then yes

[hollow narwhal]

okok

thank you

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

how can i solve algebraically

[royal basin]

did you forget the word "lim" there

[timid silo]

yes i did

[royal basin]

multiply numerator and denominator by x^-2, expand, observe you now have constant+null in both

[timid silo]

why x^-2

[tardy epoch]

It's the highest power in the denominator

[royal basin]

incorrect.

on two accounts.

one, you did not multiply the entire num and the entire denom by x^-2, and only multiplied one term in each, selectively.

two, the product of x^2 and x^-2 isn't 0.

[timid silo]

it is 1 and -1

so what is multiplying by x^-2 doing

[tardy epoch]

No?

When you multiply numbers with the same base, you add their exponents

[royal basin]

what's "it"?

[timid silo]

x^0 is 1 is it not?

-x^0 would be -1

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

.close

[obtuse pebble]

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[timid silo]

please help

[timid silo]

tag me when responding

[keen vapor]

How would you write this in exponent form?

[lost tree]

Isn't this already ur channel dxvid

[timid silo]

i rewrote it as 1/(x-sqrtx)

[keen vapor]

but thats not exponent form isnt it?

[timid silo]

it would be (x^-1) - (x^(-1/2))

and i find the integral of each of those

so it would be lnx - 2sqrt(x) wouldnt it

[viral blade]

$\f 1{a+b}\neq \f 1a +\f 1b$

No

[keen vapor]

hold on

[supple granite]

Not what I got

[lost tree]

A substitution might help

[keen vapor]

ik i just realized

[warm shale]

monikanicity

[timid silo]

oh

[viral blade]

same thing for minus

[timid silo]

so how could i do it

[viral blade]

no clue

[supple granite]

That's not equal to 1/(x-sqrtx)

[lost tree]

U = sqrt(x)

[timid silo]

ok let me try

[supple granite]

Or expand the fraction by 1/sqrtx and substitute u=sqrtx-1

[timid silo]

would i substitute it when its in exponential form?

or the original form?

[lost tree]

Do u mean with sqrt(x) written as x^½ ?

It doesn't matter how you do it

[timid silo]

when its x^(-1/2)

[lost tree]

Yeah well it doesn't matter how it is written

You can just substitute it in the original

[timid silo]

ok

[timid silo]

wait so then dx = du / (1/2sqrtx)?

[lost tree]

$dx=\frac{du}{\frac{1}{2\sqrt{x}}}$

[warm shale]

𝕾ilver𝕾oldier

[timid silo]

yes s

so now what

hi

[lost tree]

Hi r u new

Do u have a question?

Okay please read #❓how-to-get-help

[timid silo]

what do i do after?

[lost tree]

Well what do y get after making the substitution

[timid silo]

1/(x-u) du/(1/2sqrtx)

[lost tree]

Write all the x's in terms of u

[timid silo]

how

[lost tree]

u = sqrt(x), then what is x?

[timid silo]

idk

u^2?

[lost tree]

Yes

X = u²

[timid silo]

i still dont know where to go after

the du is confusing me

@lost tree

[lost tree]

Why the du is just like the dx u have here

[timid silo]

yes but it is du / (1/2sqrtx)

[lost tree]

Right you can write that as du * (1/(½))

And just move the du to the very end

[timid silo]

wuldnt it be 1/x^1/2

[lost tree]

U can simplify that just like how u wud something like $\frac{2}{\frac{5}{3}}$ or whatever

[warm shale]

𝕾ilver𝕾oldier

[timid silo]

i forgot how to do that lol

[lost tree]

As I said write everything in terms of u

[lost tree]

When u divide by a fraction, you multiply by its reciprocal

[timid silo]

ah ok

yeah im super confused

are u able to write it all out for me

[lost tree]

2 ÷ (5/3) = 2 multiplied by reciprocal of (5/3)

[timid silo]

let me send what i have so far

[lost tree]

Do u know what the reciprocal of 5/3 is

[timid silo]

yes 3/5

question 6

[lost tree]

Okay sqrt(x) is not sqrt(u)

[timid silo]

oh yeah

so its 1/ 1/2u

right

[lost tree]

Yes

[lost tree]

So here u got 1 ÷ (1/2u)

That is equal to 1 × reciprocal of (1/2u)

[timid silo]

yes so it becomes 2u

[lost tree]

Yes

[timid silo]

so then the u cancels out

and its 2/u^2

du

[lost tree]

Uhm

What u get is $\frac{1}{u^2-u}\cdot2u$

[warm shale]

𝕾ilver𝕾oldier

[timid silo]

yes

so it is basically 2u / (u^2 - u)

[lost tree]

Yes

[timid silo]

then the u cancels

[lost tree]

Yes..

