help-27

[amber valley]

That question?

[whole grove]

yeah

What is the probability that 6 is the highest number showing?

but now this one

same info as before diff question

[amber valley]

Then yes, just the prob of getting at least one 6

[whole grove]

oh wait

at least one

or exactly one

because those are diff

[amber valley]

At least, if you get two the highest is still a 6

[whole grove]

oh u right

i almost sold

lol

thanks

[amber valley]

[whole grove]

.close

[devout snow]

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[blazing patrol]

I'm working on a school paper about birth control methods, and I'm discussing the possibility that combining two less effective methods might be more effective than using just one. In this context, I'm wondering how to calculate the combined effectiveness of two percentages. For example, let's say Method A has a 92% effectiveness rate and Method B has a 70% effectiveness rate. How would I combine these two percentages to find the overall effectiveness of using both methods together?

[tough storm]

1-0.92

1-0.70

multiply the answer

then do 1-ans

[dawn scarab]

i believe you can just keep the original numbers and multiply

rather than subtract 1 at the end

correct me if im wrong though

[blazing patrol]

So the answer would be: 0,08*0,30 = 0,024
1 - 0,024 = 0,976

97,6%

[tough storm]

yup

there might be a simpler method but it hasn’t failed me yet

[blazing patrol]

That is so helpful, thank you very much

[dawn scarab]

sorry, but to my understanding, the effectiveness should be lower than either of the ones provided

at least from a conceptual basis, if you have 92%, and 70%, it should be something lower than 92

if you combined a procedure with a 100% success rate, then did one with a 70% success rate right after, you have a lower
chance versus 100 and 100

[blazing patrol]

I don’t know how to explain it in math linguistics but couldn’t be lower than the efficiency of one of them because one’s efficiency does not depend on the other.

[dawn scarab]

i guess this comes down to how you define "using both methods together?"

[tough storm]

no it would still be 100%

he’s not putting the tests into a solution

they’re separate

[dawn scarab]

then we're assuming they are independent events

got it

you can just ignore the above then

[blazing patrol]

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[surreal prism]

how does x= 1 +lambda make lambda = x-1

shouldn't it be 1-x?

or is the idea that we try to keep x positive

but then how does z = lambda = z+5/-1

[spark sleet]

If you do -1 on each side, you get lambda = x - 1

[devout snow]

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[thin pagoda]

ALLAHUAKBAR

[surreal prism]

so I need to find the direction vectors of 4x+3y+6z=12

[thin pagoda]

Use chatgpt

[surreal prism]

It's not always correct

[thin pagoda]

Email ur teacher

[surreal prism]

anyways i got to the point of x=12-3y-6z/4 y=12-3y-4x/6 and z=12-3y-4x/6

they're out

[spare crypt]

I'd probably have to see all the options but you get take differences between the 3 points to get some parallel vectors

[devout snow]

@surreal prism Has your question been resolved?

[surreal prism]

theres are all the options

[spare crypt]

yea those are correct, the others are additions so they'll definitely not be parallel

[devout snow]

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[brittle adder]

Hi struggling with this question:
A surface z(x,y) is defined by the relationship xz+z^2=x^2+2xy+y^2
Ive found the tangent plane equation to the surface point (5,1,4) to be 8x+12y-13z=0
Use the linear approximation to estimate the value of z when x=5.1 and y=1.05 <-- this is the part I need help with.

[devout snow]

@brittle adder Has your question been resolved?

[robust dust]

it just means plug in x=5.1 and y=1.05 into the plane

and get the corresponding value of z

the tangent plane is precisely the linear approximation of the function at a point

[brittle adder]

oh my gosh thanks! I was definitely way over thinking it and going in circles

.close

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[halcyon kernel]

How can i find a and x values?

[rocky sequoia]

you're not finding the x value, but for a can you see the pattern

[halcyon kernel]

ia it 5?

is*

[rocky sequoia]

not quite but on the right track

you see how in the pattern x has 'no coefficient' (coefficient of 1)

[halcyon kernel]

yeah so i need to take out 8

[rocky sequoia]

try to manipulate the numbers such that x^2 term has a coefficient of 1

[rocky sequoia]

yh that too

[halcyon kernel]

but that wouldnt change the bottom term at all

[rocky sequoia]

yes you would have to factorise for the bottom

[halcyon kernel]

alright i will give it another shot

hmm

not sure where it went wrong

.close

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[stark shale]

the solution space of a system of linear equations is preserved under linear combinations.
Is this something that emerges and I just have to accept it? i'm trying to understand this at a more intuitive level for my brain.

[stark shale]

geometryically, why would that be? it's not obvious, right?

[graceful cosmos]

Let's say we have two solutions, x and y.

That is, they obey Ax = b, and Ay = b

Wait, is this a true/false question?

Because A(x + y)
= Ax + Ay
= 2b
Isn't a solution

[stark shale]

mmmm what?

i believe in your example A is a matric and x is a vector

x is the solution, with components x1, x2

i dont think you have 2 solutions x and y in the way you wrote it... thosre are 2 different equations

it's not a true or false. have you just read the first line by any chance?

[devout snow]

@stark shale Has your question been resolved?

[graceful cosmos]

x and y are two different vectors in the solution space. However, x + y isn't, implying the solution space isn't closed under linear combinations

If b were 0, however...

[stark shale]

what i mean is

a linear combination of 2 lines doesn't change the solution of the 2 lines

it will be another line that passes through the intersection point

[devout snow]

@stark shale Has your question been resolved?

[graceful cosmos]

Did you still have a question?

[stark shale]

yea

[devout snow]

@stark shale Has your question been resolved?