And it doesn't turn into 2/u² when u do that

[timid silo]

2/u^2 - 1

[lost tree]

U can't do $\frac{2\cancel{u}}{u^2-\cancel{u}}$

[warm shale]

𝕾ilver𝕾oldier

[timid silo]

oh im forgetting so many basics

so what can u do

[lost tree]

If u r gonna cancel something like that, it must be a factor of both the numerator and denominator

[timid silo]

ohh i see

[lost tree]

Try factorising the denominator

[timid silo]

got it

ok i got it thank u so much

ill let u know if u need more help

@lost tree hey sorry this is just really confusing to me

so i rewrote it as 1/(e^z - e^-z)

could i do usub again?

[tardy epoch]

Why do you have a minus sign

[timid silo]

i mean plus

[timid silo]

accident

[tardy epoch]

Multiply top and bottom by e^z and try u sub

[timid silo]

would u be e^z?

[tardy epoch]

Yup

[timid silo]

so now the top e^z cancels out

now whats left is 1/(u^2 + e^-z)

oh then whats left?

@tardy epoch

[frosty spoke]

multiply both the top and the bottom by e^z

show what you get

[timid silo]

i got e^z / ((e^z)^2 + e^-z)

[tardy epoch]

(a+b) c = ab + bc

[tardy epoch]

Use this in the denominator

[timid silo]

how?

ill just come back to this question

[tardy epoch]

You didn't multiply e^(-z) term

You did (a+b)c = ac + b

[obtuse pebble]

@timid silo Has your question been resolved?

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[cloud nimbus]

!claim

[cloud nimbus]

how would i do this

[timid silo]

Slope of tangent to parabola is
2x+ a

Now equate it with 1

[cloud nimbus]

where did u get 2x+a

[timid silo]

Dy/dx

[cloud nimbus]

we havent learnt that yet

[timid silo]

Ok

So wait

Do u know the condition of tangency in parabola

[cloud nimbus]

the way the textbook wants us to do it, is through letting them equal each other, and then using the discriminant

[cloud nimbus]

nope

so like x^2 +ax+1-x=0

and then using the discriminant i believe

[timid silo]

So may be they want u to do by putting y = x in parabola equation

Than using the quadratic formula

[cloud nimbus]

but theres 2 varibles

[timid silo]

Of satisfying

D greater or equal to 0

[cloud nimbus]

so would the c value be 1-x?

[timid silo]

It will give u inequality in a

[timid silo]

Make equation in x²

X² + x(a-1) +1

This

[cloud nimbus]

how did u get x(a-1)

[timid silo]

By taking x common

And see

It is tangent so it must have

1 root

So just keep D equal to 0

(a-1)² - 4( 1)(1) =0

,

[cloud nimbus]

ahhh ok

got it thanks!

[timid silo]

What's ur standard?

Like class

[cloud nimbus]

its australian

but yr11 math methods

which is like adv math

[timid silo]

U are 11 yr

?

[cloud nimbus]

yep

[timid silo]

Noice

I don't even know what is quadratic formula when I was 11

[cloud nimbus]

no haha, as in year 11

im 16

[timid silo]

Ohh

Age

[cloud nimbus]

.close

[obtuse pebble]

Channel closed

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[normal breach]

Can someone help me find the standard deviation of x here? I know the formula, but I don't know how to calculate it with the given information I have

Think I got it figured out

.close

[obtuse pebble]

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[timid silo]

hello : )

[past walrus]

is this the answer?

[timid silo]

well, rationalising it should lead to the answer 2-root 3

would you like to tell me how you have done it?

[past walrus]

yea

[timid silo]

rationalizing your answer would give us...

5+3 root 3

so there's something wrong with your process and/or calculation

how have you solved this question?

[past walrus]

I used the tan(a)+tan(b)/1-tan(a)tan(b)

equation

[timid silo]

right

and tan pi/6 is tan 30

and tan pi/4 is tan 45

so i hope you substituted the correct values?

[past walrus]

I used the sines and cosines of those angles

[timid silo]

ah, it might've made your calculations a bit more difficult

well, tan 30 is 1/root 3 and tan 45 is 1

[past walrus]

i re-did it and got -1 this time

[timid silo]

💀

tell me each step please

there is no way you can get -1 for this question (´。＿。｀)

[past walrus]

question

how do u know tan30 is 1/root3?

[timid silo]

it's given in the trigonometry table's standard values

but either way tan can also be found by sin x/cos x

[past walrus]

well I mean I did the sin/cos and got root3/3

[timid silo]

well root 3/3 will also work

can you tell me your step by step procedure for solving this?

or send a picture?

[past walrus]

uh sure give me one sec

[timid silo]

it seems like you've gotten your concept down, so there might be some error with your calculations

[past walrus]

@timid silo

[timid silo]

yes, this is correct

[past walrus]

so can this be my final answer?

or does it have to be 2-root3?

[timid silo]

depends on your teacher, honesdtly

however, 3 and 3 can get cancelled

the 3 in the denominator

[obtuse pebble]

@past walrus Has your question been resolved?

[faint olive]

No, its not simplified

Mostly you will not get full marks on that

[past walrus]

so what will my answer be then

[faint olive]

@past walrus

[past walrus]

this is one of my teachers examples

would having it as 3-root3/3+roo3 also count?

[timid silo]

yeah this'll also work

your teacher didn't rationalise the denominator

[faint olive]

Weird teacher, but yes that works also

[timid silo]

but if they've given this as example , then it's fine

@faint olive old man, go study physics

[past walrus]

yea he said in the video, "there's not really anyway to simplify that anymore"

[faint olive]

Already doing that

[past walrus]

so ig its fine

[faint olive]

Laserr

[faint olive]

I am a multi-tasker

[faint olive]

Maybe

.close

[obtuse pebble]

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[upper hinge]

i have no clue as to how to start this

[thick gyro]

Write out PR @upper hinge

[obtuse pebble]

@upper hinge Has your question been resolved?

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[still badger]

Let S be an ordered set. Let A be a nonempty finite subset of S. Then A is bounded. Prove that inf A and sup A exists and are both in A. Hint: Use induction

[still badger]

Welp

I tried my best

Or lets just say i dont even know how to get started eh

[obtuse pebble]

@still badger Has your question been resolved?

[kind hawk]

they gave the hint to use induction

have you tried doing a base case

[still badger]

Base case as in which case

Im sry i just dont understand how to implement induction on such a thing

[kind hawk]

let the size of A be n

we want to do induction over n

what could the base case be

[still badger]

Then? Say size of A is 1. The inf and sup will be the only element in A? Smth like tat?

Sry for the delay btw

[kind hawk]

yes

so now imagine you have a set A of size n+1

how can you use the induction hypothesis which says that for a set of size n the inf and sup exist

[still badger]

So maybe i can use contradiction?

Ya tat makes a bit sense actually

Btw for the base case should i use size of 1 or size of 2

[still badger]

!close

.close

[obtuse pebble]

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[molten moon]

I need to get the value of the side of the square

[molten moon]

It was on my math competition and everyone is wonder how it is possible

[timid silo]

can you get a clearer image

we can try drawing hypotenuses of the triangles

or the diagonal of the square

[molten moon]

Solving it graphically was 1 point solving with equations was 5

[timid silo]

let me try it

[median dome]

draw the diagonal

of the square

[molten moon]

And what then

[median dome]

use pythagoras

[molten moon]

I need the length of the diagonal

[median dome]

yes

draw it once

and send the image

[molten moon]

I dont really want to draw on the paper

Ur not supposed to do that

[median dome]

well then draw somewhere else idk

[molten moon]

For the com

[timid silo]

what ?

wdym

[molten moon]

I need to solve it with equations

[timid silo]

well youre supposed to draw the diagrams somewhere if you wanna be able to solve problems like those

[timid silo]

even for that you need a diagram to make equations using pythagoras theorem

[molten moon]

Aight ill make a copy of the paper

[timid silo]

nice

@median dome do we bisect the base of those two triangles ?

[median dome]

what

[timid silo]

with the diagonal

[median dome]

do you mean the side with length 3?

no i dont think it will bisect

[timid silo]

yes

[obtuse pebble]

@molten moon Has your question been resolved?

[molten moon]

Diagonals

[molten moon]

<@&286206848099549185>

[dark stirrup]

Are you given the value of those angles? Or any other info? Otherwise, this can't be solved

[molten moon]

Value of angles in the middle is 90° both

The dot in the middle of the angle is kind of lost on the photocopy

[dark stirrup]

Oh both interior angles are 90?

[molten moon]

Yes

Scroll up there is a photo without diagonals drawn

[dark stirrup]

Such an annoying drawing, right angles are usually draw square instead of rounded

[molten moon]

In my country they teach it round and a dot inside

[dark stirrup]

Fair enough

Okay

Much easier problem now

You were already given answer, so I'll just explain a solution

Rotate the square so 9 and 12 are parallel to x axis.