[stone stump]

your question isnt quite clear

do you mean why we can do row operations do solve a linear system?

[stark shale]

mmmm, i understand doing row operations is like orienting the planes so that the intersection point is clear

but why, when we do linear combinations , the solution doesn't change?

@stone stump

[stone stump]

I dont understand what you mean by linear combinations

if not row operations

[stark shale]

i mean **that **,yes

that if i add 2 vectors, the 3rd vectors will span the same space of the other 2 vectors. the rank doesn't change. i'm not adding a new dimension. the 3rd vector is linearly dependent with other two

c1*vector1 + c2*vector2

@stone stump

I'm not saying it's not row operations, i'm saying my question is different than

why we can do row operations do solve a linear system?

[stone stump]

but which vectors are you trying to add

[stark shale]

i'm not sure, what kind of vectors can i choose from?

[stone stump]

well you are the one trying something so you should tell me what you are trying

[stark shale]

are you trolling me ? 😄

from another perspective but same question

https://www.desmos.com/calculator/pcu24pz62s

why if i change the parameter "a" the intersection doesnt change?

[stone stump]

well its a row operation

you are adding a multiply of the first row to the second row

[stark shale]

so, why this happens?

i'm looking for a geometric intuition if possible

[stone stump]

I cant offer you a geometric intuition. the nice thing about algebra is that we can leave the geometry behind and just think about the equations

and for those it is very clear that row operations are allowed

[stark shale]

ok, good to know

thank you! @stone stump

.close

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[modest eagle]

How could i integrate this

[devout snow]

Please don't occupy multiple help channels.

[sullen urchin]

split the equation into 3 parts

int(dx/x^5) + int(5^2-x*25^xdx) + int(dx/2)

[modest eagle]

ok i see

[sullen urchin]

then 1st and 3rd are prolly easier than 2nd

[modest eagle]

yeah how can i integrate the 2nd?

thats basically my doubt

[sullen urchin]

in 2nd just use u.v formula for integrals

[modest eagle]

ohh

i see thank youu

you mind if i ask you something else?

[sullen urchin]

int(uvdx) = u int(vdx) - int(du/dx*int(vdx)dx)

[modest eagle]

this one what method should i follow?

i think i can separate t

[sullen urchin]

yeah thats by substitution, you can do that as well

[modest eagle]

okayy i see

[sullen urchin]

but remember to take ln as u and not as v in this eqn

[modest eagle]

oh ok

[sullen urchin]

else it'll not make a complete eqn

[modest eagle]

it only happens

with ln

[sullen urchin]

yeah there's a series followed for determining u and v

its ILATE

short for Inverse Trigno

Logarithm

Algerbriac

[sullen urchin]

Trigno

[modest eagle]

this one

[sullen urchin]

yeah

[modest eagle]

ohhh

great

in my exam it will only have inverse and logs

so chill

[sullen urchin]

good

[modest eagle]

well thank you very much for the help really appreciate

[sullen urchin]

no problem

[modest eagle]

.close

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[buoyant path]

Hello, i cant get near the answer at this equasion, my brain just blocked it self. Can somebody help

[shrewd thistle]

How familiar are you with your order of operations?

[buoyant path]

How shouod I answer at this question, I know order

[restive river]

what is (a-b)^2

[shrewd thistle]

Well, in that case, start with the first item in the order

if possible

[buoyant path]

You can do this
(A-B) × (A-B) = A² -ab +ab + b²

[shrewd thistle]

since it's not, move on to the second

[buoyant path]

-ab

My bad

I need to start with (3√2-4)²

Am I right?

[shrewd thistle]

daniel's right, yeah

[buoyant path]

I got lost guys

[shrewd thistle]

What have you got so far?

[buoyant path]

(3√2-4) (3√2-4) = 9×2

To here

Yea, pr9babli is

Probably*

[shrewd thistle]

Nope.. wrong channel

you're good so far

But if you're getting confused by the square root, just change it to "x" for the time being

[buoyant path]

Wher do I need to go which channel

[shrewd thistle]

instead of 3sqrt(2), just write 3x

Remember your substitution, though.. we'll have to convert it back at the end

[buoyant path]

Sure

[shrewd thistle]

No, I'm trying to multitask and it's not working - you're good so far

[buoyant path]

Okay okay

[shrewd thistle]

So you've gotten this far

What's next?

[buoyant path]

Now I got this far 9x-24x+16

[shrewd thistle]

Good... let's leave that alone for a moment and deal with the sqrt(32)

[buoyant path]

Sure

[shrewd thistle]

Is 32 a perfect square?

[buoyant path]

No it isnt

[shrewd thistle]

Ok.. are any of its factors perfect squares?

[buoyant path]

We can use 4, because 32:4 is 8

Am I right?

[shrewd thistle]

Cool, you're already on the next step.. yeah - what would that look like?

[buoyant path]

√8*4

[shrewd thistle]

not quite

[buoyant path]

We can root 4

[shrewd thistle]

We can do better than root 4

[buoyant path]

6

[shrewd thistle]

isn't a factor of 32

[buoyant path]

16?

16*2 is 32

[shrewd thistle]

There it is!

[buoyant path]

√16*2

[shrewd thistle]

Now.. you can't do this across addition or subtraction signs, but we can do this across multiplication and division..

[buoyant path]

Yes

[shrewd thistle]

Take the square root of 16 and multiply it by the square root of 2

That allows you to take the 4 outside the radical

And hey, since sqrt(2) = x, let's just change this one while we're at it

[buoyant path]

Oke

[shrewd thistle]

What do we do next?

[buoyant path]

We need to add √2

[shrewd thistle]

You're missing something

[buoyant path]

+24√2?

[buoyant path]

Oh yea

[shrewd thistle]

What happens when we square √2?

[buoyant path]

We got 1.4

[shrewd thistle]

nope.. that's just √2

what is (√2)^2

[buoyant path]

Two

[shrewd thistle]

Bingo

[buoyant path]

9 * 2 = 18

[shrewd thistle]

Yep.. keep going

[buoyant path]

18+16-20√2
34-20√2

[shrewd thistle]

Now then.. depending on what your instructor wants, we're either done, or there's one more thing we can do

Are you supposed to use a calculator and get a decimal value for square roots?

Or are you supposed to give an exact answer?

[buoyant path]

Mi friend got it like this

[buoyant path]

Not really

[shrewd thistle]

Your friend didn't properly FOIL out (3(√2) - 4)

your friend just immediately applied the exponent.. and that's incorrect

[buoyant path]

Oke, I will tel him

On photomath it tels me again other answer

Oh no
My bad

[shrewd thistle]

looked like a *2 instead of a ^2

[elder wasp]

Plugged into mathway and got the answer

Nawh but you gotta foil it first

Then simplify

[buoyant path]

Sory for suspecting in you @shrewd thistle

[elder wasp]

The rest is easy

[buoyant path]

I didnt understand rest, but john tought me

[elder wasp]

It comes together quite nicely

[shrewd thistle]

Nah, you're good

[buoyant path]

Thanks

[elder wasp]

With the square root of 32 being 4_/2 and just adding it to 24_/2

[shrewd thistle]

lame questions at the start are lame, because I just don't know what you know

[elder wasp]

_/

4_/2

Discord is so weird

[shrewd thistle]

If you're on PC, learn the joys of charmap

[elder wasp]

Cellphone

Not

Ok

[shrewd thistle]

Sad times

[elder wasp]

Nawh i got pc

But mah dad usin

[buoyant path]

Should I close this now?

[elder wasp]

If you have the answer

[shrewd thistle]

If you're good with that problem, yes

[buoyant path]

I wil, thanks again John

.close

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[buoyant path]

.reopen

[devout snow]

✅

[buoyant path]

Oh no one mor equasi9n

(4√3 +2√8)² - (3√12 -5√2

)*

[sullen urchin]

simplify the square, using (a+b) ^2 = a^2+b^2 +2ab

then it should be good enough to act like an algebraic eqn

[buoyant path]

Tnks i'll be back if smth i'll not understand

When i get (4√3+2√8)² can i do this

(4√3 + 4√2)²

16×3 + 16×2

And at the end can I multiply √3 and √2 to get √6 and ut in i to the equasion?

Is ths right

@shrewd thistle

[devout snow]

@buoyant path Has your question been resolved?

[devout snow]

@buoyant path Has your question been resolved?

[devout snow]

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[vernal grail]

Hi, i am having problem with this ... Two masses $m_1 = 5kg$ and $m_2 = 10kg$ are connected as in the figure. The plane, inclined by $\alpha = 30$°, is rough with static friction coefficients $\mu_s = 0.5$ and dynamic $\mu_d = 0.3$. Determine whether the two masses, initially at rest, move and, if so, with what acceleration.

[woven radish]

alee

[vernal grail]

this is the figure

[ocean yacht]

draw the fbd

[vernal grail]

ok

this is correct?

[lost laurel]

what about the frictional force?

[vernal grail]

@lost laurel

like this?

[ocean yacht]

why is the normal applied to m2?

[vernal grail]

i think i fixed

[ocean yacht]

alr now we just need the frictional forces

use

fs = mu(N)

[vernal grail]

what is fs?

[ocean yacht]

static friction

[vernal grail]

I was thinking that on $m_2$, on the x axis there are no forces, while on the y axis we could say that $m_2g - T = 0$

[woven radish]

alee

[devout snow]

@vernal grail Has your question been resolved?

[devout snow]

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[onyx rune]

what have you tried?

[jagged bone]

factor

but doesnt lead anywhere

[onyx rune]

try divide numerator and denominator by x^2

[jagged bone]

then plug in 0?

will give 0

[onyx rune]

yes

no

it wont give 0

[jagged bone]

or 1

1

[onyx rune]

no

it gives -7/9

after plugging 0

[jagged bone]

wow i wrote 5x-7x

[onyx rune]

what did you get after doing this?

[jagged bone]

i get it

[onyx rune]

ok

[jagged bone]

ty

.close

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[willow quartz]

how can I isolate t to just the right side?

[willow quartz]

like why cant I do this?

[hasty saffron]

1. Complete the square
2. How did you get that

[willow quartz]

I just divided the 105 byt 5.1 and -4.9

[noble ridge]

that would also divide the -4.9 by 5.1 and vice versa

[willow quartz]

so how would I do the division then

[hasty saffron]

That's just -105/0.2 on the left.
Which gives -105=0.2(t+t²), which changes the given question

[willow quartz]

yes I understand that but I dont understand what im supposed to do instead

[smoky nimbus]

You don't need to, just move all the terms to one side and it's a quadratic

[hasty saffron]

Either use the quadratic formula or try to complete the square

[smoky nimbus]

Where you can factor or using any other method you know

[willow quartz]

yeah but like if I wanted to divide without factoring

is there a way to do that?

[smoky nimbus]

Not as easily with quadratics

The best and easiest is factoring

[willow quartz]

ok

well thanks

im gonna factor it

.close

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[dim horizon]

I need help with this question

I will post my work momentarily

I am stuck here

[scarlet sequoia]

you should do x^2(x+1)

and then

A/x + B/x^2 + C/(x+1)

[dim horizon]

Let me try

[dim horizon]

How do you get this

A/x+1 + Bx+C/x^2

why would it not be the above instead?