Then your path goes 9 right, 3 down, 12 right

In total, 21 right and 3 down

[molten moon]

Yes

[dark stirrup]

,w 21^2+3^2

[warm shale]

Results provided by WolframAlpha

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Requested by SWR#8588

[molten moon]

I just got it

Jesus

Im stupis

[dark stirrup]

Cool

All good

I was having more trouble because I did not know they were right angles

But good work

[molten moon]

Jesus christ

How did i not see that

[dark stirrup]

x=15, right?

[molten moon]

I was looking for half the time of the comp

Yes i think so

[dark stirrup]

Cool

Well yeah, hope that change in perspective helped

Glad I could help

[molten moon]

.close

[obtuse pebble]

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[halcyon owl]

explain this please

[visual apex]

What he just solved for e^k

[ruby path]

Divide 1800 on both sides, and take the 8th root on both sides

[visual apex]

Btw where do I get help

[ruby path]

Go to #❓how-to-get-help

[halcyon owl]

gotcha

.close

[obtuse pebble]

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[radiant relic]

Hey I've been trying to solve 3 a and 3 b, but I seem to be getting the wrong answer

[radiant relic]

I know I need to differentiate and sub in 0 but it doesn't give the right answer

[median dome]

,rccw

[median dome]

what are you getting for your derivative?

[radiant relic]

Sin(2) - 3t^2 + 8t

[median dome]

aha

sin(2)?

use the chain rule to differentiate sin(2x)

[radiant relic]

Chain rule?

[median dome]

yes

do you know what it is?

[radiant relic]

I don't think I've come across it

[median dome]

,w chain rule

[thick gyro]

you might have missed a lesson then

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by kheerii#8421

[median dome]

but

you should watch a video or something

to understand it properly

[radiant relic]

Yeah

[radiant relic]

No clue what that means haha

I think we just haven't covered it yet, because we have 1 teacher for stats and mechanics, and 1 for pure so the lessons are prlerfectly in sync

And this is a much later chapter to the pure stuff

[median dome]

https://www.youtube.com/watch?v=0T0QrHO56qg

[radiant relic]

Alright I'll take a look, thanks!

[median dome]

no worries

[median dome]

jbtw the derivative will be 2cos(2t) - 3t^2 + 8t

[radiant relic]

.close

[obtuse pebble]

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[blazing oyster]

guys can you arrange this equation so i can get an f(x)

[bold bane]

,rotate

[obtuse pebble]

@blazing oyster Has your question been resolved?

[obtuse pebble]

@blazing oyster Has your question been resolved?

[blazing oyster]

No sir it hasn't been resolved

[obtuse pebble]

@blazing oyster Has your question been resolved?

[obtuse pebble]

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[rich basin]

hello

[rich basin]

i am stuck on a word problem that requires me to use algebra

here is the question:
A hospital has a childrens’ A&E and an adult A&E. Over the weekend, in total, there were 186 people admitted to A&E. The mean age of the children admitted was 12, and the mean age of adults admitted was 50. The mean age of people admitted was 32,425. How many children were admitted to A&E?

i understand that one would have to do x(children) + y(adults) = 186 but from here on out I have no clue to how to solve it

[fierce lagoon]

You need to create a system of equations

if x is the number of children and y the number of adults then one equation is x + y = 186

But now you need to use info about their age to create the second

[rich basin]

yeah, that is where i am stuck

i dont know how to write an equation for the mean age of people

[fierce lagoon]

Think about it for a bit

How the number of people and their ages may affect the total age

[rich basin]

i think that the second equation would be something like:
...=32,435

i tried understanding the basics of a mean but cannot create the equation

[fierce lagoon]

Well don't worry about the mean

Just assume every child is 12 years of age

And every adult is 50 years of age

[rich basin]

alright let me think

could you give me a hint perhaps?

[fierce lagoon]

The number of children times their mean age equals to the total number of children's age

[rich basin]

so the same for adults then 50x + 12y = 32,425

?

.close

[obtuse pebble]

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[latent jasper]

I need to check if it is convergence or divergence $\int_{1}^{3} \frac1{x-1}dx$. I've done it but need to confirm if is right

[warm shale]

afeAlway

[latent jasper]

$\int_{1}^{3} \frac1{x-1}dx = ..... = \lim_{c \to 1^{+}} ln (3-1) - ln(c-1) = \lim_{c \to 1^{+}} ln (2) - ln(c-1) = ln(2) - ln(1^{+} - 1) = ln2 + \infty = \infty$

[warm shale]

afeAlway

[latent jasper]

So basically it is divergence

Am I right?

[obtuse pebble]

@latent jasper Has your question been resolved?

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[obtuse pebble]

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[mossy minnow]

How does one determine if an equation of this form has integer solutions and if so, how many?(x and e are both integers)

[mossy minnow]

Tried looking online but found nothing on something like this