[scarlet sequoia]

it could be A/(x+1) + B/x + (Cx+D)/x^2 but C would be zero anyway

if you have a factors like x^2, x^3, ..., x^n you simply do it the way I did

[dim horizon]

so x^2 is not an irreducible quadratic factor?

it's considered linear?

[scarlet sequoia]

if it was e.g. (x + 1)^2 then yours would be necessary

[dim horizon]

I'm getting confused here, is x^2 considered a linear factor?

how were you able to write it as simply B/x^2

[scarlet sequoia]

it's reducible though x^2 = x * x

I mean factorable

[dim horizon]

Ok, so there are two buckets, linear and quadratic factors, but if it's not irreducible quadratic, its linear (technically)

I see what you mean

[scarlet sequoia]

Your strategy is used when we deal with terms of the form (ax^2 + bx + c)^m and it cannot be factored further

e.g. x^2 + 2x + 2

then you'd write (Ax+B)/(x^2 + 2x + 2)

[dim horizon]

Ok sure, so I get that you would have B/x^2 + C/x+1 now

But where does the A/x come from

we only have two factors right?

[scarlet sequoia]

yes but when you do PFD and you have factor of the form (ax^2 + bx + c)^m or (ax+b)^m you do:

A1/(ax+b) + A2/(ax+b)^2 + A3/(ax+b)^3 + ... + A_(m-1)/(ax+b)^(m-1) + A_m/(ax+b)^m

(we often use next letters of alphabet instead of A1, A2, A3 ...)

it's a rule

So if it was for example x^3

You would write down A/x + B/x^2 + C/x^3

etc.

[dim horizon]

mmmm

I see what you mean

[scarlet sequoia]

Same for (ax^2 + bx + c)^m

[dim horizon]

if it was just x we could leave it alone

but because its x^2 we need A/x + A/x^2

and so on and so forth to degree n

[scarlet sequoia]

yes, you can see that the power tells us how many terms we should have for that factor

x^2 --> power is 2 ---> A/x + B/x^2 ---> 2 terms for this factor

[scarlet sequoia]

exactly

[dim horizon]

alright I'll give it another go with this in mind

[scarlet sequoia]

And always try to factor a denominator completely

[dim horizon]

I did the algebra but something seems off

[scarlet sequoia]

it should be Ax(x+1) + B(x+1) + Cx^2

remember you should multiply both sides by x^3 + x^2 = x^2(x+1)

Seems that you multiplied them by x * x^2 * (x+1)

[dim horizon]

I see my mistake

I thought it was missing x^2 as a factor but it already has x, so just x was needed not x^2

thanks for the help, I managed to solve out the problem

I think I understand it a little better

.close

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[devout snow]

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[boreal basin]

hi
when i have The Comparison Theorem for Improper Integrals at 2 ponits should i found 2 fcts that i know that they are conv or divergent?
and if both of those 2 functions are conv that mean my function is conv?

[boreal basin]

i'm try to solve + sin(x)/x^2 from 0 to inf using comparison theroem

[cinder bobcat]

you just need to find a function that is bigger than this one whose integral converges

[boreal basin]

what about 1/x^2

sin(x)/x^2 ?< 1/x^2
x^2*sin(x) ?< x^2
sin(x) ?< 1

we know that sin(x) <= 1
which mean that

sin(x)/x^2 < 1/x^2
since 1/x^2 is conv that mean that sin(x)/x^2 is also conv
like tihs?

[devout snow]

@boreal basin Has your question been resolved?

[cinder bobcat]

that would be almost perfect except 1/x^2 does not converge from 0 to infinity bc of the asymptote at x = 0

you could create a piece-wise function that covers that area if you wanted

[boreal basin]

Alright,
Also Just here, i think i can't do that right?
Since the equation is false

• I think there is error at the value of intrgral, i mean it's can't from 0,1 since it's undefined at 1 (it's will give me negative value for inside the square

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[devout snow]

Please don't occupy multiple help channels.

@hasty flame Has your question been resolved?

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[bright juniper]

In 2007 the first guy averaged to use 5mb of WAP GPRS and 30mb of internet GPRS. In 2008 the GPRS usage increased by 2x and WAP GPRS usage by 5x how many percent did internet GPRS usage increase

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[bright juniper]

<@&286206848099549185>

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@bright juniper Has your question been resolved?

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[jagged bone]

how does 35pi/6 equal tht

[crisp niche]

35 = 36 -1

[jagged bone]

what would i do with that info

[crisp niche]

do you know what a fraction is? and how to perform operations on fractions?

[jagged bone]

ik it equals 35pi/6 but how would i use 6pi-pi/6 to solve this

[crisp niche]

read the photo I gave you, there you can find formulae for such situations,

for sinus and cosinus

[crisp niche]

analogously, you may perform your calculations

[devout snow]

@jagged bone Has your question been resolved?

[jagged bone]

so i would end up with -1/2 / sqrt3/2 +1

all squared

[crisp niche]

correct)

[jagged bone]

1/7?

[crisp niche]

no

[jagged bone]

after squaring wouldnt it be 1/4 / 3/4+1

[crisp niche]

no

look:

try to continue what I ve written

[jagged bone]

-1/(sqrt3)/+1

then multpliy by itself

[devout snow]

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[jagged bone]

.reopen

[devout snow]

✅

[jagged bone]

1/3+4sqrt3 +4

which 1/7 4sqrt3

[crisp niche]

but now, you should apply conjugate of the denominator

since, it is common knowledge that we do not leave radical symbol in denominator

[jagged bone]

could we have just squared the fraction from the first step?

[crisp niche]

yes but all depends on fluence of calculations, if you're enough fluent then sure, i was doing it step by step

for you

we also use such method:

[jagged bone]

to rationalize multpliy top and bottom by 7-4sqrt3?

[crisp niche]

that can help you , to remove radical symbol in the denominator

yes

that precisely i want from you )

very well

[jagged bone]

7-4sqrt3 / 1

[crisp niche]

yes correct)

[devout snow]

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[icy zephyr]

no idea how to start this out

[dense jay]

which direction first

[icy zephyr]

wdym?

[dense jay]

its an iff
so youre proving f being surjective implies g is

and youre also then proving g being surjective implies f is

[icy zephyr]

right... ill be honest im very lost

[vale cipher]

So we have to prove it both ways, but how would we go about proving the first, being f being surjective implies g

or rather the other one, not sure which to begin with

me and @icy zephyr are working on the same problem btw @dense jay

[devout snow]

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[devout snow]

@icy zephyr Has your question been resolved?

[icy zephyr]

.close

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[restive river]

how do i do this?

[safe jasper]

uh two for slope are gradient and rate of change

i've always heard y-intercept called that but i think vertical intercept could work

[restive river]

okay!!

just need to figure out 100% for y-intercept now

[safe jasper]

try initial value?

do you get multiple attempts

[restive river]

yep that was it!!

[safe jasper]

😄

[restive river]

also this one as well

math makes my head hurt :((

[safe jasper]

what's the question with this one, i don't see it ask anything

[restive river]

oh oops one second

[safe jasper]

okay so for the rate of change for the current year, you'd find the slope between two points on the graph, by dividing the difference in their y values by the difference in their x values (rise over run). you could pick the points (0,0) and (2,16) for simplicity, so the slope would be (16 - 0) / (2 - 0) = 16 / 2 = 8

for the rate of change of the previous year, you do the same thing with two points, but this time your x values are in the left column and your y values are in the right column. taking the points (2,12) and (4,24), the rate of change is (24 - 12) / (4 - 2) = 12 / 2 = 6

and for part B, you know that in the current year each bushel costs 8 dollars and each bushel costs 6 dollars in the previous year, so a bushel of corn is 8 - 6 = 2 more dollars in the current year than in the previous year

[restive river]

woahh okay!! thank yiu so much

my math teacher isnt the best at explaining but that is explained really well

[devout snow]

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[main ravine]

i got x^3 - x

but aanswer key says am wron

[dreamy gorge]

can you send ur work

[lusty sapphire]

The problem is a definite integral, not indefinite

[main ravine]

dont i just substitute x in the antiderivative?

and ignore the 2

thats how i was taught it

[main ravine]

i was taught to just plugin x in the antiderivative

and ignore the 2

[lusty sapphire]

No you do not ignore the 2.

[main ravine]

also this , shouldnt in be D

[lusty sapphire]

The definite integral requires you consider the 2

[main ravine]

thats how i was taught it

i literally just plugin x

without doing anything

[lusty sapphire]

That's derivative of a definite integral

[main ravine]

wdym

[lusty sapphire]

This is a a derivative of a definite integral

And this is just a definite integral (no derivative included)

[main ravine]

oh

but that doesnt make sense.. how is it useful

to find the derivative

of an antiderivative

you basically get f'(x) again

[lusty sapphire]

It's one of the fundamental theorems of calculus and it can be useful in chain rule situations

for example $\frac{d}{dx}\left[\int_{e^x}^{x^3} \sin t^2 dt\right]$

[woven radish]

SWR

[main ravine]

3x^2 * sinx^3 - sin*e^2x?

[lusty sapphire]

close, but you seem to get the idea

[devout snow]

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[granite geode]

why was positive 15 used at start of numerator for the quadratic equation when its -15

[true oar]

first of all you dont need to use the quadratic formula to solve this

[granite geode]

oh?

[true oar]

you can use cross multiplication method

factors of 9: 1, 3, 9

factors of 4: 1, 2, 4

so o(3x-1)(3x-4)=0

so x=1/3, 4/3

[true oar]

quadratic formula is \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

help how do i yuse latex

[granite geode]

oh its - infront of b

i see

[true oar]

@woven radish \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

[granite geode]

😅

[true oar]

,help

[woven radish]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[granite geode]

its ok, i think i udnerstand now

Thanks!

[true oar]

,tex \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

,tex \frac{-b \pm \sqrt{b^{2}-4ac}}}{2a}

[woven radish]

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

[true oar]

,tex \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

[woven radish]

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

[true oar]

wtf

[granite geode]

XD

its ok haha

thanks for helping

[true oar]

yw

[granite geode]

its just the -b part that i forgot

[true oar]

dude i need help now lmao

why is latex doing this

[granite geode]

so its basically -(-15) yea?

[true oar]

yes

,tex \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

[granite geode]

awesome

[woven radish]

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

[dusky void]

wow

[true oar]

,tex \frac{a}{b}

[woven radish]

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

[true oar]

okay the frrac is doing weird things

[dusky void]

hha

[granite geode]

.close

[devout snow]

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[dusky void]

,tex -b\pm \sqrt{b^{2}-4ac}

[woven radish]

Wind chaser
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[devout snow]

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[restive river]

i was studying rotating functions using r cos ( theta + alpha) but i didnt know parametric equations and functions, where do i study about trigonometric parametric coordinates and equation would i start with polar

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[devout snow]

@restive river Has your question been resolved?

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[vernal blaze]

,help

[woven radish]

A brief description and guide on how to use me was sent to your DMs!
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[stark anvil]

#latex-testing

[devout snow]

@vernal blaze Has your question been resolved?

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[wise spade]

So, a guy sold 2 of item a, 5 of item b and bought 10 of item c on a market, he has 14$ left. If he would have sold 5 of item b and 4 of item c he could have bought 5 of item a, if he would have sold 3 of item b and 3 of item c he wouldve been short 21$ to buy 5 of item a

my question is, where do i start? What do i do? I gotta solve this using linear equation systems, but only recently joined this class, teacher isnt here to explain, and i hit a brickwall

[fresh elbow]

represent x as the starting money, a,b,c as the item prices

[wise spade]

i see, just solve it graphically then?

[fresh elbow]

no? i think

x-2a-5b-10c = 14

[wise spade]

then i dont see

[fresh elbow]

that's the first equation

[wise spade]

aah, wait lemme think for a second

[fresh elbow]

write out the other equations

[wise spade]

b: 5a-4b-5c=god knows..?

[fresh elbow]

oh shoot i misread

x+ 2a+5b-10c = 14

[wise spade]

aa, i was wondering

[fresh elbow]

x+5b+4c = 5a

can you find the third equation

[wise spade]

3a+3b-5c=-21?

[fresh elbow]

not exactly

where is your starting money

[wise spade]

x+3a+3b-5c=-21?

[fresh elbow]

so your statement says

if I sold 3 of a, 3 of b, and bought 5 of c, I would have -21 money

can you fix it

[wise spade]

Where do i adjust the price and where the quantity of items tho..? Currently struggeling with that, if i put -21 at a variable i would have -21 of item a, if i just add another variable at the equal i still something wrong as starting money

[fresh elbow]

ok so the left side of the equation

is the money you have after selling 3b and 3c

can you make that

the quantity of the items is the coefficient

the price of the item is a,b,c depending on the item

so if I have 3 of item a, the price would be 3a

[wise spade]

x-3a+3b+5c=god knows, i subtract a from whatever b and c equals to and we know its negative..?

teacher came back, thx for the help!

ill ask her ^^

[fresh elbow]

it would be x+3b+3c = 5a -21

[wise spade]

.close

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[tranquil bloom]

b)

[devout snow]

@tranquil bloom Has your question been resolved?

[tranquil bloom]

<@&286206848099549185>

i got that B=1+3i, and changed to cis

but the argument doesn't come out to be a nice number, so im guessing there's a better way

nvm got it

.close

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[quaint egret]

Anyone know the formula to calculate compound interest?

[tribal pier]

Depends on if it's continuously compounding or not

[fresh elbow]

if it wasn't continuously compounding it's not compound interest

[tribal pier]

It's different

Infinite vs finite

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[dry iron]

can someone explain to me why the first and last number doesn't get multiplied by 2?

[dense jay]

because they only occur in one trapezium each

the rest appear in two

[dusky void]

good

[dry iron]

wdym

[dense jay]

they only appear in the area formula for the first and last trapezia

the others appear in two trapezia each so theyre multiplied by two

[dry iron]

ooh

[dry iron]

thank you

[earnest karma]

hello

[devout snow]

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[sudden path]

how do i even start?

[feral agate]

They meant OP bisects AB at right angles, not AC

To prove that OAP is congruent to OBP, you need 3 similarities between the two triangles (equal sides or equal angles)

Try to look for as many as you can prove

[dusky void]

O and P are the centers of the circles, and OA and OB are radii of the circle centered at O, while AP and BP are radii of the circle centered at P. This implies that OA = OB and AP = BP, as radii of the same circle are equal.

Additionally, by the definition of a chord, OP is the same line segment for both circles, so we have that OP is common to both triangles
△OAP and △OBP. Hence, by the Side-Side-Side (SSS) criterion, the two triangles are congruent.

[feral agate]

!nosols

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dusky void]

wow

[devout snow]

@sudden path Has your question been resolved?

[dusky void]

why

[devout snow]

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[vital flicker]

help pls

[winter patrol]

!status

[devout snow]

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

[soft umbra]

@vital flicker

[vital flicker]

1

[winter patrol]

introduce a variable for your radii

then note that triangle ABC is a right triangle in which case pytthagors is applicable

[vital flicker]

oh okay

wait, lmao, i just figured everything out, i remember the concept, thanks for your time and sorry for the disturbance

.close

[devout snow]

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[real pecan]

Help pls

[soft umbra]

!help

[devout snow]

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

[devout snow]

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[fierce matrix]

I have three tables: https://paste.pythondiscord.com/YYIQ, and trying to figure out how to calculate crude death rate, and age-standardized death rate from Chronic Obstructive Pulmonary Disease (COPD) for Uganda and US in 2019.

[fierce matrix]

How to approach this task?

[devout snow]

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@fierce matrix Has your question been resolved?

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[devout snow]

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[warm gazelle]

How do i merge these two into x(y)?

[warm gazelle]

do I just put y inside x(t) as parameter?

these two are both dependant on the same time, i want to represent it as x(y)

Or do I first solve for t=... inside my y=y(t)

and just put that t inside x(t)

[devout snow]

@warm gazelle Has your question been resolved?

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@warm gazelle Has your question been resolved?

[spare crypt]

yea solve for t=... inside y=y(t), you'll need the quadratic formula

[warm gazelle]

Obv will get 2 answers, for t1, and t2 right?

Or am i supposed to exclude one

or nvm

i dont work with values

so idc abt excluding

cant decide from just the formula

ty

.close

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[honest solstice]

hello i was wondring what's the probability of having a shiny things in the game called apex legends so here some values to know: the chance of getting heirloom (shiny thing) in a pack is 0,5%, you only have 600 packs and there's no guarantee. i'm looking for answer but if u know what to use to get the answer i can do the math after (it's been a so long since i've done probability so idk what i should do). Sorry in advance if it's not clear, i'm not that fluent in english.

[fresh elbow]

What are you trying to find

[twin hornet]

Given the probability is consistent, independent events (probably not likely in most video games) Then it's simply 600x0.5% or 600x0.005

[topaz beacon]

are you trying to find about how many you get or the probability of you getting one?

[twin hornet]

I assumed from the first sentence the probability of getting one from 600 packs

[topaz beacon]

that wouldnt be the right formula then

it would be hard to figure out the probability of getting 1 or more, so what we can do is find the probability of not getting one

that probability is 99.5% or .995 for one pack open

assuming the events are independent you would multiply this to itself 600 times to get the probability of not winning

,w .995^600

[woven radish]

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Requested by garlicbredfries

[topaz beacon]

about 4.9% of not getting anything

therefore, there is a 95.1% chance of getting something

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[lean cipher]

Given a permutation of size n:
What is the probability of 1 not being a fixed point, but for 2 being a fixed point?

[lean cipher]

The answer is no to whoever deleted their comment just now

Anyway more formally:
If A_i means “i is a fixed point”
What is P(A_1^c and A_2)

(I calculated either 0 or (n-2)/[n(n-1)]

[hushed lance]

I’m not sure if this is right but I would say that there are n-2 choices for where 1 ends up

[lean cipher]

There are n-1 choices, hence the 0 I mentioned just now

1 could go to 2

[hushed lance]

then that probably means I don’t know what a fixed point is lol sorry

[lean cipher]

Np

[spare crypt]

the strategy is P(2 fixed) - P(1 and 2 fixed)

[lean cipher]

Just a moment, I need to think about that

[lean cipher]

This makes intuitive sense but on the formal level I’m not so sure

Doesn’t this require some form of independency?

[spare crypt]

nah, and fixed points definitely aren't independent, it's just set theory like something you'd draw a venn diagram for

[lean cipher]

Yeah I know they aren’t independent, that’s why I’m asking

[spare crypt]

P(2)=P(2 and 1) + P(2 and 1^c) so P(2 and 1^c)=P(2)-P(2 and 1)

[lean cipher]

Oh

Yeah I guess that’s a good way of proving that

[spare crypt]

yea it helps to have intuition for set theory stuff with complements for probability

[lean cipher]

Absolutely. Unfortunately (or not) I transitioned during my studies and my previous university really lacked in discrete math (which is where combinatorics was studied)

I’ll take that advice to heart though, thank you

.close

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[timid peak]

Im guessing this means without loss of generality

what is that supposed to mean?

[sonic smelt]

It does

[sonic smelt]

It means that you have several "symmetrical" cases

[heady tulip]

physics?

[sonic smelt]

Meaning you can work on just one case and then use it everywhere else

[stone stump]

if you have some numbers a,b then you can for example wlog assume that a>=b

cause the case b>=a will behave the same way

[timid peak]

no, this proof

[sonic smelt]

That is assuming swapping a and b in the statement that you are proving does not change it

[stone stump]

if x_2 in X and x_1 in A\X, then the rest of the arguments are still the same

[timid peak]

ok

[stone stump]

just with the places of x_1 and x_2 switched

[devout snow]

@timid peak Has your question been resolved?

[timid peak]

its so over

ahhhhh

why am i like this

ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

ok

ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

ok

im good now

how is the range a subset of B??

oh it is a subset if its not surjective

[timid peak]

am i stupid??

yes that must be the case

wtf why g(y)=1 if y<0

huh???

but y cant be less than 0? its 0 inclusive until infinity for domain

<@&286206848099549185>

oh

oh ma god

im blind

g's domain is all reals

but Im still kinda confused, for y, are they saying its f(x)?

but that cannot be the case because you cant have x^2 be <0

eve though f maps to all reals..

[devout snow]

@timid peak Has your question been resolved?

[timid peak]

how did they get this part?

[sand dove]

This? This is AM-GM inequality

x1 = 1, x2=x3=...=xn+1 = 1 + 1/n

[timid peak]

not sure how to get this part..

how would you expand (1+1/n)^n?

[sand dove]

As I said, AM-GM inequality

$1\cdot \Big(1+\frac 1n\Big)^n = a_1\cdot a_2\cdot ... \cdot a_{n+1}$ where $a_1 = 1$ and $a_2 = ... = a_{n+1} = 1+\frac 1n$

[woven radish]

rafilou2003

[sand dove]

And then AM-GM applied to those ak

[timid peak]

oh ok

thank you!

how did they go from (1+1/n)^n < (1+1/m)^m+1 to the AM-GM...

[sand dove]

Again, find the appropriate $a_1,...,a_{n+m}$

[woven radish]

rafilou2003

[timid peak]

no like from here to here..

[sand dove]

We WILL show

[sand dove]

They haven't shown this yet

[sand dove]

IN ORDER TO show it, they use AM-GM to find this FIRST

[timid peak]

maybe my algebra is bad

or maybe i should prob just sleep for a bit

i can’t get to this

[sand dove]

Where does this first line even come from?

Oh right you multiply both sides by 1/...

[timid peak]

yea this part

[sand dove]

How does the thing inside parentheses change?

[timid peak]

ohhh fck

its reciprocol

of (1-1/m)^m

not this is equal to..

ahh im so tired i dont know if i should nap for a bit or not

cause then I dont know if I would be able to sleep later

ahhh decisions

ok yea ill sleep for a lil bit

i cant think properly

[devout snow]

@timid peak Has your question been resolved?

[timid peak]

im reading it over again

not sure how these two are equivalent

because thats what its saying, divide 1 by (1-(1/m))^m and you get (1+ 1/(m-1))^m which proves section 6.1

i'll move ontoo other questions

[upper schooner]

You can always "shift" m to (m + 1) of course: if (m - 1) is to be a strictly positive integer, you could "replace" all m's with (m + 1) and it basically be the same thing

[timid peak]

ahhh @upper schooner youre here

yay

ok umm not sure what you mean by shift m to m+1.. m+1 would be (1+1/m)^m(1+1/m) but not sure how that can be 'transformed' to the equation required

[upper schooner]

As in, the statement could be restated, instead of them saying
[
\pqty{1 + \frac1n }^n < \pqty{1 + \frac1m}^{m + 1}
]
you could instead say (that for all $n$ and $m$) that you have
[
\pqty{1 + \frac1n }^n < \pqty{1 + \frac1{m - 1}}^m
]
(provided you have $m$ being an integer that's at least 2)

[woven radish]

@upper schooner

[timid peak]

ahh

restated, not equivalent

gotchya

[upper schooner]

Yea similar to like how in some convergence/Cauchy type problems, you could instead of choosing epsilon you can choose eps/2 or something convenient like that and it makes no real difference

[timid peak]

ahhhh

im already stuck on the first question

hm,, im trying to prove by contradiction

ill send my work so far.. but idk..

but my contradiction is not a contradiction😭

wait instead of for all n>n_0 an<=alpha it should be for all n in N

@upper schooner

[upper schooner]

hmm, do you even need to do it by contradiction at all though, might be easier to do it without (there are multiple things you'd need to consider when it comes to contradiction after all: either your sequence isn't in A, isn't increasing or doesn't converge to alpha)

Remember how the least upper bound of a set is defined and all, and note that in this case your sup isn’t in the set to begin with

[timid peak]

how can the sequence an be a subset of A, but the sup A is not in A. that means there is an x in A s.t an<x<supA for all n in N

and then the x would be the least upper bound

for {an}

[upper schooner]

For example if you took the interval [0, 1) and let your sequence be 1 - 1/n, that sequence is inside the interval, the sup is 1 and isn't in the interval, but the sequence converges to that sup

[timid peak]

but then thats a case where it should be {an} \subseteq A

[upper schooner]

That's what they ask you for here though, no(?)

[timid peak]

but they made it seem like, an is a subsequence of A

like subset of A

[upper schooner]

More that a(n) is a sequence whose terms all live in the set A

Take the set of all the terms of the sequence, and that will form a subset of A

[timid peak]

my very strong argument😭😭

[upper schooner]

hmmmm

How about we do it more directly together

Let's just for now only assume that A is a subset of the real numbers which is bounded above, but doesn't contain its sup

We're gonna cook a sequence up that would fit the properties they ask for

@timid peak pick your favourite or a random (strictly) positive number that you like

[timid peak]

ok

1

[upper schooner]

cool, that's nice

Now, (alpha - 1) is strictly less than alpha, the least upper bound of A, so (alpha - 1) is not an upper bound of A, right? What does that mean?

[timid peak]

alpha -1 must be in A or lower bound of A?

[upper schooner]

Not necessarily either of those

Think about what not being an upper bound of a set in isolation would actually mean (remember there's usually some way it's usually stated, I think we covered that before? )

[timid peak]

there exists x in A s.t alpha -1 < x

[upper schooner]

Yep, there we go, now let's bear that in mind

We also know that x <= alpha too, so that it remains in A

But actually, remembering that alpha isn't in A, we know that x < alpha, right?

[timid peak]

yes

[upper schooner]

Cool, now all of that is really nice

I'm going to make a tiny change to what we've done so far, you might see where this is likely going to go(!)

We choose a strictly positive number that we like, say 1, and note that $\alpha - 1 < \alpha = \sup(A)$.

In that case, $\alpha - 1$ is not an upper bound of $A$, and so we have some $x_1\in A$ such that $\alpha - 1 < x_1 < \alpha$ (note that $x_1 \leq \alpha$ so that $x_1 \in A$, and as $\alpha\not\in A$, we must have a strict inequality).

[woven radish]

@upper schooner

[upper schooner]

Do you have any ideas of what we could do now? And as for why I chose to label that as x1?

[timid peak]

ahh sorry I left for a little bit

back now

ok

inductive argument, since its strictly increasing and monotone, we have x1<x2.. x2<x3... so on. In which for all xn where n in N, the upper bound is alpha. Then it is proved that the sequence converges to alpha

[upper schooner]

Kinda like that sure you can then take the x1 you found, and because that's less than alpha you can find similarly some x2 in A such that x1 < x2 < alpha, then find an x3 similarly, and you notice that x1 < x2 < x3 < ... (that gives you the "subset of A" and "strictly increasing" parts that they ask you for)

Showing it converges to alpha isn't too bad

[timid peak]

ok, I will write down the proof.

yea then Ill just say |xn-alpha| < epsilon

the convergence proof

[upper schooner]

make sure you word the convergence part fine

[timid peak]

😭 the stress is at an all time high omg, my heart is pounding like i just ran a marathon or smth

[upper schooner]

Awwww well you can do it, imagine I'm cheering for you like you're doing an actual marathon or something

[timid peak]

for these type of questions, I need to claim the least upper bound or largest lower bound for the monotone subsequence (it seems for d)

and also for other questions, how would you find the limit in these recurence sequences?

some you can by inspection, some are not so easy by that method